Modeling and Analysis of Dynamic Systems, Third Edition (Solutions, Instructor Solution Manual) [3 ed.] 9781138726420, 1138726427

Citation preview

1

Review Problems 1 Evaluate the function = f ( x, y ) 23 e − x /2 cos( y − 1) for x = −0.23, y = 2.7 (a) Using the subs command. (b) By conversion into a MATLAB function. Solution

(a) >> f = sym('2*exp(-x/2)*cos(y-1)/3'); >> x = -0.23; y = 2.7; >> double(subs(f)) ans = -0.0964

(b) >> F = matlabFunction(f); >> F(-0.23,2.7) ans = -0.0964

2 Evaluate the function g ( x, y ) =sin(2 x + 1) tan( y − 12 ) for x = 0.45, y = −1.17 (a) Using the subs command. (b) By conversion into a MATLAB function. Solution

(a) >> g = sym('sin(2*x+1)*tan(y-1/2)'); >> x = 0.45; y = -1.17; >> double(subs(g)) ans = 9.5076

(b) >> G = matlabFunction(g); >> G(0.45,-1.17) ans = 9.5076

 xy + 1  Evaluate the vector function v( x, y ) =  for x 1.54, = y 2.28  = x − 2 y (a) Using the subs command. (b) By conversion into a MATLAB function. 3

2

Solution

(a) >> v = sym('[x*y+1;x-2*y]'); >> x = 1.54; y = 2.28; >> double(subs(v)) ans = 4.5112 -3.0200

(b) >> V = matlabFunction(v); >> V(1.54,2.28) ans = 4.5112 -3.0200

3x − y   y Evaluate the matrix function m( x, y ) =  −2, y = 3.35  for x =  x + 1 sin 2 y  (a) Using the subs command. (b) By conversion into a MATLAB function. 4

Solution

(a) >> m = sym('[y 3*x-y;x+1 sin(2*y)]'); >> x = -2; y = 3.35; >> double(subs(m)) ans = 3.3500 -1.0000

-9.3500 0.4048

(b) >> M = matlabFunction(m); >> M(-2,3.35) ans = 3.3500

-9.3500

-1.0000

0.4048

5 If f (t ) = e −3t /5 + t ln(t + 1) , evaluate df / dt when t = 4.4 (a) Using the subs command. (b) By conversion into a MATLAB function.

3 Solution

(a) >> f = sym('exp(-3*t/5)+t*log(t+1)'); >> df = diff(f); t = 4.4; >> double(subs(df)) ans = 2.4584

(b) >> dF = matlabFunction(df); >> dF(4.4) ans = 2.4584

( x) 23 x −1 + e − x cos x , evaluate dg / dx when x = 1.37 6 If g= (a) Using the subs command. (b) By conversion into a MATLAB function. Solution

(a) >> g = sym('2^(3*x-1)+exp(-x)*cos(x)'); >> dg = diff(g); x = 1.37; >> double(subs(dg)) ans = 17.6539

(b) >> dG = matlabFunction(dg); >> dG(1.37) ans = 17.6539

7

Solve the following initial-value problem, and evaluate the solution at x = 3.5 . ′ + xy 2 x , = ( x − 1) y= y (2) 1

Solution >> y = dsolve('(x-1)*Dy+x*y=2*x','y(2)=1','x'); >> y = matlabFunction(y); y(3.5) ans = 1.9107

4 8

Solve the following initial-value problem, and evaluate the solution at t = 2.8 .

tx + x =et , x(1) =−1

Solution

>> x = dsolve('t*Dx+x=exp(t)','x(1)=-1'); >> x = matlabFunction(x); x(2.8) ans = 4.5451

9

Plot y1 (t ) = e − t /3 cos( 12 t ) and y2 (t )= (t + 1)e − t versus 0 ≤ t ≤ 10 in the same graph. Adjust the limits on the vertical axis to −0.3 and 1.1 . Add grid and label.

Solution >> >> >> >> >>

syms t y1 = exp(-t/3)*cos(t/2); y2 = (t+1)*exp(-t); ezplot(y1,[0,10]) hold on ezplot(y2,[0,10])

1

0.8 y2

y1 0.6

0.4

0.2

0

-0.2 0

1

2

3

4

5 t

6

7

8

9

10

Figure Review1No9 10 Plot x1,2,3 (t ) = e − at /2 sin( 13 t ) , corresponding to a = 1, 2,3 , versus 0 ≤ t ≤ 5 in the same graph. Adjust the limits on the vertical axis to −0.05 and 0.3 . Add grid and label. Solution >> >> >> >> >> >> >> >> >>

syms t x1 = exp(-t/2)*sin(t/3); x2 = exp(-t)*sin(t/3); x3 = exp(-3*t/2)*sin(t/3); ezplot(x1,[0,5]) hold on ezplot(x2,[0,5]) hold on ezplot(x3,[0,5])

5

0.25

0.2 x1 0.15

x2

0.1 x3 0.05

0

-0.05

0

0.5

1

2.5 t

2

1.5

3

3.5

4

2

2.5

4.5

5

Figure Review1No10 t

11 Plot ∫ e x −t sin xdx versus −1 ≤ t ≤ 3 , add grid and label. Solution

1

>> syms t x >> integ = int(exp(x-t)*sin(x),x,1,t); >> ezplot(integ,[-1,3])

0.5

0

-0.5

-1

-1.5

-2 -1

-0.5

0

0.5

1 t

1.5

Figure Review1No11 t

12 Plot ∫ ( x + t ) 2 e− (2t − x ) dx versus 1 ≤ t ≤ 3 , add grid and label. Solution

0

>> syms t x >> integ = int((x+t)^2*exp(-(2*t-x)),x,0,t); ezplot(integ,[1,3])

3

6

1.4 1.3 1.2 1.1 1 0.9 0.8 0.7 0.6 1

1.2

1.4

1.6

1.8

2 t

2.4

2.2

2.6

2.8

3

Figure Review1No12 13 Plot z ( x, t )= (1 + cos x)e − (t −1) versus 0 ≤ x ≤ 10 for two values of t = 1.5, 2.5 in a 1× 2 tile. Add grid and title. Solution x = linspace(0,10); t = [1.5,2.5]; for i = 1:2, for j = 1:100, z(j,i) = (1+cos(x(j)))*exp(-(t(i)-1)); end end

% 100 values of z for each t

% Initiate figure subplot(1,2,1), plot(x,z(:,1)), title('t = 1.5') subplot(1,2,2), plot(x,z(:,2)), title('t = 2.5')

t = 1.5

t = 2.5 0.45

1.4

0.4

1.2

0.35 1 0.3 0.8

z

z

0.25 0.2

0.6

0.15 0.4 0.1 0.2

0

0.05

0

2

6

4

8

10

0

0

x

2

6

4 x

Figure Review1No13

8

10

7 14 Plot z ( x, t ) = sin(2.7x) cos(3.2t ) versus 0 ≤ x ≤ 5 for four values of t = 1, 1.5, 2, 2.5 in a 2 × 2 tile. Add grid and title. Solution x = linspace(0,5); t = 1:0.5:2.5; for i = 1:4, for j = 1:100, z(j,i) = sin(2.7*x(j))*cos(3.2*t(i)); end end % Initiate figure subplot(2,2,1), plot(x,z(:,1)), title('t = subplot(2,2,2), plot(x,z(:,2)), title('t = subplot(2,2,3), plot(x,z(:,3)), title('t = subplot(2,2,4), plot(x,z(:,4)), title('t =

% 100 values of z for each t

1') 1.5') 2') 2.5') t = 1.5 0.1

0.5

0.05

0

0

z

z

t=1 1

-0.05

-0.5

0

4

3

2

1

-0.1

5

2

1

x

t=2

t = 2.5

1

0.2

0.5

0.1

0 -0.5 -1

0

x

z

z

-1

3

4

5

3

4

5

0 -0.1

0

4

3

2

1

-0.2

5

0

2

1

x

x

Figure Review1No14 15 Given f ( x) = e −3 x /2 + cos( x − 1) , plot f ′( x) versus 0 ≤ x ≤ 8 . Add grid and title. Solution >> f = sym('exp(-3*x/2)+cos(x-1)'); df = matlabFunction(diff(f)); >> ezplot(df,[0,8]) 1

f'(x)

0.5

0

-0.5

-1

0

1

2

3

4 x

5

Figure Review1No15

6

7

8

8

16 Write a user-defined function with function call F=temp_conv(C) that converts the temperature from Celsius C to Fahrenheit F. Execute the function for the case of C = 27. Solution function F = temp_conv(C) F = 9*C/5+32; end >> F = temp_conv(27) F = 80.6000

17 Write a user-defined function with function call s=speed_calc(x,t0) that calculates the speed of a moving object, whose position is described by a symbolic function x, at a specified time t0. Execute the function to find the speed (at t = 2.5 ) of an object whose ) 0.75t 3 − 2t 2 + 1.4 . position is described by x(t= Solution function s = speed_calc(x,t0) v = matlabFunction(diff(x)); s = abs(v(t0)); end >> x = sym('0.75*t^3-2*t^2+1.4'); >> s = speed_calc(x,2.5) s = 4.0625

18 Write a user-defined function with function call fval = f_eval(f,a,b) where f is an anonymous function, and a and b are constants such that a < b . The function calculates the midpoint m of the interval [ a, b ] and returns the value of 12 f (a ) + f (m) + 12 f (b) . Execute f_eval

for f = xe − x , a = −3 , b = 4 .

Solution function fval = f_eval(f,a,b) m = (a + b)/2; fval = f(a)/2 + f(m) + f(b)/2; end >> f = @(x)(x*exp(-x)); >> fval = f_eval(f,-3,4)

fval = -29.7884

9

19 Repeat the example considered in this chapter involving a signal and its integral, but this time the sine wave has amplitude 0.8 and frequency 1.5 rad/sec , and the integrated signal gets amplified by a factor of 2 ; see Figure 1.22. Build the model, run the simulation and generate a plot that can be imported into a document.

Sine Wave Amplitude = 0.8 Frequency = 1.5 r/s

Scope Integrated signal amplified by 2

1 s

1

2

Out1

Gain

Integrator

Figure 1.22 Problem 19. Solution 2.5

2

1.5

Output

1 Original signal 0.5

0

-0.5

-1 0

1

2

3

4

5 t

6

7

8

9

10

Figure Review1No19

20 Create the Simulink model shown in Figure 1.23. Select the signal generator as a square wave with amplitude 0.5 and frequency 0.5 rad/sec . To flip the gain block, right click on it, then go to Rotate & Flip and choose the "Flip Block" option. Double clicking on the Sum (Commonly Used Blocks) allows control over the list of desired signs. Run simulation and generate a figure that can be imported into a document.

10

Signal generator

Sum 1 s

Scope

Integrator

Square wave Amplitude = 0.5 Frequency = 0.5 r/s

5

1

Out1

Feedback gain

Figure 1.23 Problem 20. Solution 0.15

0.1

0.05

0

-0.05

-0.1

0

1

2

5 t

4

3

6

7

8

9

10

Figure Review1No20

21 Build the Simulink model shown in Figure 1.24. Select the signal generator as a square wave with amplitude 1 and frequency 0.5 rad/sec . To flip the gain block, right click on it, then go to Rotate & Flip and choose the "Flip Block" option. Double clicking on the Sum (Commonly Used Blocks) allows control over the list of desired signs. Run simulation and generate a figure that can be imported into a document. Signal generator Sum 1 s

Square wave Amplitude = 1 Frequency = 0.5 r/s

Integrator

Scope

2

Feedback gain 1

Out1

Figure 1.24 Problem 21.

11 Solution 0.4

0.3

0.2

0.1

0

-0.1

-0.2

-0.3

-0.4

0

1

2

3

4

5 t

6

7

8

9

10

Figure Review1No21 22 Build the Simulink model shown in Figure 1.25. To flip the gain block, right click on it, then go to Rotate & Flip and choose the "Flip Block" option. Double clicking on the Sum allows control over the list of desired signs. Run simulation and generate the response plot.

Sum

Sine Wave Amplitude = 1 Frequency = 1.25 r/s

1 s

1 s

Integrator1

Integrator 2

Scope

2

Feedback gain 1

Out1

Figure 1.25 Problem 22. Solution 0.6 0.5 0.4 0.3 0.2 0.1 0 -0.1 -0.2 -0.3 -0.4

0

1

2

3

4

5 t

Figure Review1No22

6

7

8

9

10

12

23 Build the Simulink model shown in Figure 1.26. Transfer function represents the ratio of the Laplace transforms of the output and the input signals associated with a block; see Chapter 4. The Transfer Fcn block is located in the Continuous library. Run simulation and generate the response plot. Sum 1 2s+1

Sine Wave Amplitude = 1.5 Frequency = 1.25 r/s

Scope

Transfer Fcn 5

1

Out1

Gain

Figure 1.26 Problem 23. Solution 2.5 2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 -2.5

0

1

2

4

3

6

5 t

7

8

10

9

Figure Review1No23 24 Build the Simulink model shown in Figure 1.27. The Step signal has a step time of 1 and a final value of 1 , that is, it assumes a value of 1 immediately after t = 1 and a value of 0 otherwise. The Transfer Fcn block is located in the Continuous library. Run simulation and generate the response plot. s

1 s

s 2+2s+2

Step

Scope

Integrator

Transfer Fcn

2.5

Gain1

Figure 1.27 Problem 24.

4

Gain 2

1

Out1

13

Solution 1.4

1.2

1

0.8

0.6

0.4

0.2

0

0

1

2

3

4

5 t

6

7

8

9

10

Figure Review1No24 25 Build and simulate a Simulink model that integrates a sawtooth signal (amplitude 1, frequency 1 rad/sec) and displays the result, along with the sawtooth signal itself. Solution

Scope

Sawtooth Amplitude = 1 Frequency = 1 r/s

1 s

1

Out1 Integrator

Figure Model in Problem 25 5 4 3

Integrated signal

2 1 0

Sawtooth

-1 -2 -3 -4 -5

0

1

2

3

4

5 t

Figure Review1No25

6

7

8

9

10

14

26 Build the model shown in Figure 1.28, which is the Simulink model of the RLC circuit considered in this chapter and is equivalent to the Simscape model presented in Figure 1.19. Perform the simulation to confirm that both models yield the same response. Unit pulse

s 1 0 cos φ < 0

⇒

− 1 tan φ =

2nd quadrant

3

⇒

5 φ= π 6

Therefore, cos(2t + 13 π ) = sin(2t + 65 π ) . 14. 3sin ωt − cos ωt Solution = ωt D sin(ωt = + φ ) D sin ωt cos φ + D cos ωt sin φ . Comparison gives Expand 3sin ωt − cos D = 10 −1 D sin φ = sin φ < 0 ⇒ = D cos φ 3 cos φ > 0

Therefore 3sin ωt − cos = ωt

⇒

4th quadrant

− 13 tan φ =

⇒

−0.3218 rad φ=

10 sin(ωt − 0.3218) .

In Problems 15 and 16, write the expression in the form D cos(ωt + φ ) . 15. cos t − 34 sin t Solution Write cos t − 34 sin = t D cos(t += φ ) D cos t cos φ − D sin t sin φ and compare the two sides to find D sin φ = 34 D cos φ = 1

D = 54

⇒

sin φ > 0 cos φ > 0

⇒

tan φ =

3 4

1st quadrant

⇒

φ = 0.6435 rad

25

Therefore, cos t − 34 sin t =

5 4

cos(t + 0.6435) .

16. sin t − cos t Solution Write sin t − cos = φ ) D cos t cos φ − D sin t sin φ and compare the two sides to find t D cos(t += −1 D = 2 sin φ < 0 D sin φ = ⇒ −1 D cos φ = cos φ < 0

⇒

tan φ =1

3rd quadrant

⇒

φ = 54 π

2 cos(t + 54 π ) .

Therefore, sin t − cos= t

Problem Set 2.3 In Problems 1 through 8, (a) Find the Laplace transform of the given function. Use Table 2.2 when applicable. (b) Confirm the result of (a) in MATLAB. 1. eat +b , a, b = const Solution

{

}

{

}

linearity

(a)= eat eb L eat +b L=

{ }

= L eat eb

eb s−a

(b) >> syms a b t >> laplace(exp(a*t+b)) ans = -exp(b)/(a - s)

2. cos(ωt + φ ) , ω , φ = const Solution (a) Using trigonometric expansion, we find

L {cos( = ωt + φ )} L {cos ωt}cos φ − L {sin ωt}sin φ s ω s cos φ − ω sin φ = 2 cos φ − 2 sin φ = 2 2 s +ω s +ω s2 + ω 2 (b) >> syms w t p

26 >> laplace(cos(w*t+p)) ans = (s*cos(p) - w*sin(p))/(s^2 + w^2)

3. sin(ωt − φ ) , ω , φ = const Solution (a) Using trigonometric expansion, we find

L {sin( = ωt − φ )} L {sin ωt cos φ − cos= ωt sin φ } L {sin ωt} cos φ − L {cos ωt} sin φ =

ω s ω cos φ − s sin φ cos φ − 2 sin φ = 2 2 s +ω s +ω s2 + ω 2 2

(b) >> syms w t p >> laplace(sin(w*t-p)) ans = (w*cos(p) - s*sin(p))/(s^2 + w^2)

4. t 3 − 12 Solution

{

}

(a) L t 3 − 12 =

6 1 − s 4 2s

(b) >> syms t >> laplace(t^3-1/2) ans = 6/s^4 - 1/(2*s)

5. sin 2 t Solution 2 (a) Noting that sin = t

1 (1 − cos 2t ) , 2

{

}

we find

1 1 1 s  Simplify 2 L sin 2 t = L {1 − cos 2t} =  − 2 =  2 2 2 s s + 4 s ( s + 4) (b) >> syms t >> laplace(sin(t)^2)

27 ans = 2/(s*(s^2 + 4))

6. t cos ωt Solution (a) Following Eq. (2.16),

d  s  s2 − ω 2 −  2 = L {t cos ωt} =  ds  s + ω 2  ( s 2 + ω 2 ) 2 (b) >> syms t w >> simplify(laplace(t*cos(w*t))) ans = (s^2 - w^2)/(s^2 + w^2)^2

7. t cosh t Solution (a) Comparing with Eq. (2.16), we have g (t ) = cosh t so that

G ( s= ) L {cosh t= }

{

}

1 1 1 1  Simplify s + = L et + e−t=  2 2  s − 1 s + 1  s2 −1

Then,

d  s  s2 + 1 −  2 = L {t cosh t} = ds  s − 1  ( s 2 − 1) 2 (b) >> syms t >> simplify(laplace(t*cosh(t))) ans = (s^2 + 1)/(s^2 - 1)^2

8. t 2 sin( 12 t ) Solution (a) Following the general form of Eq. (2.16) with g (t ) = sin( 12 t ) and n = 2 ,

28

{

}

L= t sin( 12 t ) 2

2 d 2  12  3s − 14 =   ds 2  s 2 + 14  ( s 2 + 14 )3

(b) >> G = sym('1/(2*(s^2+1/4))'); simplify(diff(G,2)) ans = (16*(12*s^2 - 1))/(4*s^2 + 1)^3

In Problems 9 through 12, (a) Express the signal in terms of unit-step functions. (b) Find the Laplace transform of the expression in (a) using the shift on t − axis. 9. g (t ) in Figure 2.15 Solution (a) g (t ) = u (t − 1) − u (t − 2) e − s e −2 s e − s − e −2 s (b) G ( s ) = − = s s s

g (t ) 1

t 1

2

Figure 2.15 Signal in Problem 9.

10. g (t ) in Figure 2.16 Solution (a) g (t ) = −u (t ) + 2u (t − 1) − u (t − 2) (b) G ( s ) =

−1 + 2e − s − e −2 s −(1 − e − s ) 2 = s s

29

g (t ) 1

t 1

2

−1 Figure 2.16 Signal in Problem 10

t1  Solution (a) Construct g (t ) using the strategy shown in Figure PS2-3 No11, leading to g (t ) = (1 − t )u (t ) − (1 − t )u (t − 1)

(b) Rewrite the expression obtained in (a) as g (t )= u (t ) − tu (t ) + (t − 1)u (t − 1) . Then

1 1 e− s G (s) = − 2 + 2 s s s

Figure PS2-3 No11

30

t1  Solution simplify

(a) g (t= ) tu (t ) − tu (t − 1) + u (t − 1) =

tu (t ) − (t − 1)u (t − 1)

1 e− s 1 − e− s (b) G ( s ) = 2 − 2 = 2 s s s

In Problems 13 through 16, find the Laplace transform of each periodic function whose definition in one period is given.  1 if 0 < t < 12 13. h(t ) =  1 −1 if 2 < t < 1 Solution Noting P = 1 , we find

H ( s= )

1 1 − e− s

1 − s /2 1/2 − st  + e − s (1 − e− s /2 ) 2 simplify 1 − e− s /2 − st  1 − 2e e dt e dt − = = = ∫  ∫ −s −s s e s e s (1 + e − s /2 ) (1 ) (1 ) − − 1/2  0 

if 0 < t B = [1 0]; A = conv([1 1],[1 2 2]); >> [R,P,K] = residue(B,A) R = 0.5000 - 0.5000i 0.5000 + 0.5000i -1.0000 + 0.0000i P = -1.0000 + 1.0000i -1.0000 - 1.0000i -1.0000 + 0.0000i K = []

9s + 5 s ( s 2 + 4 s + 5) Solution (a) Expand as

20.

2

9s + 5 A B Cs + D As ( s 2 + 4 s + 5) + B( s 2 + 4 s + 5) + s 2 (Cs + D) = + + = s 2 ( s 2 + 4 s + 5) s s 2 s 2 + 4s + 5 s 2 ( s 2 + 4s + 5)

This leads to

A+C = 0 4A + B + D = 0 5 A + 4B = 9 5B = 5

Simultaneous solution gives A = 1, B = 1, C = −1, D = −5 so that 9s + 5 1 1 −s − 5 1 1 − 12 + 32 j − 12 − 32 j = + + = + + + s 2 ( s 2 + 4s + 5) s s 2 s 2 + 4s + 5 s s 2 s + 2 − j s + 2 + j

(b) >> B = [9 5]; A = conv([1 0 0],[1 4 5]); >> [R,P,K] = residue(B,A) R = -0.5000 + 1.5000i -0.5000 - 1.5000i 1.0000 + 0.0000i

34 1.0000 + 0.0000i P = -2.0000 -2.0000 0.0000 0.0000

+ + +

1.0000i 1.0000i 0.0000i 0.0000i

K = []

In Problems 21 through 26, find the inverse Laplace transform using the partial-fraction expansion method. s +1 s ( s + 3) Solution Expand as

21.

2

s +1 A B C As ( s + 3) + B ( s + 3) + Cs = + 2+ = s+3 s ( s + 3) s s s 2 ( s + 3)

= A+C 0 = A 92 ⇒ 3A + B = 1 ⇒ B = 13 3B = 1 C = − 92

2

2

Therefore, 2 1 − 92 s +1 9 3 = + + s 2 ( s + 3) s s 2 s + 3

L −1

⇒

2 9

+ 13 t − 92 e−3t

2s + 3 ( s + 4)( s + 1) Solution Partial-fraction expansion leads to

22.

2

2s − 3 ( As + B)( s + 1) + C ( s 2 + 4) As + B C = + = ( s 2 + 4)( s + 1) s 2 + 4 s + 1 ( s 2 + 4)( s + 1)

⇒

0 A+C = A+ B 2 = B + 4C = −3

Simultaneous solution gives A = 1, B = 1, C = −1 so that 2s − 3 s +1 −1 s 1 2 −1 = 2 + = 2 + ⋅ 2 + 2 ( s + 4)( s + 1) s + 4 s + 1 s + 4 2 s + 4 s + 1

L −1

⇒

cos 2t + 12 sin 2t − e −t

35

4s + 5 s ( s 2 + 4 s + 5) Solution Forming partial fractions,

23.

2

4s + 5 A B Cs + D ( B + C ) s 3 + ( A + 4 B + D) s 2 + (4 A + 5 B) s + 5 A = + + = s 2 ( s 2 + 4 s + 5) s 2 s s 2 + 4s + 5 s 2 ( s 2 + 4 s + 5)

Then = B+C 0 = A solve A= + 4B + D 0 = B ⇒ = 4 A + 5B 4 = C 5A = 5 D=

1 0 0 −1

Therefore 4s + 5 1 1 = − s 2 ( s 2 + 4 s + 5) s 2 ( s + 2) 2 + 12

L −1

⇒

t − e −2t sin t

s + 10 s ( s + 2 s + 5) Solution Using partial fractions,

24.

2

s + 10 A Bs + C ( A + B) s 2 + (2 A + C ) s + 5 A = + = s ( s 2 + 2 s + 5) s s 2 + 2 s + 5 s ( s 2 + 2 s + 5)

Subsequently,

= A+ B 0 = A 2 solve −2 2A + C = 1 ⇒ B= 5 A = 10 C = −3 Then s + 10 2 2s + 3 2 2( s + 1) + 1 =− =− 2 2 2 s ( s + 1) 2 + 22 s ( s + 2 s + 5) s ( s + 1) + 2

L −1

⇒

1 2 − 2e−t cos 2t − e−t sin 2t 2

36

5s + 8 s ( s + 2) 2 Solution Forming partial fractions,

25.

A2 A1 A( s + 2) 2 + A2 s + A1s ( s + 2) 5s + 8 A = ≡ + + = s ( s + 2) 2 s ( s + 2) 2 s + 2 s ( s + 2) 2

( A + A1 ) s 2 + ( 4 A + A2 + 2 A1 ) s + 4 A s ( s + 2) 2

Then,

= A 2  A + A1 0 =  A2 = 1  4 A + A2 + 2 A1 = 5 ⇒ 4 A = 8 A1 = −2  Therefore, 5s + 8 2 1 −2 ≡ + + 2 2 s ( s + 2) s+2 s ( s + 2)

L −1

⇒

2 + te −2t − 2e −2t

2 s 2 + 3s − 4 ( s + 3)( s 2 + 2 s + 2) Solution Partial-fraction expansion gives

26.

2 s 2 + 3s − 4 A Bs + C ( A + B ) s 2 + (2 A + 3B + C ) s + 2 A + 3C =+ = ( s + 3)( s 2 + 2 s + 2) s + 3 s 2 + 2 s + 2 ( s + 3)( s 2 + 2s + 2)

Comparing coefficients of like powers of s results in

= A+ B 2 = A 1 2 A + 3B + C= 3 ⇒ B= 1 2 A + 3C = −4

C= −2

Finally, 2 s 2 + 3s − 4 1 s−2 1 s +1 1 = + = + − 3⋅ 2 2 2 2 2 s + 3 ( s + 1) + 1 ( s + 3)( s + 2 s + 2) s + 3 ( s + 1) + 1 ( s + 1) 2 + 12

Term-by-term inversion yields

37

e−3t + e−t cos t − 3e−t sin t

In Problems 27 through 32, (a) Solve the IVP. (b) Confirm the result of (a) in MATLAB. 27. 23 x + 2 x = u (t ) − u (t − 1), x(0) = 0 Solution (a) Taking the Laplace transform and using the zero initial condition, yields 1 e− s ( 23 s + 2) X ( s ) = − s s Let G ( s ) =

⇒

3

3

2 2 X (s) = e− s − s ( s + 3) s ( s + 3)

1 so that by partial-fraction expansion we have g (= t) s ( s + 3)

1 (1 − e −3t ) . 3

Next,

 3   3  3 −1 x(t ) = L −1  2  − L −1  2 e− s  = L {G ( s )} − 32 L −1 e− s G ( s ) 2  s ( s + 3)   s ( s + 3) 

{

Using the shift on the t − axis for the second term, we find x(t ) = g (t ) was defined earlier, hence

3 2

}

g (t ) − 32 g (t − 1)u (t − 1) . But

x(t ) = 12 (1 − e −3t ) − 12 1 − e −3(t −1)  u (t − 1) (b) >> x = dsolve('(2/3)*Dx+2*x=heaviside(t)-heaviside(t-1)','x(0)=0'); >> pretty(x) / heaviside(t) heaviside(t - 1) exp(3) -exp(-3 t) | ------------ - ----------------------\ 2 2 exp(3 t) heaviside(t) heaviside(t - 1) exp(3 t) \ - --------------------- + ------------------------- | 2 2 /

28.  x + 2 x + x = u (t − 1) − u (t − 2), x(0) = 0, x (0) = 0 Solution

38 (a) Laplace transformation of the ODE and using the zero initial conditions, yields ( s + 1) 2 X (= s)

Let H ( s ) =

e − s − e −2 s s

⇒

X (= s)

1 1 e− s − e −2 s 2 s ( s + 1) s ( s + 1) 2

1 so that by partial-fraction expansion we have h(t ) =1 − (t + 1)e−t . Next, 2 s ( s + 1)

 1 1 −s  −1  −2 s  x(t ) = L −1  e L e L −1 e− s H ( s ) − L −1 e −2 s H ( s ) −   = 2 2  s ( s + 1)   s ( s + 1) 

{

}

{

}

By the shift on the t − axis, we find x(t ) = h(t − 1)u (t − 1) − h(t − 2)u (t − 2) . Using the expression for h(t ) defined earlier, we have x(t ) = 1 − te − (t −1)  u (t − 1) − 1 − (t − 1)e − (t −2)  u (t − 2) (b) >> x = simplify(dsolve('D2x+2*Dx+x=heaviside(t-1)-heaviside(t-2)','x(0)=0, Dx(0)=0')) x = heaviside(t - 1) - heaviside(t - 2) - heaviside(t - 2)*exp(2 - t) t*heaviside(t - 1)*exp(1 - t) + t*heaviside(t - 2)*exp(2 - t)

1  e −2t , x= 29.  x += (0) 0 , x= (0) 12 4 x Solution (a) Laplace transformation of the ODE and using the initial conditions leads to

1 1 ( s 2 + 14 s ) X ( s= )− 2 s+2

⇒

= X (s)

s+4 2 s ( s + 2)( s + 14 )

Finally, using partial-fraction expansion, we find   2 −2t 30 −t /4 s+4 x(t ) =L −1  =e − e +4 1  7  2 s ( s + 2)( s + 4 )  7

(b) >> x = simplify(dsolve('D2x + (1/4)*Dx = exp(-2*t)','x(0)=0, Dx(0)=1/2')) x = (2*exp(-2*t))/7 - (30*exp(-t/4))/7 + 4

39

30.  x + x + 12 x= δ (t − 1) , x(0)= 0 , x (0)= 12 Solution (a) Laplace transform of the ODE and using the initial conditions, we find s 2 X ( s ) − 12 + sX ( s ) + 12 X = ( s) e− s

Let H ( s ) =

⇒

X= (s)

1 1 −s 2 + e 2 2 1 s +s+ 2 s + s + 12

1 so that h(t ) = 2e −t /2 sin(t / 2) . Also, 1 s +s+ 2 2

{

}

x(t ) = L −1 H ( s )e − s + 12 L −1 { H ( s )} = h(t − 1)u (t − 1) + 12 h(t ) = 2e − (t −1)/2 sin( 12 (t − 1))u (t − 1) + e −t /2 sin(t / 2)

(b) >> x = simplify(dsolve('D2x + Dx + (1/2)*x = dirac(t-1)','x(0)=0,Dx(0)=1/2'))

x = exp(-t/2)*(sin(t/2) + 2*heaviside(t - 1)*exp(1/2)*sin(t/2 - 1/2))

31.  x + 3 x + = 2 x sin( 23 t ), x(0) = 0, x (0) = 3 Solution (a) Taking the Laplace transform of the ODE and using the given initial conditions, yields ( s 2 + 3s + 2) X (= s)

2 3

s 2 + 94

+3

⇒

X (= s)

3s 2 + 2 ( s + 1)( s + 2)( s 2 + 94 )

By partial-fraction expansion we find 3s 2 + 2 3.4615 −3.1500 −0.3115s + 0.1615 X (s) = = + + 2 s +1 s+2 ( s + 1)( s + 2)( s + 94 ) s 2 + 94 Finally, x(t ) =3.4615e −t − 3.1500e −2t − 0.3115cos( 23 t ) + 0.2423sin( 23 t ) (b)

40

>> x=simplify(dsolve('D2x+3*Dx+2*x=sin(2*t/3)','x(0)=0,Dx(0)=3')) x = (45*exp(-t))/13 - (81*cos((2*t)/3))/260 - (63*exp(-2*t))/20 + (63*sin((2*t)/3))/260

= 0, x (0) = 0 x + 2 x + 34= x e−t /2 , x(0) 32.  Solution (a) Taking the Laplace transform of the ODE and using the zero initial conditions, yields ( s 2 + 2 s + 34 ) X ( s= )

1 s + 12

⇒

X ( s= )

1 (s +

=

1 )2 ( s + 3 ) 2 2

−1 1 1 + + 2 1 1 s + 2 (s + 2 ) s + 32

Inversion yields x(t ) = −e −t /2 + te −t /2 + e−3t /2 . (b) >> x=simplify(dsolve('D2x+2*Dx+(3/4)*x=exp(-t/2)','x(0)=0,Dx(0)=0')) x = exp(-(3*t)/2)*(t*exp(t) - exp(t) + 1)

In Problems 33 through 38 decide whether the final-value theorem is applicable, and if so, find xss . s 33. X ( s ) = (4 s + 1)( s 2 + 3) Solution The poles are at − 14 , ± j 3 , two of which are located on the imaginary axis. Therefore, the FVT does not apply. 34. X ( s ) =

2 s + 13 ( s + 13 )( s 2 + s + 1)

Solution The poles are at − 13 , − 12 ±

3 2

j hence the FVT applies.

  s (2 s + 13 ) = xss lim = sX ( s )} lim  = {  0 s →0 s →0 ( s + 1 )( s 2 + s + 1)   3 35. X ( s ) =

s+2 ( s + 1)3 ( s + 3)

41 Solution The poles are at −1, −1, −1, −3 hence the FVT applies.

 s ( s + 2)  = xss lim = sX ( s )} lim  = {  0 s →0 s →0 ( s + 1)3 ( s + 3)  

36. X ( s ) =

s +1 s ( s + 3)( s + 2) 2

Solution The poles are at 0, 0, −3, −2 . Since the pole at the origin is repeated, the FVT does not apply. 37. X ( s ) =

4s 2 + 1 s (4 s 2 + 4 s + 2)

Solution The poles are at 0, − 12 ± 12 j hence the FVT applies.

 4 s 2 + 1  1 = = xss lim {sX ( s)} lim  =  s →0 s →0  4 s 2 + 4 s + 2    2 38. X ( s ) =

2 5

s +1

s ( s + 2 s + 2) 2 2

Solution The poles are at 0, −1 ± j hence the FVT applies. 2 s +1   1 = = xss lim sX ( s )} lim  2 5= {  2 s →0 s →0  ( s + 2 s + 2)    4

In Problems 39 through 42, evaluate x(0+ ) using the initial-value theorem. 39. X ( s ) =

3s 2 + 1 ( s + 1)( 23 s + 2) 2

Solution  s (3s 2 + 1)  27 = x(0+ ) lim = sX ( s )} lim  = {  s →∞ s →∞ ( s + 1)( 2 s + 2) 2   4 3

40. X ( s ) = Solution

s2 + 1 ( 12 s + 1)(9 s 2 + 6 s + 2)

42

  2 s ( s 2 + 1) = = = x(0 ) lim sX ( s )} lim  {  s →∞ s →∞ ( 1 s + 1)(9 s 2 + 6 s + 2)  2  9 +

41. X ( s ) =

s ( s + 4) 2 ( s + 1)( s + 2)( s + 3) 2

Solution

s 2 ( s + 4) 2   = = = x(0+ ) lim sX ( s )} lim  {  0 2 s →∞ s →∞  ( s + 1)( s + 2)( s + 3)    42. X ( s ) =

1 s4 + 1 4 3 2

s (2 s + 9)

Solution

 1 s 4 + 1  1 4 = x(0 ) lim = {sX ( s)} slim  =  →∞ s 2 (2 s 2 + 9) s →∞   8 +

Review Problems In Problems 1 through 4, perform the operations and express the result in rectangular form. 1. e( −1+ 2 j )/(1+ j ) Solution Noting

−1 + 2 j 1 − j 1 + 3 j × = 1+ j 1− j 2

we have e( −1+ 2 j )/(1+ j ) = e(1+3 j )/2 = e1/2 (cos 32 + j sin 32 ) = 0.1166 + 1.6446j 2. e(0.15−0.85j )

8

Solution We first write 0.15 − 0.85j = 0.8631e −1.3961j so that (0.15 − 0.85j )8 = 0.3081e −11.1690j = 0.0532 + 0.3034j

Then 0.85j ) j e(0.15−= e0.0532+0.3034 = 1.0065 + 0.3151j 8

43

3.

( 23 + 12 j ) 4

(1 + 34 j )(−1 + 13 j ) Solution ( 23 + 12 j ) 4 −0.4066 + 0.2593j (0.8333e0.6435j ) 4 0.4823e2.5740j = = = 3 1 −1.25 − 0.4167j (1 + 4 j )(−1 + 3 j ) −1.25 − 0.4167j −1.25 − 0.4167j This yields 0.2306 − 0.2843j .

(1 − 23 j ) 4 ( 23 + j )3 Solution 4.

(1 − 32 j ) 4 (1.2019e −0.5880j ) 4 1.2019e −2.3520j = = = 1.2019e −5.3004j = 3 0.9828j 3 2.9484j 2 ( 3 + j) (1.2019e ) e 5. Find all possible values of ( 3 + 13 j ) . Solution First,

2 3

+j

4/3

( 3 + 13 j )

4/3

(

= ((3 + 13 j ) 4 )1/3 = (3.0185e0.1107j ) 4

)

1/3

= (83.0123e0.4428j )1/3

The three values are generated by 0.4428 + 2kπ 0.4428 + 2kπ   83.01231/3 cos 0,1, 2 + j sin  , k= 3 3  

which yields 4.3149 + 0.6413j , − 2.7130 + 3.4160j , − 1.6018 − 4.0575j

6. Find all possible roots of the following equation:

(2 − j ) z 4 + 1 − 2 j =0 Solution The equation is re-written as z4 =

−1 + 2 j −1 + 2 j 4 + 3 j × = 2− j 2− j 5

This in turn is equivalent to = z ( 54 + 53 j )1/4 . Therefore, the four values of z are generated by

44

cos

0.6435 + 2kπ 0.6435 + 2kπ , k= 0,1, 2,3 + j sin 4 4

which gives 0.9871 + 0.1602j , − 0.1602 + 0.9871j , − 0.9871 − 0.1602j , 0.1602 − 0.9871j

In Problems 7 through 10, solve the IVP. Do not use Laplace transformation. 7. y − y tan t = t + 1 , y (0) = 1 Solution By Eq. (2.12),  eln cos t (t + 1) dt + c  =e − ln cos t [ ∫ (t + 1) cos t dt + c ] ∫  1 = [cos t + (t + 1) sin t + c ] cos t Application of the initial condition yields c = 0 , hence h =− ∫ tan tdt =ln cos t

,

y (t ) =e

− ln cos t

y (t ) =1 + (t + 1) tan t

8. y = (1 − y ) sin t , y ( 12 π ) = 2 Solution Rewrite the ODE as dy = (1 − y ) sin t dt

⇒

∫

dy = ∫ sin tdt 1− y

⇒

− ln 1 − y = − cos t + c

This implies

 e − c if 1 − y > 0 t = 1 − y kecos= , k  −c −e if 1 − y < 0 Application of the initial condition yields k = −1 , hence y (t ) = 1 + ecos t .

9.  y − y = t 2 − cos t , y (0) = 12 , y (0) = 0 Solution

45 We first consider the homogeneous equation  y− y = 0 . The characteristic equation λ 2 − 1 =0 has roots λ1,2 = ±1 so the homogeneous solution is yh = (t ) c1et + c2 e − t . Since f (t ) contains a 2nd-degree polynomial and a cosine, neither one coinciding with the independent parts in yh , no special case applies. Therefore, we choose

y p = K 2t 2 + K1t + K 0 + M 1 cos t + M 2 sin t and substitute into the ODE to obtain K 2 = −1 , K1 = 0 , K 0 = −2 , M 1 = 12 , and M 2 = 0 so that

y p =−t 2 − 2 + 12 cos t and a general solution is y= c1et + c2 e − t − t 2 − 2 + 12 cos t . Initial conditions give  y (0) = c1 + c2 − 32 =   y (0) = c1 − c2 = 0

1 2

⇒

c1 = 1 c2 = 1

⇒ y = et + e − t − t 2 − 2 + 12 cos t

10.  y + 4= y et sin t , y (0) = 0, y (0) = 101 Solution We first consider the homogeneous equation  y + 4y = 0 . The characteristic equation has roots yh (t ) c1 cos 2t + c2 sin 2t . The particular solution is λ1,2 = ±2i so the homogeneous solution is= picked as

= y p et ( K1 cos t + K 2 sin t ) and substituted into the ODE to obtain K1 = − 101 and K 2 = 102 so that y p =

1 10

et ( − cos t + 2sin t ) .

A general solution is y (t= ) c1 cos 2t + c2 sin 2t + 101 et ( − cos t + 2sin t ) . By initial conditions,

 y (0) = c1 − 101 = 0  1 1  y (0) = 2c2 + 10 = 10

⇒

c1 = 101 c2 = 0

⇒= y

1 10

t cos 2t + e ( − cos t + 2sin t ) 

11. Express the signal in Figure 2.19 in terms of unit-step functions, and subsequently find its Laplace transform using the shift on t − axis. Solution The signal g (t ) is reconstructed as g (t )= u (t ) − u (t − 1) + (−2t + 3)u (t − 1) + (−2t + 3)u (t − 32 )

46 Term-by-term Laplace transformation using the shift on t − axis yields G (s) =

1 1 − s  2 1  − s 2 −3 s /2 − e + − 2 + e − 2 e s s s s  s

1 2 − s 2 −3 s /2 . This simplifies to G ( s ) = − e − 2e s s2 s

g (t ) 1

t 1

3 2

Figure 2.19 Problem 11. 12. Find the Laplace transform of the periodic function in Figure 2.20. Solution Noting P = 32 , we find

G (s) =

3/2 1 e − st g (t )dt −3 s /2 ∫ 1− e 0

The integral is evaluated as 3/2

1

3/2

0

0

1

− st − st − st ∫ e g (t )dt = ∫ e dt + ∫ e (−2t + 3)dt =

1 2 (1 − e − s ) + 2 (e −3 s /2 − e − s ) s s

Simplifying and substituting back, we find 1 2 −3 s /2 − s + 2 (e −e ) s s G ( s) = 1 − e −3 s /2

47

g (t ) 1

t 1

3 2

5 2

3

Figure 2.20 Problem 12. 13. Find the response x(t ) of a system governed by 3 x + 7 x + 2 x = tu (t − 1) , x(0) = 0, x (0) = 0

Solution Taking the Laplace transform of the ODE, using the zero ICs, we find −s

H (s)  1 1  ⇒ X (s) =  2 2 + 2  s (3s + 7 s + 2) s (3s + 7 s + 2) 

−s

e e (3s + 7 s + 2) X ( s ) = 2 + s s 2

Inverse Laplace transforms of the two terms in H ( s ) are 

 1 1 1 −2 t + 95 e − t /3 − 74  =− 2t 20 e s (3 s + 7 s + 2)  

L −1 

2

2



 1 =  s (3s + 7 s + 2) 

L −1 

2

1 10

e −2t − 53 e − t /3 + 12

so that 1 1 −2 t h(t ) = L −1 { H ( s )} =+ + 56 e − t /3 − 54 2t 20 e

Finally, by the shift on t − axis, we have x(t ) = L −1 { H ( s )e − s } = h(t − 1)u (t − 1) =  12 (t − 1) + 201 e −2(t −1) + 65 e − (t −1)/3 − 54  u (t − 1)

14. Evaluate the convolution t ∗ u (t − a ) . Solution By definition of convolution,

  −s e  

48

t

∫ u (t − a)(t − t )dt

t ∗ u (t − a )=

0

0 if τ < a Since u (τ − a ) = , the above reduces to  1 if τ > a t

t ∗ u (t − a ) =

∫ (t − t )dt =

1 2

(t − a ) 2

a

15. Evaluate the convolution ( g ∗ h)(t ) where g (t ) is shown in Figure 2.21 and h(t ) = sin t . Solution By definition of convolution, t

( g ∗ h= )(t )

∫ g (t ) sin(t − t )dt

(*)

0

The analytical description of g (t ) is 1 − t g (t ) =  0

if if

0 < t ≤1 t ≥1

Since g (t ) exhibits different behaviors over 0 < t ≤ 1 and t ≥ 1 , the two time intervals must be examined separately. o If 0 < t ≤ 1 , then only the first segment of g (t ) contributes to the solution, and Eq. (*) gives t

t ∫ (1 − t ) sin(t − t )d=

sin t − cos t − t + 1

0

o

If t ≥ 1 , then the definite integral in Eq. (*) can be decomposed as t

1

t

0

0

1

)dt ∫ g (t ) sin(t − t )dt + ∫ g (t ) sin(t − t )dt ∫ g (t ) sin(t − t=

In the first integral g (ττ ) = 1 − , while in the second integral g (τ ) = 0 . Therefore,

49 t

1

0

0

∫ g (t ) sin(t − t )dt =∫ g (t ) sin(t − t )dt =sin t − cos t − sin(t − 1) In summary,  sin t − cos t − t + 1 if 0 < t ≤ 1 ( g ∗ h)(t ) =  sin t − cos t − sin(t − 1) if t ≥ 1

g (t ) 1

t 1

Figure 2.21 Problem 15.

16. (a) Using partial-fraction expansion, find

 4 s 3 + 8s 2 + 2  L  2 2   s (4 s + 1)  −1

(b)

Confirm the result of (a) in MATLAB.

Solution (a) Partial-fraction expansion results in 4 s 3 + 8s 2 + 2 A B Cs + D 2 s ≡ 2+ + 2 = 2+ 2 1 2 2 s (4 s + 1) s s 4s + 1 s s +4

 4 s 3 + 8s 2 + 2  1 Therefore L −1  2 2 = 2t + cos( 2 t ) . s (4 s 1) +   (b) >> syms s >> ilaplace((4*s^3+8*s^2+2)/s^2/(4*s^2+1))

50

ans = 2*t + cos(t/2)

17. (a) Find

 e − s (4 s 2 + 12 s + 5)  L   2  2 s (4 s + 8s + 5)  −1

(b) Confirm the result of (a) in MATLAB. Solution (a) Let 4 s 2 + 12 s + 5 Partial-fraction expansion 2 1 = H (s) = + 2 2 2 s (4 s + 8s + 5) 4( s + 1) + 1 2 s so that −1 = h(t ) L = {H ( s)} e−t sin( 12 t ) + 12

Shift on t − axis yields

 e − s (4 s 2 + 12 s + 5)  −1 = e − s H ( s ) e − (t −1) sin( 12 (t − 1))u (t − 1) + 12 u (t − 1)  L = 2  2 s (4 s + 8s + 5) 

L −1 

{

}

(b) >> syms s >> ilaplace(exp(-s)*(4*s^2+12*s+5)/(2*s)/(4*s^2+8*s+5)) ans = heaviside(t - 1)/2 + heaviside(t - 1)*exp(1 - t)*sin(t/2 - 1/2)

18. Consider

s 2 + 3s + 1 X (s) = s ( s + 1) 2 (a) Using the final-value theorem, if applicable, evaluate xss . (b) Confirm the result of (a) by evaluating lim x(t ) . t →∞

Solution (a) Poles of X ( s ) are at 0, −1, −1 so that the FVT is applicable:

51

s 2 + 3s + 1 = xss lim = sX ( s )} lim= 1 { s →0 s →0 ( s + 1) 2 (b) Since x(= t ) te − t + 1 , we find lim x(t ) = 1 , confirming the result of (a). t →∞

19. Consider

2(25s 2 − 30 s + 113) X (s) = s (25s 2 + 10 s + 226) (a) Using the initial-value theorem, evaluate x(0+ ) . (b) Confirm the result of (a) by evaluating lim x(t ) . t →0

Solution (a) 2(25s 2 − 30 s + 113) = x(0 ) lim = = 2 {sX ( s)} lim s →∞ s →0 25s 2 + 10 s + 226 +

(b) Since x= (t ) e −0.2t (cos 3t − sin 3t ) + 1 , we find lim x(t ) = 2 , confirming the result of (a). t →0

20. Consider

0.2s + 0.3 s (3s 2 + 2.5s + 0.5) (a) Evaluate xss using the final-value theorem. (b) Confirm the result of (a) by evaluating lim x (t ) . X (s) =

t →∞

Solution (a) s (0.2s + 0.3) = = = 0 xss lim {s [ sX (s)]} lim s →0 s →0 3s 2 + 2.5s + 0.5

(b) Since x(t ) = 54 e − t /2 − 75 e − t /3 + 53 , we have= x (t ) t ) lim lim x (= t →0

x →∞

{

7 15

7 15

e − t /3 − 52 e − t /2 and

}

e − t /3 − 52 e − t /2= 0

52

Problem Set 3.1 In Problems 1 through 8, perform the indicated operations, if defined, for the vectors and matrices below.

 0 −2 1  1  −3 1 1 1     A=   , w= 0  0 4 5 , v =  1 −3 3  , B = −2     −1  2 1 2    1. wT A Solution wT A = [ −2 −3 −1]

2. BAw Solution 1 BAw =   −8 3. ( ABT ) v Solution  5    T ( AB ) v=  −9  −31  

4. (B − vwT ) A Solution 7 3 5 (B − vwT ) A = 10 −13 20    5. Aw − BT v Solution 2    T Aw − B v = 5  9   

53 6. A(BA + vwT ) Solution Operation is undefined! A is 3 × 3 , but BA + vwT is 2 × 3 . 7. vT Bw Solution vT Bw = 6 8. ( A 2 + BT B)w Solution  16    2 T ( A + B B)w = −19   −33  

In Problems 9 through 12, find the rank by inspecting the row-echelon form of the matrix.

 3 −2 0  9. A =  −1 2 4   4 −4 −3 Solution  3 −2 0   −1 2 4   −1 2 4   −1 2 4   −1 2 4  →  3 −2 0  →  0 4 12  →  0 1 3  = A=         REF( A)  4 −4 −3  4 −4 −3  0 4 13  0 0 1 

2 10. A =  3 1 Solution  2 −1 A = 3 3 1 −2

3 ⇒ rank( A) =

−1 1  3 2  −2 0 

1  1 −2 0  1 −2 0  1 −2 0    2  →  3 3 2  → 0 9 2  → 0 9 2  =REF( A) 0   2 −1 1  0 3 1  0 0 13 

⇒ rank( A) =3

54 1 3 11. A =  0  1 Solution 1 −1  3 −2 A = 0 2  1 1

−1 −1 −2 0 2 4 1 3 −1 0 4 3

0 1  1  1

0  1 −1 −1 0  1 −1 −1 0  1 −1 −1 0  1  0 1 3 1  0 1 3 1  0 1 3 1  → → → =REF( A) 1  0 2 4 1  0 0 −2 −1 0 0 −2 −1        1  0 −2 −4 −1 0 0 2 1  0 0 0 0 

Therefore, rank( A) = 3 .

1 −1 3 2 12. A =  2 3  4 1 Solution 1 3  A= 2  4

2 0 −2 2

0 1  1  1

−1 2 2 0 3 −2 1 2

0  1 −1 2 1  0 5 −6 → 1  0 5 −6   1  0 5 −6

0  1 −1 2 1  0 5 −6 → 1  0 0 0   1  0 0 0

0 1  = REF( A) 0  0

Therefore, rank( A) = 2 . 13. Determine the value of a such that rank( A) = 2 .  2 1 0 −2  1 3 −1 4   A= 1 −2 1 −6     5 a −1 0  Solution 3 −1 4  1 3 −1 4   2 1 0 −2  1 3 −1 4  1 1 3 −1 4   2 1 0 −2  0   −5 2 −10  0 −5 2 −10       A= → → → 1 −2 1 −6  1 −2 1 −6  0 5 −2 10  0 0 0 0          0   5 a −1 0   5 a −1 0  0 a − 15 4 −20  0 a − 5 0 Therefore, a = 5 .

55

14. Given 0 1 0 = A = 0 0 1  , B  −1 −2 −1

0    0  1  2

find the rank of

B AB A 2B    Solution 0   12   1   1 2 AB =  2  , A B= − 2  − 1  − 1   2  2

⇒

0 0  2 1 B AB A B  =   0 2 1 − 1 2 2

1  2  − 12  − 12 

The rank of this matrix is readily seen to be 3.

In Problems 15 through 18, evaluate the determinant. −1 6 15. 0 1 Solution −1 2 6 0 0 1 1 −3

2 0 1

3 1 1 −3 5 −1 −3 0 1 3 1 −1 3 1 −1 2 3 2 3 1 1 −3 using the 4th row = − 0 1 −3 − 3 6 1 −3 + 6 0 1 =−(18) − 3(34) + (−41) =−161 5 −1 1 5 −1 0 5 −1 0 1 5 0 1

0 0 0 b 16. a −1 0 0 Solution 0 0 −3

−3 −2 0 1 , a, b = parameters a 2 0 2a −2

0 0 −3 0 b 0 1 = 2= a 0 b 0 6a 2b a −1 a 2 a −1 a 0 0 0 2a

56

s −1 0 0 0 s 0 −1 17. , s = parameter 1 2 s +1 0 0 1 3 s+2 Solution s −1 0 0 −1 −1 s 0 0 0 −1 using the first row 0 s 0 = s 2 s +1 0 + 1 s +1 0 1 2 s +1 0 1 3 s+2 0 3 s+2 0 1 3 s+2

= s ( s 3 + 3s 2 + 3s − 5) + (−3) = s 4 + 3s 3 + 3s 2 − 5s − 3 5 −2 2 4 18. 0 0 0 0 0 0 Solution 5 -2 0 2 4 0 0 0 -3 0 0 0 0 0 1

0 0 0 0 0 10 −3 0 0 0 −1 2 1 3 5 0 0 0 10 block upper-triangular (24)(33) 792 = = 0 0 -1 3

2×2 and 3×3

2 5

In Problems 19 through 24, find the inverse of the matrix.  cos a sin a  19. A = =  , a parameter  − sin a cos a  Solution

cos a A = cos 2 αα + sin 2 = 1 . The adjoint matrix is formed as adj( A) =   sin a − sin  cos αα A −1 =  cos   sin αα

− sin a  . Finally, cos a 

57

1 0 2  = 20. A 0 2 −1 0 0 3  Solution Upper-triangular matrix; A = 6 . Then, 6 0 −4  6 0 −4  1   −1 = = adj( A) 0 3 = 1 , A 0 3 1   6 0 0 2  0 0 2 

1 0 − 23   1 1  0 2 6  0 0 1  3  

1 −1 0  21. A =  3 0 0   2 0 −2  Solution A = −6 . Then, 0 −2 0  6 −2 0  , adj( A) =   0 −2 3

22. A

0 −2 0   0 1   −1 A =6 −2 0  =  −1 −6 0 −2 3  0

1 3 1 3 1 3

0   0  − 12 

1  a + 1 0 = a 2  , a parameter  0  1 0 a + 2 

Solution A = a 3 + 3a 2 + a = a (a 2 + 3a + 1) . Next,

0 −a   a (a + 2)   = adj( A)  2 a 2 + 3a + 1 −2(a + 1)   −a 0 a (a + 1)   Finally,

−a  0  a (a + 2)   1 = A −1 a 2 + 3a + 1 −2(a + 1)   2 2 a (a + 3a + 1)  0 a (a + 1)   −a

58

0 0   a  a+3 0  , a = parameter 23. A =  0  a + 1 a + 2 3(a + 1)  Solution Lower-triangular matrix; A = 3a (a + 1)(a + 3) . Then, 0 0   3(a + 1)(a + 3)  = adj( A)  0 3a (a + 1) 0   −(a + 1)(a + 3) − a (a + 2) a (a + 3)  Finally,

A −1

24. A

0 0   3(a + 1)(a + 3) 1  0 3a (a + 1) 0  3a (a + 1)(a + 3)   −(a + 1)(a + 3) −a (a + 2) a (a + 3) 

 − L1 sin θ1 − L2 sin (θ1 + θ 2 ) − L2 sin (θ1 + θ 2 )  =  , L1 , L2 , θ1 , θ 2 parameters  L1 cos θ1 + L2 cos (θ1 + θ 2 ) L2 cos (θ1 + θ 2 ) 

Solution A −1 =

1 L1 L2 sin θ 2

L2 cos (θ1 + θ 2 ) L2 sin (θ1 + θ 2 )      − L1 cos θ1 − L2 cos (θ1 + θ 2 ) − L1 sin θ1 − L2 sin (θ1 + θ 2 ) 

25. Prove that the inverse of a (non-singular) symmetric matrix is symmetric. Solution Suppose A is symmetric so that AT = A . Then

(= A −1 )T (= AT ) −1 A −1 This proves that A −1 is symmetric, as asserted. 26. Given A n×n and scalar k ≠ 0 , show that (kA) −1 =

1 −1 A . k

Solution By definition, we have (kA) −1 =

1 adj(kA) . But by one of the determinant properties listed kA

earlier, kA = k n A . It can also be shown that adj(kA) = k n −1adj( A) . As a result,

59

= (kA) −1

1 1 1 1 1 −1 = adj(kA) = k n −1adj( A) = adj( A) A n kA k A k A k

Problem Set 3.2 In Problems 1 through 6, solve the linear system Ax = b using Gauss elimination.

 2 2 −1 1. A =  3 1 4  , b =  −1 2 5  Solution  2 2 −1 1    Swap 1st and 3 1 4 − 2   3rd→ rows  −1 2 5 −8 

1   −2     −8 −1 2 5 −8     3 1 4 −2   2 2 −1 1 

More EROs

→

 −1 2 5 −8     0 7 19 −26  51 51   0 0 − 7 7 

 −1 0 4.2  10.4      2. A = −4.6   2.3 −3 1.5  , b =    5 3.2 1   −4.8 Solution  −1 0 4.2 10.4  −1 0 4.2 10.4    EROs    2.3 −3 1.5 −4.6  →  0 −3 11.16 19.32   5 3.2 1 −4.8  0 0 33.9040 67.9040 

Back substitution

1.4 −2.1  3  4.10      1  , b= 3. A =  11.9   1.5 −3.2   0.75 2.5 −2  −4.75 Solution  3 1.4 −2.1 4.10   3 1.4 −2.1 4.10    EROs   1 11.9  → 0 −3.9 2.05 9.85   1.5 −3.2 0 −2 −4.75 −0.3449 −0.3449  0.75 2.5 0

⇒

Back substitution

⇒

−2    x= 1   2

Back substitution

⇒

3   x= −2  1  

1   x= −1   −1

60

 3 0 −2  0      4. A = 1 −4 −3 , b = 1     2 3 2  4  Solution  3 0 −2 0  1 −4 −3 1    Swap 1st two rows   → −3  0 12 7 1 −4 −3 1  and more  EROs  2 3 2 4  0 0 1.5833 4.75  2 −1 0 3   1 3 1 −4   , b 5. A = 2 0 3 5    −1 −2 4 6  Solution  2 −1 0 3 10    EROs  1 3 1 −4 −14  → 2 0 3 5 9     −1 −2 4 6 11 

Back substitution

⇒

2   x =−  2 3  

 10  −14       9   11 

 2 −1 0 3 10    −5.5 −19  1  0 3.5  0 0 4.7143 3.5714 2.4286    0 1.5152 3.0303   0 0

 −2 4 3 1   3 2 −6 −1  , b 6. A =  1 −2 −3 2     4 8 9 −3 Solution −2 4 3 1 7  1   EROs   3 2 −6 −1 −6  → 0  1 −2 −3 2 1  0     4 8 9 −3 −3 0

Back substitution

⇒

7 −6      1  −3

−2 −3 2 1   8 3 −7 −9  0 −3 5 9   0 0 28 56 

Back substitution

⇒

In Problems 7 through 10, solve the linear system Ax = b using (a) The inverse of the coefficient matrix. (b) The "\" operator in MATLAB.

−1    12  x= 1 3  2 

1 −2    x=    −1  2 

61

 2 1 −2   4a      2  , b = a  , a parameter = 7. A  3 0 =    −1 2 −1 0 Solution  −4 −3 2  4a   a  1      −1 1 −4 −10   a =  0  (a) x= A b=  −19  6 −5 −3   0  −a  (b) >> syms a >> A = [2 1 -2;3 0 2;-1 2 -1]; b = [4*a;a;0]; x = A\b x = a 0 -a

 a − 1 2a + 1 −1   5a      parameter −2a −1  , b = 8. A =  −4a  , a =  1    −1 2a + 1 3a + 1 7 a + 2  Solution 1 − 6a 2 −6a 2 − 7 a − 2 −4a − 1   5a  −1    (a) x= A −1b= −3a a − 2   −4a = 3a 2 − 2a − 2 3  6a + 2a + 2  1 −a (2a + 1) −2a 2 − 1 7 a + 2  

1    2  1   

(b) >> syms a >> A = [a-1 2*a+1 -1;1 -2*a -1;-1 2*a+1 3*a+1]; b = [5*a;-4*a;7*a+2]; x = A\b x = 1 2 1

9. A

0 2 2 0 1 3 0 0  2 2 = , b  0 0 − 12 1     0 0 1 −1

0  1     3 − 2   2 

62 Solution  − 43  1  14 −1 (a) = x A= b −1  0  0

0  0   −1 0 0   1  =  0 −2 −2  − 32   0 −2 −1  2  1

0

−1   1   1 −1

(b) >> A = [2 2 0 0;1/2 3/2 0 0;0 0 -1/2 1;0 0 1 -1]; b = [0;1;-3/2;2]; x = A\b x = -1 1 1 -1

3 2 0 0   −1   1 4 −1 0   1 2 3  , b  − 6  10. A = = 1  1 0 32 1  − 2     0 0 1 −1  2  Solution 5 2 2   −1   − 103  9 − 152 −3 −3  − 16  −2  4 −1 (a) = x A= b =  −2 −3 −3 − 12  11  43  4  −2 −3 52   2   3

−1 1     1 −1

(b) >> A = [3 2 0 0;1/2 4/3 -1 0;1 0 3/2 1;0 0 1 -1]; b = [-1;-1/6;-1/2;2]; >> x = A\b x = -1.0000 1.0000 1.0000 -1.0000

In Problems 11 through 14, (a) Solve the linear system Ax = b using Cramer’s rule. (b) Repeat (a) in MATLAB.  1 5 0 13     11. A = 0  −1 2 1  , b =    −2 5 3 1

63 Solution (a) The determinant of the coefficient matrix is 1 5 0

D= −1 2 1 = 6 −2 5 3 Since D ≠ 0 , we proceed by evaluating 13 5 0 1

13 0 1 5 13 D1 = 0 2 1= 18 , D2 = −1 0 1 = 12 , D3 = −1 2 0 = −6 1 5 3 −2 1 3 −2 5 1 Subsequently, 18 12 −6 x1 = = 3 , x2 = = 2 , x3 = = −1 6 6 6 (b)

>> >> >> >> >>

A = [1 5 0;-1 2 1;-2 5 3]; b = [13;0;1]; d = det(A); A1=A; A1(:,1)=b; d1=det(A1); A2=A; A2(:,2)=b; d2=det(A2); A3=A; A3(:,3)=b; d3=det(A3); x1=d1/d; x2=d2/d; x3=d3/d; x=[x1;x2;x3]

x = 3 2 -1

 2 12 −1 19      12. A=  1 1 −2  , b=  26  −8  1 −1 1    2  2 Solution (a) The determinant of the coefficient matrix is 2 12 −1 9 D =1 1 −2 =− 4 1 −1 12 2 Since D ≠ 0 , we find 19 12 −1

D1 =26

1

−8 −1

−2 =−9

,

1 2

2 19 −1 27 D2 =1 26 −2 =− 2 1 −8 12 2

,

2 D3 =1 1 2

so that x1 =

D1 = 4 D

,

x2 =

D2 = 6 D

,

x3 =

D3 = −8 D

1 2

1

19 26 = 18

−1 −8

64 (b) >> A = [2 1/2 -1;1 1 -2;1/2 -1 1/2]; b = [19;26;-8]; d = det(A); >> A1=A; A1(:,1)=b; d1=det(A1); >> A2=A; A2(:,2)=b; d2=det(A2); >> A3=A; A3(:,3)=b; d3=det(A3); >> x1=d1/d; x2=d2/d; x3=d3/d; x=[x1;x2;x3] x = 4.0000 6.0000 -8.0000

s2 + s + 1 − 1  F  3 = 3 13. A  , = b  =  , s, F parameters 1 s + 13  0  − 3 Solution (a) The determinant of the coefficient matrix is

s 2 + s + 13 D= − 13

− 13 s+

1 3

=s 3 + 43 s 2 + 23 s

We proceed by evaluating

F − 13 D1 == ( s + 13 ) F , 0 s + 13

s 2 + s + 13 F 1 D2 = = 3 F − 13 0

so that

= x1

1 ( s + 13 ) F 3F = , x 2 3 3 4 2 2 4 2 s +3s +3s s + 3 s + 23 s

(b) >> >> >> >> >>

syms s F A = [s^2+s+1/3 -1/3;-1/3 s+1/3]; b = [F;0]; d = det(A); A1=A; A1(:,1)=b; d1=det(A1); A2=A; A2(:,2)=b; d2=det(A2); x1=d1/d; x2=d2/d; x=[x1;x2]

x = (F*(3*s + 1))/(s*(3*s^2 + 4*s + 2)) F/(s*(3*s^2 + 4*s + 2))

2 0 0 3  1 − 4 1 0 2 3  , b = 14. A =  −1 0 32 1    0 1 1 0

7  − 8   3  3 − 4   12 

65 Solution (a) The determinant of the coefficient matrix is 3 D=

1 2

2 − 34

−1 0

0 0

0 0 1 0 9 = − 3 1 2 2 1 1

Since D ≠ 0 , we proceed by evaluating

7

2

0 0 3 7 0 0 −8 −8 −4 1 1 0 1 0 9 3 3 2 3 D1 = = − , D = = −9 2 −3 3 −3 0 2 1 −1 4 23 1 2 4 1 0 1 1 0 12 1 1 2 3

2

7

0 3 2 0 7 −8 −4 −4 1 1 0 9 1 −38 9 2 3 3 2 3 = = = − D3 = D , 4 −3 −3 3 −1 0 4 1 4 −1 0 2 4 2 0 0 12 1 0 0 1 12 Finally, x1 == 1 , x2 2 , x3 = − 12 , x4 = 1 (b) >> >> >> >> >> >> >>

A = [3 2 0 0;1/2 -4/3 1 0;-1 0 3/2 1;0 0 1 1]; b = [7;-8/3;-3/4;1/2]; d = det(A); A1=A; A1(:,1)=b; d1=det(A1); A2=A; A2(:,2)=b; d2=det(A2); A3=A; A3(:,3)=b; d3=det(A3); A4=A; A4(:,4)=b; d4=det(A4); x1=d1/d; x2=d2/d; x3=d3/d; x4=d4/d; x=[x1;x2;x3;x4]

x = 1.0000 2.0000 -0.5000 1.0000

In Problems 15 through 17, solve the homogeneous linear system Ax = 0 , where the components of x are x1 , x2 , ... , xn .

 −1 3 4  A  2 −2 1  15.=  1 1 5 

66 Solution Since the coefficient matrix is singular, the system has infinitely many solutions. The rowechelon from of the system is obtained as  2 −2 1 0    0 4 9 0  0 0 0 0  The last row being entirely zero suggests that there is one free variable. The second row yields 4 x2 + 9 x3 = 0 . Choosing x3 as the free variable, this yields x2 = − 94 x3 . Using this information

x . Therefore, in the first row, we find x1 = − 11 4 3  x1    x=  x2 = x   3

− 11  4  9  − 4  x3  1   

−11   For example, letting x3 = 4 generates one of the infinitely many solutions, as  −9  .  4     2 0 1 16. A =  −1 1 3  2 −2 3 Solution Since the coefficient matrix is non-singular, A = 18 , the system only has a trivial solution.  5 −3 2  17. A =  0 2 4   −3 2 6  Solution Since the coefficient matrix is non-singular, A = 68 , the system only has a trivial solution. 18. Determine a so that the following system has a non-trivial solution:  2 −1 0 0   x1  0  1 a 0 0         x2  = 0   0 0 2a 1   x3  0   1  0 0 −1 2   x4  0  Solution Since the determinant of the coefficient matrix is (2a + 1)(a + 1) , the system has a non-trivial solution for both cases of a = −1 or a = − 12 .

67

Problem Set 3.3 In Problems 1 through 8, (a) Find the eigenvalues and eigenvectors of the matrix. (b) Confirm the results of (a) in MATLAB. 2 4  1. A =    0 −3 Solution (a) Upper-triangular matrix; λ ( A)= 2, −3 . For λ1 = 2 we solve

[ A − 2I ] v1 =0

0 4  0  EROs 0 1  0  1  ⇒  v1 =  ⇒  v1 =  ⇒ v1 =    0 −5 0  0 0  0  0 

For λ2 = −3 we solve

[ A + 3I ] v 2 =0

5 4 0  4 ⇒  v 2 =  ⇒ v 2 =   0 0  0  −5

(b) >> A = [2 4;0 -3]; [V,D] = eig(A) V = 1.0000 0

-0.6247 0.7809

D = 2 0

0 -3

 5 −2  2. A =   2 0  Solution (a) λ ( A) = 1, 4 . For λ1 = 1 we solve

[ A − I ] v1 =0 For λ2 = 4 we solve

 4 −2  0  ⇒  v1 =    2 −1 0 

 2 −1 0  ⇒  v1 =  ⇒  0 0  0 

EROs

1  v1 =  2 

68

[ A − 4I ] v 2 =0

1 −2  0  v2 = ⇒      2 −4  0 

1 −2  0  v2 = ⇒    ⇒  0 0  0 

EROs

2  v2 =   1 

(b) >> A = [5 -2;2 0]; [V,D] = eig(A) V = 0.8944 0.4472

0.4472 0.8944

D = 4 0

0 1

0 3 3. A =   3 0 Solution (a) λ ( A) = ±3 . For λ1 = 3 we solve

[ A − 3I ] v1 =0

 −3 3  0  ⇒  v =   1   3 −3 0 

1 −1 0  ⇒  v =   ⇒ 1  0 0  0 

EROs

1 v1 =  1

For λ2 = −3 we solve

[ A + 3I ] v 2 =0

 3 3 0  ⇒  v2 =     3 3 0 

(b) >> A = [0 3;3 0]; [V,D] = eig(A) V = -0.7071 0.7071

0.7071 0.7071

D = -3 0

0 3

1 1 0  4. A =  3 3 0  0 0 −2 

1 1  0  1 ⇒  v2 =   ⇒ v2 =    0 0  0  −1

EROs

69 Solution (a) Block diagonal matrix; λ (= A) 0, 4, −2 . For λ1 = 0 we solve

1 1 0  0      0 ⇒  3 3 0  v1 = [ A ] v1 = 0  0  0 0 −2   

1 1 0  0  1       ⇒ 0 0 0  v1 = 0  ⇒ v1 = −1   0 0 0 1  0   

EROs

For λ2 = 4 we solve

 −3 1 0  0    [ A − 4I ] v 2 =0 ⇒  3 −1 0  v 2 =0 0   0 0 −6   

 −3 1 0  0  1        ⇒  0 0 0 v 2 = 0  ⇒ v 2 = 3  0  0   0 0 1     

EROs

For λ3 = −2 we solve

3 1 0 0    [ A + 2I ] v3 =0 ⇒ 3 5 0 v3 =0 0  0 0 0   

1 0 0  0  0        ⇒ 0 1 0  v 3 =0  ⇒ v 3 =0      0 0 0  0  1 

EROs

(b) >> A = [1 1 0;3 3 0;0 0 -2]; [V,D] = eig(A) V = -0.7071 0.7071 0

-0.3162 -0.9487 0

0 0 1.0000

-0.0000 0 0

0 4.0000 0

0 0 -2.0000

D =

 −1 0 0  5. A =  1 4 0   2 −1 1  Solution (a) Lower-triangular matrix; λ ( A) = −1, 4,1 . For λ1 = −1 we solve

70

0 0 0 0    [ A + I ] v1 =0 ⇒ 1 5 0  v1 =0    2 −1 2  0  For λ2 = 4 we solve

0 0 0  0  −10        ⇒ 1 5 0  v1 =0  ⇒ v1 = 2  0   11  0 −11 2     

EROs

 −5 0 0  0     1 0 0 v = 0 [ A − 4I ] v 2 =⇒   2 0  0   2 −1 −3   For λ3 = 1 we solve  −2 0 0  0    [ A − I ] v3 =0 ⇒  1 3 0 v3 =0    2 −1 0  0 

1 0 0  0  0     ⇒ 0 0 0  v 2 = 0  ⇒ v 2 = −3 0  1 0 1 3    

EROs

1 0 0  0  0        ⇒ 0 1 0  v 3 =0  ⇒ v 3 =0      0 0 0  0  1 

EROs

(b) >> A = [-1 0 0;1 4 0;2 -1 1]; [V,D] = eig(A) V = 0 0 1.0000

0 0.9487 -0.3162

0.6667 -0.1333 -0.7333

D = 1 0 0

0 4 0

0 0 -1

 3 −4 −2  6. A =  −1 4 1   2 −6 −1 Solution (a) λ ( A) = 3, 2,1 . For λ1 = 3 we solve  0 −4 −2  [ A − 3I ] v1 = 0 ⇒  −1 1 1  v1 =  2 −6 −4  For λ2 = 2 we solve

0    0  0   

0 2 1  ⇒ 1 1 0  v1 = 0 0 0 

EROs

0    0  ⇒ v1 = 0   

 −1   1 −2   

71

 1 −4 −2  0    [ A − 2I ] v 2 = 0 ⇒  −1 2 1  v 2 = 0 0   2 −6 −3  

1 0 0  0  0       ⇒  0 2 1  v 2 = 0  ⇒ v 2 =  1  0  −2  0 0 0     

EROs

For λ3 = 1 we solve

 2 −4 −2  0      [ A − I ] v3 = 0 ⇒  −1 3 1  v3 = 0    2 −6 −2  0 

 −1 0 1  0  1        ⇒  0 1 0  v 3 = 0  ⇒ v 3 = 0       0 0 0  0  1 

EROs

(b) >> A = [3 -4 -2;-1 4 1;2 -6 -1]; [V,D] = eig(A) V = -0.4082 0.4082 -0.8165

-0.0000 0.4472 -0.8944

0.7071 0.0000 0.7071

3.0000 0 0

0 2.0000 0

0 0 1.0000

D =

 1 −2 4  7. A =  −1 2 2   −5 10 −2  Solution (a) λ ( = A) 3, 0, −2 . For λ1 = 3 we solve  −2 −2 4  [ A − 3I ] v1 = 0 ⇒  −1 −1 2  v1 =  −5 10 −5

0    0  0   

1 1 −2  ⇒ 0 0 0  v1 = 0 1 −1

EROs

0    0  ⇒ v1 = 0   

1  1 1 

For λ2 = 0 we solve

 1 −2 4  0    [ A ] v 2 = 0 ⇒  −1 2 2  v 2 = 0    −5 10 −2  0 

1 −2 0  0  2        ⇒  0 0 1  v 2 = 0  ⇒ v 2 = 1    0   0 0 0  0   

EROs

72 For λ3 = −2 we solve

 3 −2 4  0    [ A + 2I ] v3 = 0 ⇒  −1 4 2 v3 = 0 0   −5 10 0    (b)

1 0 2  0  2       ⇒  0 1 1  v 3 = 0  ⇒ v 3 =  1  0  −1 0 0 0     

EROs

>> A = [1 -2 4;-1 2 2;-5 10 -2]; [V,D] = eig(A) V = -0.5774 -0.5774 -0.5774

-0.8944 -0.4472 0.0000

-0.8165 -0.4082 0.4082

3.0000 0 0

0 -0.0000 0

0 0 -2.0000

D =

 0 2 −2 0   −1 −5 2 −2   8. A =   0 −3 0 −3    1 2 −2 −1 Solution (a) λ ( A) =−2, −1, 0, −3 . For λ1 = −2 we solve  2 2 −2 0  0   −1 −3 2 −2     v1 =0  [ A + 2I ] v1 =0 ⇒  0 −3 2 −3 0    0   1 2 −2 1 

1 EROs  0 ⇒  0  0

1 0 1 0

0 1 0 0

0 0  1     0  0  −1 v1 =  ⇒ v1 =  1 0  0  0   1  0

For λ2 = −1 we solve  1 2 −2 0  0  −2   −1 −4 2 −2      0  1  v2 = ⇒ v = [ A + I ] v 2 =0 ⇒      2 0 −3 1 −3 0  0    −1  1 2 −2 0  0  For λ3 = 0 we solve

73  0 2 −2 0  0   −1 −5 2 −2  0      A v 0 v = ⇒ = [ ] 3   ⇒  0 −3 0 −3 3 0     1 2 −2 −1 0  For λ4 = −3 we solve

1 −1   v 3 =  −1  1 

 3 2 −2 0  0  0   −1 −2 2 −2  0      1    v4 = [ A + 3I ] v 4 =0 ⇒    ⇒ v4 =   0 −3 3 −3 0  1    0   1 2 −2 2  0  (b) >> A = [0 2 -2 0;-1 -5 2 -2;0 -3 0 -3;1 2 -2 -1]; [V,D] = eig(A) V = 0.5774 -0.5774 0.0000 0.5774

-0.8165 0.4082 -0.0000 -0.4082

-0.5000 0.5000 0.5000 -0.5000

0.0000 0.7071 0.7071 0.0000

-2.0000 0 0 0

0 -1.0000 0 0

0 0 0.0000 0

0 0 0 -3.0000

D =

In Problems 9 through 16, find the eigenvalues, eigenvectors, algebraic and geometric multiplicity of each eigenvalue, and decide whether the matrix is defective or not. Then transform the matrix into either a diagonal or a Jordan matrix, whichever applicable.  2 −1 9. A =   6 7  Solution >> A = [2 -1;6 7]; [V,D] = eig(A) V = -0.4472 0.8944

0.3162 -0.9487

4.0000 0

0 5.0000

% A is not defective

D =

% Eigenvalues are distinct, hence each has AM=1=GM

74

>> V\A*V ans = 4.0000 0.0000

0 5.0000

% A is transformed into a diagonal matrix D

0 −1 10. A =   1 2  Solution >> A = [0 -1;1 2]; [V,D] = eig(A) V = -0.7071 0.7071

-0.7071 0.7071

1.0000 0

0 1.0000

% Indicates that A is defective

D =

% AM(1)=2 but GM(1)=1

Switching to "jordan", we find >> [V,J] = jordan(A) V = -1 1

1 0

1 0

1 1

J =

>> V\A*V ans = 1 0

1 1

% A is transformed into Jordan matrix J

75

3 0 0 11. A = 0 3 1  5 0 0  Solution >> A = [3 0 0;0 3 1;5 0 0]; [V,D] = eig(A) V = 0 1.0000 0

0 -0.3162 0.9487

0.0000 -1.0000 0.0000

% 1st and 3rd columns are the same % thus, there is a generalized eigenvector

D = 3 0 0

0 0 0

0 0 3

% Eigenvalues are 3, 3, 0 % AM(3)=2, AM(0)=1, GM(3)=1, GM(0)=1

Matrix A is defective. Therefore, we switch to jordan: >> [V,J] = jordan(A) V = 0 0.5556 -1.6667

0 1.6667 0

1.0000 -0.5556 1.6667

J = 0 0 0

0 3 0

0 1 3

% V\A*V = J (Jordan matrix)

 4 1 0 12. A =  0 4 1   −1 −1 3 Solution >> A = [4 1 0;0 4 1;-1 -1 3]; [V,D] = eig(A) V = 0.5774 + 0.0000i -0.5774 + 0.0000i 0.5774 + 0.0000i

0.5774 + 0.0000i -0.0000 + 0.5774i -0.5774 + 0.0000i

0.5774 + 0.0000i -0.0000 - 0.5774i -0.5774 - 0.0000i

76

D = 3.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i

0.0000 + 0.0000i 4.0000 + 1.0000i 0.0000 + 0.0000i

0.0000 + 0.0000i 0.0000 + 0.0000i 4.0000 - 1.0000i

% A is not defective because it has distinct eigenvalues 3, 4 ± i >> V\A*V ans = 3.0000 - 0.0000i 0.0000 - 0.0000i -0.0000 + 0.0000i

-0.0000 - 0.0000i 4.0000 + 1.0000i -0.0000 - 0.0000i

-0.0000 + 0.0000i -0.0000 + 0.0000i 4.0000 - 1.0000i

This, as expected, is the diagonal matrix D.

 −2 3 1  13. A =  −1 2 1   −1 1 0  Solution Use the eig command to find eigenvalues and eigenvectors of matrix A . >> A = [-2 3 1;-1 2 1;-1 1 0]; [V,D] = eig(A) V = -0.7071 -0.7071 0.0000

0.7071 -0.0000 0.7071

0.5774 0.5774 -0.5774

1.0000 0 0

0 -1.0000 0

0 0 0.0000

D = % Eigenvalues are 1, -1, 0 (distinct) % AM=1 for all and GM=1 automatically % V\A*V = D (diagonal matrix)

2 2 1 14. A =  0 1 0   0 0 1  Solution >> A = [2 2 1;0 1 0;0 0 1]; [V,D] = eig(A)

77 V = 1.0000 0 0

-0.8944 0.4472 0

-0.7071 0 0.7071

D = 2 0 0

0 1 0

0 0 1

AM(2)=1, AM(1)=2. But since the columns of V are linearly independent, we conclude that GM(1)=2. We know GM(2)=1 automatically.

>> D = V\A*V D = 2 0 0

0 1 0

0 0 1

% As expected, A is transformed into a diagonal D

1 0 0  15. A = 0 −2 −1 0 1 −2  Solution Use the eig command to find eigenvalues and eigenvectors of matrix A . >> A = [1 0 0;0 -2 -1;0 1 -2]; [V,D] = eig(A) V = 0.0000 + 0.0000i 0.7071 + 0.0000i 0.0000 - 0.7071i

0.0000 + 0.0000i 0.7071 + 0.0000i 0.0000 + 0.7071i

1.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i

-2.0000 + 1.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i

0.0000 + 0.0000i -2.0000 - 1.0000i 0.0000 + 0.0000i

0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 + 0.0000i

D =

% λ(A)= -2 ± i, 1 % A is not defective % V\A*V = D diagonal

78

5 0 0 16. A =  2 5 1   0 0 6  Solution >> A = [5 0 0;2 5 1;0 0 6];[V,D] = eig(A) V = 0 1.0000 0

0.0000 -1.0000 0

0 0.7071 0.7071

D = 5 0 0

0 5 0

0 0 6

The first two columns of V represent the same eigenvector, hence A is defective. Swith to "jordan":

>> [V,J] = jordan(A) V = 0 1.0000 1.0000

0 -1.0000 0

-0.5000 -1.0000 0

J = 6 0 0

0 5 0

0 1 5

0 5 0

0 1 5

>> V\A*V ans = 6 0 0

% A is transformed into a Jordan matrix J

17. Prove that if A 2×2 has a repeated eigenvalue λ , then A must be defective. Solution If the eigenvalue is repeated, we have AM(λ ) = 2 . To determine the GM we must solve the eigenvalue problem (A − λ I )v = 0 . But A − λ I is 2 × 2 and singular. Therefore, it must have

79 exactly one zero row in its REF. This means GM(λ ) = 1 , which implies that A has a generalized eigenvector and is therefore defective.

18. Prove that a singular matrix must have at least one zero eigenvalue. Solution The determinant of a matrix is the product of its eigenvalues. If the determinant is zero, then at least one of the eigenvalues of the matrix must be zero. 19. Show that if the eigenvalues of A are λ1 , λ2 , ... , λn , then the eigenvalues of A − α I are

λ1 − α , λ2 − α , ... , λn − α . Solution ( A − α I ) v =Av − λ v . But Av = λ v . Insert into the last equation to obtain ( A − αα I ) v =Av − v =λ v − α v =(λ − α ) v This describes the eigenvalue problem associated with A − α I , and proof is complete.

20. Show that the eigenvalues of A and AT are the same. Solution It suffices to show that any eigenvalue of A is also an eigenvalue of AT . If λ is an eigenvalue T 0 . But a matrix A − λ I and its transpose [ A − λ I ] =AT − λ I of A , it must satisfy A − λ I = have the same determinant, which implies AT − λ I = 0 . This is the eigenvalue problem associated with AT . Therefore, λ must be an eigenvalue of AT .

Review Problems 1. Find the value of a such that the following two matrices have the same determinant.  1 3 0    −2 a 0  ,  0 0 4 

4 0 0     0 −3 2a  1    0 2 −1

Solution Equating the two determinants, 4(a + 6) = 4(3 − a ) , yields a =

−3 2

.

2. Determine whether the following vectors are linearly independent.

3   −1 , 4  

5    4  , 2   

2   5 −2   

80 Solution Assemble the vectors so that they form the rows of a matrix:  3 −1 4  5 4 2     2 5 −2  The determinant of this matrix is zero, which implies the rows are linearly dependent.

3. Prove that the product of two symmetric matrices is not necessarily symmetric. Solution T BA ≠ AB . Let A and B be symmetric so that AT = A and BT = B . But ( AB= ) BT A= T

4. If A is m × m and symmetric, and B is a general m × n matrix, show that BT AB is n × n and symmetric. Solution

(

)

T

T T BT AB is clearly n × n . Also BT= AB B= A B BT AB , and the proof is complete.

5. Given

0 1 0 A = = 0 0 1  , C  2 0  −2 −1 −2 

1 3

find the rank of

CT 

AT CT

( AT ) 2 CT 

Solution CT 

AT CT

2  T 2 T C  0 ( A )= 1 3

−2 3 5 3 −2 3

4 3

 0 3 

6. Determine a such that rank( A) = 3 , where 7 −2 1  3  −1 4 −3 2   A=  1 a − 4 5 3   12 1 5 7 Solution Using EROs, we find

⇒ = rank 3

81  −1 −2 1  7 3  −1 4 −3 2    −1    −3 2  EROs  0 a 4 2 5  EROs  0 A = → → 0  1 a − 4 5 3  0 19 −11 7       12 1 5  0 7  0 40 −20 19 

4 a 0 0

−3 2 −38 − 11 a −80 − 20 a

 5  −95 +5  a  −200 + 19  a 2

For the rank to be 3, the ratios of the last two entries in the third column and those in the fourth column must be the same:

−38 − 11a −95 + 7 a = −80 − 20a −200 + 19a

⇒

69a 2 − 138a = 0 ⇒ a = 0, 2

Inserting a = 0 reveals that rank( A) = 4 , hence rejected. It is easily verified that a = 2 is the only acceptable value. 7. Find the inverse of the rotation matrix

 cos θ R =  − sin θ  0

sin θ cos θ 0

0 0  1 

Solution Noting R = 1 , we find −1

= R

cos θ 1 = adj(R )  sin θ R  0

− sin θ cos θ 0

0 0  1 

8. Find the value(s) of a for which the following system only has a trivial solution.  23 1 −2   x1  0       0   a −2 1   x2  =    1 1 −5  x3  0   

Solution The homogeneous system has a trivial solution if and only if the coefficient matrix is nonsingular. Since A= 3a + 3 , we must have a ≠ −1 . 9. Solve the following linear system using Cramer’s rule; a is a parameter.  x1   3 0 −1 2   2a + 4  x   2 2 0 −2   2      = = , A  , x  2 = , b  Ax b=   −1 3 1 −1  x3   2a − 2     x4   2 −1 2 3   2a 

82 Solution The determinant of the coefficient matrix is 3 0 −1 2 2 2 0 −2 = −84 D= −1 3 1 −1 2 −1 2 3 Since D ≠ 0 , we proceed by evaluating 2a + 4 0 −1 2 3 2a + 4 −1

2 2 0 D1 = 2a − 2 3 1 2a

−1

2

−2

2

2 2 0 = −84 , D2 = −1 −1 2a − 2 1

−2

3

3

3 0 2a + 4 2 −2 2 2 2 = 84 D3 = −1 3 2a − 2 −1 2 −1 2a 3

2

,

2a

2

−1

= −84a

3 0 −1 2a + 4 2 2 0 2 = −84a D4 = −1 3 1 2a − 2 2 −1 2 2a

Finally,

x1 = 1 , x2 = a , x3 = −1 , x4 = a 10. Solve the linear system in Problem 9 using the inverse of the coefficient matrix. Solution We find A −1 as −8   −4 −22 12  −18 6 −30 6  1 1   A −1 = = adj( A) A −84  28 −14 0 −28    −22 26 −18 −2  The solution is then obtained as −8  2a + 4   −4 −22 12  −18 6 −30 6   2  1   −1   = = x A= b  −84  28 −14 0 −28  2a − 2     −22 26 −18 −2   2a 

1   a   −1  a 

11. Show that any matrix with distinct eigenvalues is non-defective. Solution Since the eigenvalues are assumed distinct, each has AM = 1 . But GM ≤ AM so that each eigenvalue also has GM = 1 . Therefore, there is no generalized eigenvector and the matrix is non-defective.

83

 −3 −2 0  = A  0 −1 0  . 12. (a) Find all eigenvalues and eigenvectors of  1 −1 −2  (b) Confirm the results of (a) in MATLAB. Solution (a) A is block lower triangular, hence its eigenvalues are those of the upper-left corner block and the single block of −2 . Therefore, λ =−1, −2, −3 and the corresponding eigenvectors are found as 1 0  1       v1 = −1 , v 2 = 0  , v 3 = 0 2 1  −1       (b) >> A = [-3 -2 0;0 -1 0;1 -1 -2]; [V,D] = eig(A) V = 0 0 1.0000

0.7071 0 -0.7071

-0.4082 0.4082 -0.8165

D = -2 0 0

0 -3 0

0 0 -1

13. Find the eigenvalues and the algebraic and geometric multiplicity of each, and decide whether the matrix is defective. 1 0 0 0  −2 3 0 0   A=  1 0 −1 1     0 2 0 −1 Solution A is block lower triangular, comprised of two 2 × 2 blocks, one block is lower triangular, the other upper triangular. Therefore λ =−1, −1,1,3 . For the two eigenvalues that occur only once, the AM and GM are both 1. Let us inspect λ = −1 as follows.

( A + I= )v 0

Elementary

⇒

row operations

1 0  0  0

0 0 0 1

0 0 0 0

0  a  0   b  =  1  c   0  d 

0    0    0  0 

84 It is clear that there is only one free variable, which means only one independent eigenvector can be found for this eigenvalue. Therefore, AM = 2 while GM = 1. This implies there is a generalized eigenvector, hence the matrix is defective. 14. Prove that the eigenvalues of a matrix are preserved under a similarity transformation, that is, if S −1AS = B , then the eigenvalues of A and B are the same. Solution Let λ be an eigenvalue of B so that λ I - B = 0 . Substitute for B :

λI − B = 0 ⇒

λ I − S −1AS = S −1 (λ I − A)S = S −1 λ I − A S = 0

0 , which means λ is an eigenvalue of A . But S −1 S = 1 . Therefore λ I − A =

Find the modal matrix and use it to transform A into a diagonal or a Jordan matrix:

15.

 −1 0  A= 0  0

4 −1 2  1 1 4  0 −1 0   0 2 −5

Solution >> A = [-1 4 -1 2;0 1 1 4;0 0 -1 0;0 0 2 -5]; [V,D] = eig(A) V = 1.0000 0 0 0

0.8944 0.4472 0 0

0.1374 -0.5494 0 0.8242

1.0000 -0.0000 0.0000 0.0000

D = -1 0 0 0

0 1 0 0

0 0 -5 0

0 0 0 -1

The first and last columns of V are the same, thus we must switch to jordan: V = -0.0833 0.3333 0 -0.5000

2.8333 1.4167 0 0

-6.0000 0 0 0

-2.7500 -1.5000 1.0000 0.5000

85

J = -5 0 0 0

0 1 0 0

0 0 -1 0

0 0 1 -1

16. Show that the rotation matrix R of Problem 7 is orthogonal. Then, verify that each of its eigenvalues has an absolute value of 1. Solution It is readily observed that cos θ − sin θ 0  = R −1 = sin θ cos θ 0  RT  0 0 1 

0: Therefore, R is orthogonal. The three eigenvalues of R are the roots of R − λ I = 1, cos θ ± j sin θ Each of the three clearly has an absolute value of 1. 17. Prove that if k = const and matrix A has eigenvalues λ1 , ... , λn with corresponding eigenvectors v1 , ... , v n , then the eigenvalues of kA are k λ1 , ... , k λn with eigenvectors

v1 , ... , v n . Solution Multiply both sides of Av = λ v by k to get ( kA ) v = ( k λ ) v . This is the eigenvalue problem for

kA so that the eigenvalues are k λ1 , ... , k λn and the corresponding eigenvectors are v1 , ... , v n . In Problems 18 through 20, decide whether the statement is true or false. 18. If n ≥ 3 , then every n × n matrix with any repeated eigenvalues is defective. Solution False 19. The rank of a singular 4 × 4 matrix is 3. Solution False 20. If all eigenvalues of a matrix are real, then the matrix is symmetric. Solution False

86

Problem Set 4.1 In Problems 1 through 8, assuming general initial conditions, express the system model in (a) Configuration form. (b) Standard, second-order matrix form.  1  x1 + x1 + 13 ( x1 − x2 ) = 0 1.  2 x2 + 12 x2 − 13 ( x1 − x2 ) = e −2t /3   Solution (a) The generalized coordinates are q1 = x1 and q2 = x2 . Then  q1 = −2q1 − 23 (q1 − q2 ) = f1 (q1 , q2 , q1 , q2 , t )  − 12 q2 + 13 (q1 − q2 ) + e −2t /3 = f 2 (q1 , q2 , q1 , q2 , t ) q2 = (b) The second-order matrix form is obtained as

x1  1 0   x1   13 − 13   x1   0   12 0     −2t /3       + 0 1     +  1 1    = 0 1     x2   2   x2    − 3 3   x2  e x + x + 2( x1 − x2 ) = e−2t  1  2.  33 1 1 0  5 x2 − 2( x1 − x2 ) = Solution (a) The second equation is algebraic, hence may be used to eliminate one of the two variables: x1 + x1 + 136 x1 = e−2t . Therefore, there is x into the first equation, we find 13  Inserting x2 = 10 13 1 only one generalized coordinate q1 = x1 and the configuration form is derived as

q1 = −3q1 − 18 q + 3e−2t = f (q1 , q1 , t ) . 13 1 (b) The second-order matrix form happens to be scalar ( n = 1 )

= M  13 = , C [1]= , K  136 = , x x= e−2t 1 , f x + k x + cx1 − k2 ( x2 − x1 ) = F1 (t ) m  3.  1 1 1 1 ; Mechanical system in Figure 4.3 x2 + k2 ( x2 − x1 ) = F2 (t )  m2  Solution (a) The generalized coordinates are q1 = x1 and q2 = x2 . Then

87 1   q1 = m [ −k1q1 − cq1 + k2 (q2 − q1 ) + F1 (t ) ] = f1 (q1 , q2 , q1 , q2 , t )  1  q = 1 [ −k (q − q ) + F (t ) ] = f (q , q , q , q , t ) 2 2 1 2 2 1 2 1 2  2 m2

x2

x1 k1

F2(t)

F1(t) m2

m1

Figure 4.2 Mechanical system in Problem 3.

(b) The second-order matrix form is obtained as

x1   c 0   x1   k1 + k2  m1 0     +   0 m       +  −k 2   x2   0 0   x 2  2 

−k2   x1   F1   =   k2   x2   F2 

x + c x + k x − k ( x − x ) − c2 ( x2 − x1 ) = F (t )  m  4.  1 1 1 1 1 1 2 2 1 ; Mechanical system in Figure 4.3    m x + k ( x − x ) + c ( x − x ) = 0 2 2 2 2 1 2 2 1 

x1

x2 k2

k1 F (t ) c1

m1

m2 c2

Figure 4.3 Mechanical system in Problem 4. Solution (a) The generalized coordinates are q1 = x1 and q2 = x2 . Then   q1 =   q =  2

1 [ −c1q1 − k1q1 + k2 (q2 − q1 ) + c2 (q2 − q1 ) + F (t )] = f1 (q1, q2 , q1, q2 , t ) m1 1 [ −k2 (q2 − q1 ) − c2 (q2 − q1 )] = f 2 (q1, q2 , q1, q2 , t ) m2

88 (b) The second-order matrix form is obtained as

x1  c1 + c2  m1 0    +   0 m    2   x2   −c2 

−c2   x1   k1 + k2  + c2   x2   −k2

−k2   x1   F (t )   =   k2   x2   0 

θ1 + θ1 + 3 (θ1 − θ 2 ) = e−t /2 sin t 2 5.    3 0 θ 2 + θ 2 − 2 (θ1 − θ 2 ) = Solution (a) The generalized coordinates are q1 = q1 and q2 = q 2 . Then  q1 = −q1 − 32 (q1 − q2 ) + e −t /2 sin t =f1 (q1 , q2 , q1 , q2 , t )  − q2 + 32 (q1 − q2 ) =f 2 (q1 , q2 , q1 , q2 , t ) q2 = (b) The second-order matrix form is obtained as 1 0  θ1  1 0  θ1   32 0 1     + 0 1     +  − 3   θ 2    θ 2   2

e −t sin t   − 32  θ1   =  3     0   2   θ 2  

 mx1 + kx1 − 12 k ( x2 − x1 ) − 2c( x2 − x1 ) = F1 (t )  2 1 6.  mx2 − 3 k ( x3 − x2 ) + 2 k ( x2 − x1 ) + 2c( x2 − x1 ) = 0 2mx + kx + 2 k ( x − x ) + cx = F2 (t ) 3 3 3 2 3 3  Solution (a) The generalized coordinates are q1 = x1 , q2 = x2 , and q3 = x3 . Then  q1 =   =  q2    q3 = 

1  −kq1 + 12 k (q2 − q1 ) + 2c(q2 − q1 ) + F1  = f1 (q1 , q2 , q3 , q1 , q2 , q3 , t ) m 1 2  k (q3 − q2 ) − 12 k (q2 − q1 ) − 2c(q2 −= q1 )  f 2 (q1 , q2 , q3 , q1 , q2 , q3 , t ) m 3 1  −kq3 − 23 k (q3 − q2 ) − cq3 + F2  = f3 (q1 , q2 , q3 , q1 , q2 , q3 , t ) 2m 

(b) The second-order matrix form is obtained as x1   2c −2c 0   x1   32 k  m 0 0             1  0 m 0   x2  +  −2c 2c 0   x2  +  − 2 k  0 0 2m    x3   0 0 c   x3   0

− 12 k 7 k 6 2 −3k

0   x1   F1      − 23 k   x2  = 0  5 k   x3   F2  3

89  12 mx1 + 2kx1 − k ( x2 − x1 ) − 12 c( x2 − x1 ) = F1 (t )  1 7.  2mx2 − k ( x3 − x2 ) + k ( x2 − x1 ) + 2 c( x2 − x1 ) − c( x3 − x2 ) = F2 (t ) mx + 2kx + k ( x − x ) + c( x − x ) = 0 3 3 2 3 2  3

Solution (a) The generalized coordinates are q1 = x1 , q2 = x2 , and q3 = x3 . Then 2  1 q1 = m  −2kq1 + k (q2 − q1 ) + 2 c(q2 − q1 ) + F1  = f1 (q1 , q2 , q3 , q1 , q2 , q3 , t )  1   k (q3 − q2 ) − k (q2 − q1 ) − 12 c(q2 − q1 ) + c(q3 − q2 ) += = F2  f 2 (q1 , q2 , q3 , q1 , q2 , q3 , t )  q2 2 m  1  f3 (q1 , q2 , q3 , q1 , q2 , q3 , t )  q3 =m [ −2kq3 − k (q3 − q2 ) − c(q3 − q2 ) ] = 

(b) The second-order matrix form is obtained as  12 m 0 0    x1   12 c − 12 c 0   x1   3k     1   x2  +  − 2 c 23 c −c   x2  +  − k  0 2m 0     0 m    x3   0 c   x3   0 −c  0

−k 2k −k

0   x1   F1 (t )      − k   x2  = F2 (t )  3k   x3   0 

  x= − 14 x + F (t ) 8.  x 2ϕ = 2ϕ + ϕ − 

Solution (a) The generalized coordinates are q1 = x and q2 = ϕ . Using the expression for x given by the first equation in the second equation, we have  x= − 14 x + F    1  2ϕ = 2ϕ + ϕ + 4 x − F

The configuration form is then found as  − 14 q1 + F = f1 (q1 , q2 , q1 , q2 , t )  q1 =  1 1 1  q2 =q2 + 2 q2 + 8 q1 − 2 F =f 2 (q1 , q2 , q1 , q2 , t )

(b) The second-order matrix form is obtained as x   14 1 0    0 2  ϕ +  − 1      4

0   x  0 0   x   F   +  =   −2  ϕ  0 −1 ϕ  − F 

90 Problem Set 4.2 In Problems 1 through 8, find a suitable set of state variables, derive the state-variable equations, and form the state equation. 1. mx + bx=  e −2t /3 , m, b = const > 0 Solution The ODE is second-order in x , therefore two initial conditions, x(0) and x (0) , are needed. Thus, there are two state variables: x1 = x , x2 = x . The state-variable equations are derived as    x1 = x2  x = 1  −bx2 + e −2t /3    2 m

The state equation is then formed as

0 1  0  x1   x  x = Ax + Bu , x = B = e −2t /3 ,  =   , A= 0 − b  1 , u =  x x  2   m  m

x +  x + 12 x + 2 x = cos t 2. 13  Solution The ODE is third-order in x , therefore three initial conditions, x(0) , x (0) , x(0) , are needed, and x . The state-variable equations are then there are three state variables: x1 = x , x2 = x , x3 =  derived as  x = x 2  1  x2 = x3  1     x3 = 3  − x3 − 2 x2 − 2 x1 + cos t 

The state equation is subsequently formed as 0 1  x1     x = Ax + Bu , x = 0  x2  , A = 0 x   −6 − 3  3 2 

0 0   0  , u = 1 , B= cos t   −3  3

x1 + 14 x1 + 32 ( x1 − x2 ) = F1 (t )   3.  2 F2 (t ) 2 x2 + x2 + 3 ( x2 − x1 ) = Solution The first ODE is second-order in x1 , hence two initial conditions, x1 (0) , x1 (0) , are needed. The second one is first-order in x2 so that only x2 (0) is needed. Therefore, there are a total of three state variables: x1 = x1 , x2 = x2 , x3 = x1 . As a result,

91

 x1 = x3  1 2  x2 = 2  − x2 − 3 ( x2 − x1 ) + F2 (t )   − 14 x3 − 23 ( x1 − x2 ) + F1 (t )  x3 =

F  Noting that u =  1  , the state equation is obtained as  F2  0 0 1   x1  0 0   F1   1     5 1 x = Ax + B3×2u 2×1 , x =  x2  , A =    3 −6 0  , B= 0 2  , u =  F2  x  2 2 1 −    − 1 0    3 3 3 4  

  x + 52 x + x = 2y 4.  1 F (t ) 2 y + 2 y + x = Solution The first ODE is second-order in x , hence two initial conditions, x(0) , x (0) , are needed. The second one is first-order in y so that only y (0) is needed. Therefore, there are a total of three state variables: x1 = x , x2 = y , x3 = x . The state-variable equations are  x1 = x3  1 1  x2 = 2  − 4 x2 − x1 + F (t )   − 52 x3 − x1 + 2 x2  x3 = The state equation is then formed as 0 0  x1   1   1 x = Ax + Bu , x =  x2  , A = − 2 − 8     x3   −1 2

1  0   1 0  , B= F (t ) 2 , u = 2 − 5   0 

x1 + 2 x1 + 2( x1 − x2 ) = F (t )   5.  1 0  3 x2 − 2( x1 − x2 ) = Solution Because the second equation is algebraic, variables x1 and x2 are linearly dependent, hence cannot be selected as state variables. Solve the second equation for x2 to find x2 = 76 x1 and insert into the first equation to obtain  x1 + 2 x1 + 72 x1 = F (t ) . We now have a second-order ODE in x1 and can proceed as usual. There are two state variables here: x1 = x1 and x2 = x1 . Note that x2 is the standard notation for a state variable and should not be confused with x2 in the original system. Consequently,

92

 x1 = x2  −2 x2 − 72 x1 + F (t )  x2 =

The state equation is

0  x1  x = Ax + Bu , x =   , A=  2  x1  − 7

1 0  , B= F (t )    , u= −2  1 

 z1 + z1 + 12 k ( z1 − z2 ) = e −t /3   6.  , k = const 1 0   z2 − 2 k ( z1 − z2 ) = Solution Because the second equation is algebraic, variables z1 and z2 are linearly dependent, hence

cannot be selected as state variables. Solve the second equation for z2 to find z2 =

k z k +2 1

and

z1 + z1 + k +k 2 z1 = e −t /3 . We now have a second-order ODE insert into the first equation to obtain  in z1 and can proceed as usual. There are two state variables here: x1 = z1 and x2 = z1 . Then,  x1 = x2  − x2 − k +k 2 x1 + e−t /3  x2 = The state equation is

 0  z1  x = Ax + Bu , x =   , A= − k  z1   k +2

1 0  e−t /3 , B= , u=    −1 1 

x + 4 x1 − 2 x2 + x1 − x2 = F1 (t ) 2  7.  1 x2 − 2 x1 + 2 x2 − x1 + x2 = F2 (t )   Solution The two second-order ODEs require a total of four initial conditions for complete solution, hence there are four state variables. We choose them as x1 = x1 , x2 = x2 , x3 = x1 , and x4 = x2 . Then,  x1 = x3  x = x  2 4  1   x3 = 2 [ −4 x3 + 2 x4 − x1 + x2 + F1 (t ) ]  x4 = 2 x3 − 2 x4 + x1 − x2 + F2 (t )

F  Noting that u =  1  , the state equation is obtained as  F2 

93

x = Ax + B 4×2u 2×1

0 1 0  x1  0  0 0 x  0   0 0 0 0 1  F1 (t )   2    , u= = , x= , B   , A=    − 12 12 −2 1   12 0   F2 (t )   x3       x4   1 −1 2 −2  0 1

 mL12θ1 + (mgL1 + kL22 )θ1 − kL22θ 2 = 0 8.  , m, k , g , L1 , L2 = const 2 2 2 0 mL1θ2 + (mgL1 + kL2 )θ 2 − kL2θ1 = Solution The two second-order ODEs require a total of four initial conditions for complete solution, hence there are four state variables. We choose them as x1 = θ1 , x2 = θ 2 , x3 = θ1 , and x4 = θ2 . Then,  x1 = x3  x = x 4  2  1  x3 = 2 mL1   1  x4 = 2 mL1 

(

)

(

)

 − mgL1 + kL22 x1 + kL22 x2     − mgL1 + kL22 x2 + kL22 x1   

The state equation is obtained as

0   0  x1   x   mgL + kL2  2 1 2 x Ax = = , x  =  , A − 2 mL x  1  3  2  x4  kL2   mL12

0 0 kL22 mL12 −

mgL1 + kL22 mL12

1 0 0 1   0 0   0 0 

In Problems 9 and 10, derive the state-variable equations (in vector form) for the given nonlinear system model.

x + 13 x 2 + 2 x = e−2t /3 9. 2  Solution The ODE is second-order in x , therefore two initial conditions, x(0) and x (0) , are needed, and there are two state variables: x1 = x , x2 = x . The state-variable equations are derived as   x1 = x2  − 16 x22 − x1 + 12 e −2t /3   x2 =

94

In vector form,

x = f (x, u, t )

where  x1  x = = , u  x2 

x2    1 2  − 6 x2 − x1 + u 

−2t /3 1 e= , f 2

  x + 1 x x = x23 10.  1 3 1 1   x2= x1 + 2sin t Solution The first ODE is second-order in x1 , hence two initial conditions, x1 (0) , x1 (0) , are needed. The

second one is first-order in x2 so that only x2 (0) is needed. Therefore, there are a total of three state variables: x1 = x1 , x2 = x2 , x3 = x1 . Then,

 x = x  1 3  x2= x1 + 2sin t  − 13 x3 x3 + x23  x3 =

In vector form,

x = f (x, u, t )

where

 x1    t, f = = x  x2=  , u 2sin x   3

  x3   x1 + u    1 3 − 3 x3 x3 + x2 

Problems 11 through 14 are concerned with the stability of systems. A linear dynamic system is stable if the homogeneous solution of its mathematical model, subjected to the prescribed initial conditions, decays. More practically, a linear system is stable if the eigenvalues of its state matrix all have negative real parts, that is, they all lie in the left half-plane. 11. Decide whether the system in Problem 1 is stable. Solution The state matrix is formed as 0 1  A= b 0 − m  The eigenvalues of A are 0, − mb . Since one of these is always 0, we conclude that the system is unstable regardless of the values of m and b .

95

12. Decide whether the system in Problem 5 is stable. Solution The state matrix is 1 0 A= 2   − 7 −2  Since its eigenvalues are −0.1548, − 1.8452 , the system is stable.

13. Decide whether the system in Problem 4 is stable. Solution The state matrix is 0 0 1   1  1 A= − 2 − 8 0   −1 2 − 2  5  Since the eigenvalues are 0.1620 ± 1.1397j , − 0.8490 , the system is unstable. 14. Find the range of values of b for which a system described by  y − (b − 1) y − y = F (t ) is stable. Solution Selection of x1 = y and x2 = y as state variables leads to the state matrix 1  0 = A   1 b − 1

eigenvalues

⇒

l (= A)

b − 1 ± (b − 1) 2 + 4 2

However, since (b − 1) 2 + 4 > b − 1 for all values of b , at least one of the two eigenvalues will always have a positive real part. Therefore, the system is unstable for all values of b .

In Problems 15 through 18, find the state-space form of the mathematical model

  x1 + 3x1 + 32 ( x1 − x2 ) = 0 15.  , outputs are x1 and x1 . 3 F (t ) 2 x2 + x2 − 2 ( x1 − x2 ) = Solution Choosing state variables x1 = x1 , x2 = x2 , x3 = x1 , the state-variable equations are obtained as

96  x1 = x3  5 3 1  x2 =2  − 2 x2 + 2 x1 + F   −3 x3 − 32 x1 + 32 x2  x3 =

⇒

0  3 x = 4  3 − 2

0 − 54 3 2

1  x1  0       0  x + 1  u , x = x2  , u =F   −3 0   x3 

The outputs are x1 and x1 so that the output equation is written as

 x1   x1  1 0 0    0  = y =     x2  +   u  x3  0 0 1  2×3  x  0   3   z + 3 ( z − z ) + 2 z1 + z1 = F (t ) 16.  1 43 1 2 , outputs are z1 and z2 .  z + ( z − z ) = 0   2 4 2 1 Solution With state variables x1 = z1 , x2 = z2 , x3 = z1 , we find

 x1 = x3  − 34 x2 + 34 x1  x2 =  x = 7 3  3 − 4 x1 + 4 x2 − 2 x3 + F

0  3 ⇒ x =  4 − 7  4

0 − 34 3 4

1  x1  0       0  x + 0  u , x = F (t )  x2  , u =   −2  1   x3 

The outputs are z1 and z2 , thus

 x1   x1  1 0 0    0  = y =     x2  +   u  x2  0 1 0  2×3  x  0   3 x1 + 9( x1 − x3 ) − 0.8( x2 − x1 ) − 2( x2 − x1 ) = F (t )  2   0 x2 + 0.8( x2 − x1 ) + 2( x2 − x1 ) = 17.   , outputs are x2 and x2 . 2 x 0.45( x − x ) = 1 3  3 Solution x . Using this, eliminate x3 in the The third equation is algebraic. Solving for x3 yields x3 = 0.45 2.45 1 first equation, while the second equation remains as is. Then x1 + 9.35x1 − 0.8( x2 − x1 ) − 2 x2 = F (t ) 2   x2 + 0.8( x2 − x1 ) + 2( x2 − x1 ) = 0  

Select state variables as x1 = x1 , x2 = x2 , x3 = x1 , x4 = x2 , noting that x3 is not the same as x3 in the original model. The state equation is formed as

97

0 1 0   x1   x1   0 0  x  x   0    0 0 1  0  2  2   x = u , x= F (t ) x+  =   , u=  −4.675 1 −0.4 0.4  1   x3   x1      −2 0.8 −0.8  2 0  x4   x2  The output equation is

 x1     x2  0 1 0 0   x2  0  = y =     +  u  x4  0 0 0 1  2×4  x3  0   x4    x1 + 52 ( x1 − x3 ) − 53 ( x2 − x1 ) − 12 ( x2 − x1 ) = 0  18.   , outputs are x2 and x3 . x2 + 53 ( x2 − x1 ) + 12 ( x2 − x1 ) = F (t ) = 3  x3 5 ( x1 − x3 )

Solution There are five state variables: x1 = x1 , x2 = x2 , x3 = x3 , x4 = x1 , x5 = x2 . The state-variable equations are then obtained as

 x1 = x4  x = x 5  2 3  x3 5 x1 − 53 x3 =  x = − 109 x1 + 12 x2 + 52 x3 − 53 x4 + 53 x5  4 − 53 x5 + 53 x4 − 12 x2 + 12 x1 + F (t )  x5 = Consequently, the state equation = x Ax + Bu is formed as

 0  0  = A  53  9  − 10  1  2

0

0

1

0

0

0

0

− 53 2 5

0 − 53

0

3 5

1 2 − 12

0 1   0= , B  3 5  − 53  5×5

System outputs are x2 and x3 , hence the output equation is

0  0    = 0  , u F (t )   0  1  5×1

98

 x1  x  2  x2  0 1 0 0 0    0  = y =     x3  +   u  x3  0 0 1 0 0  2×5  x  0  2×1  4  x5  19. A dynamic system is governed by 4  y + 8 y + 3 y = f (t ) , where f and y are the system input and output, respectively. Derive the state-space form of the decoupled system. Solution Selecting the state variables as x1 = y and x2 = y , the state equation is formed as

0 x = − 3  4

1 0  x1   y  x + 1u = Ax + Bu , x =  =   , u =f (t )  −2   x2   y  4

The output equation is

x  y = x1 = [1 0]  1  + 0 ⋅ u = Cx + Du  x2 

 −2 −2  λ ( A) = − 12 , − 32 , V =   . The transformed (via x = Vx ) state-space form is 1 3   −81   − 12 0   = x +  1  u  x  3   0 − 2   8  Decoupled     y =[ −2 −2] x + 0 ⋅ u 20. A dynamic system model is derived as

x1 − x1 − 3 x1 + x2 = f (t )  4   1 0  x2 + 3 x1 + 2 x2 = Where the input is f and the outputs are x2 and x1 . (a) Find the state-space form. (b) Derive the state-space form of the decoupled system. Solution (a) There are three state variables x1 = x1 , x2 = x2 , x3 = x1 , which lead to

99  x1 = x3  − 13 x1 − 2 x2  x2 =  x = 1 [ x + 3 x − x + f ] 1 2  3 4 3

x Ax + Bu where The resulting state equation is = 0  1 A= − 3  3  4 The output equation is

0 −2 − 14

0 1  x1       0 , B = f (t ) , x =  x2  0 , u = x  1 1  3 4 4

 x1   x2  0 1 0    0  = y =     x2  +   u  x3  0 0 1  2×3  x  0  2×1  3 (b) % Define state-space form matrices A,B,C,D >> A = [0 0 1;-1/3 -2 0;3/4 -1/4 1/4]; >> B = [0;0;1/4]; C = [0 1 0;0 0 1]; D = [0;0]; >> [V,Dtilda] = eig(A); % Solve eigenvalue problem for A >> Btilda = V\B; Ctilda = C*V; % Define transformed matrices B,C >> dec_sys = ss(Dtilda,Btilda,Ctilda,D) dec_sys = a = x1 x2 x3

x1 1.016 0 0

x2 0 -0.7885 0

x3 0 0 -1.977

x2 -0.2112 -0.6052

x3 0.9887 0.1338

% Decoupled

b = x1 x2 x3

u1 -0.1996 -0.184 -0.02369

c = y1 y2

x1 0.07732 -0.7104

d = u1 y1 0 y2 0 Continuous-time state-space model.

Problem Set 4.3 In Problems 1 through 8, find all possible input-output equations.

100

  x1 + x1 + 13 ( x1 − x2 ) = f (t ) 1.  , f (t ) = input, x1 , x2 = outputs 1 0  x2 − 3 ( x1 − x2 ) = Solution Since there are two outputs and one input, we expect two I/O equations. Assuming zero initial conditions, Laplace transformation of the governing equations yields 2 F ( s) ( s + s + 13 ) X1 ( s ) − 13 X 2 ( s ) =  1 1 − 3 X1 ( s) + ( s + 3 ) X 2 ( s) = 0 

Since both x1 and x2 are the outputs, we solve the above system once for X1 ( s ) and a second time for X 2 ( s ) by means of Cramer’s rule: − 13

F (s) X1 ( s) =

s + 13

0 s +s+ − 13 2

1 3

− 13 s + 13

=

( s + 13 ) F ( s ) s + 3

4 s2 3

cross multiply

⇒

+ s 2 3

(3s 3 + 4 s 2 + 2 s ) X1 ( s ) = (3s + 1) F ( s )

s 2 + s + 13 X 2 (s) =

F (s) 1 F (s) 0 3 = 3 2 4 s2 + 2 s 1 1 s + s +s+ 3 −3 3 3 1 1 s+ 3 −3 − 13

cross multiply

⇒

(3s 3 + 4 s 2= + 2s) X 2 ( s) F ( s)

Interpretation of these two equations in time domain results in the two desired I/O equations:

3 x1 + 4  x1 + 2 x1 = 3 f + f

,

3 x2 + 4  x2 + 2 x2 = f

  x1 + 2 x1 + 53 ( x1 − x2 ) = f1 (t ) , f1 (t ), f 2 (t ) = inputs, x2 = output 2.  x2 + x2 − 53 ( x1 − x2 ) = f 2 (t ) 3 Solution Two I/O equations are expected. Laplace transformation of the original system leads to 2 3 3  F1 ( s )  ( s + 2s + 5 ) X1 ( s) − 5 X 2 ( s) =  3 2 3 F2 ( s )  − 5 X 1 ( s ) + (3s + s + 5 ) X 2 ( s ) =

101 Since x2 is the output, we solve the above system for X 2 ( s ) via Cramer’s rule:

= X 2 ( s)

s 2 + 2 s + 53 F1 ( s ) 3 − 53 F2 ( s ) s 2 + 2 s + 53 5 = + F s F1 ( s ) ( ) 2 ∆( s) ∆( s) − 53 s 2 + 2 s + 53 − 53

3s 2 + s + 53

s 2 + 95 s . When f1 is the input, we set f 2 = 0 so that F2 ( s ) = 0 in the where ∆( s ) = 3s 4 + 7 s 3 + 22 5 above equation, and 3 X 2 (s) 5 = F1 ( s ) ∆( s )

⇒

s 2 + 95 s ) X 2 (= s) (3s 4 + 7 s 3 + 22 5

3 F (s) 5 1

When f 2 is the input, we set f1 = 0 so that F1 ( s ) = 0 , and 2 X 2 ( s ) s + 2 s + 53 = ∆( s) F2 ( s )

⇒

s 2 + 95 s ) X 2 ( s ) = ( s 2 + 2 s + 53 ) F2 ( s ) (3s 4 + 7 s 3 + 22 5

Transforming the data back to the time domain, the two I/O equations are

 3x2(4) + 7 x2 + 22 x + 9 x =f 2 + 2 f2 + 53 f 2 5 2 5 2

3  3 x2(4) + 7 x2 + 22 x + 9 x = f , 5 2 5 2 5 1

x1 + x1 + 13 ( x1 − x2 ) = f1 (t )   , f1 (t ) , f 2 (t ) = inputs, x1 = output 3.  1 x2 − 3 ( x1 − x2 ) = f 2 (t )   Solution Two input-output equations are expected. Laplace transformation of the original system leads to 2 1 1  F1 ( s ) ( s + s + 3 ) X 1 ( s ) − 3 X 2 ( s ) =  1 2 1 F2 ( s )   − 3 X1 ( s) + ( s + 3 ) X 2 ( s) =

Since x1 is the output, we solve the above system for X 1 ( s ) via Cramer’s rule: F1 ( s ) = X1 ( s)

− 13

F2 ( s ) s 2 + 13 ( s 2 + 13 ) F1 ( s ) 13 F2 ( s ) = + ∆( s) ∆( s) − 13 s 2 + s + 13 − 13

s 2 + 13

102

where ∆( s ) = s 4 + s 3 + 23 s 2 + 13 s . When f1 is the input, we set f 2 = 0 so that F2 ( s ) = 0 in the above equation, and 2 X1 ( s ) s + 13 = F1 ( s ) ∆( s)

⇒

( s 4 + s 3 + 23 s 2 + 13 s ) X 1 ( s ) = ( s 2 + 13 ) F1 ( s )

When f 2 is the input, we set f1 = 0 so that F1 ( s ) = 0 , and 1 X1 ( s) 3 = F2 ( s ) ∆( s )

⇒

( s 4 + s 3 + 23 s 2 + 13 s ) X= 1 (s)

1 3

F2 ( s )

Transforming the data back to the time domain, the two I/O equations are

x1(4) +  x1 + 23  x1 + 13 x1 = f1 + 13 f1 ,

1 f x1(4) +  x1 + 23  x1 + 13 x1 = 3 2

x + x + 2( x1 − x2 ) = 0   , f (t ) = input, x1 = output 4.  1 1 x2 + x2 − 2( x1 − x2 ) = f (t )   Solution One I/O equation is expected. Laplace transformation of the original system leads to  ( s 2 + s + 2) X1 ( s ) − 2 X 2 ( s ) = 0  2 F ( s) −2 X1 ( s ) + ( s + s + 2) X 2 ( s ) = Since x1 is the output, we solve the above system for X 1 ( s ) via Cramer’s rule: 0 X1 ( s) =

−2

F (s) s 2 + s + 2 2 F (s) = 4 2 s + 2 s 3 + 5s 2 + 4 s s +s+2 −2 −2

s2 + s + 2

The I/O equation is therefore obtained as

x1(4) + 2 x1 + 5 x1 + 4 x1 = 2f

103

θ + 2 θ + 1 (θ − θ ) = u (t ) 5.  1 31 1 3 1 2 , u (t ) = input, θ 2 = output ( ) 0 θ θ θ − − =  2 1 2 3  Solution The second equation is algebraic. Solving for θ1 in the second equation, we have θ1 = 4θ 2 . u (t ) . Inserting into the first equation yields the I/O equation as 4θ + 8 θ + θ = 2

3 2

2

x + x + 3( x1 − x2 ) = f1 (t )   6.  1 1 , f1 (t ) , f 2 (t ) = inputs, x1 , x2 = outputs x2 − 3( x1 − x2 ) = f 2 (t )   Solution Four input-output equations are expected. Laplace transformation of the original system leads to ( s 2 + s + 3) X 1 ( s ) − 3 X 2 ( s ) = F1 ( s )  2 F2 ( s )  −3 X1 ( s ) + ( s + 3) X 2 ( s ) = Solving the above system for X 1 ( s ) via Cramer’s rule, we find F1 ( s ) = X1 ( s)

−3

F2 ( s ) s 2 + 3 3F2 ( s ) ( s 2 + 3) F1 ( s ) = 4 3 + s + s + 6 s 2 + 3s s 4 + s 3 + 6 s 2 + 3 s −3 s2 + s + 3 −3

(a)

s2 + 3

Solving the system for X 2 ( s ) via Cramer’s rule, we find

X 2 (s) =

s 2 + s + 3 F1 ( s ) F2 ( s ) −3 3F1 ( s ) ( s 2 + s + 3) F2 ( s ) = + s 4 + s 3 + 6 s 2 + 3s s 4 + s 3 + 6 s 2 + 3 s s2 + s + 3 −3 −3

s2 + 3

When f1 is the input, we set F2 ( s ) = 0 in Eqns. (a) and (b), and two of the I/O equations are obtained as

X1 ( s) =

= X 2 (s)

( s 2 + 3) F1 ( s ) s 4 + s 3 + 6 s 2 + 3s 3F1 ( s ) s + s 3 + 6 s 2 + 3s 4

⇒

⇒

x1(4) +  x1 + 6  x1 + 3x1 =  f1 + 3 f1

x2(4) = +  x2 + 6  x2 + 3x2 3 f1

(b)

104

When f 2 is the input, we set F1 ( s ) = 0 in Eqns. (a) and (b), and the other two I/O equations are obtained as

= X1 ( s)

X 2 (s) =

3F2 ( s ) s 4 + s 3 + 6 s 2 + 3s

( s 2 + s + 3) F2 ( s ) s 4 + s 3 + 6 s 2 + 3s

⇒

⇒

x1(4)= +  x1 + 6  x1 + 3x1 3 f 2

x2(4) +  x2 + 6  x2 + 3x2 =  f 2 + f2 + 3 f 2

 1 q + q + 2(q1 − q2 ) + 13 q1 = v(t ) 7.  2 1 1 , v(t ) = input, q1 , q2 = outputs 0  q2 − 2(q1 − q2 ) = Solution There are two I/O equations. Proceeding as always, 2 V (s) ( 12 s + s + 73 )Q1 ( s ) − 2Q2 ( s ) =  0  −2Q1 ( s ) + ( s + 2)Q2 ( s ) =

Solve for Q1 ( s ) and Q2 ( s ) separately: V (s) Q1 ( s ) =

0 1 s2 2

−2 s+2

+ s + 73 −2

−2 s+2

=

1 s3 2

+ s + 73 V ( s ) 0 −2 = 1 s2 + s + 7 −2 2 3 s+2 −2

( s + 2)V ( s ) s + 32 )Q1 ( s ) =( s + 2)V ( s ) ⇒ ( 12 s 3 + 2 s 2 + 11 3 2 11 2 + 2s + 3 s + 3

1 s2 2

Q2 ( s ) =

2V ( s ) 11 s + 2 )Q ( s ) 2V ( s ) ⇒ ( 12 s 3 + 2 s 2 += 2 3 3 1 s 3 + 2 s 2 + 11 s + 2 2 3 3

In time-domain, the two I/O equations are described by 1  q + 2q1 + 11 q + 2 q 2 1 3 1 3 1

=+ v 2v ,

1  q 2 2

+ 2q2 + 11 q + 32 q2 = 2v 3 2

105   x1 + x1 − x3 − ( x2 − x1 ) − 12 ( x2 − x1 ) = f (t )  1 8.   , x2 + x2 − x1 + 2 ( x2 − x1 ) = 0  x= x − x  3 1 3

f (t ) = input, x2 , x3 = outputs

Solution Two input-output equations are expected. Laplace transformation of the original system leads to ( s 2 + s + 3 ) X 1 ( s ) − ( s + 1 ) X 2 ( s ) − X 3 ( s ) = F ( s) 2 2  2 1 1 0 =  −( s + 2 ) X 1 ( s ) + ( s + s + 2 ) X 2 ( s )  0 − X1 ( s) + ( s + 1) X 3 ( s ) = 

Solving for X 2 ( s ) and X 3 ( s ) via Cramer’s rule:

= X 2 ( s)

s 2 + s + 32 F ( s ) −1 −( s + 12 ) 0 0 −1 0 s +1 ( s + 12 )( s + 1) F ( s ) = s 5 + 3s 4 + 4 s 3 + 2 s 2 + 12 s s 2 + s + 23 −( s + 12 ) −1 −( s + 12 ) −1

s 2 + s + 12 0

0 s +1

s 2 + s + 32

−( s + 12 )

F ( s)

−( s + 12 ) s + s + 12 0 −1 0 0 ( s 2 + s + 12 ) F ( s ) = s 5 + 3s 4 + 4 s 3 + 2 s 2 + 12 s −1 s 2 + s + 32 −( s + 12 ) 2

= X 3 (s)

−( s + 12 )

s 2 + s + 12

−1

0 s +1

0

In time-domain, the two I/O equations are described by

x2(5) + 3 x2(4) + 4 x2 + 2  x2 + 12 x2 = f + 32 f + 12 f x (5) + 3 x (4) + 4 x + 2  x + 1 x = f + f + 1 f 3

3

3

3

2 3

2

9. A mechanical system model is derived as mx + bx + kx = f (t ) where m = 2 , b = 12 , k = 5 , and applied force f (t ) = 52 e−t , all in consistent physical units. The system is subjected to zero initial conditions. Assuming x is the output, find the transfer function.

106 Solution The transfer function is independent of the specific nature of the input, and is found as

X (s) 1 1 = = 2 2 F ( s ) ms + bs + k 2 s + 12 s + 5 10. In Problem 9, find the transfer function if x is the output. Solution The transfer function is independent of the specific nature of the input. It is readily seen that

X (s) 1 1 = = F ( s ) ms 2 + bs + k 2 s 2 + 12 s + 5 Noting that L { x} = sX ( s ) − x(0) = sX ( s ) found as L { x} = L {f}

due to zero initial conditions, the transfer function is sX ( s ) s = 2 F ( s ) 2 s + 12 s + 5

In Problems 11 through 14, a system model, as well as its inputs and outputs are provided. Find the appropriate transfer function or matrix. x + 0.4x1 + 9( x1 − x2 ) = f (t )   , 11.  1 x2 + 0.8x2 + 9( x2 − x1 ) = 0 0.6 Solution Laplace transformation yields

f (t ) = input, x1 = output

 ( s 2 + 0.4s + 9) X1 ( s ) − 9 X 2 ( s ) = F ( s)  2 0 −9 X1 ( s ) + (0.6s + 0.8s + 9) X 2 ( s ) = Solving for X1 ( s ) results in −9

F (s) = X1 ( s)

0 0.6s + 0.8s + 9 0.6s 2 + 0.8s + 9 = F (s) 0.6s 4 + 1.04s 3 + 14.72s 2 + 10.8s −9 s 2 + 0.4s + 9 2

−9

0.6s 2 + 0.8s + 9

Since the system is SISO, the only transfer function is

107

X1 ( s) 0.6s 2 + 0.8s + 9 = F ( s ) 0.6s 4 + 1.04s 3 + 14.72s 2 + 10.8s  q + q + 1 (q − q ) + q1 = v(t ) 12.  1 11 2 1 2 , v(t ) = input, q1 = output 0  q2 − 2 (q1 − q2 ) = Solution Laplace transformation of the original system leads to

( s 2 + s + 3 )Q1 ( s ) − 1 Q2 ( s ) = V (s) 2 2  1 1 0  − 2 Q1 ( s ) + ( s + 2 )Q2 ( s ) = Solve this system of algebraic equations for Q1 ( s ) via Cramer’s rule:

= Q1 ( s )

V ( s ) − 12 0 s + 12 ( s + 12 )V ( s ) = s 3 + 23 s 2 + 2 s + 12 s 2 + s + 32 − 12 − 12 s + 12

Since the system is SISO, there is only a single transfer function s + 12 Q1 ( s ) = V ( s ) s 3 + 32 s 2 + 2 s + 12

x + x − ( x2 − x1 ) − ( x2 − x1 ) = f1   , f1 (t ) , f 2 (t ) = inputs, x1 , x2 = outputs 13.  1 1 x2 + ( x2 − x1 ) + ( x2 − x1 ) = f2 2  Solution We expect a 2 × 2 transfer matrix with the following structure:  X1 ( s)  X1 ( s)   F ( s ) F ( s ) 1 2  G ( s ) G ( s ) = F2 0= F1 0   11  12 = G ( s ) =    X 2 (s) G21 ( s ) G22 ( s )  2×2  X 2 ( s )   F (s) F2 ( s ) F 0  = F2 0= 1  1  To find the four transfer functions listed above, we take the Laplace transform of the governing equations, with zero initial conditions, ( s 2 + s + 2) X1 ( s ) − ( s + 1) X 2 ( s ) = F1 ( s )  2 F2 ( s )  −( s + 1) X 1 ( s ) + (2 s + s + 1) X 2 ( s ) =

108

Solving for X1 ( s ) , we find

F1 ( s ) = X1 ( s)

−( s + 1)

F2 ( s ) 2 s 2 + s + 1 2 s 2 + s + 1 s +1 = F1 ( s ) + F2 ( s ) ∆( s) ∆( s) ∆( s)

(a)

where ∆( s )=

s2 + s + 2 −( s + 1)

−( s + 1)

= 2 s 4 + 3s 3 + 5s 2 + s + 1 2s + s + 1 2

Solving for X 2 ( s ) , yields

= X 2 (s)

s 2 + s + 2 F1 ( s ) −( s + 1) F2 ( s ) s 2 + s + 2 s +1 = F2 ( s ) + F1 ( s ) ∆( s) ∆( s) ∆( s)

where ∆( s ) is as before. The four transfer functions are then determined as follows.

X1 ( s) X1 ( s) s +1 2s 2 + s + 1 , = = F1 ( s ) F 0= F2 ( s ) F 0 ∆( s ) ∆( s) = 2

1

X 2 (s) X 2 (s) s +1 s2 + s + 2 = = , F1 ( s ) F 0= ∆( s) F2 ( s ) F 0 ∆( s) = 2

1

Ultimately, the transfer matrix is formed as

 2s 2 + s + 1  ∆( s) G (s) =   s +1   ∆( s )

   2 s + s + 2  ∆( s )  s +1 ∆( s)

x1 + x1 − x3 − 2( x2 − x1 ) = f1 (t )    x2 + 2( x2 − x1 ) = f 2 (t ) 14.   , f1 (t ) , f 2 (t ) = inputs, x1 , x2 = outputs  x= x − x  3 1 3 Solution Taking the Laplace transform, we find

(b)

109  ( s 2 + 2 s + 1) X 1 ( s ) − 2 sX 2 ( s ) − X 3 ( s ) = F1 ( s )   2 F2 ( s ) −2 sX1 ( s ) + ( s + 2 s ) X 2 ( s ) =  − X ( s ) + ( s + 1) X ( s ) = 0 3   1

Solving separately for X1 ( s ) and X 2 ( s ) , yields −2 s

F1 ( s )

−1

F2 ( s ) s + 2 s 2

= X1 ( s)

0 s +1 0 0 ( s + 1)( s 2 + 2 s ) 2 s ( s + 1) = + F s F2 ( s ) ( ) 1 5 4 3 2 5 s + 5s + 5s + 2 s s + 5s 4 + 5s 3 + 2 s 2 −1 s 2 + 2 s + 1 −2 s −2 s −1

= X 2 (s)

s 2 + 2s 0

0 s +1

s 2 + 2 s + 1 F1 ( s ) −1 −2 s F2 ( s ) 0 −1 s +1 0 s 3 + 3s 2 + 3s 2s 2 + 2s = + F s F2 ( s ) ( ) 1 s 5 + 5s 4 + 5s 3 + 2 s 2 s 5 + 5s 4 + 5s 3 + 2 s 2 −1 s 2 + 2 s + 1 −2 s −2 s −1

s 2 + 2s 0

0 s +1

Note that we do not cancel terms involving s from the numerator and denominator in each fraction because valuable information about the system will be lost otherwise. The entries of the 2 × 2 transfer matrix G ( s ) are finally formed as

= G11 ( s )

X1 ( s) X1 ( s) ( s + 1)( s 2 + 2 s ) 2 s ( s + 1) = 5 G12 ( s ) = ,= 4 3 2 5 F1 ( s ) F 0= F2 ( s ) F 0 s + 5s 4 + 5s 3 + 2 s 2 s + 5s + 5s + 2 s 2

G21 ( s ) =

1

X 2 (s) X 2 (s) s 2 + 2s s 3 + 3s 2 + 3s G s , ( ) = 5 = = 22 F1 ( s ) F 0= F2 ( s ) F 0 s 5 + 5s 4 + 5s 3 + 2 s 2 s + 5s 4 + 5s 3 + 2 s 2 2

15. The governing equation for an electric circuit is derived as

1

110 t

L

di 1 + Ri + ∫ i (t )dt = v(t ) dt C0

where L , R , and C are the inductance, resistance, and capacitance, all constants, i (t ) is the current and v(t ) is the applied voltage. Initial conditions are assumed to be zero. If v(t ) and i (t ) are the system input and output, respectively, find the transfer function. Solution Laplace transformation of the governing equation yields

1    Ls + R + =  I (s) V (s) Cs  

Transfer function

⇒

1 I (s) Cs = = 2 V ( s ) Ls + R + 1 LCs + RCs + 1 Cs

16. Electric charge q and electric current i are related via i = dq / dt . In Problem 15, find the transfer function if v(t ) and q (t ) are the system input and output, respectively. Solution 1 Noting i = dq / dt , the governing equation can be rewritten as Lq + Rq + q = v(t ) . With zero C initial conditions, Laplace transformation yields

1  2  Ls + Rs + =  Q( s) V ( s) C 

Transfer function

⇒

1 Q( s) C = = 2 1 V ( s ) Ls 2 + Rs + LCs + RCs + 1 C

17. The I/O equation for a dynamic system is given as 3 x + 2 x + 4 x = f (t ) where f and x denote the input and output, respectively. (a) Find the system’s transfer function. (b) Assuming f (t ) is the unit-impulse, find the expression for X ( s ) using (a). (c) Find the steady-state value xss via the final-value theorem. Solution X (s) 1 = 2 (a) F ( s ) 3s + 2 s + 4 1 (b) Knowing F ( s ) = 1 , from (a) we find X ( s ) = 2 . 3s + 2 s + 4 (c) Since the poles of X ( s ) are −0.3333 ± 1.1055j , the final-value theorem applies, and s   xss lim sX ( s )} lim  = = = {  0 2 s →0 s →0  3s + 2 s + 4 

18. The mathematical model of a dynamic system is given as

111

  x1 + 2 x1 + 12 ( x1 − x2 ) = f (t )  1 x2 + x2 − 2 ( x1 − x2 ) = 0 3 where f is the input and x1 and x2 are the outputs. (a) Find the appropriate transfer functions. (b) Assuming f (t ) is the unit-impulse, find the expressions for X1 ( s ) and X 2 ( s ) using (a). (c) Find the steady-state values of x1 and x2 using the final-value theorem. Solution (a) Laplace transformation of the model yields 2 1 1  F (s) ( s + 2 s + 2 ) X 1 ( s ) − 2 X 2 ( s ) =  0 − 12 X 1 ( s ) + (3s 2 + s + 12 ) X 2 ( s ) =  

Solving for X1 ( s ) and X 2 ( s ) yields − 12

F (s)

0 3s 2 + s + 12 3s 2 + s + 12 = F (s) 3s 4 + 7 s 3 + 4 s 2 + 32 s − 12 s 2 + 2 s + 12

= X1 ( s)

− 12

= X 2 (s)

3s 2 + s + 12

s 2 + 2 s + 12 F ( s ) 1 − 12 0 2 = F (s) 4 3 2 1 1 3s + 7 s + 4 s 2 + 32 s s + 2s + 2 −2 − 12

3s 2 + s + 12

The two transfer functions are 3s 2 + s + 12 X1 ( s) = 4 F ( s ) 3s + 7 s 3 + 4 s 2 + 32 s

,

1 X 2 (s) 2 = 4 F ( s ) 3s + 7 s 3 + 4 s 2 + 32 s

,

X 2 (s) =

(b) Knowing F ( s ) = 1 , from (a) we find X1 ( s) =

3s 2 + s + 12 3s 4 + 7 s 3 + 4 s 2 + 32 s

1 2

3s 4 + 7 s 3 + 4 s 2 + 32 s

(c) Poles of X1 ( s ) and X 2 ( s ) are located at the origin (simple pole) and in the left half-plane,

112 thus the final-value theorem applies, and 2    3s + s + 12  = = = x1ss lim sX s ( ) lim { }  3  1 s →0 s → 0 3s + 7 s 2 + 4 s + 3  2   1     2 = = = x2 ss lim sX 2 ( s )} lim  3 {  2 3 s →0 s → 0 3s + 7 s + 4 s +    2

1 3 1 3

19. The state-space representation of a system model is described as

x Ax + Bu =  y Cx + Du = where

x  0 1 0  = x  1= , A  = , B  = , C   −2 −1 1   x2 

, D [0 1] =

0= , u f

Find (a) The I/O equation. (b) The transfer function. Solution (a) The state-variable equations are

 x1 = x2  −2 x1 − x2 + f  x2 = Take the Laplace transform and solve for X 2 ( s ) , because x2 is the output, to find

0  sX1 ( s ) − X 2 ( s ) =  F ( s)  2 X1 ( s ) + ( s + 1) X 2 ( s ) =

⇒

sF ( s ) X 2 (s) = 2 s +s+2

x2 + x2 + 2 x2 = f . This yields the I/O equation as  (b) The transfer function is

X 2 (s) s = 2 . F (s) s + s + 2

20. The state-space representation of a system model is described as

x Ax + Bu =  y Cx + Du = where

113

0 x  = x  1=  , A  1  x2  − 2

1 0 1 0  0 = , B = , C  = , D =     , u f −3 1  0 1  0 

Find the transfer matrix. Solution The state-variable equations are

 x1 = x2  − 12 x1 − 3 x2 + f  x2 = Laplace transformation yields

0  sX1 ( s ) − X 2 ( s ) = 1 F (s)  2 X1 ( s ) + ( s + 3) X 2 ( s ) = The output equation indicates that x1 and x2 are the outputs. Solving the above system for

X1 ( s ) and X 2 ( s ) , we find = X1 ( s)

−1 0 F ( s) s + 3 1 = 2 F ( s) s −1 s + 3s + 12 1 s+3 2

s = X 2 (s)

0 1 F (s) s 2 = F (s) 2 s −1 s + 3s + 12 1 s+3 2

Then, G= 11 ( s )

X1 ( s) 1 = 2 , F ( s ) s + 3s + 12

G= 21 ( s )

X 2 (s) s = 2 F ( s ) s + 3s + 12

Finally,

1    s 2 + 3s + 1  2 G (s) =    s  2  1  s + 3s + 2 

Problem Set 4.4 In Problem 1 through 6, find the state-space form directly from the input-output equation.

114 1. 12  y + y + y = u Solution Multiply by 2 to obtain  y + 2 y + 2 y = 2u . Comparing with Eq. (4.14) we have n = 2 , a1 = 2 ,

a2 = 2 , b0 = 0 , b1 = 2 , b2 = 0 . By Eq. (4.17), x  0 1 0  = x  1= = , B =  , A    , u u  −2 −2  1   x2  By Eq. (4.18) we= have C

2] , D [0=

0 . The state-space form is therefore obtained as

 0 1 0  = x +  u  x     −2 −2  1     y = [ 0 2] x  2. 3 y + y + 2 y = 3u + 2u Solution Divide by 3 to rewrite in standard form, as  y + 13 y + 23 y =u + 23 u . Comparing with Eq. (4.14) we find n = 3 , a1 = 0 , a2 = 13 , a3 = 23 , b0 = 0 , b1 = 1 , b2 = 0 , b3 = 23 . By Eq. (4.17),

 x1    = x  x2=  , A    x3 

0  0  2 − 3

1 0 − 13

0  1=  , B 0 

0 =  0  , u u 1 

2 0 1 , D 0 = By Eq. (4.18) we have C = . The state-space form is therefore obtained as 3 

 0 1 0 0    = 0 1  x + 0  u  x  0  1  2  1   − 3 − 3 0    y =  2 0 1 x 3   3.  y + 52  y + y + 2 y =  u + 2u Solution Compare with Eq. (4.14): n = 3 , a1 = 52 , a2 = 1 , a3 = 2 , b0 = 1 , b1 = 0 , b2 = 2 , b3 = 0 . Then, Eqns. (4.17) and (4.18) yield

115

  =  x     y =  4.

2 3

 y + y + 32 y =

Solution Multiply by

3 2

0 1 0  0    1  x + 0  u 0 0 2  1   −2 −1 − 5   −2 1 − 52  x + 1 ⋅ u

1u  + u + u 3

to get  y + 32 y + 94 y = 12 u + 23 u + 23 u and compare with Eq. (4.14) to deduce n = 2 ,

a1 = 32 , a2 = 94 , b0 = 12 , b1 = 32 , b2 = 32 . The state equation is then obtained as 0 = x  9 − 4 Equation (4.18) yields C =  83

3 4

and D =

1  0  3x +  u − 2 1  1 2

so that the output equation= is y  83

3 x + 12 ⋅ u 4

The state-space form is  = x    y = 

5. 2 y (4) +  y + 2 y + y =  u + 2u + u Solution y + y + 12 y= Divide by 2 to get y (4) + 12 

0 − 9  4  83

1  0  3x +  u − 2 1  3 x + 12 ⋅ u 4

1  u + u + 12 u 2

with Eq. (4.14): n = 4 , a1 = 0 , a2 = 12 ,

a3 = 1 , a4 = 12 , b0 = 0 , b1 = 12 , b2 = 1 , b3 = 0 , b4 = 12 . Then, Eqns. (4.17) and (4.18) yield  1 0  0 0 x  =  0 0   1   − 2 −1     y =  12 0 1

0 1 0 − 12 1x 2

0 0   0  0  x + u 1 0     0  1 

.

116

y + 14 y + y = 4u 6. y (4) +  Solution Compare with Eq. (4.14): n = 4 , a1 = 0 , a2 = 1 , a3 = 14 , a4 = 1 , b0 = 0 , b1 = 0 , b2 = 0 , b3 = 0 ,

b4 = 4 . The state-space form is then formed as  1 0  0 0 x  =  0 0   1  −1 − 4     y = [ 4 0 0

0 1 0 −1

0 0   0  0 x +  u 1 0     0  1 

0] x

7. Derive the state-space representation in controller canonical form as described by Eqns. (4.19) and (4.20). Solution , x2 v ( n−2) , ... , = , xn v , the state-variable x1 v ( n−1)= xn−1 v= Choosing the state variables as= equations are formed as x1 = −a1 x1 − a2 x2 + ⋅⋅⋅ − an xn + u x2 = x1 ... xn−1 = xn−2 xn = xn−1

x Ax + Bu where Thus, the corresponding state equation is =  −a1 −a2  1 0  A  ... = ...  0  0  0 0

... −an−2 ... 0 ... ... ... 1 ... 0

−an−1 −an  0 0   , B ... ... =  0 0  1 0 

The output equation is written as

= y b0v ( n ) + b1v ( n−1) + ⋅⋅⋅ + bn−1v + bn v = b0 x1 + b1 x1 + ⋅⋅⋅ + bn−1 xn−1 + bn xn

1 0   = ... , u u   0  0 

117 Since the output equation cannot contain x1 , we substitute for x1 using the last relation in the state-variable equations above. The result is

= y b0 (−a1 x1 − a2 x2 − ⋅⋅⋅ − an xn + u ) + b1 x1 + ⋅⋅⋅ + bn−1 xn−1 + bn xn = (−b0 a1 + b1 ) x1 + (−b0 a2 + b2 ) x2 + ⋅⋅⋅ + (−b0 an + bn ) xn + b0u Finally, the output equation is expressed as = y Cx + Du where

C= b0 [ −b0 a1 + b1 −b0 a2 + b2 ... −b0 an + bn ]1×n , D =

8. A system’s I/O equation is provided as 5 6

y (4) +  y + 16  y + 12 y =  u + u + 13 u

(a) Find the state-space representation in controller canonical form. (b) Confirm the results of (a) in MATLAB. Solution y + 15  y + 53 y = 65  u + 65 u + 52 u . Comparing with Eq. (4.14), (a) Multiply by 65 to get y (4) + 65 

n = 4 , a1 = 65 , a2 = 15 , a3 = 0 , a4 = 53 , b0 = 0 , b1 = 56 , b2 = 0 , b3 = 65 , b4 =

2 5

y , x2 =  y , x3 = y and x4 = y , Eqns. (4.19) and (4.20) yield Selecting the state variables as x1 =   − 65  1 = A  0  0

− 15 0 1 0

0 − 53   0 0 , B = 0 0  1 0

1  0  =  , C 6 5 0    0 

0

6 5

2 , D 0 = 5

(b) The transfer function is directly obtained from the I/O equation, as Y (s) = U (s)

s 3 + s + 13 5 6

s 4 + s 3 + 16 s 2 + 12

>> Num = [1 0 1 1/3]; Den = [5/6 1 1/6 0 1/2]; >> [A,B,C,D] = tf2ss(Num,Den) A = -1.2000 1.0000 0 0

-0.2000 0 1.0000 0

0 0 0 1.0000

-0.6000 0 0 0

118

B = 1 0 0 0 C = 1.2000

0

1.2000

0.4000

D = 0

In Problems 9 through 12, given the transfer function Y ( s ) / U ( s ) , find (a) The input-output equation. (b) The state-space form directly from the input-output equation in (a).

9.

s + 34 s 3 + 13 s + 1

Solution (a) Using the given transfer function, s + 34 Y (s) = 3 ⇒ (s 3 + 13 s + 1)Y ( s ) =( s + 34 )U ( s ) 1 U (s) s + 3 s + 1

Transforming back to time domain, we find  y + 13 y + y = u + 34 u . (b) Compare with Eq. (4.14), and subsequently use Eqns. (4.17) and (4.18) to find

  =  x      = y 10.

 0 1 0 0      0 0 1  x + 0  u 1   1   −1 − 3 0  3  4 1 0  x + 0 ⋅ u

1 s2 + 1 5 2 3 s + 5 s + 101

Solution (a) Using the given transfer function, 1 s2 + 1 Y (s) = 25 1 U ( s ) s + 53 s + 10

1 )Y ( s )= ( 1 s 2 + 1)U ( s ) ⇒ (s 2 + 53 s + 10 5

119

1 y = 1u  + u . Transforming back to time domain, we have  y + 53 y + 10 5

(b) Compare with Eq. (4.14), and use Eqns. (4.17) and (4.18) to find  = x    =  y

 0 − 1  10 49  50

1 0  3 x +  u − 5 1  3  1 − 25  x + 5 ⋅u

s3 + 2s + 1 2 s3 + s 2 + s + 1 3 Solution (a) Using the given transfer function,

11.

Y ( s) = U (s)

s3 + 2s + 1 ⇒ ( 23 s 3 + s 2 + s + 1)Y ( s )= ( s 3 + 2 s + 1)U ( s ) 2 s3 + s 2 + s + 1 3

Transforming back to time domain, we have

2  y+ 3

 y + y + y =  u + 2u + u .

(b) Multiply by 32 to get  y + 32  y + 32 y + 32 y = 32  u + 3u + 32 u . Compare with Eq. (4.14), and use Eqns. (4.17) and (4.18) to find  0   =  x  0  − 3  2     y = − 3  4 

1 0 − 23 3 4

0  0   1  x + 0  u 1  − 23  − 94  x + 32 ⋅ u

s3 + 2s 2 s 4 + s 2 + 3s + 2 Solution (a) Using the given transfer function, 12.

Y (s) s3 + 2s = ⇒ (2 s 4 + s 2 + 3s + 2)Y ( s ) = ( s 3 + 2 s )U ( s ) 4 2 U ( s ) 2 s + s + 3s + 2

Transforming back to time domain, we have 2 y (4) +  y + 3 y + 2 y =  u + 2u .

120

(b) Divide by 2, compare with Eq. (4.14), then use Eqns. (4.17) and (4.18) to find    x =       =  y

1 0 0 0  0 0  3  −1 − 2

− 12

1 0 0

1  x + 0⋅u 2

0 1 0

0 0    0 0 x +  u 1 0     0  1 

13. The state-variable equations and the output equation for a dynamic system are given as

 x1 = x2 , = y  − x2 − x1 + u  x2 =

1x +x 2 2 1

Find the transfer function (or matrix) by determining the Laplace transforms of x1 and x2 in the state-variable equations and using them in the Laplace transform of the output equation. Solution Laplace transformation of state-variable equations yields

0  sX1 ( s ) − X 2 ( s ) =  U (s)  X1 ( s ) + ( s + 1) X 2 ( s ) = Using Cramer’s rule,

= X1 ( s)

1 s = U (s) , X 2 (s) 2 U (s) 2 s + s +1 s + s +1

Insert into the Laplace transform of the output equation:

Y (s) =

1 2

X1 ( s) + X 2 ( s) =

1 2

s + s +1 2

U (s) +

s U (s) s + s +1 2

Finally, the transfer function is found as

s + 12 Y (s) G = (s) = 2 U (s) s + s + 1 14. Repeat Problem 13 for

 x1 = x2 ,  − 12 x2 − x1 + 12 u  x2 =

x  y =  1  x2 

121 Solution Laplace transformation of state-variable equations yields

0  sX1 ( s ) − X 2 ( s ) =  1 1 U (s)  X1 ( s) + ( s + 2 ) X 2 ( s) = 2 Using Cramer’s rule, = X1 ( s)

1 1s 2 2 = U ( s ) , X ( s ) U (s) 2 2 2 1 s + 2 s +1 s + 12 s + 1

Insert into the Laplace transform of the output equation: 1   2 U (s)   2 1  s + 2 s +1  Y( s) =   1s   2  s 2 + 1 s + 1U ( s)    2

The transfer matrix is then obtained as 1   2  2 1   s + 2 s + 1 G (s) =   1s   2  s 2 + 1 s + 1   2

In Problems 15 through 19, the matrices in the state-space form of a system model are given. (a) Find the transfer function (or transfer matrix) using Eq. (4.21). (b) Verify the result of (a) using the ss2tf command.

1  0 0 = = , B = , C  12 = 1 , D 0 15. A     1 2 − − 2   7  Solution (a) The system is SISO, hence there is only one transfer function. Since G ( s ) is 1× 1 , it is simply denoted by G ( s ) . With D = 0 , Eq. (4.21) reduces to G= ( s ) C( sI − A ) −1 B . Then, 7 s + 1 7  0  1 7(2 s + 1) 2s + 1 C( sI − A) −1 B = 12 1 2 =2 G ( s) =  −14 7 s   2  = 2 7 s + s + 14     7 s + s + 14 s + 17 s + 2 (b) >> A = [0 1;-2 -1/7]; B = [0;2]; C = [1/2 1]; D = 0; [n,d] = ss2tf(A,B,C,D) n = 0 d =

2

1

% Num = 2s+1;

122 1.0000

0.1429

2.0000

% Den = s2+(1/7)s+2

0 0 1 = = , B = 1] , D 13 16. A   2  , C [0 =  −1 −2  3 Solution (a) The system is SISO, hence there is only one transfer function. Since G ( s ) is 1× 1 , it is simply denoted by G ( s ) . Eq. (4.21) yields  s + 2 1  0  1 1 s 2 + 4s + 1 ) C( sI − A) −1 B + D= [ 0 1] 2 G ( s= + =   2 s + 2 s + 1  −1 s   3  3 3( s 2 + 2s + 1) (b) >> A = [0 1 ;-1 -2]; B = [0;2/3]; C = [0 1]; D = 1/3; >> [n,d] = ss2tf(A,B,C,D) n = 0.3333

1.3333

0.3333

d = 1

2

1

0 0 1 0 1 0 0      17. A  0 0 1= , 0 , = = = B C   , D 02×1    0 0 1    − 2 − 1 −1 1 3 2  3  Solution = (a) The output equation is y C2×3x3×1 + D2×1u , hence there are two outputs and one input. Therefore, G ( s ) is 2 ×1 . By Eq. (4.21),

G ( s ) =C( sI − A) −1 B + D 3s 2 + 3s + 1 3( s + 1) 3  0    1 0 0  1 3s ( s + 1) 3s   0  =  −2   3 2 0 0 1  3s + 3s + s + 2  1 −( s + 2) 3s 2   2   −2 s 1 3 =   3 2 2(3s + 3s + s + 2)  s 2  (b) >> A = [0 1 0;0 0 1;-2/3 -1/3 -1]; B = [0;0;1/2]; C = [1 0 0;0 0 1]; >> D = [0;0]; [n,d] = ss2tf(A,B,C,D) n = 0 0

0 0.5000

0 0

0.5000 0

1.0000

1.0000

0.3333

0.6667

d =

123

0 0  0 1 0  12 0  0 1 0      18. A B C D = = = = 0 0 1 , 1 0 , ,   0 0 1      0 0    1    −2 −1 −2  0 4  Solution = (a) The output equation is y C2×3x3×1 + D2×2u 2×1 , hence there are two outputs and two inputs. Therefore, G ( s ) is 2 × 2 . By Eq. (4.21),

G ( s ) =C( sI − A ) −1 B + D  s 2 + 2s + 1  3 2  s + 2s + s + 2 0 1 0  −2 0 0 1   3 2    s + 2s + s + 2  −2s  3  s + 2s 2 + s + 2  s 3 + 4 s 2 + 5s + 2  3 2( s + 2 s 2 + s + 2) =  −( s + 2)  3 2  s + 2 s + s + 2

1 2 s +1 s 2 s +1 −1 2 s +1

  s3 + 2s 2 + s + 2  0 0   s   12 0  1 0   +  0 0 s 3 + 2 s 2 + s + 2    1 0 4    2  s  3 2 s + 2s + s + 2  1

 s  4( s 3 + 2 s 2 + s + 2)   s2  4( s 3 + 2 s 2 + s + 2) 

(b) >> A = [0 1 0;0 0 1;-2 -1 -2]; B = [0 0;1 0;0 1/4]; C = [0 1 0;0 0 1]; >> D = [1/2 0;0 0]; >> [n1,d1] = ss2tf(A,B,C,D,1) % Input 1 n1 = 0.5000 0

2.0000 0

2.5000 -1.0000

1.0000 -2.0000

1.0000

2.0000

1.0000

2.0000

% 0.5s3+2s2+2.5s+1 % -(s+2)

d1 =

>> [n2,d2] = ss2tf(A,B,C,D,2)

% Input 2

n2 = 0 0

0 0.2500

0.2500 0

0 0

1.0000

2.0000

1.0000

2.0000

d2 =

% 0.25s % 0.25s2

124

0 1 0 0 0  1 0 0  0 0     = 1  , B = 1 0  , C  = , D  19. A  0 0=  0 1 0 0 1     −1 0 −1 0 2  Solution = (a) The output equation is y C2×3x3×1 + D2×2u 2×1 , hence there are two outputs and two inputs. Therefore, G ( s ) is 2 × 2 . By Eq. (4.21),

G ( s ) =C( sI − A) −1 B + D  s ( s + 1) 1  0 0  s +1 1 0 0    1  + 0 0  1 ( 1) 1 0 =  − s s + s    3 2  0 1  0 1 0  s + s + 1   2  0 2   −1 s    −s 2  s +1   s3 + s 2 + 1 3 2 s + s +1   =  s ( s + 1) s3 + s 2 + 2s + 1   3 2  s3 + s 2 + 1   s + s +1 (b) >> A = [0 1 0;0 0 1;-1 0 -1]; B = [0 0;1 0;0 2]; C = [1 0 0;0 1 0]; >> D = [0 0;0 1]; >> [n1,d1] = ss2tf(A,B,C,D,1) % Input 1 n1 = 0 0

0 1

1 1

1 0

% s+1 % s2+s

d1 = 1.0000

1.0000

-0.0000

>> [n2,d2] = ss2tf(A,B,C,D,2)

1.0000 % Input 2

n1 = 0 1.0000

0 1.0000

0 2.0000

2.0000 1.0000

1.0000

1.0000

-0.0000

1.0000

% 2 % s3+s2+2s+1

d1 =

20. A dynamic system model has state variables x1 and x2 and input u . The two I/O equations are derived as  x1 + x1 + 2 x1 =u + 13 u ,  x2 + x2 + 2 x2 =2u Find the state-space representation for this model and subsequently the transfer matrix.

125 Solution Using Eqns. (4.17) and (4.18), the first I/O equation generates

0 1 0  = A  = , B = , C  13 = 1 , D 0    −2 −1 1  The second I/O equation gives

0 1 0  = A  = , B =   , C  −2 −1 1 

[2

= 0] , D 0

As expected, A and B are the same in both cases. This implies that the system model statespace form is x Ax + Bu =  y Cx + Du = where  13 1  0 1 0  0 = A  = , B = , C = , D    1  0  −2 −1     2 0 The transfer matrix is therefore 2 ×1 and obtained as G= ( s ) C( sI − A) −1 B  3s + 1   13 1   s + 1 1 0   3( s 2 + s + 2)  1  = =  2  −2 s  1   2 0   s s 2 + + 2       2   s +s+2  The two individual transfer functions in G ( s ) are clearly valid as observed by the given I/O equations.

126 Problem Set 4.5 1. Reduce the block diagram in Figure 4.25 by moving the constant gain block K to the right of the summing junction. Subsequently, find the transfer function Y ( s ) / U ( s ) .

U (s)

+

K

Σ

Y (s)

G (s) −

H (s) Figure 4.25 Problem 1. Solution Based on the strategy in Figure 4.15, the block diagram reduces to what is shown in Figure PS45 No1. This is a negative feedback loop, hence

Y (s) KG KG = = U ( s ) 1 + KG H 1 + GH K U (s)

+

Σ

−

K

Y ( s)

G

H/K Figure PS4-5 No1

2. Consider the block diagram in Figure 4.26. Use the block diagram reduction steps listed below to find Y ( s ) / U ( s ) . (a) Replace the single loop containing G1 ( s ) and H ( s ) with a single equivalent block. (b) Move the gain block K to the right of the summing junction. (c) Replace the resulting loop with a single equivalent block. U (s)

K

+

Σ−

−

G1 ( s )

G2 ( s )

H ( s)

Figure 4.26 Problem 2. Solution Details are shown in Figure PS4-5No2.

Y (s)

127

U (s)

K

Σ

G1 1 + HG1

Σ

K

+ −

Y (s)

G2

(a)

U (s)

+

−

(b)

G1G2 1 + HG1

Y (s)

1/ K

U (s) (c )

KG1G2 1 + HG1 + G1G2

Y ( s)

Figure PS4-5 No2 3. For the block diagram in Figure 4.27, find the transfer function Y ( s ) / U ( s ) using (a) Block diagram reduction techniques. (b) Mason’s rule. H (s) G2 ( s ) U (s)

+

Σ

G1 ( s )

G3 ( s )

G4 ( s )

+

+

Σ

Y (s)

+

Figure 4.27 Problem 3. Solution (a) The parallel connection involving G2 and G3 is replaced with a single block G2 + G3 . This block, together with G1 and G4 form a series combination, hence replaced with G1 (G2 + G3 )G4 . This new block and H are in a parallel connection. The final block is G1 (G2 + G3 )G4 + H . (b) There are no loops and three forward paths with gains H , G1G2G4 and G1G3G4 . All paths are also coupled, thus the special case applies and yields

Y (s) = H + G1G2G4 + G1G3G4 U (s) This agrees with the result of (a). 4. Using Mason’s rule, find Y ( s ) / U ( s ) for the block diagram in Figure 4.16, Example 4.21. Solution There is one loop with gain −G1H1 and two forward paths with gains G1G2 K and G1H 2 K . Since all paths are coupled, special Mason’s rule gives

128

Y ( s ) G1G2 K + G1H 2 K G1K (G2 + H 2 ) = = U (s) 1 + G1H1 1 + G1H1 5. Find Y ( s ) / U ( s ) in Figure 4.28 using Mason’s rule. Solution There are three loops with gains G1 H1 , − KG1G2 H 3 and −G2 H 2 , as well as one forward path with gain KG1G2 . Since the three loops and the forward path are coupled, the special case of Mason’s rule applies. Therefore, KG1G2 Y (s) = U ( s ) 1 + KG1G2 H 3 − G1H1 + G2 H 2

6. Use block diagram reduction techniques listed below to find the overall transfer function for the diagram in Figure 4.28. (a) Move the summing junction of the negative feedback loop containing H 2 ( s ) outside of the psitive loop containing H1 ( s ) . (b) In the ensuing diagram, replace the loop containing H1 ( s ) with a single block. (c) Similarly, replace the two remaining loops with single equivalent blocks to obtain one single block whose input is U ( s ) and whose output is Y ( s ) . H1 ( s ) U ( s)

+

Σ

−

K

+

Σ

+

G1 ( s )

+

Σ

−

G2 ( s )

Y (s)

H 2 (s) H 3 (s)

Figure 4.28 Problems 5 and 6. Solution Details are shown in Figure PS4-5No6. The overall transfer function is readily obtained as KG1G2 Y (s) = U ( s ) 1 + KG1G2 H 3 − G1H1 + G2 H 2

129 H1 ( s ) U (s)

+

Σ

−

K

+

Σ

+ −

Σ

+ G2 ( s)

G1 ( s )

Y (s)

H 2 ( s) G1 ( s ) (a )

U (s)

H 3 ( s)

+

Σ

−

K

+

Σ

−

G1 ( s) 1 − G1 ( s ) H1 ( s )

G2 ( s )

Y ( s)

H 2 ( s) G1 ( s ) (b)

U (s)

H 3 ( s)

+

Σ

−

(c )

K

G1 ( s )G2 ( s ) 1 − G1 ( s ) H1 ( s ) + G2 ( s ) H 2 ( s)

Y (s)

H 3 ( s)

U ( s)

KG1 ( s)G2 ( s) 1 − G1 ( s ) H1 ( s ) + G2 ( s ) H 2 ( s) + KG1 ( s)G2 ( s) H 3 ( s)

Y (s)

(d )

Figure PS4-5No6 7. In Figure 4.29, find the transfer function Y ( s ) / U ( s ) using (a) Block diagram reduction steps. (b) Mason’s rule. Solution (a) The negative feedback loop is replaced with a single block of

G1 . This, together with 1 + G1H1

130

K and G2 form a series connection, hence replaced with

KG1G2 . This block and H 2 are in a 1 + G1H1

parallel connection, thus the transfer function is KG1G2 Y = + H2 U 1 + G1H1

(b) Since not all paths are coupled, general case of Mason’s rule will be applied. There are two forward paths and one negative feedback loop. The quantities in Eq. (4.25) are:

F1 = H 2 , F2 = KG1G2 , D = 1 + G1H1 D1 = same as D when restricted to the bottom portion= 1 + G1H1 D2 = same as D when restricted to the top portion = 1 Then KG1G2 Y ( s ) H 2 (1 + G1H1 ) + KG1G2 = = H2 + U (s) 1 + G1H1 1 + G1H1

U ( s)

+

H 2 (s)

Σ

Y ( s)

+ K

+

Σ

−

G1 ( s )

G2 ( s )

H1 ( s )

Figure 4.29 Problem 7.

8. Find the overall transfer function in Figure 4.30 using Mason’s rule. Solution There are two loops with gains G2 H 2 and −G2G3 H 3 , and one forward path with gain G1G2G3 . All paths are coupled hence Mason’s rule (special case) applies: G1G2G3 Y (s) = U ( s ) 1 − G2 H 2 + G2G3 H 3

131 G1 ( s )

U (s)

+ −

Σ

G2 ( s )

+

G3 ( s )

Y (s)

H 2 (s) H 3 (s)

Figure 4.30 Problem 8. 9. For the block diagram in Figure 4.31, find Y ( s ) / U ( s ) using (a) Block diagram reduction. (b) Mason’s rule. U (s)

+ −

Σ

G1 ( s )

+ −

Σ−

G2 ( s )

G3 ( s )

Y (s)

H 2 (s) H1 ( s )

Figure 4.31 Problem 9. Solution (a) The negative loop on the left is replaced with

G1 . The negative loop containing G2 1 + G1H1

G2 . This block and G3 are in a series and part of a negative 1 + G2 H 2 feedback, which is replaced with

and H 2 is replaced with

G2G3 G2G3 1 + G2 H 2 = G2G3 1 + G2 H 2 + G2G3 1+ 1 + G2 H 2

This block and

G1 are in a series, hence 1 + G1H1 G1G2G3 Y (s) = U ( s ) (1 + G1H1 )(1 + G2 H 2 + G2G3 )

(b) General case of Mason’s rule applies because the loop on the left is not coupled with the other two. There is one forward path and three negative feedback loops. The quantities in Eq. (4.25) are: F1 = G1G2G3 ,

D= 1 + G1H1 + G2 H 2 + G2G3 + G1H1G2 H 2 + G1H1G2G3   Single-loop gains

Gain product of non-touching two-loops

132

D1 = same as D when restricted to the portion not touching the forward path = 1 Therefore, G1G2G3 Y (s) = U ( s ) 1 + G1H1 + G2 H 2 + G2G3 + G1H1G2 H 2 + G1H1G2G3

Of course, this agrees with the result of (a).

10. Consider the block diagram in Figure 4.32 where V ( s ) represents an external disturbance. (a) Find the transfer function Y ( s ) / U ( s ) by setting V ( s ) = 0 . (b) Find Y ( s ) / V ( s ) by setting U ( s ) = 0 . Solution (a) When V ( s ) = 0 , the forward path going from U ( s ) to Y ( s ) has a gain of G1G2 , while the two loops have gains −G2 H 2 and −G1G2 H1 . Using the special case of Mason’s rule, G1G2 Y (s) = U ( s ) 1 + G2 H 2 + G1G2 H1

(b) When U ( s ) = 0 , the forward path goes from V ( s ) to Y ( s ) and has gain G2 . The two loops have gains −G2 H 2 and −G1G2 H1 . Mason’s rule (special case) yields G2 Y (s) = V ( s ) 1 + G2 H 2 + G1G2 H1 External disturbance

V (s) U (s)

+

Σ

−

G1 ( s )

+

Σ

+ −

G2 ( s )

Y (s)

H 2 ( s) H1 ( s )

Figure 4.32 Problem 10.

11. The state-variable equations and output equation of a system are given as

 x1 = x2 , = y 2 x1 + x2  1 x − x + 2u x = − 2 1 2 2 

133 where u and y are the input and output, respectively. (a) Construct the block diagram. (b) Find the transfer function Y ( s ) / U ( s ) directly from the block diagram. (c) Find the transfer function Y ( s ) / U ( s ) from the state-space form and compare with (b). Solution (a) Details are shown in Figure PS4-5No11.

U ( s)

2

+ −

Σ −

x2

1 s

x2

1 s

x1

2

+

x2 +

Σ

Y (s)

x2 1 x 2 1

1 2

Figure PS4-5No11 (b) There are two forward paths with gains

1 4 1 2 and , and two loops with gains − and − 2 . 2 s s 2s s

All paths are coupled and by the special case of Mason’s rule we have 4 2 + 2 Y (s) 4s + 8 s s = = 2 1 1 U ( s) 1 + + 2s + 2s + 1 2 s 2s (c) In state-space form, we have 1 0 0 = A  1 = , B =   , C [ 2 1] 2  − 2 −1 Therefore, 4( s + 2) Y (s) = C( sI − A) −1 B = 2 U (s) 2s + 2s + 1 This agrees with the result of (b). 12. The state-space representation of a system model is given as x Ax + Bu =  y Cx + Du = with 1  0 x  0  = = , B   = , C 1 12  = , D 0= , u u x  1  ,= A   2 − 2 − x 3    2 3  (a) Contruct the block diagram. (b) Find the transfer function Y ( s ) / U ( s ) directly from the block diagram. (c) Find the transfer function Y ( s ) / U ( s ) from the state-space form and compare with (b). Solution (a) The state-space form reveals the following information:

134

 x1 = x2 ,  −2 x1 − 23 x2 + 3u  x2 =

y= x1 + 12 x2

Proceeding as always, the block diagram is constructed and shown in Figure PS4-5No12. 1 2

U (s)

3

+ −

Σ −

2x1

2 3

x2

x2

x2

1 s

1 s

x1

+

Σ

+ Y (s)

2 3

2

Figure PS4-5No12 (b) Two forward paths with gains

2 3 3 2 − and , and two loops with gains and . All − 3s 2s s2 s2

paths are coupled and by the special case of Mason’s rule we have

3 3 + 3(3s + 6) Y (s) s 2 2s = = U ( s ) 1 + 2 + 2 2(3s 2 + 2 s + 6) 3s s 2 (c) In state-space form, we have

1  0 0 = = , B = A    , C 1 2  3  −2 − 3  Therefore,

1 2

Y (s) 9( s + 2) = C( sI − A) −1 B = 2 U (s) 2(3s + 2 s + 6)

This agrees with the result of (b).

In problems 13 through 16, the I/O equation of a dynamic system is provided. (a) Find the state-space model. (b) Construct the block diagram based on the information in (a). (c) Determine the overall transfer function directly from the block diagram in (b). 1 u 13. 3  y + y + 2 y = 3 Solution (a) With x1 = y and x2 = y , we find

135

 x1 = x2 ,  1   1  x2 = 3  −2 x1 − x2 + 3 u 

y = x2

Therefore, the matrices in the state-space representation are

0 = A  2 − 3

1 0  = , B =  1 1 , C − 3 9

1] , D [0 =

0

(b) The block diagram is constructed as shown in Figure PS4-5No13. y = x2 U (s)

+

1 9

−

x2

1 s

Σ −

1 s

x2

x1

1 3 2 3

Figure PS4-5No13 (c) Noting that x2 is the output, there is one forward path with gain gains −

1 , and two loops with 9s

2 1 and − 2 . All paths are coupled and by the special case of Mason’s rule we have 3s 3s 1 Y (s) s 9s = = U ( s) 1 + 1 + 2 3(3s 2 + s + 2) 2 3s 3s

14. 23  y + 2 y + y = 13 u + 2u Solution (a) With x1 = y and x2 = y , we find

 x1 = x2 , = y 2 x1 + 13 x2  3 = − − + x ( x 2 x u ) 1 2  2 2 (b) The block diagram can then be constructed as shown in Figure PS4-5No14. 1 3

U (s)

3 2

+ −

Σ −

x2

x2

1 s

1 s

3 3 2

Figure PS4-5No14

x1

2

+

+

Σ

Y ( s)

136

3 3 3 1 and 2 , and two loops with gains − and − 2 . All s 2s 2s s paths are coupled and by the special case of Mason’s rule we have 1 3 + 2 Y (s) s+6 2 s s = = 2 U (s) 1 + 3 + 3 2s + 6s + 3 s 2s 2 (c) Two forward paths with gains

15. 2 y +  y + y =  u + 2u Solution y , we find (Section 4.4) (a) With x1 = y , x2 = y and x3 =  0 1 0  0   0  , C = 1  − 14 1 − 14  , D = A= 2  0 0 1  , B=   − 1 0 − 1    1    2 2 The above can then be expressed as  x1 = x2  x3 , y= − 14 x1 + x2 − 14 x3 + 12 u  x2 =  x = 1 1  3 − 2 x3 − 2 x1 + u

(b) The block diagram can then be constructed as shown in Figure PS4-5No15. Direct link from input to output

1 2

x3

U ( s)

+ −

Σ −

x3

1 s

x3

1 4

1 s

x2

1 s

x1

1 4

− +

+ −

Σ

Y (s)

1 2 1 2

Figure PS4-5No15 (c) Because not all paths are coupled, the general case of Mason’s rule applies as follows: Four −1 1 1 −1 forward paths with gains F1 = , F2 = , F3 = 3 and F4 = 2 , and two loops with gains 2 4s 4s s 1 1 and − 3 . − 2s 2s

D =+ 1

1 1 + 3 , D1 =D , D2 == 1 D3 =D4 2s 2s

− s 2 + 4s − 1 Y ( s ) ∑ i =1 Fi Di 12 D + F2 + F3 + F4 1 s3 + 2s = = = + = 2 2(2 s 3 + s 2 + 1) 2 s 3 + s 2 + 1 U (s) D D 4

137

16.  y + 12 y + 2 y = 12  u +u Solution y , we find (Section 4.4) (a) With x1 = y , x2 = y and x3 =  0 1 0 0    0 , C = 1 0 − 14 0  , D = A= 0 1 , B = 2 0    −2 − 1 0    1    2  The above can then be expressed as  x1 = x2  x3 , y= − 14 x2 + 12 u  x2 =  x = 1  3 − 2 x2 − 2 x1 + u

(b) The block diagram can then be constructed as shown in Figure PS4-5No16. Direct link from input to output

1 2 1 4

U (s)

+ −

Σ −

x3

x3

1 s

1 s

x2

1 s

−

Σ

+ Y (s)

x1

1 2

2

Figure PS4-5No16 (c) Because not all paths are coupled, general case of Mason’s rule applies as follows: There are −1 2 1 1 two forward paths with gains F1 = and F2 = 2 , and two loops with gains − 2 and − 3 . 2 4s s 2s

1 2 + D , D2 = , D1 = 1 2s 2 s3 1 2 − 2 Y ( s ) ∑ i =1 Fi Di 12 D + F2 1 s3 + 2 4s = == + = 2 1 + 1 + 2 2s3 + s + 4 U (s) D D 2s 2 s3 D= 1+

17. A system is described by its transfer function s + 13 Y (s) = U ( s ) s (2 s 2 + s + 2) (a) Find the I/O equation. (b) Find the state-space model from the I/O equation. (c) Build a block diagram based on the information in (b). Find the transfer function directly from the block diagram and compare with the given transfer function. Solution (a) The input-output equation is derived as 2 y +  y + 2 y = u + 13 u .

138

y , we find (b) With x1 = y , x2 = y and x3 =   x1 = x2  x2 x3 , = y =  x = 1 (− x − 2 x + u ) 3 2  3 2

1x 6 1

+ 12 x2

(c) The block diagram can then be constructed as shown in Figure PS4-5No17. There are two 1 1 1 1 forward paths with gains 2 and 3 , and two loops with gains − and − 2 . All paths are 2s 6s s 2s coupled and by the special case of Mason’s rule we have 1 1 + 3 2 3s + 1 Y (s) 6s s = 2= 1 1 U (s) 1 + + 3s (2 s 2 + s + 2) 2 2s s This agrees with the original transfer function. 1 2

U ( s)

+ −

Σ −

x3

1 s

x3

1 s

x2

1 s

x1

1 6

+

+

Σ

Y (s)

1 2

Figure PS4-5No17 18. Repeat Problem 17 for s 2 + 13 Y (s) = U ( s) 2s 2 + s + 2

Solution (a) The input-output equation is derived as 2  y + y + 2 y = u + 13 u . (b) With x1 = y and x2 = y , we find (Section 4.4)

1  0 0 1  − 13 − 14  , D = A=  −1 − 1  , B = 1  , C = 2    2 The above can then be expressed as  x1 = x2 , y= − 13 x1 − 14 x2 + 12 u  1 x +u x = − x − 1 2 2  2 (c) Because not all paths are coupled, the general case of Mason’s rule applies as follows: 1 1 1 1 Three forward paths with gains F1 = , F2 = − 2 and F3 = − , and two loops with gains − 4s 2 2s 3s and −

1 . s2

139

1 1 + , D1 =D , D2 == 1 D3 2s s 2 1 1 3 − 2− 1 D+F +F F D s 2 + 13 Y ( s ) ∑ i =1 i i 2 1 2 3 4s = = = = + 3s U (s) D D 2 1 + 1 + 1 2s 2 + s + 2 2s s 2 D =+ 1

This agrees with the original transfer function. Direct link from input to output

1 2 1 4

U ( s)

+ −

Σ −

x2

1 s

x1

1 s

x2

− −

1 3

+

Σ

Y (s)

1 2

Figure PS4-5No18

19. The block diagram representation of a system model is presented in Figure 4.33, where x1 and x2 denote the two state variables. (a) Derive the state-space form directly from the block diagram. (b) Find the transfer function directly from the block diagram. U (s)

2

+ −

Σ

−

1 s

X 2 (s)

+

Σ

1 s

−

1 3

X1 ( s)

1 2

3

Figure 4.33 Problem 19. Solution (a) The left and right summing junctions respectively yield  sX 2 ( s ) = − 13 X 2 ( s ) − 3 X1 ( s ) + 2U ( s )  − 12 X1 ( s ) + X 2 ( s )  sX1 ( s ) = The output is clearly Y ( s ) = X1 ( s ) . Therefore,

 x1 = − 12 x1 + x2 ,  −3 x1 − 13 x2 + 2u  x2 = The state-space form is then obtained as x Ax + Bu =  y Cx + Du = with

y = x1

Y (s)

140

− 1 1  x  0 = x  1  ,= A  2 = , B , C [1 0] = , D 0= , u u  2 = 1 − 3 − x      2 3  (b) To find the transfer function, we must use the general case of Mason’s rule. The pertinent quantities are 2 F1 = 2 s

,

1 D =+

1 1 3 + + 2+ 3s 2 s s  Single-loop gains

1 6s 2

1 D1 =

,

Non-touching two-loops

The transfer function is then found as

2 ⋅1 2 Y ( s ) F1D1 12 s = = = 2 1 1 3 1 U (s) D 1 + + + 2 + 2 6 s + 5s + 19 3s 2 s s 6s 20. Repeat Problem 19 for the block diagram in Figure 4.34, where x1 , x2 and x3 denote the state variables. 1 3

2 U (s)

+ −

Σ

−

1 s

X 3 (s)

1 s

X 2 (s)

1 s

X1 ( s)

−

+

Σ +

Y ( s)

2 3 1 2

Figure 4.34 Problem 20. Solution − 23 X 3 ( s ) − 12 X1 ( s ) + U ( s ) . The next two (a) The left summing junction yields sX 3 ( s ) = integrators give sX 2 ( s ) = X 3 ( s ) and sX1 ( s ) = X 2 ( s ) . The right summing junction yields

Y ( s ) =X1 ( s ) − 2 X 3 ( s ) + 13 U ( s ) . Therefore,  x1 = x2  ,  x2 =x3  x = 2 1  3 − 3 x3 − 2 x1 + u

y =x1 − 2 x3 + 13 u

The state-space form is then obtained as

x Ax + Bu =  y Cx + Du = with

141

0 1 0  x1  0      x = x2  , A = 0 0 1  , B =0  , C =[1 0 −2] , D =13 , u =u x  − 1 0 − 2  1   3 3  2 (b) To find the transfer function, we must use the general case of Mason’s rule. The pertinent quantities are 1 2 1 F1 = , F2 = − , F3 =3 , 3 s s The transfer function is then found as

1+ D=

2 1 + 3 3s 2 s

,

1 , D3 = 1 D1 = D , D2 =

2 1 − + 3 D + F2 + F3 1 Y (s) 6 s 3 − 32 s 2 + 21 s s = = + = U (s) D 3 1+ 2 + 1 3(6 s 3 + 4 s 2 + 3) 3s 2 s 3 1 3

Problem Set 4.6 In Problems 1 through 10, the mathematical model of a nonlinear system is provided. Using the procedure outlined in this section, find the operating point and derive the linearized model. 1.  x + 2 x = + x3 u (t ) , = u (t ) unit-step , = x(0) 12 , = x (0) 0 Solution The 4-step procedure is followed as listed below: Step 1: The operating point satisfies x 3 = 1 so that x = 1 . Step 2: The nonlinearity f ( x) = x 3 is linearized about x = 1 as f ( x) = x3 ≅ f (1) + 3 x 2 

x =1

∆x = 1 + 3∆x

Step 3: Substitute x = 1 + ∆x and the linear approximation into the original equation: (1 + ∆x) + 2(1 + ∆x) + [1 + 3 = ∆x] u (t )

⇒

∆ x + 2∆x += 3∆x 0

Step 4: Adjust the initial conditions:

∆x(0) = x(0) − 1 = − 12 , ∆x (0) = x (0) = 0 Therefore, the linearized model is obtained as

∆ x + 2∆x + 3∆x =0 , ∆x(0) =− 12 , ∆x (0) =0 1 + sin t , x (0) = x + 13 x3 = 1 , x (0) = 1 2.  3

142 Solution The 4-step procedure is followed as listed below: Step 1: The operating point satisfies 13 x 3 = 13 so that x = 1 . Step 2: The nonlinearity f ( x) = x 3 is linearized about x = 1 as f ( x) = x3 ≅ f (1) + 3 x 2 

x =1

∆x = 1 + 3∆x

Step 3: Substitute x = 1 + ∆x and the linear approximation into the original equation: (1 + ∆x) + 13 [1 + 3∆x] =

1 3

+ sin t

⇒

∆ x + ∆x = sin t

Step 4: Adjust the initial conditions: ∆x(0) = x(0) − 1 = 0 , ∆x (0) = x (0) = 1

Therefore, the linearized model is obtained as ∆ x + ∆x = sin t , ∆x(0) = 0 , ∆x (0) =1

 x + x + 3 x x =13 + cos t , x(0) =− 12 , x (0) =0 Solution We follow the 4-step procedure outlined in the section. First note that

3.

1 2

 3 x 2 if x ≥ 0 f ( x) = 3x x =  2 −3 x if x < 0

Differentiate

⇒

 6 x if x ≥ 0 f ′( x) =  −6 x if x < 0

Step 1: The operating point x satisfies 3x x = 13 . 1 Case (1): x < 0 leads to −3x 2 = , which has no real solution. 3

Case (2): x > 0 leads to 3x 2 = 13 so that x = ± 13 . But since x > 0 , the only valid solution is

x = 13 . Therefore, the operating point is x = 13 . Step 2: Linearizing f ( x) = 3 x x about the operating point, we find

f ( x)= 3 x x ≅ f ( 13 ) + [ 6 x ]x = 1 ⋅ ∆x = 3

1 3

+ 2∆x

Step 3: Insert x =∆x + 13 and the linear approximation into the original equation:

143 1 2

∆ x + ∆x +  13 + 2∆x  =

1 3

+ cos t

⇒

1 2

∆ x + ∆x + 2∆x = cos t

Step 4: Initial conditions are adjusted: ∆x(0) =x(0) − 13 =− 65 , ∆x (0) = x (0) = 0 . Therefore, the linearized model is 1 2

∆ x + ∆x + 2∆x =cos t , ∆x(0) =− 65 , ∆x (0) =0

x + x + x x = 18 + sin( 12 t ) , x(0) = 0 , x (0) = 1 4.  Solution First note that

 x x if x ≥ 0 f ( x) = x x =   x − x if x < 0 Step 1: The operating point must satisfy x

⇒

 3 x if x ≥ 0 2 f ′( x) = 3  2 − x if x < 0

x = 18 .

1 Case (1): x < 0 yields x − x = , which has no real solution. 8

Case (2): x > 0 leads to x x = 18 , thus x = 14 . Thus, the operating point is x = 14 . Step 2: Linearization of f ( x) = x x about x =

1 4

yields

f ( x) ≅ 18 +  23 x  1 ∆x = x= 4

Step 3: Substitute x =

1 4

1 8

+ 34 ∆x

+ ∆x and the linear approximation just obtained, into the original model:

∆ x + ∆x +  18 + 34 ∆x  =

1 8

+ sin( 12 t )

⇒

∆ x + ∆x + 34 ∆x = sin( 12 t )

Step 4: Adjust the initial conditions: ∆x(0) =x(0) − 14 =− 14 , ∆x (0) = x (0) = 1 . Therefore, the linearized model is described as

∆ x + ∆x + 43 ∆x =sin( 12 t ) , ∆x(0) =− 14 , ∆x (0) =1  3 x 1 , f ( x) = 5.  x + 23 x + f ( x) =+ 1 e −t /3 , x(0) = 1 , x (0) =  3  −3 x

if

x≥0

if

x≤0

144 Solution Step 1: The operating point satisfies f ( x ) = 1 . Case (1): x < 0 yields −3 x = 1 , which has no real solution. Case (2): x > 0 leads to 3 x = 1 , thus x = 19 . Thus, the operating point is x = 19 . Step 2: Linearization of f ( x) about x = f ( x) ≅ f

Step 3: Substitute x =

1 9

1 9

yields

( 19 ) + 2 3 x

x=

∆x = 1 + 92 ∆x 1 9

+ ∆x and the linear approximation just obtained, into the original model:

∆ x + 32 ∆x + 1 + 92 ∆x = 1 + e−t /3

⇒

Step 4: Adjust the initial conditions: ∆x(0)= x(0) − 19= linearized model is described as

∆ x + 32 ∆x + 92 ∆x = e−t /3 8 9

, ∆x (0) = x (0) =

∆ x + 23 ∆x + 92 ∆= x e −t /3 , ∆x(0) =

6.

1 3

 x + x + f ( x) =3 + sin t , x(0) =0 , x (0) =−1 ,

8 9

, ∆x (0) =

1 3

. Therefore, the

1 3

 1 (1 − e − x /2 ) if x ≥ 0 f ( x) =  2 − x /2 1 ) if x < 0 − 2 (1 − e

Solution Step 1: The operating point x satisfies f ( x ) = 3 . Due to the nature of f ( x) , we need to inspect two cases: Case (1): x > 0 leads to Case (2): x < 0 leads to

1 (1 − e − x /2 ) = 3 , which has no solution 2 − 12 (1 − e− x /2 ) = 3 , which implies e− x /2

because e− x /2 > 0 . 1 . Since this is = 7 , thus x = ln 49

1 a negative value, the solution is accepted and therefore the operating point is x = ln 49 .

Step 2: Linearization of f ( x) is done via f ( x) ≅ f ( x ) + f ′( x ) ⋅ ∆x . But since x < 0 , the branch of f ( x) corresponding to x < 0 is of interest only. Therefore, f ( x) ≅  − 12 (1 − e− x /2 ) 

+  − 14 e− x /2  1 = x ln= x ln 491 49

∆x = 3 − 74 ∆x

Step 3: Substituting the linear approximation, as well as x =∆x + ln 12 , into the original system:

145

1 ∆ x + ∆x + (3 − 74 3

∆x) = 3 + sin t

⇒

1 ∆ x + ∆x − 74 3

∆x = sin t

1 = ln 49 Step 4: The initial conditions are adjusted as follows: ∆x(0) = x(0) − ln 49 , ∆x (0) = x (0) = −1 . In summary, the linearized model is obtained as 1 ∆ x + ∆x − 74 3

∆x =sin t , ∆x(0) =ln 49 , ∆x (0) =−1

 x 1 x − x x1 (0) = 34 = 7.  1 2 −21 1 −t , x2 (0) = 1 2 2 x2 + 1 + e   x= Solution Step 1: The operating point satisfies  12 x2 − x1 = 0  ⇒  −1 2 1 0 x + =   2

x2 = −2 x1 = −1

Therefore, the operating point is (−1, −2) . Step 2: Linearize f ( x2 ) = 2 x2−1 about the operating point: f ( x2 ) ≅

2 x2

 2 1 + − 2  ∆x2 =−1 − ∆x2 2 x2 =−2  x2  x2 =−2

∆x2 − 2 , x1 =∆x1 − 1 , into the Step 3: Substitute the linear approximation, together with x2 = original system: 1  ∆x1 = 2 (∆x2 − 2) − (∆x1 − 1)  −1 − 12 ∆x2 + 1 + e−t ∆x2 =

1  ∆x1 = 2 ∆x2 − ∆x1  − 12 ∆x2 + e−t ∆x2 =

simplify

⇒

Step 4: Adjust the initial conditions:

∆x1 (0) = x1 (0) += 1 74 ∆x2 (0) = x2 (0) += 2 3 In conclusion, the linearized model is described as 1  ∆x1 = 2 ∆x2 − ∆x1  −t 1 ∆x2 =− 2 ∆x2 + e

,

7 ∆x1 (0) = 4 3 ∆x2 (0) =

146

 x1 = x2  8.  3 1   x2 =− x1 − x1 x1 − x2 − 2 + sin( 3 t )

,

x1 (0) = 1 x2 (0) = −1

Solution Step 1: The operating point ( x1 , x2 ) satisfies Second equation  0 = x2 ⇒ x1 + x1 x1 + 2 = 0  − x1 − x1 x1 − x23 − 2  0 = 0 , which has no real solution. Case (1): x > 0 leads to x12 + x1 + 2 =

Case (2): x < 0 yields − x12 + x1 + 2 =− ( x1 − 2 )( x1 + 1) =0 , hence x1= 2, −1 . Since x < 0 , we have x1 = −1 . Therefore, the operating point is (−1, 0) . Step 2: The two nonlinear elements f = x1 x1 and g = x23 must be linearized about (−1, 0) . We first note that  x12 if x1 ≥ 0 f = x1 x1 =  2 − x1 if x1 < 0

Differentiate

⇒

 2 x1 if x1 ≥ 0 f′=  −2 x1 if x1 < 0

Then, f ≅ f (−1) + [ −2 x1 ]o.p. ∆x1 =−1 + 2∆x1

g ≅ g (0) + 3 x22 

o.p.

∆x2 = 0

Step 3: Substitute x1 = −1 + ∆x1 , x2 = 0 + ∆x2 , and the linear approximations into the original equation: ∆x2 ∆x1 =   1  ∆x2 =−(−1 + ∆x1 ) − (−1 + 2∆x1 ) − 2 + sin( 3 t )

⇒

∆x2  ∆x1 =   1 ∆x2 =−3∆x1 + sin( 3 t )

Step 4: Using x1 (0) = −1 + ∆x1 (0) , x2 (0) = ∆x2 (0) , we find the linear model as ∆x2 ∆x1 (0) = 2  ∆x1 = ,   1 ∆x2 (0) = −1 ∆x2 =−3∆x1 + sin( 3 t )

147

 x =− x2 − x1 x1 − 2 + sin 2t x1 (0) = 32 , 9.  1 x2 (0) = 0  x2 = x1 − x2 − 2 Solution Step 1: The operating point ( x1 , x2 ) satisfies  x2 + x1 x1 + 2 = 0  0  x1 − x2 − 2 =

Add

⇒

x1 + x1 x1 = 0

Case (1): x1 ≥ 0 leads to x1 + x12 = 0 with solutions x1= 0, −1 , hence x1 = 0 is accepted and then

x2 = −2 . Therefore, the operating point is (0, −2) . Case (2): x1 ≤ 0 leads to x1 − x12 = 0 with solutions x1 = 0,1 , hence x1 = 0 again. Therefore, the only operating point is (0, −2) . Step 2: Linearization of x1 x1 about (0, −2) yields

x1 x1 ≅ x1 x1 + [ 2 x1 ]x =0 ⋅ ∆x1 = 0 1

∆x2 − 2 , and the linear approximation, into the original Step 3: Substitute x1 = ∆x1 and x2 = system to obtain −(∆x2 − 2) − 2 + sin 2t  ∆x1 =  ∆x2 =∆x1 − (∆x2 − 2) − 2

Simplify

⇒

−∆x2 + sin 2t  ∆x1 =  ∆x2 =∆x1 − ∆x2

Step 4: Initial conditions are adjusted as:

∆x1 (0)= x1 (0)=

2 3

, ∆x2 (0)= x2 (0) + 2= 2

In summary, the linearized model is

 ∆x1 = −∆x2 + sin 2t ,  ∆x2 = ∆x1 − ∆x2  − x1 + 3 x23 x1 (0) = 0  x = 10.  1 , 1 − t /3 x2 (0) = 1   x2 = x1 + 24 + 3 e Solution Step 1: The operating point satisfies

2 ∆x1 (0) = 3 ∆x2 (0) = 2

148 3 x23 − x1 = 0  ⇒  0   x1 + 24 =

x2 = −2 x1 = −24

Therefore, the operating point is (−24, −2) . Step 2: Linearize f ( x2 ) = x23 about the operating point: f ( x2 ) ≅  x23 

+ 3 x22  ∆x2 =−8 + 12∆x2 x2 = −2  x2 = −2

∆x2 − 2 , x1 =∆x1 − 24 , into the Step 3: Substitute the linear approximation, together with x2 = original system:  ∆x1 =−(∆x1 − 24) + 3(−8 + 12∆x2 )  ( x1 − 24) + 24 + 13 e−t /3 ∆x2 =∆

simplify

⇒

∆x1 =−∆x1 + 36∆x2  1 − t /3  ∆x2 =∆x1 + 3 e

Step 4: Adjust the initial conditions: ∆x1 (0)= x1 (0) + 24= 24 , ∆x2 (0)= x2 (0) + 2= 3 In conclusion, the linearized model is described as

∆x1 = −∆x1 + 36∆x2  1 − t /3  ∆x2 =∆x1 + 3 e

,

∆x1 (0) = 24 ∆x2 (0) = 3

In Problems 11 through 14, a nonlinear model is given. (a) Obtain the linear state-space form in Simulink. (b) Derive the linearized model analytically to confirm the result of (a).

x + x + x x =1 + sin t , x(0) = 0 , x (0) =1 11.  Solution (a) Choosing state variables x1 = x and x2 = x , we find x (0) = 0  x1 = x2 , 1 (a)  x2 (0) = 1  x2 =− x2 − x1 x1 + 1 + sin t Build the model shown in Figure PS4-6No11 for the nonlinear system in Eq. (a) and save it as 'Problem11'. Double-clicking on an integrator block allows us to input the appropriate initial condition for the output signal of that block. Since the first integrator has x2 as its output, we use x2 (0) = 1 . For the second integrator, x1 (0) = 0 . The nonlinear element x1 x1 is handled by the function block from the Simulink User-Defined Functions library, where u is used as the

149 input variable name. The linearization input point is chosen at the output signal of the summing junction, the linearization output point at the output signal x1.

Figure PS4-6No11 The linear model is extracted from the nonlinear system in Eq. (a) as follows. Note that the model, 'Problem11', must be open before we can proceed. Using the procedure outlined in this section, we find the operating point and the state-space model as follows. >> sys='Problem11'; >> load_system(sys); >> opspec = operspec(sys); % Specify the properties of the first state variable. >> opspec.States(1).SteadyState = 1; >> opspec.States(1).x = 0; % Initial value >> opspec.States(1).Min = 0; % Minimum value % Do the same for the second state variable >> opspec.States(2).SteadyState = 1; >> opspec.States(2).x = 1; >> opspec.States(2).Min = 0; % Find the operating point based on the specifications listed above. >> [op,opreport] = findop(sys,opspec); Operating Point Search Report: --------------------------------Operating Report for the Model Problem11. (Time-Varying Components Evaluated at time t=0) Operating point specifications were successfully met. States: ---------(1.) Problem11/Integrator 1 x: 0 dx: 0 (0) (2.) Problem11/Integrator 2 x: 1 dx: 0 (0) Inputs: None ----------

150 Outputs: ---------(1.) Problem11/Out1 = x1 y: 1

[-Inf Inf]

% Get the linearization annotations >> IO=getlinio(sys); % Extract the linear state-space model >> LIN = linearize('Problem11',op,IO)

a = Integrator 1 Integrator 2

Integrator 1 -1 1

Integrator 1 Integrator 2

Sum1 1 0

Integrator 2 -2 0

% Controller canonical form

b =

c = x1

Integrator 1 0

x1

Sum1 0

Integrator 2 1

d =

Continuous-time state-space model.

Note that o the state matrix is in the controller canonical form, and o the input in the linear model is comprised of the time-varying portion of the input in the original nonlinear system. In the current case, the input for the nonlinear system is 1 + sin t . This implies that the input for the linear model is simply sin t . (b) The operating point is (1, 0) and the linearized model is ∆ x + ∆x + 2∆x =sin t ∆x and ∆x2 = ∆x , To derive the controller canonical form, choose the state variables as ∆x1 = and the state-variable equations for the linearized model are obtained as ∆x1 = −∆x1 − 2∆x2 + sin t  ∆x1  ∆x2 = In vector form,

DD x A x + Bu =  =  y CDx + Du where

(b)

151

 Dx   −1 −2  1  = , A  = , B  = , C Dx  1  =  1 0 0  Dx2  This confirms the set of results realized in Simulink.

, D [0 1] =

0= , u sin t

3 12.  x + x= + x u (t ) , = u (t ) unit-step , = x(0) 1 , = x (0) 1 Solution (a) The Simulink model is built as shown in Figure PS4-6No12 and saved as 'Problem12'. Employing the procedure outlined in this section, we find the operating point and the state-space model as follows.

>> >> >> >> >> >> >> >> >> >>

sys='Problem12'; load_system(sys); opspec = operspec(sys); opspec.States(1).SteadyState = 1; opspec.States(1).x = 1; opspec.States(1).Min = 0; opspec.States(2).SteadyState = 1; opspec.States(2).x = 1; opspec.States(2).Min = 0; [op,opreport] = findop(sys,opspec);

Operating Point Search Report: --------------------------------Operating Report for the Model Problem12. (Time-Varying Components Evaluated at time t=0) Operating point specifications were successfully met. States: ---------(1.) Problem12/Integrator 1 x: 0 dx: 9.93e-09 (0) (2.) Problem12/Integrator 2 x: 1 dx: 0 (0) Inputs: None ---------Outputs: ---------(1.) Problem12/Out1 = x1 y: 1

[-Inf Inf]

>> IO=getlinio(sys); >> LIN = linearize('Problem12',op,IO) LIN = LIN = a = Integrator 1

Integrator 1 -1e-10

Integrator 2 -1

152 Integrator 2

1

0

b = Empty matrix: 2-by-0 c = x1

Integrator 1 0

Integrator 2 1

d = Empty matrix: 1-by-0 Continuous-time state-space model.

IC: x2(0) = 1 Unit step at t = 0

1 s

IC: x1(0) = 1 1 s

x2

Integrator 1

x1

Integrator 2

1 Out1 = x1

u^3 Nonlinear element Scope

Figure PS4-6No12

(b) The 4-step procedure is followed as listed below: Step 1: The operating point satisfies x = 1 . Step 2: The nonlinearity f ( x ) = x 3 is linearized about x = 0 as f ( x ) ≅ f (0) + 3 x 2 

x =0

∆x =0

Step 3: Substitute x = 1 + ∆x and the linear approximation into the original equation: (1 + ∆x) + 0 + [1 += ∆x] u (t )

⇒

Letting x1 = ∆x and x2 = ∆x , the above reduces to

 x1 = − x2   x2 = x1 The matrices in the state-space form are then identified as

∆ x= + ∆x 0

153

0 −1 0  = A  = , B =   , C 1 0  0

1] , D [0 =

0

This agrees with (a).

13.  x + x x + x = 12 + cos( 12 t ) , x(0) = 12 , x (0) = 0 Solution (a) The Simulink model is built as shown in Figure PS4-6No13 and saved as 'Problem13'. Noting that cos = 2t sin ( 2t + 12 π ) , we generate cos 2t using the Sine Wave block from the Sources library by setting the amplitude to 1 , frequency to 2 rad/sec , and phase to 12 π . Employing the procedure outlined in this section, we find the operating point and the state-space model as follows. >> >> >> >> >> >> >> >> >> >>

sys='Problem13'; load_system(sys); opspec = operspec(sys); opspec.States(1).SteadyState = 1; opspec.States(1).x = 1/2; opspec.States(1).Min = 0; opspec.States(2).SteadyState = 1; opspec.States(2).x = 0; opspec.States(2).Min = 0; [op,opreport] = findop(sys,opspec);

Operating Point Search Report: --------------------------------Operating Report for the Model Problem13. (Time-Varying Components Evaluated at time t=0) Operating point specifications were successfully met. States: ---------(1.) Problem13/Integrator 1 x: 0 dx: 1.06e-09 (0) (2.) Problem13/Integrator 2 x: 1.5 dx: 0 (0) Inputs: None ---------Outputs: ---------(1.) Problem13/Out1 = x1 y: 1.5

[-Inf Inf]

>> IO=getlinio(sys); >> LIN = linearize('Problem13',op,IO)

154 LIN = a = Integrator 1 -1e-05 1

Integrator 1 Integrator 2

Integrator 2 -1 0

b = Sum1 1 0

Integrator 1 Integrator 2 c = x1

Integrator 1 0

Integrator 2 1

d = x1

Sum1 0

Continuous-time state-space model.

cos(t/2)

IC: x2(0) = 0 1 s

1/2 Constant

IC: x1(0) = 1/2 1 s

x2

Integrator 1

x1

Integrator 2

1 Out1 = x1

u*abs(u) Nonlinear element Scope

Figure PS4-6No13

(b) The 4-step procedure is followed as listed below: Step 1: The operating point satisfies x = 12 . Step 2: The nonlinearity f ( x ) = x x is linearized about x = 0 as

f ( x ) ≅ f (0) + [ 2 x ]x =0 ∆x =0 Step 3: Substitute x =

1 2

+ ∆x and the linear approximation into the original equation:

( 12 + ∆x) + 0 + [ 12 + ∆x] =

1 2

+ cos( 12 t )

Letting x1 = ∆x and x2 = ∆x , the above reduces to

⇒

∆ x + ∆x = cos( 12 t )

155

 x1 = − x2 + cos( 12 t )   x2 = x1 The matrices in the state-space form are then identified as

0 −1 1  = A  = , B  = , C  1 0  0 

, D [0 1] =

0= , u cos( 12 t )

This agrees with (a). 14. 2  x + x3 =81 + sin t , x(0) =1 , x (0) =1 (a) The Simulink model is built as shown in Figure PS4-6No14 and saved as 'Problem14'. Employing the procedure outlined in this section, we find the operating point and the state-space model as follows. >> >> >> >> >> >> >> >> >> >>

sys='Problem14'; load_system(sys); opspec = operspec(sys); opspec.States(1).SteadyState = 1; opspec.States(1).x = 1; opspec.States(1).Min = 0; opspec.States(2).SteadyState = 1; opspec.States(2).x = 1; opspec.States(2).Min = 0; [op,opreport] = findop(sys,opspec);

Operating Point Search Report: --------------------------------Operating Report for the Model Problem14. (Time-Varying Components Evaluated at time t=0) Operating point specifications were successfully met. States: ---------(1.) Problem14/Integrator 1 x: 0 dx: -2.33e-09 (0) (2.) Problem14/Integrator 2 x: 0.5 dx: 0 (0) Inputs: None ---------Outputs: ---------(1.) Problem14/Out1 = x1 y: 0.5

[-Inf Inf]

>> IO=getlinio(sys); >> LIN = linearize('Problem14',op,IO)

156 LIN = a = Integrator 1 Integrator 2

Integrator 1 0 1

Integrator 2 -0.375 0

b = Integrator 1 Integrator 2

Sum1 0.5 0

c = x1

Integrator 1 0

Integrator 2 1

d = x1

Sum1 0

Continuous-time state-space model.

sin(t)

IC: x2(0) = 1

1/8

1/2

Constant

Gain

1 s

IC: x1(0) = 1 1 s

x2

x1

Integrator 2

Integrator 1

1 Out1 = x1

u^3 Nonlinear element

Scope

Figure PS4-6No14 (b) The 4-step procedure is followed as listed below: Step 1: The operating point satisfies x = 12 . Step 2: The nonlinearity f ( x) = x 3 is linearized about x = f ( x) ≅ f ( 12 ) + 3 x 2  Step 3: Substitute x =

1 2

x = 12

∆x =

1 2

as

1 8

+ 34 ∆x

+ ∆x and the linear approximation into the original equation:

2( 12 + ∆x) + [ 18 + 34 ∆x] = 18 + sin t

Letting x1 = ∆x and x2 = ∆x , the above reduces to

⇒

2∆ x + 34 ∆x = sin t

157

 x1 =12 (− 34 x2 + sin t )   x2 = x1 The matrices in the state-space form are then identified as

0 − 83   12  = A  = , B , C   = 0 1 0 

, D [0 1] =

0= , u sin t

This agrees with (a). Review Problems 1. The mathematical model of a dynamic system is derived as x1 + x1 + x1 − 3x2 = 0     1 f (t ) 2 x2 − x1 − 4 x2 = (a) If f (t ) is the input, and x1 and x1 are the outputs, obtain the state-space form. (b) Determine if the system is stable (Problem Set 4.2) by examining the state-space form. (c) Find the transfer matrix directly from the given model. (d) Confirm the result of (c) using ss2tf. Solution (a) There are three state variables: x1 = x1 , x2 = x2 , x3 = x1 . The state-variable equations are  x1 = x3  1 1  x2 = 2  x1 + 4 x2 + f   − x3 − x1 + 3 x2  x3 = The state equation is then formed as

 x1  0 0 1 0    1 1 0 , B= 1 , u = f (t ) x = Ax + Bu , x =  x2  , A = 2 8  2 x   −1 3 −1  0   3 The output equation is

1 0 0  0  y= Cx + Du , C = 0 0 1  , D = 0      (b) >> A = [0 0 1;1/2 1/8 0;-1 3 -1]; eig(A) ans =

158 0.7640 + 0.0000i -0.8195 + 1.2065i -0.8195 - 1.2065i

System is unstable because one of the eigenvalues of the state matrix lies in the right half-plane. (c) Laplace transformation yields 2 0  ( s + s + 1) X1 − 3 X 2 =  1 F (s) − X1 + (2 s − 4 ) X 2 =

Solve for X1 :

−3 F 2 s − 14 X1 3F 12 = = ⇒ 3 2 3 F 8s + 7 s 2 + 7 s − 13 2 s + 74 s + 74 s − 134 s2 + s + 1 −3 0

X1 =

−1

2 s − 14

The transfer function corresponding to x is

sX1 12s = 3 F 8s + 7 s 2 + 7 s − 13 Therefore, the transfer matrix is 12    8s 3 + 7 s 2 + 7 s − 13  G (s) =   12 s    8s 3 + 7 s 2 + 7 s − 13  (d) >> A = [0 0 1;1/2 1/8 0;-1 3 -1]; B = [0;1/2;0]; >> C = [1 0 0;0 0 1]; D = [0;0]; >> [NUM,DEN] = ss2tf(A,B,C,D,1) % option '1' may be omitted NUM = 0 0

0 0

0 1.5000

1.5000 0

1.0000

0.8750

0.8750

-1.6250

DEN =

159 2. A system is described by its governing equations  12  x1 + x1 − 2( x2 − x1 ) = f (t )  x2 + 2( x2 − x1 ) = 0   where f is the input, while x2 and x2 are the outputs. (a) Obtain the state-space form. (b) Find the transfer matrix directly from the state-space form. (c) Find the transfer matrix using the governing equations, and compare with (b). Solution (a) There are four state variables: x1 = x1 , x2 = x2 , x3 = x1 , x4 = x2 . The state-variable equations are

 x1 = x3  x = x  2 4  −6 x1 + 4 x2 + 2 f  x3 =  = x4 2 x1 − 2 x2 The state equation is then formed as

 x1  0 0 x  0 0  2  x = Ax + Bu , x =   , A=  −6 4 x  3   x4   2 −2

1 0 0   0  0 1   , u= , B= f (t ) 2 0 0    0 0 0 

The output equation is

 x2  0 1 0 0 0 y= C2×4 x 4×1 + Du , C = , D=  =   0 0 0 0 1     x4  (b) >> A = [0 0 1 0;0 0 0 1;-6 4 0 0;2 -2 0 0]; B = [0;0;2;0]; >> C = [0 1 0 0;0 0 0 1]; D = [0;0]; >> [nG, dG] = ss2tf(A,B,C,D) nG = 0 0

0 0

0 0

0 4

4 0

dG = 1.0000

-0.0000

8.0000

-0.0000

4.0000

160 The transfer matrix is 2 ×1 because there are two outputs and one input: 4    s 4 + 8s 2 + 4  G (s) =   4s    s 4 + 8s 2 + 4 

(c) Laplace transformation yields ( 1 s 2 + 3) X1 − 2 X 2 = F 2  2 0 −2 X1 + ( s + 2) X 2 = Solve for X 2 : 1 s2 2

= X2

+3 F

−2

1 s2 2

0 = +3 −2

−2

1 s4 2

2F + 4s 2 + 2

X2 ⇒= F

1 s4 2

2 + 4s 2 + 2

s2 + 2

Since the other output is x2 , the second transfer function is simply

sX 2 2s = 4 1 F s + 4s 2 + 2 2 Results clearly agree with (b).

3. A dynamic system with input f and output x is modeled as 3   x + x += kx f (t ) , = k const > 0 4 (a) Find the state-space form, and determine the value(s) of k for which the system is stable. (b) Find the transfer function directly from the state-space form. (c) Find the transfer function directly from the given model, and compare with (b). Solution (a) There are two state variables: x1 = x , x2 = x . The state-variable equations are

 x1 = x2  4  x2 = 3 (− x2 − kx1 + f )

161 The state equation is then formed as

 0  x1  x = Ax + Bu , x =   , A= − 4 k  x2   3

1  0 , B= f (t )  4 , u = 4 − 3 3

The output is y= x= x2 , thus

y= Cx + Du , C = 0 [0 1] , D = The eigenvalues of the state matrix are λ =− 23 ±

4 9

− 43 k . Since k > 0 ,

4 9

− 43 k
0 . (b) Since the system is SISO, there is one transfer function: −1

−1   0  s = 2    4 4 s + 3 3 3 s + s + k 4

 s C( sI − A) B = G (s) = [0 1]  4 k 3 −1

(c) Laplace transformation of the model yields ( 34 s 2 + s + k ) X= (s) F (s)

X (s) = F (s)

⇒

3 s2 4

1 +s+k

Since x is the output, the transfer function is sX ( s ) s = 2 F ( s ) 34 s + s + k 4. The I/O equation of a system is derived as 2 y + 12  y + y =  u + 13 u + u (a) Find the state-space representation. (b) Find the transfer function from the state-space form. Solution (a) State-space form is found as

0  x1   x = Ax + Bu , x =   , A= 0  x2  − 1  2

 14 y =+ Cx Du , C =

1 0  0   0  , u = 0 1  , B= u   1  0 − 14  1 6

1 − 81  , D = 2

2 3

for all k .

162 (b) Using the state-space form,

−1

G ( s= ) C( sI − A) B + D=

 14

1 6

 s −1 0    1 − 8   0 s −1  1 0 s + 1 4 2

−1

0 3 0  + 1= 6 s + 2 s + 6   2 12 s 3 + 3s 2 + 6 1 

5. A system’s transfer function is given as

Y (s) s 2 (2s + 1) = 3 2 U ( s ) 13 s + s + s + 2 (a) Find the state-space form. (b) Find the transfer function from (a). Solution y +  y + y + 2 y = 2 u + u .The state-space form is obtained as (a) The I/O equation is 13 

 x1  0 1 0 0     0 , u = u x = Ax + Bu , x =  x2  , A =  0 0 1 , B=   x   −6 −3 −3 1   3

y= Cx + Du , C = 6 [ −36 −18 −15] , D =

(b) Using the state-space form,

G ( s ) =C( sI − A) −1 B + D  s −1 0  =[ −36 −18 −15] 0 s −1  6 3 s + 3

−1

0  2 0  + 6 = 3s (2 s + 1)   s 3 + 3s 2 + 3s + 6 1 

6. Find the input-output equation of a SISO system whose state-space form is given as x Ax + Bu =  y Cx + Du = where 1  0 0 = A  1 = , B , u u= , C  23 13 = , D 0   3 = 1 − − 2  4 2 Solution We will first find the transfer function, as

163

−1

G ( s )= C( sI − A) B + D=

s

 23

1 3 1 4

−1

−1   0  2( s + 2)   3 = 1 s + 2   2  4s 2 + 2s + 1

As a result, the I/O equation is 4  y + 2 y + y = 2u + 4u .

7. Repeat Problem 6 for

0 1 0  A= u , C= [ 2 −1] , D =2  −1 −4  , B = 3 , u =     Solution We will first find the transfer function, as −1

 s −1  0  2 s 2 + 5s + 8 + = G ( s )= C( sI − A) B + D= [ 2 −1]  2    s 2 + 4s + 1 1 s + 4   3  −1

As a result, the I/O equation is  y + 4 y + y = 2u + 5u + 8u .

8. Find all possible input-output equations for a system with state-space form x Ax + Bu =  y Cx + Du = where 1 0  0 1 0  0 = A  1 = = , B  1 = , u u= , C  , D     0 2 0  − 2 −1 3 Solution We will first find the transfer matrix, as

1 0   s G ( s= ) C( sI − A) −1 B + D=   1 0 2   2

2   3   −1   0  0  2s 2 + 2s + 1   + = 4s s + 1  13  0     2 3   2s + 2s + 1 

The system has one input u and two outputs y1 and y2 . As a result, the two I/O equations are obtained as

2  y1 += 2 y1 + y1

2u 3

,

2 y2 + = 2 y 2 + y2

4 u 3

164 In Problems 9 through 12, given the block diagram representation of the system, find the overall transfer function. 9. Figure 4.38

U (s)

k1

+

Σ Σ

1 s

− +

1 s

k2

Y ( s)

k2

Y ( s)

k3

+ k4 Figure 4.38 Problem 9. Solution Based on Figure Reviwe4No9, we write

s 2 X ( s ) =k1U ( s ) − [k3 sX ( s ) + k4 X ( s )] which yields

k1 X (s) = 2 U ( s ) s + k3 s + k4 Since Y ( s ) = k2 X ( s ) , the transfer function is found as

kk Y (s) = 2 1 2 U ( s ) s + k3 s + k4

U (s)

k1

+

Σ Σ

s 2 X (s)

− +

1 s

sX ( s )

k3

+ k4 Figure Review4No9

10. Figure 4.39

1 s

X ( s)

165

G4 ( s ) H 2 (s) +

U (s)

Σ

−

G1 ( s )

G2 ( s )

+

Σ

+ −

H1 ( s )

G3 ( s )

+

Σ

+

H 3 (s)

Figure 4.39 Problem 10. Solution There are three forward paths and two loops: Forward path gain F1 = G4

Loop gain −G1H1

F2 = H 2G3

−G3 H 3

F3 = G1G2G3 In the general case of Mason’s rule, we have

D = 1 − [ −G1H1 − G3 H 3 ] + G1G3 H1H 3

D1 Same = as D when restricted to portion not touching 1st path D D2 = Same as D when restricted to portion not touching 2nd path = 1 + G1H1 as D when restricted to portion not touching 3rd path 1 D3 Same = Then, Mason’s rule yields

Y ( s ) G4 [1 + G1H1 + G3 H 3 + G1G3 H1H 3 ] + (1 + G1H1 ) H 2G3 + G1G2G3 = U (s) 1 + G1H1 + G3 H 3 + G1G3 H1H 3

11. Figure 4.40

Y ( s)

166 G4 ( s ) H 2 (s)

+

U (s)

Σ

G1 ( s )

−

+

Σ

+

+

G2 ( s )

Σ

+

G3 ( s )

Σ

+ Y (s)

+ H1 ( s )

H 3 (s)

Figure 4.40 Problem 11. Solution There are two forward paths and four loops, as shown in Figure Review4No11: Forward path 12456

Gain F1 = G1G2G3

Loop 2482

Gain −G1H1

1236

F2 = G4

5675

G3 H 3

4594

G2 H 2 −G4 H 3 H 2 H1

236759482

3

G4 ( s ) H 2 (s) U (s)

1

+

Σ

−

G1 ( s ) 2 H1 ( s )

+

Σ

+

4

G2 ( s )

9

+

Σ +

5

G3 ( s )

H 3 (s)

8

+

Σ

+

6

Y (s)

7

Figure Review4No11

In the general case of Mason’s rule, we have

D = 1 − [ −G1H1 + G2 H 2 + G3 H 3 − G4 H 3 H 2 H1 ] − G1H1G3 H 3

D1 Same = as D when restricted to portion not touching 1st path 1 D2 = Same as D when restricted to portion not touching 2nd path = 1 − G2 H 2 Then, Mason’s rule yields

167

G1G2G3 + (1 − G2 H 2 )G4 Y (s) = U ( s ) 1 − [ −G1H1 + G2 H 2 + G3 H 3 − G4 H 3 H 2 H1 ] − G1H1G3 H 3

12. Figure 4.41 G4 ( s )

U (s)

+

Σ

−

G1 ( s )

k

H1 ( s )

+

Σ

−

G2 ( s )

G3 ( s )

+

Σ

H 2 ( s)

Figure 4.41 Problem 12. Solution There are two forward paths and two loops: Forward path gain F1 = G4

Loop gain −G1H1

F2 = kG1G2G3

−G2 H 2

In the general case of Mason’s rule, we have

D = 1 − [ −G1H1 − G2 H 2 ] + G1H1G2 H 2

D1 Same = as D when restricted to portion not touching 1st path D D2

Same = as D when restricted to portion not touching 2nd path 1

Then, Mason’s rule yields

Y ( s ) G4 [1 + G1H1 + G2 H 2 + G1H1G2 H 2 ] + kG1G2G3 = U (s) 1 + G1H1 + G2 H 2 + G1H1G2 H 2

In Problems 13 and 14, given the block diagram, find the system’s I/O equation. 13. Figure 4.42

+ Y ( s)

168 1 3

+

4 9

U (s)

+ −

Σ

−

1 s

1 s

1 s

Σ

+ −

Y (s)

1 3

2 3

Figure 4.42 Problem 13.

Solution Because not all paths are coupled, the general case of Mason’s rule applies as follows: There are 4 −1 1 2 three forward paths with gains F1 = , F2 = 2 and F3 = 3 , and two loops with gains − 2 3 3s 9s 3s and −

1 . s3

2 1 + 3 , D1 = D , D2 = 1, D3 = 1 2 3s s 4 1 3 − 3 1D+F +F 2 F D Y ( s ) ∑ i =1 i i 3 s3 + 2s 1 9s 2 3 s 3 == = + = U (s) D D 3 1 + 2 + 1 3s 3 + 2 s + 3 3s 2 s 3 D= 1+

Subsequently, the I/O equation is formed as 3 y + 2 y + 3 y = u + 2u .

14. Figure 4.43 1 2 1 4

− U (s)

+ −

Σ

−

1 s

1 s

1 s

1 2 1 2

Figure 4.43 Problem 14.

1 4

+ −

+

Σ

Y ( s)

169

Solution Because not all paths are coupled, the general case of Mason’s rule applies as follows: Four 1 −1 −1 1 forward paths with gains F1 = , F2 = , F3 = 3 and F4 = 2 , and two loops with gains 2 4s s 4s 1 1 and − 3 . − 2s 2s

D =+ 1

1 1 + 3 , D1 =D , D2 == 1 D3 =D4 2s 2s

− s 2 + 4s − 1 Y ( s ) ∑ i =1 Fi Di 12 D + F2 + F3 + F4 1 s3 + 2s = = = + = 2 2(2 s 3 + s 2 + 1) 2 s 3 + s 2 + 1 U (s) D D Therefore, the I/O equation is formed as 2 y +  y + y =  u + 2u . 4

In Problems 15 and 16, given the state-space representation of the system model, construct the appropriate block diagram, and directly use it to find the transfer matrix.

x Ax + Bu = 15.  y Cx + Du = where  x1    = , A x  x2  = x   3

0  0 − 1  2

1 0   0 1 = , B −1 − 12 

0 0 = , D 0= , u u   , C [1 2 0] = 1 

Solution The block diagram is shown in Figure Review4No15.

Figure Review4No15

170

There are two forward paths with gains

−

1 2 1 1 and 2 , and three loops with gains − , − 2 , and 3 2s s s s

1 . All paths are coupled. By Mason’s rule, we have 2s 3 1 2 + Y (s) 2(2 s + 1) s3 s 2 = = U (s) 1 + 1 + 1 + 1 2s3 + s 2 + 2s + 1 2 3 2s s 2s

Note that in the above transfer function, the numerator and denominator have 2 s + 1 in common which can algebraically be cancelled to yield

2 . However, this – known as pole-zero s2 + 1

cancellation – must be avoided because important information about the system may be lost.

x Ax + Bu = 16.  y Cx + Du = where x  0 1 0  1 0  0  = = , B  = , C  = , D = x  1= , A     , u u  −1 −3 1  0 2  0   x2  Solution The block diagram is shown in Figure Review4No16.

Figure Review4No16

Y1 ( s ) , we U (s) 1 1 3 note that there is one forward path with gain 2 and two feedback loops with gains − , − 2 . s s s There are two outputs and one input, hence the transfer matrix G ( s ) is 2 ×1 . To find

171 By Mason’s rule,

1 Y1 ( s ) 1 s 2= (s) = G11 = 2 3 1 U (s) 1 + + s + 3s + 1 2 s s

Y2 ( s ) 2 , we note that there is one forward path with gain and two feedback loops with U (s) s 1 3 gains − , − 2 . By Mason’s rule, s s 2 Y2 ( s ) 2s s= (s) = G21 = 2 U ( s ) 1 + 3 + 1 s + 3s + 1 s s2 To find

 G (s)  Therefore, the transfer matrix is G ( s ) =  11  . G21 ( s ) 

17. A system is atable if the poles of the overall transfer function lie in the left half-plane; these poles are the same as the eigenvalues of the state matrix. Consider the block diagram representation of a system shown in Figure 4.44, where k > 0 is a parameter. Determine the range of values of k for which the system is stable. U (s)

1 2

+ −

Σ

−

1 2s + 1

1 s

1 3

Y ( s)

ks + 1 3

Figure 4.44 Problem 17. Solution The overall transfer function is formed by Mason’s rule, as 1 1 Y (s) 6 s (2 s + 1) 6 ( s) = G = = 2 1 3 ks + U ( s) 1 + (k + 2) s + 2 s + 3 + 2 s + 1 s (2 s + 1)

The poles of G ( s ) are s1,2 =−1 ± 1 − 3(k + 2) . Since k > 0 , both poles are complex with negative real parts. This implies that the system is stable for all k > 0 .

172

18. Repeat Problem 17 for the block diagram in Figure 4.45, where k > 0 is a parameter. 2

U (s)

+ −

Σ

−

2 s +1

1 s

+

+

Σ

Y (s)

s+k k

Figure 4.45 Problem 18. Solution The overall transfer function is formed by Mason’s rule, as

2 4 + 4s + 2 Y (s) s ( s + 1) s + 1 ( s) = G = = 2 U ( s ) 1 + 2( s + k ) + 2k 3s + (1 + 2k ) s + 2k s +1 s ( s + 1)

−(1 + 2k ) ± (1 + 2k ) 2 − 24k . But (1 + 2k ) 2 − 24k < 1 + 2k for all 6 k > 0 . Therefore, both poles lie in the left half-plane for all values of k > 0 . This implies that the system is stable for all k > 0 .

The poles of G ( s ) are s1,2 =

19. Derive the linearized model for the nonlinear system described by x1 (0) = −1 = x1 x1 x1 + x2 − 1 + sin t ,  x2 (0) = 0 − x1 − x2 − 1  x2 = Solution Step 1: The operating point ( x1 , x2 ) satisfies 0= x1 x1 + x2 − 1 Add ⇒ x1 x1 − x1 − 2 = 0  − x1 − x2 − 1 0 = Case (1): x1 > 0 leads to x12 − x1 − 2 = ( x1 − 2)( x1 + 1) = 0 , hence x1= 2, −1 . Since x1 > 0 , then

we have x1 = 2 so that x2 = −3 . Therefore, the operating point is (2, −3) .

0 , which has no real solution. Case (2): x1 < 0 yields − x12 − x1 − 2 = Step 2: The nonlinear element f = x1 x1 must be linearized about (2, −3) . We first note that

173  x12 if x1 ≥ 0 f = x1 x1 =  2 − x1 if x1 < 0

Differentiate

⇒

 2 x1 if x1 ≥ 0 f′=  −2 x1 if x1 < 0

Then, f ≅ f (2) + [ 2 x1 ]o.p. ∆x1 = 4 + 4∆x1

Step 3: Substitute x1 = 2 + ∆x1 , x2 = −3 + ∆x2 , and the linear approximations into the original equation:

∆x1 = 4 + 4∆x1 − 3 + ∆x2 − 1 + sin t   ∆x2 = −2 − ∆x1 + 3 − ∆x2 − 1

∆x1 = 4∆x1 + ∆x2 + sin t   ∆x2 = −∆x1 − ∆x2

⇒

Step 4: Using x1 (0) = 2 + ∆x1 (0) , x2 (0) = −3 + ∆x2 (0) , we find the linear model as

∆x1 (0) = −3 ∆x1 = 4∆x1 + ∆x2 + sin t ,  ∆x2 (0) = 3  ∆x2 = −∆x1 − ∆x2

20. Repeat Problem 19 for x2 + 2sin t  1  x=  3   x2 =x1 + 3 x2 − 1

,

x1 (0) = 1 x2 (0) = 0

Solution Step 1: The operating point ( x1 , x2 ) satisfies   0 = x2  3  0 =x1 + 3 x2 − 1

⇒

x1 = 1

Therefore, the operating point is (1, 0) . Step 2: The nonlinear element f = x13 must be linearized about (1, 0) : f ≅ f (1) + 3 x12 

o.p.

∆x1 = 1 + 3∆x1

174 Step 3: Substitute x1 = 1 + ∆x1 , x2 = ∆x2 , and the linear approximations into the original equation: ∆x2 + 2sin t ∆x1 =   ∆x2 = 1 + 3∆x1 + 3∆x2 − 1

∆x2 + 2sin t ∆x1 =   ∆x2 = 3∆x1 + 3∆x2

⇒

Step 4: Using x1 (0) = 1 + ∆x1 (0) , x2 (0) = ∆x2 (0) , we find the linear model as ∆x2 + 2sin t ∆x1 =   ∆x2 = 3∆x1 + 3∆x2

,

∆x1 (0) = 0 ∆x2 (0) = 0

175

Problem Set 5.1 1.

If the 50-kg block in Figure 5.12 is released from rest at A, determine its kinetic energy and velocity after it slides 5 m down the plane. Assume that the plane is smooth.

Figure 5.12 Problem 1.

Solution Choosing point B as the datum and applying the principle of conservation of energy gives TB + VB = TA + VA where VB = 0 and TA = 0 1 2

2 a mv = mgh = 50(9.81)(5sin 60 = ) 2123.93 J B

The velocity at point B is = vB

2TB = m

2(2123.93) = 9.22 m/s 50

Figure PS5-1 No1 2. Repeat Problem 1 if the coefficient of kinetic friction between the block and the plane is μk = 0.15.

Solution From the free-body diagram shown below, we obtain the friction force μ k mg cos= 60 (0.15)(50)(9.81)(cos 60 ) 36.79 N = Ff μ= = kN Applying the principle of work and energy gives where TA = 0 TB − TA = Wmg + WF f 1 = mvB2 50(9.81)(5sin 60 ) − 36.79(5) = 1939.98 J 2

The velocity at point B is = vB

2TB = m

2(1939.98) = 8.81 m/s 50

Figure PS5-1 No2 3.

The ball in Figure 5.13 has a mass of 6 kg and is fixed to a rod having a negligible mass. Assume that the ball is released from rest when θ = 60°. a. Determine the gravitational potential energy of the ball when θ = 60° and 30°. The datum is shown in Figure 5.13. b. Determine the kinetic energy and the velocity of the ball when θ = 30°.

176

Figure 5.13 Problem 3.

Solution a.

With the chosen datum, the gravitational potential energy of the ball when θ = 60 is V1 =mgh1 =(6)(9.81)(1 − cos 60 )=29.43 J and the one when θ = 30 is V2 =mgh2 =(6)(9.81)(1 − cos 30 )=7.89 J

b.

Applying the principle of conservation of energy, we obtain the kinetic energy of the ball when θ = 30 T1 + V1 = T2 + V2 where T1 = 0

T2 = V1 − V2 = 29.43 − 7.89 = 21.54 J The velocity of the ball when θ = 30 is 

= v2

2T2 = m

2(21.54) = 2.68 m/s 6

Figure PS5-1 No3 4.

The 6-kg slender rod in Figure 5.14 is released from rest when θ = 60°. a. Determine the gravitational potential energy of the rod when θ = 60° and 30°. The datum is shown in Figure 5.14. b. Determine the kinetic energy and the angular velocity of the rod when θ = 30°. The mass moment of inertia of the slender rod about the fixed point O is IO = mL2/3, where L is the length of the rod.

Figure 5.14 Problem 4.

Solution a.

With the chosen datum, the gravitational potential energy of the rod when θ = 60 is V1 =mgh1 =(6)(9.81)(1 − 12 cos 60 )=44.15 J and the one when θ = 30 is V2 =mgh2 =(6)(9.81)(1 − 12 cos 30 )=33.37 J

b.

Applying the principle of conservation of energy, we obtain the kinetic energy of the rod when θ = 30

177

T1 + V1 = T2 + V2

where T1 = 0

T2 = V1 − V2 = 44.15 − 33.37 = 10.78 J Note that T2 =

1 I ω2 , 2 O 2

where

= IO

1 3

2 mL =

2 1 (6)(1 = ) 3

2 kg ⋅ m 2

The angular velocity of the rod when θ = 30 is

= ω2

2T2 = IO

2(10.78) = 3.28 rad/s 2

Figure PS5-1 No4 5.

Determine the elastic potential energy of the system shown in Figure 5.15 if the 10-kg block moves downward a distance of 0.05 m. Assume that the block is originally in static equilibrium.

Figure 5.15 Problem 5.

Solution Since the block is originally in static equilibrium, the spring is compressed and the deformation is mg 10(9.81) ∆x= = = 0.20 m k 500 If the block moves a distance of 0.05 m, the total deformation of the spring is 0.25 m. Thus, the elastic potential energy is 2 = Ve 12= kx 2 12 (500)(0.25 = ) 15.63 J 6.

If the disk in Figure 5.16 rotates in the clockwise direction by 10°, determine the elastic potential energy of the system. Assume that the springs are originally undeformed.

Figure 5.16 Problem 6.

178

Solution If the disk rotates in the clockwise direction by 10°, the two torsional springs will be twisted in the opposite directions. The angular deformation of each spring is 10°. The total elastic potential energy is

 10π  = Ve 2= ( 12 Kθ2 ) 2 ( 12 ) (400)=   12.18 J  180  2

7.

Determine the equivalent spring constant for the system shown in Figure 5.17.

Figure 5.17 Problem 7.

Solution

For the three springs in parallel, we have keq1 = k2 + k3 + k4 . They are connected with another two springs in series. Thus, the equivalent spring constant for the system is 1 1 1 1 1 1 1 = + + = + + keq k1 keq1 k5 k1 k2 + k3 + k4 k5 keq =

8.

k1 (k2 + k3 + k4 )k5 (k1 + k5 )(k2 + k3 + k4 ) + k1k5

Determine the equivalent spring constant for the system shown in Figure 5.18.

Figure 5.18 Problem 8.

Solution For the two springs in series, kk 1 1 1 or keq1 = 1 2 = + k1 + k2 keq1 k1 k2

For the two springs in parallel, we have keq2= k3 + k4 . keq1 and keq2 are connected in parallel. Thus, the equivalent spring constant for the system is k1k2 + k3 + k 4 k1 + k2 Derive the spring constant expression for the axially loaded bar shown in Figure 5.19. Assume that the cross-sectional area is A and the modulus of elasticity of the material is E. keq = keq1 + keq2 =

9.

Figure 5.19 Problem 9.

Solution The force-deflection relation of an axially loaded bar is x =

L f . Therefore, keq= EA

f EA . = x L

179

10. The uniform circular shaft in Figure 5.20 acts as a torsional spring. Assume that the elastic shear modulus of the material is G. Derive the equivalent spring constant corresponding to a pair of torques applied at two free ends.

Figure 5.20 Problem 10.

Solution For a shaft, we have L τ GJ where the polar moment of inertia for a cylinder is J = 321 πd 4 . Thus, 32 L θ= τ Gπd 4 The equivalent spring stiffness is τ Gπd 4 K eq = = θ 32 L θ=

11. A rod is made of two uniform sections, as shown in Figure 5.21. The two sections are made of the same material, and the modulus of elasticity of the rod material is E. The areas for the two sections are A1 and A2, respectively. Derive the equivalent spring constant corresponding to a tensile force applied at the free end.

Figure 5.21 Problem 11.

Solution The rod with two uniform sections can be treated as two axial springs in series. Using the result obtained in Problem 9, the equivalent spring stiffness for each section is EA ki = i Li where

i = 1, 2 . Thus, the equivalent spring stiffness for the rod is EA1 A2 L L 1 1 1 = + = 1 + 2 or keq = L1 A2 + L2 A1 keq k1 k2 EA1 EA2

12. Derive the spring constant expression of the fixed-fixed beam in Figure 5.22. The Young’s modulus of the material is E and the moment of inertia of cross-sectional area is I. Assume that the force f and the deflection x are at the center of the beam.

Figure 5.22 Problem 12.

Solution For a fixed-fixed beam with a load at midspan, the force-deflection relation is x = spring stiffness is keq=

f 192 EI . = x L3

L3 f . Therefore, the equivalent 192 EI

180

13. Derive the spring constant expression of the simply supported beam in Figure 5.23. The Young’s modulus of the material is E and the moment of inertia of cross-sectional area is I. Assume that the force f and the deflection x are at the center of the beam.

Figure 5.23 Problem 13.

Solution For a simply supported beam with a load at midspan, the force-deflection relation is x = equivalent spring stiffness is keq=

L3 f . Therefore, the 48EI

f 48 EI . = x L3

14. Derive the spring constant expression of the simply supported beam in Figure 5.24. The Young’s modulus of the material is E and the moment of inertia of cross-sectional area is I. Assume that the applied load f is anywhere between the supports.

Figure 5.24 Problem 14.

Solution For a simply supported beam with a load applied anywhere between supports, the force-deflection relation is a 2b2 x= f 3EIL where L is the length of the beam and L = a+b. Therefore, the equivalent spring stiffness is f 3EI (a + b) keq= = x a 2b2 15. Determine the equivalent damping coefficient for the system shown in Figure 5.25.

Figure 5.25 Problem 15.

Solution For the two dampers in series,

bb 1 1 1 or beq1 = 1 2 = + b1 + b2 beq1 b1 b2 beq1 and b3 are connected in parallel. Thus, the equivalent damping coefficient for the system is beq = beq1 + b3 =

b1b2 + b3 b1 + b2

16. Determine the equivalent damping coefficient for the system shown in Figure 5.26.

Figure 5.26 Problem 16.

181

Solution For the two dampers in parallel,

beq1= b1 + b2 beq1 and b3 are connected in series. Thus, the equivalent damping coefficient for the system is 1 1 1 1 1 bb +b b or beq = 1 3 2 3 = + = + beq beq1 b3 b1 + b2 b3 b1 + b2 + b3

Problem Set 5.2 1.

For the system shown in Figure 5.38, the input is the force f and the output is the displacement x of the mass. a. Draw the necessary free-body diagram and derive the differential equation of motion. b. Using the differential equation obtained in Part (a), determine the transfer function. Assume initial conditions x(0) = 0 and x (0) = 0 . c. Using the differential equation obtained in Part (a), determine the state-space representation.

Figure 5.38 Problem 1.

Solution a.

The free-body diagram is shown in the figure below. Due to gravity, the spring is compressed by δ st when the mass is in static equilibrium and mg = kδ st . Choosing the static equilibrium as the origin of the coordinate x and applying Newton’s second law in the x-direction gives f (t ) + ↓ x : f + mg − k ( x + δ st ) − bx =mx or mx + bx + kx =

b.

c.

Figure PS5-2 No1 Taking the Laplace transform of both sides of the preceding equation with zero initial conditions results in X ( s) 1 = 2 F ( s ) ms + bs + k The state, the input, and the output are  x1   x  = x = , u f ,= y x   =  x2   x 

 0  x1   = k    x 2   −  m

1  0  x1     + b   1 u −   x2    m m x  = y [1 0]  1  + 0 ⋅ u  x2 

2.

Repeat Problem 1 for the system shown in Figure 5.39.

182

Figure 5.39 Problem 2.

Solution a.

The free-body diagram is shown in the figure below. Due to gravity, the spring is stretched by δ st when the mass

is in static equilibrium and mg sin 30 = kδ st . Choosing the static equilibrium as the origin of the coordinate x and applying Newton’s second law in the x-direction gives +: f + mg sin 30 − k ( x + δ st ) − bx = mx mx + bx + kx = f (t )

b. c.

Figure PS5-2 No2 Refer to the solution to Part (b) in Problem 1. Refer to the solution to Part (c) in Problem 1.

3.

Repeat Problem 1 for the system shown in Figure 5.40.

Figure 5.40 Problem 3.

Solution a.

The free-body diagram is shown in the figure below. Due to gravity, the spring is stretched by δ st when the mass is in static equilibrium and mg = kδ st . Choosing the static equilibrium as the origin of the coordinate x and applying Newton’s second law in the x-direction gives + ↓ x : f + mg − k ( x + δst ) − bx =mx mx + bx + kx = f (t )

b.

Figure PS5-2 No3 Refer to the solution to Part (b) in Problem 1.

183

c. 4.

Refer to the solution to Part (c) in Problem 1. Repeat Problem 1 for the system shown in Figure 5.41.

Figure 5.41 Problem 4.

Solution a.

b. c. 5.

The free-body diagram is shown in the figure below. Choosing the static equilibrium as the origin of the coordinate x and applying Newton’s second law in the x-direction gives + → x : f − kx − bx =mx mx + bx + kx = f (t )

Figure PS5-2 No4 Refer to the solution to Part (b) in Problem 1. Refer to the solution to Part (c) in Problem 1. For Problems 1 through 4, use Simulink to construct block diagrams to find the displacement output x(t) of the system subjected to an applied force f(t) = 10u(t), where u(t) is the unit-step function. The parameter values are m = 1 kg, b = 2 N·s/m, and k = 5 N/m. Assume zero initial conditions.

Solution Since the systems in Problem 1 through 4 have the same mathematical model, we can use the same Simulink block diagram to find the displacement output.

Figure PS5-2 No5a

184

6.

Figure PS5-2 No5b The system shown in Figure 5.42 simulates a machine supported by rubbers, which are approximated as four same spring-damper units. The input is the force f and the output is the displacement x of the mass. The parameter values are m = 500 kg, b = 250 N·s/m, and k = 200,000 N/m. a. Draw the necessary free-body diagram and derive the differential equation of motion. b. Determine the transfer function form. Assume zero initial conditions. c. Determine the state-space representation. d. Find the transfer function from the state-space form and compare with the result obtained in Part (b).

Figure 5.42 Problem 6.

Solution a.

b.

c.

d.

The free-body diagram is shown in the figure below. Note that four same spring-damper units are connected in parallel. Therefore, the equivalent spring stiffness is 4k and the equivalent damping coefficient is 4b. Let us choose the static equilibrium as the origin of the coordinate x. Then, the static spring force will be cancelled by the weight. Applying Newton’s second law in the x-direction gives + ↑ x : f − 4kx − 4bx =mx mx + 4bx + 4kx = f (t ) or 500  x + 1000 x + 800000 x = f (t )

Figure PS5-2 No6 Taking the Laplace transform of both sides of the preceding equation with zero initial conditions results in 1 X ( s) = F ( s ) 500s 2 + 1000s + 800000 The state, the input, and the output are  x1   x  = x = , u f ,= y x   =  x2   x  The state equation and the output equation in matrix form are 0  1   x1    x1   0 x  u , = = + y [1 0]  1  + 0 ⋅ u 1         x 2   −1600 −2   x2    x2   500  Using the results obtained in Part (c), we can derive the transfer function X ( s) Y ( s) = =C( sI − A ) −1 B + D F ( s) U ( s)

=

 0   s + 2 1 1 s + 2 s + 1600  −1600 s   500 

  0] 2 [1=   1 

1 500 s + 2 s + 1600 2

185

7.

which is the same as the one obtained in Part (b). Repeat Problem 6 for the system shown in Figure 5.43.

Figure 5.43 Problem 7.

Solution a.

b.

c.

d.

The free-body diagram is shown in the figure below. Choosing the static equilibrium as the origin of the coordinate x and applying Newton’s second law in the x-direction gives + → x : f − 2kx − 2bx =mx mx + 2bx + 2kx = f (t ) or 20  x + 250 x + 800 x = f (t )

Figure PS5-2 No7 Taking the Laplace transform of both sides of the preceding equation with zero initial conditions results in X (s) 1 = F ( s ) 20 s 2 + 250 s + 800 The state, the input, and the output are  x1   x  , u f ,= y x = x =   =  x2   x  The state equation and the output equation in matrix form are 1   x1   0   x1   0 x  = u, = y [1 0]  1  + 0 ⋅ u     +     x2   x2   −40 −12.5  x2  0.05 Using the results obtained in Part (c), we can derive the transfer function −1

−1   0  s 0]      40 s + 12.5 0.05  s + 12.5 1  0  1 0.05 = [1= 0] 2     2 s  0.05 s + 12.5s + 40 s + 12.5s + 40  −40 which is the same as the one obtained in Part (b). X (s) Y (s) = = C( sI − A) −1 B + D= F (s) U (s)

8.

[1

The system shown in Figure 5.44 simulates a vehicle traveling on a rough road. The input is the displacement z. a. Draw the necessary free-body diagram and derive the differential equation of motion. b. Assuming zero initial conditions, determine the transfer function for two different cases of output: (1) displacement x and (2) velocity x . c. Determine the state-space representation for two different cases of output: (1) displacement x and (2) velocity x .

186

Figure 5.44 Problem 8.

Solution a.

We choose the static equilibrium as the origin of the coordinate x. Then, the static spring force will be cancelled by the weight. Assume that x > z > 0 . Applying Newton’s second law in the x-direction gives + ↑ x : − k ( x − z ) − b ( x − z ) =mx

mx + bx + kx = bz + kz

b.

c.

Figure PS5-2 No8 Assuming zero initial conditions and taking Laplace transform of the above differential equation gives ( ms 2 + bs + k ) X ( s) = ( bs + k ) Z ( s) If the displacement x is the output, the transfer function is X ( s) bs + k = 2 Z ( s ) ms + bs + k If the velocity x is the output, the transfer function is V ( s ) sX ( s ) bs 2 + ks = = Z ( s ) Z ( s ) ms 2 + bs + k Rewrite the transfer function X ( s ) Z ( s ) as

X ( s) X ( s) W ( s)  b k 1 = =  s+  Z ( s) W ( s) Z ( s)  m m  s2 + b s + k m m where

X ( s) b k s+ = W ( s) m m 1 W ( s) = Z ( s) s 2 + b s + k m m Thus in the time-domain, we have b k w + w m m b k  = w − w − w + z m m T Select the state and the = input as x [ = w w ] , u z . The state-space equations with the displacement x as the output are 1   0  x1    x1   0  k b   x1  = = k b  +   1 u y  m m   x  + 0 ⋅ u    x x − −   2  2  2   m  m If the velocity x is the output, following the same procedure gives the output equation  bk b2 k   x1  b y= − − +  + u  2 m 2 m   x2  m  m = x

187

9.

Repeat Problem 8 for the system shown in Figure 5.45, where the cam and follower impart a displacement z to the lower end of the system.

Figure 5.45 Problem 9.

Solution a.

We choose the static equilibrium as the origin of the coordinate x. Then, the static spring force will be cancelled by the weight Assume that x > z > 0 . The free-body diagram is shown in the figure below. Applying Newton’s second law in the x-direction gives + ↑ x : − k1 x − bx − k2 ( x − z ) =mx mx + bx + ( k1 + k2 ) x = k2 z

b.

Figure PS5-2 No9 Assuming zero initial conditions and taking Laplace transform of the above differential equation give k2 Z ( s ) ( ms 2 + bs + k1 + k2 ) X ( s) = If the displacement x is the output, the transfer function is k2 X ( s) = 2 Z ( s ) ms + bs + k1 + k2 If the velocity x is the output, the transfer function is k2 s V ( s ) sX ( s ) = = Z ( s ) Z ( s ) ms 2 + bs + k1 + k2

c.

The state and the input are x =

T x x ] , u [=

z . The state-space equations with the displacement x as the output is

 0  x1   = k1 + k2    x 2   − m 

1  0  x1     + b   k u −   x2   2  m m x  = y [1 0]  1  + 0 ⋅ u  x2  If the velocity x is the output, the output equation is x  = y [0 1]  1  + 0 ⋅ u  x2 

188

10. For the system shown in Figure 5.46, the input is the force f and the outputs are the displacements x1 and x2 of the masses. a. Draw the necessary free-body diagrams and derive the differential equations of motion. b. Write the differential equations of motion in the second-order matrix form. c. Using the differential equations obtained in Part (a), determine the state-space representation.

Figure 5.46 Problem 10.

Solution a.

We choose the displacements of the two masses x1 and x2 as the generalized coordinates. The static equilibrium positions of m1 and m2 are set as the coordinate origins. Assume that x2 > x1 > 0 . Applying Newton’s second law in the x-direction gives + ↓ x : ∑ Fx =max Mass 1: −k1 x1 + k2 ( x2 − x1 ) + b ( x2 − x1 ) = m1  x1

Mass 2: f − k2 ( x2 − x1 ) − b ( x2 − x1 ) = m2  x2 Rearranging the equations, we have m1  x1 + bx1 − bx2 + ( k1 + k2 ) x1 − k2 x2 = 0

m2  x2 − bx1 + bx2 − k2 x1 + k2 x2 = f

b.

c.

Figure PS5-2 No10 The differential equations can be expressed in second-order matrix form as x1   b −b   x1   k1 + k2 −k2   x1   0   m1 0    +   =    0 m     +  k2   x2   f   −b b   x2   −k2  2   x2  The state, the input, and the output are

189

 x1   x1  x  x  x   2  2 , u f ,= = x = y  1   =   x2   x3   x1   x4   x 2  The state-space representation in matrix form is 0 1 0   0  0   0 0 1   x1     x1   0  x1  0   x        k1 + k2 k2 b b   x2   1 0 0 0   x2   0   2 − +  0 u y    −   =  +  u  m1 m1 m1 m1   x3    0 1 0 0  x3  0  x3   1  x4   k2  x4  k b b  x    − 2 −   4   m2   m2 m2 m2   m2

11. Repeat Problem 10 for the system shown in Figure 5.47.

Figure 5.47 Problem 11.

Solution a.

We choose the displacements of the two masses x1 and x2 as the generalized coordinates. The static equilibrium positions of m1 and m2 are set as the coordinate origins. Assume that x1 > x2 > 0 . Applying Newton’s second law in the x-direction gives ∑ Fx = ma x +: Mass 1: f − k1 ( x1 − x2 ) − b1 ( x1 − x 2 ) = m1  x1

Mass 2: k1 ( x1 − x2 ) + b1 ( x1 − x2 ) − k2 x2 − b2 x2 = m2  x2 Rearranging the equations, we have

m1  x1 + b1 x1 − b1 x 2 + k1 x1 − k1 x2 = f    m2 x2 − b1 x1 + (b1 + b2 ) x2 − k1 x1 + (k1 + k2 ) x2 = 0

b.

Figure PS5-2 No11 The differential equations can be expressed in second-order matrix form as x1   b1 −b1   x1   k1 −k1   x1   f   m1 0    +     0 m      x  +  −k k + k   x  = x − b b + b  2 2  1 1 2   2   1 1 2   2  0

190

c.

The state, the input, and the output are  x1   x1  x  x  x   2  2 , u f ,= = x = y  1   =  x x  x2   3  1  x4   x 2  The state-space representation in matrix form is 0 1 0   0 0  0 0 1   x1     x1   0  x1  0  x        k1 k1 b b1   x2   1 0 0 0   x2   0   2 − 1 +  1 u y    −   =  +  u  m1 m1 m1   x3    0 1 0 0  x3  0  x3   m1 m 1  x4   k1  x4  k +k b1 b + b  x    − 1 2 − 1 2   4   0   m2 m2 m2   m2

12. Repeat Problem 10 for the system shown in Figure 5.48.

Figure 5.48 Problem 12.

Solution a.

We choose the displacements of the two masses x1 and x2 as the generalized coordinates. The static equilibrium positions of m1 and m2 are set as the coordinate origins. Assume that x2 > x1 > 0 . Applying Newton’s second law in the x-direction gives + → x : ∑ Fx =ma x Mass 1: −k1 x1 − b1 x1 + k2 ( x2 − x1 ) = m1  x1 Mass 2: f − k2 ( x2 − x1 ) = m2  x2 Rearranging the equations, we have

m1  x1 + b1 x1 + ( k1 + k2 ) x1 − k2 x2 = 0

m2  x2 − k2 x1 + k2 x2 = f

b.

c.

Figure PS5-2 No12 The differential equations can be expressed in second-order matrix form as x1  b1 0  x1   k1 + k2 −k2   x1   0   m1 0    +  =    0 m     + k2   x2   f   0 0  x 2   −k2 2   x2   The state, the input, and the output are  x1   x1  x  x  x   2  2 = , u f ,= x = y  1   =  x2   x3   x1   x4   x 2  The state-space representation in matrix form is

191

 0  x1   0     x 2   k1 + k2 =   − m1  x3    x 4   k2   m2 13.

0 0 k2 m1 −

k2 m2

1 0 −

b1 m1 0

0  0  1   x1     x1    0      x   1 0 0 0   x2   0  0  2 = +  0 u y   +  u  0 1 0 0  x3  0   x3   1   x4   x    0  4   m2  

For Problems 10 through 12, use MATLAB commands to define the systems in the state-space form and then convert to the transfer function form. Assume that the displacements of the two masses, x1 and x2, are the outputs, and all initial conditions are zero. The masses are m1 = 5 kg and m2 = 15 kg. The spring constants are k1 = 7.5 kN/m and k2 = 15 kN/m. The viscous damping coefficients are b1 = b2 = b = 250 N·s/m.

Solution For the system in Problem 10, the MATLAB session is % define the system parameters m1 = 5; m2 = 15; b1 = 250; b2 = 250; b = 250; k1 = 7500; k2 = 15000; % define the system in the state-space form A = [0 0 1 0; 0 0 0 1; -(k1+k2)/m1 k2/m1 -b/m1 b/m1; k2/m2 -k2/m2 b/m2 -b/m2]; B = [0; 0; 0; 1/m2]; C = [1 0 0 0; 0 1 0 0]; D = zeros(2,1); % convert to the transfer function form sys_ss = ss(A,B,C,D); sys_tf = tf(sys_ss); The command tf returns two transfer functions X1 ( s) 3.333s + 200 = F ( s ) s 4 + 66.67 s 3 + 5500 s 2 + 2.5 × 104 s + 1.5 × 106

X 2 (s) 0.06667 s 2 + 3.333s + 300 = 4 F (s) s + 66.67 s 3 + 5500 s 2 + 2.5 × 104 s + 1.5 × 106 For the systems in Problems 11 through 12, we only need to modify the state-space matrices. % define the system in Problem 11 in the state-space form A = [0 0 1 0; 0 0 0 1; -k1/m1 k1/m1 -b1/m1 b1/m1; k1/m2 -(k1+k2)/m2 b1/m2 -(b1+b2)/m2]; B = [0; 0; 1/m1; 0]; C = [1 0 0 0; 0 1 0 0]; D = zeros(2,1); The command tf returns two transfer functions for the system in Problem 11

192

X1 ( s) 0.2 s 2 + 6.667 s + 300 = 4 3 F ( s ) s + 83.33s + 3833s 2 + 7.5 × 104 s + 1.5 × 106 X 2 (s) 3.333s + 100 = 4 3 F (s) s + 83.33s + 3833s 2 + 7.5 × 104 s + 1.5 × 106 % define the system in Problem 12 in the state-space form A = [0 0 1 0; 0 0 0 1; -(k1+k2)/m1 k2/m1 -b1/m1 0; k2/m2 -k2/m2 0 0]; B = [0; 0; 0; 1/m2]; C = [1 0 0 0; 0 1 0 0]; D = zeros(2,1); The command tf returns two transfer functions for the system in Problem 11 X1 ( s) 200 = 4 3 2 F ( s ) s + 50 s + 5500 s + 5 × 104 s + 1.5 × 106

X 2 (s) 0.06667 s 2 + 3.333s + 300 = 4 F (s) s + 50 s 3 + 5500 s 2 + 5 × 104 s + 1.5 × 106 14. For the system in Figure 5.49, the inputs are the forces f1 and f2 applied to the masses and the outputs are the displacements x1 and x2 of the masses. a. Draw the necessary free-body diagrams and derive the differential equations of motion. b. Write the differential equations of motion in the second-order matrix form. c. Using the differential equations obtained in Part (a), determine the state-space representation.

Figure 5.49 Problem 14.

Solution a.

We choose the displacements of the two masses x1 and x2 as the generalized coordinates. The static equilibrium positions of m1 and m2 are set as the coordinate origins. Assume that x2 > x1 > 0 . Applying Newton’s second law in the x-direction gives + → x : ∑ Fx =ma x Mass 1: f1 − k1 x1 + k2 ( x2 − x1 ) + b2 ( x2 − x1 ) = m1  x1

Mass 2: f 2 − k3 x2 − k2 ( x2 − x1 ) − b2 ( x2 − x1 ) = m2  x2 Rearranging the equations, we have m1  x1 + b2 x1 − b2 x2 + (k1 + k2 ) x1 − k2 x2 = f1

m2  x2 − b2 x1 + b2 x2 − k2 x1 + (k2 + k3 ) x2 = f2

Figure PS5-2 No14

193

b.

c.

The differential equations can be expressed in second-order matrix form as −k2   x1   f1  x1   b2 −b2   x1   k1 + k2  m1 0    +   +   =    0 m     k2 + k3   x2   f 2  2   x2    −b2 b2   x2   −k2 The state, the input, and the output are  x1   x1  x  x  f  x   2  2 = x = u  1  ,= y  1    ,=  f2   x2   x3   x1   x4   x 2  The state-space representation in matrix form is 0 1 0  0   0 0  0  x   x 0 0 1 0   1   x1   1  0  x   x       u k + k k b b 1 0 0 0 1      x2  0 0  u1   2  2 1 2 2 2 − 2 0  1 y   +  =   −  +   m1 m1 m1 m1   x3   m1 0 1 0 0  x3  0 0 u2   x3    u2   x4   k2  x4  k +k 1  b2 b  x   − 2 3 − 2  4  0   m2  m2 m2 m2    m2

15. Repeat Problem 14 for the system shown in Figure 5.50.

Figure 5.50 Problem 15.

Solution a.

We choose the displacements of the two masses x1 and x2 as the generalized coordinates. The static equilibrium positions of m1 and m2 are set as the coordinate origins. Assume that x2 > x1 > 0 .

Figure PS5-2 No15 Applying Newton’s second law in the x-direction gives + → x : ∑ Fx =ma x

m1  x1 Mass 1: f1 − k1 x1 − b1 x1 + k2 ( x2 − x1 ) =

m2 x2 Mass 2: f 2 − k2 ( x2 − x1 ) − k3 x2 = Rearranging the equations, we have m1  x1 + b1 x1 + (k1 + k2 ) x1 − k2 x2 = f1 b.

c.

m2  x2 − k2 x1 + (k2 + k3 ) x2 = f2 The differential equations can be expressed in second-order matrix form as −k2   x1   f1  x1  b1 0   x1   k1 + k2  m1 0    +   +   =    0 m     k2 + k3   x2   f 2   0 0   x2   −k2  2   x2  The state, the input, and the output are

194

 x1   x1  x  x  f  x   2  2 = x = u  1  ,= y  1    ,=   f2   x2   x3   x1   x4   x 2  The state-space representation in matrix form is 0 1 0 0   0 0    0 0 1   x1  0 0   x1   0  x1    x        x   1  u  k1 + k2 k2 b 1 0 0 0  x2  0 0  u1   2 − 1 0  2  +  = 0  1 y    −  +    m1 m1 m1 0 1 0 0  x3  0 0 u2    x3   m1  x3    u2   x    x4   k2  x4  k +k 1  − 2 3 0 0  4   0   m2  m2  m2  

16.

For Problems 14 and 15, use MATLAB commands to define the systems in the state-space form and then convert to the transfer function form. Assume that the displacements of the two masses, x1 and x2, are the outputs, and all initial conditions are zero. The masses are m1 = 5 kg and m2 = 15 kg. The spring constants are k1 = 7.5 kN/m, k2 = 15 kN/m, and k3 = 30 kN/m. The viscous damping coefficients are b1 = b2 = 250 N·s/m.

Solution For the system in Problem 14, the MATLAB session is % define the system parameters m1 = 5; m2 = 15; b1 = 250; b2 = 250; k1 = 7500; k2 = 15000; k3 = 30000; % define the system in the state-space form A = [0 0 1 0; 0 0 0 1; -(k1+k2)/m1 k2/m1 -b2/m1 b2/m1; k2/m2 -(k2+k3)/m2 b2/m2 -b2/m2]; B = [0 0; 0 0; 1/m1 0; 0 1/m2]; C = [1 0 0 0; 0 1 0 0]; D = zeros(2,2); % convert to the transfer function form sys_ss = ss(A,B,C,D); sys_tf = tf(sys_ss); The command tf returns four transfer functions 2  3.333s + 200  X1 ( s ) X 1 ( s )  0.2 s + 3.333s + 600   F (s) F (s)   3.333s + 200 0.06667 s 2 + 3.333s + 300  2  1 =   X 2 (s) X 2 (s)  s 4 + 66.67 s 3 + 7500 s 2 + 1.25 × 105 s + 1.05 × 107    F1 ( s ) F2 ( s )  For the systems in Problem 15, we only need to modify the state-space matrices. % define the system in Problem 15 in the state-space form A = [0 0 1 0; 0 0 0 1; -(k1+k2)/m1 k2/m1 -b2/m1 b2/m1; k2/m2 -(k2+k3)/m2 b2/m2 -b2/m2]; B = [0 0; 0 0; 1/m1 0; 0 1/m2]; C = [1 0 0 0; 0 1 0 0]; D = zeros(2,2);

195

The command tf returns four transfer functions for the system in Problem 15 2  200  X1 ( s ) X 1 ( s )  0.2 s + 600   F (s) F (s)   200 0.06667 s 2 + 3.333s + 300  2  1 =   X 2 ( s ) X 2 ( s )  s 4 + 50 s 3 + 7500 s 2 + 1.5 × 105 s + 1.05 × 107    F1 ( s ) F2 ( s )  17. For the system in Figure 5.51, the input is the force f and the outputs are the displacement x1 of the mass and the displacement x2 of the massless junction A. a. Draw the necessary free-body diagrams and derive the differential equations of motion. Determine the number of degrees of freedom and the order of the system. b. Write the differential equations of motion in the second-order matrix form. c. Using the differential equation obtained in Part (a), determine the state-space representation.

Figure 5.51 Problem 17.

Solution a.

One massless junction is included in this system. We choose the displacements of the mass and the junction A as the generalized coordinates, which are denoted by x1 and x2 . Thus, this is a two-degree-of-freedom system. Assume that x2 > x1 > 0 .

Figure PS5-2 No17 Applying Newton's second law gives Mass:

b.

c.

+ → x : ∑ Fx =ma x −k1 x1 − b1 x1 + k2 ( x2 − x1 ) + b2 ( x 2 − x1 ) = mx1 + → x : ∑= = Fx ma 0 Massless junction: x   k ( x x ) b ( x x ) f = 0 − − − − + A: 2 2 1 2 2 1 The equations can be rearranged as follows. Thus, this is a third-order system. mx1 + (b1 + b2 ) x1 − b2 x 2 + ( k1 + k2 ) x1 − k2 x2 = 0 b2 x 2 − b2 x1 − k2 x1 + k2 x2 = f The differential equations can be expressed in second-order matrix form as x1  b1 + b2 −b2   x1   k1 + k2 −k2   x1   0   m1 0   +  +  =    0 0   b2   x 2   −k2 k2   x2   f     x2   −b2 The state, the input, and the output are  x1   x1  x      x = y  1 = x2   x2  = , u f ,=  x2   x   x   3  1 The state-space representation in matrix form is 0 1   x1   0   x1   x1   0 1 0 0   0         k b k2 b2 1 x = + 1b u y   x 2  =−   x2  +   u  2 2  2  2 0 1 0  x  0  x   − k m  x  1 m   0 b m −  3  3  1 1  3  

196

18. Repeat Problem 17 for the system in Figure 5.52, the input is the displacement z and the outputs are the displacements x1 and x2.

Figure 5.52 Problem 18.

Solution a.

Assume that x1 > x2 > z > 0 .

Figure PS5-2 No18 Applying Newton's second law give Mass:

+ ↑ x : ∑ Fx =ma x

mx1 − k1 ( x1 − x2 ) − b ( x1 − x 2 ) =

Massless junction

A:

+ ↑ x : ∑= Fx ma = 0 x

k1 ( x1 − x2 ) + b ( x1 − x 2 ) − k2 ( x2 − z ) = 0

The equations can be rearranged as

b.

c.

mx1 + bx1 − bx 2 + k1 x1 − k1 x2 = 0   bx2 − bx1 − k1 x1 + ( k1 + k2 ) x2 = k2 z The differential equations can be expressed in second-order matrix form as x1   b −b   x1   k1 −k1   x1   0   m 0   +  +     =  0 0       x2   −b b   x 2   −k1 k1 + k2   x2  k2 z  The state, the input, and the output are  x1   x1  x      = x = x2   x2  = , u z,= y  1  x2   x   x   3  1 The state-space representation in matrix form is 0 1   x1   0   x1   0  x1     1 0 0   0  x  +  k b  u  x = k b − ( k + k ) b 1 y   2  1 1 2 2   x2  +   u   2  =  0 1 0  x  0  x   0  x  k m   0 − k m  3  2  3  2   3

197

Problem Set 5.3 1.

Consider the rotational system shown in Figure 5.61. The system consists of a massless shaft and a uniform thin disk of mass m and radius r. The disk is constrained to rotate about a fixed longitudinal axis along the shaft. The shaft is equivalent to a torsional spring of stiffness K. Draw the necessary free-body diagram and derive the differential equation of motion.

Figure 5.61 Problem 1.

Solution The free-body diagram of the disk is shown below. Applying the moment equation to the fixed point O gives I Oα +↶: ∑ M O =  τ − Kθ − Bθ = θ I  O

where the mass moment of inertia for a thin disk about its center of gravity is I O = 12 mr 2 . Thus, we have 1 mr 2 θ + Bθ + Kθ = τ 2

Figure PS5-3 No1 2.

Repeat Problem 1 for the system shown in Figure 5.62.

Figure 5.62 Problem 2.

Solution The free-body diagram of the disk is shown below. Applying the moment equation to the fixed point O gives I Oα +↶: ∑ M O = τ − 2 Kθ − 2 Bθ = I  θ O

where the mass moment of inertia for a thin disk about its center of gravity is I O = 12 mr 2 . Thus, we have 1  Kθ = mr 2 θ + 2 Bθ+2 τ 2

Figure PS5-3 No2 3.

Consider the torsional mass–spring system in Figure 5.63. The mass moments of inertia of the two disks are I1 and I2, respectively. The massless torsional springs represent the elasticity of the shafts.

198

a. b. c.

Draw the necessary free-body diagrams and derive the differential equations of motion. Provide the equations in the second-order matrix form. Determine the transfer functions Θ1(s)/T(s) and Θ2(s)/T(s). All the initial conditions are assumed to be zero. Determine the state-space representation with the angular displacements θ1 and θ2 as the outputs.

Figure 5.63 Problem 3.

Solution

a. We choose the angular displacements θ1 and θ2 as the generalized coordinates. Assume that θ1 > θ2 > 0 . The free-body diagrams are shown in the figure below. Applying the moment equation about the fixed points O1 and O2, respectively, gives I Oα +↶: ∑ M O = τ − K θ − K (θ − θ )= I  θ 1 1

2

1

2

1 1

K 2 (θ1 − θ2 )= I 2 θ2

Rearranging the equations results in

I1 θ1 + ( K1 + K 2 )θ1 − K 2 θ2 =τ I  θ − K θ + K θ =0 2 2

2 1

which can be written in the second-order matrix form θ1   K1 + K 2  I1 0     0 I    +  − K θ 2   2 2 

2 2

− K 2   θ1   τ   =   K 2  θ 2  0 

Figure PS5-3 No3 b.

All the initial conditions are assumed to be zero. Taking Laplace transform of the above differential equations gives  I1s 2 + K1 + K 2 − K 2   Θ1 ( s )  T ( s )    = . I 2 s 2 + K 2  Θ 2 ( s )   0  − K2  Using Cramer’s rule, we can solve for Θ1 ( s ) Τ( s ) and Θ2 ( s ) Τ( s ) ,

Θ1 ( s ) I2 s2 + K2 I2 s2 + K2 = = , T( s ) ( I1s 2 + K1 + K 2 )( I 2 s 2 + K 2 ) − K 22 I1 I 2 s 4 + ( I1 K 2 + I 2 K1 + I 2 K 2 ) s 2 + K1 K 2 Θ2 ( s) K2 K2 = = . T( s ) ( I1s 2 + K1 + K 2 )( I 2 s 2 + K 2 ) − K 22 I1 I 2 s 4 + ( I1 K 2 + I 2 K1 + I 2 K 2 ) s 2 + K1 K 2 c.

The state, the input, and the output are  x1   θ1   x  θ  θ   2  2 = , u τ,= x = y  1    = x θ θ2   3  1   x4  θ2  The state-space representation in matrix form is

199

0   0  x1    x   K1 + K 2  2 =   −  I1 x  3   x4   K 2  I2 

1 0 0 0 1   x1    0   x    K2 0 0  2  +  1  u I1   x3   I   x   1  K − 2 0 0   4   0  I2   x1     1 0 0 0   x2   0  = y   +  u  0 1 0 0  x3  0  x4 

4.

0 0

Repeat Problem 3 for the system shown in Figure 5.64, where the torsional viscous damper represents the fluid coupling. The input is the angular displacement φ at the end of the shaft.

Figure 5.64 Problem 4.

Solution a.

Assume that θ1 > θ2 > φ > 0 . The free-body diagrams are shown in the figure below.

Figure PS5-3 No4 Applying the moment equation about the fixed points O1 and O2, respectively, gives +↶:

∑ MO = I Oα

− K1θ1 − B1θ 1 − K 3 ( θ1 − θ2 ) = I1 θ1 K (θ − θ ) − K (θ − φ ) = I  θ 3

Rearranging the equations results in

1

2

2

2

2 2

I1 θ1 + B1θ 1 + ( K1 + K 3 ) θ1 − K 3θ2 = 0 I  θ − K θ + (K + K )θ = Kφ 2 2

3 1

2

3

which can be written in the second-order matrix form θ1   B1 0  θ 1   K1 + K 3  I1 0    0 I    +  0 0    +  − K   θ 2   3 2  θ 2   b.

2

2

− K 3   θ1   0   =   K 2 + K 3  θ2   K 2φ 

All the initial conditions are assumed to be zero. Taking Laplace transform of the above differential equations gives  I1s 2 + B1s + K1 + K 3   Θ1 ( s )   0  − K3 =    2 − K3 I 2 s + K 2 + K 3   Θ2 ( s )   K 2 Φ ( s )   Using Cramer’s rule, we can solve for Θ1 ( s ) Φ ( s ) and Θ2 ( s ) Φ ( s ) ,

200

Θ1 ( s ) K 2 K 3 = Φ( s) ∆( s )

where ∆( s )=

c.

(I s 1

2

2 Θ2 ( s ) K 2 ( I1s + B1s + K1 + K 3 ) = Φ( s) ∆( s )

+ B1s + K1 + K 3 )( I 2 s 2 + K 2 + K 3 ) − K 32

The state, the input, and the output are  x1   θ1   x  θ  θ   2  2 , u φ ,= = x = y  1    = θ 2   x3   θ1        x4   θ 2  The state-space representation in matrix form is 0 0 1 0   0  0 0 0 1   x1     x1    x1    0       x2    K3 B1 1 0 0 0  x 2   K1 + K 3    x2   0  0  = − +  0 u y    −  +  u I1 I1 I1 0 1 0 0  x3  0  x3     x3   K     x4   2  K2 + K3  x 4   K 3  x4   0 0 −  I 2    I I 2 2  

5.

Consider the pendulum system shown in Figure 5.65. The system consists of a bob of mass m and a uniform rod of mass M and length L. The pendulum pivots at the joint O. Draw the necessary free-body diagram and derive the differential equation of motion. Assume small angles for θ.

Figure 5.65 Problem 5.

Solution For the pendulum system, the free-body diagram is shown in the figure below, where Rx and Ry are the x and y components of the reaction force at the joint O, respectively. Note that the system rotates about a fixed axis through the point O.

Figure PS5-3 No5 Applying moment equation about the fixed point O gives

+ :

∑M

= IOα − 12 L sin θ·Mg − L sin θ·mg − Bθ= I O θ The system consists of a point mass and a slender rod. The total mass moment of inertia about point O is O

201

= I O I O _ mass + I O _ rod

where I O _ mass = mL2 and IO_rod can be obtained using the parallel-axis theorem,

I O _ rod = I C _ rod + Md 2 = Then I= O

( m + 13 M ) L2 . Substituting I

form, we obtain

O

1 12

ML2 + M ( 12 L ) = 2

1 3

ML2

into the equation and rearranging it in the input–output differential equation

( m + 13 M ) L2θ + Bθ + ( m + 12 M ) gL sin θ = 0

For small angles θ, sin θ ≈ θ. The equation of motion becomes m + 13 M L2 θ + Bθ + m + 12 M gLθ = 0

(

6.

)

(

)

Repeat Problem 5 for the systems shown in Figure 5.66, where (a) the mass of the rod is neglected and (b) no bob is attached to the rod.

Figure 5.66 Problem 6.

Solution a.

The free-body diagram of the system in (a) is shown below, where the mass of the rod is neglected. Applying moment equation about the fixed point O gives

+ :

∑M

O

= IOα

− L sin θ·mg − Bθ= I O θ

where the mass moment of inertia about point O is I O = mL2 . Therefore, the differential equation of motion is mL2 θ + Bθ + mgL sin θ = 0 For small angles θ, sin θ ≈ θ. The equation of motion becomes mL2 θ + Bθ + mgLθ = 0

b.

Figure PS5-3 No6a The free-body diagram of the system in (b) is shown below, where no bob is attached to the rod. Applying moment equation about the fixed point O gives

+ :

∑M

= IOα  − L sin θ·Mg − Bθ= I O θ 1 2

O

202

where the mass moment of inertia about point O can be obtained using the parallel-axis theorem, 1 ML2 + M 1 L IO = I C + Md 2 = ( 2 ) =13 ML2 12 2

Thus, the differential equation of motion is 1 ML2 θ + Bθ + 12 MgL sin θ = 0 3

For small angles θ, sin θ ≈ θ. The equation of motion becomes 1 ML2 θ + Bθ + 12 MgLθ = 0 3

Figure PS5-3 No6b 7.

The system shown in Figure 5.67 consists of a uniform rod of mass M and length L and a translational spring of stiffness k at the rod’s left tip. The friction at the joint O is modeled as a damper with coefficient of torsional viscous damping B. The input is the force f and the output is the angle θ. The position θ = 0 corresponds to the static equilibrium position when f = 0.

a. b. c.

Figure 5.67 Problem 7. Draw the necessary free-body diagram and derive the differential equation of motion for small angles θ. Using the linearized differential equation obtained in Part (a), determine the transfer function Θ(s)/F(s).  Assume that the initial conditions are θ(0) = 0 and θ(0) = 0. Using the differential equation obtained in Part (a), determine the state-space representation.

Solution a.

At equilibrium, we have

0 +↷: ∑ M O = L ⋅ kδ st = 0 δ st = 0 Applying the moment equation to the fixed point O gives I Oα +↷: ∑ M O = 1 2

I O θ − 12 L cos θ ⋅ f k + 12 L cos θ ⋅ f − Bθ = where, for small angular motions, the spring force can be approximated as f k = k ( 12 Lθ) . I= Note that the mass moment of inertia about point O is I= O C

1 12

Thus, the differential equation of motion for small angles θ is 1  1 kL2 θ= 1 Lf ML2 θ+Bθ+ 12

4

2

ML2 .

203

b.

c.

8.

Figure PS5-3 No7 Taking Laplace transform of the above linearized differential equation gives Θ( s ) 6L = 2 2 F ( s ) ML s + 12 Bs + 3kL2 The state, the input, and the output are  x1  θ  , u f ,= y θ = x =   =  x2   θ  The state equation and the output equation in matrix form are 1   0  0   x1    x1   x   = y [1 0]  1  + 0 ⋅ u 12 B    +  6  u=    3k  −  x2   x2  −  x2   M  ML  ML2  Repeat Problem 7 for the system shown in Figure 5.68. When θ = 0 and f = 0, the springs are at their free lengths.

Figure 5.68 Problem 8.

Solution a.

Applying the moment equation to the fixed point O gives I Oα +↷: ∑ M O =

L cos θ ⋅ f + L sin θ ⋅ mg + 12 L sin θ ⋅ Mg − 12 L cos θ ⋅ (2 f k ) = I O θ where, for small angular motions, the spring force can be approximated as f k = k ( 12 Lθ)

Figure PS5-3 No8 Note that the mass moment of inertia about point O is I O = I O _ mass + I O _ rod = mL2 + 13 ML2 Thus, the differential equation of motion for small angles θ is (m + 1 M ) L2 θ+ 1 kL2 θ − (m + 1 M ) gLθ=fL 3

b.

2

2

Taking Laplace transform of the above linearized differential equation gives Θ( s ) 1 = F ( s ) (m + 13 M ) Ls 2 + 12 kL − (m + 12 M ) g

204

c.

The state, the input, and the output are  x1  θ  , u f ,= y θ = x =   =  x2   θ  The state equation and the output equation in matrix form are 0 1 0     x1   1   x1  +  u 1 = − kL + ( m + M ) g 1    2   2  0   x2    x2   1   (m + 13 M ) L  + ( ) m M L 3   y =

9.

[1

x  0]  1  + 0 ⋅ u  x2 

Example 5.4 Part (d) shows how one can represent a linear system in Simulink based on the differential equation of the system. A linear system can also be represented in transfer function or state-space form. The corresponding blocks in Simulink are Transfer Fcn and State-Space, respectively. Consider Problem 7 and construct a Simulink block diagram to find the output θ(t) of the system, which is represented using (a) the linearized differential equation of motion, (b) the transfer function, and (c) the state-space form obtained in Problem 7. The parameter values are M = 0.8 kg, L = 0.6 m, k = 100 N/m, B = 0.4 N·s/m, and g = 9.81 m/s2. The input force f is the unit-impulse function, which has a magnitude of 20 N and a time duration of 0.01 s.

Solution The linearized differential equation of motion obtained in Problem 7 is   0.024θ+0.4θ+9θ=0.3 f The transfer function obtained in Problem 7 is Θ( s ) 3.6 = F ( s ) 0.288s 2 + 4.8s + 108 The state-space equations obtained in Problem 7 is 1   x1   0   x1   0 x  u = y [1 0]  1  + 0 ⋅ u =     +     x2   −375 −16.67   x2  12.5  x2  The Simulink block diagrams constructed based on the differential equation, the transfer function, the state-space form are shown below. Run simulations and the same response in Plot d is obtained.

Figure PS5-3 No9a

Figure PS5-3 No9b

Figure PS5-3 No9c

205

0.07 0.06 0.05 0.04

(rad)

0.03 0.02 0.01 0 -0.01 -0.02 0

0.2

0.4

0.6

0.8

1

Time (s)

Figure PS5-3 No9d 10.

Repeat Problem 9 using the linearized differential equation of motion, the transfer function, and the state-space form obtained in Problem 8. The parameter values are m = 0.2 kg, M = 0.8 kg, L = 0.6 m, k = 100 N/m, and g = 9.81 m/s2. The input force f is the unit-impulse function, which has a magnitude of 20 N and a time duration of 0.01 s.

Solution The linearized differential equation of motion obtained in Problem 8 is  0.168θ+14.47θ=0.6f The transfer function obtained in Problem 8 is Θ( s ) 1 = F ( s ) 0.28s 2 + 24.114 The state-space equations obtained in Problem 7 is 1   x1   0   x1   0 x  = y [1 0]  1  + 0 ⋅ u      x  + 3.57  u =  x 86.12 0 −  2    x2   2  The Simulink block diagrams constructed based on the differential equation, the transfer function, the state-space form are similar to those in Problem 9. Run simulations and the same response as shown in the figure below is obtained. 0.08

0.06

0.04

(rad)

0.02

0

-0.02

-0.04

-0.06

-0.08 0

0.5

1

1.5

2

Time (s)

Figure PS5-3 No10 11. Consider the system shown in Figure 5.69, in which the motion of the rod is small angular rotation. When θ = 0, the springs are at their free lengths. a. Determine the mass moment of inertia of the rod about point O. Assume that a > b. b. Draw the necessary free-body diagram and derive the differential equation of motion for small angles θ.

206

Figure 5.69 Problem 11.

Solution a.

Applying the parallel axis theorem gives the mass moment of inertia of the rod about point O

b.

I O = I C + Md 2 = 121 M ( a + b) 2 + M ( a − 12 ( a + b) ) = 13 M ( a 2 + b2 − ab) Applying the moment equation to the fixed point O gives I Oα +↶: ∑ M O = 1 −a cos θ ⋅ f + ( a − ( a + b)) sin θ ⋅ Mg − b cos θ ⋅ f = I  θ 2

k1

k2

2

O

where sin θ ≈ θ , cos θ ≈ 1 , and the spring forces are f k 1 = k1aθ and f k 2 = k2 bθ . Thus, the differential equation of motion for small angles θ is 2 2 2 1 1  2 3 M ( a + b − ab)θ+a k1θ + b k 2 θ − 2 ( a − b ) Mgθ=0

Figure PS5-3 No11 12. Consider the system shown in Figure 5.70, in which a lever arm has a spring–damper combination on the other side. When θ = 0, the system is in static equilibrium. a. Assuming that the lever arm can be approximated as a uniform slender rod, determine the mass moment of inertia of the rod about point O. b. Draw the necessary free-body diagram and derive the differential equation of motion for small angles θ.

Figure 5.70 Problem 12.

Solution a.

Applying the parallel axis theorem gives the mass moment of inertia of the rod about point O 2 2 7 1 1 1 IO = I C + Md 2 = 12 ML + M ( 2 L − 4 L ) = 48 ML 2

b.

At equilibrium, we have

207

0 +↷: ∑ M O = 1 4

L ⋅ Mg − 43 L ⋅ kδ st = 0 Mg = 3kδ st

Figure PS5-3 No12 Applying the moment equation to the fixed point O gives I Oα +↷: ∑ M O =

− 43 L cos θ ⋅ f k − 43 L cos θ ⋅ f b + 14 L cos θ ⋅ Mg = I Oθ where sin θ ≈ θ , cos θ ≈ 1 , the spring force = f k k ( 43 Lθ + δst ) and the damping force  f = b( 3 Lθ) b

4

Thus, the differential equation of motion for small angles θ is 2 2 2 7 9 9 3 1 0 48 ML θ + 16 bL θ + 16 kL θ + 4 kδst L − 4 MgL = Substituting the static equilibrium condition gives 2 2 2 7 9 9 48 ML θ + 16 bL θ + 16 kL θ=0 13. Consider the two-degree-of-freedom system shown in Figure 5.71, in which two simple inverted pendulums are connected by a translational spring of stiffness k. Each pendulum consists of a point mass m concentrated at the tip of a massless rope of length L. θ1 and θ2 are the angular displacements of the pendulums. When θ1 = 0 and θ2 = 0, the spring is at its free length. a. Draw the necessary free-body diagrams and derive the differential equations of motion. Assume small angles for θ1 and θ2. Provide the equations in the second-order matrix form. b. Using the differential equations obtained in Part (a), determine the state-space representation with the angular velocities θ 1 and θ 2 as the outputs.

Figure 5.71 Problem 13.

Solution a.

Assume that θ1 > θ2 > 0 . The free-body diagrams are shown in the figure below. Applying the moment equation about the fixed points O1 and O2, respectively, gives I Oα +↷: ∑ M O = − L sin θ ⋅ mg − L cos θ ⋅ f = I  θ 1

1

k

O1 1

− L sin θ 2 ⋅ mg + L cos θ 2 ⋅ f k = I O2  θ2 I= mL2 . Assuming small angles, we have cosθ ≈ 1 and where I= O1 O2

208

f k= kL ( sinθ1 − sin θ 2 ) ≈ kL ( θ1 − θ 2 )

Rearranging the equations results in

mL2 θ1 + kL2 (θ1 − θ 2 ) + mgLθ1 = 0 2 2 mL θ 2 − kL (θ1 − θ 2 ) + mgLθ 2 = 0 which can be written in the second-order matrix form  mL2 θ1   kL2 + mgL −kL2   θ1  0  0    +          = θ 2   −kL2 mL2   kL2 + mgL  θ 2  0   0

Figure PS5-3 No13 b.

The state and the output are  x1     x2  = x =   x3    x4   The state-space representation in matrix form is 0 0    x  1 0 0  x    2   −mg − kL k  =  x m  3   mL  x4   −mg − kL k  m mL

 θ1    θ 2  y    ,=  θ1    θ 2  

 θ 1       θ 2  

1 0 0 1   x1      x2  0 0   x  3   x  0 0  4  

 x1    0 0 1 0    x2  y=   0 0 0 1   x3    x4  

14. Repeat Problem 13 for the system shown in Figure 5.72, in which each pendulum is a uniform slender rod of mass m and length L. The inputs are the torques τ1 and τ2 applied to the pivots O1 and O2, respectively. When θ1 = 0, θ2 = 0, τ1 = 0, and τ2 = 0, the spring is at its free length.

Figure 5.72 Problem 14.

Solution a.

Assume that θ1 > θ2 > 0 . The free-body diagrams are shown in the figure below. Applying the moment equation about the fixed points O1 and O2, respectively, gives I Oα +↶: ∑ M O = τ + 1 L sin θ ⋅ mg − 2 L cos θ ⋅ f = I  θ 1

2

1

3

1

k

O1 1

τ 2 + 12 L sin θ 2 ⋅ mg + 32 L cos θ 2 ⋅ f k = I O2  θ2

209

I= where I= O1 O2

1 3

mL2 . Assuming small angles, we have cosθ ≈ 1 and = f k k ( 32 L ) ( sinθ1 − sin θ 2 ) ≈ k ( 32 L ) ( θ1 − θ 2 )

Rearranging the equations results in 1 3

mL2 θ1 + 94 kL2 (θ1 − θ 2 ) − 12 mgLθ1 = τ1

θ 2 − 94 kL2 (θ1 − θ 2 ) − 12 mgLθ 2 = τ2 mL2 which can be written in the second-order matrix form  13 mL2 − 94 kL2   θ1   τ1  0    θ1   94 kL2 − 12 mgL +          = 1 4 mL2   kL2 − 12 mgL  θ 2   τ 2  θ 2   − 94 kL2 3 9  0 1 3

Figure PS5-3 No14 b.

The state, the input, and the output are  x1  x   2 = x =   x3    x4  

 θ1  θ   θ  τ   2 u  1  ,= y  1    ,=  τ2  θ2   θ1    θ2   The state-space representation in matrix form is 0  0 0 1 0  0    x  x1      x1  0 0  0 0 0 1 1  x   x      2   9mg − 8kL 0 0 1 0  x2  0 0  u1    u1   2  3 4k y     0   0 0   +  2 =  +   x u 3m 0 0 0 1   x3  0 0 u2   x3   6mL  2   3   mL  x4    4k 9mg − 8kL 3   x    x4   0 0  4   0 2   3m 6mL mL   

15. Consider the system shown in Figure 5.73, in which a uniform sphere of mass m and radius r rolls along an inclined plane of 30°. A translational spring of stiffness k is attached to the sphere. Assuming that there is no slipping between the sphere and the surface, draw the necessary free-body diagram and derive the differential equation of motion.

Figure 5.73 Problem 15.

Solution The free-body diagram of the system is shown in the figure below, where the normal force N and the friction force f are reaction forces at the contact point. Assuming that the sphere rolls down the incline, the spring is in compression and fk is the spring force. When the sphere is at the static equilibrium position, we have fk = kδst, where δst is the static deformation of the spring. Then, + ↶:

∑M

IC

=0

210

r sin 30°·mg − r·k δst = 0 0.5mg = k δst We choose the static equilibrium position as the origin. Because of no slipping, the contact point is the IC. Applying the moment equation to the IC gives + ↶:

∑M = IC

I IC α

r sin 30°·mg − r·k ( x + δst= ) I IC θ

where I IC =I C + md 2 =25 mr 2 + mr 2 =75 mr 2 . Introducing the static equilibrium condition, 0.5 mg = kδst, and the assumption of no slipping, x = rθ, we obtain the differential equation of motion 7 mr 2 θ + kr 2 θ = 0 5

Figure PS5-3 No15 16. Consider the system shown in Figure 5.74. A uniform solid cylinder of mass m, radius R, and length L is fitted with a frictionless axle along the cylinder’s long axis. A spring of stiffness k is attached to a bracket connected to the axle. Assume that the cylinder rolls without slipping on a horizontal surface. Draw the necessary free-body diagram and derive the differential equation of motion.

Figure 5.74 Problem 16.

Solution The free-body diagram of the system is shown, where the normal force N and the friction force f are reaction forces at the contact point. When the cylinder is at the static equilibrium position, δst = 0, where δst is the static deformation of the spring. We choose the static equilibrium position as the origin. Because the cylinder rolls without slipping, the contact point is the IC. Applying the moment equation to the IC gives + ↷:

∑M = IC

I IC α

− R·kx =I IC θ

I C + md 2 =12 mR 2 + mR 2 =23 mR 2 . Introducing the assumption of no slipping, x = Rθ, we obtain the where I IC = differential equation of motion 3 2

mR 2 θ + kR 2 θ = 0

Figure PS5-3 No16

211

Problem Set 5.4 1.

For the pulley system in Example 5.14, draw the free-body diagram and kinematic diagram, and derive the equation of motion using the force/moment approach.

Figure 5.82 A pulley system.

Solution The free-body diagram and the kinematic diagram are shown below.

Figure PS5-4 No1 At static equilibrium, we have

∑

+ : MO = 0 r·mg − r·k δst = 0 or mg = k δst where δst is the static deformation of the spring. Assume that the block and the pulley are displaced in their positive directions. The spring force is fk = k(x + δst). For the block (translation only), applying the force equation in the x direction gives +↓ x:

∑F

+ :

∑M

=maCx mg − T = mx For the pulley (rotation only), applying the moment equation about the fixed point O gives x

= IOα r·T − r·k ( x + δst )= I O θ Combining the two equations and eliminating the unknown T results in r ( mg − mx) − rk ( x + δst )= I O θ Introducing the geometric constraint, x = rθ, and the static equilibrium condition, mg = kδst, the equation becomes ( I O + mr 2 ) θ + kr 2 θ = 0 2.

O

The double pulley system shown in Figure 5.89 has an inner radius of r1 and an outer radius of r2. The mass moment of inertia of the pulley about point O is IO. A translational spring of stiffness k and a block of mass m are suspended by cables wrapped around the pulley as shown. Draw the free-body diagram and kinematic diagram, and derive the equation of motion using the force/moment approach.

212

Figure 5.89 Problem 2.

Solution The free-body diagram and the kinematic diagram are shown below. At static equilibrium, we have f k = kδst and

0 +↶: ∑ M O =

r1 ⋅ mg − r2 ⋅ kδst = 0 where δst is the static deformation of the spring. Choose the equilibrium as the origin. Assume that the block and the double pulley are displaced in their positive directions. Note that x = r1θ and the spring force= f k k ( r2 θ + δst ) .

Figure PS5-4 No2 For the block (translation only), applying the force equation in the x-direction gives + ↓ x : ∑ Fx =maCx mg − T = mx = mr  θ 1

For the double pulley (rotation only), applying the moment equation about the fixed point O yields I Oα +↶: ∑ M O = r ⋅T − r ⋅ k (r θ + δ ) = I  θ 1

2

2

st

O

Combining the two equations and eliminating the unknown T results in r1 ( mg − mr1 θ ) − r2 k ( r2 θ + δst ) = I O θ Introducing the static equilibrium condition, r1mg = r2 kδst , the equation becomes 0 ( I + mr 2 ) θ + kr 2θ = O

3.

1

2

Consider the mechanical system shown in Figure 5.90. A disk-shaft system is connected to a block of mass m through a translational spring of stiffness k. The elasticity of the shaft and the fluid coupling are modeled as a torsional spring of stiffness K and a torsional viscous damper of damping coefficient B, respectively. The radius of the disk is r and its mass moment of inertia about point O is IO. Assume that the friction between the block and the horizontal surface cannot be ignored and is modeled as a translational viscous damper of damping coefficient b. The input to the system is the force f. Draw the necessary free-body diagram and the kinematic diagram, and derive the equations of motion.

213

Figure 5.90 Problem 3.

Solution This is a mixed system, where the mass block undergoes translational motion along x-direction and the motion of the disk is pure rotation about point O that is fixed. Choose the displacement of the block x and the angular displacement of the disk θ as the generalized coordinates. Assuming that the block and the disk are displaced in their positive directions and rθ > x > 0, the spring is in tension and the magnitude of the spring force is fk = k(rθ − x). The free-body diagram and the kinematic diagram of the system are shown in the figure blow, where W is the weight of the disk and N is the normal force supporting the block.

Figure PS5-4 No3 For the block (translation only), applying the force equation in the x direction gives

+← x:

∑F

=maCx k ( rθ − x ) − bx = mx For the disk (rotation only), applying the moment equation about the fixed point O gives x

∑

Rearranging the two equations gives

M O= I O α + ↷: − K θ − Bθ − r·k ( rθ − x ) + r· f= I O θ

mx + bx + kx − krθ =0

I O θ + Bθ + ( K + kr 2 )θ − krx = rf

4.

Consider the mechanical system shown in Figure 5.91, where the motion of the rod is small angular rotation. When θ = 0 and f = 0, the deformation of each spring is zero and the system is at static equilibrium. Assume that the friction between the block of mass m1 and the horizontal surface cannot be ignored and is modeled as a translational viscous damper of damping coefficient b. a. Assuming that a > c > 0, determine the mass moment of inertia of the rod about the pivot point O. b. Draw the necessary free-body diagram and the kinematic diagram, and derive the equations of motion for small angles.

214

Figure 5.91 Problem 4.

Solution a.

The mass moment of inertia of the rod about point O can be found by applying the parallel axis theorem

I O = I C + Md 2 = 121 M ( a + c ) 2 + M ( a − 12 ( a + c ) ) = 13 M ( a 2 + c 2 − ac ) Choose the displacements of the blocks, x1 and x2, and the angular displacement of the rod θ as the generalized coordinates. Assume that the blocks and the rod are displaced in their positive directions, x1 > aθ > 0, and cθ > x2 > 0. Both springs are in tension and the magnitudes are = f k 1 k1 ( x1 − aθ) 2

b.

f k 2 k 2 ( cθ − x2 ) = The free-body diagram and the kinematic diagram of the system are shown in the figure below.

Figure PS5-4 No4 For the blocks (translation only), applying the force equation in the x direction gives Mass 1: + → x :

∑

Mass 2: + ← x :

∑F

Fx =maCx f − k1 ( x1 − aθ) − bx1 =m1 x1

=maCx  k2 ( cθ − x2 ) = m2 x2 For the rod (rotation only), applying the moment equation about the fixed point O gives + ↷:

x

∑M = O

IOα

a cos θ⋅ k1 ( x1 − aθ) + ( a − ( a + c )) sin θ⋅ Mg − c cos θ⋅ k2 ( cθ − x2 )= I O θ 1 2

where sin θ ≈ θ and cos θ ≈ 1 for small angles. Substituting the mass moment of inertia IO and rearranging the three equations give

215

m1 x1 + bx1 + k1 x1 − k1aθ = f m2  x2 + k 2 x2 − k 2 cθ = 0 1 3

5.

M ( a 2 + c 2 − ac ) θ + ( k1a 2 + k2 c 2 − 12 ( a − c ) Mg )θ − k1ax1 − k2 cx2 = 0

Consider the mechanical system shown in Figure 5.92, in which a simple pendulum is pivoted on a cart of mass m and is constrained to rotate in a vertical plane. The pendulum consists of a point mass M concentrated at the tip of a massless rod of length L. Assume that the friction between the cart and the horizontal surface cannot be ignored. Denote the displacement of the cart as x and the angular displacement of the pendulum as θ. Draw the necessary free-body diagram and kinematic diagram, and derive the equations of motion for small angles.

Figure 5.92 Problem 5.

Solution The free-body diagram and the kinematic diagram are shown in the figure below.

Figure PS5-4 No5 Applying the force equation to the whole system along the x-direction gives 2

+ → x : ∑ Fx =∑ mi ( aCi ) x i =1

−bx − kx = mx + Mx + ML θ ⋅ cos θ − MLθ 2 ⋅ sin θ Applying the moment equation to the pendulum about the pivot point P results in +↷: ∑ M P = I C α + M eff _ ma L sin θ ⋅ Mg = L cos θ ⋅ Mx + L ⋅ MLθ C

Rearranging the two equations gives

(m + M )  x + ML θ cos θ − MLθ 2 sin θ + bx + kx = 0 2 MLx cos θ + ML θ − MgL sin θ = 0

For small angular motions, we have cos θ ≈ 1 , sin θ ≈ θ , θ 2 θ ≈ 0 . Thus, (m + M )  x + ML θ + bx + kx = 0 2 MLx + ML θ − MgLθ = 0 6.

Consider the mechanical system shown in Figure 5.93. The inputs are the force f1 applied to the cart and the force f2 applied at the tip of the rod. The outputs are the displacement x of the cart and the angular displacement θ of the pendulum.

216

a. b.

Draw the free-body diagram and kinematic diagram, and derive the equations of motion for small angles. Using the differential equation obtained in Part (a), determine the state-space representation.

Figure 5.93 Problem 6.

Solution a.

The free-body diagram and the kinematic diagram are shown in the figure below.

Figure PS5-4 No6 Applying the force equation to the whole system along the x-direction gives 2

+ → x : ∑ Fx =∑ mi ( aCi ) x i =1

f1 − kx + f 2 cos θ = mx + Mx + M L2 θ ⋅ cos θ − M L2 θ 2 ⋅ sin θ Applying the moment equation to the pendulum about the pivot point P results in +↷: ∑ M P = I C α + M eff _ ma C

L 2

where

sin θ ⋅ Mg + Lf 2= I Cθ + L2 cos θ ⋅ Mx + L2 ⋅ M L2 θ

I C = 121 ML2 for a uniform thin rod. Rearranging the two equations gives

( m + M )  x + 12 ML θ cos θ − 12 MLθ 2 sin θ + kx = f1 + f 2 cos θ 1 MLx cos θ + 1 ML2 θ − 1 MgL sin θ = Lf 2

3

2

For small angular motions, we have cos θ ≈ 1 , sin θ ≈ θ , θ 2 θ ≈ 0 . Thus, ( m + M )  x + 12 ML θ + kx = f1 + f 2 2 1 1 MLx + ML θ − 1 MgLθ = Lf 2

b.

3

The state, the input, and the output are specified as

2

2

2

217

 x1   x   x  θ  f  x  2   = x = u  1 = y     =  θ   f2   x3   x   x4   θ  We then take the time derivative of each state variable. For the first two, x1= x= x3 x = θ= x 2

4

To find x3 and x 4 , we rewrite the two linearized differential equations as

x   f1 + f 2 − kx   m + M 12 ML    =   1 ML 1 ML2   1 3  2   θ   Lf 2 + 2 MgLθ  θ can be solved using Cramer’s rule, from which x and  −4kx − 3Mgθ + 4 f1 − 2 f 2  x= M + 4m 6 Mkx + 6 ( M + m ) Mgθ − 6 Mf1 + ( 6 M + 12m )  θ= ML( M + 4m ) Thus, the state-space equation in matrix form is 0 0 1 0 0     x  x1     0 0 0 1 0  1   x    2    x2   4k 3Mg 4 0 0   +  −   − x M + 4m  x3   M + 4m   3   M + 4m  x 4     6k 6( M + m ) g 6 x 0 0  4   −  L( M + 4m ) L( M + 4m )   L( M + 4m )

f2

     f1  2 −   M + 4m   f 2  6 M + 12m  ML( M + 4m )  0 0

 x1    1 0 0 0    x 2   0 0   f1  y  =  +   0 1 0 0  x3  0 0  f 2   x4 

7.

For the mechanical system in Problem 2, use the energy method to derive the equation of motion.

Solution The static equilibrium position and the deformed positions are shown in the figure below. At equilibrium, we have r1mg = r2 kδst . Assuming that the block and the pulley move along the direction shown gives T =T +T = 1 mx 2 + 1 I θ 2 = 1 mr 2 θ 2 + 1 I θ 2 block

pulley

2

2

O

2

1

2

O

V= Vg + Ve = −mgr1θ + k ( r2 θ + δst ) where the static equilibrium position is chosen as the datum for the gravitational potential energy. Because of the static equilibrium condition, the total potential energy becomes = V 12 kr22 θ2 + 12 kδst2 The total energy of the system is T= + V 12 mr12 θ 2 + 12 I O θ 2 + 12 kr22 θ2 + 12 kδst2 for which the time derivative is d  + I θθ  + kr 2 θθ (T + V=) mr12θθ O 2 dt Applying the principle of conservation of energy gives 1 2

( mr

2 1

+ I O )  θ + kr22 θ = 0

2

218

Figure PS5-4 No7 8.

For the mechanical system in Problem 11 of Problem Set 5.3, use the energy method to derive the equation of motion.

Solution

Figure PS5-4 No8 The system is only subjected to the gravitational and spring forces, and it is a conservative system. When θ = 0, the springs are at their free length. Assuming that the rod rotates along the positive direction by a small angle θ gives T = 12 I O θ 2 V = Vg + Ve = Mg ⋅ 12 ( a − b) cos θ + 12 k1 ( aθ) 2 + 12 k2 (bθ) 2

The total energy of the system is T += V 12 I O θ 2 + 12 Mg ( a − b) cos θ + 12 k1a 2 θ2 + 12 k2b2 θ2 for which the time derivative is d  + 1 Mg ( a − b) sin θθ + k a 2 θθ + k b 2 θθ (T + V )= I Oθθ 1 2 2 dt Applying the principle of conservation of energy gives I Oθ + 12 Mg ( a − b) sin θ + k1a 2 θ + k2b2 θ=0 where the mass moment of inertia of the rod about pivot O is = I O 13 M ( a 2 + b2 − ab) 9.

Repeat Problem 13 of Problem Set 5.3 using Lagrange’s equations.

Solution The system is only subjected to the gravitational and spring forces, and it is a conservative system.

219

Figure PS5-4 No9 The kinetic energy of the system is

T = Tball1 + Tball2 = 12 mL2 θ 12 + 12 mL2 θ 22

Assume that θ1 > θ2 > 0 . Then, the spring is under elongation and the elastic potential energy is

= Ve 12 k ( L sin θ1 − L sin θ 2 ) 2 Using the datum defined in the figure, we can obtain the potential energy Vg = −mgh1 − mgh2 = −mgL cos θ1 − mgL cos θ 2 The total potential energy is V = Ve + Vg = 12 kL2 (sin θ1 − sin θ 2 ) 2 − mgL(cos θ1 + cos θ 2 ) Apply Lagrange's equations d  ∂T  ∂T ∂V + = 0 i = 1, 2  − dt  ∂qi  ∂qi ∂qi For i = 1 , q1 = θ1 , q1 = θ 1 ,

∂T = mL2 θ 1 ∂θ 1

d  ∂T  2 θ1   = mL  dt  ∂θ 1  ∂T =0 ∂θ1

∂V = kL2 (sin θ1 − sin θ 2 ) cos θ1 + mgL sin θ1 ∂θ1 Substituting into Lagrange's equation results in d  ∂T  ∂T ∂V + = mL2 θ1 + kL2 (sin θ1 − sin θ 2 ) cos θ1 + mgL sin θ1= 0  − dt  ∂θ 1  ∂θ1 ∂θ1 ∂T Similarly, for i = 2 , q2 = θ 2 , q2 = θ 2 , = mL2 θ 2 ∂θ 2

d  ∂T  dt  ∂θ 2

 2 θ2  = mL  

∂T =0 ∂θ 2

∂V = kL2 (sin θ1 − sin θ 2 )(− cos θ 2 ) + mgL sin θ 2 ∂θ 2 Substituting into Lagrange's equation gives d  ∂T  ∂T ∂V + = mL2 θ 2 − kL2 (sin θ1 − sin θ 2 ) cos θ 2 + mgL sin θ 2= 0  − dt  ∂θ 2  ∂θ 2 ∂θ 2 For small motions, the two differential equations of motion after linearization are

220

mL2θ1 + kL2 (θ1 − θ 2 ) + mgLθ1 = 0 2 2 mL θ 2 − kL (θ1 − θ 2 ) + mgLθ 2 = 0 10. Repeat Problem 14 of Problem Set 5.3 using Lagrange’s equations.

Solution

Note that the system is subjected to external torques, τ1 and τ2, which are nonconservative.

Figure PS5-4 No10 The kinetic energy of the system is

I= where I= O1 O2

1 3

T = Trod1 + Trod2 = 12 I O1 θ 12 + 12 I O2 θ 22

ML2 . Assume that θ1 > θ2 > 0 . Then, the spring is under elongation and the elastic potential

energy is

= Ve 12 k ( 32 L sin θ1 − 32 L sin θ 2 ) 2 Using the datum defined in the figure, we can obtain the potential energy Vg = mgh1 + mgh2 = mg ( 12 L cos θ1 ) + mg ( 12 L cos θ 2 ) The total potential energy is V = Ve + Vg = 92 kL2 (sin θ1 − sin θ 2 ) 2 + 12 mgL(cos θ1 + cos θ 2 ) Apply Lagrange's equations d  ∂T  ∂T ∂V + = Qi i= 1, 2  − dt  ∂qi  ∂qi ∂qi ∂T 1 2  = 3 mL θ1 ∂θ 1

For i = 1 , q1 = θ1 , q1 = θ 1 ,

d  ∂T  1 2   = mL θ1 dt  ∂θ 1  3 ∂T =0 ∂θ1 ∂V 4 2 = kL (sin θ1 − sin θ 2 ) cos θ1 − 12 mgL sin θ1 ∂θ1 9

∂r j

2

∑F

Q1 =

j =1

j

∂θ1

=( τ1k )

∂ ∂ ( θ1k ) + ( τ2k ) ( θ2k ) =τ1 ∂θ1 ∂θ1

Substituting into Lagrange's equation results in d  ∂T  ∂T ∂V 1 2 4 2 + = mL θ1 + 9 kL (sin θ1 − sin θ 2 ) cos θ1 − 12 mgL sin θ= τ1  − 1 dt  ∂θ 1  ∂θ1 ∂θ1 3 ∂T 1 2  Similarly, for i = 2 , q2 = θ 2 , q2 = θ 2 , = 3 mL θ 2 ∂θ 2

221

d  ∂T  dt  ∂θ 2

 1 2  = 3 mL θ 2  ∂T =0 ∂θ 2

∂V = ∂θ 2

4 9

∂r j

2

∑F

Q2 =

j =1

kL2 (sin θ1 − sin θ 2 )(− cos θ 2 ) − 12 mgL sin θ 2

j

∂θ2

=( τ1k )

∂ ∂ ( θ1k ) + ( τ2k ) ( θ2k ) =τ2 ∂θ2 ∂θ2

Substituting into Lagrange's equation gives d  ∂T  ∂T ∂V 1 2 4 2 + = mL θ 2 − 9 kL (sin θ1 − sin θ 2 ) cos θ 2 − 12 mgL sin θ= τ2  − 2 dt  ∂θ 2  ∂θ 2 ∂θ 2 3 For small motions, the two differential equations of motion after linearization are 1 mL2 θ1 + 94 kL2 (θ1 − θ 2 ) − 12 mgLθ1 = τ1 3 1 3

mL2 θ 2 − 94 kL2 (θ1 − θ 2 ) − 12 mgLθ 2 = τ2

11. Repeat Problem 5 using Lagrange’s equations.

Solution The system is subjected to a damping force, and it is a nonconservative system.

Figure PS5-4 No11 The kinetic energy of the system is

T = Tblock + Tball = where

1 2

mx 2 + 12 Mv 2

 2 + 2 x ( Lθ)  cos θ v2 = x 2 + ( Lθ)

Thus, we have

T =12 ( m + M ) x 2 + 12 ML2 θ 2 + MLx θ cos θ Using the datum defined in the figure, we can obtain the potential energy V= Ve + Vg=

1 2

kx 2 + Mgh=

1 2

kx 2 + MgL cos θ

We then apply Lagrange's equations

For i = 1 , q1 = x , q1 = x ,

d  ∂T  ∂T ∂V + = Qi i= 1, 2  − dt  ∂qi  ∂qi ∂qi ∂T = ( m + M ) x + MLθ cos θ ∂x d  ∂T   ( m + M )  x + ML θcosθ − MLθsinθ ⋅ θ  = dt  ∂x  ∂T =0 ∂x ∂V = kx ∂x

222

∂r ∂ F = −bx Q1 = ( −bxi ) ( xi ) = ∂x ∂x Substituting into Lagrange's equation results in d  ∂T  ∂T ∂V + = −bx (m + M )  x + ML θ cos θ − MLθ 2 sin θ + kx =  − dt  ∂x  ∂x ∂x ∂T Similarly, for i = 2 , q2 = θ , q2 = θ , = ML2 θ + MLx cos θ ∂θ d  ∂T  ML2θ + MLx cos θ − MLx sin θ ⋅ θ  = dt  ∂θ  ∂T  θ = MLx θsin ∂θ ∂V = − MgL sin θ ∂θ

∂r ∂ Q2 = F = 0 ( −bxi ) ( xi ) = ∂θ ∂θ

Substituting into Lagrange's equation gives d  ∂T  ∂T ∂V + = ML2 θ + MLx cos θ − MgL sin θ= 0  − dt  ∂θ  ∂θ ∂θ For small motions, the equations of motion after linearization are (m + M )  x + ML θ + bx + kx = 0 2 ML θ + MLx − MgLθ = 0 12. Repeat Problem 6 using Lagrange’s equations.

Solution The system is subjected to two external forces, f1 and f2, which are nonconservative.

Figure PS5-4 No12 The kinetic energy of the system is 1 = T T= mx 2 + 12 MvC2 + 12 I C θ 2 cart + Trod 2 where I C = 121 ML2 and 2   ( x + L2 θcosθ) vC2 = + L2 θsinθ

Thus, we have

T =12 (m + M ) x 2 + 16 ML2 θ 2 + 12 MLxθ cos θ Using the datum defined in the figure, we can obtain the potential energy V= Ve + Vg=

We then apply Lagrange's equations

1 2

kx 2 + Mgh=

1 2

kx 2 + 12 MgL cos θ

223

d  ∂T  dt  ∂qi

d  ∂T  dt  ∂x

∂r j

2

F ∑= ∂x j

j =1

=

i= 1, 2

∂T (m + M ) x + 12 MLθ cos θ = ∂x

For i = 1 , q1 = x , q1 = x ,

= Q1

 ∂T ∂V + = Qi −  ∂qi ∂qi

  (m + M )  x + 12 ML θcosθ − 12 MLθsinθ ⋅ θ =  ∂T =0 ∂x ∂V = kx ∂x

∂ ∂x

∂ ∂x

( f1i ) ( xi ) + ( f 2 cos θi + f 2 sin θj) ( xi + L sin θi − L cos θj)

( f1i ) i + ( f 2 cos θi + f 2 sin θj) i=

f1 + f 2 cos θ

Substituting into Lagrange's equation results in d  ∂T  ∂T ∂V + = (m + M )  x + 12 ML θ cos θ − 12 MLθ 2 sin θ + kx = f1 + f 2 cos θ  − dt  ∂x  ∂x ∂x ∂T Similarly, for i = 2 , q2 = θ , q2 = θ , = 13 ML2 θ + 12 MLx cos θ ∂θ d  ∂T  1 2 1 ML θ + 2 MLx cos θ − 12 MLx sin θ ⋅ θ  = dt  ∂θ  3 ∂T = − 12 MLxθ sin θ ∂θ ∂V = − 12 MgL sin θ ∂θ

∂r j

2

Q2 =

F ∑= ∂θ j

j =1

=

∂ ∂θ

∂ ∂θ

( f1i ) ( xi ) + ( f 2 cos θi + f 2 sin θj) ( xi + L sin θi − L cos θj)

θj) ( f1i ) 0 + ( f 2 cos θi + f 2 sin θj)( L cos θi + L sin=

(

)

2 f 2 L cos 2 θ + sin= f2 L θ

Substituting into Lagrange's equation gives d  ∂T  ∂T ∂V 1 2 1 + = ML θ + 2 MLx cos θ − 12 MgL sin= θ f2 L  − dt  ∂θ  ∂θ ∂θ 3 For small motions, the equations of motion after linearization are (m + M )  x + 12 ML θ + kx = f1 + f 2 1 3

ML2 θ + 12 MLx − 12 MgLθ = f2 L

13. A robot arm consists of rigid links connected by joints allowing the relative motion of neighboring links. The dynamic model for a robot arm can be derived using Lagrange’s equations d  ∂T  ∂T ∂V − + = t i , i =… 1, 2, , n, dt  ∂θ i  ∂θi ∂θi where θi is the angular displacement of the ith joint, τi is the torque applied to the ith joint, and n is the total number of joints. Consider a single-link planar robot arm as shown in Figure 5.94. Use Lagrange’s equations to derive the dynamic model of the robot arm. Assume that the motion of the robot arm is constrained in a vertical plane, and the joint angle varies between 0° and 360°.

224

Figure 5.94 Problem 13.

Solution The kinetic energy of the robot arm is

2 2 = T 12= I O θ 2 12 ( 13 mL = )θ 16 mL2 θ 2 Using the datum defined in the figure, we can obtain the potential energy V= V= mgh = mg ⋅ 12 L cos θ g

Figure PS5-4 No13 We then apply Lagrange's equations d  ∂T  ∂T ∂V − + = t dt  ∂θ  ∂θ ∂θ

where ∂T 1 2  = mL θ ∂θ 3

d  ∂T  1 2   = mL θ dt  ∂θ  3 ∂T =0 ∂θ ∂V = − 12 mgL sin θ ∂θ Substituting into Lagrange's equation results in d  ∂T  ∂T ∂V + =  − dt  ∂θ  ∂θ ∂θ

1 3

mL2θ − 12 mgL sin = θ τ

14. Repeat Problem 13 for a two-link planar robot arm as shown in Figure 5.95. Assume that the motion of the robot arm is constrained in a horizontal plane, and the joint angles vary between 0° and 360°.

Figure 5.95 Problem 14.

Solution The motion of the robot arm is assumed to be constrained in a horizontal plane. Thus, the gravitational potential energy 0 ) and the Lagrange’s equations can be rewritten as does not change ( ∂V ∂θi = d  ∂T  ∂T = ti , i =… 1, 2, , n,  − dt  ∂θ i  ∂θi

225

The kinematic diagram is shown below.

Figure PS5-4 No14 The kinetic energy of the robot arm is

= T Tlink1 + Tlink2 Note that link 1 rotates about a fixed axis through the point O, and 2 2 = Tlink1 12= I1O θ 12 12 ( 13 mL = )θ1 16 mL2 θ 12 The link 2 undergoes general plane motion, and Tlink2 =12 mv22C + 12 I 2C ω 22 =12 mv22C + 12 ( 121 mL2 )(θ 1 + θ 2 ) 2 where v2C is the velocity at the center of gravity of link 2. Denote the velocity at the end of link 1 as v1E and the relative velocity of the center of gravity of link 2 with respect to the end of link 1 as vr, we can obtain 2 2 v22C =vr2 + v12E − 2vr v1E cos ( π − θ2 ) = 12 L ( θ 1 + θ 2 ) + ( Lθ 1 ) + 2  12 L ( θ 1 + θ 2 )  ( Lθ 1 ) cos θ2 using the law of cosine. Thus, the total kinetic energy is 2 T= 16 mL2 θ 12 + 12 m  14 L2 ( θ 1 + θ 2 ) + L2 θ 12 + L2 θ 1 ( θ 1 + θ 2 ) cos θ2  + 241 mL2 (θ 1 + θ 2 ) 2   =

2 3

mL2 θ 12 + 16 mL2 ( θ 1 + θ 2 ) + 12 mL2 ( θ 12 + θ 1θ 2 ) cos θ2 2

We then apply Lagrange's equations. For i = 1 , ∂T 4 2  1 2   = 3 mL θ1 + 3 mL θ1 + θ2 + 12 mL2 2θ 1 + θ 2 cos θ2 ∂θ

(

)

(

)

1

d  ∂T   =  dt  ∂θ 1 

4 3

 +  mL2 θ1 + 13 mL2 ( θ1 +  θ2 ) + 12 mL2 ( 2θ θ2 ) cos θ2 − 12 mL2 ( 2θ 1 + θ 2 ) sin θ2 ⋅ θ 2 1

∂T =0 ∂θ1 Substituting into Lagrange's equation results in d  ∂T  ∂T  − dt  ∂θ 1  ∂θ1

=

4 3

 +  θ1 + 13 mL2 ( θ1 +  θ2 ) + 12 mL2 ( 2θ θ2 ) cos θ2 − 12 mL2 ( 2θ 1 + θ 2 ) sin θ2 ⋅ θ 2 mL2 1

θ1 + ( 13 + 12 cos θ2 )mL2 θ2 − 12 mL2 ( 2θ 1θ 2 + θ 22 ) sin θ2 = τ1 = ( 53 + cos θ2 )mL2 Similarly, for i = 2 ,

∂T = ∂θ 2 d  ∂T   =  dt  ∂θ 2 

1 3

1 3

mL2 ( θ 1 + θ 2 ) + 12 mL2 θ 1 cos θ2

mL2 ( θ1 +  θ2 ) + 12 mL2θ1 cos θ 2 − 12 mL2 θ 1 sin θ2 ⋅ θ 2

∂T = − 12 mL2 ( θ 12 + θ 1θ 2 ) sin θ2 ∂θ2 Substituting into Lagrange's equation gives

226

d  ∂T  ∂T  − dt  ∂θ 2  ∂θ2 =

1 3

mL2 ( θ1 +  θ2 ) + 12 mL2 θ1 cos θ2 − 12 mL2 θ 1 sin θ2 ⋅ θ 2 + 12 mL2 ( θ 12 + θ 1θ 2 ) sin θ2

= ( 13 + 12 cos θ2 )mL2 θ1 + 13 mL2 θ2 + 12 mL2 θ 12 sin θ2 = τ2 Thus, the two differential equations of motion are ( 53 + cos θ2 )mL2θ1 + ( 13 + 12 cos θ2 )mL2θ2 − 12 mL2 2θ 1θ 2 + θ 22 sin θ2 = τ1

(

)

( + cos θ2 )mL  θ1 + mL  θ2 + mL θ sin θ2 = τ2 1 3

1 2

2

1 3

2

1 2

2 2 1

Problem Set 5.5 1.

Repeat Example 5.18, and determine a mathematical model for the simple one-degree-of-freedom system shown in Figure 5.96a in the form of a differential equation of motion in θ2.

Solution As shown in Example 5.18, the differential equation of motion in θ1 is

  r12 θ1 =τ1  I C1 + 2 I C2   r2   Introducing the geometric constraint θ2 θ1 =r1 r2 , the above differential equation can be rewritten as    r2  r12 θ2  = τ1  I C1 + 2 I C2    r2    r1  which can be simplified as

 r22  r θ2 = 2 τ1  2 I C1 + I C2   r1  r1  If the gears have negligible inertia or zero angular acceleration, and if the energy loss due to the friction between the gear teeth can be neglected, we have τ1 θ2 r1 = = τ 2 θ1 r2 Thus, the dynamics seen from the output side is  r22  θ2 =τ2  2 I C1 + I C2    r1  2.

Repeat Example 5.19, and determine a mathematical model for the single-link robot arm shown in Figure 5.97 in the form of a differential equation of motion in the load variable θ.

Solution As shown in Example 5.19, the differential equation of motion in terms of the angular displacement of the motor is θm + ( Bm + N 2 B )θ m =τ m ( I m + N 2 I ) By the geometry of the gears, θ r1 = = N θ m r2 the above differential equation can be rewritten in the load variable θ,  θ θ τm ( I m + N 2 I ) + ( Bm + N 2 B ) = N N which can be simplified as  1    1  1  2 I m + I  θ +  2 Bm + B  θ= N τ m N  N  If the gears have negligible inertia or zero angular acceleration, and if the energy loss due to the friction between the

227

gear teeth can be neglected, we have

τm r1 θ = = = N τ θ m r2

Thus, the dynamics seen from the output side is  1    1  1  2 I m + I  θ +  2 Bm + B  θ= N τ N  N  3.

Consider the one-degree-of-freedom system shown in Figure 5.99. The system consists of two gears of mass moments of inertia I1 and I2 and radii r1 and r2, respectively. The applied torque on gear 1 is τ1. Assume that the gears are connected with flexible shafts, which can be approximated as two torsional springs of stiffnesses K1 and K2, respectively. a. Draw the necessary free-body diagrams, and derive the differential equation of motion in θ1. b. Using the differential equation obtained in Part (a), determine the transfer function Θ2(s)/T1(s). c. Using the differential equation obtained in Part (a), determine the state-space representation with θ2 as the output.

Figure 5.99 Problem 3.

Solution a.

The free-body diagram is shown in the figure below.

Figure PS5-5 No3 Applying the moment equation to each gear gives I Oα +↶: ∑ M O =

I1 θ1 Gear 1: τ1 − K1θ1 − r1 F =  −I θ Gear 2: K θ − r F = 2 2

2

2 2

Combining the two equations and eliminating F yields r τ 1 − K1θ1 − 1 ( I 2θ2 + K 2 θ2 ) = I1 θ1 r2 By the geometry of the gears, θ2 = ( r1 r2 )θ1 . Substituting it into the previous differential equation gives

  r12    r12 I + I θ + K + τ1  1 2 2  1  1 2 K 2  θ1 = r2  r2   

228

b.

c.

Assuming zero initial conditions and taking Laplace transform of the above differential equation, we have Θ1 ( s ) 1 = Τ1 ( s )   r12  2  r12  I1 + 2 I 2  s +  K1 + 2 K 2  r2  r2    Applying the geometric constraint gives r1 Θ2 ( s ) r1 Θ1 ( s ) r2 = = 2 Τ1 ( s ) r2 Τ1 ( s )   r1  2  r12  I1 + 2 I 2  s +  K1 + 2 K 2  r r 2 2     The state, the input, and the output are  x1  θ1  = x = y θ2 , u τ1 ,=   =  x2  θ1  The state equation and the output equation in matrix form are 0 1 0     x1   2  x1    u 2 2 = r    r2 K1 + r1 K 2  + 0   x2   2 2 2   x 2   − 2 2  r2 I1 + r1 I 2   r2 I1 + r1 I 2 

r y 1 =  r2 4.

x  0  1  + 0 ⋅ u   x2 

Consider the gear–train system shown in Figure 5.100. The system consists of a rotational cylinder and a pair of gears. The gear ratio is N = r1/r2. The applied torque on the cylinder is τa. Assume that the gears are connected with flexible shafts, which can be approximated as two torsional springs of stiffnesses K1 and K2, respectively. a. Draw the necessary free-body diagrams, and derive the differential equations of motion. b. Using the differential equation obtained in Part (a), determine the state-space representation. Use θa, θ1, ωa, and ω1 as the state variables, and use θ2 and ω2 as the output variables.

Figure 5.100 Problem 4.

Solution a.

Assume that θa > θ1 > 0 . The free-body diagram is shown in the figure below.

229

Figure PS5-5 No4 Applying the moment equation to each gear gives IOα +↶: ∑ M O =

I θa Gear 1: τ a − K1 ( θ a − θ1 ) =

I1 θ1 Gear 2: K1 ( θ a − θ1 ) − r1 F = Gear 3: −r2 F + K 2 θ 2 = − I 2 θ2 Combining the last two equations and eliminating F yields r K1 ( θ a − θ1 ) − 1 ( I 2 I1 θ2 + K 2θ2 ) = θ1 r2

= θ2 By the geometry of the gears, we have

(I

1

r1 r2 ) θ1 (=

Nθ1 . Substituting it into the previous equation gives

+ N I 2 )  θ1 − K1θ a + ( K1 + N 2 K 2 ) θ1 = 0 2

The differential equations of motion for the system are θ a + K1θ a − K1θ1 = τa I 2  0 ( I1 + N I 2 ) θ1 − K1θa + ( K1 + N 2 K 2 ) θ1 = b.

The state, the input, and the output are

 x1  x   2 = x =   x3    x4   The state equation in matrix form is 0    x  1 0  x    K1  2 =    −  x I  3   x4   K1  2  I1 + N I 2

 θa  θ  θ   1 u τ,= y  2   ,= ω ω 2   a   ω1  

1 0 0 0 0 1   x1      0   x2    K1 0 0   +  1  u I   x3   I  2  x    K + N K2 0 0   4   0  − 1 2 I1 + N I 2  By the geometry of the gears, we have θ 2 = Nθ1 and ω2 = Nω1 . Thus, the output equation in matrix form is 0 N = y  0 0

5.

0

 x1  x  0 0 0   2  +  u  0 N   x3  0   x4   

A three-degree-of-freedom gear–train system is shown in Figure 5.101, which consists of four gears of moments of inertia I1, I2, I3, and I4. Gears 2 and 3 are meshed and their radii are r2 and r3, respectively. Gears 1 and 2 are connected by a relatively long shaft, and gears 3 and 4 are connected in the same way. The shafts are assumed to be flexible, and can be approximated by torsional springs. The applied torque and load torque are τa and τl on gear 1 and gear 4, respectively. The gears are assumed to be rigid and have no backlash. Derive the differential equations of motion.

230

Figure 5.101 Problem 5.

Solution Assume that θ1 > θ 2 > 0 and θ3 > θ 4 > 0 . The free-body diagram is shown in the figure below.

Figure PS5-5 No5 Applying the moment equation to each gear gives IOα +↶: ∑ M O =

I1 θ1 Gear 1: τ a − K1 ( θ1 − θ 2 ) =

I 2 θ2 Gear 2: K1 ( θ1 − θ 2 ) − r2 F = − I 3 θ3 Gear 3: −r3 F + K 2 ( θ3 − θ 4 ) = θ4 Gear 4: − τl − K 2 ( θ3 − θ 4 ) =− I 4 Combining the equations of gears 2 and 3 and eliminating F yields r r K1 ( θ1 − θ 2 ) − 2 I 3 θ3 − 2 K 2 ( θ3 − θ 4 ) = I 2 θ2 r3 r3

By the geometry of the gears, we have θ3 = ( r2 r3 ) θ2 . Substituting it into the previous differential equation gives

   r22  r2 r θ 2 − K1θ1 +  K1 + 22 K 2  θ 2 − 2 K 2 θ 4 = 0  I 2 + 2 I 3   r3 r3  r3    Similarly, the equation of gear 4 becomes r I 4 θ4 − 2 K 2θ2 + K 2θ4 = τl r3 Combining with the equation of gear 1 gives the differential equations of motion for the system I1 θ1 + K1θ1 − K1θ 2 = τa    r22  r2 r θ 2 − K1θ1 +  K1 + 22 K 2  θ 2 − 2 K 2 θ 4 = 0  I 2 + 2 I 3   r3 r3  r3    r θ4 − 2 K 2θ2 + K 2θ4 = τl I 4 r3

6.

Repeat Problem 5. Assume that the shaft connecting gears 1 and 2 is relatively short and rigid.

Solution

Assuming that the shaft connecting gears 1 and 2 is relatively short and rigid, we have θ1 = θ 2 . Assume that

θ3 > θ 4 > 0 . The free-body diagram is shown in the figure below. Applying the moment equation to each gear gives

IOα +↶: ∑ M O =  Gears 1 and 2: τ a − r2 F = ( I1 +I 2 )θ 2

231

I 2 θ2 Gear 2: K1 ( θ1 − θ 2 ) − r2 F =

− I 3 θ3 Gear 3: −r3 F + K 2 ( θ3 − θ 4 ) = θ4 Gear 4: − τl − K 2 ( θ3 − θ 4 ) =− I 4 Combining the first two equations and eliminating F yields r r τ a − 2 I 3 θ3 − 2 K 2 ( θ3 − θ 4 ) = ( I1 + I 2 )  θ2 r3 r3 By the geometry of the gears, θ3 = ( r2 r3 )θ2 . Substituting it into the above equation and the equation of gear 4 gives

 r22  r2 r θ 2 + 22 K 2 θ 2 − 2 K 2 θ 4 = τa  I1 + I 2 + 2 I 3   r3 r r 3 3   r I 4 θ4 − 2 K 2θ2 + K 2θ4 = τl r3

Figure PS5-5 No6

Problem Set 5.6 (Note: All MATLAB figures are given at the end of Problem Set 5.6.) 1.

Consider the mass–spring–damper system in Problem 6 of Problem Set 5.2. Assume that the force acting on the mass block is a unit-impulse function with a magnitude of 100 N and a duration of 0.1 sec. The parameter values are m = 500 kg, b = 250 N·s/m, and k = 200,000 N/m. a. Build a Simulink model based on the differential equation of motion of the system and find the displacement output x(t). b. Build a Simscape model of the physical system and find the displacement output x(t).

Solution The differential equation of the system is x= mx + 4bx + 4kx = f or  2.

1 m

( f − 4bx − 4kx)

Repeat Problem 1 for the mass–spring–damper system shown in Figure 5.117, in which the origin of the coordinate x is set at equilibrium. Assume x(0) = 0.1 m and x (0) = 0 m/s. The parameter values are m = 20 kg, b = 125 N⋅s/m, and k = 400 N/m.

Figure 5.117 Problem 2.

Solution The differential equation of the system is

232

x= mx + bx + 2kx = 0 or  3.

1 m

( −bx − 2kx )

Consider the two-degree-of-freedom mass–spring system shown in Figure 5.118. The parameter values are m1 = m2 = 5 kg, k1 = 2000 N/m, and k2 = 4000 N/m. Assume that initially x(0) = [0 0]T and x (0) = [1 0] . T

a. b.

Figure 5.118 Problem 3. Build a Simulink model based on the differential equations of motion of the system and find the displacement outputs x1(t) and x2(t). Build a Simscape model of the physical system and find the displacement outputs x1(t) and x2(t).

Solution The differential equations of motion of the system is

m1 x1 + (k1 + k2 ) x1 − k2 x2 = 0 m2 x2 − k2 x1 + k2 x2 = 0 or

 x1 =  = x2 4.

[ −(k1 + k2 ) x1 + k2 x2 ] 1 k x − k2 x2 ] m [ 2 1

1 m1

2

Repeat Problem 3 for the two-degree-of-freedom quarter-car model in Example 5.5. Assume that the surface of the road can be approximated as a sine wave z = Z0sin(2πvt/L), where Z0 = 0.01 m, L = 10 m, and the speed v = 20 km/h. If the car moves at a speed of 100 km/h, rerun the simulations and compare the results with those obtained in the case of 20 km/h. Ignore the control force f for both cases.

Solution The differential equations of motion of the system is m1 x1 + b1 x1 − b1 x2 + k1 x1 − k1 x2 = 0

m2  x2 − b1 x1 + b1 x2 − k1 x1 + ( k1 + k2 ) x2 = k2 z or

 x1 = m1 [ −b1 x1 + b1 x2 − k1 x1 + k1 x2 ] 1

 = x2 5.

1 m2

[b1 x1 − b1 x2 + k1 x1 − (k1 + k2 ) x2 + k2 z ]

Consider the disk–shaft system in Problem 2 of Problem Set 5.3. The system is approximated as a single-degree-of-freedom rotational mass–spring system, where m = 10 kg, r = 0.05 m, B = 1 N⋅m⋅s/rad, and K = 1000 N⋅m/rad. a. Assume that a torque τ = 50u(t) N⋅m is acting on the disk, which is initially at rest. Build a Simscape model of the physical system and find the angular displacement output θ(t). b. Assuming that the external torque is τ = 0 and the initial angular displacement is θ(0) = 0.1 rad, find the angular displacement output θ(t).

Solution The differential equations of motion of the system is 1 mr 2 θ + 2 Bθ + 2 Kθ = τ 2 or

 θ=

2 mr 2

( τ − 2Bθ − 2Kθ )=

(

80 τ − 2θ − 2000θ

)

233

6.

Consider the pendulum-bob system in Problem 5 of Problem Set 5.3. The parameter values are m = 0.1 kg, M = 1.2 kg, L = 0.6 m, and B = 0.25 N⋅s/m. The initial angular displacement is θ(0) = 0.1 rad and the initial angular velocity is θ = 0.1 rad/s . a. Build a Simulink block diagram based on the nonlinear mathematical model of the system and find the angular displacement output θ(t). b. Build a Simscape model of the nonlinear physical system and find the angular displacement output θ(t).

Solution The nonlinear mathematical model of the system is m + 13 M L2 θ + Bθ + m + 12 M gL sin θ = 0

(

)

(

)

θ + 0.25θ + 4.12sin θ = 0 0.18 or

 = θ

1 0.18

( −0.25θ − 4.12sin θ)

234

MATLAB Figures Problem 1

Figure PS5-6 No1a Simulink block diagram. Mechanical Translational Reference2

f(x) = 0 Solver Configuration

Mechanical Translational Reference1

Mass

C V PS S P Ideal Translational PS-Simulink Motion Sensor Converter

C

Group 1

R

R

R

Translational Spring

C

C

C

C

r1

C

C

R

R

R

R

R

r3

C

Translational Damper

R

Simulink-PS Converter1

S Ideal Force Source

C

Signal Builder1

S PS

Mechanical Translational Reference

Figure PS5-6 No1b Simscape block diagram. 10

2.5

-4

2

1.5

1

Displacement x (m)

Signal 1

R

0.5

0

-0.5 -1

-1.5 -2

-2.5 0

0.5

1

1.5

2

2.5

3

3.5

Time (s)

Figure PS5-6 No1c

4

4.5

5

Scope

235

Problem 2

Figure PS5-6 No2a Simulink block diagram.

Translational Mechanical Translational Reference

Spring

Mechanical Translational Reference1

R

C

R

Translational Damper

R

C

C

Translational Solver Configuration

C V

R

Spring1

P Ideal Translational Motion Sensor

Mass

f(x) = 0

Figure PS5-6 No2b Simscape block diagram. 0.1

0.08

Displacement x (m)

0.06

0.04

0.02

0

-0.02 0

0.2

0.4

0.6

0.8

1

1.2

1.4

Time (s)

Figure PS5-6 No2c

1.6

1.8

2

PS S PS-Simulink Converter

Scope

236

Problem 3

Figure PS5-6 No3a Simulink block diagram. Solver Configuration

Translational

Spring k1

Spring k2

R

Translational

PS-Simulink Converter

Ideal Translational Motion Sensor

Ideal Translational Motion Sensor1

Mechanical Translational Reference2

S PS

C V P Mechanical Translational Reference1

R

PS-Simulink Converter1 x2

x1

Scope

Figure PS5-6 No3b Simscape block diagram.

S PS

C

R

R

C V P

C Mechanical Translational Reference

Mass m2

Mass m1

f(x) = 0

Scope1

237

0.05

0.04

0.04

0.03

0.03

0.02

0.02

0.01

2

1

(m)

(m)

0.05

0.01

0

Displacement x

Displacement x

0

-0.01

-0.02

-0.01

-0.02

-0.03

-0.03

-0.04

-0.04

-0.05

-0.05 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

0.1

0.2

0.3

Time (s)

0.4

0.5

Time (s)

Figure PS5-6 No3c Problem 4

Figure PS5-6 No4a Simulink block diagram.

0.6

0.7

0.8

0.9

1

238

Mechanical Translational Reference1

C V P Ideal Translational Motion Sensor

Mass 1

x1

R

C

Solver Configuration

Scope

Translational Spring 1 Mechanical Translational Reference2

R

Mass 2

C

R

Translational Damper

Translational Spring 2

R

R

S

C

R

Fcn velocity z_dot (Z0 = 0.01 m L = 10 m v = 20 or 100 km/h)

PS-Simulink Converter

C

f(x) = 0

PS S

C V

x2

P Ideal Translational Motion Sensor1

Ideal Translational Velocity Source

PS S PS-Simulink Converter1

Scope1

S PS

Z0*2*pi*v/L*cos(2*pi*v/L*u) Clock

Mechanical Translational Reference

Simulink-PS Converter1

Figure PS5-6 No4b Simscape block diagram. 0.015

0.02

0.015 0.01 0.01 0.005

1

2

(m)

(m)

0.005

0

Displacement x

-0.005

-0.01

-0.005

-0.01 -0.015 -0.015 -0.02

-0.025

-0.02 1

0

2

3

4

5

6

7

8

9

10

1

0

2

3

4

5

6

7

8

9

10

6

7

8

9

10

Time (s)

Time (s)

Figure PS5-6 No4c: v = 20 km/h 10

-3

10

6

-3

8

6 4

4

(m)

2 2

1

(m)

2

0

Displacement x

0

Displacement x

Displacement x

0

-2

-2

-4

-4 -6

-6

-8 0

1

2

3

4

5

6

7

8

9

10

0

1

2

Time (s)

3

4

5

Time (s)

Figure PS5-6 No4d: v = 100 km/h

239

Problem 5 S PS

A

S R

Step1

C

Simulink-PS Converter

Mechanical Rotational Reference

R

R

Rotational

PS-Simulink Converter

C

Rotational

Spring

Spring1

Inertia

R

C

PS S

W

C Ideal Rotational Motion Sensor

Ideal Torque Source C

f(x) = 0

R

R

Rotational Damper

C

Rotational Damper1

Solver Configuration

Figure PS5-6 No5a Simscape block diagram. Set the external torque τ as 50 Nm and the initial angular displacement θ(0) as 0 rad. 0

(rad)

-0.02

Angular Displacement

-0.04

-0.06

-0.08

-0.1

-0.12 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

Time (s)

Figure PS5-6 No5b Angular displacement output. Set the external torque τ as 0 and the initial angular displacement θ(0) as 0.1 rad. 0.1

(rad)

0.08

0.06

Angular Displacement

0.04

0.02

0

-0.02

-0.04

-0.06 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

Time (s)

Figure PS5-6 No5c Angular displacement output.

Mechanical Rotational Reference1

Scope1

240

Problem 6

Figure PS5-6 No6a Simulink block diagram.

θ + 0.25θ = −4.12sin θ , where −4.12sin θ is considered as a torque applied to the Rewrite the equation as 0.18 system. Mechanical Rotational Reference

f(x) = 0

C

Mechanical Rotational Reference2

Rotational Damper

PS S

R

Solver Configuration

R S PS

R

S Ideal Torque Source

Simulink-PS Converter

Inertia

PS-Simulink Converter

A W

C

C Ideal Rotational Motion Sensor

-4.12*sin(u) Fcn

Figure PS5-6 No6b Simscape block diagram. Run either of the simulations returns the curve shown below. 0.12

0.1

(rad)

0.08

0.06

Angular Displacement

0.04

0.02

0

-0.02

-0.04

-0.06

-0.08 0

1

2

3

4

5

6

7

Time (s)

Figure PS5-6 No6c

8

9

10

Mechanical Rotational Reference1

Scope1

241

Review Problems 1.

Determine the equivalent spring constant for the system shown in Figure 5.119.

Figure 5.119 Problem 1.

Solution The equivalent spring constant for the middle part is 1 1 1 kk = + or keq1 = 2 3 keq1 k2 k3 k 2 + k3 The equivalent spring constant for the system is

keq = k1 + keq1 + k4 = 2.

k1k2 + k1k3 + k2 k3 + k2 k4 + k3 k4 k 2 + k3

Determine the equivalent spring constant for the system shown in Figure 5.120.

Figure 5.120 Problem 2.

Solution The equivalent spring constant for the bottom part is k k +k k 1 1 1 or keq1 = 2 3 2 4 = + k 2 + k3 + k 4 keq1 k2 k3 + k4 The equivalent spring constant for the system is

k k +k k +k k +k k +k k keq =k1 + keq1 = 1 2 1 3 1 4 2 3 2 4 k 2 + k3 + k 4 3.

Consider the system shown in Figure 5.121, in which a mass–spring system is hung from the middle of a massless beam. Assume that the beam can be modeled as a spring and the equivalent stiffness at the midspan is 192EIA/L3, where E is the modulus of elasticity of beam material and IA is the area moment of inertia about the beam’s longitudinal axis.

a. b.

Figure 5.121 Problem 3. Derive the differential equation of motion for the system. Using the differential equation obtained in Part (a), determine the transfer function X(s)/F(s). Assume that the initial conditions are x(0) = 0 and x (0) = 0 .

242

c.

Using the differential equation obtained in Part (a), determine the state-space representation. Assume that the output is the displacement x of the mass.

Solution a.

The system is equivalent to the mass-spring system shown in the figure below, where the spring kb represents the flexibility of the beam and kb = 192EIA/L3. The equivalent spring stiffness of the system is 1 1 1 L3 1 = + = + keq k b k 192 EI A k or

192 EI A k L3k + 192 EI A Choose the static equilibrium as the origin of the coordinate x. The equation of motion for the system is mx + keq x = f (t ) keq =

b.

c.

Figure Review5 No3 Taking the Laplace transform of both sides of the preceding equation with zero initial conditions results in X ( s) 1 = 2 F ( s ) ms + keq The state, the input, and the output are  x1   x  , u f ,= y x = x =   =  x2   x  The state equation and the output equation in matrix form are 1  0 0 x   x1    x1     = y [1 0]  1  + 0 ⋅ u =    keq  + 1 u     0  x2   x2   x2   −  m  m

4.

An accelerometer attached to an object can be modeled as a mass–damper–spring system, as shown in Figure 5.122. Denote the displacement of the mass relative to the object, the absolute displacement of the mass, and the absolute displacement of the object as x(t), y(t), and z(t), respectively, where x(t) = y(t) – z(t) and x(t) is measured electronically. a. Draw the necessary free-body diagram and derive the differential equation in terms of x(t). b. Using the differential equation obtained in Part (a), determine the transfer function X(s)/Z(s). Assume that the initial conditions are x(0) = 0 and x (0) = 0 . c. Using the differential equation obtained in Part (a), determine the state-space representation. The input is the absolute displacement of the object z(t) and the output is the displacement of the mass relative to the object x(t).

243

Figure 5.122 Problem 4.

Solution a.

b.

c.

Choose the static equilibrium as the origin of the coordinate y and assume that y > z > 0 . The free-body diagram of the mass is shown below.

Figure Review5 No4 Applying Newton’s second law in the y-direction gives  + ↑ y : − k ( y − z ) − b ( y − z ) =my Note that x(t) = y(t) – z(t). Thus, the equation can be rewritten as −mz − kx − bx= m(  x +  z ) or mx + bx + kx = Assuming zero initial conditions and taking Laplace transform of the above differential equation gives X ( s) −ms 2 = 2 Z ( s ) ms + bs + k Rewrite the transfer function X ( s ) Z ( s ) as 1 X ( s) X ( s) W ( s) = = −s2 b k Z ( s) W ( s) Z ( s) 2 s + s+ m m where X ( s) = −s2 W ( s) W ( s) 1 = Z ( s) s2 + b s + k m m Thus in the time-domain, we have  x = −w b k  = w − w − w + z m m Select the state and the input as  x1   w  = x = , u z   =  x2   w  The state equation is 1   0  x1    x1   0 = k b  +  u   −   x2   1   x 2   − m  m and the output equation is  k b   x1  = y    − 1⋅ u  m m   x2 

244

5.

Consider a quarter-car model shown in Figure 5.123, where m1 is the mass of the seats including passengers, m2 is the mass of one-fourth of the car body, and m3 is the mass of the wheel–tire–axle assembly. The spring k1 represents the elasticity of the seat supports, k2 represents the elasticity of the suspension, and k3 represents the elasticity of the tire. z(t) is the displacement input due to the surface of the road. Draw the necessary free-body diagrams and derive the differential equations of motion. Write the differential equations of motion in the second-order matrix form.

Solution Choose the displacements of the three masses x1 , x2 , and x3 as the generalized coordinates. The static equilibrium positions of m1 , m2 , and m3 are set as the coordinate origins. Assume that x1 > x2 > x3 > z > 0 . The free-body diagram is shown below.

Figure 5.123 Problem 5.

Figure Review5 No5

Applying Newton’s second law in the x-direction gives + ↑ x : ∑ Fx =ma x

Mass 1: −k1 ( x1 − x2 ) − b1 ( x1 − x 2 ) = m1  x1

Mass 2: k1 ( x1 − x2 ) + b1 ( x1 − x 2 ) − k2 ( x2 − x3 ) − b2 ( x 2 − x3 ) = m2  x2 Mass 3: k2 ( x2 − x3 ) + b2 ( x 2 − x3 ) − k3 ( x3 − z ) = m3  x3 Rearranging the equations into

m1  x1 + b1 x1 − b1 x 2 + k1 x1 − k1 x2 = 0

m2  x2 − b1 x1 + ( b1 + b2 ) x 2 − b2 x3 − k1 x1 + ( k1 + k2 ) x2 − k2 x3 = 0 m3  x3 − b2 x 2 + b2 x3 − k2 x2 + ( k2 + k3 ) x3 = k3 z

The differential equations can be expressed in second-order matrix form as 0    x1   b1 −b1 0   x1   k1 −k1  m1 0       0 m   0   x + −b b + b −b2  x 2  + −k1 k1 + k2 2   2  1 1 2     0 m3    x3  b2   −b2 −k2   0  x3   0  0 6.

0   x1   0      −k2   x2  = 0      k2 + k3    x3   k3 z 

The system shown in Figure 5.124 consists of a uniform rod of mass m and length L and a translational spring of stiffness k at the rod’s tip. The friction at the joint O is modeled as a damper with coefficient of torsional viscous damping B. The input is the force f and the output is the angle θ. The position θ = 0 corresponds to the static equilibrium position when f = 0.

245

a. b.

Draw the necessary free-body diagram and derive the differential equation of motion for small angles θ. Using the linearized differential equation obtained in Part (b), determine the transfer function Θ(s)/F(s).  = 0. Assume that the initial conditions are θ(0) = 0 and θ(0)

c.

Using the differential equation obtained in Part (b), determine the state-space representation.

Figure 5.124 Problem 6.

Solution a.

At equilibrium, we have

0 +↶: ∑ M O =

kδ st = 0 δ st = 0 Applying the moment equation to the fixed point O gives I Oα +↶: ∑ M O =

− L cos θ ⋅ f k − 12 L sin θ ⋅ mg − Bθ + 12 L cos θ ⋅ f = I O θ where, for small angular motions, the spring force can be approximated as

f k = kLθ Note that the mass moment of inertia about point O is 1 mL2 + m( 1 L ) 2 = 1 mL2 IO = I C + md 2 = 12 2 3 Thus, the differential equation of motion for small angles θ is 1 3

b.

c.

1 mL2 θ + Bθ + kL2 θ + 12 mgLθ = 2 fL

Figure Review5 No6 Taking Laplace transform of the above linearized differential equation gives Θ( s ) 3L = F ( s ) 2mL2 s 2 + 6 Bs + 6kL2 + 3mgL The state, the input, and the output are  x1  θ  = x = , u f ,= y θ   =  x2   θ  The state equation and the output equation in matrix form are

246

 x1  =    x2 

7.

0   −6kL − 3mg  2mL 

  0   x1    y + 3B   3  u = − 2   x2   mL   2mL  1

[1

x  0]  1  + 0 ⋅ u  x2 

Consider the system shown in Figure 5.125. Assume that a cylinder of mass m rolls without slipping. Draw the necessary free-body diagram and derive the differential equation of motion for small angles θ.

Figure 5.125 Problem 7.

Solution The free-body diagram is shown in the figure below. Because the cylinder rolls without slipping, the contact point is the IC.

Figure Review5 No7 Choose the static equilibrium position as the origin. Applying the moment equation to the IC gives I ICα +↷: ∑ M IC = − (a + r ) ⋅ f − (a + r ) ⋅ f = I θ k1

k2

IC

where for a cylinder, I IC =I C + md 2 =12 mr 2 + mr 2 =23 mr 2

For small angles θ , the spring forces are f k 1 = k1aθ and f k 2 = k2 aθ . Thus, the differential equation of motion is 3 mr 2 θ + a ( a + r )( k + k ) θ = 0 2

8.

1

2

The pulley of mass M shown in Figure 5.126 has a radius of r. The mass moment of inertia of the pulley about the point O is IO. A translational spring of stiffness k and a block of mass m are connected to the pulley as shown. Assume that the pulley rolls without slipping. Derive the equation of motion using (a) the force/moment approach, and (b) the energy approach.

Figure 5.126 Problem 8.

Solution The free-body diagram and the kinematic diagram are shown in the figure below. Because the pulley rolls without f k k= δst mg , where δst is the static deformation slipping, the contact point is the IC. At static equilibrium, we have = of the spring. We choose the equilibrium as the origin. Assume that the block and the pulley are displaced in their f k k ( rθ + δst ) . positive directions. Note that x = rθ and the spring force=

247

Figure Review5 No8 For the block (translation only), applying the force equation in the x-direction gives + ↓ x : ∑ Fx =maCx mg − T = mx = mr θ For the pulley, applying the moment equation about the IC yields I ICα +↷: ∑ M IC =

r ⋅ T − r ⋅ k ( rθ + δst ) =I ICθ

where I IC = I O + Md = I O + Mr . Combining the two equations and eliminating the unknown T results in r ( mg − mr θ ) − rk ( rθ + δ ) = ( I + Mr 2 )  θ 2

2

st

O

Introducing the static equilibrium condition, mg = kδst , the equation becomes 0 ( I + Mr 2 + mr 2 ) θ + kr 2θ = O

9.

Consider a half-car model shown in Figure 5.127, in which IC is the mass moment of inertia of the car body about the pitch axis, mb is the mass of the car body, mf is the mass the front wheel−tire−axle assembly, and mr is the mass of the rear wheel−tire−axle assembly. Each of the front and rear wheel−tire−axle assemblies is represented by a mass–spring–damper system. The input is the force f, and the car undergoes vertical and pitch motion. Derive the equations of motion using the force/moment approach.

Figure 5.127 Problem 9.

Solution Choose xf , xr , xb , and θ as the generalized coordinates. The static equilibrium positions of mf , mr , and mb are set as the coordinate origins. Assume that xb − aθ > xf > 0 , xb + bθ > xr > 0 , and θ > 0 . The free-body diagram and the kinematic diagram are shown in the figure below.

248

Figure Review5 No9 Applying the force equations to mf , mr , and the rod mb gives

+ ↑ x : ∑ Fx =maCx ksf ( xb − aθ − xf ) + bsf ( x b − aθ − xf ) − kf xf = mf  xf  k ( x + bθ − x ) + b ( x + bθ − x ) − k x = m  x sr

b

r

sr

b

r

r r

r r

f − ksf ( xb − aθ − xf ) − bsf ( x b − aθ − xf ) − ksr ( xb + bθ − xr ) − bsr ( x b + bθ − x r ) = mb  xb Applying the moment equation to the rod gives IC α +↶: ∑ M C =  ksf ( xb − aθ − xf )a cos θ + bsf ( x b − aθ − xf )a cos θ − ksr ( xb + bθ − xr )b cos θ − bsr ( x b + bθ − x r )b cos θ + fc cos θ = I C θ For small angular motions, we have cos θ ≈ 1 . The equations can be written in second-order matrix form 0 bsf a −bsf xf   bsf  mf 0 0 0      xf  0 m      0 0 xr 0 bsr −bsr −bsr b   x r  r   +    0 0 mb 0    bsf + bsr −bsf a + bsr b   x b  xb   −bsf −bsr       2 2    0 0 0 I C   θ   bsf a −bsr b −bsf a + bsr b bsf a + bsr b   θ   ksf + kf  0 +  −ksf   ksf a

0 ksr + k r −ksr −ksr b

−ksf −ksr

  xf   0  −ksr b   xr   0    =   ksf + ksr −ksf a + ksr b   xb   f      −ksf a + ksr b ksf a 2 + ksr b2   θ   fc  ksf a

10. Consider the mechanical system shown in Figure 5.128, where a uniform rod of mass m and length L is attached to a massless rigid link of equal length. Assume that the system is constrained to move in a vertical plane. Denote the angular displacement of the link as θ1, and the angular displacement of the rod as θ2. Derive the equations of motion for small angles using (a) the force/moment approach, and (b) the energy approach.

Figure 5.128 Problem 10.

Solution (a) The free-body diagram and the kinematic diagram are shown in the figure below.

249

Figure Review5 No10a Applying the force equations to the rod gives + → x : ∑ Fx =maCx  cos θ − mLθ 2 sin θ − m( 1 L)θ 2 sin θ −T sin= θ1 mL θ1 cos θ1 + m( 12 L)θ 2 2 1 1 2 2 2

+ ↑ y : ∑ Fy =maCy  sin θ + mLθ 2 cos θ + m( 1 L)θ 2 cos θ T cos θ1= − mg mL θ1 sin θ1 + m( 12 L)θ 2 2 1 1 2 2 2 Eliminating the unknown force T yields

mL θ1 + 12 mL θ2 cos ( θ2 − θ1 ) − 12 mLθ 22 sin ( θ2 − θ1 ) + mg sin θ1 = 0 Applying the moment equation to the rod about the O1 results in +↶: ∑ M O1 =I C α + M eff _ maC

 − 12 L sin θ2 ⋅ mg = I C θ2 + 12 L cos ( θ2 − θ1 ) ⋅ mL θ1 + 12 L sin ( θ2 − θ1 ) ⋅ mLθ 12 + 12 L ⋅ m( 12 L)θ 2 where I C = 121 mL2 . Rearranging the equation gives 1  1  1 2 1 0 3 mLθ 2 + 2 mL cos ( θ 2 − θ1 ) θ1 + 2 mL sin ( θ 2 − θ1 ) θ1 + 2 mg sin θ 2 = For small angular motions, we have

mL θ1 + 12 mL θ2 + mgθ1 = 0 1 θ + 1 mL θ + 1 mgθ = 0 mL 1

2

3

2

2

2

(b) The system is only subjected to the gravitational forces, and it is a conservative system. The rod undergoes general plane motion, and T =12 mvC2 + 12 I Cω 2 =12 mv 2 + 12 121 mL2 θ 22

(

)

where v is the velocity at the center of gravity of the rod, and 2 2 v 2 = 12 Lθ 2 + Lθ 1 + 2 12 Lθ 2 Lθ 1 cos ( θ2 − θ1 )

(

) ( )

(

)( )

using the law of cosine. Thus, the kinetic energy is 2 2 = T 12 m  12 Lθ 2 + Lθ 1 + 2 12 Lθ 2 Lθ 1 cos ( θ2 − θ1 )  + 12 ( 121 mL2 ) θ 22   22 22 2  1 1 1 = mL θ + mL θ + mL θ θ cos ( θ − θ )

(

2

) ( )

1

6

2

2

(

1 2

)( ) 2

1

Using the datum defined in the figure, we can obtain the potential energy V= Vg = −mgh = −mg ( L cos θ1 + 12 L cos θ2 )

Figure Review5 No10b

250

We then apply Lagrange's equations. For i = 1 , ∂T = mL2 θ 1 + 12 mL2 θ 2 cos ( θ2 − θ1 )  ∂θ1

d  ∂T  2 2 2 1 1   =   mL θ1 + 2 mL θ2 cos ( θ2 − θ1 ) − 2 mL θ2 sin ( θ2 − θ1 ) ⋅ ( θ2 − θ1 ) dt  ∂θ 1 

∂T =− 12 mL2 θ 1θ 2 sin ( θ2 − θ1 ) ⋅ ( −1) ∂θ1 ∂V = mgL sin θ1 ∂θ1 Substituting into Lagrange's equation results in d  ∂T  ∂T ∂V 0 + = mL2θ1 + 12 mL2θ2 cos ( θ2 − θ1 ) − 12 mL2 θ 22 sin ( θ2 − θ1 ) + mgL sin θ=  − 1 dt  ∂θ 1  ∂θ1 ∂θ1 Similarly, for i = 2 ,

∂T 1 2  1 2  = mL θ2 + 2 mL θ1 cos ( θ2 − θ1 ) ∂θ 2 3 d  ∂T   =  dt  ∂θ 2 

1 3

mL2θ2 + 12 mL2θ1 cos ( θ2 − θ1 ) − 12 mL2 θ 1 sin ( θ2 − θ1 ) ⋅ ( θ 2 − θ 1 )

∂T = − 12 mL2 θ 1θ 2 sin ( θ2 − θ1 ) ∂θ2 ∂V 1 = mgL sin θ2 ∂θ2 2 Substituting into Lagrange's equation gives d  ∂T  ∂T ∂V 1 2 1 2 + = 3 mL θ2 + 2 mL θ1 cos ( θ2 − θ1 ) + 12 mL2 θ 12 sin ( θ2 − θ1 ) + 12 mgL sin= θ2 0   − dt  ∂θ2  ∂θ2 ∂θ2 Thus, the two differential equations of motion are mL θ1 + 12 mL θ2 cos ( θ2 − θ1 ) − 12 mLθ 22 sin ( θ2 − θ1 ) + mg sin θ1 = 0 2 1 1 1 1    mLθ + mLθ cos ( θ − θ ) + mLθ sin ( θ − θ ) + mg sin θ = 0 3

2

2

For small angular motions, we have

1

2

1

1

2

2

1

2

2

mL θ1 + 12 mL θ2 + mgθ1 = 0 1 1 1   mLθ + mLθ + mgθ = 0 2

1

3

2

2

2

11. Consider the mechanical system shown in Figure 5.129, in which a block of mass m is connected to a spring of stiffness k and it can slide without friction inside a vertical tube. Assume that the tube pivots at the joint O, and it can be approximated as a uniform rod of mass M and length L. The unstretched length of the spring is L/2. Denote the angular displacement of the tube as θ, and the relative displacement of the block with respect to the tube as x. Derive the nonlinear equations of motion using (a) the energy approach, and (b) the force/moment approach.

Figure 5.129 Problem 11.

251

Solution (a) The system is a conservative system. The tube undergoes general plane motion, and Ttube 12= I Oω 2 12 13 ML2 θ 2 =

(

)

The block rotates together with the tube and at the same time slides in the tube, and the kinetic energy is = Tblock 12= mvC2 12 m( x 2 θ 2 + x 2 ) The total kinetic energy is

T = 16 ML2 θ 2 + 12 mx 2 θ 2 + 12 mx 2 Using the datum defined in the figure, we can obtain the potential energy V= Vg + Ve = − Mg L2 cos θ − mgx cos θ + 12 k ( x − L2 ) 2

Figure Review5 No11a We then apply Lagrange's equations. For i = 1 , ∂T = 13 ML2 θ + mx 2 θ ∂θ d  ∂T  1 2 2   = ML θ + mx θ + 2mxx dt  ∂θ  3

∂T =0 ∂θ ∂V = Mg L2 sinθ + mgx sin θ ∂θ Substituting into Lagrange's equation results in d  ∂T  ∂T ∂V 1 2 + = ML θ + mx 2 θ + 2mxx + Mg L2 sinθ + mgx sin= θ 0  − dt  ∂θ  ∂θ ∂θ 3 Similarly, for i = 2 , ∂T = mx ∂x d  ∂T    = mx dt  ∂x  ∂T = mxθ 2 ∂x ∂V = −mg cos θ + k ( x − L2 ) ∂x Substituting into Lagrange's equation gives d  ∂T  ∂T ∂V + = mx − mxθ 2 − mg cos θ + k ( x − L2 ) = 0  − dt  ∂x  ∂x ∂x Thus, the two differential equations of motion are 1 ML2 θ + mx 2 θ + 2mxx + Mg L2 sinθ + mgx sin θ = 0 3 mx − mxθ 2 − mg cos θ + k ( x − L2 ) = 0

252

(b) The free-body diagram and the kinematic diagram are shown in the figure below. Note that the tube and the block rotate together about pivot O, and at the same time the block moves relatively inside the tube.

Figure Review5 No11b Applying the force equation to the block gives + → x : ∑ Fx =maCx − f k + mg cos θ = max

f k k ( x − L2 ) . To find the acceleration ax, it is more convenient to use the polar coordinate where the spring force =      x . Taking system. The velocity components of the block in the polar coordinate system are vθ = xθ uθ and vx = xu time derivatives gives     vθ = xθ uθ + x θuθ + xθ uθ     x = vx  xux + xu     Note that uθ = −θ ux and ux = θ uθ . Thus,      u vθ = xθ uθ + x θuθ − xθθ x     = vx  xux + xθuθ and the acceleration component along the x direction is  a =  x − xθθ x

Substituting it into the force equation gives −k ( x − L2 ) + mg cos θ = mx − mxθ 2 Applying the moment equation to the tube and the block about the O results in +↶: ∑ M O=

∑I

C

α + ∑ M eff _ maC

− L2 sin θ ⋅ Mg − x sin θ ⋅ mg= I C  θ + L2 ⋅ M L2  θ + x ⋅ maθ where I C = 121 mL2 and the acceleration component of the block along the θ direction is a = xθ + x θ + xθ θ

Substituting them into the moment equation gives  θ + 14 ML2 θ + xm( xθ + x θ + xθ) − L2 sin θ ⋅ Mg − x sin θ ⋅ mg = 121 ML2 Rearranging the above two equations gives mx − mxθ 2 + k ( x − L2 ) − mg cos θ = 0 1 3

ML2 θ + 2mxxθ + mx 2 θ + 12 MgL sin θ + mgx sin θ = 0

12. A rack and pinion is a pair of gears that convert rotational motion into translation. As shown in Figure 5.130, a torque τ is applied to the shaft. The pinion rotates and causes the rack to translate. The mass moment of inertia of the pinion is I and the mass of the rack is m. Draw the free-body diagram and derive the differential equation of motion.

253

Figure 5.130 Problem 12.

Solution The free-body diagram is shown in the figure below.

Figure Review5 No12 Applying the moment equation to the pinion gives I Oα +↷: ∑ M O = τ − rF = I θ Applying the force equation to the track gives + ← x : ∑ Fx =ma x F = mx

Combining the two equations and eliminating F yields I θ + mrx = τ Introducing the geometric constrain x = rθ gives the differential equation of motion τ ( I + mr 2 ) θ = 13.

Consider the mass–spring–damper system shown in Problem 9 of Problem Set 5.2, in which the cam and follower impart a displacement z(t) in the form of a periodic sawtooth function (see Figure 5.131) to the lower end of the system. The values of the system parameters are m = 12 kg, b = 200 N⋅s/m, k1 = 4000 N/m, and k2 = 2000 N/m. a. Build a Simulink model based on the differential equation of motion of the system and find the displacement output x(t). b. Build a Simscape model of the physical system and find the displacement output x(t).

Figure 5.131 Problem 13.

Solution The differential equation of motion of the system is mx + bx + ( k1 + k2 ) x = k2 z or

 = x

1 m

 k2 z − bx − ( k1 + k2 ) x 

254

Figure Review5 No13a Simulink block diagram.

C R

Spring

C

Translational Solver Configuration

R

Mechanical Translational Reference Mechanical Translational Reference1

Translational Damper

f(x) = 0 R

Signal Builder velocity

PS-Simulink Converter

x Scope

C

S

Mechanical Translational Reference2

Group 1 Signal 1

PS S

R

C

Spring1

Ideal Translational Velocity Source

P Ideal Translational Motion Sensor Mass

R

Translational

C V

S PS Simulink-PS Converter

Figure Review5 No13b Simscape block diagram. 10

10

-3

8

Displacement x (m)

6

4

2

0

-2 0

1

2

3

4

5

6

Time (s)

Figure Review5 No13c 14. Case Study Consider the ball and beam system [?] shown in Figure 5.132, in which I and Ib are the mass moments of inertia of the beam and the ball, respectively, m is the mass of the ball, and r is the radius of the ball. The beam is made to

255

rotate in a vertical plane by applying a torque τ at the center of rotation and the ball is free to roll (with one degree-of-freedom) along the beam. It is assumed that the ball remains in contact with the beam and that the rolling occurs without slipping. The system parameters are I = 0.02 kg·m2, Ib = 2×10−6 kg·m2, m = 0.05 kg, and r = 0.01 m. a. Choosing the beam angle θ and the ball position x as generalized coordinates, derive the nonlinear equations of motion of the system using the Lagrange’s equations. b. Choosing θ, θ , x, and x as state variables and τ as the input, find the state variable equations of the nonlinear system. c. Assuming small angular motion, find the state-space form for the linearized system (with θ and x as outputs). d.

Build two Simulink block diagrams, one using the differential equations obtained in Part (a) and the other using the state-space form found in Part (c), compare the beam angle θ(t) and the ball displacement x(t) when the torque applied to the system is a pulse function, τ(t) = 1 Nm for 1 ≤ t ≤ 2 s.

Figure 5.132 Problem 14.

Solution a.

The kinetic energy of the ball and beam system is = T Tbeam + Tball Note that the beam rotates about a fixed axis through the center, and Tbeam = 12 Iθ 2 The motion of the ball is more complicated than that of the beam. As shown in the figure below, the ball rotates together with the beam about the pivot, and it rolls without slipping along the beam. The kinetic energy of the ball consists of three parts 1 Tball = ( I b + mx 2 )θ 2 + 12 mx 2 + 12 I b α 2 2 Note that αr = x due to the geometric constraint.

Figure Review5 No14a Thus, the total kinetic energy is T=

1 2

I ( I + I b + mx 2 )θ 2 + 12 (m + b2 ) x 2 r

The total potential energy is

= V V= mgx sin θ ball We then apply Lagrange's equations. For i = 1 , ∂T = ( I + I b + mx 2 )θ  ∂θ d  ∂T  2     = ( I + I b + mx )θ + 2mxxθ dt  ∂θ  ∂T =0 ∂θ ∂V = mgx cos θ ∂θ

256

F Q1 =

∂r ∂ = τ ( τk ) ( θk ) = ∂θ ∂θ

Substituting into Lagrange's equation results in

( I + I b + mx 2 )θ + 2mxxθ + mgx cos θ = τ

Similarly, for i = 2 ,

I ∂T = (m + b2 ) x ∂x r Ib d  ∂T  x  =  (m + 2 )  dt  ∂x  r ∂T = mxθ 2 ∂x ∂V = mg sin θ ∂θ ∂r = Q2 F= 0 ∂x Substituting into Lagrange's equation gives Ib )  0 x − mxθ 2 + mg sin θ = r2 Thus, the two differential equations of motion are ( I + I b + mx 2 )θ + 2mxxθ + mgx cos θ = τ (m +

Ib )  0 x − mxθ 2 + mg sin θ = 2 r Writing the above nonlinear equations in the matrix form.  I + I b + mx 2 0      θ  =  τ − 2mxxθ − mgx cos θ      I 2 b  x   mxθ − mg sin θ  0 m + 2    r   Applying the Cramer’s rule gives I  θ= (m + b2 )(τ − 2mxxθ − mgx cos θ) r  x = ( I + I b + mx 2 )(mxθ 2 − mg sin θ) Choose θ, θ , x, and x as state variables and τ as the input. The state variable equations are (m +

b.

x1 = x2 x2 = (m +

Ib )(u − 2mx2 x3 x4 − mgx3 cos x1 ) r2

x3 = x4 x4 = ( I + I b + mx32 )(mx22 x3 − mg sin x1 ) c.

Linearize the nonlinear equations about the equilibrium, θ ≈ 0 and x ≈ 0. Note that the nonlinear terms xxθ ≈ 0 , x 2 ≈ 0 , xθ 2 ≈ 0 , cosθ ≈ 1, and sinθ ≈ θ. The linearized model of the system is

I  θ= (m + b2 )(τ − mgx) r  x =( I + I b )(−mgθ) and the state-space form is

257

 x1   x   2 =    x3   x4 

0    0   0   − mg ( I + I b )

0  0   x1    I I    0 −mg (m + b2 ) 0   x2   m + b2   +  r r u x3     0 0 1 0   x    0 0 0   4   0 

1

0

 x1     y1  1 0 0 0   x2  0  =     +  u  y2  0 0 1 0   x3  0   x4 

d.

The Simulink block diagram for the nonlinear system is shown below, in which two Fcn blocks are used to θ and x . According to the order of the input signals, the expressions in the two Fcn blocks are construct 

θ : (m+Ib/r^2)*(u(5)-m*g*u(3)*cos(u(1))-2*m*u(3)*u(4)*u(2)) for  for x : (I+Ib+m*u(3)*u(3))*(-m*g*sin(u(1))+m*u(3)*u(2)*u(2))

Figure Review5 No14b The Simulink block diagram for the linearized system can be built in the same way, only with different expressions defined in the two Fcn blocks: θ : (m+Ib/r^2)*(u(5)-m*g*u(3)) for  for x : (I+Ib)*(-m*g*u(1)) The nonlinear and linear responses of the beam angle and the ball displacement are shown in the figure below. For a small torque input, the nonlinear and linearized models give almost the same results. 0

0.7

Nonliear

Nonliear Linear

0.6

Linear

-0.01

-0.02

ball displacement x (m)

(rad)

0.5

Beam angle

0.4

0.3

0.2

0.1

-0.03

-0.04

-0.05

-0.06

-0.07

0

-0.08 0

2

4

6

8

10

0

Time (s)

2

4

6

Time (s)

Figure Review5 No14c

8

10

258

Problem Set 6.1 1.

Determine the equivalent resistance Req for the circuit shown in Figure 6.8.

Figure 6.8 Problem 1.

Solution The equivalent resistance for the parallel-connected resistors R1 and R3 is RR 1 1 1 or Req1 = 1 3 = + R1 + R3 Req1 R1 R3 They are connected with the resistor R2 in series. The equivalent resistance for the circuit is RR R R + R2 R3 + R1 R3 Req =R2 + Req1 =R2 + 1 3 = 1 2 R1 + R3 R1 + R3 2.

Determine the equivalent resistance Req for the circuit shown in Figure 6.9.

Figure 6.9 Problem 2.

Solution The equivalent resistance for the series-connected resistors R2 and R3 is Req1 = R2 + R3 They are connected with the resistor R1 in parallel. The equivalent resistance for the circuit is R R + R1 R3 1 1 1 1 1 or Req = 1 2 = + = + R1 + R2 + R3 Req Req1 R1 R2 + R3 R1 3.

Determine the equivalent resistance Req for the circuit shown in Figure 6.10. Assume that all resistors have the same resistance of R.

Figure 6.10 Problem 3.

259

Solution The equivalent resistance for the parallel-connected resistors is R 1 1 1 1 1 1 1 3 or Req1 = = + + = + + = 3 Req1 R2 R3 R4 R R R R They are connected with the resistor R1 and R5 in series. The equivalent resistance for the circuit is Req = R1 + Req1 + R5 = R + 13 R + R = 73 R 4.

Determine the equivalent resistance Req for the circuit shown in Figure 6.11. Assume that all resistors have the same resistance of R.

Figure 6.11 Problem 4.

Solution The resistors R3, R4, and R5 are connected in series and the equivalent resistance is Req1 = R3 + R4 + R5 = R + R + R = 3R They are connected with R1 and R2 in parallel. The equivalent resistance for the circuit is 3 1 1 1 1 1 1 1 7 or Req2 = R = + + = + + = 7 Req R1 R2 Req1 R R 3R 3R 5.

A potentiometer is a variable resistor with three terminals. Figure 6.12a shows a potentiometer connected to a voltage source. The two end terminals are labeled as 1 and 2, and the adjustable terminal is labeled as 3. The potentiometer acts as a voltage divider, and the total resistance is separated into two parts as shown in Figure 6.12b, where R13 is the resistance between terminal 1 and terminal 3, and R32 is the resistance between terminal 3 and terminal 2. Determine the relationship between the input voltage vi and the output voltage vo.

Figure 6.12 Problem 5. (a) Potentiometer and (b) voltage divider.

Solution Denoting the current in the circuit to be i, we obtain the input voltage = vi i ( R13 + R32 ) and the output voltage

vo = iR32 Thus the relationship between the input voltage vi and the output voltage vo is vo R32 = vi R13 + R32

260

6.

The output voltage of a voltage divider is not fixed but varies according to the load. a. Find the output voltage vo in Figure 6.13 for two different values of load resistance: (1) RL = 10 kΩ and (2) RL = 100 kΩ. b. Is it possible to get an output voltage vo of greater than 9 V? If yes, determine the minimum value of the load resistance.

Figure 6.13 Problem 6.

Solution a.

If = RL 10 kΩ , then the equivalent resistance for the parallel-connected resistors is

1 1 1 1 1 19 90 = + = + = or = Req kΩ Req R2 RL 9 10 90 19 Refer to the result in Problem 5. The output voltage is Req 90 19 = vo = vi = (10 ) 8.26 V R1 + Req 1 + 90 19 If= RL 100 kΩ , following the same procedure gives

1 1 1 1 1 109 900 or= Req kΩ = + =+ = 109 Req R2 RL 9 100 900 and

vo = b.

Req 900 109 vi = = (10 ) 8.92 V R1 + Req 1 + 900 109

If the output voltage vo > 9 V , then we have

Req Req = vi (10) > 9 V R1 + Req 1 + Req Solving for Req gives Req > 9 kΩ . However,

1 1 1 1 1 1 = + =+ > Req R2 RL 9 RL 9 which implies that Req < 9 kΩ no matter how big the load resistance is. Therefore, for the given voltage divider, it is impossible to get an output voltage vo of greater than 9 V. 7.

Consider an inductive divider shown in Figure 6.14. For an alternating current (AC) input vi, prove that the output L2 voltage of the inductive voltage divider is vo = vi . L1 + L2

Solution Denoting the current in the circuit to be i, we obtain the input voltage di di di vi = L1 + L2 = ( L1 + L2 ) dt dt dt The output voltage is vo = L2 di dt . Thus the relationship between the input voltage vi and the output voltage vo is

vo L2 L2 vi or vo = = L1 + L2 vi L1 + L2

261

Figure 6.14 Problem 7. 8.

Consider a capacitive divider shown in Figure 6.15. For an AC input vi, prove that the output voltage of the C1 capacitive voltage divider is vo = vi . C1 + C2

Figure 6.15 Problem 8.

Solution Denoting the current in the circuit to be i, we obtain the input voltage  1 1 1 1  vi = ∫ idt + idt =  +  ∫ idt ∫ C1 C2  C1 C2  and the output voltage

vo =

1 idt C2 ∫

Thus the relationship between the input voltage vi and the output voltage vo is

vo 1 C2 C1 C1 vi or vo = = = C1 + C2 vi 1 C1 + 1 C2 C2 + C1 9.

Consider a circuit of two inductors, L1 and L2, in series. Prove that the equivalent inductance of the circuit is Leq = L1 + L2.

Solution As shown in the figure below, for two inductors in series, we have di di di v = v1 + v2 = L1 + L2 = ( L1 + L2 ) dt dt dt di For the equivalent circuit, we have v = Leq . Thus, dt Leq= L1 + L2

Figure PS6-1 No9

262

10. Consider a circuit of two inductors, L1 and L2, in parallel. Prove that the equivalent inductance of the circuit is 1 1 1 . = + Leq L1 L2

Solution As shown in the figure below, for two inductors in parallel, we have 1 1  1 1 i = i1 + i2 = ∫ vdt + ∫ vdt =  +  ∫ vdt L1 L2  L1 L2  For the equivalent circuit, we have i =

1 vdt . Therefore, Leq ∫ 1 1 1 = + Leq L1 L2

Figure PS6-1 No10 11. Consider a circuit of two capacitors, C1 and C2, in series. Prove that the equivalent capacitance of the circuit is 1 1 1 . = + Ceq C1 C2

Solution As shown in the figure below, for two capacitors in series, we have  1 1 1 1  v = v1 + v2 = ∫ idt + idt =  +  ∫ idt ∫ C1 C2  C1 C2  For the equivalent circuit, we have v =

1 idt . Therefore, Ceq ∫ 1 1 1 = + Ceq C1 C2

Figure PS6-1 No11 12. Consider a circuit of two capacitors, C1 and C2, in parallel. Prove that the equivalent capacitance of the circuit is Ceq = C1 + C2.

Solution As shown in the figure below, for two capacitors in parallel, we have dv dv dv i = i1 + i2 = C1 + C2 = ( C1 + C2 ) dt dt dt dv For the equivalent circuit, we have i = Ceq . Therefore, dt Ceq= C1 + C2

263

Figure PS6-1 No12 13. The current through an inductor of 5 mH is shown in Figure 6.16. Find the voltage across the inductor. What is the energy stored in the inductor when (1) t = 4 s and (2) t = 6 s?

Figure 6.16 Problem 13.

Solution The voltage across the inductor is

=  2 L 10 mV 0 ≤ t ≤ 4 di = v L=  dt −4 L =−20 mV 4 < t ≤ 6 The energy stored in the inductor is

= E (t ) which gives

E (t = 4) =

1 2

E (t = 6) =

1 2

Li 2 (t ) − 12 Li 2 (0)

( 5 ×10 ) (8) − 0 = 0.16 J ( 5 ×10 ) ( 0 ) − 0 = 0 J −3

2

−3

1 2

2

14. The voltage across a capacitor of 500 μF is shown in Figure 6.17. Find the current through the capacitor. What is the energy stored in the capacitor when (1) t = 1 s and (2) t = 4 s?

Solution The current through the capacitor is

3C 1.5 mA 0 ≤ t ≤ 1 dv= = i C=  0 3 θL and applying the moment equation to the rotor and the load gives K t ia − K m (θ m − θ L ) − Bm (θ m − θ L ) = I m θm

(

)

− τ L + K m ( θm − θ L ) + Bm θ m − θ L − BL θ L − K L θ L = I L θL where K t ia = τ m . Assuming all the initial conditions to be zero and taking the Laplace transform, we have

( La s + Ra ) I a ( s ) + K e sΘm ( s ) = Va ( s )

( I m s 2 + Bm s + K m )Θm ( s ) − ( Bm s + K m )ΘL ( s ) − K t I a ( s ) = 0 ( I L s 2 +Bm s + BL s + K m + K L )ΘL ( s ) − ( Bm s + K m )Θm ( s ) = −Τ L ( s ) Applying Cramer’s rule gives ( L s + Ra )( I m s 2 + Bm s + K m ) + K t K e s K ( B s + Km ) ΘL ( s ) = − a TL ( s ) + t m Va ( s ) ∆ (s) ∆ (s) where

2 ∆ ( s ) = ( La s + Ra ) ( I m s 2 + Bm s + K m )( I L s 2 + Bm s + BL s + K m + K L ) − ( Bm s + K m )    + K t K e s( I L s 2 + Bm s + BL s + K m + K L )

284

Thus, the transfer functions Ω L ( s ) TL ( s ) and Ω L ( s ) Va ( s ) are

Ω L ( s ) sΘ L ( s ) s( La s + Ra )( I m s 2 + Bm s + K m ) + K t K e s 2 = = − TL ( s ) TL ( s ) ∆ (s)

Ω L ( s ) sΘL ( s ) sK t ( Bm s + K m ) = = Va ( s ) Va ( s ) ∆ (s) b.

c.

A block diagram representing the dynamics of the armature-controlled motor is shown below.

Figure PS6-4 No5 Refer to the differential equations in Part (a). Let x1 ia= x2 θ m= x3 θ L= x4 θ m= x5 θ L = and = u1 v= u2 τ L a The state variable equations are di 1 1 x1 = a = va − Ra ia − K e θ m = ( − Ra x1 − K e x4 + u1 ) dt La La x= θ= x

(

)

2

m

4

x= θ= x5 3 L

4 θ= x= m

1  K t ia − K m (θ m − θ L ) − Bm (θ m − θ L )  Im 

1 ( K t x1 − K m x2 + K m x3 − Bm x4 + Bm x5 ) Im

=

(

)

1  x5 =  θL = − τ L + K m ( θ m − θ L ) + Bm θ m − θ L − BL θ L − K L θ L  I  L

1  −u2 + K m x2 − ( K m + K L ) x3 + Bm x4 − ( Bm + BL ) x5  IL  Assuming the angular velocity ω L to be the output gives = y ω= θ= x =

L

L

5

Thus, the state equation and the output equation in matrix form are

285

 x1   x   2   x3   x   4  x5 

 Ra − L  a  0   0   Kt   Im  0 

0

0

−

0

0

1

0

0

0

Km Im

−

−

Km Im

Km IL

−

Km + KL IL

Ke La

Bm Im

Bm IL

 1  L   a   0   x1  0   x2       1   x3  + 0  x    4  Bm   x5  0 Im    Bm + BL  −  0 IL   0

    0     u1  0    u2   0   1 −  IL  0

 x1  x   2  u  = y [0 0 0 0 1]  x3  + [0 0]  1  u2  x   4  x5  6.

A more complicated model of the field-controlled motor is shown in Figure 6.52, in which the rotor is connected to an inertial load through a flexible and damped shaft. Km and Bm represent the torsional stiffness and the torsional viscous damping of the shaft, respectively. The mass moments of inertia of the motor and the load are Im and IL, respectively. a. Assuming zero initial conditions, derive the transfer functions ΩL(s)/Vf(s) and ΩL(s)/TL(s). b. Assuming the angular velocity ωL to be the output, draw a block diagram to represent the dynamics of the field-controlled motor. c. Determine the state-space form.

Figure 6.52 Problem 6.

Solution a.

Applying Kirchhoff’s current law to the field circuit and the moment equation to the mechanical part gives di Lf f + Rf if = vf dt I m θ m + Bm ( θ m − θ L ) + K m ( θ m − θ L ) = τ m = K t if

I L θ L + ( BL + Bm )θ L − Bm θ m + ( K L + K m )θ L − K m θ m = −τ L The above three equations are the system differential equations of the field-controlled DC motor shown in Figure 6.52. Assuming all the initial conditions to be zero and taking the Laplace transform, we have Lf sI f ( s ) + Rf I f ( s ) = Vf ( s )

286

(I

(I

L

m

s 2 + Bm s + K m ) Θm ( s ) − ( Bm s + K m ) ΘL ( s ) = K t I f ( s)

s 2 + BL s + Bm s + K L + K m ) ΘL ( s ) − ( Bm s + K m ) Θm ( s ) = −TL ( s )

Applying Cramer’s rule gives

− ( Lf s + Rf ) ( I m s 2 + Bm s + K m )

= ΘL ( s )

∆ (s)

TL ( s ) +

K t ( Bm s + K m ) Vf ( s ) ∆ (s)

where

2 ∆ ( s ) = ( Lf s + Rf ) ( I m s 2 + Bm s + K m )( I L s 2 + BL s + Bm s + K L + K m ) − ( Bm s + K m )    Thus, the transfer functions Ω L ( s ) TL ( s ) and Ω L ( s ) Vf ( s ) are 2 Ω L ( s ) sΘ L ( s ) − s ( Lf s + Rf ) ( I m s + Bm s + K m ) = = TL ( s ) TL ( s ) ∆ (s)

Ω L ( s ) sΘ L ( s ) K t s ( Bm s + K m ) = = Vf ( s ) Vf ( s ) ∆ (s) b.

A block diagram representing the dynamics of the field-controlled motor is shown below. v_f

1

i_f

Kt

tau_m

Lf.s+Rf

1 s

1/Im

theta_m_dot

theta_m

1 s

Km

Bm

tau_L

1 s

1/IL

theta_L_dot

1 s

theta_L

BL

KL

c.

Figure PS6-4 No6 Refer to the differential equations in Part (a). Let x1 if= x2 θm= x3 θ L= x4 θ m= x5 θ L = and = u1 v= u2 τ L f The state variable equations are di 1 1 x1 = f = ( vf − Rf if ) = ( − Rf x1 + u1 ) dt Lf Lf  x= θ= x 2

m

4

x= θ= x5 3 L

x 4 = θm =

1 1  K t if − K m ( θ m − θ L ) − Bm ( θ m − θ L ) = ( K t x1 − K m x2 + K m x3 − Bm x4 + Bm x5 )   Im Im 1  − τ L + K m θ m − ( K L + K m )θ L + Bm θ m − ( BL + Bm )θ L  x5 =  θL = IL  =

1 [ −u2 + K m x2 − ( K L + K m )x3 + Bm x4 − ( BL + Bm )x5 ] IL

287

Assuming the angular velocity ω L to be the output gives = y ω= θ= x5 L L Thus, the state equation and the output equation in matrix form are  Rf  1 0 0 0 − L 0  L  f   f    0 0 1 0  x1  0   x1  0  x     x2    2       0 0 0 1  x3  0   x3  + 0  x    x    4   Kt  4  Km Km Bm Bm − −   x5  0  x5   I I I I I m m m m m      Km K L + K m Bm BL + Bm  − − 0  0 IL IL IL IL     x1  x   2  u  = y [0 0 0 0 1]  x3  + [0 0]  1  u2  x  4    x5 

    0    u  0  1  u2   0   1 −  IL  0

Problem Set 6.5 1.

Reconsider the RC circuit shown in Figure 6.24. Use the impedance method to determine the transfer function I(s)/Va(s) and the input–output differential equation relating vC and va. Assume that all the initial conditions are zero.

Solution Replacing the passive elements with their impedance representations gives the circuit in the s domain.

Figure PS6-5 No1 Note

Z1 = R

1 Cs Applying Kirchhoff’s voltage law to the equivalent impedance circuit yields Z1 I ( s ) + Z 2 ( s ) I ( s ) = Va ( s ) Thus, the transfer function relating the input voltage va and the output current i is I ( s) Cs 1 = = Va ( s ) Z1 + Z 2 ( s ) RCs + 1 Note that the current is related to the output voltage by VC ( s ) = Z 2 ( s ) I ( s ) Thus, we have VC ( s ) Z 2 ( s ) I ( s ) 1 = = Va ( s ) Va ( s ) RCs + 1 Z 2 ( s) =

288

By transforming VC ( s ) Va ( s ) from the s domain to the time domain with the assumption of zero initial conditions, we obtain the differential equation of the system RCvC + vC = va 2.

Reconsider the RL circuit shown in Figure 6.25. Use the impedance method to determine the transfer function VL(s)/Va(s) and the input–output differential equation relating iL and va. Assume that all the initial conditions are zero.

Solution Replacing the passive elements with their impedance representations gives the circuit in the s domain as shown in the figure below, where Z1 = R

Z 2 ( s ) = Ls

Figure PS6-5 No2 Applying Kirchhoff’s voltage law to the equivalent impedance circuit yields Z1 I L ( s ) + Z 2 ( s ) I L ( s ) = Va ( s ) Thus, the transfer function relating the input voltage va and the output current iL is I L ( s) 1 1 = = Va ( s ) Z1 + Z 2 ( s ) R + Ls Note that the current is related to the output voltage by VL ( s ) = Z 2 ( s ) I L ( s ) Thus, we have

VL ( s ) Z 2 ( s ) I L Ls = = Va ( s ) Va ( s ) R + Ls Also, by transforming I L ( s ) Va ( s ) from the s domain to the time domain with the assumption of zero initial conditions, we obtain the differential equation of the system di L L + RiL = va dt 3.

Reconsider the RLC circuit shown in Figure 6.26. Use the impedance method to determine the input–output differential equation relating vo and va. Assume that all the initial conditions are zero.

Solution Replacing the passive elements with their impedance representations gives the circuit in the s domain as shown in the figure below.

Figure PS6-5 No3

289

Note that

Z1 = R 1 1 Ls or Z 2 ( s ) = = Cs + Z 2 (s) Ls LCs 2 + 1 Applying Kirchhoff’s voltage law in the equivalent impedance circuit yields Z1 I ( s ) + Z 2 ( s ) I ( s ) = Va ( s ) Thus, the transfer function relating the input voltage va and the current i is I (s) 1 LCs 2 + 1 = = Va ( s ) Z1 + Z 2 ( s ) RLCs 2 + Ls + R Note that the current is related to the output voltage by Vo ( s ) = Z 2 ( s ) I ( s ) Thus, we have Vo ( s ) Z 2 ( s ) I ( s ) Ls = = 2 Va ( s ) Va ( s ) RLCs + Ls + R By transforming Vo ( s ) Va ( s ) from the s domain to the time domain with the assumption of zero initial conditions, we obtain the differential equation of the system RLCvo + Lvo + Rvo = Lva 4.

Reconsider the RLC circuit shown in Figure 6.27. Use the impedance method to determine the input–output differential equation relating i and va. Assume that all the initial conditions are zero.

Solution Replacing the passive elements with their impedance representations gives the circuit in the figure below, where RCs + 1 1 = Z1 ( s ) =R + Cs Cs Z 2 ( s ) = Ls

Figure PS6-5 No4 Applying Kirchhoff’s current law to node 1 in the equivalent impedance circuit yields = I (s) I R (s) + I L (s) where V (s) IR ( s ) = a Z1

IL ( s ) =

Va ( s ) Z2

Thus, we have

 1 1 I (s) =  + Z Z  1 2 which gives the transfer function

 LCs 2 + RCs + 1 Va ( s )  Va ( s ) = 2 RLCs + Ls 

s domain as shown in the

290

I ( s ) LCs 2 + RCs + 1 = Va ( s ) RLCs 2 + Ls By transforming I ( s ) Va ( s ) from the s domain to the time domain with the assumption of zero initial conditions, we obtain the differential equation of the system d 2i di RLC 2 + L = LCva + RCva + va dt dt 5.

Reconsider the RLC circuit shown in Figure 6.28. Use the impedance method to determine the input–output differential equation relating vo and va. Assume that all the initial conditions are zero.

Solution Replacing the passive elements with their impedance representations gives the circuit in the s domain as shown in the figure below, we have 1 1 1 RLs or Z1 ( s ) = = + Z1 ( s ) Ls R Ls + R

1 Cs Applying Kirchhoff’s voltage law to the equivalent impedance circuit yields Z1 ( s ) I ( s ) + Z 2 ( s ) I ( s ) = Va ( s ) Thus, we have 1 I (s) LCs 2 + RCs = = Va ( s ) Z1 ( s ) + Z 2 ( s ) RLCs 2 + Ls + R Note that the current is related to the output voltage by Vo ( s ) = Z 2 ( s ) I ( s ) Thus, the transfer function relating the input voltage va and the output voltage vo, Vo ( s ) Z 2 ( s ) I ( s ) Ls + R = = Va ( s ) Va ( s ) RLCs 2 + Ls + R Z 2 (s) =

By transforming Vo ( s ) Va ( s ) from the s domain to the time domain with the assumption of zero initial conditions, we obtain the differential equation of the system RLCvo + Lvo + Rvo = Lva + Rva

Figure PS6-5 No5 6.

Reconsider the RLC circuit shown in Figure 6.29. Use the impedance method to determine the input–output differential equation relating vo and va. Assume that all initial conditions are zero.

Solution Replacing the passive elements with their impedance representations gives the circuit in the s domain, where Z1 ( s= ) R1 + Ls R2 1 1 = + Cs or Z 2 ( s ) = Z 2 ( s ) R2 R2Cs + 1

291

Figure PS6-5 No6 Applying Kirchhoff’s voltage law yields

Z1 ( s ) I ( s ) + Z 2 I ( s ) = Va ( s ) Thus, we have

R2Cs + 1 I ( s) 1 = = 2 Va ( s ) Z1 ( s ) + Z 2 ( s ) R2 LCs + ( R1 R2C + L) s + R1 + R2 Note that the output voltage is Vo ( s ) = Z 2 ( s ) I ( s ) Thus, the transfer function relating the input voltage va and the output voltage vo is Vo ( s ) Z 2 ( s ) I ( s ) R2 = = 2 Va ( s ) Va ( s ) R2 LCs + ( R1 R2C + L) s + R1 + R2 Transforming Vo ( s ) Va ( s ) from the s domain to the time domain with the assumption of zero initial conditions gives R2 LCvo + ( R1 R2 C + L)vo + ( R1 + R2 )vo = R2 va . 7.

Reconsider the RLC circuit shown in Figure 6.30. Use the impedance method to determine the input–output differential equation relating vo and va. Assume that all initial conditions are zero.

Solution Replacing the passive elements with their impedance representations gives the circuit in the s domain, where Z1 ( s= ) R1 + L1 s

RLs 1 1 1 or Z 2 ( s ) = 2 2 = + L2 s + R2 Z 2 ( s ) R2 L2 s

Figure PS6-5 No7 Applying Kirchhoff’s voltage law yields

Z1 ( s ) I ( s ) + Z 2 I ( s ) = Va ( s ) Thus, we have

L2 s + R2 I (s) 1 = = 2 Va ( s ) Z1 ( s ) + Z 2 ( s ) L1 L2 s + R1 L2 s + R2 ( L1 + L2 ) s + R1 R2 Note that the output voltage is Vo ( s ) = Z 2 ( s ) I ( s ) Thus, the transfer function relating the input voltage va and the output voltage vo is Vo ( s ) Z 2 ( s ) I ( s ) R2 L2 s = = Va ( s ) Va ( s ) L1 L2 s 2 + R1 L2 s + R2 ( L1 + L2 ) s + R1 R2

292

Transforming

8.

from the s domain to the time domain with the assumption of zero initial conditions gives

Reconsider the op-amp circuit shown in Figure 6.41. Use the impedance method to determine the differential equation relating the input voltage vi and the output voltage vo.

Solution Replacing the passive elements with their impedance representations gives the op-amp circuit in the s domain as shown in the figure below, we have and

Figure PS6-5 No8 Because the current drawn by the op-amp is very small, applying Kirchhoff’s current law to node 1 yields

where

. Thus, we have

which gives the transfer function relating the input voltage vi and the output voltage vo,

Transforming

from the s domain to the time domain with the assumption of zero initial conditions gives

Problem Set 6.6 (Note: All MATLAB figures are given at the end of Problem Set 6.6.) 1.

Consider the RL circuit shown in Figure 6.25 (Problem Set 6.2, Problem 2), in which R = 35 Ω and L = 10 H. When the switch is closed at 0 second, the circuit is driven by a 6V DC voltage source. Assume that all initial conditions are zero. a. Build a Simscape model of the physical system and find the loop current iL(t) and the voltage across the inductor vL(t). b. Build a Simulink model of the system based on the differential equation relating iL and va, and find the loop current iL(t). c. Build a Simulink model of the system based on the transfer function VL(s)/Va(s), and find the voltage across the inductor vL(t).

Solution The differential equation relating iL and va is or The transfer function VL(s)/Va(s) is

293

2.

Consider the parallel RLC circuit shown in Example 6.2, in which R = 2 Ω, L = 1 H, and C = 0.5 F. The circuit is driven by a controlled current source ia(t) = 10u(t), where u(t) is a unit-step function. a. Build a Simscape model of the physical system and find the voltage across the capacitor vC(t) and the current through the inductor iL(t). b. Refer to the results obtained in Example 6.2. Build a Simulink model of the system based on the transfer function VC(s)/Ia(s) and find the voltage across the capacitor vC(t). c. Refer to the results obtained in Example 6.2. Build a Simulink model of the system based on the transfer function IL(s)/Ia(s) and find the current through the inductor iL(t).

Solution The transfer function VC(s)/Ia(s) is The transfer function IL(s)/Ia(s) is

3.

A simple low-pass filter can be realized by an RLC circuit (see Figure 6.21 in Example 6.4), which passes signals with frequencies lower than a certain cutoff frequency and attenuates signals with frequencies higher than the cutoff frequency. Assume that the parameter values are R = 500 Ω, L = 100 mH, and C = 10 µF. The circuit is connected to an AC voltage source, which has an amplitude of 1 V and a varying frequency. a. Build a Simscape model of the physical system and find the output voltage vo(t) when the frequency of the input voltage is 500, 1000, and 2500 rad/s. b. Build a Simulink model of the system based on the transfer function Vo(s)/Va(s), and verify the results obtained in Part (a).

Solution The transfer function Vo(s)/Va(s) is

4.

A simple band-pass filter can be realized by an RLC circuit (see Figure 6.73), which passes frequencies within a certain range and attenuates frequencies outside that range. Assume that the parameter values are R = 500 Ω, L = 100 mH, and C = 10 µF. The circuit is connected to an AC voltage source, which has an amplitude of 1 V and a varying frequency. a. Build a Simscape model of the physical system and find the output voltage vo(t) when the frequency of the input voltage is 1000, 800, and 1200 rad/s. b. Derive the transfer function Vo(s)/Va(s), build a Simulink model of the system based on this transfer function, and verify the results obtained in Part (a).

Figure 6.73 Problem 4.

Solution The transfer function Vo(s)/Va(s) is

5.

Consider the op-amp integrator in Figure 6.35. Assume that the parameter values are R = 1 MΩ and C = 1 µF. Build a Simscape model of the op-amp circuit and find the output voltage vo(t) when the input voltage is vi = −0.1 V.

294

6.

Consider the op-amp circuit shown in Figure 6.74, in which the parameter values are C1 = 0.8 µF, R1 = 10 kΩ, C2 = 80 pF, and R2 = 100 kΩ. The circuit is connected to an AC voltage source, which has an amplitude of 1 V and a frequency of 200 Hz. Build a Simscape model of the op-amp circuit and find the output voltage vo(t).

Figure 6.74 Problem 6.

295

MATLAB Figures Problem 1 PS

S PS Clock

Simulink-PS Converter

-

+

+

Resistor

Switch

I -

iL(t)

PS S

PS-Simulink Converter

Current Sensor

Scope

DC Voltage Source -

f(x) = 0

+

Inductor

Solver Configuration

+

Electrical Reference V

Voltage Sensor PS S

vL(t)

PS-Simulink Converter1

Scope1

Figure PS6-6 No1a

Figure PS6-6 No1b

Figure PS6-6 No1c 0.18

6

0.16 5 0.14

(A) L

L

(A)

4 0.12

3

Inductor Voltage v

Inductor Current i

0.1

0.08

0.06

2

1

0.04 0 0.02

-1

0 0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

0

Time (s)

0.5

1

1.5

2

2.5

Time (s)

Figure PS6-6 No1d

3

3.5

4

4.5

5

296

+

Problem 2

I

-

Current Sensor

PS S

Step

+

+

Scope

V

-

iL(t)

PS-Simulink Converter Voltage Sensor

Capacitor

-

-

Inductor

-

Resistor

S PS

+

+

Controlled Current Source

Simulink-PS Converter PS S

vC(t)

PS-Simulink Converter1

f(x) = 0

Scope1

Electrical Reference

Solver Configuration

Figure PS6-6 No2a

Figure PS6-6 No2b

14

8

12

6

10

(A)

10

L

C

(V)

Figure PS6-6 No2c

8

Inductor Current i

Capacitor Voltage v

4

2

0

-2

6

4

2

-4

0 0

5

10

15

0

Time (s)

5

10

Time (s)

Figure PS6-6 No2d

15

297

Problem 3 -

+

Resistor

V

-

-

Voltage Sensor

-

Capacitor

AC Voltage Source

+

+

+

Inductor

vC(t)

f(x) = 0

PS S

Solver Configuration

PS-Simulink Converter

Electrical Reference

Scope

Figure PS6-6 No3a

Figure PS6-6 No3b 500 rad/s

1000 rad/s

2

6

1.5

4

1

Output voltage (V)

0

0

-2

-0.5

-4

-1

-6

-1.5 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0

0.01

0.02

0.03

0.04

Time (s)

0.05

Time (s) 2000 rad/s 1

0.8

0.6

0.4

0.2

Output Voltage (V)

Output voltage (V)

2 0.5

0

-0.2

-0.4

-0.6

-0.8 -1 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

Time (s)

Figure PS6-6 No3c

0.08

0.09

0.1

0.06

0.07

0.08

0.09

0.1

298

Problem 4 -

+

Inductor

Voltage Sensor

V

-

-

Capacitor

-

AC Voltage Source

+

+

+

Resistor

PS S PS-Simulink Converter1

f(x) = 0 Solver Configuration

Scope1

Electrical Reference

Figure PS6-6 No4a

Figure PS6-6 No4b 1000 rad/s

1

Output Voltage (V)

0.5

0

-0.5

-1

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

Time (s) 1200 rad/s 1

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

Output Voltage (V)

Output Voltage (V)

800 rad/s 1

0

-0.2

-0.4

0

-0.2

-0.4

-0.6

-0.6

-0.8

-0.8 -1

-1 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

0

0.01

0.02

0.03

0.04

0.05

Time (s)

Time (s)

Figure PS6-6 No4c

0.06

0.07

0.08

0.09

0.1

299

Problem 5 +

Capacitor -

-

+

f(x) = 0

-

+ Resistor

Solver Configuration

+

Op-Amp

Voltage Sensor

V

-

DC Voltage Source

PS S

PS-Simulink Converter

Electrical Reference

Figure PS6-6 No5a

0.06

0.05

Output Voltage (V)

0.04

0.03

0.02

0.01

0 0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

Time (s)

Figure PS6-6 No5b

0.4

vo(t)

0.45

0.5

Scope

300

Problem 6 -

+

Capacitor1 -

+

Solver Configuration

Resistor1 -

+

f(x) = 0

+

-

+

Resistor

Capacitor

+

Op-Amp

Voltage Sensor

-

V

AC Voltage Source

PS S PS-Simulink Converter

Electrical Reference

Figure PS6-6 No6a

10

Output Voltage (V)

5

0

-5

-10

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

Time (s)

Figure PS6-6 No6b

Review Problems 1.

Determine the equivalent resistance Req for the circuit shown in Figure 6.75.

Figure 6.75 Problem 1.

Solution For the top path, the equivalent resistances for the two parallel-connected circuits are 3R 1 1 4 + = or 4 3R R 3R

vo(t) Scope

301

1 1 1 + =or 3R 4 R 12 R 3R They are connected in series. Thus, we have

3R 15 R + 3R = 4 4 Similarly, for the bottom path, the equivalent resistance for the parallel-connected resistors, 5R and 20R, is 1 1 1 + =or 4R 5R 20 R 4 R They are connected with the resistor, R, in series. Thus, we have R + 4R = 5R The equivalent resistor for the entire circuit is 15 4 1 7 R + = or 7 15 R 5 R 15 R 2.

Find R13 and R32 for the voltage divider shown in Figure 6.76 so that the current is limited to 2 A when vi = 110 V and vo = 100 V.

Figure 6.76 Problem 2.

Solution The current is 2 A and vo = 100 V. Thus, the resistor R32 is R32 =

vo 100 = = 50 Ω . i 2

For a voltage divider, we have

vo R32 100 V 10 = = = vi R13 + R32 110 V 11 Thus, the resistor R13= 3.

R32 = 5Ω. 10

Consider the second-order RC circuit shown in Figure 6.77. Assume that all the initial conditions are zero. a. Use the node or loop method to derive the input–output differential equation relating vo and va and find the transfer function Vo(s)/Va(s). b. Use the impedance method to determine the transfer function Vo(s)/Va(s), and compare with the result obtained in Part (a).

Figure 6.77 Problem 3.

302

Solution a.

All currents entering or leaving a node are labeled as shown in the figure below.

Figure Review6 No3a Applying Kirchhoff’s current law gives

iR1 = iC1 + iR2 = iC1 + iC2 va − v1 = C1v1 + C2 vo R1 Applying Kirchhoff’s voltage law gives

= v1 R2iR2 + vo = v1 R2C2 vo + vo Combining the above two differential equations gives  va − R2 C= R1C1 ( R2 C2 vo + vo ) + R1C2 vo 2 vo − vo which can be rearranged and gives the differential equation relating vo and va, R1 R2C1C2 vo + ( R1C1 + R1C2 + R2C2 )vo + vo = va Taking Laplace transform gives Vo ( s ) 1 = Va ( s ) R1 R2C1C2 s 2 + ( R1C1 + R1C2 + R2C2 ) s + 1 b.

Replacing the passive elements with their impedance representations gives the circuit in the s domain as shown in the figure below, where Z1 = R1 1 Z 2 ( s) = C1s 1 Z 3 ( s= ) R2 + C2 s

Figure Review6 No3b The impedances Z2 and Z3 are connected in parallel, and we have R2C2 s + 1 1 1 1 = + or Z 4 ( s ) = Z 4 ( s) Z 2 ( s) Z3 ( s) R2C1C2 s 2 + (C1 + C2 ) s Applying Kirchhoff’s current law to node 1 yields I= ( s ) I1 ( s ) + I 2 ( s )

Va ( s ) V ( s ) V1 ( s ) V1 ( s ) = 1 + = Z1 + Z 4 ( s ) Z 2 ( s ) Z 3 ( s ) Z 4 ( s ) where

= V1 ( s ) R2 I 2 ( s ) + Vo ( s )

303

The output voltage is

V = V= o ( s) C2 ( s )

1 I 2 ( s) C2 s

Thus, we have

Va ( s ) R C s +1 = 2 2 Vo ( s ) Z1 + Z 4 ( s ) Z 4 ( s) which gives the transfer function relating the input voltage va and the output voltage vo, Vo ( s ) Z 4 ( s) = Va ( s ) [ Z1 + Z 4 ( s )] ( R2C2 s + 1) Substituting the expressions of the impedances results in Vo ( s ) 1 = 2 Va ( s ) R1 R2C1C2 s + ( R1C1 + R1C2 + R2C2 ) s + 1 which is same as the one obtained in Part (a). 4.

Repeat Problem 3 for the RLC circuit shown in Figure 6.78. Assume zero initial conditions. a. Use the node or loop method to derive the input–output differential equation relating i and va, and find the transfer function I(s)/Va(s). b. Use the impedance method to determine the transfer function I(s)/Va(s), and compare with the result obtained in Part (a).

Figure 6.78 Problem 4.

Solution a.

All currents entering or leaving a node are labeled as shown in the figure below.

Figure Review6 No4a Applying Kirchhoff’s voltage law to the two loops shown in the figure gives 1 RCi1 + ∫ i1dt = va C di RLi2 + L 2 = va dt which can be written in the s-domain as Cs 1   Va ( s ) Va ( s ) or I1 ( s ) =  RC +  I1 ( s ) = RCCs + 1 Cs  

304

( RL + Ls ) I 2 ( s ) = Va ( s ) or I 2 ( s ) =

1 Va ( s ) RL + Ls

Applying Kirchhoff’s current law to node 1 gives

i= i1 + i2 or in the s-domain,

I= ( s ) I1 ( s ) + I 2 ( s ) = I (s)

Cs 1 Va ( s ) + Va ( s ) RC Cs + 1 RL + Ls

( RCCs + 1)( RL + Ls ) I ( s ) =[( RL + Ls )Cs + ( RCCs + 1)]Va ( s ) Thus, we have

b.

I (s) LCs 2 + ( RLC + RCC ) s + 1 = Va ( s ) RC LCs 2 + ( RC RLC + L) s + RL

Taking inverse Laplace transform gives the differential equation relating i and va, d 2i di RC LC 2 + ( RC RLC + L) + RLi = LCva + ( RLC + RCC )va + va dt dt Replacing the passive elements with their impedance representations gives the circuit in the s domain as shown in the figure below, where 1 ) RC + Z1 ( s= Cs Z 2 ( s= ) RL + Ls

Figure Review6 No4b The impedances Z1 and Z2 are connected in parallel, and we have 1 1 1 = + Z ( s ) Z1 ( s ) Z 2 ( s ) or ( RL + Ls )( RCCs + 1) Z ( s) = LCs 2 + ( RLC + RCC ) s + 1 The relation between the current in the loop and the voltage applied is Va ( s ) = Z ( s ) I ( s ) which gives the transfer function relating the input voltage va and the output current i, LCs 2 + ( RL C + RC C ) s + 1 I (s) 1 = = Va ( s ) Z ( s ) RC LCs 2 + ( RC RL C + L) s + RL which is same as the one obtained in Part (a). 5.

Consider the RLC circuit shown in Figure 6.79, and assume zero initial conditions. a. Use the node or loop method to derive the input–output differential equation relating vo and va, and find the transfer function Vo(s)/Va(s). b. Use the impedance method to determine the transfer function Vo(s)/Va(s), and compare with the result obtained in Part (a).

305

Figure 6.79 Problem 5.

Solution a.

All currents entering or leaving node 1 and node 2 are labeled as shown in the figure below.

Figure Review6 No5a At node 1, applying Kirchhoff’s current law gives iC − iR1 − iR2 = 0 v1 v1 − v2 d = C ( va − v1 ) − − 0 dt R1 R2 which can be rearranged and gives the first differential equation  1 1  1 R1 R2Cva Cv1 +  +  v1 − v2 = Cva or R1 R2Cv1 + ( R1 + R2 ) v1 − R1v2 = R2  R1 R2  Applying Kirchhoff’s current law to node 2 gives

iR2 − iL = 0 v1 − v2 1 − ∫ v2 dt = 0 R2 L v1 − v2 1 − v2 = 0 R2 L which can be rearranged and gives the second equation 1 1 1 − v1 + v2 + v2 = 0 or − Lv1 + Lv2 + R2 v2 = 0 R2 R2 L In second-order matrix form, we have  R1 R2C 0   v1   R1 + R2 − R1   v1   R1 R2Cva   +  =    −L R2   v2   0  L   v2   0  Assume that all the initial conditions are zero. Taking Laplace transform gives − R1  V1 ( s )   R1 R2CsVa ( s )   R1 R2Cs + R1 + R2  =   − Ls Ls + R2  V2 ( s )   0   Applying Cramer’s rule gives V2 ( s ) Vo ( s ) R1 LCs 2 = = 2 Va ( s ) Va ( s ) R1 LCs + ( R1 R2C + L ) s + ( R1 + R2 ) b.

Replacing the passive elements with their impedance representations gives the circuit in the s domain as shown in the figure below, we have

306

1 Cs Z 2 = R1 Z 3 ( s= ) R2 + Ls The impedances Z2 and Z3 are connected in parallel, and we have R Ls + R1 R2 1 1 1 = + or Z 4 ( s ) = 1 Ls + R1 + R2 Z 4 ( s) Z 2 Z3 ( s) Z1 ( s ) =

Figure Review6 No5b Applying Kirchhoff’s current law to node 1 yields I= I R1 ( s ) + I R2 ( s ) C ( s)

Va ( s ) V ( s ) V ( s ) V1 ( s ) = 1 + 1 = Z1 ( s ) + Z 4 ( s ) Z2 Z3 ( s) Z 4 ( s) where

= V1 ( s ) R2 I R 2 ( s ) + Vo ( s ) The output voltage is

V= V= LsI R 2 ( s ) o ( s) L ( s) Thus, we have

Va ( s ) 1  R2  =  + 1Vo ( s ) Z1 ( s ) + Z 4 ( s ) Z 4 ( s )  Ls  which gives the transfer function relating the input voltage va and the output voltage vo, Vo ( s ) LsZ 4 ( s ) = Va ( s ) [ Z1 ( s ) + Z 4 ( s )] ( Ls + R2 ) Substituting the expressions of the impedances yields the same result as the one obtained in Part (a). 6.

Consider the circuit shown in the Figure 6.77 (Review Problems, Problem 3). a. Determine a suitable set of state variables and obtain the state-space representation. b. Find the transfer function directly from the state-space form and compare with the result obtained in Problem 3.

Solution a.

Refer to Figure 6.77 in Problem 3. Note that the circuit has two independent energy storage elements. This implies that two states are needed, and they are = x1 v= x2 vC2 C1 , Their time derivatives are dvC1 1 = x1 = iC1 dt C1

= x 2

dvC2 1 = iC2 dt C2

where

 1 v −v v −v 1  1 1 iC1 = iR1 − iR2 =a C1 − C1 C2 = −  +  x1 + x2 + u R1 R2 R2 R1  R1 R2 

307

vC1 − vC2 1 1 = x1 − x2 R2 R2 R2 Thus, the complete set of two state-variable equations is  1 1  1 1 − + x1 = x2 + u  x1 + R2C1 R1C1  R1C1 R2C1  = iR2 = iC2

= x 2

1 1 x1 − x2 R2C2 R2C2

The output equation is

b.

= y v= vC2= x2 o The state equation and the output equation can be written in matrix form as follows:   1 1  1  +  1  −     x  1    R1C1 R2C1  R2C1   x1    = +       R1C1  u   x x 1 1  2  2  0  −   R2C2 R2C2   x  = y [0 1]  1  + 0 ⋅ u  x2  The state-space form can be converted to a transfer function using Y (s) −1 =C ( sI − A ) B + D. U (s) Substituting A, B, C, and D matrices obtained in Part (a) gives −1

  1 1  1  + s +    1   − R1C1 R2C1  R2C1   Vo ( s ) 1   R1C1  1] [0= 2     Va ( s ) R1R2C1C2 s + ( R1C1 + R1C2 + R2C2 ) s + 1 1 1 s+ −    0  R C R C  2 2 2 2  which is the same as the transfer function obtained in Problem 3. 7.

Repeat Problem 6 for the circuit shown in Figure 6.78 (Review Problems, Problem 4).

Solution a.

Refer to Figure 6.78 in Problem 4. Note that the circuit has two independent energy storage elements. This implies that two states are needed, and they are = x1 v= x2 iL C, Their time derivatives are dvC 1 = x1 = iC dt C diL 1 = x 2 = vL dt L where va − vC u − x1 = = iC iR= C RC RC

vL =va − vR L =va − RL iR L =va − RL iL =u − RL x2 Thus, the complete set of two state-variable equations is 1 1 x1 = − x1 + u RCC RCC R 1 x 2 = − L x2 + u L L The output equation is

308

v −v 1 1 − y == i iC + iL =a C + iL = x1 + x2 + u RC RC RC The state equation and the output equation can be written in matrix form as follows:  1   1  − 0    1 x  1  x1  x  R C RCC  1 +  C u , = y =− 1  1  + ⋅u      x x RL   2   1   2   RC   x2  RC − 0   L  L  b.

The state-space form can be converted to a transfer function using Y (s) −1 =C ( sI − A ) B + D. U (s) Substituting A, B, C, and D matrices obtained in Part (a) gives −1

I ( s) Va ( s )

1    1  s+ 0     1  LCs 2 + ( RLC + RCC ) s + 1 RCC R C 1   C + − 1  = R   1  RC RC LCs 2 + ( RC RLC + L) s + RL  RC  0 s+ L   L   L 

which is the same as the transfer function obtained in Problem 4. 8.

Repeat Problem 6 for the circuit shown in Figure 6.79 (Review Problems, Problem 5).

Solution a.

Refer to Figure 6.79 in Problem 5. Note that the circuit has two independent energy storage elements. This implies that two states are needed, and they are = x1 v= x2 iL C, Their time derivatives are dvC 1 = x1 = iC dt C diL 1 = x 2 = vL dt L where v −v 1 1 iC = iR1 + iR2 = a C + iL = u − x1 + x2 R1 R1 R1

vL = va − vC − vR2 = va − vC − R2iL = u − x1 − R2 x2 Thus, the complete set of two state-variable equations is 1 1 1 − x1 = x1 + x2 + u R1C C R1C R 1 1 x 2 = − x1 − 2 x2 + u L L L The output equation is y =vL =u − x1 − R2 x2 The state equation and the output equation can be written in matrix form as follows: 1   1  1  − x   x1   R1C C   x1   R1C   + u y = [ −1 − R2 ]  1  + 1 ⋅ u =     R   x2   1   x2   x2   1 − 2  − L  L  L b.

The state-space form can be converted to a transfer function using Y (s) −1 =C ( sI − A ) B + D. U (s) Substituting A, B, C, and D matrices obtained in Part (a) gives

309

−1

1 1   1   − s + R C  RC Vo ( s ) R1LCs 2 C 1   1  +1 = = [ −1 − R2 ]  Va ( s ) R1LCs 2 + ( R1R2C + L ) s + ( R1 + R2 ) R2   1   1 + s  L L   L  which is the same as the transfer function obtained in Problem 5. 9.

The op-amp circuit shown in Figure 6.80 is an active low-pass filter. Derive the input–output differential equation relating the output voltage vo(t) and the input voltage vi(t). Assuming zero initial conditions, find the transfer function Vo(s)/Vi(s) directly from the input–output equation.

Figure 6.80 Problem 9.

Solution Note that the current drawn by the op-amp is very small, i.e., i− ≈ i+ ≈ 0 . Applying Kirchhoff’s current law gives

iR1 = iR2 + iC d ( v − − vo ) vi − v− v− − vo = +C R1 R2 dt where v− ≈ v+ = 0 . Thus, we have

vi −vo = − Cvo R1 R2 which can be rearranged as

1 1 vo = − vi R2C R1C Assume that all the initial conditions are zero. Taking Laplace transform gives Vo ( s ) − R2 = Vi ( s ) R1 R2Cs + R1 vo +

10. Repeat Problem 9 for the op-amp circuit shown in Figure 6.81, which represents an active band-pass filter.

Figure 6.81 Problem 10.

Solution Note that the current drawn by the op-amp is very small, i.e., i− ≈ i+ ≈ 0 . Applying Kirchhoff’s current law to node 1 gives iR1 = iR2 + iC1 + iC2

310

and applying Kirchhoff’s current law to node 2 gives

where

. Thus, we have or

Substituting the second equation into the first one to eliminate the terms associated with v1 or its derivatives yields

Taking the time derivative and rearranging gives

Assume that all the initial conditions are zero. Taking Laplace transform gives

Figure Review6 No10 11.

Consider the RLC circuit shown in Figure 6.30 (Problem Set 6.2, Problem 15), where R1 = 100 Ω, L = 20 H, R2 = 400 Ω, and C = 1/120 F. The circuit is driven by a 100 V DC voltage source. a. Build a Simscape model of the physical system and find the output voltage vo(t). b. Build a Simulink model of the system based on the state-space form and find the output voltage vo(t).

Solution The Simscape and Simulink block diagrams are shown on the next page. Running either of the block diagrams gives the same voltage response curve shown below.

Figure Review6 No11a

311

Simscape block diagram:

Figure Review6 No11b Simulink block diagram:

Figure Review6 No11c 12. Case Study Consider the DC motor–driven wheeled mobile robot shown in Figure 6.82a, in which m is the mass of the wheeled mobile robot, r is the radius of the driving wheel, and τ is the torque delivered to the wheeled mobile robot by the DC motor. For simplicity, the motion is restricted to one spatial dimension. Figure 6.82b shows the simplified drive system, including the equivalent electrical circuit of the DC motor, the rotor of the DC motor, the gears, and the driving wheel. The motor parameter values are armature inductance La = 0.001 H, resistance Ra = 2.6 Ω, back emf constant Ke = 0.008 V⋅s/rad, and torque constant Kt = 0.008 N⋅m/A. The mass moment of inertia of the motor can be negligible. The gear ratio N = θ/θm = τm/τ = 1/3.7. The wheel and axle mechanism converts the rotational motion to translation, and the wheel radius r = 0.00635 m. The mass of the cart m = 0.455 kg.

a. b. c. d.

Figure 6.82 Problem 12. Derive the equations of motion of the system. Choose the armature current ia, the robot displacement x, and the robot velocity the state-space form of the system. Assuming zero initial conditions. Find the transfer function X(s)/Va(s).

as state variables and find

Following Figure 6.45, build a Simulink block diagram using the differential equations obtained in Part (a) and find the displacement output x(t) when the voltage applied to the DC motor is a pulse function, va(t) = 1 V for 1 ≤ t ≤ 2 s.

312

e.

Build a Simscape model of the wheeled mobile robot and find the displacement output x(t) when the voltage applied to the DC motor is a pulse function, va(t) = 1 V for 1 ≤ t ≤ 2 s.

Solution a.

Applying Kirchhoff’s voltage law gives

where

. The gear ratio is

Also, using the geometric constraint,

, which implies that

, we have

Applying Newton’s second law along x-direction gives where the driving force generated by the motor is

Note that the gear ratio is

where

. Therefore, we have

The equations of motion of the system are

b.

Let The state variable equations are

The displacement x is assumed to be the output, i.e., Thus, the state equation and the output equation in matrix form are

c.

Assuming all the initial conditions to be zero and taking the Laplace transform for the equations in Part (a), we have

313

( La s + Ra ) I a ( s ) +

Ke s X ( s) = Va ( s ) Nr

Kt I a ( s ) − ms 2 X ( s ) = 0 Nr Eliminating Ia(s) and solving X(s)/Va(s) yield X (s) K t Nr = Va ( s ) mN 2 r 2 La s 3 + mN 2 r 2 Ra s 2 + K t K e s d.

Simulink block diagram

e.

Simscape block diagram

Figure Review6 No12a

Mechanical Translational Reference

Gear Box

Inductor

+

S

Resistor

A

O

P

Wheel and Axle

+

Solver Configuration

+

-

R

f(x) = 0

-

C V

P Ideal Translational Motion Sensor

Mass

PS S PS-Simulink Converter

C

-

Controlled Voltage Source

Rotational Electromechanical Converter

R

Group 1 Signal 1

Signal Builder

S PS Simulink-PS Converter

Electrical Reference

Mechanical Rotational Reference

Figure Review6 No12b Running either of the block diagrams gives the displacement response as shown in the figure below.

Figure Review6 No12c

Scope

314

Problem Set 7.1 1. A car has an internal volume of 2.8 m3. If the sun heats the car from a temperature of 20°C to a temperature of 35°C, what will the pressure inside the car be? Assume the pressure is initially 1 atm.

Solution   This is a constant-volume process, where p1 1= atm 101325 Pa. We = T1 20 = C 293 K,= T2 35 = C 308 K, and= have T1 T2 = p1 p2 Thus, the pressure inside the car will be T2 308 = p2 = p1 = (101325 ) 106512 Pa = 1.07 atm T1 293

2.

Find the pneumatic capacitance of dry air in a rigid container with volume 15 ft3 for an isothermal process. Assume the air is initially at ambient temperature of 76°F.

Solution The capacitance of a container is

C=

V nRgT

For an isothermal process, n = 1, and we have V 15 = = = 1.63 × 10−5 slug⋅ft2/lb C RgT 1716.6 ( 76 + 460 ) 3.

Figure 7.3 shows a pneumatic system, in which the pneumatic capacitances of the two rigid containers are C1 and C2, respectively. Dry air at a constant temperature passes through a valve of linear resistance R1 into container 1. The pressure pi at the inlet of the valve is constant, and it is greater than p1. The air flows from container 1 to container 2 through a valve of linear resistance R2. Derive the differential equations in terms of the pressures p1 and p2. Write the equations in second-order matrix form.

Figure 7.3 Problem 3.

Solution Applying the law of conservation of mass to container 1 gives dm = qmi − qmo dt where dp dm dm dp1 = = C1 1 dt dp1 dt dt p − p1 qmi = i R1 p − p2 qmo = 1 R2 Substituting these expressions results in dp  1 p pi 1  C1 1 +  +  p1 − 2 = dt  R1 R2  R2 R1 Applying the law of conservation of mass to container 2 gives

315

dm = qmi − qmo = qmi dt where

dp dm dm dp2 = = C2 2 dt dp2 dt dt p1 − p2 qmi = R2 Substituting these expressions results in

dp2 p1 p2 − + = 0 dt R2 R2 The system differential equations in second-order matrix form is  dp1   1 + 1  − 1   pi     R2   p1    C1 0   dt   R1 R2   =  R1   0 C   dp  +  1 1   p2    2  2   −  0  dt   R2 R2  C2

4.

Figure 7.4 shows a pneumatic piston, which can exert a force in one direction and serve as a translational actuator. The displacement of the piston is x, the total mass of the piston and the load is m, and b is the damping coefficient. Use subscript i to denote the chamber index and i = 1, 2. As shown in Figure 7.4, pi is the absolute pressure in the chamber, qmi is the mass flow rate at the port, and Ai is the effective area of the piston. The two chambers have the same initial volume of V0. Assuming ideal gas and isothermal process, R is the specific gas constant and T is the absolute gas temperature. Derive the dynamic equations of the pressure change in the pneumatic chambers and the equation of motion of the mass block.

Figure 7.4 Problem 4.

Solution Applying the law of conservation of mass to chamber 1 gives dm = qm1 dt where dm d (ρV ) dV dρ = = ρ +V dt dt dt dt Note that for ideal gas, p ρ= 1 RT and also, the volume of the chamber is V= V0 + A1 x Substituting these expressions results in p1 1 dp1 q= A1 x + (V0 + A1 x ) m1 RT RT dt Applying the law of conservation of mass to chamber 2 gives dm = qm2 dt

316

where

dm d (ρV ) dV dρ = = ρ +V dt dt dt dt

Note that for ideal gas,

ρ=

p2 RT

and also, the volume of the chamber is

V= V0 − A2 x Substituting these expressions results in

p 1 dp2 qm2 = − 2 A2 x + (V0 − A2 x ) RT RT dt For the piston, applying Newton’s second law gives p1 A1 − p2 A2 − bx = mx Thus, the system differential equations are dp1 A1 x RT = qm1 − p1 dt V0 + A1 x V0 + A1 x dp2 A2 x RT = qm2 + p2 dt V0 − A2 x V0 − A2 x mx + bx = p1 A1 − p2 A2

Problem Set 7.2 1.

Derive the capacitance of the tank shown in Figure 7.13. There is an opening at the top of the tank at height H.

Figure 7.13 Problem 1.

Solution

The hydraulic capacitance can be derived using C = A( h ) g . The cross-sectional area is

A(= h ) π(R − h tan α) 2 where

tan α =

R−r H

Figure PS7-2 No1 Thus, we have

A( h ) =

π(RH − Rh + rh ) 2 H2

and

C=

π(RH − Rh + rh ) 2 H 2g

317

2.

Derive the capacitance of the tank shown in Figure 7.14.

Figure 7.14 Problem 2.

Solution The hydraulic capacitance can be derived using C = A( h ) g , where A( h ) = bL . The length L is L −L H −h h L= L1 + h tan α = L1 + h 2 1 = L1 + L2 H H H Thus, we have L bH − L1bh + L2 bh C= 1 Hg

Figure PS7-2 No2 3.

Consider the rectangular tank in Figure 7.15a and the pyramid tank in Figure 7.15b. The volume flow rate into each tank through a pipe is qi. The liquid leaves each tank through a valve of linear resistance R. The density ρ of the liquid is constant. a. Derive the dynamic model of the liquid height h for each tank. b. For each tank, write the differential equation in terms of hydraulic capacitance and hydraulic resistance. Compare the models for the two single-tank systems.

Figure 7.15 Problem 3.

Solution a.

Applying the law of conservation of mass to the rectangular tank gives dm = qmi − qmo dt dm d dh where = = ( ρbLh ) ρbL dt dt dt qmi = ρqi

= qmo

pa + ρgh − pa ρgh = R R

318

Substituting these expressions into the law of conservation of mass gives dh ρg ρbL + h = ρqi dt R Applying the law of conservation of mass to the pyramid tank gives dm = qmi − qmo dt where the expressions of qmi and qmo are the same as the ones given above. Note that dm d dV dh = = ( ρV ) ρ dt dt dh dt As shown in the figure below, the volume of the liquid in the pyramid tank is V = 13 a 2 ( h + y ) − 13 r 2 y where r= a − 2h tan α a = y −h 2 tan α Substituting r and y into the expression of V and simplifying, we have ( a − 2h tan α)3 a3 = V − 6 tan α 6 tan α and dV = ( a − 2h tan α) 2 dh Substituting these expressions into the law of conservation of mass gives dh ρg ρ( a − 2h tan α) 2 + h= ρqi dt R

b.

Figure PS7-2 No3 The hydraulic capacitance for the rectangular tank is A ( h ) bL = C = g g Thus, the differential equation can be written as dh ρg dh R ρgC + h= ρqi or RC + h = qi dt R dt g The hydraulic capacitance for the pyramid tank is A ( h ) r 2 ( a − 2h tan α) 2 = = = C g g g Thus, the differential equation can be written as dh ρg dh R ρgC + h= ρqi or RC + h = qi dt R dt g The differential equations in terms of hydraulic capacitance and hydraulic resistance for the two single-tank systems are the same.

319

4.

Consider the single-tank liquid-level system shown in Figure 7.16, where the volume flow rate into the tank through a pipe is qi. The pump is connected to the bottom of the tank through a valve of linear resistance R1. The pressure of the fluid increases by Δp when crossing the pump. The liquid leaves the tank through a valve of linear resistance R2. Derive the differential equation relating the liquid height h and the volume flow rate qi at the inlet. The tank’s cross-sectional area A is constant. The density ρ of the liquid is constant.

Figure 7.16 Problem 4.

Solution Applying the law of conservation of mass to the tank gives dm = qmi − qmo dt For constant fluid density and constant cross-sectional area, we have dm d dh = = ( ρAh ) ρA dt dt dt The mass flow rate into the tank is p + ∆p − ( pa + ρgh ) ∆p − ρgh ρqi + a ρqi + qmi = = R1 R1 The mass flow rate out of the tank is pa + ρgh − pa ρgh = qmo = R2 R2 Substituting these expressions into the law of conservation of mass gives ∆p − ρgh ρgh dh − ρA = ρqi + dt R1 R2 Rearranging the equation gives dh  ρg ρg  ∆p ρA +  +  h =ρqi + dt  R1 R2  R1 5.

Consider the two-tank liquid-level system shown in Figure 7.17. The liquid is pumped into tanks 1 and 2 through valves of linear resistances R1 and R2, respectively. The pressure of the fluid increases by Δp when crossing the pump. The cross-sectional areas of the two tanks are A1 and A2, respectively. The liquid flows from tank 1 to tank 2 through a valve of linear resistance R3 and leaves tank 2 through a valve of linear resistance R4. The density ρ of the liquid is constant. Derive the differential equations in terms of the liquid heights h1 and h2. Write the equations in second-order matrix form.

Figure 7.17 Problem 5.

320

Solution Applying the law of conservation of mass to tank 1 gives dm = qmi − qmo dt where dh dm d = = ( ρA1h1 ) ρA1 1 dt dt dt pa + ∆p − pa ∆p = qmi = R1 R1

= qmo

( pa + ρgh1 ) − ( pa + ρgh2 )

= R3

ρgh1 − ρgh2 R3

Substituting these expressions results in

dh1 ρg ∆p ρg + h1 − h2 = dt R3 R3 R1 Apply the law of conservation of mass to tank 2, where dh dm d = = ( ρA2 h2 ) ρA2 2 dt dt dt ( pa + ∆p ) − pa ( pa + ρgh1 ) − ( pa + ρgh2 ) ∆p ρgh1 − ρgh2 qmi = + = + R2 R3 R2 R3 ρA1

qmo =

( pa + ρgh2 ) − pa

= R4

ρgh2 R4

Substituting these expressions results in

 ρg ρg  dh2 ρg ∆p − h1 +  +  h2 = dt R3 R R R2 4   3 The system differential equations in second-order matrix form are ρg   ∆p   dh1   ρg −   R3  h1   R1  ρA1 0   dt   R3   = +      0 ρA  dh  ρg ρg ρg   h2   ∆p  2  2  − +  dt   R3 R3 R4   R2  ρA2

6.

Figure 7.18 shows a liquid-level system, in which two tanks have cross-sectional areas A1 and A2, respectively. The volume flow rate into tank 1 is qi. A pump is connected to the bottom of tank 1 and the pressure of the fluid increases by Δp when crossing the pump. Tank 2 is located higher than tank 1 and the vertical distance between the two tanks is H. The liquid is pumped from tank 1 to tank 2 through a valve of linear resistance R1 and leaves tank 2 through a valve of linear resistance R2. The density ρ of the liquid is constant. Derive the differential equations in terms of the liquid heights h1 and h2. Write the equations in second-order matrix form.

Figure 7.18 Problem 6.

321

Solution Applying the law of conservation of mass to tank 1 gives dm = qmi − qmo dt where dh dm d = = ( ρA1h1 ) ρA1 1 dt dt dt qmi = ρqi

( pa + ρgh1 + ∆p ) − ( pa + ρgh2 + ρgH )

ρgh1 + ∆p − ρgh2 − ρgH = R1 R1 Substituting these expressions results in dh ρg ρg ∆p ρ g ρA1 1 + h1 − h2 = ρqi − + H dt R1 R1 R1 R1 Applying the law of conservation of mass to tank 2 gives dm = qmi − qmo dt where dh dm d = = ( ρA2 h2 ) ρA2 2 dt dt dt p + ρ gh + ∆ p − p + ρ gh + ρ gH ( a ) ( a ) ρgh1 + ∆p − ρgh2 − ρgH 1 2 = qmi = R1 R1 qmo =

= qmo

pa + ρgh2 ) − pa (= R2

ρgh2 R2

Substituting these expressions results in

ρA2

 ρg ρg  dh2 ρg ∆p ρ g − h1 +  +  h2 = − H dt R1 R1 R1  R1 R2 

The system differential equations in second-order matrix form are ρg  ∆p ρ g    dh1   ρg ρqi − − + H      R1 R1  h1   R1 R1  ρA1 0   dt    + =      0 ρA   dh   ρg ρg ρg   h2   ∆p ρg 2  2   H − + −  R1 R1   dt   R1 R1 R2  7.

Consider the single-tank liquid-level system shown in Figure 7.19, where the volume flow rate into the tank through a pipe is qi. The liquid leaves the tank through an orifice of area Ao. Denote Cd as the discharge coefficient, which is the ratio of the actual mass flow rate to the theoretical one, and lies in the range of 0 < Cd < 1 because of friction effects. Derive the differential equation relating the liquid height h and the volume flow rate qi at the inlet. The tank’s cross-sectional area is constant. The density ρ of the liquid is constant.

Figure 7.19 Problem 7.

322

Solution Applying the law of conservation of mass to the tank gives dm = qmi − qmo dt where dm d dh = = ( ρAh ) ρA dt dt dt qmi = ρqi

qmo = ρ(vAo )Cd v is used to denote the velocity of the liquid flowing through the orifice, and v = 2 gh according to the principle of conversation energy. Substituting these expressions results in dh A + 2 gh AoCd = qi dt 8.

Figure 7.20 shows a hydraulic system of two interconnected tanks, which have the same cross-sectional area of A. A pump is connected to tank 1. Assume that the relationship between the voltage applied to the pump and the mass flow rate into tank 1 is linear, i.e., qmi = kpva, where kp is called the pump constant and can be obtained by experimental measurements. Tank 1 is connected to tank 2, which is connected to a reservoir. The liquid leaves each tank through an outlet of area Ao at the bottom. Derive the differential equations in terms of the liquid heights h1 and h2.

Figure 7.20 Problem 8.

Solution Applying the law of conservation of mass to tank 1 gives dm = qmi − qmo dt where dh dm d = = ( ρAh1 ) ρA 1 dt dt dt qmi = k p va

qmo = ρ 2 gh1 Ao Substituting these expressions results in

323

dh1 + ρ 2 gh1 Ao = k p va dt Apply the law of conservation of mass to tank 2, where dh dm d = = ( ρAh2 ) ρA 2 dt dt dt qmi = ρ 2 gh1 Ao ρA

qmo = ρ 2 gh2 Ao Substituting these expressions results in

ρA

dh2 − ρ 2 gh1 Ao + ρ 2 gh2 Ao = 0 dt

Thus, the system differential equations are

ρA

ρA

dh1 + ρ 2 gh1 Ao = k p va dt

dh2 − ρ 2 gh1 Ao + ρ 2 gh2 Ao = 0 dt

Problem Set 7.3 1.

Consider heat transfer through an insulated frame wall of a house. The thermal conductivity of the wall is 0.055 W/(m·°C). The wall is 0.15 m thick and has an area of 15 m2. The inside air temperature is 20°C and the heat transfer coefficient for convection between the wall and the inside air is 2.6 W/(m·°C). On the outside of the wall, the heat transfer coefficient for convection between the wall and the outside air is 10.4 W/(m·°C) and the outside air temperature is –10°C. Determine the heat flow rate through the wall.

Solution The heat transfer through the wall can be represented using a thermal circuit with three thermal resistances connected in series. Two modes of heat transfer, conduction and convection, are involved. The corresponding thermal resistances are 1 1 = R1 = = 2.56 × 10−2  C·s/J h1 A 2.6 × 15 L 0.15 R= = = 1.82 × 10−1  C·s/J 2 kA 0.055 × 15 1 1 = = = 6.41 × 10−3  C·s/J R3 h2 A 10.4 × 15 The total thermal resistance is 3

R = eq

R ∑= i

2.14 × 10−1 ° C·s/J

i =1

Thus, the heat flow rate through the wall is

qh = 2.

20 − (−10) ∆T = = 140.19 W R eq 2.14 × 10−1

Consider heat transfer through a double-pane window as shown in Figure 7.31a. Two layers of glass with thermal conductivity k1 are separated by a layer of stagnant air with thermal conductivity k2. The inner surface of the window is at temperature T1 and exposed to room air with heat transfer coefficient h1. The outer surface of the wall is at temperature T2 and exposed to air with heat transfer coefficient h2. Assume that k1 = 0.95 W/(m·°C), k2 = 0.0285 W/(m·°C), h1 = h2 = 10 W/(m2·°C), T1 = 20°C, and T2 = 35°C. The thickness of each glass layer is 4 mm, the thickness of the air layer is 8 mm, and the cross-sectional area of the window is 1.5 m2. a. Determine the heat flow rate through the double-pane window. b. Determine the temperature distribution through the double-pane window. c. Repeat Parts (a) and (b) for the single-pane glass window shown in Figure 7.31b.

324

Figure 7.31 Problem 2.

Solution a.

The heat transfer through the double-pane glass window can be represented using a thermal circuit with five thermal resistances connected in series. From right to left, the corresponding thermal resistances are 1 1 R1 = = = 6.67 × 10−2 °C⋅s/J (outer air) h2 A 10 × 1.5 L1 0.004 R= R= = = 2.81 × 10−3 °C⋅s/J (glass) 2 4 k1 A 0.95 × 1.5 L2 0.008 R3 = = = 1.87 × 10−1 °C⋅s/J (stagnant air) k2 A 0.0285 × 1.5 1 1 R5 = = = 6.67 × 10−2 °C⋅s/J (room air) h1 A 10 × 1.5

= Req

5

R ∑= i =1

i

3.26 × 10−1 °C⋅s/J

Thus, the heat flow rate through the double-pane glass window is ∆T 35 − 20 = qh = = 46.01 W Req 3.26 × 10−1 b.

From right to left, the heat flow rate through each layer is Outer air: = qh h2 A ( T1 − T3 )

k1 A (T3 − T4 ) L1 k2 A Stagnant air: = qh (T4 − T5 ) L2 k1 A qh = Glass: (T5 − T6 ) L1 = qh Glass:

Room air: = qh h1 A ( T6 − T2 )

With the given temperatures T1 = 35°C and T2 = 20°C, we have T3 = 31.93°C, T4 = 31.80°C, T5 = 23.20°C, and T6 = 23.07°C. The temperature distribution through the double-pane glass window is shown below.

c.

Figure PS7-3 No2a For the single-pane glass window, the corresponding thermal resistances from right to left are 1 1 = R1 = = 6.67 × 10−2 °C⋅s/J (outer air) h2 A 10 × 1.5

325

°C⋅s/J (glass) °C⋅s/J (room air) °C⋅s/J Thus, the heat flow rate through the single-pane glass window is W The heat flow rate through each layer is Outer air: Glass: Room air: With the given temperatures T1 = 35°C and T2 = 20°C, we have T3 = 27.80°C and T4 = 27.19°C. The temperature distribution through the single-pane glass window is shown below.

Figure PS7-3 No2b 3.

The junction of a thermocouple can be approximated as a sphere with a diameter of 1 mm. As shown in Figure 7.32, the thermocouple is used to measure the temperature of a gas stream. For the junction, the density is ρ = 8500 kg/m3, the specific heat capacity is c = 320 J/(kg·°C), and the thermal conductivity is k = 40 W/(m·°C). The temperature of the gas Tf is 120°C and the initial temperature of the sphere T0 is 25°C. The heat transfer coefficient between the gas and the junction is h = 70 W/(m2·°C).

a. b. c.

Figure 7.32 Problem 3. Determine if the junction’s temperature can be considered uniform. Derive the differential equation relating the junction’s temperature T(t) and the gas’s temperature Tf. Using the differential equation obtained in Part (b), construct a Simulink block diagram to find out how long it will take the thermocouple to read 99% of the initial temperature difference.

Solution a.

The characteristic length of the junction is

The Biot number of the junction is

Thus, the junction can be treated as a lump-temperature system, and its temperature can be considered uniform.

326

b.

Applying the law of conservation of energy to the junction gives

where

The thermal capacitance of the junction is J/°C and the thermal resistance due to convection is °C⋅s/J Thus, the dynamic model of the junction’s temperature is or

c.

where T(0) = 25°C and Tf =120°C. The Simulink block diagram is shown in Plot (a).

Figure PS7-3 No3a Based on the response shown in Plot (b), we can estimate the time it will take the thermocouple to read 99% of the initial temperature difference. As seen from the plot, it takes about 30 seconds for the thermocouple to read 119.05°C (= 0.99× (120−25)+25).

Figure PS7-3 No3b 4.

Figure 7.33 shows a thin-walled glass of milk, which is taken out of the refrigerator at a uniform temperature of 3°C and is placed in a large pan filled with hot water at 60°C. Assume that the assumption of the lumped system analysis is applicable because the milk is stirred constantly, so that its temperature is uniform at all times. The glass container is cylindrical in shape with a radius of 3 cm and a height of 6 cm. The estimated parameters of the

327

milk are density ρ = 1035 kg/m3, specific heat capacity c = 3980 J/(kg·°C), and thermal conductivity k = 0.56 W/(m·°C). The heat transfer coefficient between the water and the glass is h = 250 W/(m2·°C). a. Derive the differential equation relating the milk’s temperature T(t) and the water temperature. b.

Using the differential equation obtained in Part (a), construct a Simulink block diagram. How long will it take for the milk to warm up from 3°C to 58°C?

Figure 7.33 Problem 4.

Solution a.

Applying the law of conservation of energy to the glass of milk gives

where

For the glass of milk, we have m3 m2 The thermal capacitance of the glass of milk is J/°C and the thermal resistance due to convection is °C⋅s/J Thus, the dynamic model of the chicken’s temperature is or where T(0) = 3°C and Tf = 60°C. The Simulink block diagram is similar to the one shown in Problem 3. The response of the temperature b. of the glass of milk is shown in Figure PS7-3 No4 on the next page. As seen from the plot, it takes 563 seconds for the milk to warm up from 3 °C to 58 °C. 5.

The room shown in Figure 7.34 has a heater with heat flow rate input of q0. The thermal capacitances of the heater and the room air are C1 and C2, respectively. The thermal resistances of the heater–air interface and the room wall–ambient air interface are R1 and R2, respectively. The temperatures of the heater and the room air are T1 and T2, respectively. The temperature outside the room is T0, which is assumed to be constant. a. Derive the differential equations relating the temperatures T1, T2, the input q0, and the outside temperature T0. b. Using the differential equations obtained in Part (a), determine the state-space form of the system. Assume the temperatures T1 and T2 as the outputs.

328

70 o

58 C 60

o

C)

50

Temperature (

40

30

20

10 563 seconds 0 0

200

400

600

800

Time (s)

Figure PS7-3 No4

Figure 7.34 Problem 5.

Solution a.

Applying the law of conservation of energy to the heater, we have dU = qhi − qho dt where dT dU = C1 1 dt dt qhi = q0

qho =

T1 − T2 R1

Substituting these expressions gives

C1

dT1 T −T = q0 − 1 2 dt R1

dT1 1 1 + T1 − T2 = q0 dt R1 R1 Applying the law of conservation of energy to the room air, we have dU = qhi − qho dt where dT dU = C2 2 dt dt C1

1000

329

qhi =

T1 − T2 R1

qho =

T2 − T0 R2

Substituting these expressions gives

dT2 T1 − T2 T2 − T0 − C= 2 R1 R2 dt  1 dT2 1 1  1 − T1 +  +  T2 =T0 R1 R R R dt 2  2  1 The system of differential equations can be written in second-order matrix form as follows: 1   dT1   1 −  q0   R    R C1 0   dt   1 1     T1  = +      1   0 C     T T d T 1 1 1  2  2 2 0  −  +   R2   dt   R1  R1 R2   To represent a thermal system in the state-space form, the temperature of each thermal capacitance is often chosen as a state variable. As specified, the state, the input, and the output are q0   x1   T1   T1  x = u   ,= y  . =    ,= T x T  2  2 T2   0 C2

b.

The state-variable equations are

dT1 1 1 1 x1 = = − T1 + T2 + q0 dt R1C1 R1C1 C1  1 dT 1 1  1 x2 = 2 = T1 −  + T0  T2 + dt R1C2 R2C2  R1C2 R2C2  or in matrix form

1  − R C  x1   1 1 =     x2   1  R1C2

 1  C x    1 + 1    1 1    x2   − +  0   R1C2 R2 C2   1 R1C1

 0  u   1  1  u2   R2 C2 

The output equation is

 y1  1 0   x1  0 0   u1  =     +     y2  0 1   x2  0 0  u2  6.

For the three-room house shown in Figure 7.35, all rooms are perfectly square and have the same dimensions. An air conditioner produces an equal amount of heat flow qho out of each room. The temperature outside the house is To. Assume that there is no heat flow through the floors or ceilings. The thermal resistances through the inner walls and the outer walls are Ri and Ro, respectively. The thermal capacitance of each room is C. Derive the differential equations for this system.

Figure 7.35 Problem 6.

330

Solution Due to symmetry, the temperatures in the rooms on the left and on the right are the same, and they are assumed to be T1. Assume that the temperature in the room in the middle is T2. It is obvious that T1 > T2 since there are heat flow into the room through the walls on the three sides instead of on the two sides.

Figure PS7-3 No6 Applying the law of conservation of energy to the room either on the left or the right, we have

where

Substituting these expressions gives

Applying the law of conservation of energy to room in the middle, we have

where

Substituting these expressions gives

Problem Set 7.4 1.

Dry air at a constant temperature of 25°C passes through a valve out of a rigid cubic container of 1.5 m on each side (see Figure 7.42). The pressure po at the outlet of the valve is constant, and it is less than p. The valve resistance is approximately linear, and R = 1000 Pa·s/kg. Assume the process is isothermal. a. Develop a mathematical model of the pressure p in the container. b.

Construct a Simulink block diagram to find the output p(t) of the pneumatic system if the pressure inside the container initially is 2 atm and the pressure at the outlet is 1 atm.

331

Figure 7.42 Problem 1.

Solution a.

Applying the law of conservation of mass gives

where

The mathematical model of the pressure

in the container is

For an isothermal process, the pneumatic capacitance of the container is kg⋅m2/N b.

The Simulink block diagram is constructed based on the above differential equation, where kPa, and kg⋅m2/N. Double-click the Integrator block to set Pa⋅s/kg , the initial pressure to be 202650 Pa (2 atm). The response of the system is shown in the Plot (b).

Figure PS7-4 No1a

Figure PS7-4 No1b

332

2.

Figure 7.43 shows a liquid-level system in which two tanks have hydraulic capacitances C1 and C2, respectively. The volume flow rate into tank 1 is qi. The liquid flows from tank 1 to tank 2 through a valve of linear resistance R1 and leaves tank 2 through a valve of linear resistance R2. The density ρ of the liquid is constant. a. Derive the differential equations in terms of the liquid heights h1 and h2. Write the equations in second-order matrix form. b. Assume that the volume flow rate qi is the input and the liquid heights h1 and h2 are the outputs. Determine the state-space form of the system. c.

Construct a Simulink block diagram to find the outputs h1(t) and h2(t) of the liquid-level system. Assume ρ = 1000 kg/m3, g = 9.81 m/s2, C1 = 0.15 kg·m2/N, C2 = 0.25 kg·m2/N, R1 = R2 = 100 N·s/(kg·m2), and initial liquid heights h1(0) = 1 m and h2(0) = 0 m. The volume flow rate qi is a step function with a magnitude of 0 before t = 0 s and a magnitude of 0.25 m3 /s after t = 0 s.

Figure 7.43 Problem 2.

Solution a.

Applying the law of conservation of mass to tank 1 gives

where

Substituting these expressions results in

Applying the law of conservation of mass to tank 2 gives

where

Substituting these expressions results in

The system differential equations in second-order matrix form is

333

b.

The state vector, the input, and the output vector are

The state-variable equations are

or in matrix form

The output equation is

c.

The Simulink block diagram constructed based on the differential equations is given below. The response is shown in Plot(b).

Figure PS7-4 No2a

334

Figure PS7-4 No2b 3.

A chicken is taken out of the oven at a uniform temperature of 150°C and is left out in the open air at the room temperature of 25°C. Assume that the chicken can be approximated as a lumped model. The estimated parameters are mass m = 1.75 kg, heat transfer surface area A = 0.3 m2, specific heat capacity c = 3220 J/(kg·°C), and heat transfer coefficient h = 15 W/(m2·°C). a. Derive the differential equation relating the chicken’s temperature T(t) and the room temperature. b.

Using the differential equation obtained in Part (a), construct a Simulink block diagram and find the temperature of the chicken.

c. d.

Build a Simscape model of the system and find the temperature of the chicken. Assume that the chicken can be served only if its temperature is higher than 80°C. Based on the simulation results obtained in Parts (b) and (c), can the chicken be left at the room temperature of 25°C for 1 hour?

Solution a.

Applying the law of conservation of energy to the chicken gives

where

The thermal capacitance of the chicken is J/°C and the thermal resistance due to convection is °C⋅s/J Thus, the dynamic model of the chicken’s temperature is

where b.

°C and

°C.

The Simulink block diagram constructed based on the differential equation is similar to the one given in Problem Set 7.3, Problem 3.

335

c.

d.

The Simscape block diagram is given below.

Figure PS7-4 No3a The response of the temperature of the chicken is shown below. As seen from the plot, the chicken has a temperature below 80°C after it is left at the root temperature of 25°C for one hour. Thus, the chicken cannot be served.

Figure PS7-4 No3b

Review Problems 1.

Dry air at a constant temperature of T passes through a valve into a rigid spherical container (see Figure 7.44). The pressure at the inlet is pi and the one at the outlet is pa. The linear resistances of the two valves at the inlet and the outlet are R1 and R2, respectively. Assume that the process is isothermal. a. Develop a mathematical model of the pressure p in the container. b. Denote the volume flow rate at the outlet as qo. Determine the transfer function relating pi and p for this pneumatic system if qo = 0.

Figure 7.44 Problem 1.

336

Solution Applying the law of conservation of mass gives

dm = qmi − qmo dt dp pi − p p − pa = − C dt R1 R2 For an isothermal process, the pneumatic capacitance of the spherical container is V 4πr 3 C = = Rair T 3Rair T The mathematical model of the pressure p in the container is 4πr 3 dp  1 1  1 1 + +  p = pi + pa 3Rair T dt  R1 R2  R1 R2 2.

A single-tank liquid-level system is shown in Figure 7.45, in which water flows into the tank at a volume flow rate qi and out of the tank through two valves at points 1 and 2. The linear resistances of the two valves are R1 and R2, respectively. Assuming h > h1, derive the differential equation relating the liquid height h and the volume flow rate qi at the inlet. The cross-sectional area of the tank A is constant. The density ρ of the liquid is constant.

Figure 7.45 Problem 2.

Solution Applying the law of conservation of mass to the tank gives dm = qmi − qmo dt where dm d dh = = ( ρAh ) ρA dt dt dt qmi = ρqi

pa + ρg ( h − h1 ) − pa pa + ρg ( h − h2 ) − pa ρg ( h − h1 ) ρg ( h − h2 ) qmo = + = + R1 R2 R1 R2 Substituting these expressions into the law of conservation of mass gives ρgh1 ρgh2 dh  ρg ρg  ρA +  +  h =ρqi + + dt  R1 R2  R1 R2

3.

A watermelon is taken out of the refrigerator at a uniform temperature of 3°C and is exposed to 32°C air. Assume that the watermelon can be approximated as a sphere and the temperature of the watermelon is uniform. The estimated parameters are the density ρ = 120 kg/m3, diameter D = 35 cm, specific heat capacity c = 4200 J/(kg·°C), and heat transfer coefficient h = 15 W/(m2·°C). a. Derive the differential equation relating the watermelon’s temperature T(t) and the air temperature.

337

b.

Using the differential equation obtained in Part (a), construct a Simulink block diagram and find the temperature of the watermelon.

c. d.

Build a Simscape model of the system. Based on the simulation results obtained in Parts (b) and (c), how long will it take before the watermelon is warmed up to 20°C?

Solution a.

Applying the law of conservation of energy to the watermelon gives

where

The thermal capacitance of the watermelon is J/°C and the thermal resistance due to convection is °C⋅s/J Thus, the dynamic model of the watermelon’s temperature is or

a.

b.

where T(0) = 3°C and Tf = 32°C. The Simulink block diagram is given below.

Figure Review7 No3a The Simscape block diagram is given below.

Figure Review7 No3b Running either of the block diagrams yields the same response curve as shown in the figure below.

338

c. 4.

Figure Review7 No3c It will take about 1741 seconds (or 29 minutes) before the watermelon is warmed up to 20°C. Case Study Figure 7.46 represents the temperature dynamics of two adjacent objects, where the thermal capacitances of the objects are C1 and C2, respectively. Assume that the temperatures of both objects are uniform, and they are T1 and T2, respectively. The heat flow rate into object 1 is q0, and the temperature surrounding object 2 is T0. There are two modes of heat transfer involved, conduction between the objects and convection between object 2 and the air. The corresponding thermal resistances are R1 and R2, respectively. a. Derive the differential equations relating the temperatures T1, T2, the input q0, and the outside temperature T0. b.

Build a Simscape model of the physical system and find the temperature outputs T1(t) and T2(t). Use default values for the blocks of Thermal Mass (mass = 1 kg, specific heat = 447 J⋅K/kg, and initial temperature = 300 K), Conductive Heat Transfer (area = 1×10−4 m2, thickness = 0.1 m, and thermal conductivity = 401 W/(m⋅K)), and Convective Heat Transfer (area = 1×10−4 m2 and heat transfer coefficient = 20 W/(m2⋅K). Assume that the heat flow rate is q0 = 400 J/s and the surrounding temperature is T0 = 298 K.

c.

Build a Simulink block diagram based on the differential equaitons obtained in Part (a) and find the temperature outputs T1(t) and T2(t).

Figure 7.46 Problem 4.

Solution a.

The differential equations relating the temperatures T1, T2, the input q0, and the outside temperature T0 are

The equations can be rearranged as

339

b.

The Simscape block diagram corresponding to the physical system is shown in the figure below. Two Thermal Mass blocks are used to represent objects 1 and 2. A Conductive Heat Transfer block and a Convective Heat Transfer block are used to represent the two modes of heat transfer involved, conduction between the objects and convection between object 2 and the air. An Ideal Heat Flow Source block is included to represent the heat flow rate input q0, and an Ideal Temperature Source block is used to represent the surrounding temperature T0.

Figure Review7 No4a c.

The Simulink block diagram built based on the differential equaitons obtained in Part (a) is shown the figure below. Based on the default values used in Simscape modeling, the following system parameters can be determined,

Figure Review7 No4b Run either simulation to generate the figure below, showing the resulting temperature outputs T1(t) and T2(t) of two adjacent objects.

340

700 T 1

650

T 2

600

Temperature (K)

550

500

450

400

350

300 0

100

200

300

Time (s)

Figure Review7 No4c

400

500

341

Problem Set 8.2 1. Find the transient and steady-state responses of a system modeled as  (0) 1 2  x+ = 3 x e −t , x= (0) 0 , x=

Solution Characteristic equation 2λ 2 + 3λ = 0 yields λ= 0, − 32 , hence xc (t )= A + Be−3t /2 . The particular solution is picked as x p (t ) = Ke−t . Substitution into the ODE gives K = −1 so that x p (t ) = −e−t . A general solution is then x(t ) = A + Be −3t /2 − e −t . Applying initial conditions, we find A = 1 and

B = 0 so that x(t ) = 1 − e−t . Therefore, xtr = −e−t and xss = 1 . Consider a first-order system with time constant τ and zero initial condition. Find the system’s unit-impulse response for τ = 14 and τ = 34 , plot the two curves versus 0 ≤ t ≤ 2 in the same graph, and comment. Solution When τ = 14 , the unit-impulse response is

2.

x1 (t ) = 4e −4t

When τ = 34 , it is x2 (t ) = 43 e −4t /3 >> t = linspace(0,2); >> x1 = 4.*exp(-4.*t); x2 = (4/3).*exp(-4.*t/3); >> plot(t,x1,t,x2) % Figure PS8-2No2

As expected, the response x1 associated with the smaller time constant τ = state faster.

1 4

reaches steady-

4

3.5

Unit-impulse response

3

x1

2.5

2

1.5

x2

1

0.5

0

0

0.2

0.4

0.6

0.8

1 t

1.2

Figure PS8-2No2

1.4

1.6

1.8

2

342

Consider a first-order system with time constant τ and zero initial condition. Find the system’s unit-step response for τ = 13 and τ = 23 , plot the two curves versus 0 ≤ t ≤ 2 in the same graph, and comment. Solution When τ = 13 , the unit-step response is

3.

x1 (t ) = 1 − e −3t

When τ = 23 , it is x2 (t ) = 1 − e −3t /2 >> t = linspace(0,2); >> x1 = 1.-exp(-3.*t); x2 = 1.-exp(-3.*t/2); >> plot(t,x1,t,x2) % Figure PS8-2No3

As expected, the response x1 associated with the smaller time constant τ = 13 reaches steadystate faster. 1 0.9 0.8

Unit-step response

0.7

x1

0.6 0.5 0.4 0.3 x2

0.2 0.1 0

0

0.2

0.4

0.6

0.8

1 t

1.2

1.4

1.6

1.8

2

Figure PS8-2No3 4. The temperature of a steel sphere with initial temperature T0 submerged in water with temperature Ti is governed by the initial-value problem  + T T ,= RCT = T (0) T0 i

where R is the thermal resistance due to convection and C is the thermal capacitance of the sphere. Find the temperature of the sphere at any time, as well as its steady-state value. Solution The time constant is τ = RC . The input Ti is a constant, thus modeled as a step with magnitude Ti . The response is then determined as

343

T= (t ) T0e−t /( RC ) + Ti 1 − e−t /( RC )  The steady-state value of the steel temperature is Tss = Ti , as expected.

5. Repeat Problem 4 for the case when the temperature of the water tank increases linearly with time at a rate of r . Solution (a) t T + T = rt . The input is a ramp function with magnitude r , hence the response is

(

)

) e −t /tt T (t= Ta + r t − t 1 − e−t /   

)

(

The zero-state response is r t − t 1 − e−t /t  , while the zero-input response is e −t /t Ta .   (b) The steady-state temperature is r (t − t ) .

6. A single-tank liquid-level system with inflow rate qi as its input and liquid level h as its  + gh Rq (t= output is modeled as RAh = ), h(0) 0 , where R, A, g = const . If the inflow rate i

is a unit-step, find the liquid level in terms of the physical parameters. Also find the steadystate value of the liquid level. Solution Rewriting the model in standard form, we find RA  R = h+h q= i (t ), h(0) 0 g g so that the time constant is τ = RA / g . Since qi is the unit-step, the input in Eq. (a) is a step function with magnitude R / g . Therefore, the response is h= (t ) The steady-state response is hss = R / g .

(

R 1 − e − gt / RA g

)

(a)

344 7. The equation of motion of the mechanical system in Figure 8.4 is by + k ( y − x) = 0 where x and y are the input and the output, respectively, and b, k = const . Assuming zero initial condition, find the response when x is a (a) unit-impulse, (b) unit-step. x k b

y

Figure 8.4 Problem 7. Solution (a) Rewriting the model in standard form, we find b y = + y x, y= (0) 0 k

so that the time constant is τ = b / k . When x is the unit-impulse, we have y (t ) = bk e− kt / b . (b) When x is the unit-step, the response is y (t ) = 1 − e − kt / b .

8. The equation of motion for the torsional mechanical system in Figure 8.5 is derived as Jθ + Bθ = T (t ) , where J , B = const , θ is the angular displacement, and T is a constant applied torque. Express the model as first-order in angular velocity ω = θ . Assuming ω (0) = ω0 , find ω (t ) . Also identify the transient and steady-state responses. Solution J T T , and in standard form ω + ω = so that the time The model is expressed as J ω + Bω = B B constant is τ = J / B . Since T = const , the input is interpreted as a step function with magnitude T / B , and T (t ) e − Bt / J ω0 + 1 − e − Bt / J ω= B

(

Therefore,= ωtr (t ) e− Bt / J ω0 −

T − Bt / J T and ωss (t ) = . e B B

)

345

θ

B

J

T

Figure 8.5 Problem 8. 9. The governing equation for an RC circuit driven by an applied voltage

v(t )

is derived as

 + V v(t )= RCV = , V (0) V0

where V is the voltage across the capacitor and V0 is the capacitor voltage at the initial time. Assume that the applied voltage is a ramp with slope of 12 . Find the steady-state error between the input and the system response. Solution The time constant is τ = RC and the input is v(t ) = 12 ur (t ) . The response is obtained as

(

)

= V (t ) V0e−t /( RC ) + 12 t − RC 1 − e −t /( RC )   

Therefore, V= ss

1 (t 2

− RC ) and ess = 12 RC .

10. Find the unit-ramp response of the RL circuit in Example 8.3. Solution L di 1 The governing equation of the circuit is + i = va (t ) where va = ur . The time constant is R dt R τ = L / R and the input is interpreted as a ramp function with slope 1/ R . As a result, we find

i (t ) =

1 L − ( R / L )t   t − 1 − e  R R 

11. A first-order dynamic system is modeled as = y + 3 y

f (t ) = , y (0) 1

Assuming the input f (t ) is a step function with magnitude 0.8 , find yss . Solution 1 f (t ) so that τ = 1 and the input is interpreted as a step of In standard form, we have 13 y + y = 3 3

346 magnitude

0.8 3

. The response is

(

y (t ) =e −3t + 0.8 1 − e −3t 3 The steady-state response is yss =

0.8 3

)

.

12. A first-order dynamic system is modeled as 1 2

 + 5w g (t ) = w = , w(0)

2 3

If the input g (t ) is a ramp function with a slope of 52 , find the steady-state response and the steady-state error. Solution 1 g (t ) so that τ = 1 and the input is interpreted as a ramp In standard form, we have 101 w + w = 10 5 of slope

1 2

. The response is ) w(t=

The steady-state response is

1 (t 2

2 e −10t 3

(

)

1 1 − e −10t  + 12 t − 10  

1 ) , while the steady-state error is e = 1 ⋅ 1 = 1 . − 10 ss 2 10 20

Problem Set 8.3 1. Show that the free response of an overdamped, second-order system has a steady-state value of zero. Solution In this case, the two poles are given by s1 = −ζωn + ωn ζ 2 − 1 , s2 = −ζωn − ωn ζ 2 − 1 . Noting that ζ 2 − 1 < ζ , we conclude that s1 < 0 . It is also clear that s2 < 0 . Therefore, the two terms in Eq. (8.15) contain exponential decays and x(t ) → 0 as t → ∞ .

In Problems 2 through 5, for each given system model, (a) Identify the damping type and find the free response. (b) Plot the free response using the initial command. 2. 3 x+2 = x + x 0 ,= x(0) 0 ,= x (0)

1 3

347 Solution (a) In standard form,  x + 23 x + 13 x = 0 implies ωn2 = 13 and 2ζωn = 23 . Therefore, ωn = and ζ =

1 3

, hence the system is underdamped; ωd = ωn 1 − ζ 2 = x(t ) =

1 2

e −t /3 sin(t

2 3

1 3

r/s

r/s . Free response is

2 ) 3

(b) >> A=[0 1;-1/3 -2/3]; C=[1 0]; sys=ss(A,[],C,[]); >> x0=[0;1/3]; >> initial(sys,x0) % Figure PS8-3No2 Response to Initial Conditions 0.3

0.25

Amplitude

0.2

0.15

0.1

0.05

0

-0.05

0

5

10

15

20

25

Time (seconds)

Figure PS8-3No2

3.  x + 3 x= + 4x 0 , = x(0) 52 , = x (0) 0 Solution (a) Comparing with the standard form, we find ωn2 = 4 and 2ζωn = 3 . Therefore, ωn = 2 r/s and ζ = 34 , hence the system is underdamped; ωd = ωn 1 − ζ 2 = by = x(t ) e −3t /2  52 cos 

7 2

t+

6 5 7

(b) >> A=[0 1;-4 -3]; C=[1 0]; sys=ss(A,[],C,[]); >> x0=[2/5;0]; >> initial(sys,x0) % Figure PS8-3No3

sin

7 2

7 2

t 

r/s . Free response is given

348

Response to Initial Conditions

0.4

0.35

0.3

0.25

Amplitude

0.2

0.15

0.1

0.05

0

-0.05

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

Time (seconds)

Figure PS8-3No3 4. 4  x + 8 x + 3 x = 0 , x(0) = 1 , x (0) = −1 Solution (a) Comparing  x + 2 x + 34 x = 0 with the standard form, we find ωn2 = Therefore, ωn =

3 2

r/s and ζ =

Therefore x(t ) = − 52 e −t /2 + 12 e

2 3 −3t /2

3 4

and 2ζωn = 2 .

, hence the system is overdamped. Then s1,2 = − 12 , − 32 .

.

(b) >> A=[0 1;-3/4 -2]; C=[1 0]; sys=ss(A,[],C,[]);x0=[1;-1]; >> initial(sys,x0) % Figure PS8-3No4

Response to Initial Conditions 1

0.9

0.8

0.7

Amplitude

0.6

0.5

0.4

0.3

0.2

0.1

0

0

1

2

3

4

5 Time (seconds)

Figure PS8-3No4

6

7

8

9

10

349

5. 4  x+4 = x + x 0 ,= x(0) 13 ,= x (0) 1 Solution (a) Comparing  x + x + 14 x = 0 with the standard form, we find ωn2 =

1 4

and 2ζωn = 1 . Therefore,

ωn = 12 r/s and ζ = 1 , hence the system is critically damped. Therefore = x(t ) e −t /2  13 + 76 t 

(b) >> A=[0 1;-1/4 -1]; C=[1 0]; sys=ss(A,[],C,[]); >> x0=[1/3;1]; >> initial(sys,x0) % Figure PS8-3No5

Response to Initial Conditions 1

0.9

0.8

0.7

Amplitude

0.6

0.5

0.4

0.3

0.2

0.1

0

0

2

4

6

10

8

12

14

16

18

Time (seconds)

Figure PS8-3No5 6.

A dynamic system is modeled as 4  x + cx= + 5x 0 , = x(0) 1 , = x (0) 0

Plot (in the same graph) the system’s free response associated with c = 4,5 , and comment. Solution >> >> >> >> >> >>

A=@(c)([0 1;-5/4 -c/4]); % Define A as a function of c C=[1 0]; x0=[1;0]; sys1=ss(A(4),[],C,[]); sys2=ss(A(5),[],C,[]); initial(sys1,x0) % Figure PS8-3No6 hold on initial(sys2,x0)

350

It is readily observed that the response associated with c = 5 stabilizes faster due to increased damping. Response to Initial Conditions

1

0.8

0.6

Amplitude

0.4

0.2 c=5 0

c=4

-0.2

-0.4

0

2

4

6

8

10

12

Time (seconds)

Figure PS8-3No6 7. Show that the impulse response of an underdamped, second-order system has a steady-state value of zero. Solution The impulse response is given as   ζω x + x + A = x(t ) e −ζωnt  x0 cos ωd t + n 0 0 sin ωd t  ωd  

The trigonometric functions cos ωd t and sin ωd t oscillate between −1 and 1, and their coefficients are finite. Since ζωn > 0 , the exponential decay term will dominate the response for large t , causing it to approach zero as t → ∞ .

8. Show that the response of an underdamped, second-order system to an impulsive input ) , a const > 0 , and zero initial conditions, is described by δ (t − a=

 1 −ζωn (t −a )  sin ωd (t − a )  u (t − a )  e  ωd  where u (t ) is the unit-step function. Hint: Use shift on t − axis, Section 2.3. Use this result to find the response x(t ) of

351  x + x + 0.89x= δ (t − 1) , x(0)= 0 , x (0)= 0

Solution Noting L {δ (t − a )} = e − as , the response under consideration is provided by  1 e − as  −1  −1 − as G ( s ) = , x(t ) L= e G s = L ( )   2 2 ( s + ζωn ) 2 + ωd2  ( s + ζωn ) + ωd 

{

}

Then, by Eq. (2.18) in Section 2.3,

{

}

L −1 e− as G ( s ) =g (t − a )u (t − a ) where g (t ) = L −1{G ( s )} is given by Eq. (8.18) with A = 1 and zero initial conditions, as g (t ) =

1

ωd

e −ζωnt sin ωd t

Finally,

 1  x(t ) = g (t − a )u (t − a ) =  e−ζωn (t −a ) sin ωd (t − a)  u (t − a)  ωd  Comparing the model with standard form, we have ωn2 = 0.89 and 2ζωn = 1 so that

ωn = 0.9434 r/s , ζ = 0.53 , and ωd = ωn 1 − ζ 2 = 0.8 r/s . The response is then obtained as 1.25e − (t −1)/2 sin(0.8(t − 1))  u (t − 1)  

In Problems 9 and 10, assuming zero initial conditions, (a) Find the response x(t ) in closed form. (b) (c)

Plot the response using the impulse command. Plot the response through the simulation of the Simulink model of the system.

9.  x + 4 x + x = 5δ (t ) Solution (a) ωn2 = 1 and 2ζωn = 4 so that ωn = 1 r/s , ζ = 2 . The poles are s1,2 =−2 ± 3 and the (overdamped system) impulse response is = x(t )

5 2 3

e( −2+ 

3)t

− e( −2−

3)t 



352 (b)

With state variables x1 = x , x2 = x , the state-space matrices are found as

0 1 0  = = , B = 0] , D 0 A    , C [1 =  −1 −4  5  >> A=[0 1;-1 -4]; B=[0;5]; C=[1 0]; >> sys=ss(A,B,C,[]); >> impulse(sys) % Figure PS8-3No9 Impulse Response 1.4

1.2

Amplitude

1

0.8

0.6

0.4

0.2

0

0

5

10

15

20

25

Time (seconds)

Figure PS8-3No9 (c) The model is built as in Figure 8.15 with the state-space matrices defined in (b) and initial conditions specified as matrix B . The impulse response agrees with that in Figure PS83No9.

10. 2  x + 2 x + 3 x = 0.6δ (t ) Solution 3 (a) In standard form, we have  x + x + 32 x = δ (t ) . Therefore, ωn2 = 2

ωn =

3 2

r/s , ζ =

1 6

, and ωd = ωn 1 − ζ 2 = x(t ) =

(b)

5 2

3 2

and 2ζωn = 1 so that

r/s . The (underdamped system) response is

0.6 −t /2 e sin 5

5 2

t

With state variables x1 = x , x2 = x , the state-space matrices are found as

353

 0 1 0 = A  3 = , B = 0] , D 0   , C [1= 0.3  − 2 −1 >> A=[0 1;-3/2 -1]; B=[0;0.3]; C=[1 0]; >> sys=ss(A,B,C,[]); >> impulse(sys) % Figure PS8-3No10

Impulse Response 0.16

0.14

0.12

0.1

Amplitude

0.08

0.06

0.04

0.02

0

-0.02

-0.04

0

2

4

6

8

10

12

14

Time (seconds)

Figure PS8-3No10 (c) The model is built as in Figure 8.15 with the state-space matrices defined in (b) and initial conditions specified as matrix B . The impulse response agrees with that in Figure PS83No10. 11. The governing equation for an RLC circuit driven by the applied voltage va (t ) is derived as L

di 1 + Ri + ∫ i dt = va (t ) dt C

where L = 4 H , R= 4 Ω and C = 12 F . (a) Write the governing equation in terms of the electric charge q . (b)

Assuming zero initial conditions, plot q and i versus t (same figure) when the applied voltage va is a unit-impulse.

Solution (a) Using i = q , the governing equation is expressed as

354 1 Lq + Rq= + q va (t ) C

(b)

⇒

and rewrite

 + 12 q q + q=

1 4

va

We first obtain the state-space form as

 0 =  x  1  − 2   y = 1 0   

>> >> >> >>

Substitute

1 0  x + 1u  −1 4 0 x 1 

q  q  x = where=    , u = δ (t ) q   i 

A = [0 1;-1/2 -1]; B = [0;1/4]; C = eye(2); sys = ss(A,B,C,[]); [y,t] = impulse(sys); plot(t,y) % Figure PS8-3No11

0.25

0.2

Impulse response

0.15

0.1 Electric charge 0.05

0 Current -0.05

-0.1

0

2

4

8

6

10

12

14

t

Figure PS8-3No11

12. Consider the RLC circuit in Problem 11. (a) Write the governing equation in terms of the electric charge q . (b)

Assuming initial conditions are q (0) = 0 , q (0) = 1 , plot q and i versus t (same figure) when the applied voltage va is a unit-impulse.

355 Solution (a) Using i = q , the governing equation is expressed as 1 Lq + Rq= + q va (t ) C

(b)

⇒

and rewrite

 + 12 q q + q=

1 4

va

We first obtain the state-space form as

 0 =  x  1  − 2   y = 1 0   

1 0  x + 1u  −1 4

q  q (0)  0  = x = x(0) = where  , u δ (t ),=    q  q (0)  1 

0 x 1 

A = [0 1;-1/2 -1]; B = [0;1/4]; C = eye(2); x0=[0;1]; sys=ss(A,B,C,[]); [x_impulse,t]=impulse(sys); x_initial=initial(sys,x0,t); % Use the same t used for impulse x_total=x_impulse+x_initial; plot(t,x_total) % Figure PS8-3No12

1.4

1.2

1

0.8

Impulse response

>> >> >> >> >> >> >>

Substitute

0.6 Electric charge 0.4

0.2

0 Current -0.2

-0.4

0

2

4

8

6 t

Figure PS8-3No12

10

12

14

356 13.

Consider a single-degree-of-freedom mechanical system as in Figure 8.6 of this section, where = m 3= kg , k 20 N/m . Assuming zero initial conditions, plot (in a single figure) the response x to a unit-impulse force for two cases of c = 2 N-sec/m and c = 4 N-sec/m , and discuss the results. Solution >> >> >> >> >>

A=@(c)([0 1;-20/3 -c/3]); % Define A as a function of c B=[0;1/3];C=[1 0]; sys1=ss(A(2),B,C,[]); % System with c=2 sys2=ss(A(4),B,C,[]); % System with c=4 impulse(sys1,sys2) % Figure PS8-3No13

As expected, the system response associated with larger damping stabilizes at zero at a faster rate. Impulse Response 0.12

0.1

0.08

0.06 c=2

Amplitude

0.04

c=4

0.02

0

-0.02

-0.04

-0.06

-0.08

0

2

4

6

8

10

12

14

16

18

Time (seconds)

Figure PS8-3No13

14. Consider a single-degree-of-freedom mechanical system as in Figure 8.6, where the applied force is 10δ (t ) = and m 9= kg , c 6 N-sec/m, = k 5 N/m . Assume x(0) = 0.5 , x (0) = 0 . (a) Find the response x in closed form. (b) Plot the response x using the initial and impulse commands. Also plot the closedform response of (a) to confirm the result. Solution 10 (a) In standard form, we have  x + 23 x + 95 x = δ (t ) . Therefore, ωn2 = 95 and 2ζωn = 23 so that 9

ωn =

5 3

r/s , ζ=

1 5

< 1 , and ωd =

2 3

r/s . The (underdamped) system response is

= x(t ) e −t /3 0.5cos( 23 t ) + 11.5 sin( 32 t )  6

357

(b) >> >> >> >> >> >>

A = [0 1;-5/9 -2/3]; B = [0;10/9]; C = [1 0]; sys=ss(A,B,C,[]); x0=[0.5;0]; [x_impulse,t]=impulse(sys); % Impulse response [x_initial,t]=initial(sys,x0,t); % Response to ICs x_total=x_impulse+x_initial; % Total response plot(t,x_total) % Figure PS8-3No14

1.2

1

0.8

Response

0.6 0.5

0.4

0.2

0

-0.2

-0.4

0

2

4

6

8

10

12

14

16

18

t

Figure PS8-3No14 The following code produces exactly the same figure as above. x=@(t)(exp(-t/3)*(.5*cos(2*t/3)+(11.5/6)*sin(2*t/3))); ezplot(x,[0,18])

In Problems 15 and 16, u (t ) denotes the unit-step. Assuming zero initial conditions, (a) Find the response x(t ) in closed form. (b)

Plot the response using the step command.

15. 9  x + 12 x + 13 x = u (t ) Solution 1 u (t ) . Therefore, (a) In standard form, we have  x + 43 x + 13 x= ωn2 = 139 and 2ζωn = 9 9

ωn =

13 3

r/s , = ζ

2 13

< 1 , and ωd = 1 r/s . The (underdamped) system response is 1 1 − e −2t /3 cos t + 2 sin t  x(t ) = ( ) 13  3

4 3

so that

358

(b)

The transfer function

1 can be used to generate the step response: 9 s + 12 s + 13 2

>> n=1; d=[9 12 13]; sys=tf(n,d); >> step(sys) % Figure PS8-3No15

Step Response 0.09

0.08 1/13

0.07

Amplitude

0.06

0.05

0.04

0.03

0.02

0.01

0

0

1

2

3

4

5

6

7

8

9

10

Time (seconds)

Figure PS8-3No15

16. 3 x + 12 x + 10 x = 10u (t ) Solution 10 (a) In standard form, we have  x + 4 x + 103 x = u (t ) . Therefore, ωn2 = 103 and 2ζωn = 4 so that 3

ωn =

10 3

r/s ,= ζ

2 3 10

> 1 . The (overdamped) system response is

5  1 ( −2+ 1+ x(t ) = e 6  −2 + 23  (b)

The transfer function

2 3

)t −

1 −2 −

 −2− )t (  e 2 3

2 3

 

10 can be used to generate the step response: 3s + 12 s + 10 2

>> n=10; d=[3 12 10]; sys=tf(n,d); >> step(sys) % Figure PS8-3No16

359

Step Response 1

0.9

0.8

0.7

Amplitude

0.6

0.5

0.4

0.3

0.2

0.1

0

0

1

2

3

4

5

6

7

Time (seconds)

Figure PS8-3No16

In Problems 17 through 20, u (t ) denotes the unit-step. (a) Plot the response x using the step and initial commands, showing the step response, the initial response and the total response in one figure. Verify the steady-state value of x . (b) Plot the response x using the lsim command. 5  (0) 17. 14  x + x += 2u (t ), x= (0) 0, x= 4 x Solution

1 2

(a) A=[0 1;-5 -4]; B=[0;8]; C=[1 0]; sys=ss(A,B,C,[]); x0=[0;1/2]; [x_step,t]=step(sys); x_initial=initial(sys,x0,t); x_total=x_step+x_initial; plot(t,x_step) hold on plot(t,x_initial) hold on plot(t,x_total) % Complete Figure PS8-3No17

The steady-state value

A 8 = = 1.6 is confirmed in Figure PS8-3No17. ωn2 5

360

1.8

1.6

1.4

1.2 Step response 1 Total response 0.8

0.6

0.4

0.2 Initial response 0

-0.2

0

0.25

0.5

0.75

1

1.25

1.5

1.75

2

2.25

2.5

2.75

3

3.25

Figure PS8-3No17

(b) t=linspace(0,3.5); u=0*t+1; % Input data u must be a matrix of real values lsim(sys,u,t,x0) % Produces the same response as in Figure PS8-3No17

 (0) 1 18.  x + x += 4 x 3u (t ), x= (0) 0, x= Solution

(a) A=[0 1;-4 -1]; B=[0;3]; C=[1 0]; sys=ss(A,B,C,[]); x0=[0;1]; [x_step,t]=step(sys); x_initial=initial(sys,x0,t); x_total=x_step+x_initial; plot(t,x_step) hold on plot(t,x_initial) hold on plot(t,x_total) % Complete Figure PS8-3No18

The steady-state value

A 3 = = 0.75 is confirmed in Figure PS8-3No18. ωn2 4

3.5

361

1.2 Total response

1 Step response 0.8

0.6

0.4 Initial response 0.2

0

-0.2

0

2

1

3

4

5

6

7

8

9

10

Figure PS8-3No18

(b) t=linspace(0,10); u=0*t+1; % Input data u must be a matrix of real values lsim(sys,u,t,x0) % Produces the same response as in Figure PS8-3No18

19. 9  x + 12 x + = 4 x 0.6u (t ), x(0) =

2 3

, x (0) = 0

Solution (a) A=[0 1;-4/9 -4/3]; B=[0;0.6/9]; C=[1 0]; sys=ss(A,B,C,[]); x0=[2/3;0]; [x_step,t]=step(sys); x_initial=initial(sys,x0,t); x_total=x_step+x_initial; plot(t,x_step) hold on plot(t,x_initial) hold on plot(t,x_total) % Figure PS8-3No19

A The steady-state value = 2

ωn

0.6 9 4 9

= 0.15 is confirmed in Figure PS8-3No19.

362

0.7

0.6

0.5

0.4

0.3 Total response Initial response 0.2 0.15 0.1 Step response 0

0

2

4

6

8

10

12

Figure PS8-3No19

(b) t=linspace(0,14); u=0*t+1; % Input data u must be a matrix of real values lsim(sys,u,t,x0) % Produces the same response as in Figure PS8-3No19

20. 6  x + 11x + 3 x = u (t ), x(0) = 1, x (0) = − 12 Solution (a) A=[0 1;-1/2 -11/6]; B=[0;1/6]; C=[1 0]; sys=ss(A,B,C,[]); x0=[1;-1/2]; [x_step,t]=step(sys); x_initial=initial(sys,x0,t); x_total=x_step+x_initial; plot(t,x_step) hold on plot(t,x_initial) hold on plot(t,x_total) % Figure PS8-3No20

The steady-state value

A = ωn2

1 6 3 6

=

1 is confirmed in Figure PS8-3No20. 3

14

363

1 0.9 0.8 0.7 0.6 0.5 Total response 0.4 Initial response 0.3 0.2 Step response

0.1 0

0

2

4

6

8

10

12

14

16

18

20

Figure PS8-3No20 (b) t=linspace(0,20); u=0*t+1; % Input data u must be a matrix of real values lsim(sys,u,t,x0) % Produces the same response as in Figure PS8-2No20

21.

The mechanical system in Figure 8.19, where all parameter values are in consistent physical units, is subject to zero initial conditions. Plot two response curves for x2 : one corresponding to f1 as the only input, and another to f 2 only. Finally, plot the total response x2 . x1

x2 k2 = 2

k1 = 5

f 2 = u (t )

f1 = 5u (t ) m1 = 0.7

b = 0.6

m2 = 1

Figure 8.19 Problems 21 and 22. Solution The equations of motion are

x1 + 5 x1 − 2( x2 − x1 ) − 0.6( x2 − x1 ) = f1 0.7  x2 + 2( x2 − x1 ) + 0.6( x2 − x1 ) = f2   Selecting state variables = x1 x1= , x2 x= x= x2 , the state equation is formed as 2 , x3 1 , x4

364

0 1  0  0 0 0  x = 7 0.6 2  − 0.7 − 0.7 0.7  −2 0.6  2

0  0  0 1  x+ 5 0.6   0.7 0.7   −0.6  0

 x1  x  u (t )   2 x=   , u=   u (t )   x3   x4 

0 0  u , 0  1

The following script produces a pair of plots, one showing the contribution of f1 to x2 (labeled x21 ), the other reflecting the contribution of f 2 to x2 (labeled x22 ). Since the inputs are step functions, and initial conditions are zero, we will use the step command. A=[0 0 1 0;0 0 0 1;-7/0.7 2/0.7 -0.6/0.7 0.6/0.7;2 -2 0.6 -0.6]; B=[0 0;0 0;5/0.7 0;0 1]; C=[0 1 0 0]; % x2 is to be plotted sys=ss(A,B,C,[]); [x2,t]=step(sys); subplot(1,2,1), plot(t,x2(:,:,1)), title('Contribution of f1 to x2') subplot(1,2,2), plot(t,x2(:,:,2)), title('Contribution of f2 to x2') % Figure PS8-3No21a x2=x2(:,:,1)+x2(:,:,2); plot(t,x2) % Figure PS8-3No21b

Contribution of f1 to x2

Contribution of f2 to x2 1.4

2

1.3 1.8 1.2 1.6

1.1

x 21

1

1.4

x 22

0.9 1.2 0.8 0.7

1

0.6 0.8 0.5 0.6

0.4 0.3

0.4

0.2 0.2 0.1 0

0

5

10

15

20

25

30

35

40

0

0

Figure PS8-3No21a

5

10

15

20

25

30

35

40

365

3.5 3.25

x2

3 2.75 2.5

Response

2.25 2 1.75 1.5 1.25 1 0.75 0.5 0.25 0 0

5

10

15

20 t

25

30

35

40

Figure PS8-3No21b

22.

The mechanical system in Figure 8.19, where all parameter values are in consistent physical units, is subject to initial conditions x1 (0) = 1, x2 (0) = 1, x1 (0) = −1, x2 (0) = 1 . Plot the response x2 . Solution The equations of motion are

x1 + 5 x1 − 2( x2 − x1 ) − 0.6( x2 − x1 ) = f1 0.7  x2 + 2( x2 − x1 ) + 0.6( x2 − x1 ) = f2   Selecting state variables = x1 x1= , x2 x= x= x2 , the state equation is formed as 2 , x3 1 , x4

0 1  0  0 0 0  x = 7 0.6 2  − 0.7 − 0.7 0.7  −2 0.6  2

0  0  0 1   x + 0.6  5  0.7 0.7   −0.6  0

0 0  u , 0  1

 x1  x  u (t )   2 x=   , u=   u (t )   x3   x4 

A=[0 0 1 0;0 0 0 1;-7/0.7 2/0.7 -0.6/0.7 0.6/0.7;2 -2 0.6 -0.6]; B=[0 0;0 0;5/0.7 0;0 1]; C=[0 1 0 0]; % x2 is to be plotted x0=[1;1;-1;1]; sys=ss(A,B,C,[]); [x_step,t]=step(sys,10); % Total step response = step response to f1 + step response to f2 x2_step=x_step(:,:,1)+x_step(:,:,2); x2_initial=initial(sys,x0,t); % Response to initial conditions x2_total=x2_step+x2_initial; plot(t,x2_total) % Figure PS8-3No22

366

2.5 2.4 2.3 2.2 2.1 2

Response x2

1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1

0

5

10

15

20 t

25

30

35

40

Figure PS8-3No22

Reconsider Problem 22. Plot the response x2 by simulating the Simulink model of the system using the state-space block. Solution The equations of motion are 23.

x1 + 5 x1 − 2( x2 − x1 ) − 0.6( x2 − x1 ) = f1 0.7  x2 + 2( x2 − x1 ) + 0.6( x2 − x1 ) = f2   Selecting state variables = x1 x1= , x2 x= x= x2 , the state-space form is obtained as 2 , x3 1 , x4 0 1  0  0 0 0  x = 7 0.6 2  − 0.7 − 0.7 0.7  −2 0.6  2

0  0  0 1   x + 0.6  5  0.7 0.7   −0.6  0

= y

[0

0 0  u , 0  1

1   u (t )  1 x0 =   , u=   u (t )  −1  1 

1 0 0] x + [ 0 0] u

The model is shown in Figure PS8-3No23a. Double-click on the state-space block to enter all information on pertinent matrices, vectors and initial conditions. Simulate the model.

367

Step

x' = Ax+Bu y = Cx+Du

Vector Concatenate

1

Output x(t)

State-Space

Step1

Scope

Figure PS8-3No23a plot(tout,yout)

2.5

x

2

2

1.5

1

0

1

2

3

4

5 t

6

7

8

9

10

Figure PS8-3No23b This agrees with the result of Problem 22 once the range of t is adjusted.

24.

In the mechanical system in Figure 8.20, the parameter values in consistent physical units are = m 1,= k1 10,= k2 10,= b1 5,= b2 8 The system is subject to initial conditions x1 (0) = 0, x2 (0) = 0, x1 (0) = −1 and unit-step applied forces. Plot the response x1 .

368 x1

x2

k1

k2

f1

f2 m

b2

b1

Figure 8.20 Problems 24, 25 and 26. Solution The equations of motion are

f1  mx1 + k1 x1 + b1 x1 − k2 ( x2 − x1 ) − b2 ( x2 − x1 ) =  f2 k2 ( x2 − x1 ) + b2 ( x2 − x1 ) = Selecting state variables = x1 x= x= x1 , the state equation is formed as 1 , x2 2 , x3

0   k2 x =  b2  − k1 m

0 − k2 b2

0

0 1   1 x+ 0  1 − b1   m m 

0  f 1  1 , b2      f2  1  m

 x1    x=  x2  x   3

k1=10; k2=10; m=1; b1=5; b2=8; A=[0 0 1;k2/b2 -k2/b2 1;-k1/m 0 -b1/m]; B=[0 0;0 1/b2;1/m 1/m]; C=[1 0 0]; % x1 is to be plotted x0=[0;0;-1]; sys=ss(A,B,C,[]); [x_step,t]=step(sys); % Total step response = step response to f1 + step response to f2 x1_step=x_step(:,:,1)+x_step(:,:,2); x1_initial=initial(sys,x0,t); % Response to initial conditions x1_total=x1_step+x1_initial; plot(t,x1_total) % Figure PS8-3No24 0.25

0.2

0.15

x

1

0.1

0.05

0

-0.05

-0.1

0

0.5

1

1.5 t

Figure PS8-3No24

2

2.5

369

25.

Reconsider Problem 24. Plot the response x1 by simulating the Simulink model of the system using the state-space block. Solution The equations of motion are

f1  mx1 + k1 x1 + b1 x1 − k2 ( x2 − x1 ) − b2 ( x2 − x1 ) =  f2 k2 ( x2 − x1 ) + b2 ( x2 − x1 ) = Selecting state variables = x1 x= x= x1 , the state-space form is derived as 1 , x2 2 , x3

0 0  − k k2  2 x = b b 2  2  − k1 0 m

0 1   1 x+ 0  1 − b1   m m 

0  1 u , b2   1 m 

0 0] x + [ 0 0] u

[1

= y

0 u (t )    x0 = 0 , u=   u (t )  −1  

The model is shown in Figure PS8-3No25a. Double-click on the state-space block to enter all information on pertinent matrices, vectors and initial conditions. Simulate the model.

Step

x' = Ax+Bu y = Cx+Du

Vector Concatenate

1

Output x(t)

State-Space

Step1

Scope

Figure PS8-3No25a plot(tout,yout) 0.25

0.2

0.15

x

1

0.1

0.05

0

-0.05

-0.1

0

1

2

3

4

5 t

6

Figure PS8-3No25b

7

8

9

10

370

This agrees with the result of Problem 24 once the range of t is adjusted.

Reconsider Problem 24 but assume f1 = 0 and f 2 is a unit-impulsive force. All parameter values remain unchanged. Assuming zero initial conditions, plot the response x2 . Solution The equations of motion are 26.

0  mx1 + k1 x1 + b1 x1 − k2 ( x2 − x1 ) − b2 ( x2 − x1 ) =  f2 k2 ( x2 − x1 ) + b2 ( x2 − x1 ) = Selecting state variables = x1 x= x= x1 , the state-space form is derived as 1 , x2 2 , x3

0  k x  b2 =  2  − k1 m

0 1    1  x +  b1  δ (t )   12  − b1   m  m 

0 − k2 b2

0

y = [ 0 1 0] x k1=10; k2=10; k3=20; m=1; b1=5; b2=8; A=[0 0 1;k2/b2 -k2/b2 1;-k1/m 0 -b1/m]; B=[0;1/b2;1/m]; C=[0 1 0]; % x2 is to be plotted sys=ss(A,B,C,[]); [x2_impulse,t]=impulse(sys); plot(t,x2_impulse) % Figure PS8-3No26

0.25

0.2

x

2

0.15

0.1

0.05

0

0

0.5

1

1.5

2

2.5 t

Figure PS8-3No26

3

3.5

4

4.5

5

371

In Problems 27 through 30, the governing equations and initial conditions of a dynamic system are provided. Plot the specified output(s), together with the input(s), by using the lsim command. Note that u (t ) denotes the unit-step function.

x + 4 x += 5x 27. 8

3 − t /3 e sin t , 5

x(0) =

2 3

, x (0) = 0 , 0 ≤ t ≤ 15

Output : x(t )

Solution A=[0 1;-5/8 -1/2]; B=[0;3/40]; C=[1 0]; sys=ss(A,B,C,[]); t=linspace(0,15); u=exp(-t/3).*sin(t); x0=[2/3;0]; lsim(sys,u,t,x0) % Figure PS8-3No27 Linear Simulation Results 0.7

0.6

0.5

0.4

Amplitude

0.3 Output x(t)

Input

0.2

0.1

0

-0.1

-0.2

-0.3

0

5

10 Time (seconds)

Figure PS8-3No27 28. 64  x + 16 x + 17 x= u (t − 1), x(0)= 0 , x (0)= 1 , 0 ≤ t ≤ 40 Output : x(t )

Solution A=[0 1;-17/64 -16/64]; B=[0;1/64]; C=[1 0]; sys=ss(A,B,C,[]); t=linspace(0,40); u=heaviside(t-1); x0=[0;1]; lsim(sys,u,t,x0) % Figure PS8-3No28

15

372 Linear Simulation Results 1.5 Output x(t) Input u(t-1) 1

Amplitude

0.5

0

-0.5

-1

0

5

10

15

20

25

30

35

40

Time (seconds)

Figure PS8-3No28

x1 + 53 x1 − 32 x2 − 3( x2 − x1 ) = u (t ) = x1 (0) 0,= x1 (0) 1   , , 0 ≤ t ≤ 20 29.  sin t = x2 + 23 ( x2 − x1 ) + 3( x2 − x1 ) = x2 (0) 1,= x2 (0) 0   Outputs : x1 (t ), x2 (t ) Solution A=[0 0 1 0;0 0 0 1;-5/3 2/3 -3 3;2/3 -2/3 3 -3]; B=[0 0;0 0;1 0;0 1]; C=[1 0 0 0;0 1 0 0];sys=ss(A,B,C,[]); t=linspace(0,20); u=[heaviside(t);sin(t)]; x0=[0;1;1;0]; lsim(sys,u,t,x0) % Figure PS8-3No29 Linear Simulation Results 4 3 Output x 1

To: Out(1)

2 1 0

Inputs

Amplitude

-1 -2 6 5

To: Out(2)

4 Output x 2

3 2 1 0 -1 -2

0

2

4

6

8

10 Time (seconds)

Figure PS8-3No29

12

14

16

18

20

373

2  x1 + 2 x1 + 3x1 − 32 ( x2 − x1 ) − 4( x2 − x1 ) = e−t /4 sin t = x1 (0) 1,= x1 (0) 0 , , 0 ≤ t ≤ 15 30.  − t /3 3 (0) 1 x =   ( ) 4( ) x x x x e − + − = 2  2 2 1 2 1 Outputs : x1 (t ), x2 (t ) Solution >> >> >> >> >> >>

A=[0 0 1;3/8 -3/8 1;-1 0 -3/2]; B=[0 0;0 1/4;1/2 1/2]; C=[1 0 0;0 1 0]; sys=ss(A,B,C,[]); t=linspace(0,15); u=[exp(-t/4).*sin(t);exp(-t/3)]; x0=[1;1;0]; lsim(sys,u,t,x0) % Figure PS8-3No30 Linear Simulation Results 1

Output x 1

To: Out(1)

0.5

Amplitude

0

-0.5 1.5

To: Out(2)

1

Output x 2

0.5

Inputs

0

-0.5

0

5

10 Time (seconds)

Figure PS8-3No30 Problem Set 8.4 In Problems 1 through 6, find the frequency response of the given system. 1. 13 x + 2 x = 2.6sin 2t Solution In standard form, we have frequency response is xss= (t )

1 6

x + x = 1.3sin 2t so that τ = 16 , F0 = 1.3 , and ω = 2 r/s . The

1.3 1+ (

)

1 2 3

1 1 sin(2t − tan −= 1.2333sin(2t − 0.3218) 3)

15

374

2. y + 5 y = 14.5sin( 12 t ) Solution In standard form, we have frequency response is

(t ) yss=

1 5

y + y = 2.9sin( 12 t ) so that τ = 15 , F0 = 2.9 , and ω =

2.9 1 + ( 101 )

2

1 2

r/s . The

1 sin( 12 t − tan −1= 2.8856sin( 12 t − 0.0997) 10 )

3. 4  x + 12 x + 13 x = 40sin( 13 t ) Solution The transfer function is 1 G (s) = 2 4 s + 12 s + 13

FRF

⇒

1 G ( jω ) ω = = 2 = 0.0723 − 0.0230j 13 − 4ω + 12ω j ω = 1 3

1 3

indicating a fourth quadrant location. The phase is therefore calculated as φ = −0.3080 rad . The magnitude is 0.0759 . Noting that F0 = 40 , we find 0.3080) 3.0348sin( 13 t − 0.3080) xss (t ) F0 G ( jω ) sin( ωt + φ ) 40(0.0759) sin( 13 t −= = =

4. 2  x + 5 x + 8 x = 25sin 3t Solution The transfer function is 1 G (s) = 2 2 s + 5s + 8

FRF

⇒

−10 − 15 j 1 = G ( jω ) ω =3 = −10 + 15 j 325

indicating a third quadrant location. The phase is therefore calculated as in Figure PS8-4No4, 1 . Noting that F0 = 25 , we which yields φ = − 12 π − 0.5880 = −2.1588 rad . The magnitude is 325 find = xss (t ) F0 G ( jω ) sin(= ωt + φ )

25 sin(3t − = 2.1588) 1.3868sin(3t − 2.1588) 325

375

Figure PS8-4No4 5. 4  x + 2 x + 10 x = 28.8sin(2t ) Solution The transfer function is 1 G ( s ) =2 4 s + 2 s + 10

FRF

⇒

1 −3 − 2 j G ( jω ) ω = 2 = = −6 + 4 j 26

indicating a third quadrant location. The phase is calculated as shown in Figure PS8-4No5. This leads to φ = − 12 π − 0.9828 = −2.5536 rad . 3 − 26

0

0.5880 rad

0.9828 2 − 26

Figure PS8-4No5 The magnitude is 0.1387 . Noting that F0 = 28.8 , we find 2.5536) 3.9938sin(2t − 2.5536) = xss (t ) F0 G ( jω ) sin( = ωt + φ ) 28.8(0.1387) sin(2t −=

6. 10  x + 0.8x + 20 x = 35sin t Solution The transfer function is 1 G ( s) = 2 10 s + 0.8s + 20

FRF

⇒

1 10 − 0.8j G ( jω ) ω =1 = = 10 + 0.8j 100.64

376 indicating a fourth quadrant location. The phase is calculated as φ = − tan −1 0.8 −0.0798 rad . 10 = 1 The magnitude is 100.64 = 0.0997 . Noting that F0 = 35 , we find = ωt + φ ) 35(0.0997) sin(t − 0.0798) = 3.4889sin(t − 0.0798) xss (t ) F0 G ( jω ) sin(=

7. Consider the mechanical system in Figure 8.28. Assuming m = 12 kg , b = 20 N-sec/m , k = 200 N/m , F0 = 47 N , and ω = 5 rad/sec , find the system’s frequency response.

k

b

f (t ) = F0 sin ωt

m

x k

Figure 8.28 Problem 7. Solution The transfer function is G (s) =

1 12s + 20s + 200 2

FRF

⇒

G ( jω ) ω =5 =

1 −100 − 100 j = −100 + 100 j 20000

indicating a third quadrant location. The phase is calculated as φ = − 12 π − 14 π = − 43 π rad . The 1 magnitude is 20000 = 0.0071 . Noting that F0 = 47 , we find = xss (t ) F0 G ( jω ) sin( = ωt + φ ) 47(0.0071) sin(5 = t − 34 π ) 0.3323sin(5t − 34 π )

8. Find the frequency response of the undamped mechanical system in Figure 8.29, assuming m = 10 kg , k = 150 N/m , F0 = 115 N , and ω = 4 rad/sec . Solution The transfer function is FRF 1 G (s) = ⇒ 10 s 2 + 150

1 G ( jω ) ω = 4 = − 10

377

k f (t ) = F0 sin ωt

m

x

Figure 8.29 Problem 8. Since this is a real, negative number, its magnitude is 101 and its phase is − 12 π . Noting that F0 = 115 , we find 115 sin(4t = = xss (t ) F0 G ( jω ) sin(ω= t +φ) − 12 π ) 11.5sin(4t − 12 π ) 10

9. Show that the frequency response of an RLC circuit driven by a sinusoidal applied voltage v(t ) = V0 sin ωt is described by = iss (t )

ωCV0

(1 − LCω ) 2

2

RCω   cos  ωt − tan −1  1 − LCω 2   + ( RCω ) 2

Solution Using KVL, the governing equation of the circuit is derived as L

di 1 + Ri + ∫ i (t ) dt = V (s) dt C

so that the transfer function and FRF are

1 I (s) Cs =2 G (s) = = V ( s ) Ls + R + 1 LCs + RCs + 1 Cs

FRF

⇒

ωCj G ( jω ) = 1 − LCω 2 + RCω j

The magnitude and phase of FRF are found as

G ( jω= )

ωC

(1 − LCω ) 2

2

+ ( RCω ) 2

φ 90a − tan −1 , =

RCω 1 − LCω 2

Finally, the frequency response is = iss

ωCV0

(1 − LCω ) 2

2

RCω   sin  ωt + 90a − tan −1  1 − LCω 2   + ( RCω ) 2

378

which, of course, may be written as = iss

ωCV0

(1 − LCω ) 2

2

RCω   cos  ωt − tan −1  1 − LCω 2   + ( RCω ) 2

In Problems 10 through 12, draw the Bode plot and identify the corner frequency, as well as the asymptotic approximations of magnitude for low-frequency and high-frequency ranges. 10. G ( s ) =

2 5

2s + 5

Solution Rewrite as 2 1 G (s) = ⋅2 25 5 s + 1

FRF

⇒

2 1 G ( jω ) = ⋅2 25 5 ω j + 1

so that τ = 52 . The magnitude is

2 1 G ( jω ) = ⋅ 25 1 + ( 52 ω ) 2

Logarithmic magnitude

⇒

20log G ( jω )

20 log

2 1 dB + 20 log 25 1 + ( 52 ω ) 2

Since the first term is 20 log 252 = −21.9382 , the magnitude (dB) is obtained by lowering the standard magnitude curve by 21.9382 dB . The low-frequency magnitude is −21.9382 dB and the corner frequency is= ω 1/= τ 52 r/s . The high-frequency magnitude is approximated by 20 log 252 − 20 log( 52 ω ) dB . For instance, at ω = 100 r/s , magnitude is roughly −53.9794 dB . These facts are confirmed by Figure PS8-4No10.

>> sys=tf(2/5,[2 5]); >> bode(sys) % Figure PS8-4No10

379

Bode Diagram -20 -21.9382 -25

Magnitude (dB)

-30 -35 -40 -45 -50 -55

Phase (deg)

-60 0

-45

Corner frequency = 5/2 -90 -1 10

0

1

10

2

10

10

Frequency (rad/s)

Figure PS8-4No10

11. G ( s ) =

4 3s + 23

Solution Rewrite as 1 G ( s) = 6⋅ 9 2 s +1

FRF

⇒

1 G ( jω ) = 6⋅ 9 2 ω j +1

so that τ = 92 . The magnitude is

1 G ( jω ) = 6⋅ 1 + ( 92 ω ) 2

Logarithmic magnitude

⇒

20log G ( jω )

20 log 6 + 20 log

1 1 + ( 92 ω ) 2

dB

Since the first term is 20 log 6 = 15.5630 , the magnitude (dB) is obtained by raising the standard magnitude curve by 15.5630 dB . The low-frequency magnitude is 15.5630 dB and the corner frequency is= ω 1/= τ 92 r/s . The high-frequency magnitude is approximated by 20 log 6 − 20 log( 92 ω ) dB . For instance, at ω = 100 r/s , magnitude is roughly −37.5012 dB . These facts are confirmed through Figure PS8-4No11.

>> sys=tf(4,[3 2/3]); >> bode(sys) % Figure PS8-4No11

380

Bode Diagram

20 15.5630

Magnitude (dB)

10 0 -10 -20 -30

Phase (deg)

-40 0

-45

-90 -2 10

0

-1

10

Corner freq = 0.22

10

1

10

2

10

Frequency (rad/s)

Figure PS8-4No11

12. G ( s ) =

2.5 5s + 34

Solution Rewrite as 10 1 G (s) = ⋅ 20 3 3 s +1

so that τ =

20 3

FRF

⇒

10 1 G ( jω ) = ⋅ 20 3 3 ω j +1

. The magnitude is

10 1 G ( jω ) = ⋅ 3 1 + ( 203 ω ) 2

Logarithmic magnitude

⇒

20log G ( jω )

20 log

10 1 + 20 log dB 3 1 + ( 203 ω ) 2

Since the first term is 20 log 103 = 10.4576 , the magnitude (dB) is obtained by raising the standard magnitude curve by 10.4576 dB . The low-frequency magnitude is 10.4576 dB and the corner frequency is ω= 1/ τ= 203= 0.15 r/s . The high-frequency magnitude is approximated by 20 log 103 − 20 log( 203 ω ) dB . For instance, at ω = 100 r/s , magnitude is roughly −46.0206 dB . These facts are confirmed through Figure PS8-4No12. >> sys=tf(2.5,[5 3/4]); >> bode(sys) % Figure PS8-4No12

381

Bode Diagram 20 10.4576 10

Magnitude (dB)

0 -10 -20 -30 -40

Phase (deg)

-50 0

-45

-90 -2 10

-1

10

0

10

Corner freq = 0.15

1

10

2

10

Frequency (rad/s)

Figure PS8-4No12

13. The Bode plot of a first-order system indicates that the corner frequency is 2.25 rad/sec and the approximate low-frequency logarithmic magnitude is −23.10 dB . (a) Find the transfer function. (b) Using the results of (a), find the approximate value of the logarithmic gain at ω = 100 rad/sec . Solution 1 (a) Let the generic first-order transfer function be G ( s ) = K . The corner frequency is τ s +1 2.25 = 1/ τ , thus τ = 94 . The asymptotic approximation for the low-frequency range is given by 20 log K = −23.10 which implies K = 0.07 . Therefore, the transfer function is found as G (s) =

0.07 s +1

4 9

(b) The asymptotic approximation for the high-frequency range is given by 20 log K − 20 log τω . Inserting for K and τ from (a) and ω = 100 rad/sec , we find −56.0544 dB .

14. Repeat Problem 13 when the corner frequency is 0.45 rad/sec and the approximate lowfrequency logarithmic magnitude is 56.25 dB .

382 Solution 1 . The corner frequency is τ s +1 0.45 = 1/ τ , thus τ = 209 . The asymptotic approximation for the low-frequency range is given by 20 log K = 56.25 which implies K = 649.3816 . Therefore, the transfer function is

(a) Let the generic first-order transfer function be G ( s ) = K

G (s) =

649.3816 20 9 s +1

(b) The asymptotic approximation for the high-frequency range is given by 20 log K − 20 log τω . Inserting for K and τ from (a) and ω = 100 rad/sec , we find 9.3142 dB .

15. The Bode plot of a first-order system indicates that the approximate low-frequency and highfrequency (at ω = 100 rad/sec ) logarithmic magnitudes are 8.34 dB and −48.53 dB , respectively. Find the system’s transfer function. Solution 1 Let the generic first-order transfer function be G ( s ) = K . The asymptotic approximation τ s +1 for the low-frequency range is given by 20 log K = 8.34 which implies K = 2.6122 . The asymptotic approximation for the high-frequency range is 20 log K − 20 log τω . Using K = 2.6122 and ω = 100 rad/sec , this gives 20 log 2.6122 − 20 log100τ = −48.53

⇒

τ= 7.06

Therefore, the transfer function is G (s) =

16. Show that the magnitude G ( jω ) =

2.6122 7.06s + 1

1 2

attains a maximum when

1 − (ω / ωn ) 2  + (2ζω / ωn ) 2 ω = 1 − 2ζ 2 ωn

Solution Maximum G ( jω ) is achieved when

(1 − r 2 ) 2 + (2ζ r ) 2 , where r = ω / ωn , is minimum, which

is equivalent to minimizing (1 − r 2 ) 2 + (2ζ r ) 2 . Differentiate with respect to r and set equal to zero:

383 2(1 − r 2 )(−2r ) + 2(2ζ r )(2ζ )= 0 Solving for r , we find r= 1 − 2ζ 2

r ≠0

⇒

− (1 − r 2 ) + 2ζ 2= 0

ω = 1 − 2ζ 2 ωn

⇒

In Problems 17 through 19, draw the Bode plot, identify the resonant frequency and the peak magnitude (dB), if applicable, and find the approximate low-frequency logarithmic magnitude. 17. G ( s ) =

5 s + 2s + 9 2

Solution Rewrite G ( s ) to extract the standard, 2nd-order transfer function 5 9 ⋅ 2 G (s) = 9 s + 2s + 9

1 ωn = 3 r/s, ζ = 3

⇒

Since 20 log 95 = −5.1055 dB , the dB plot for the current problem is obtained by lowering the standard plot by 5.1055 dB . Since ζ < 0.7071 , the magnitude curve has a peak which occurs at

ωr = ωn 1 − 2ζ 2

ω=n 3,= ζ 1/3

= 2.6458 r/s

The peak is measured as

G ( jω ) standard max

1 = 2ζ 1 − ζ 2

1.5910 =

⇒

20 log1.5910 = 4.0334 dB

ζ =1/3

−1.0721 dB. But since the curve is lowered by 5.1055 dB , the peak is 4.0334 − 5.1055 = >> sys=tf(5,[1 2 9]); >> bode(sys) % Figure PS8-4No17

384

Bode Diagram 10 0 -5.1

-1.07

Magnitude (dB)

-10 -20 -30 -40 -50 -60 -70 -80 0

Phase (deg)

-45

-90

-135

resonant frequency 2.64

-180 -1 10

0

1

10

10

2

10

Frequency (rad/s)

Figure PS8-4No17

18. G ( s ) =

48 s + 2.6s + 6.76 2

Solution Rewrite G ( s ) to extract the standard, 2nd-order transfer function G ( s ) = 7.1006 ⋅

6.76 s + 2.6s + 6.76 2

⇒

ωn = 2.6 r/s, ζ = 0.5

Since 20 log 7.1006 = 17.0259 dB , the dB plot for the current problem is obtained by raising the standard plot by 17.0259 dB . Since ζ < 0.7071 , the magnitude curve has a peak which occurs at

ωr = ωn 1 − 2ζ 2

= ωn 2.6, = ζ 0.5

= 1.8385 r/s

The peak is measured as

G ( jω ) standard max

1 = 2ζ 1 − ζ 2

= 1.1547

⇒

20 log1.1547 = 1.2494 dB

ζ = 0.5

But since the curve is raised by 17.0259 dB , the peak is 1.2494 + 17.0259 = 18.2753 dB. >> sys=tf(48,[1 2.6 6.76]); >> bode(sys) % Figure PS8-4No18

385

Bode Diagram 30 18.27 17.02

20

Magnitude (dB)

10 0 -10 -20 -30 -40 -50 -60 0

Phase (deg)

-45

-90

-135 1.84 -180 -1 10

0

1

10

10

2

10

Frequency (rad/s)

Figure PS8-4No18

19. G ( s ) =

3 2 s + 8.64 s + 11.52 2

Solution Rewrite G ( s ) to extract the standard, 2nd-order transfer function G ( s ) = 0.2604 ⋅

5.76 s + 4.32 s + 5.76 2

⇒

ωn = 2.4 r/s, ζ = 0.9

Since 20 log 0.2604 = −11.6872 dB , the dB plot for the current problem is obtained by lowering the standard plot by 11.6872 dB . Since ζ > 0.7071 , the magnitude curve does not have a peak. >> sys=tf(3,[2 8.64 11.52]); >> bode(sys) % Figure PS8-4No19

386

Bode Diagram 0

Magnitude (dB)

-10 -11.69 -20 -30 -40 -50 -60 -70 -80 0

Phase (deg)

-45

-90

-135

-180 -1 10

2

1

0

10

10

10

Frequency (rad/s)

Figure PS8-4No19

20. Show that in the Bode plot of the standard, second-order transfer function, the asymptotic approximation of the logarithmic magnitude for the high-frequency range is given by ω −40 log ωn which represents a straight line with a slope of −40 dB/decade. Solution The logarithmic magnitude for the standard, second-order transfer function is 20 log

1 2

1 − (ω / ωn ) 2  + (2ζω / ωn ) 2

2 = −10 log  1 − (ω / ωn ) 2  + (2ζω / ωn ) 2  dB  

2

In the high-frequency range, we have ω / ωn >> 1 so that 1 − (ω / ωn ) 2  ≅ (ω / ωn ) 4 . But since ω / ωn >> 1 , the term (2ζω / ωn ) 2 is negligible relative to (ω / ωn ) 4 . As a result, 2

2 2 4 1 − (ω / ωn )  + (2ζω / ωn ) ≅ (ω / ωn )

Therefore, −10 log  1 − (ω / ωn ) 2  + (2ζω / ωn ) 2  ≅ −10 log(ω / ωn ) 4 = −40 log ω / ωn   2

387 Problem Set 8.5 In Problems 1 through 3, find e At , where t is scalar, using (a) The expm command. (b) The inverse Laplace-transform approach.

0 4  1. A =   1 3  Solution (a) >> A=[0 4;1 3]; syms t >> simplify(expm(A*t)) ans = [ (exp(-t)*(exp(5*t) + 4))/5, (4*exp(-t)*(exp(5*t) - 1))/5] [ (exp(-t)*(exp(5*t) - 1))/5, (exp(-t)*(4*exp(5*t) + 1))/5]

e

At

 1 e−t (e5t + 4) =  5 − t 5t  15 e (e − 1)

4 e − t (e5t − 1)  5  1 e − t (4e5t + 1)  5

(b) >> simplify(ilaplace(inv(s*eye(2)-A))) ans = [ (exp(-t)*(exp(5*t) + 4))/5, (4*exp(-t)*(exp(5*t) - 1))/5] [ (exp(-t)*(exp(5*t) - 1))/5, (exp(-t)*(4*exp(5*t) + 1))/5]

 −2 1  2. A =    0 2 Solution (a) >> A=[-2 1;0 2]; syms t >> simplify(expm(A*t)) ans = [ exp(-2*t), sinh(2*t)/2] [ 0, exp(2*t)]

e

At

e−2t =  0

(b) >> simplify(ilaplace(inv(s*eye(2)-A)))

1 sinh 2t  2  2t

e



388

ans = [ exp(-2*t), sinh(2*t)/2] [ 0, exp(2*t)]

1 1 0    3. A = 1 1 0  0 0 1  Solution (a) >> A=[1 1 0;1 1 0;0 0 1]; syms t >> simplify(expm(A*t)) ans = [ exp(2*t)/2 + 1/2, exp(2*t)/2 - 1/2, 0] [ exp(2*t)/2 - 1/2, exp(2*t)/2 + 1/2, 0] [ 0, 0, exp(t)]

e At

 1 (1 + e2t )  2 2t = 12 (e − 1)  0 

1 (e 2t − 1) 2 1 (1 + e 2t ) 2

0

0  0  et 

(b) >> simplify(ilaplace(inv(s*eye(3)-A))) ans = [ exp(2*t)/2 + 1/2, exp(2*t)/2 - 1/2, 0] [ exp(2*t)/2 - 1/2, exp(2*t)/2 + 1/2, 0] [ 0, 0, exp(t)]

4.

Consider the state equation subjected to an initial state vector described by 0  3 −1 0  x =  1  −4 3  x + 1  u , x(0) =     − 2  where

 12 if 0 < t < 2 = u unit-pulse =  0 otherwise Find the state vector x using the formal-solution approach. Solution Due to the nature of the unit pulse, the response x must be determined for two different time intervals: 0 < t < 2 and t > 2 .

389

For 0 < t < 2 , the integration variable τ runs from 0 to t and during this time, the input is simply 1: 2 t

x1 (t ) = e x0 + ∫ e A (t −t ) B ⋅ 12 dt At

0

For t > 2 , the integration variable τ runs from 0 to t > 2 , so the integral can be decomposed as t

2

0

0

1/2

t

0

A ( t −ttt ) A (t − ) A (t − ) = )d ∫ e Bu (tttttt ∫ e B u( ) d + ∫ e B u( ) d

In the first integral, the input is

1 2

2

, and it is zero in the second integral. Therefore, 2

e At x0 + ∫ e A (t −t ) B ⋅ 12 dt x 2 (t ) = 0

A=[3 -1;-4 3]; B=[0;1]; x0=[0;-1/2]; syms t tau % 0 < t < 2 x1=simplify(expm(A*t)*x0+int(expm(A*(t-tau))*B*(1/2),tau,0,t)) x1 = exp(5*t)/10 - 1/10 - exp(5*t)/5 - 3/10

 1 (e5t − 1)  = x1  10  , 0 2 1 1 1 − 5 e − 4 e − 20 e  It is readily verified that at t = 2 , the two response segments x1 and x 2 agree.

390 In Problems 5 through 8, find the state vector via the formal-solution approach.

1   1 -1 -1 + u , u = unit-step , x(0) =   x 5. x =    1  -4 1  0 2 Solution >> A=[1 -1;-4 1]; B=[1;1/2]; x0=[-1;0]; syms t tau >> x=simplify(expm(A*t)*x0+int(expm(A*(t-tau))*B,tau,0,t)) x = 1/2 - (3*exp(3*t))/8 - (9*exp(-t))/8 (3*exp(3*t))/4 - (9*exp(-t))/4 + 3/2

 1 − 1 (3e3t + 9e−t )  x =  2 8 3t −t  3 1  2 + 4 (3e − 9e ) 

 5 1 1 0  sin t , x(0) = 6. x =    −4 1 x +  −1 u , u =     1  Solution >> A=[5 1;-4 1]; B=[1;-1]; x0=[0;1]; syms t tau >> x=simplify(expm(A*t)*x0+int(expm(A*(t-tau))*B*sin(tau),tau,0,t)) x = exp(3*t)/25 - cos(t)/25 - (11*sin(t))/50 + (11*t*exp(3*t))/10 (51*exp(3*t))/50 - cos(t)/50 + (7*sin(t))/50 - (11*t*exp(3*t))/5

 1 e3t + 11 te3t − 1 (2 cos t + 11sin t )  x =  25 3t 10 3t 50  51 11 1  50 e − 5 te − 50 (cos t − 7 sin t ) 

0 1  0  1 7. x =  3  x +   u , u = unit-ramp , x(0) =   1  1 0 - 2  >> A=[0 1;0 -3/2]; B=[0;1]; x0=[1;1]; syms t tau >> x=simplify(expm(A*t)*x0+int(expm(A*(t-tau))*B*tau,tau,0,t)) x = t^2/3 - (26*exp(-(3*t)/2))/27 - (4*t)/9 + 53/27 (2*t)/3 + (13*exp(-(3*t)/2))/9 - 4/9

 1 t 2 − 26 e −3t /2 − 4 t + 53  27 9 27 x = 3  13 −3t /2 2 4 e t + −   9 3 9

391

0 1 0 0     x  0 1 0  x +  0  u ,= u e−t /3 sin t , x= (0) 8.= 1    −1 −2 −3 4 Solution

0    1    0 

>> A=[1 0 0;0 1 0;-1 -2 -3]; B=[0;0;1/4]; x0=[0;1;0]; syms t tau >> x=simplify(expm(A*t)*x0+int(expm(A*(t-tau))*B*exp(tau/3)*sin(tau),tau,0,t)) x = 0 exp(t) (155*exp(-3*t))/292 - exp(t)/2 - (3*exp(-t/3)*(3*cos(t) - 8*sin(t)))/292

  0   x= et   155 −3t 1 t 3 −t /3   292 e − 2 e − 292 e (3cos t − 8sin t ) 

In Problems 9 and 10, the state-space form of a system model is specified. Using the formal-solution approach, find the output y (t ) .

1  x = Ax + Bu 0 1 0 C x = , A  = , B = , 1 = 1 , u sin = t , 9.  [ ]    0  1 =  y Cx  −2 −2  −1 − 2  Solution >> A=[0 2;-2 -4]; B=[1;-1/2]; C=[1 1]; x0=[0;-1]; syms t tau >> x=simplify(expm(A*t)*x0+int(expm(A*(t-tau))*B*sin(tau),tau,0,t)); >> y=simplify(C*x) y = sin(t)/5 - cos(t)/10 - (9*exp(-2*t))/10

1 cos t − 9 e −2t y =15 sin t − 10 10

2 0 2 x Ax + Bu = = , A  0 = 0 1  , B 10.  y = Cx   0 0 4 

0 0  0  − t /2   e     =  , C 1= 1 0 1 0 , = , u x [ ]   0  0   1      0 1  1 

392 Solution >> >> >> >>

A=[2 0 2;0 0 1;0 0 4]; B=[0 0;1 0;0 1]; C=[1 1 0]; x0=[0;0;1]; syms t tau x=simplify(expm(A*t)*x0+int(expm(A*(t-tau))*B*[exp(-tau/2);1],tau,0,t)); y=simplify(C*x)

y = (25*exp(4*t))/16 - (3*exp(2*t))/2 - 2*exp(-t/2) - t/4 + 31/16

= y

25 4t e 16

31 − 23 e 2t − 2e −t /2 − 14 t + 16

Problem Set 8.6 In Problems 1 through 6, plot the specified output using the RK4 method. 1. 3 x + x += x3 43 , x(0) = 13 , x (0) = 0 , 0 ≤ t ≤ 20 , output : x Solution With state variables x1 = x and x2 = x , the nonlinear state-variable equations are obtained as

 x1 = x2  3 4 1  x2 = 3 (− x2 − x1 + 3 )

,

x1 (0) = 13 x2 (0) = 0

In vector form,

x  x f (t , x= = ) , x  1= , f  x2  >> >> >> >> >>

x2    13  x = , 1    , 0 ≤ t ≤ 20 0 3 4  3 (− x2 − x1 + 3 )  0 

t = linspace(0,20); x0 = [1/3;0]; f = @(t,x)([x(2);(-x(2)-x(1)^3+4/3)/3]); x = RK4System(f, t, x0); x2 = x(2,:); % Extract x2 from the state vector plot(t,x2) % Figure PS8-6No1

393

0.6

0.5

0.4

0.3

Output x

2

0.2

0.1

0

-0.1

-0.2

-0.3

-0.4

0

2

4

6

8

10

12

14

16

18

20

Figure PS8-6No1 2. 2 x + 9  x + 10 x + 3 x 2 =e −2t /5 , x(0) =0 , x (0) =− 14 ,  x(0) =0 , 0 ≤ t ≤ 15 , output : x Solution Choosing state variables x1 = x , x2 = x , and x3 =  x , the nonlinear state-variable equations are obtained as

 x x= = x1 (0) 0 2  1 , x2 (0) = − 14  x2 = x3  −2t /5 2 1 x3 (0) = 0 )  x3 =2 (−9 x3 − 10 x2 − 3 x1 + e The problem is formulated as

  x2  x1  0        1 x = f (t , x) , x =  x2  , f =  x3  , x0 = − 4  , 0 ≤ t ≤ 15   1   −2t /5  2 )  0   x3   2 (−9 x3 − 10 x2 − 3 x1 + e >> >> >> >> >>

t = linspace(0,15); x0 = [0;-1/4;0]; f = @(t,x)([x(2);x(3);(-9*x(3)-10*x(2)-3*x(1)^2+exp(-2*t/5))/2]); x = RK4System(f, t, x0); x1 = x(1,:); % Extract x1 from the state vector plot(t,x1) % Figure PS8-6No2

394

0.02

0

-0.02

Output x

1

-0.04

-0.06

-0.08

-0.1

-0.12

-0.14

-0.16

10

5

0

15

t

Figure PS8-6No2

3.  x + 2 x + x x = 1 + sin t , x(0) = 1 , x (0) = 0 , 0 ≤ t ≤ 10 , output : x Solution With state variables x1 = x and x2 = x , the nonlinear state-variable equations are obtained as  x1 = x2   x2 =−2 x2 − x1 x1 + 1 + sin t

,

x1 (0) = 1 x2 (0) = 0

The problem is formulated as x  = x f (t , x= ) , x  1= , f  x2  >> >> >> >> >>

x2   1  , x0   , 0 ≤ t ≤ 10  = 0  −2 x2 − x1 x1 + 1 + sin t 

t = linspace(0,10); x0 = [1;0]; f = @(t,x)([x(2);-2*x(2)-x(1)*sqrt(abs(x(1)))+1+sin(t)]); x = RK4System(f, t, x0); x2 = x(2,:); % Extract x2 from the state vector plot(t,x2) % Figure PS8-6No3

395

0.6

0.4

0.2

Output x

2

0

-0.2

-0.4

-0.6

-0.8

0

1

2

3

4

5 t

6

7

8

9

10

Figure PS8-6No3 4. 4  x + 3 x + 2 x x =12 + sin t , x(0) =1 , x (0) =−1 , 0 ≤ t ≤ 15 , output : x Solution With state variables x1 = x and x2 = x , the nonlinear state-variable equations are obtained as  x1 = x2  1 1  x2 = 4 (−3 x2 − 2 x1 x1 + 2 + sin t )

,

x1 (0) = 1 x2 (0) = −1

The problem is formulated as x2   x  1 x f (t ,= x) , x  1= = , x0   , 0 ≤ t ≤ 15  , f 1 = 1 −1  x2   4 (−3 x2 − 2 x1 x1 + 2 + sin t )  >> >> >> >> >>

t = linspace(0,15); x0 = [1;-1]; f = @(t,x)([x(2);(-3*x(2)-2*x(1)*abs(x(1))+(1/2)+sin(t))/4]); x = RK4System(f, t, x0); x1 = x(1,:); % Extract x1 from the state vector plot(t,x1) % Figure PS8-6No4

396

1

0.9

0.8

Output x

1

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

5

10

15

t

Figure PS8-6No4  x1 = x2 5.  3 1 2 − t /3   x2 =− 3 x1 − x1 x1 − x2 − 1 + 3 e

,

x1 (0) = 1 , 0 ≤ t ≤ 10 , output : x1 x2 (0) = − 12

Solution The problem is formulated as

x  x f (t ,= x) , x  1= = , f  x2  >> >> >> >> >>

x2   1  x = ,  1   1  , 0 ≤ t ≤ 10 0 3 2 − t /3 − 2  − 3 x1 − x1 x1 − x2 − 1 + 3 e 

t = linspace(0,10); x0 = [1;-1/2]; f = @(t,x)([x(2);-x(1)/3-x(1)*abs(x(1))-x(2)^3-1+2*exp(-t/3)/3]); x = RK4System(f, t, x0); x1 = x(1,:); % Extract x1 from the state vector plot(t,x1) % Figure PS8-6No5

397

1

0.5

Output x

1

0

-0.5

-1

-1.5

0

2

1

3

4

5 t

7

6

9

8

10

Figure PS8-6No5  x = −2 x1 − x23 + e−t cos t 6.  1  x2= x1 + 4

x1 (0) = 0 , 0 ≤ t ≤ 5 , outputs : x1 , x2 x2 (0) = 23

,

Solution In vector form,

x  x f (t , x= = ) , x  1= , f  x2 

t = linspace(0,5); x0 = [0;2/3]; f = @(t,x)([-2*x(1)-x(2)^3+exp(-t)*cos(t);x(1)+4]); x = RK4System(f, t, x0); plot(t,x) % Figure PS8-6No6 3

2

x 1

2

0

Output

>> >> >> >>

−2 x1 − x23 + e−t cos t  0  , x0  2  , 0 ≤ t ≤ 5  = x1 + 4 3  

-1

-2

x

1

-3

-4

-5

-6

0

0.5

1

1.5

2

2.5

t

Figure PS8-6No6

3

3.5

4

4.5

5

398

7.

A nonlinear dynamic system is described by x1 (0) = 0  x1 = − x2 − x1 x1 − 1 , , 0 ≤ t ≤ 10  x2 (0) = − 13  x2 = x1 − x2 − 1

(a) Use the Simulink model to plot x2 (t ) . (b) Derive the linearized model analytically, build its Simulink model where the linear model is represented by the State-Space block, and use it to plot the time variations of the variable that is compatible with x2 (t ) . Compare the plots generated in (a) and (b) and comment. Solution (a) The model is built and shown in Figure PS8-6No7-1. Run the simulation, followed by >> plot(tout,yout)

% Figure PS8-6No7-2

1 1

Constant1 Constant x1

1 s

Summing junction

Summing junction1

Integrator 1 IC: x1(0) = 0

x2

1 s

Scope

Integrator 2 IC: x2(0) = -1/3

1

u*abs(u)

Output x2(t)

Fcn

Figure PS8-6No7-1 -0.3

-0.4

-0.5

-0.6

x

2

-0.7

-0.8

-0.9

-1

-1.1

-1.2

-1.3

0

1

2

3

4

5 t

6

Figure PS8-6No7-2

7

8

9

10

399

(b) The operating point is (0, −1) . The linearized model is derived as ∆x1 = −∆x2   ∆x2 = ∆x1 − ∆x2

,

∆x1 (0) = 0 , 0 ≤ t ≤ 10 2 ∆x2 (0) = 3

The model in Figure PS8-6No7-3 is built using the state-space matrices and initial state: 0 −1 0  = A  = , B ,C  0 = 1 −1  

,D [0 1]=

0  0,= x0  2  3

Figure PS8-6No7-3 Simulate this model. The output will be ∆x2 by design. But in order to get access to the second variable compatible with x2 in the nonlinear model, we realize that ∆x2 − 1 =x2 . Therefore, the command that will generate the desired plot is >> plot(tout,yout-1)

Embedding the plot generated in (a) into this new figure yields Figure PS8-6No7-4. As expected, the nonlinear and linear responses agree near the second coordinate of the operating point. -0.3

-0.4

-0.5

-0.6

x

2

-0.7

-0.8

-0.9 Operating point x 2 = -1

-1 Nonlinear -1.1 Linear -1.2

-1.3

0

1

2

3

4

5 t

6

Figure PS8-6No7-4

7

8

9

10

400

8. Repeat Problem 7 when is the output. Solution (a) The model is built and shown in Figure PS8-6No8-1. Run the simulation, followed by >> plot(tout,yout)

% Figure PS8-6No8-2

Figure PS8-6No8-1

Figure PS8-6No8-2

(b) The operating point is

. The linearized model is derived as

401 ∆x1 = −∆x2   ∆x2 = ∆x1 − ∆x2

,

∆x1 (0) = 0 , 0 ≤ t ≤ 10 2 ∆x2 (0) = 3

The model in Figure PS8-6No8-3 is built using the state-space matrices and initial state: 0 −1 0  = A  = , B ,C  0 = 1 −1  

[1

0  0]= , D 0,= x0  2  3

Figure PS8-6No8-3 Simulate this model. The output will be ∆x1 by design. This happens to be compatible with x1 because ∆x1 = x1 . Therefore, the command that will generate the desired plot is >> plot(tout,yout)

Embedding the plot generated in (a) into this new figure yields Figure PS8-6No8-4. As expected, the nonlinear and linear responses agree near the second coordinate of the operating point. 0.1

0.05

0 Operating point x1 = 0 -0.05

x

1

-0.1

-0.15

-0.2 Nonlinear -0.25 Linear -0.3

-0.35

-0.4

0

1

2

3

4

5 t

6

Figure PS8-6No8-4

7

8

9

10

402

9.

A nonlinear dynamic system is governed by x2 − x1  x= 1  x2 2 x2−1 + 1 =

x1 (0) = 23 x2 (0) = − 12

,

, 0 ≤ t ≤ 10

(a) Use the Simulink model to plot x2 (t ) . (b) Derive the linearized model analytically, and use the initial command to plot the time variations of the variable that is compatible with x2 (t ) . Compare the plots generated in (a) and (b) and comment. Note that the initial command will automatically determine the most suitable time duration for the response curve. Solution (a) The model is built and shown in Figure PS8-6No9-1. Run the simulation, followed by >> plot(tout,yout)

% Figure PS8-6No9-2 1

Out x2

Scope x2 IC: x2(0) = -1/2 1

Constant

x2

1 s

Summing junction

1 s

Integrator 1

Summing junction1

2/u

Fcn

Figure PS8-6No9-1

Integrator 2 IC: x1(0) = 2/3

x1

Scope x1

403

-0.5

x

2

-1

-1.5

-2

0

1

2

3

5 t

4

6

7

8

9

10

Figure PS8-6No9-2 (b) The operating point is (−2, −2) . The linearized model is derived as

∆x1 = ∆x2 − ∆x1   1  ∆x2 =− 2 ∆x2

,

8 ∆x1 (0) = 3 , 0 ≤ t ≤ 10 3 ∆x2 (0) = 2

But in order to get access to the second variable compatible with x2 in the nonlinear model, we realize that ∆x2 − 2 = x2 . >> A = [-1 1;0 -1/2]; C = [0 1]; sys = ss(A,[],C,[]); >> x0 = [8/3;3/2]; >> [init_resp,t] = initial(sys,x0); % Lower by 2 units for compatibility with x2 and complete Figure PS8-6No9-3 >> plot(t,init_resp-2)

Embedding the plot generated in (a) into this new figure yields Figure PS8-6No9-3. As expected, the nonlinear and linear responses agree near the second coordinate of the operating point.

404

-0.5

x

2

-1

Linear -1.5

Nonlinear

Operating point x 2 = -2 -2 0

1

2

3

4

5

6

7

8

9 t

10

11

12

13

14

15

16

17

18

Figure PS8-6No9-3

10.

The mathematical model of a nonlinear system is given below, where u (t ) is the unit-step input.  x1 = x2  − x1 − x1 x1 − x2 + u (t )  x2 =

,

x1 (0) = 0.4 , 0 ≤ t ≤ 10 x2 (0) = −1

(a) Plot x2 (t ) by simulating the Simulink model. (b) Derive the linearized model analytically, and use the initial command to plot the time variations of the variable that is compatible with x2 (t ) . Compare the plots generated in (a) and (b) and comment. (c) Find the linear model using the linearization procedure in Simulink. Plot the response x2 (t ) to confirm the result of (b). Solution (a) The model is built and shown in Figure PS8-6No10-1. Run the simulation, followed by >> plot(tout,yout)

% Initiate Figure PS8-6No10-2

405

1

Out x2

Scope x2 IC: x2(0) = -1 x2

1 s

Step Summing junction1

x1

1 s

Integrator 1

Integrator 2 IC: x1(0) = 0.4

Scope x1

u*(1+abs(u))

Fcn1

Figure PS8-6No10-1

0.6 Linear 0.4 Nonlinear 0.2 Operating point x2

x

2

0

-0.2

-0.4

-0.6

-0.8

-1

0

1

2

3

4

5 t

6

7

8

9

Figure PS8-6No10-2

(b) The operating point is (0.6180, 0) . The linearized model is derived as

 ∆x1 =∆x2  ∆x2 = −2.2360∆x1 − ∆x2

,

∆x1 (0) =−0.2180 , 0 ≤ t ≤ 10 ∆x2 (0) = −1

Noting ∆x2 = x2 , we write the following script to plot the linearized response x2 . >> >> >> >>

A = [0 1;-2.2360 -1]; C = [0 1]; sys = ss(A,[],C,[]); x0 = [-0.2180;-1]; [init_resp,t] = initial(sys,x0); plot(t,init_resp) % Complete Figure PS8-6No10-2

10

406

The linear and nonlinear responses for x2 agree at the second coordinate of the operating point. (c) Refer to the model displayed in Figure PS8-6No10-3. >> sys='ProbPS86No103'; >> load_system(sys); >> opspec = operspec(sys); % Specify the properties of the first state variable >> opspec.States(1).SteadyState = 1; >> opspec.States(1).x = 0.4; % Initial value >> opspec.States(1).Min = 0; % Minimum value % Specify the properties of the second state variable >> opspec.States(2).SteadyState = 1; >> opspec.States(2).x = -1; % Initial value >> opspec.States(2).Min = 0; % Minimum value % Find the operating point based on the above specifications >> [op,opreport] = findop(sys,opspec);

Operating Point Search Report: --------------------------------Operating Report for the Model ProbPS86No103. (Time-Varying Components Evaluated at time t=0) Operating point specifications were successfully met. States: ---------(1.) ProbPS86No103/Integrator 1 x: 0 dx: -2.22e-13 (0) (2.) ProbPS86No103/Integrator 2 x: 0.618 dx: 0 (0) Inputs: None ---------Outputs: ---------(1.) ProbPS86No103/Out x2 y: 0 [-Inf Inf]

% Get the linearization annotations >> IO = getlinio(sys); % Extract the linear state-space model >> LIN = linearize('ProbPS86No103',op,IO) LIN =

407

a = Integrator 1 Integrator 2

Integrator 1 -1 1

Integrator 2 -2.236 0

b = Integrator 1 Integrator 2

Summing junc 1 0

c = Integrator 1

Integrator 1 1

Integrator 2 0

d = Integrator 1

Summing junc 0

Continuous-time state-space model.

1

Out x2

Scope x2 IC: x2(0) = -1 1 s

Step Summing junction1

x2

1 s

Integrator 1

x1

Integrator 2 IC: x1(0) = 0.4

u*(1+abs(u))

Fcn1

Figure PS8-6No10-3 (linear analysis points inserted) >> >> >> >>

A = [-1 -2.2360;1 0]; C = [1 0]; sys = ss(A,[],C,[]); x0 = [-1;-0.2180]; [init_resp,t] = initial(sys,x0); plot(t,init_resp) % Figure PS8-6No10-4

The plot confirms the result of (b).

408

0.6

0.4

0.2

Linearized x

2

0

-0.2

-0.4

-0.6

-0.8

-1

0

1

3

2

4

5 t

6

7

8

10

9

Figure PS8-6No10-4 11.

Consider the nonlinear model x += ax x e −2t /3 sin t , = x(0) 0.3 , 0 ≤ t ≤ 15 where a is a parameter. Use the RK4 method to plot the response x(t ) for a = 0.2, 0.8, 1.5 versus t in the same graph. Solution x0=0.3; t=linspace(0,15); f1=@(t,x)(-0.2*x*abs(x)+exp(-2*t/3)*sin(t)); f2=@(t,x)(-0.8*x*abs(x)+exp(-2*t/3)*sin(t)); f3=@(t,x)(-1.5*x*abs(x)+exp(-2*t/3)*sin(t)); x1=RK4System(f1,t,x0); x2=RK4System(f2,t,x0); x3=RK4System(f3,t,x0); plot(t,x1,t,x2,t,x3) % Figure PS8-6No11 0.9

0.8

0.7 a = 0.2 0.6

0.5

x(t)

>> >> >> >> >> >> >> >>

a = 0.8 0.4 a = 1.5 0.3

0.2

0.1

0

0

5

10 t

Figure PS8-6No11

15

409

12.

The pendulum system in Figure 8.35 consists of a uniform thin rod of length l and a concentrated mass m at its tip. The friction at the pivot causes the system to be damped. When the angular displacement θ is not very small, the system is described by the nonlinear model ml 2θ + 0.24θ + 12 mg l sin θ = 0 g Assume, in consistent physical units, that= ml 2 1.5, = 6.8 . Two sets of initial conditions l are to be considered: (1) θ (0) = 10 , θ(0) = 0 , and (2) θ (0) = 20 , θ(0) = 0 . Using the RK4 method plot the two angular displacements θ1 and θ 2 corresponding to the two sets of initial conditions versus 0 ≤ t ≤ 10 in the same graph. Angle measures must be converted to radians.

m Figure 8.35 Problem 12. Solution Dividing the equation of motion by ml 2 , we have

θ +

0.24  12 mg l sin θ = 0 θ+ ml 2 ml 2

⇒

θ + 0.16θ + 3.4sin θ = 0

There are two state variables: , x2 θ . The state-variable equations are formed as = x1 θ=

x1 = x2 x2 = −0.16x2 − 3.4sin x1 The two sets of initial conditions are

x1 (0) = 10(π /180) radians x2 (0) = 0 In vector form,

,

x1 (0) = 20(π /180) radians x2 (0) = 0

410

x2 x    (t , x) , x  1 = , f (t , x)  = x f=  , x0  x2  −0.16x2 − 3.4sin x1  >> >> >> >> >> >> >> >>

t = linspace(0,10); x0 = [10*pi/180;0]; f = @(t,x)([x(2);-0.16*x(2)-3.4*sin(x(1))]); xRK4 = RK4System(f,t,x0); theta1 = xRK4(1,:); plot(t,theta1) % Initiate Figure PS8-6No12 hold on x0 = [20*pi/180;0]; xRK4 = RK4System(f,t,x0); theta2 = xRK4(1,:); plot(t,theta2) % Complete figure

0.4

0.3 Initial 20 degrees

Angular displacement

0.2

0.1

0

Initial 10 degrees

-0.1

-0.2

-0.3

-0.4

0

1

2

3

4

5 t

6

7

8

9

10

Figure PS8-6No12

13.

A nonlinear dynamic system model is derived as x + 2= x3 u (t ) , x(0) = 0 , 0≤t ≤3

where u (t ) is the unit-step function. (a) Build the Simulink model and use it to plot the response x(t ) . (b) Derive the linearized model analytically. Build a Simulink model and use it to plot the time variations of the variable in the linear model that is compatible with x(t ) . Compare the plots generated in (a) and (b) and comment. Solution (a) The model is built and shown in Figure PS8-6No13-1. Run the simulation, followed by >> plot(tout,yout)

% Figure PS8-6No13-2

411

Figure PS8-6No13-1

0.8

0.7

0.6

x(t)

0.5

0.4

0.3

0.2

0.1

0

0

1

0.5

1.5 Time

3

2.5

2

Figure PS8-6No13-2 (b) The operating point is x = ( 12 ) . The linearized model is derived as 1/3

∆x + 6 ( 12 )

2/3

∆x =0 , ∆x(0) =− ( 12 )

1/3

, 0≤t ≤3

The model is built as in Figure PS8-6No13-3, where the gain is K = 6 ( 12 )

2/3

.

412

Figure PS8-6No13-3 Simulate this model. The output is ∆x by design. However, in order to get access to the variable compatible with x in the nonlinear model, we recognize that ∆x + ( 12 )

1/3

=x . Therefore, the

command that will generate the desired plot is >> plot(tout,yout+(1/2)^(1/3))

Embedding the plot generated in (a) into this new figure yields Figure PS8-6No13-4. As expected, the nonlinear and linear responses agree near the operating point. 0.8 0.7937 Operating point

0.7 Linear 0.6

Nonlinear

x(t)

0.5

0.4

0.3

0.2

0.1

0

0

0.5

1

1.5 Time

2

2.5

3

Figure PS8-6No13-4 14. The nonlinear spring in the mechanical system shown in Figure 8.36 is identified by the spring force fspr = x x . The system is subjected to a unit-step externally applied force u (t ) , and initial conditions x(0) = 0 , x (0) = 1 .

413 (a) Derive the state-variable equations in vector form. (b) Using the RK4 method, plot the displacement x(t ) versus 0 ≤ t ≤ 10 sec.

x fspr = x x f (t ) = u (t ) m = 1 kg

c = 1 N.s/m

Figure 8.36 Problem 14. Solution (a) The system’s motion is described by  x + x += x x u (t ) , = x(0) 0 , = x (0) 1

With state variables x1 = x and x2 = x , the nonlinear state-variable equations are obtained as

 x1 = x2  − x2 − x1 x1 + u (t )  x2 =

,

x1 (0) = 0 x2 (0) = 1

In vector form,

x2   x  0  x f (t , x= = ) , x  1= , x0   , 0 ≤ t ≤ 10 , f  = 1   x2  − x2 − x1 x1 + 1 (b) t = linspace(0,10); x0 = [0;1]; f = @(t,x)([x(2);-x(2)-x(1)*abs(x(1))+1]); x = RK4System(f, t, x0); x1 = x(1,:); % Extract x1 from the state vector plot(t,x1) % Figure PS8-6No14 1.4

1.2

1

Response to a unit-step force

>> >> >> >> >>

0.8

0.6

0.4

0.2

0

0

1

2

3

4

5 Time (sec)

6

Figure PS8-6No14

7

8

9

10

414

Review Problems 1. A linear system is modeled as 2x  + 3 x= 5

5 u (t ) + ur (t ) 3

, x(0)= 0

where u (t ) and ur (t ) denote the unit-step and unit-ramp functions, respectively. (a) Express the response x(t ) in closed form. (b) Plot the response x(t ) using the lsim command. Solution 2 (a) In standard form, 152 x += . Since the system is linear, we will x 95 u (t ) + 13 ur (t ) so that τ = 15 use superposition to find its total response. The response in x(= t)

For

2 15

2 15

5 x + x = u (t ) is given by 9

5 (1 − e −15t /2 ) 9

1 u (t ) , the response is x + x = 3 r

x(t ) = 13 t − 152 (1 − e−15t /2 )  Superposition yields x(t ) = 95 (1 − e −15t /2 ) + 13 t − 152 (1 − e −15t /2 )  . (b) Rewrite the model as x = − 152 x + 25 u (t ) + 52 ur (t ) . The following script generates the 6 response. A = -15/2; B = 1; C = 1; sys = ss(A,B,C,[]); x0 = 0; t = linspace(0,10); u = 25/6.+5*t/2; lsim(sys,u,t,x0) % Figure Review8No1

Linear Simulation Results 30

25

20

Amplitude

>> >> >> >>

15

Input 10

5 Response

0

0

1

2

3

4

5

6

Time (seconds)

Figure Review8No1

7

8

9

10

415

2. The model of a second-order system is given as 4  x + 4 x + 54 x = 0 , x(0) = 0.5 , x (0) = −1 (a) Identify the damping type and express the response x(t ) in closed form. (b) Plot the response x(t ) using the initial command. Solution 5 2 < 1 (underdamped), ω = 1 r/s , and (a) We have ωn = r/s, ζ = d 4 4 5

= x(t ) e−t /2  12 cos 14 t − 3sin 14 t 

(b) Using x1 = x and x2 = x , the state-space matrices are found as

 0 = A  5  − 16

1 0 0.5 = , B  = , C [1 0]= , D 0,= x0    −1 0  −1 

>> A=[0 1;-5/16 -1]; C=[1 0]; x0=[0.5;-1]; >> sys=ss(A,[],C,[]); >> initial(sys,x0) % Figure Review8No2 Response to Initial Conditions 0.5

0.4

0.3

Amplitude

0.2

0.1

0

-0.1

-0.2

-0.3

-0.4

0

2

4

6

8

10

Time (seconds)

Figure Review8No2

3. A second-order dynamic system is modeled as 9  x + 6 x + 10 x= 14δ (t ) , x(0) = 0 , x (0) = − 14 9

12

14

416 (a) Find the response x(t ) in closed form. Plot the response x(t ) using the impulse and initial commands. Also plot the impulse and initial responses in the same figure. Solution

(b)

10 3 1 r/s , and (a) We have ωn = r/s, ζ = < 1 (underdamped), ωd = 9 9 10

 1 14  − t /3 − 4 + 9 = x(t ) e=  1 sin( 19 t )   9 

47 − t /3 e sin( 19 t ) 4

(b) Using x1 = x and x2 = x , the state-space matrices are found as

1  0 0 0 A  10 B C x = = = = = = δ , , 1 0 , D 0, u , ] [  1  14  0 2 − 4   − 81 − 3  9 A=[0 1;-10/81 -2/3]; B=[0;14/9]; C=[1 0]; x0=[0;-1/4]; sys=ss(A,B,C,[]); [yinit,t]=initial(sys,x0); [yimp,t]=impulse(sys); plot(t,yinit,t,yimp,t,yinit+yimp) % Figure Review8No3

2

1.5

Impulse response

1

Response

>> >> >> >> >>

Total response

0.5

0

Initial response

-0.5

0

5

10

15 t

Figure Review8No3

20

25

417

4.

The mathematical model of a dynamic system is described by 4  x + 4 x += 5x

10 u (t ) 3 r

, x= (0) 0 , x= (0)

2 3

where ur (t ) is the unit-ramp. Plot the response x(t ) by (a) Using the lsim command over 0 ≤ t ≤ 5 . (b) Simulating the Simulink model of the system. Solution (a) Using x1 = x and x2 = x , the state-space matrices are found as

1 0  0 B  10  ,= C [1 0] ,= D 0,= u ur , = x0  2  ,=  −1  12  3

0 = A  5 − 4

>> A=[0 1;-5/4 -1]; B=[0;10/12]; C=[1 0]; x0=[0;2/3]; sys=ss(A,B,C,[]); >> t=linspace(0,5); u=t; lsim(sys,u,t,x0) % Figure Review8No4-1 Linear Simulation Results 5

4.5

4

3.5

Amplitude

3

2.5

2

1.5 Input

Response

1

0.5

0

0

0.5

1

1.5

2.5

2

3.5

3

4

4.5

5

Time (seconds)

Figure Review8No4-1 (b) The model is built as in Figure Review8No4-2, where the state-space matrices and initial state are defined in (a). Simulation leads to a response plot that exactly agrees with that in (a). x' = Ax+Bu y = Cx+Du

Ramp

1

Out1 State-Space

Scope

Figure Review8No4-2

418

5.

A dynamic system model is given as

 2  x1 + x1 − 13 ( x2 − x1 ) = u (t ) = , x1 (0) 1,= x2 (0) 0,= x1 (0)  5 1 u ( t )  x2 + x2 + 3 ( x2 − x1 ) = 2

1 5

where u (t ) is the unit-step. Plot x1 versus 0 ≤ t ≤ 15 by (a) Using the lsim command. (b) Simulating the Simulink model of the system. Solution (a) With = x1 x= x= x1 , the state-space matrices are obtained as 1 , x2 2 , x3 0 0  1 4 A=  3 −3 − 1 1  6 6 >> >> >> >>

 1  0    5 0, x0 0  u, C = [1 0 0] , D == , B = 2, u = 1  1 − 12  2 5  1 0

A=[0 0 1;1/3 -4/3 0;-1/6 1/6 -1/2]; B=[0;5/2;1/2]; C=[1 0 0]; sys=ss(A,B,C,[]); t=linspace(0,15); u=0*t+1.; x0=[1;0;1/5]; lsim(sys,u,t,x0) % Figure Review8No5-1

Linear Simulation Results 7

6

Amplitude

5

4

3

2

1

0

10

5

15

Time (seconds)

Figure Review8No5-1 (b) The model is built as in Figure Review8No5-2, where the state-space matrices and initial state are defined in (a). Simulation leads to a response plot that exactly agrees with that in (a).

419

Figure Review8No5-2 6.

The governing equations for a system are derived as

x1 + x1 + 3x1 − 2 x2 = 10δ (t )  3 , x1 (0) 1,= x2 (0) 0,= x1 (0) 0 =   x 3 x 3 x 0 − + = 1 2  2 where δ (t ) is the unit-impulse. Plot x1 using the impulse and initial commands. Solution With = x1 x= x= x1 , the state-space matrices are obtained as 1 , x2 2 , x3

0 0 1  0 1        δ, C = 0, x0 = A= [1 0 0] , D = 0   3 −3 0  , B =  0 , u = 0  10 2 1  −1   − 3    3  3 A=[0 0 1;3 -3 0;-1 2/3 -1/3]; B=[0;0;10/3]; C=[1 0 0]; sys=ss(A,B,C,[]); x0=[0;1;0]; [imp_res,t]=impulse(sys); % Same array of time for both data [init_res,t]=initial(sys,x0,t); plot(t,init_res+imp_res) % Figure Review8No6 4

3.5

3

2.5

1

2

Response x

>> >> >> >> >>

1.5

1

0.5

0

-0.5

-1

0

5

15

10 t

Figure Review8No6

20

420

7.

The mechanical system in Figure 8.37, where all parameter values are in consistent physical units, is subject to initial conditions x1 (0) = 1, x2 (0) = 1, x1 (0) = −1, x2 (0) = 1 . Plot the response x1 versus 0 ≤ t ≤ 25 . x1

x2 k2 = 1

k1 = 3 f = 15δ (t ) m1 = 0.8

b = 0.6

m2 = 1.2

Figure 8.37 Problem 7. Solution The equations of motion are

x1 + 3x1 − ( x2 − x1 ) − 0.6( x2 − x1 ) = 15δ (t )  0.8  x2 + x2 − x1 + 0.6( x2 − x1 ) = 0 1.2 Selecting state variables = x1 x1= , x2 x= x= x2 , the state equation is formed as 2 , x3 1 , x4

 0   0 x = 4  − 0.8  1  1.2

0 0

1 0

1 0.8 1 − 1.2

− 0.6 0.8 0.6 1.2

0  0    1   0 u , + x 0.6  15   0.8 0.8    0.6 − 1.2 0 

 x1  x   2 δ (t ) x=   , u= x 3    x4 

The following script produces the plot for response x1 .

A=[0 0 1 0;0 0 0 1;-4/0.8 1/0.8 -0.6/0.8 0.6/0.8;1/1.2 -1/1.2 0.6/1.2 0.6/1.2]; B=[0;0;15/0.8;0]; C=[1 0 0 0]; % x1 is to be plotted x0=[1;1;-1;1]; sys=ss(A,B,C,[]); [x1_imp,t]=impulse(sys,25); x1_initial=initial(sys,x0,t); % Response to initial conditions x1_total=x1_imp+x1_initial; plot(t,x1_total) % Figure Review8No7

421

8

7

6

5

Response x

1

4

3

2

1

0

-1

-2

0

5

10

15

20

25

t

Figure Review8No7

8. The Bode plot of a first-order system indicates that the corner frequency is 0.35 rad/sec and high-frequency (at ω = 100 rad/sec ) logarithmic magnitude is −37.45 dB . (a) Find the system’s transfer function. (b) Using the results of (a), find the approximate low-frequency logarithmic magnitude. Solution 1 (a) Let the generic first-order transfer function be G ( s ) = K . The corner frequency is τ s +1 1/ τ = 0.35 which implies τ = 207 . The asymptotic approximation for the high-frequency range is 20 log K − 20 log τω . Using τ = 207 and ω = 100 rad/sec , this gives −37.45 20 log K − 20 log 2000 7 =

⇒

K= 3.8321

Therefore, the transfer function is G (s) =

3.8321 20 7 s +1

(b) The approximate low-frequency logarithmic magnitude is = 20 log K 20 = log 3.8321 11.6687 dB

422 9. Determine the damping ratio associated with a second-order system in the standard form of Eq. (8.32) that corresponds to a maximum (peak) logarithmic magnitude of 15.22 dB . Solution We have 20 log G ( jω ) max = 15.22 dB , hence G ( jω ) max = 5.7677 . Therefore, G ( jω ) max =

1 = 5.7677 2ζ 1 − ζ 2

Solving this equation yields ζ = 0.0870, 0.9962 . However, for the magnitude curve to attain a maximum peak, we must have ζ < 0.7071 . Therefore, the only acceptable value here is ζ = 0.0870 .

10. The Bode plot of a second-order system indicates that the approximate low-frequency magnitude is 2.28 dB , the approximate high-frequency ( ω = 1000 r/s ) magnitude is −106 dB , and the logarithmic gain attains a peak value of 4.98 dB . Find the system’s transfer function. Solution ωn2 Let the generic second-order transfer function be G ( s ) = K 2 . The asymptotic s + 2ζωn s + ωn2 approximation for the low-frequency range is given by 20 log K = 2.28 which implies ω K = 1.3002 . The approximation for the high-frequency range is 20 log K − 40 log . Using ωn K = 1.3002 and ω = 1000 r/s , this gives ωn = 1.9634 r/s . The peak logarithmic gain is given as 4.98 dB . This peak includes the effect of 20 log K = 2.28 . To find the maximum peak associated with the standard second-order transfer function, 4.98 dB − 2.28 dB = 2.7 dB . Then,

1 = = 20 log G ( jω ) max 20 log 2.7 2ζ 1 − ζ 2 Solving for ζ we find ζ = 0.3997 to be the only acceptable value (less than 0.7071 ). Therefore, the transfer function is 1.96342 5.0121 G ( s ) 1.3002 = 2 2 2 s + 2(0.3997)(1.9634) s + 1.9634 s + 1.5695s + 3.8549

11.

Consider

 x1 =− x1 x1 + ax2 − 23 + sin t ,  − x1 − 12 x2 − 1  x2 =

x1 (0) = 23 x2 (0) = 1

423 where a is a parameter. Using the RK4 method, plot x1 versus 0 ≤ t ≤ 10 for the two cases a = 1.25 , a = 2.5 in the same graph. Solution >> >> >> >> >> >>

t=linspace(0,10); x0=[2/3;1]; f1=@(t,x)([-x(1)*abs(x(1))+1.25*x(2)-2/3+sin(t);-x(1)-x(2)/2-1]); f2=@(t,x)([-x(1)*abs(x(1))+2.5*x(2)-2/3+sin(t);-x(1)-x(2)/2-1]); x1=RK4System(f1,t,x0); x11=x1(1,:); x2=RK4System(f2,t,x0); x12=x2(1,:); plot(t,x11,t,x12) % Figure Review8No11

1

a = 2.5 0.5

a = 1.25

x

1

0

-0.5

-1

-1.5

0

1

2

3

4

5 t

6

7

8

Figure Review8No11

12.

The nonlinear state-variable equations for a dynamic system are derived as −2 x2 + 3.3sin t x1 (0) = 0.7  x1 = ,  3 x2 (0) = 0  x2 =x1 − 3 x2 + 5

Plot x1 (t ) versus 0 ≤ t ≤ 10 by (a) Using the RK4 method. (b) Simulating the Simulink model of the system. Solution (a) >> >> >> >>

t=linspace(0,10); x0=[0.7;0]; f=@(t,x)([-2*x(2)+3.3*sin(t);x(1)^3-3*x(2)+5]); x=RK4System(f,t,x0); x1=x(1,:); plot(t,x1) % Figure Review8No12-1

9

10

424

1

0.5

0

x

1

-0.5

-1

-1.5

-2

-2.5 0

1

5 t

4

3

2

6

8

7

9

10

Figure Review8No12-1

(b) The Simulink model is built as shown in Figure Review8No12-2. Simulation will result in the exact same plot as in (a). 1

Out1 = x1 5

1 s

3.3sin(t)

Integrator 1

Constant

Scope

IC: x1(0) = 0.7

IC: x2(0) = 0

u^3

1 s

Nonlinear function

Integrator 2

x1

3

Gain 2

Gain1

Figure Review8No12-2

x2

425

Problem Set 9.1 1.

A lightly damped single-degree-of-freedom system is subjected to free vibration. The response of the system is shown in Figure 9.5. Estimate the value of the viscous damping ratio ζ.

Figure 9.5 Problem 1.

Solution Note that the first and fifth peak displacements are about 0.62 m and 0.09 m, respectively. Thus, the logarithmic decrement is 1 x1 1 0.62 = δ = ln ln= 0.48 4 x5 4 0.09 The damping ratio is δ 0.48 = = 0.08 ζ = 2 2 2 ( 2π ) + 0.482 ( 2π ) + δ Since the damping is very small, it can also be estimated as δ 0.48 ζ≈ = = 0.08 2π 2π 2.

An underdamped single-degree-of-freedom vibrating system is viscously damped. It is observed that the maximum displacement amplitude during the third cycle is 60% of the first. Calculate the damping ratio ζ and determine the maximum displacement amplitude during the fourth cycle as a fraction of the first.

Solution It is observed that the maximum displacement amplitude during the third cycle is 60% of the first. Thus, the logarithmic decrement is 1 x1 1 1 = δ = ln = ln 0.26 2 x3 2 0.6 The damping ratio is δ 0.26 = = 0.04 ζ = 2 2 2 ( 2π ) + δ ( 2π ) + 0.262 During the fourth cycle, the maximum displacement amplitude is 32

x4 x4 x3 x2  x3  32 = =  = 0.6 = 46.48%  x1 x3 x2 x1  x1  of the first.

426

3.

Figure 9.6 is the free response of a single-degree-of-freedom system subjected to Coulomb damping. The parameters of the system include the mass (m = 40 kg) and the spring stiffness (k = 2000 N/m). Estimate the value of the kinetic friction coefficient μk.

Figure 9.6 Problem 3.

Solution Note that the first, second, and third peak displacements are about 0.4 m, 0.24 m, and 0.08 m, respectively. Thus, the decay per cycle is 4∆ =0.16 m The kinetic friction coefficient is Ff k ∆ 2000 ( 0.16 4 ) = = = 0.20 μk = mg mg 40 ( 9.81) 4.

A Coulomb damped vibrating system consists of a mass of 8 kg and a spring of stiffness 5000 N/m. The kinetic friction coefficient μk is 0.1. The initial conditions are x0 = 0.02 m and v0 = 0 m/s. a. Determine the decay per cycle. b. Determine the position when the oscillation stops.

Solution a.

b.

The magnitude of the Coulomb damping force is = Ff m= 0.1(8)(9.81) = 7.85 N k mg and F 7.85 ∆= f= = 1.6 × 10−3 m k 5000 Thus, the decay per cycle is = 4(1.6 × 10−3= 4∆ ) 6.4 × 10−3 m Note that the displacement magnitude is reduced by 2Δ after every half-cycle. The motion stops at the end of the half-cycle for which the displacement magnitude is smaller than 2Δ. This can be expressed mathematically as x0 − n (2∆ ) < 2∆ where n denotes the number of half-cycles before stopping. Solving for n and substituting the appropriate values, we have x 0.02 n= > 0 −1 = − 1 5.25 2∆ 2(1.6 × 10−3 ) The smallest integer satisfying the inequality is n = 6. Thus, the oscillation stops after six half-cycles, or three cycles, with x(t = 6T / 2) = x0 − 6(2∆) = 0.02 − 12(1.6 × 10−3 ) = 8 × 10−4 m

427

Problem Set 9.2 1.

Figure 9.13 shows the experimental data for the frequency response of a single-degree-of-freedom system. Use the half-power bandwidth method to estimate the damping ratio ζ of the system.

Figure 9.13 Problem 1.

Solution Note that the frequencies corresponding to half-power points, f1 and f2, are about 3.8 Hz and 4.5 Hz, respectively. For light damping, the natural frequency is f n ≈ 12 ( f1 + f 2 ) = 4.15 Hz Thus, the damping ratio is

ζ≈ 2.

f −f ∆ω ∆f 4.5 − 3.8 = = 2 1= = 0.08 2ω n 2 f n 2 fn 2 ( 4.15)

A viscously damped single-degree-of-freedom system is subjected to a harmonic force excitation. It is observed that the amplitude of the frequency response X/xst reaches the maximum when the driving frequency is 150 rpm. The peak value is 40. Assume the system to be lightly damped. a. Determine the damping ratio ζ. b. Determine the bandwidth of the system.

Solution a.

For small damping, the damping ratio can be estimated using the peak amplitude of the frequency response, i.e., 1 1 ζ≈ = = 0.0125 2Q 2 ( 40 )

b.

Since the system is lightly damped, the peak occurs in the immediate neighborhood of ω/ωn = 1, i.e., ωn= ≈ ω 150 = rpm 5π rad/s . Thus, the bandwidth of the system is

∆= ω 2ζω= 2 ( 0.0125 )( 5π= ) 0.39 rad/s n

3.

An industrial machine of mass M = 450 kg is supported by a spring with a static deflection xst = 0.5 cm. If the machine has a rotating unbalance me = 0.25 kg·m, determine the dynamic amplitude X at 500 rpm. Assume damping to be negligible.

Solution The stiffness of the spring is

k =

Mg 450(9.81) = = 882,900 N/m xst 0.5 × 10−2

428

and the natural frequency of the system is

k 882,900 = = 44.29 rad/s M 450 Given the driving frequency of 500 rpm, the frequency ratio is ω 500(2π ) = = 1.18 ωn 60(44.29) Neglecting the damping and solving the dynamic amplitude, we have = ωn

X=

4.

( me

M )( ω/ωn )

1 − ( ω/ωn )

2

2

=

(0.25 / 450)(1.182 ) 1 − 1.182

= 2 × 10−3 m

Tires must be balanced so that no periodic forces develop during operation. Figure 9.14 shows a tire with an eccentric mass because of uneven wear. The parameters are given as follows: the mass of the tire is M = 11.75 kg, the unbalanced mass is m = 0.1 kg, the radius of the tire is r = 22.5 cm, and the eccentric distance is e = 15 cm. Assume that the stiffness of the tire is 120 kN/m. Neglect the damping of the system. Determine the amplitude of the steady-state response of the tire caused by mass unbalance when the car moves at (a) 40 km/h and (b) 80 km/h.

Figure 9.14 Problem 4.

Solution a.

The natural frequency of the tire is

120 × 103 k = = 101.06 rad/s 11.75 M When the car moves at 40 km/h, the frequency ratio is ω vr 40000 = = = 0.49 ωn ωn 3600 ( 22.5 × 10−2 ) (101.06 ) ωn =

Neglect the damping of the system. The dimensionless ratio is

( ω ωn ) MX 0.492 = = = 0.32 me 1 − ( ω ωn )2 1 − 0.492 2

which gives the amplitude of the steady-state response of the tire as −2 0.32me 0.32 ( 0.1) (15 × 10 ) X = = = 4.09 × 10−4 m M 11.75 b. When the car moves at 80 km/h, the frequency ratio is 0.98. Repeating the calculation in Part (a) gives = X 3.10 × 10−2 m. 5.

Reconsider Example 9.5, in which the mathematical model of a vehicle is given by an ordinary differential equation mx + bx + kx = bz + kz with m = 3000 kg, b = 2000 N⋅s/m, and k = 50 kN/m. It is observed that the base excitation due to the roughness of the road surface is also related to the speed of the vehicle, i.e., z = 0.01sin(0.2πvt). Determine the transmissibility and the dynamic amplitude X of the vehicle when it moves at a speed of (a) 25 km/h and (b) 105 km/h.

Solution

429

a.

The natural frequency and the damping ratio of the vehicle system are

k = m

= ωn

50 × 103 = 4.08 rad/s 3000

and

= ζ

b 2000 = = 0.08 2 mk 2 3000 ( 50 × 103 )

When the vehicle moves at a speed of 25 km/h, the frequency of the harmonic excitation is 0.2π ( 25000 ) ω 0.2π = = v = 4.36 rad/s 3600 and the frequency ratio is ω 4.36 = = 1.07 ω n 4.08 Thus, the transmissibility is

1 +  2 ( 0.08 )(1.07 )  1 + ( 2ζω/ω n ) X = = = 4.52 2 2 2 Z0 1 − ( ω/ω n )2  + ( 2ζω/ω n )2 1 − 1.072 ) + ( 2 ( 0.08 )(1.07 ) ) (   Thus, the dynamic amplitude of the vehicle is= X 4.52 × 10−2 m. When the vehicle moves at a speed of 105 km/h, the frequency of the harmonic excitation is 0.2π (105000 ) = ω 0.2π = v = 18.33 rad/s 3600 and the frequency ratio= is ω ω n 18.33 = 4.08 4.49 . Repeating the calculation in Part (a) gives X = 6 × 10−4 m. 2

2

b.

6.

Precision instruments must be placed on rubber mounts, which act as springs and dampers, to reduce the effects of base vibration. Consider a precision instrument of mass 110 kg mounted on a rubber block. For the entire assembly, the spring stiffness is 250 kN/m and the damping ratio is 0.10. Assume that the base undergoes vibration, and the displacement of the base is expressed as y(t) = Y0sin(ωt). Determine the dynamic amplitude of the system if the acceleration amplitude of the base excitation is 0.15 m/s2 and the excitation frequency is (a) 8 Hz and (b) 24 Hz.

Solution a.

The expression for the acceleration of the base is  y (t ) = −ω 2Y0 sin ωt 2 2 where ω Y0 = 0.15m/s . For an excitation frequency of 8 Hz, the displacement amplitude of the base is 0.15 0.15 = Y0 = = 5.94 × 10−5 m 2 ω2 [8(2π)] The frequency ratio is

ω = ωn

8(2π) = 1.05 250000 / 110

and the transmissibility is

1 + [ 2(0.1)(1.05) ] 1 + ( 2ζω/ωn ) = = 4.37 2 2 (1 − 1.052 ) 2 + ( 2(0.1)(1.05) ) 1 − ( ω/ωn )2  + ( 2ζω/ωn )2   Thus, the dynamic amplitude of the system is X = 4.37 (5.94 × 10−5) = 2.60 × 10−4 m. For an excitation frequency of 24 Hz, the displacement amplitude of the base is 0.15 0.15 6.60 × 10−6 m = Y0 = = 2 2 ω 24(2π) [ ] X Y0

b.

The frequency ratio is

2

2

430

ω 24(2π) = = 3.16 ωn 250000 / 110 Repeating the calculation in Part (a) gives X = 8.60 × 10−7 m.

Problem Set 9.3 1.

A 25-kg instrument is suspended from a ceiling by four springs, each of which has a stiffness of 800 N/m. The ceiling vibrates with a frequency of 2 Hz and amplitude of 0.05 mm due to the air conditioning compressor and chiller mounted on the roof. Neglecting the damping, determine the maximum displacement amplitude of the instrument.

Solution The natural frequency of the instrument is

k 4(800) = = 11.31 rad/s m 25 When the ceiling vibrates with a frequency of 2 Hz, the frequency ratio is ω 2(2π) = = 1.11 ωn 11.31 Thus, the transmissibility is X 1 1 = = = 4.31 2 Z 0 1 − ( ω/ωn ) 1 − 1.112 = ωn

and the maximum displacement amplitude of the instrument is X = 4.31(0.05 × 10−3) = 2.15 × 10−4 m. 2.

Consider a 10-kg instrument placed on a floor that vibrates with a frequency of 3000 rpm and amplitude of 2 mm due to nearby machinery. A vibration isolator is designed to protect the instrument from the vibration of the floor. Assume that the damping ratio of the isolator is 0.05 and the maximum allowable acceleration amplitude of the instrument is 2g. a. Determine the stiffness of the isolator. b. Determine the maximum displacement amplitude of the instrument.

Solution a.

Assume that the expression for the displacement of the instrument is x(t) = Xsin(ωt+φ). The acceleration amplitude of the instrument is A = Xω 2 , where the maximum allowable acceleration amplitude is 2g. When the floor vibrates with a frequency of 3000 rpm and amplitude of 2 mm, the transmissibility is 2 ( 9.81) X A = = = 0.10 2 Z 0 ω Z 0 ( 3000 × 2π 60 )2 ( 0.002 ) Assume that the damping ratio of the isolator is 0.05. The transmissibility can be given by

1 + ( 2ζω/ωn ) X = = 2 Z0 1 − ( ω/ωn )2  + ( 2ζω/ωn )2   which is equivalent to 2

4

1 +  2 ( 0.05 )( ω/ωn )  = 0.10 2 1 − ω/ωn )2  +  2 ( 0.05 )( ω/ωn )  2   (   2

2

 ω   ω  0   − 0.99   − 99 =  ωn   ωn  Solving for ( ω ω n ) gives 2

( ω ωn )

2

= 10.46 or ( ω ωn ) = −9.47 2

The negative one is not valid. Thus, the frequency ratio ω ω n is

= ωn Thus, the stiffness of the isolator is

10.46 or 3.23, and the natural frequency is

ω 3000(2π) = = 97.26 rad/s 3.23 60(3.23)

431

b.

3.

2 = k ω= 97.262 (= 10 ) 95 kN/m nm The maximum displacement amplitude of the instrument is 2 ( 9.81) A = = = X 1.99 × 10−4 m 2 2 ω ( 3000 × 2π 60 )

A 9000-kg air conditioning compressor mounted on a roof is supported by four springs. The static deformation of each spring is 4 cm. Neglecting the damping, determine the force transmissibility when the compressor works at 60 Hz.

Solution The equivalent spring stiffness is

mg 9000(9.81) = = 2207.25 kN 0.04 ∆

= keq Thus, the natural frequency of the system is

keq 2207.25 × 103 = = 15.66 rad/s m 9000 Neglecting the damping, the force transmissibility when the compressor works at 60 Hz is FT 1 1 = = = 0.17% 2 F0 1 − ( ω ωn )2  60(2π)  1−    15.66  = ωn

4.

Consider the single-degree-of-freedom system shown in Figure 9.8. The excitation force due to the rotating unbalance is meω2sin(ωt). Assume that the system has the following parameters: m = 5 kg, M = 100 kg, e = 0.1 m, k = 5,000 N/m, and b = 200 N·s/m. a. Determine the force transmitted to the support when the system runs at the rotating speed ω = 1.5ωn. b. Determine the force transmissibility. c. The spring is replaced to decrease the force transmissibility at the same excitation frequency to 20%. Determine the stiffness of the new spring.

Solution a.

The natural frequency of the system is

= ωn

k = M

5000 = 7.07 rad/s 100

and the damping ratio of the system is

b 200 = = 0.14 2 mk 2 100 ( 5000 )

= ζ

When the system runs at the rotating speed ω = 1.5ωn, the dimensionless ratio is

( ω ωn ) 1.52 = = 1.71 2 2 2 1 − ( ω ωn )2  +  2ζ ( ω ωn )  2 1 − 1.52  +  2 ( 0.14 )(1.5 )      Thus, the dynamic amplitude is 1.71me 1.71( 5 )( 0.1) X = = = 8.6 × 10−3 m M 100 The force transmitted to the support is 2

MX = me

FT = kX 1 + ( 2ζω ωn ) = 5000 ( 8.6 × 10−3 ) 1 +  2 ( 0.14 )(1.5 )  = 46.64 N The force transmissibility is 2

b.

2

432

c.

gives

If the force transmissibility is decreased to 20%, solving for

or

,

which is not valid. Thus, the new natural frequency is rad/s and the stiffness of the new spring is kN/ m 5.

A rotating machine has a mass of 6 kg and a natural frequency of 5 Hz. Due to a rotating unbalanced mass, the machine is subjected to a harmonic disturbance force meω2sin(ωt). When the machine operates at a frequency of 3.5 Hz, the amplitude of the disturbance force is 40 N. a. Design a vibration absorber assuming that the maximum allowable displacement of the absorber is 5 cm. b.

Using MATLAB, write an m-file to plot X1/me versus ω. Use the decibel units to show the vertical axis.

Solution a.

The natural frequency of the absorber is required to be the same as the disturbance frequency, rad/s The maximum allowable displacement amplitude of the absorber is 5 cm. Thus, the spring stiffness of the absorber is N/m and the mass of the absorber is kg

b.

The following is the MATLAB session. m = 6; %main mass, kg wn = 2*pi*5; %natural frequency of the main mass, rad/s F1 = 40; %disturbance force, N wf = 2*pi*3.5; %disturbance frequency, rad/s me = F1/wf^2; %rotating unbalanced mass X2 = 0.05; %maximum displacement amplitude of the absorber, m w2 = wf; %natural frequency of the absorber, rad/s k2 = F1/X2; %stiffness of the absorber, N/m m2 = k2/w2^2; %mass of the absorber, kg u = m2/m; v = w2/wn; w = 0:0.1:50; r = w./wn; X1xst = (v^2-r.^2)./((v^2-r.^2).*(1+u*v^2-r.^2)-u*v^4); xst = F1/m; X1me = X1xst*xst/me; X1me_log = 20*log10(X1me); plot(w,X1me_log); xlabel('\omega (rad/s)'); ylabel('X_1/me (dB)'); grid on;

%ratio X1/xst %ratio X1/me %decibel units

433

100

80

40

1

X /me (dB)

60

20

3.5 Hz

0

-20 0

10

20

30

40

50

(rad/s)

Figure PS9-3 No5 6.

The pendulum in Figure 9.19 is known as a tuned mass damper. It is mounted in a building, which is simplified as a block of mass m1 supported by a spring of stiffness k. The mass of the pendulum is m2 and the length l is tunable. The tunable pendulum is used to control the vibration of the building under extreme wind loads. Assume that the force due to gusty winds can be modeled as a harmonic force F1sin(ωt), and the excitation frequency ω is very close to the natural frequency of the building. Prove that the forced vibration of the building can be eliminated when the length of the pendulum is tuned such that ω = g l , where g is the gravitational acceleration.

Figure 9.19 Problem 6.

Solution Assume that x2 > x1 > 0. The free-body diagram of the building and the pendulum is shown in the figure below.

Figure PS9-3 No6 Applying Newton’s second law along x-direction gives + → x : ∑ Fx =maCx

Building: F1 sin(ωt ) − kx1 + T sin φ = m1  x1  Pendulum: −T sin φ = m2 x2 Note that for small angles, Tcosφ = m2g. Thus, we have

434

The above two differential equations can be rewritten as

Comparing the equations with (9.65),

the term of m2g/l is similar to k2. The tunable pendulum can only be effective when its own natural frequency is the same as the excitation frequency, i.e.,

Problem Set 9.4 1.

Consider the two-degree-of-freedom mass–spring system shown in Figure 9.22. Assume that m1 = m, m2 = 2m, and k1 = k2 = k, where m = 5 kg and k = 2000 N/m. a. Derive the equations of motion and express them in matrix form. b. Solve the associated eigenvalue problem by hand. Plot the two modes and explain the nature of the mode shapes. c.

Solve the associated eigenvalue problem using MATLAB.

Figure 9.22 Problem 1.

Solution a.

The differential equations of motion in matrix form are

b.

where kg, , and The mass and stiffness matrices are

N/m.

Substituting M and K into the frequency equation

, we have

which gives the two eigenvalues , Thus, the two natural frequencies are rad/s

435

rad/s To determine the eigenvectors or modal vectors, we insert

(r = 1, 2) into the equation

. For

r = 1, we have

Substituting

Assuming

and dividing by k results in

leads to

. In a similar way, we can find the other modal vector. ,

The modal vectors represent physically the shape of the natural modes as shown in the figure below. In mode 1, rad/s. In mode 2, the two the two masses move in the same direction and oscillate with a frequency of masses move in the opposite directions with a frequency of

rad/s.

Figure PS9-4 No1 c.

2.

The following is the MATLAB session. m = 5; k = 2000; M = diag([m 2*m]); K = [2*k -k; -k k]; [U,Omega2] = eig(K,M); omegar = diag(sqrt(Omega2));

% % % %

mass matrix stiffness matrix solve the eigenvalue problem extract the natural frequencies

A three-story building can be modeled as a three-degree-of-freedom system as shown in Figure 9.23, in which the horizontal members are rigid and the columns are massless beams acting as springs. Assume that m1 = 1200 kg, m2 = 2400 kg, m3 = 3000 kg, k1 = 500 kN/m, k2 = 1000 kN/m, and k3 = 1500 kN/m.

a.

Figure 9.23 Problem 2. Derive the differential equations for the horizontal motion of the masses.

436

b. c.

Solve the associated eigenvalue problem by hand. Plot the three modes and explain the nature of the mode shapes. Solve the associated eigenvalue problem using MATLAB.

Solution a.

The differential equations of motion in matrix form are

b.

where m1 = 1200 kg, m2 = 2400 kg, m3 = 3600 kg, k1 = 500 kN/m, k2 = 1000 kN/m, and k3 = 1500 kN/m. The mass and stiffness matrices are

where m = 1200 kg and k = 500 kN/m. Substituting M and K into the frequency equation

Dividing the above equation by k and letting

gives

, we have

The three roots are ,

,

Thus, the three natural frequencies are rad/s rad/s rad/s To determine the eigenvectors or modal vectors, we insert For

(

) into the equation

, we have

Substituting

Assuming

and dividing by

leads to

and

results in

. Thus, the first modal vector is

.

437

In a similar way, we can find the other two modal vectors.

The modal vectors represent physically the shape of the natural modes as shown in the figure below. In mode 1, all floors move in the same direction and oscillate with a frequency of 11.16 rad/s. In mode 2, all floors oscillate with a frequency of 23.29 rad/s. The floors 2 and 3 move in the same direction, while floor 1 moves in the opposite direction. In mode 3, the floors 1 and 3 move in the same direction, while floor 2 moves in the opposite direction. All masses oscillate with a frequency of 32.68 rad/s.

Figure PS9-4 No2 c.

3.

The following is the MATLAB session. M = diag([1200 2400 3600]); K = [500 -500 0; -500 1500 -1000; 0 -1000 2500]*1000; [U,Omega2] = eig(K,M); omegar = diag(sqrt(Omega2));

% mass matrix % stiffness matrix

% solve the eigenvalue problem % extract the natural frequencies

Consider the two-degree-of-freedom mass–spring system in Figure 9.22. Normalize the modal vectors X1 and X2 obtained in Part (b) of Problem 1 to a set of orthonormal modal vectors satisfying Equations 9.96 and 9.97.

Solution Write the first modal vector in the form

, where

nonzero scaling constant. Substituting

into Eq. (9.96) gives

is orthonormal. Inserting

in Problem 1 yields

if

Thus,

and

and

is the modal vector after normalization and

is a

438

0.1650   = X α= X1  1  0.2939  Following the same procedure yields the other orthonormal modal vector,  0.4157   = X 0.4157 = X2  2   −0.1167  4.

Consider the three-story building system in Figure 9.23. Normalize the modal vectors X1, X2, and X3 obtained in Part (b) of Problem 2 to a set of orthonormal modal vectors satisfying Equations 9.96 and 9.97.

Solution

 = αX , where X  is the modal vector after normalization and α is a Write the first modal vector in the form X 1 1 1  into Eq. (9.96) gives nonzero scaling constant. Substituting X 1

 T MX  = X 1 1

M ( αX 1 ) ( αX 1 ) = T

2 T α= X1 MX1 1

 is orthonormal. Inserting X and M in Problem 2 yields if X 1 1

 1   2  T α 2 X= α MX 1 1  0.7009  0.3417 

T

0 0  1  1200  0   0.7009  2799α 2 2400 0   =  = 1  0 0 3600  0.3417 

Thus, α = 0.0189 and

0.0189   X X1 0.0132  = α= 1  0.0065  Following the same procedure yields the other two orthonormal modal vectors,  0.0198   −0.0091     X 2 = 0.0198X 2 =  −0.0060 = = X3  0.0143   , X3 0.0143  −0.0111  −0.0106  5.

Consider the two-degree-of-freedom mass–spring system in Figure 9.22. Determine the response of the system subjected to initial excitations x(0) = [0 0]T and x (0) = [1 0] by modal analysis. Note that the natural frequencies T

were found in Problem 1 and the modal vectors were normalized in Problem 3.

Solution From Problems 1 and 3, we obtained the two natural frequencies ω1 = 9.36 rad/s, ω2 = 30.20 rad/s and the two orthonormal modal vectors  = 0.1650  , X  =  0.4157  X 1 2 0.2939   −0.1167      The modal responses are  T Mx(0) cos(ω t ) + 1 X  T Mx (0) sin(ω t ) r = X qr (t ) = 1, 2 r r r r ωr where  5 0  1  1 q1 (t ) = [0.1650 0.2939]     sin ( 9.36t ) 0.0881sin ( 9.36t ) 9.36 0 10  0 

 5 0  1  1 q2 (t ) = 0.0688sin ( 30.20t ) [0.4157 −0.1167]     sin ( 30.20t ) = 30.20 0 10  0  The response of the system to the given initial excitations is

439

 q (t ) + X  q (t ) = x(t ) X 1 1 2 2 0.1650   0.4157  = 0.0881sin ( 9.36t )   + 0.0688sin ( 30.20t )  −0.1167  0.2939     0.0145 0.0286     =   sin ( 9.36t ) +  −0.0080  sin ( 30.20t ) 0.0259     6.

Consider the three-story building system in Figure 9.23. Determine the response of the system subjected to initial excitations x(0) = [0.01 0 0]T and x (0) = [ 0 0 0] by modal analysis. Note that the natural frequencies were found T

in Problem 2 and the modal vectors were normalized in Problem 4.

Solution From Problems 2 and 4, we obtained the natural frequencies ω1 = 11.16 rad/s, ω2 = 23.29 rad/s, ω3 = 32.68 rad/s and the orthonormal modal vectors  0.0198  0.0189   −0.0091  =  −0.0060  , X  = 0.0132  , X  =  0.0143  X 2 3 1        −0.0111  0.0065   −0.0106  The modal responses are

 T Mx(0) cos(ω t ) + 1 X  T Mx (0) sin(ω t ) r = qr (t ) = X 1, 2,3 r r r r ωr where

T = = q1 (t ) X 0.2268cos (11.16t ) 1 Mx(0) cos(ω1t ) T = = q2 (t ) X 0.2376 cos ( 23.29t ) 2 Mx(0) cos(ω 2 t )

 T Mx(0) cos(ω t ) = −0.1092 cos ( 32.68t ) q3 (t ) = X 3 3 The response of the system to the given initial excitations is  q (t ) + X  q (t ) + X  q (t ) x(t ) = X 1 1 2 2 3 3  0.0043  0.0047   0.0010      = 0.0030  cos (11.16t ) +  −0.0014  cos ( 23.29t ) +  −0.0016  cos ( 32.68t )  0.0015   −0.0026   0.0012  7.

Consider the two-degree-of-freedom mass–spring system in Figure 9.22. Determine the response of the system if a harmonic force F(t) = 2 cos(3t) is applied to mass 2. Note that the natural frequencies were solved in Problem 1 and the modal vectors were normalized in Problem 3.

Solution The force vector can be written as

 0   0  0  = f =   =    cos(3t )  F (t )  2 cos(3t )  2  From Problems 1 and 3, we obtained the two natural frequencies ω1 = 9.36 rad/s, ω2 = 30.20 rad/s and the two orthonormal modal vectors  = 0.1650  , X  =  0.4157  X 1 2 0.2939   −0.1167      The modal responses are  Tf X r 0 cos(ωt ) r 1, 2 = qr (t ) = 2 ωr − ω2 where

440

0  0.2939]   2  cos(3t ) 0.0075cos(3t ) = q1 (t ) = 2 2 9.36 − 3 0  [0.4157 −0.1167]   2  cos(3t ) = q2 (t ) = −2.5846 × 10−4 cos(3t ) 2 2 30.20 − 3 The response of the system to the given harmonic excitation is  q (t ) + X  q (t ) x(t ) X = 1 1 2 2

[0.1650

0.1650   0.4157  = 0.0075cos(3t )  − 2.5846 × 10−4 cos(3t )    0.2939   −0.1167   0.0011 =  cos(3t ) 0.0022  8.

Consider the three-story building system in Figure 9.23. Determine the response of the system if a harmonic force F(t) = 15 sin(0.1t) is applied to the top story. Note that the natural frequencies were found in Problem 2 and the modal vectors were normalized in Problem 4.

Solution The force vector can be written as

 F (t )  15sin(0.1t )  15       = f = 0   0=   0  sin(0.1t )  0    0 0       From Problems 2 and 4, we obtained the natural frequencies ω1 = 11.16 rad/s, ω2 = 23.29 rad/s, ω3 = 32.68 rad/s and the orthonormal modal vectors  0.0198  0.0189   −0.0091        X1 = 0.0132  , X 2 =  −0.0060  , X3 =  0.0143   −0.0111  0.0065   −0.0106  The modal responses are  Tf X r 0 = qr (t ) = sin(ωt ) r 1, 2,3 2 ωr − ω2 where 15 [0.0189 0.0132 0.0065]  0  0   sin(0.1t ) 0.0023sin(0.1t ) = q1 (t ) = 2 2 11.16 − 0.1 15 [0.0198 −0.0060 −0.0111]  0  0   sin(0.1 q2 (t ) t ) 4.4755 × 10−4 sin(0.1t ) = = 23.292 − 0.12 15   − − 0.0091 0.0143 0.0106 [ ] 0  0   sin(0.1t ) = −1.2781× 10−4 sin(0.1t ) q3 (t ) = 32.682 − 0.12 The response of the system to the given harmonic excitation is

441

9.

Solve Problem 5 using MATLAB.

Solution The following is the MATLAB session written based on Eqs. (9.106) and (9.114). % eigenvalue problem m = 5; k = 2000; M = diag([m 2*m]); % mass matrix K = [2*k -k; -k k]; % stiffness matrix [U,Omega2] = eig(K,M); % solve the eigenvalue problem omegar = diag(sqrt(Omega2)); % extract the natural frequencies % modal response to initial excitations x0 = [0 0]'; % initial displacement xv0 = [1 0]'; % initial velocity q0 = U'*M*x0; % initial modal displacement qv0 = U'*M*xv0; % initial modal velocity figure(1); t = 0:0.001:1; q = zeros(length(t),2); for i = 1:2 subplot(2,1,i); q(:,i) = q0(i)*cos(omegar(i)*t)+qv0(i)/omegar(i)*sin(omegar(i)*t); plot(t,q(:,i),'b-'); end % system response to initial excitations x = zeros(2,length(t)); for i = 1:length(t) x(:,i) = U*q(i,:)'; end figure(2); for i = 1:2 subplot(2,1,i); plot(t,x(i,:),'b-'); end

Figure PS9-4 No9a Modal responses. 10.

Solve Problem 6 using MATLAB.

Solution

Figure PS9-4 No9b System responses.

442

The MATLAB session is similar to the one given in Problem 9. % eigenvalue problem M = diag([1200 2400 3600]); % mass matrix K = [500 -500 0; % stiffness matrix -500 1500 -1000; 0 -1000 2500]*1000; [U,Omega2] = eig(K,M); % solve the eigenvalue problem omegar = diag(sqrt(Omega2)); % extract the natural frequencies % modal response to initial excitations x0 = [0.01 0 0]'; % initial xv0 = zeros(3,1); % initial q0 = U'*M*x0; % initial qv0 = U'*M*xv0; % initial

displacement velocity modal displacement modal velocity

figure(1); t = 0:0.001:1; q = zeros(length(t),3); for i = 1:3 subplot(3,1,i); q(:,i) = q0(i)*cos(omegar(i)*t)+qv0(i)/omegar(i)*sin(omegar(i)*t); plot(t,q(:,i),'b-'); end % system response to initial excitations x = zeros(3,length(t)); for i = 1:length(t) x(:,i) = U*q(i,:)'; end figure(2); for i = 1:3 subplot(3,1,i); plot(t,x(i,:),'b-'); end

11.

Figure PS9-4 No10a Modal responses. Solve Problem 7 using MATLAB.

Figure PS9-4 No10b System responses.

Solution The following is the MATLAB session written based on Eqs. (9.121) and (9.122).

443

% eigenvalue problem m = 5; k = 2000; M = diag([m 2*m]); K = [2*k -k; -k k]; [U,Omega2] = eig(K,M); omegar = diag(sqrt(Omega2));

% % % %

mass matrix stiffness matrix solve the eigenvalue problem extract the natural frequencies

% modal response to external excitations f0 = [0 2]'; omega = 3; figure(1); t = 0:0.01:10; q = zeros(length(t),2); for i = 1:2 subplot(2,1,i); q(:,i) = U(:,i)'*f0/(omegar(i)^2-omega^2)*cos(omega*t); plot(t,q(:,i),'b-'); end % system response to external excitations x = zeros(2,length(t)); for i = 1:length(t) x(:,i) = U*q(i,:)'; end figure(2); for i = 1:2 subplot(2,1,i); plot(t,x(i,:),'b-'); end

Figure PS9-4 No11a Modal responses. 12.

Solve Problem 8 using MATLAB.

Solution The MATLAB session is similar to the one given in Problem 11. % eigenvalue problem

Figure PS9-4 No11b System responses.

444

M = diag([1200 2400 3600]); K = [500 -500 0; -500 1500 -1000; 0 -1000 2500]*1000; [U,Omega2] = eig(K,M); omegar = diag(sqrt(Omega2));

% solve the eigenvalue problem % extract the natural frequencies

% modal response to external excitations f0 = [15 0 0]'; omega = 0.1; figure(1); t = 0:0.1:100; q = zeros(length(t),3); for i = 1:3 subplot(3,1,i); q(:,i) = U(:,i)'*f0/(omegar(i)^2-omega^2)*sin(omega*t); plot(t,q(:,i),'b-'); end % system response to external excitations x = zeros(3,length(t)); for i = 1:length(t) x(:,i) = U*q(i,:)'; end figure(2); for i = 1:3 subplot(3,1,i); plot(t,x(i,:),'b-'); end 10

-3

1

0

20

0 10

40

60

80

0

100

0

-3

10

5

2

0

x

q

2

1

20

40

60

80

100

20

40

60

80

100

20

40

60

80

100

-5

0 -5

-1 20

0 10

2

40

60

80

0

100

-4

10

2

3

0

x

3

-4

-1

-5

q

10

1

x

q

1

5

-5

0 -2

-2 0

20

40

60

80

100

0

Time (s)

Figure PS9-4 No12a Modal responses.

Figure PS9-4 No12b System responses.

Problem Set 9.5 1.

For a single-degree-of-freedom system, denote the Fourier transforms of the displacement response and excitation force as X(jω) and F(jω), respectively. The expression of the imaginary part of the frequency response function X(jω)/F(jω) is given by Equation 9.124. Prove that the imaginary part reaches a peak at resonance.

Solution

445

The expression of the imaginary part of the frequency response function

and

Let

is given by

. When the imaginary part reaches a peak, we have

Multiplying

, where

to both the numerator and the denominator yields

Rearranging the terms in the numerator gives

Thus, at resonance, i.e., 2.

or

, we have

and the imaginary part reaches a peak.

Accelerations are often measured in vibration testing. For a single-degree-of-freedom system, denote the Fourier transforms of the acceleration response and excitation force as A(jω) and F(jω), respectively. The expression of the frequency response function A(jω)/F(jω) is given by

Using MATLAB, write an m-file to plot the magnitude, phase, real part, and imaginary part of the frequency response versus ω/ωn. Assume that m = 60 kg, b = 25 N·s/m, and k = 2000 N/m.

Solution The frequency response function

can be rewritten as

The following is the MATLAB session. % define the system m = 60; b = 25; k = 2000; wn = sqrt(k/m); zeta = b/(2*sqrt(m*k)); % calculate the magnitude, phase, real part, and imaginary part r = linspace(0.01,2,150); G = (-1/m*r.^2)./(1-r.^2+j*2*zeta.*r); Gmag = abs(G); Gpha = angle(G); Gre = real(G); Gim = imag(G); % plot the magnitude and phase figure(1);

446

subplot(2,1,1); plot(r,Gmag,'b-'); ylabel('Magnitude'); grid on; subplot(2,1,2); plot(r,Gpha,'b-'); ylabel('Phase'); xlabel('\omega/\omega_n'); grid on; % plot the real and imaginary figure(2); subplot(2,1,1); plot(r,Gre,'b-'); ylabel('Real Part'); grid on; subplot(2,1,2); plot(r,Gim,'b-'); ylabel('Imaginary Part'); xlabel('\omega/\omega_n'); grid on; 0.3

0.2

Real Part

Magnitude

0.2

0.1

0

-0.2 0

0.5

1

1.5

2

0

0.5

0

0.5

1

1.5

2

1.5

2

0.3

Imaginary Part

Phase

4

2

0

0.2

0.1

0 0

0.5

1

/

1.5 n

Figure PS9-5 No2a Magnitude and phase. 3.

0

2

1

/

n

Figure PS9-5 No2b Real and imaginary parts.

Rods, beams, plates, and so on are continuous systems, which have an infinite number of degrees of freedom and an infinite number of modes. For simplicity, assume that a cantilever beam is approximated as a single-degree-of-freedom mass–damper–spring system, for which the natural frequency is close to the first mode of the beam. The parameters of the cantilever beam are length L = 0.5 m, width b = 0.025 m, thickness h = 0.005 m, density ρ = 7850 kg/m3, and Young’s modulus E = 210 × 109 N/m2. a. It is known that the equivalent mass for the beam is meq = m/3, where m is the actual mass of the beam. Determine the equivalent stiffness keq for the beam. Calculate the natural frequency of the equivalent single-degree-of-freedom system. b. Figure 9.29 is the measured frequency response of the cantilever beam for the first mode. Determine the natural frequency and the damping ratio based on the given information in the plot. c. Compare the frequencies obtained in Parts (a) and (b). What is the error if the cantilever beam is approximated as a single-degree-of-freedom system?

447

Figure 9.29 Problem 3.

Solution a.

Refer to Example 5.3. The equivalent stiffness for a cantilever beam is

Ebh 3 = keq = 4 L3

) ( 210 ×10 ) ( 0.025)( 0.005 = 3

9

4 ( 0.5)

3

1312.5 N/m

The equivalent mass for the beam is m ρ bhL 7850 ( 0.025)( 0.005)( 0.5) meq= = = = 0.1635 kg 3 3 3 Thus, the natural frequency of the equivalent single-degree-of-freedom system is

= ωn b.

keq = meq

1312.5 = 89.60 rad/s 0.1635

From Figure 9.29, we have

f n ≈ 16.7 Hz and applying the method of half-power bandwidth gives f − f 16.95 − 16.45 ∆ω ∆f ζ ≈ = = 2 1= = 0.0150 2ω n 2 f n 2 fn 2 (16.7 ) c.

From Part (a), we have

= f n ω= 14.3 Hz n ( 2π )

Comparing it with the one obtained in Part (b) gives the error 14.3 − 16.7 e= = −14.37% 16.7 if the cantilever beam is approximated as a single-degree-of-freedom system. 4.

Figure 9.30 shows the magnitude of an experimentally determined frequency response. Estimate the number of degrees of freedom of the system and its natural frequencies.

Solution From the figure, we can determine that it is a three-degree-of-freedom system and the natural frequencies can be approximated as the frequencies corresponding to the peaks. They are f1 ≈ 15 Hz, f 2 ≈ 100 Hz, and f 3 ≈ 300 Hz.

448

Figure 9.30 Problem 4.

Review Problems 1.

Consider the single-degree-of-freedom system shown in Figure 5.42. a. Determine the undamped natural frequency ωn, the damping ratio ζ, and the damped natural frequency ωd. b. Assume f(t) = 0. Find the free vibration response of the system subjected to the initial conditions x(0) = 0 m . and c. d.

Write a MATLAB m-file to plot the system’s response obtained in Part (b). Construct a Simulink block diagram based on the differential equation of motion of the system and find the free vibration response.

e.

Build a Simscape model of the physical system and find the free vibration response.

Solution a.

As shown in Figure 5.42, m = 500 kg. Four same spring-damping units are connected in parallel, where b = 250 N⋅s/m and k = 200 kN/m. Thus, the equivalent damping is N⋅s/m kN/m The differential equation of motion of the system is We have rad/s

b.

rad/s . The free vibration response of the system subjected to the initial conditions x(0) = 0 m and

Assume is where

449

Thus, the free vibration response is c.

The following is the MATLAB session. % define system parameters m = 500; beq = 1000; keq = 800e3; % calculate natural frequency, damping ratio, and damped natural frequency wn = sqrt(keq/m); zeta = beq/(2*sqrt(keq*m)); wd = wn*sqrt(1-zeta^2); % define initial conditions x0 = 0; v0 = 1; % calculate amplitude and phase A = sqrt(x0^2+((zeta*wn*x0+v0)/wd)^2); phi = atan2((zeta*wn*x0+v0)/wd,x0); % plot response t = 0:0.01:5; x = A*exp(-zeta*wn*t).*cos(wd*t-phi); plot(t,x,'b-'); xlabel('Time (s)'); ylabel('Free Vibration Response (m)');

Figure Review9 No1a d.

The Simulink block diagram is shown in Plot (b). Set the initial velocity (by double-clicking the Integrator block) to 1 and the initial displacement (by double-clicking the Integrator1 block) to 0. The response of the system is the same as the one shown in Plot (a).

450

e.

Figure Review9 No1b The Simscape block diagram is shown in Plot (c), which returns the same curve in Plot (a). To define the nonzero initial velocity, double-click on the mass, damper, and spring blocks, type 1 for the Initial velocity, and choose the unit as m/s. The response of the system is the same as the one shown in Plot (a).

Figure Review9 No1c 2.

Consider the single-degree-of-freedom system shown in Figure 5.42. . Determine the dynamic amplitude X a. Assume f(t) = 500sin(60t) and initial conditions x(0) = 0 and and the static deflection xst of the system. Find the forced vibration response of the system. b. c.

Write a MATLAB m-file to plot the system’s response obtained in Part (a). Construct a Simulink block diagram based on the differential equation of motion of the system and find the forced vibration response.

d. e.

Build a Simscape model of the physical system and find the forced vibration response. Assume that the driving frequency varies from 0 to 100 rad/s. Write a MATLAB m-file to plot the dimensionless ratio X/xst versus the frequency ratio ω/ωn.

Solution a.

Refer to Part (a) in Review Problem 1. The differential equation of motion of the system is where x(0) = 0 and

. Note that the driving frequency is 60 rad/s and the amplitude of the harmonic

excitation is 500 N. Refer to Problem 1, where we found that of the system is amplitude

rad/s and

. The dynamic

451

m

The static deflection

of the system is m

The steady-state response of the system is where

m and

Thus, we have b.

The following is the MATLAB session. m = 500; beq = 1000; keq = 800e3; wn = sqrt(keq/m); zeta = beq/(2*sqrt(keq*m)); F0 = 500; w = 60; xst = F0/k; X = F0/k/sqrt((1-(w/wn)^2)^2+(2*zeta*w/wn)^2); phi = -atan2(2*zeta*w/wn,1-(w/wn)^2); t = 0:0.001:1; x = X*sin(w*t+phi); plot(t,x,'b-'); xlabel('Time (s)'); ylabel('Forced Vibration Response (m)');

Figure Review9 No2a c.

The Simulink block diagram is shown in Plot (b) and the response of the system is shown in Plot (c). Note this is the total response of the system, and the steady-state response is the same as the one shown in Plot (a).

452

Figure Review9 No2b

d.

e.

Figure Review9 No2c The Simscape block diagram is shown in Plot (d), which returns the same response shown in Plot (c).

Figure Review9 No2d The following is the MATLAB session. w = linspace(0,100,500); r = w./wn; Xxst = 1./sqrt((1-r.^2).^2+(2*zeta.*r).^2); plot(r,Xxst,'b-'); xlabel('\omega/\omega_n (rad/s)'); ylabel('X/x_{st}');

453

20

X/x

st

15

10

5

0 0

1

0.5

1.5

/

n

2

2.5

(rad/s)

Figure Review9 No2d 3.

A machine can be considered as a rigid mass with a rotating unbalanced mass. To reduce the vibration, the machine is mounted on a support system with a stiffness of 10 kN/m. Assume that the total mass is M = 10 kg and the unbalanced mass is m = 0.5 kg. Determine the range of the damping ratio of the support system so that the vibration amplitude will not exceed 5% of the rotating mass’s eccentricity when the machine operates at a speed of 300 rpm.

Solution For a system with rotating unbalance, the dimensionless ratio

( ω ωn )

MX = me

2

2

1 − ( ω ω n )2  +  2ζ ( ω ω n )  2     The vibration amplitude will not exceed 5% of the rotating mass’s eccentricity, i.e.,

m ( ω ωn )

X = e

2

≤ 5% 2 2 2   +  2ζ ω ωn )  M 1 − ( ω ωn )    ( where M = 10 kg, m = 0.5 kg, k = 10 kN/m, and ω = 300 rpm = 31.42 rad/s. The natural frequency of the system is = ωn = k M 31.62 rad/s. The above inequality can be rewritten as 2

1 − ( ω ωn )2  +  2ζ ( ω ωn )  2 ≥    

20m ( ω ωn )

2

M

Solving for the damping ratio gives

400m 2 ( ω ωn ) 2 2 − 1 − ( ω ωn )  2   M 0.50 = 2 ( ω ωn ) 4

ζ≥ 4.

As shown in Figure 9.31, a machine of a mass M = 150 kg is mounted on a fixed-fixed steel beam with negligible mass. The parameters of the beam are given as follows: Young’s modulus of the beam E = 210 GPa, width b = 0.75 m, and thickness h = 3 cm. The rotating unbalance in the machine is me = 0.015 kg⋅m. The machine runs at a speed of 2400 rpm and the maximum allowable displacement is 4 mm. Determine if it is ok to design the length of the beam as 2 m. Assume damping to be negligible.

454

Figure 9.31 Problem 4.

Solution The system shown in Figure 9.31 can be simplified as a single-degree-of-freedom mass-spring system with rotating unbalance. Neglect the damping. The maximum allowable displacement of the machine is 4 mm, i.e.,

me ( ω ω n ) 0.015 ( ω ω n ) = ≤ 4 × 10−3 m 2 2 M 1 − ( ω ωn ) 150 1 − ( ω ω n ) 2

= X

2

( ω ωn ) 2 1 − ( ω ωn ) 2

≤ 40

Solving for the frequency ratio ω/ωn gives

ω 40 ω 40 ≤ = 0.9877 or ≥ = 1.0127 ωn 41 ωn 39 The machine runs at a speed of 2400 rpm. Thus, the natural frequency of the machine should be 2400(2π/60) 2400(2π/60) = 248.18 rad/s = ωn ≥ 254.46 rad/s or ω n ≤ 1.0127 0.9877 Note that the equivalent spring stiffness for a fixed-fixed beam is 16 Ebh3 16(210 × 109 )(0.75)(0.03)3 = keq = = 8505 kN/m. 23 L3 Then, the natural frequency of the equivalent system is keq 8505000 ωn = = = 238.12 rad/s 150 M which is less than 248.18 rad/s. Therefore, it is ok to design the length of the beam as 2 m. 5.

A 80-kg machine is placed on a floor that vibrates with a frequency of 50 Hz. The maximum acceleration of the floor is 12.5 cm/s2. A vibration isolator consisting of four parallel-connected springs is designed to protect the machine from the vibration of the floor. Assume that the damping ratio of the isolator is 0.1 and the maximum allowable acceleration is 2.5 cm/s2. Determine the stiffness of each spring.

Solution The maximum allowable accelerations of the machine and the floor are 2.5 cm/s2 and 12.5 cm/s2, respectively. Thus, the displacement transmissibility is X Xω2 2.5 × 10−2 = = = 0.2 2 Z0 Z0ω 12.5 × 10−2 The damping ratio of the isolator is assumed to be 0.1 and we have

X = Z0

1 + ( 2ζω ωn ) = 2 1 − ω ωn )2  + ( 2ζω ωn )2  (  2

1 + ( 0.2ω ωn ) = 0.2 2 1 − ω ωn )2  + ( 0.2ω ωn )2  (  2

which is equivalent to

( ω ωn )

4

− 2.96 ( ω ωn ) − 24 = 0 2

Solving for ( ω ω n ) gives 2

( ω ωn )

2

= 6.60 or ( ω ωn ) = −3.64 2

455

The negative one is not valid. Thus, the frequency ratio natural frequency is

is

. For the excitation frequency of 50 Hz, the rad/s

Thus, the equivalent spring stiffness of the system is kN/m For each spring, the stiffness is 6.

kN/m.

Many vibration measuring instruments consist of a case containing a mass–damper–spring system, as shown in Figure 9.32. The displacement of the mass relative to the case is measured electrically. Denote the displacement of the mass, the displacement of the case, and the displacement of the mass relative to the case as x(t), z(t), and y(t), respectively, where y(t) = x(t) − z(t). Assume harmonic excitation, z(t) = Z0sin(ωt). a. Show that the amplitude Y of y(t) is given by

b.

Using MATLAB, write an m-file to plot the frequency response Y/Z0 versus ω/ωn for different values of damping ratio: ζ = 0.1, 0.25, 0.5, 0.7, and 1.0.

Figure 9.32 Problem 6.

Solution a.

For the mass-damper-spring system, the differential equation of motion is

which can be rewritten as The system’s frequency response function is

The harmonic motion of the case is given as z(t) = Z0sin(ωt). The steady-state response of the system is also harmonic, y(t) = Ysin(ωt), where the dynamic amplitude Y is

Dividing the numerator and denominator by k gives

b.

The following is the MATLAB session. zeta = [0.1 0.25 0.5 0.7 1.0];

456

r = 0:0.01:2.5; figure; for i = 1:length(zeta) G = sqrt(r.^2)./sqrt((1-r.^2).^2+(2*zeta(i)*r).^2); plot(r,G); hold on; end grid; xlabel('\omega/\omega_n'); ylabel('Y/Z_0'); 5

= 0.1 4

Y/Z

0

3

= 0.25 2 = 0.5 = 0.7 1

= 1.0 0 0

0.5

1

1.5

/

2

2.5

n

Figure Review9 No6 7.

A 110-kg rotating machine is mounted on a vibration isolator consisting of four parallel-connected springs. The machine operates at a speed of 3000 rpm. Determine the stiffness of each spring so that the force transmitted to the ground is reduced by 90%. Assume that all the springs are identical and damping is negligible.

Solution The force transmitted to the ground is reduced by 90%. This implies that FT/F0 = 0.1. Neglecting the damping, we have FT 1 = = 0.1 F0 1 − ( ω/ω )2 n

Solving for the frequency ratio yields ω / ω n = 11 . The machine operates at a speed of 3000 rpm. Thus, the natural frequency of the system is

ω= n

ω 3000(2π / 60) = = 94.72 rad/s 11 11

and the stiffness of each spring is

= k 8.

mω2n 110(94.72) 2 = = 247 kN/m 4 4

Consider the single-degree-of-freedom system shown in Figure 9.15a, where m = 10 kg. When the mass is in equilibrium, the static deformation of the spring is 2 mm. When the mass is allowed to vibrate freely, the maximum displacement amplitude during the fourth cycle is 20% of the first. Assume that a harmonic excitation force is applied to the system. a. Determine the minimum allowable driving frequency if the force transmitted to the ground is less than the excitation force. b. If the allowable force transmissibility is 10%, determine the stiffness of the spring.

Solution a.

The static deformation of the spring is 2 mm. Thus, the stiffness of the spring is

457

k =

mg 10 ( 9.81) = = 49.05 kN/m δst 2 × 10−3

and the natural frequency of the system is ωn =

= 49050 10 70.04 rad/s. When the mass is allowed to vibrate freely, the maximum displacement amplitude during the fourth cycle is 20% of the first. Thus, the logarithmic decrement and the damping ratio are 1 x1 1 1 = = δ = ln ln 0.54 3 x4 3 0.2

δ = 2 ( 2π ) + δ2

= ζ

0.54 = 0.09 2 ( 2π ) + 0.542

If the force transmitted to the ground is less than the excitation force, we have

FT = F0

1 + ( 2ζω ω n )

2

2 2

1 − ( ω ω n )  +  2ζ ( ω ω n )     

0 

Thus, the frequency ratio ω ω n is greater than 2 . This result can also be obtained by observing the transmissibility plot in Figure 9.12. The minimum allowable driving frequency is ω > 2ω = 2 ( 70.04 = ) 99.05 rad/s n b.

If the allowable force transmissibility is 10%, we have

1 + ( 2ζω ωn ) = 10% 2 1 − ( ω ωn )2  +  2ζ ( ω ωn )  2     2

FT = F0 which is equivalent to

( ω ωn )

4

− 5.21( ω ωn ) − 99 = 0 2

Solving for ( ω ω n ) gives ( ω ωn ) = 12.89 or ( ω ωn ) = −7.68 , which is not valid. Thus, 2

2

2

= ωn 99.05 = 12.89 27.59 rad/s

2 = k ω= 27.592 (= 10 ) 7612 N/m nm

9.

Consider the vibration absorber shown in Figure 9.16. It is known that two new resonant frequencies in the neighborhood of the excitation frequency are created. Denote the two new natural frequencies as ωn1 and ωn2. Assume ν = 1. Determine ωn1/ω1 and ωn2/ω1 for the following cases: μ = 0.05, 0.1, 0.15, 0.2, and 0.25.

Solution For the vibration absorber in Figure 9.16, the equation of motion is x1   k1 + k2 −k2   x1   F1   m1 0    +  =   sin(ωt )  0 m    k2   x2   0  2   x2    −k2 The natural frequencies can be determined by solving characteristic equation k1 + k2 − m1ω 2 −k2 = ( k1 + k2 − m1ω 2 )( k2 − m2 ω 2 ) − k22 = 0 −k2 k2 − m2 ω 2 Introducing the notations μ = m2 m1 and ν = ω 2 ω1 , we can rewrite the above equation as

 ν 2 − ( ω ω1 )2  1 + μν 2 − ( ω ω1 )2  − μν 4 = 0    The details can be found in Section 9.3.2. When ν = 1 , we have 1 − ( ω ω1 )2  1 + μ − ( ω ω1 )2=  −μ   

( ω ω1 ) − ( 2 + μ )( ω ω1= ) +1 4

2

0

458

which has two roots of

, corresponding to

and

. The results for the different values of

are listed below. 0.05 0.89 1.12

Table Review9 No9 0.1 0.15 0.85 0.82 1.17

1.21

0.2 0.80

0.5 0.70

1.25

1.41

10. A 75-kg table saw is driven by a motor, which runs at a constant speed of 1000 rpm and produces a 150-N force. Assume that the stiffness provided by the table legs is 750 kN/m and the damping is negligible. a. Determine the dynamic amplitude of the table. b. Design a vibration absorber to reduce the table oscillation to zero. Assume that the maximum allowable displacement of the absorber is 2 mm. What is the value of the mass ratio μ?

Solution a.

The natural frequency of the table saw is

The table saw runs at a constant speed of 1000 rpm, or 104.72 rad/s. Assume that the damping is negligible. The dynamic amplitude of the table is

b.

The natural frequency of the absorber is required to be the same as the excitation frequency,

The maximum allowable displacement amplitude of the absorber is 2 mm. Considering the magnitude only, the spring stiffness of the absorber can be obtained as

Thus, the mass of the absorber is

The value of the mass ratio is μ = m2/m1 = 6.84/75 = 0.09. 11. Consider the linearized model of the double pendulum system in Example 5.15. Assume m = 1 kg and L = 1 m. a. Use the modal analysis approach to determine the response of the system to the initial excitations θ1(0) = 0.05 , and . rad, θ2(0) = 0.1 rad, b. c.

Write a MATLAB m-file to plot the response of the system. Construct a Simulink block diagram based on the linearized model and find the response of the system.

Solution a.

Refer to Example 5.15. The linearized model of the double pendulum system is

where m = 1 kg and L = 1 m. Solving the eigenvalue problem

gives the natural frequencies rad/s,

rad/s

459

and the modal vectors , Normalizing the modal vectors gives , For θ1(0) = 0.05 rad, θ2(0) = 0.1 rad,

, and

, the modal responses are

where

The response of the system to the given initial excitations is

b.

Refer to Problem 9 in Problem Set 9.4.

Figure Review9 No11a Modal responses. c.

Solving for the accelerations

and

Figure Review9 No11b System responses.

using the linearized model of the system gives

460

based on which a Simulink block diagram is constructed and shown below. Run the simulation. The same responses as in Figure Review9 No11b can be obtained.

Figure Review9 No11c Simulink block diagram. 12. Consider the two-degree-of-freedom mass–spring system in Figure 5.118. a. Use the modal analysis approach to determine the response of the system if a harmonic force f = 40sin(5πt) N is applied to mass 1. Write a MATLAB m-file to plot the response of the system.

b.

Solution a.

Refer to Figure 5.118 (Problem 3 in Problem Set 5.6). The differential equations of motion in matrix form is

where m1 = m2 = 5 kg, k1 = 2000 N/m, and k2 = 4000 N/m. Solving the eigenvalue problem

gives the natural frequencies rad/s, and the normalized modal vectors , The force vector can be written as

The modal responses are

where

The response of the system to the given harmonic excitation is

b.

Refer to Problem 11 in Problem Set 9.4.

rad/s

461

0.04

0.01

0.02

0.005 1

0

x

q

1

0 -0.02

-0.005 -0.01

-0.04 0

0.2

0.4

0.6

1

0.8

0.02

0.02

0.01

0.01

0.2

0.4

0

0.2

0.4

0.6

0.8

1

0.6

0.8

1

2

0

x

q

2

0

0

-0.01

-0.01 -0.02

-0.02 0

0.2

0.4

0.6

1

0.8

Time (s)

Time (s)

Figure Review9 No12a Modal responses.

Figure Review9 No12b System responses.

13. Vibration tests are conducted to estimate parameters of a single-degree-of-freedom mass–damper–spring system. a. The frequency response of the system is shown in Figure 9.33. Determine the system’s parameters including the mass m, the stiffness k, and the damping b.

Figure 9.33 Problem 13(a). b.

The free vibration response of the system is given in Figure 9.34, which shows the recorded displacement response of the first three cycles. Using the value of the mass found in Part (a), determine the stiffness k and the damping b.

462

Figure 9.34 Problem 13(b).

Solution a.

For small damping, the natural frequency can be estimated using the peak of the frequency response, i.e., ω n ≈ 20 rad/s. Applying the half-power bandwidth method gives the damping ratio ∆ω 20.9 − 18.9 = 0.05 ζ ≈= 2ω n 2 ( 20 ) Note that the magnitude of the frequency response function X ( jω ) F ( jω ) is

X ( jω ) = F ( jω )

1k 2

1 − ( ω ω n )2  +  2ζ ( ω ω n )  2     The DC gain (i.e., the value when ω = 0 ) is 1 k . From Figure 9.33, we have

20log10 (1 k ) = −116.9 dB which gives the stiffness = k 10 = 700 kN/m. Thus, the mass and the damping are k 700000 = = 1750 kg m = 2 ωn 202 116.9 20

= = = b 2ζ mk 2 ( 0.05) 1750 ( 700000 ) 3500 N⋅s/m b.

From Figure 9.34, the logarithmic decrement is x1 0.0232 = δ ln= ln = 0.3168 x2 0.0169 and the damping ratio is δ 0.3168 = ζ = = 0.05 2 2 ( 2π ) + δ2 ( 2π ) + 0.31682 The damped natural frequency is

ω= d

2π 2π = = 19.95 rad/s Td 0.315

and the undamped natural frequency is

= ωn Thus, the stiffness is

ωd = 1 − ζ2

19.95 = 19.98 rad/s 1 − 0.052

463

kN/m and the damping is N⋅s/m 14. Case Study Consider a small airplane shown in Figure 9.35. Assume that it can be approximated as a four-degrees-of-freedom system. The values of the physical parameters are: m1 = m3 = 250 kg, m2 = 1000 kg, m4 = 75 kg, k1 = k2 = 600 kN/m (wings), k3 = 150 kN/m (landing-gear system), and k4 = 350 kN/m (tires). In addition, it is assumed that the damping coefficient for each mode is 0.05. It is desired to investigate the landing dynamics of this airplane. This may be done by examining both the response of the system to a set of initial conditions, and also the response to a harmonic input. a. Derive the differential equations of motion for the vertical motion of the airplane. b.

Determine the natural frequencies and mode shapes. Discuss the mode shapes as descriptively as possible.

c.

Model the initial touchdown event with the following set of conditions: z(0) = [0 0 0 0]T m and m/s. Plot the motion of each of the masses for the 2 seconds following touchdown.

d.

The airplane has a normal landing speed of 120 km/hr. The runway has a contour that can be modeled by a sine wave with a wavelength l of 30 m and an amplitude Y0 of 0.15 m as shown. Assume that the plane lands at a speed v of 120 km/h and that it remains in contact with the runway after touchdown so that steady state conditions are established. The input motion to the tires is given by

. Using the modal analysis approach, solve the forced vibration problem for the steady-state amplitudes of response relative to the runway. Discuss your results.

Figure 9.35 Problem 14.

Solution a.

Neglecting the damping, the differential equations of motion for the vertical motion of the airplane is

where m1 = m3 = 250 kg, m2 = 1000 kg, m4 = 75 kg, k1 = k2 = 600 kN/m, k3 = 150 kN/m, and k4 = 350 kN/m. b.

Defining K and M matrices, and calling eig command gives the natural frequencies and mode shapes. [U,Omega2] = eig(K,M); Omega = diag(sqrt(Omega2)); The natural frequencies are 8.3069, 48.9898, 60.1446, 82.0383 rad/s, or 1.322, 7.797, 9.572, and 13.057 Hz. The modal matrix is

464

and each mode shape is shown in the figure below.

Figure Review9 No14a Modal shapes. c.

Below is the MATLAB session. zeta = 0.05; Omegad = Omega*sqrt(1-zeta^2); z0 = zeros(4,1); zv0 = ones(4,1)*(-0.5); % modal responses q0 = U'*M*z0; qv0 = U'*M*zv0; C = sqrt(q0.^2+((zeta*Omega.*q0+qv0)./Omegad).^2); phi = atan2(zeta*Omega.*q0+qv0,Omegad.*q0); figure; t = 0:0.001:2; q = zeros(length(t),4); for i = 1:4 subplot(2,2,i); q(:,i) = C(i)*exp(-zeta*Omega(i)*t).*cos(Omegad(i)*t-phi(i)); plot(t,q(:,i),'b-'); end % actual responses z(t) z = zeros(4,length(t)); for i = 1:length(t) z(:,i) = U*q(i,:)'; end figure; for i = 1:4 subplot(2,2,i); plot(t,z(i,:),'b-'); end

465

Figure Review9 No14b Modal responses for initial excitation.

Figure Review9 No14c Actual responses for initial excitation. d.

Below is the MATLAB session. zeta = 0.05; omegad = omegar*sqrt(1-zeta^2); v = 120e3/3600; L = 30; y0 = 0.15; omega = 2*pi*v/L; % modal responses Y0 = [0 0 0 3.5e5*y0]'; Yc0 = [zeta zeta zeta zeta]'.*y0; Grabs = 1./sqrt((1-(omega./omegar).^2).^2+(2*zeta*omega./omegar).^2); Grpha = atan2(2*zeta*omega./omegar,1-(omega./omegar).^2); figure; t = 0:0.001:2; q = zeros(length(t),4); for i = 1:4

466

subplot(2,2,i); q1(:,i) = U(:,i)'*Y0/(omegar(i)^2)*Grabs(i)*sin(omega*t-Grpha(i)); q2(:,i) = U(:,i)'*Yc0/(omegar(i)^2)*Grabs(i)*cos(omega*t-Grpha(i)); q(:,i) = q1(:,i)+ q2(:,i); plot(t,q(:,i),'b-'); end

4

10

2

0

0

2

20

q

q

1

% actual responses z(t) z = zeros(4,length(t)); for i = 1:length(t) z(:,i) = U*q(i,:)'; end figure; for i = 1:4 subplot(2,2,i); plot(t,z(i,:),'b-'); end 10

-23

-2

-10

-4

-20 0

0.5

1

1.5

2

0

0.5

1

1.5

2

1.5

2

Time (s) 1

0.1

0.5

0

0

4

0.2

q

q

3

Time (s)

-0.1

-0.5 -1

-0.2 0

0.5

1

1.5

2

0

0.5

Time (s)

1

Time (s)

Figure Review9 No14d Modal responses for external excitation. 1

0.5

2

0

z

z

1

0.5

0

-0.5 -1

-0.5 0

0.5

1

1.5

2

0

0.5

1

1.5

2

1.5

2

Time (s) 0.4

0.5

0.2

0

0

4

1

z

z

3

Time (s)

-0.5

-0.2

-1

-0.4 0

0.5

1

Time (s)

1.5

2

0

0.5

1

Time (s)

Figure Review9 No14e Actual responses for external excitation.

467

Problem Set 10.1 1.

Draw a block diagram for the feedback control of a liquid-level system, which consists of a valve with a control knob (0–100%) and a liquid-level sensor. Clearly label essential components and signals.

Solution

Figure PS10-1 No1 2.

Draw a block diagram for the feedback control of a single-link robot arm system, which consists of a DC motor to produce the driving force and an encoder to measure the joint angle. Clearly label essential components and signals.

Solution

Figure PS10-1 No2 3.

Determine the transfer functions U(s)/E(s), Y(s)/R(s), and E(s)/R(s) in Figure 10.4.

Figure 10.4 Problem 3.

Solution By block algebra, the equivalent transfer function for the forward path is kI   10   10kp s + 10kI G1 ( s ) =  kp + s   s 2 + 2 s + 10  = 3 2  s + 2 s + 10 s   The transfer function between R( s ) and Y ( s ) is

10kp s + 10kI G1 ( s ) Y (s) = = 3 2 R( s ) 1 + G1 ( s ) s + 2 s + (10 + 10kp ) s + 10kI The transfer function between R( s ) and E ( s ) is

E (s) Y (s) s 3 + 2 s 2 + 10 s = 1− = 3 2 R( s) R ( s ) s + 2 s + (10 + 10kp ) s + 10kI

468

4.

The block diagram in Figure 10.5 represents a rocket attitude control system. Determine the transfer function Θ(s)/Θr(s).

Figure 10.5 Problem 4.

Solution By block algebra, the transfer function for the forward path is

and the transfer function for the feedback path is The transfer function between Θ(s)/Θr(s) is

Consider the control system in Example 10.2. Build a Simulink block diagram to simulate reference tracking 5. control, in which the signal R(s) is a sine wave with a magnitude of 0.1 m and a frequency of 2 rad/s. Show the actual position response and the reference signal in the same scope.

Solution

Figure PS10-1 No5a Simulink block diagram.

Figure PS10-1 No5b Responses.

469

6.

Reconsider the control system in Example 10.2. a. Convert the transfer function G(s) = Y(s)/U(s) to a differential equation of y(t). b.

Using the differential equation obtained in Part (a) to represent the plant, build a Simulink block diagram to simulate regulation control, in which the reference signal R(s) is zero. Assume that the initial m/s. conditions are y(0) = 0.1 m and

Solution a.

The transfer function

in Example 10.2 can be written as

which is a differential equation of

.

b.

Figure PS10-1 No6a Simulink block diagram.

Figure PS10-1 No6b Responses.

Problem Set 10.2 1.

The transfer function of a dynamic system is given by

a.

Using Routh’s stability criterion, determine the stability of the system.

b.

Using MATLAB, solve for the poles of the system and verify the result obtained in Part (a).

Solution a.

The characteristic equation of the closed-loop system is . The Routh array can be formed as follows. Since the elements in the first column are not all positive, the system is unstable.

470

b.

2.

>> num = [1 1]; >> den = [4 5 2 1 6]; >> G = tf(num,den); >> pole(G) The command pole returns the poles of the system: have positive real parts, and thus the system is unstable.

and

. There are two poles that

The transfer function of a dynamic system is given by

a. b. c.

Using Routh’s stability criterion, determine the stability of the open-loop system. Suppose that a negative unity feedback is applied to this open-loop system. Using Routh’s stability criterion, determine the stability of the resulting closed-loop system. Using MATLAB, solve for the poles of the open-loop and closed-loop systems and verify the results obtained in Parts (a) and (b).

Solution a.

b.

For the open-loop system, the characteristic equation is , where two coefficients are negative. According to the first condition of Routh’s stability criterion (all the coefficients of the characteristic polynomial are positive), the open-loop system is unstable. If a negative unity feedback is applied to this open-loop system, the closed-loop transfer function is

Thus, the characteristic equation of the closed-loop system is formed as follows.

c.

. The Routh array can be

Because the elements in the first column of the Routh array are all positive, the closed-loop system is stable. >> num = [20 50]; >> den = [1 10 -5 -30]; >> G = tf(num,den); >> pole(G) >> clp = feedback(G,1); >> pole(clp)

471

The command pole returns the poles of the open-loop system ( −10.20 , −1.62 , 1.82) and those of the closed-loop system ( −8.51 , −0.74 ± 1.34 j ). There is one open-loop pole that is positive real, and thus the open-loop system is unstable. For the closed-loop system, all poles have negative real parts, and thus the system is stable. 3.

)

(

The unit-step response for a second-order system Y ( s ) / U ( s ) = ω 2n / s 2 + 2ζω n s + ω n2 is given by

  ζ = y (t ) 1 − e −ζωn t cos ( ωd t ) + sin ( ωd t ) .   1 − ζ2   Prove that the relationship between the overshoot Mp and the damping ratio is

M p = e −pζ /

1−ζ 2

.

Solution Differentiating y (t ) with respect to t , we obtain

    ζ ζ = y (t ) ζω n e − ζωn t  cos ωd t + sin ωd t  − e − ζωn t  −ωd sin ωd t + ωd cos ωd t      1 − ζ2 1 − ζ2     = ζω n e − ζωn t cos ωd t + ζω n e − ζωn t

ζ 1 − ζ2

sin ωd t + e − ζωn t ωd sin ωd t −e − ζωn t

ζω n 1 − ζ 2 1 − ζ2

cos ωd t

  ζ = e − ζωn t  ζω n + ωd  sin ωd t   1 − ζ2   Note that the time derivative y (t ) equals zero when y (t ) reaches its maximum value at the peak time tp ,   ζ  ζω n + ωd  sin ωd tp = 0   1 − ζ2   This occurs when sin ωd tp = 0 . Therefore, we have e

− ζω n tp

ω d tp = π and

tp =

π ωd

The maximum value of y is πζ

−   2 ζ  cos π + y ( tp ) = 1− e sin π  = 1 + e 1−ζ 2   1− ζ   − ζω n π ωd

The overshoot M p can be determined by computing y (tp ) − 1 , which is M p = e − πζ 4.

(

.

)

Consider a second-order system Y ( s ) / U ( s ) = ω n2 / s 2 + 2ζω n s + ω n2 , which has two poles at −4±4j. a. b.

Determine the undamped natural frequency ωn, damping ratio ζ, and damped natural frequency ωd of the system. Estimate the rise time tr, overshoot Mp, peak time tp, and 2% settling time ts in the unit-step response for the system.

Solution a.

1− ζ 2

For a second-order system with two poles at −4±4j, the undamped natural frequency ωn, damping ratio ζ, and damped natural frequency ωd are

ωn = Re 2 + Im 2 = (−4) 2 + 42 = 4 2 rad/s

472

| Re | 4 = = ωn 4 2

ζ= sin = θ

ωd = ω n = 1 − ζ2 4 2 1 − b.

( ) 2 2

2

2 2

= 4 rad/s or ωd = Im = 4 rad/s

The rise time tr can be solved from the equation ωn tr ≈ 1.12 − 0.078ζ + 2.230ζ 2 2 2 2 1.12 − 0.078ζ + 2.230ζ 2 1.12 − 0.078( 2 ) + 2.230( 2 ) = = 0.44 s 5 5 The overshoot Mp and peak time tp are

tr ≈

( ) e= −p (

−pζ / 1−ζ 2

M= p =e

2 )/ 2

1−

2 2

2

4.32%

p p = = 0.79 s ωd 4

tp= The 2% settling time ts is

(

ln ∆ 1 − ζ − ts = ζωn

5.

2

)

 ln  0.02 1 −  = − 

( ) 2 2

2

( ) (5) 2 2

(

   = 1.20 s

)

Consider a second-order system Y ( s ) / U ( s ) = ω 2n / s 2 + 2ζω n s + ω n2 . Sketch the allowable region of the poles in the s-plane if the requirements for the system’s unit-step response are Mp ≤ 10% and rise time tr ≤ 0.2 s.

Solution

For Mp ≤ 10 %, we have

= ζ

| ln M p |

| ln 0.1 | ≥ = 0.59 2 p + (ln M p ) p + (ln 0.1) 2 2

2

The corresponding angle θ is

= θ sin −1 ζ ≥ sin −1 0.59 = 36° The approximated relationship between tr, the natural frequency ωn, and the damping ratio ζ for a second-order system is given by ωn ≈

1.12 − 0.078ζ + 2.230ζ 2 tr

The natural frequency ωn increases with increasing damping ratio ζ. For time tr ≤ 0.2 s and the smallest possible damping ζ = 0.59, we have

ωn ≥

1.12 − 0.078(0.59) + 2.230(0.59) 2 = 9.4 rad/s 0.2

Figure PS10-2 No5

473

The area to the left of the gray boundary shown in the figure above is the allowable region for the poles in the s-plane so that the two performance requirements are met. 6.

)

(

Consider a second-order system Y ( s ) / U ( s ) = ω 2n / s 2 + 2ζω n s + ω n2 . Sketch the allowable region of the poles in the s-plane if the requirements for the system’s unit-step response are Mp ≤ 25%, 5% settling time ts ≤ 2 s, and rise time tr ≤ 0.5 s.

Solution

For Mp ≤ 25 %, we have

= ζ

| ln M p | p + (ln M p ) 2

2

| ln 0.25 | = 0.40 2 p + (ln 0.25) 2

≥

The corresponding angle θ is

= θ sin −1 ζ ≥ sin −1 0.40 = 24° The relationship between ts, the natural frequency ωn, and the damping ratio ζ for a second-order system is given by ζωn = −

(

)

ln ∆ 1 − ζ 2 ts

Note that ζωn increases with increasing damping ratio ζ. Thus, for 5% settling time ts ≤ 2 s and the smallest possible damping ζ = 0.40, we have

ζωn ≥ −

(

ln 0.05 1 − 0.402

) = 1.54

2 The approximated relationship between tr, the natural frequency ωn, and the damping ratio ζ for a second-order system is given by ωn ≈

1.12 − 0.078ζ + 2.230ζ 2 tr

The natural frequency ωn increases with increasing damping ratio ζ. For time tr ≤ 0.5 s and the smallest possible damping ζ = 0.40, we have

1.12 − 0.078(0.40) + 2.230(0.40) 2 = 2.89 rad/s 0.5 The area to the left of the gray boundary shown in the figure below is the allowable region of the poles in the s -plane so that the three performance requirements are met. ωn ≥

Figure PS10-2 No6

474

7.

Consider a second-order system a.

b.

.

Write a MATLAB m-file to plot the magnitude of the system’s frequency response function for the following cases: ωn = 2 rad/s and ζ = 0.01, 0.1, 0.5, and 1. Summarize the effects of the damping ratio on the frequency-domain performance. Repeat Part (a) for the following cases: ζ = 0.1 and ωn = 1, 2, and 6 rad/s. Summarize the effects of the natural frequency on the frequency-domain performance.

Solution a.

b.

As shown in the figure below, the resonant peak is related to the damping of the system. The smaller the damping, the higher the resonant peak. wn = 2; zeta = [0.01 0.1 0.5 1]; for i = 1:length(zeta) num = [wn^2]; den = [1 2*zeta(i)*wn wn^2]; sys = tf(num,den); bode(sys); hold on; end legend('\zeta = 0.01','\zeta = 0.1','\zeta = 0.5','\zeta = 1');

Figure PS10-2 No7 As shown in the figure below, the bandwidth is related to the natural frequency of the system. The bandwidth increases when the natural frequency increases. zeta = 0.1; wn = [1 2 6]; for i = 1:length(wn) num = [wn(i)*wn(i)]; den = [1 2*zeta*wn(i) wn(i)*wn(i)]; sys = tf(num,den); bode(sys); hold on; end legend('\omega_n = 1','\omega_n = 2','\omega_n = 6');

475

Bode Diagram 20

ωn increases

Magnitude (dB)

0 -20 ωn = 1

-40

ωn = 2

-60

ωn = 6

Phase (deg)

-80 0

ωn increases

-45 -90 -135 -180 -1

10

0

10

1

10

2

10

Frequency (rad/sec)

Figure PS10-2 No8 8.

The unit-step response of a dynamic system is shown in Figure 10.13. The maximum value of the response is 1.1 and its steady-state value is 0.95. The response reaches 10% and 90% of its steady-state value at 0.048 s and 0.21s,

(

)

respectively. Find the transfer function of the system if it can be approximated as a s 2 + 2ζωn s + ω2n .

Figure 10.13 Problem 8.

Solution Using the maximum value and the steady-state value, we can find the overshoot, i.e., ymax − yss 1.1 − 0.95 = Mp = = 15.79% yss 0.95 from which the damping ratio can be determined, and | ln M p | | ln 0.1579 | = ζ = = 0.51 2 2 2 p + (ln M p ) p + (ln 0.1579) 2 The response reaches 10% and 90% of its steady-state value at 0.048 s and 0.21 s, respectively. Thus, the rise time tr = 0.21 – 0.048 = 0.162 s, which gives the undamped natural frequency

476

To determine the value of a, we apply the final-value theorem to find the steady-state value of the unit-step response, i.e.,

which gives Therefore, the transfer function of the system is 99.81/(s2+10.46s+105.06).

Problem Set 10.3 1.

Consider the feedback system shown in Figure 10.26. a. Using Routh’s stability criterion, determine the range of the control gain K for which the closed-loop system is stable. b.

Use MATLAB commands to find the unit-step responses for open-loop and closed-loop control. Assume that the control gain is K = 30. Compare the open- and closed-loop responses.

Figure 10.26 Problem 1.

Solution a.

The closed-loop transfer function

is

The closed-loop characteristic equation is

, for which we can construct the Routh

array

All elements in the first column of Routh array must be positive. This leads to b.

Below is the MATLAB session. figure(1); olp = tf([1 1],conv([1 -2 0],[1 5])); step(olp); figure(2); K = 30; clp = feedback(K*olp,1); step(clp);

.

477

10

14

24

2.5

12 2

Closed-loop Response

Open-loop Response

10

8

6

4

1.5

1

0.5 2

0

0 0

5

10

15

20

25

30

0

Time (seconds)

4

2

6

8

Time (seconds)

Figure PS10-3 No1 2.

Consider the feedback system shown in Figure 10.27. Using Routh’s stability criterion, determine the range of the control gain K for which the closed-loop system is stable.

Figure 10.27 Problem 2.

Solution The closed-loop transfer function Y ( s ) R ( s ) is

1 K 2 Y ( s) Ks + 6 K ( s − 1)( s + 4) = = 4 s + 2 s + 10s 3 + 23s 2 + ( K − 10) s + 2 K − 24 1 R( s ) 1 + K ( s 2 − 1)( s + 4) s + 6 The closed-loop characteristic equation is s 4 + 10s 3 + 23s 2 + ( K − 10) s + 2 K − 24 = 0 , for which we can construct the Routh array s4 : 1 23 2 K − 24

s 3 : 10

K − 10

0

240 − K 10(23) − ( K − 10) = s2 : 2 K − 24 10 10 5K − 0.1K 2 (24 − 0.1K )( K − 10) − 10(2 K − 24) = s1 : 0 24 − 0.1K 24 − 0.1K s 0 : 2 K − 24 In order to make the closed-loop system stable, all elements in the first column of Routh array must be positive, K < 240  24 − 0.1K > 0 →  2  5K − 0.1K > 0 → 0.1K 2 − 5K < 0 → 0 < K < 50  24 − 0.1 K  → K > 12  2 K − 24 > 0 which leads to 12 < K < 50 . 3.

Reconsider Example 10.7. Using the final-value theorem, verify the steady-state errors to a unit-step input for open-loop and closed-loop control without and with disturbance.

478

Solution For the open-loop system, we have Y ( s ) = G ( s )U ( s ) + G ( s ) D ( s ) = KG ( s ) R ( s ) + G ( s ) D ( s ) where K = 2.5 . The steady-state value of y (t ) is 4 1 10 = yss lim = sY ( s ) lim sKG= ( s ) R ( s ) lim s ( 2.5)= lim = 1 (without disturbance) s →0 s →0 s →0 s → 0 s + 10 s s + 10 = yss lim = sY ( s ) lim s [ KG ( s ) R ( s ) + G ( s ) D ( s )] s →0

s →0

(with disturbance) 4 1 4  −1   10 − 4  = lim s ( 2.5) + = lim = 0.6    s →0 s + 10 s s + 10  s   s →0 s + 10  Therefore, the steady-state error is 0 without disturbance or 0.4 with disturbance. For the closed-loop system, KG ( s ) G( s) = Y ( s) R( s ) + D( s) 1 + KG ( s ) 1 + KG ( s ) where K = 2.5 . The steady-state value of y (t ) is

yss lim sY ( s ) lim s = = s →0

s →0

50 ( 4 ) 1 200 KG ( s ) R ( s ) lim s = = lim= 0.95 (without disturbance) 0 0 s s → → 1 + KG ( s ) s + 10 + 4 ( 50 ) s s + 210

 KG ( s )  G( s) = = yss lim sY ( s ) lim s  R( s ) + D( s)  s →0 s →0 1 + KG ( s )  1 + KG ( s )   50 ( 4 ) 1 4 200 − 4  1  = lim s  + −   lim = 0.93 = → s →0 s 0 s + 210  s + 10 + 4 ( 50 ) s s + 10 + 4 ( 50 )  s   Therefore, the steady-state error is 0.05 without disturbance or 0.07 with disturbance. 4.

(with disturbance)

A stable system can be classified as a system type, which is defined as the degree of the polynomial for which the steady-state error is a nonzero finite constant. For instance, if the error to a step input, which is a polynomial of zero degree, is a nonzero finite constant, then such a system is called type 0, and so on. Consider the system in Figure 10.28. a. Compute the following steady-state errors for: (1) a unit-step reference input, (2) a unit-ramp reference input u = t, and (3) a parabolic reference input u = 0.5t2. b. Determine the system type.

Figure 10.28 Problem 4.

Solution a.

The closed-loop transfer function Y ( s ) R ( s ) is

Y (s) 10 s + 2 = 3 R( s ) s + 8s 2 + 35s + 2 E (s) Y (s) s 3 + 8s 2 + 25s = 1− = R( s) R( s ) s 3 + 8s 2 + 35s + 2 For a unit step reference input,

E (s) 1 s 3 + 8s 2 + 25s ( s) lim s 3 ess lim sE ( s ) lim sR= = = = 0 s →0 s →0 R( s ) s →0 s s + 8s 2 + 35s + 2 For a unit ramp reference input u = t , 1 s 3 + 8s 2 + 25s 25 E (s) lim ( ) lim ( ) lim e= sE = s sR s = s = = 12.5 ss s →0 s →0 R( s ) s →0 s 2 s 3 + 8s 2 + 35s + 2 2

479

For a parabolic reference input

,

b.

The steady-state error is a nonzero finite constant to a ramp input, which is a polynomial of degree one. Therefore, the system is type 1.

5.

Reconsider Example 10.9. Build Simulink block diagrams to simulate open-loop and closed-loop control with parameter variations. Verify the steady-state response values yss obtained in Example 10.9.

Solution The Simulink block diagram for open-loop control is shown below, where k = 40 without uncertainty or 50 with uncertainty. The steady-state response value yss is 1 without uncertainty or 0.8 with uncertainty.

Figure PS10-3 No5a The Simulink block diagram for closed-loop control is shown below. The steady-state response value yss is about 0.98 without uncertainty or 0.97 with uncertainty.

Figure PS10-3 No5b Simulation results for open-loop control and closed-loop control are shown in Plot (c) and (d), respectively.

Figure PS10-3 No5c 6.

Figure PS10-3 No5d

Consider the feedback control system shown in Figure 10.27. a. Compute the sensitivity of the closed-loop transfer function to changes in the parameter τ. b. Compute the sensitivity of the closed-loop transfer function to changes in the parameter K. c.

Assuming τ = 1 and K = 1, use MATLAB to plot the magnitude of each of the sensitivity functions for s = jω. Use the logarithmic scale for the y-axis. Comment on the effect of parameter variations in τ and K for different driving frequencies ω.

480

Figure 10.29 Problem 6.

Solution a.

The closed-loop transfer function is

The sensitivity of the closed-loop transfer function

to changes in the parameter

b.

The sensitivity of the closed-loop transfer function

to changes in the parameter

c.

Below is the MATLAB session. tau = 1; K = 1; w = 0:0.01:10; S1 = abs(-tau*j*w./(tau*j*w+1+K)); S2 = abs(-K./(tau*j*w+1+K)); semilogy(w,S1,'b-',w,S2,'r-'); axis([0 10 0.001 10]); xlabel('Frequency (rad/s)'); ylabel('Sensitivities'); legend('Sensitivity 1','Sensitivity 2');

Figure PS10-3 No6 The plot indicates that • The steady-state response is not affected by variation in because response is dependent on

because

when

.

is

is

when

. The steady-state

481

•

The closed-loop transfer function is more sensitive to variation in than 1 rad/s.

when the driving frequency

is greater

Problem Set 10.4 1.

Figure 10.40 shows a negative feedback control system. a. Design a P controller such that the damping ratio of the closed-loop system is 0.5. b. Estimate the rise time, overshoot, and 2% settling time in the unit-step response for the closed-loop system.

Figure 10.40 Problem 1.

Solution a.

The closed-loop transfer function

is

which is a second-order system. The closed-loop characteristic equation is closed-loop natural frequency and damping ratio as

and

. Denoting the

, we have

As a requirement, the damping ratio of the closed-loop system is 0.5. Substituting the value of equation set yields b.

rad/s and

into the previous

.

The rise time, overshoot, and 1% settling time are s %

s 2.

Consider the negative feedback control system shown in Figure 10.41. a. Design a P controller such that the maximum overshoot in the response to a unit-step reference input is less than 15%, the 2% settling time is less than 1 s, and the rise time is less than 0.2 s. b.

Use MATLAB to plot the unit-step response of the closed-loop system. Find the overshoot, 2% settling time, and rise time. If the time-domain specifications exceed the requirements, fine-tune and reduce them to be approximately the specified values or less.

Figure 10.41 Problem 2.

Solution a.

The closed-loop transfer function

is

482

which is a second-order system. The closed-loop characteristic equation is Denoting the closed-loop natural frequency and damping ratio as

The requirement of overshoot

and

, we have

implies that

Let ζ = 0.6. For 2% settling time ts < 1 s, we have rad/s For rise time tr < 0.2 s, we have rad/s Also, from the closed-loop characteristic equation, we have rad/s In order to satisfy all three conditions, pick

b.

rad/s. Thus, the control gain can be determined as

Below is the MATLAB session. sys = tf([5],[1 12 5]); kp = 19; clp = feedback(kp*sys,1); step(clp);

Figure PS10-4 No2 3.

Consider the feedback control system shown in Figure 10.40. a. Design a PD controller such that the closed-loop poles are at p1,2 = −6±8j. Use MATLAB to plot the unit-step response of the closed-loop system. Find the rise time, overshoot, b. peak time, and 1% settling time.

483

Figure 10.42 Problem 3.

Solution a.

The closed-loop transfer function

is

which is a second-order system. The closed-loop characteristic equation is Denoting the closed-loop natural frequency and damping ratio as

and

, we have

. This implies that

As a requirement, the closed-loop poles are located at

rad/s

where

b.

or 2. Substituting the values of

and

into the previous equation set yields

Below is the MATLAB session. sys = tf([8],[1 4 0]); kp = 12.5; kd = 1; sysc = tf([kd kp],[1]); clp = feedback(sysc*sys,1); step(clp);

Figure PS10-4 No3

484

4.

Consider the feedback control system shown in Figure 10.43. a. Find the values for kp and TD such that the maximum overshoot in the response to a unit-step reference input is less than 5% and the 2% settling time is less than 0.5 s. b. Compute the steady-state error of the closed-loop system to a unit-step reference input. c.

Verify the results in Parts (a) and (b) using MATLAB by plotting the unit-step response of the closed-loop system. If the maximum overshoot and the settling time exceed the requirements, fine-tune and reduce them to be approximately the specified values or less.

Figure 10.43 Problem 4.

Solution a.

The closed-loop transfer function

which is a second-order system. The closed-loop characteristic equation is Denoting the natural frequency and the damping ratio of the closed-loop system as

implies that

The requirement of overshoot

Pick ζ = 0.8. Thus, for 2% settling time ts < 0.5 s, we have rad/s Choosing

and

yields

b.

For a unit step reference input,

c.

Below is the MATLAB session. sys = tf([1],[0.1 0.8 1]); kp = 12.225; Td = 0.0851; sysc = tf([kp*Td kp],[1]); clp = feedback(sysc*sys,1); step(clp);

and

.

and

, we have

485

Figure PS10-4 No4 5.

Consider the feedback control system shown in Figure 10.44a. a. If the desired closed-loop poles are located within the shaded regions shown in Figure 10.44b, determine the corresponding ranges of ωn and ζ of the closed-loop system. b. Design a PI controller such that the closed-loop poles are at p1,2 = −9 ± 12 j. c. Compute the steady-state errors of the plant and the closed-loop system to a unit-step reference input. Verify the results in Part (c) using MATLAB by plotting the unit-step responses of the plant and the d. closed-loop system.

Figure 10.44 Problem 5.

Solution a.

Note that the semicircles in Figure 10.44(b) indicate lines of constant natural frequencies lines indicate constant damping ratios . Thus, for the shaded regions, we have

and the diagonal

or which implies b.

because

The closed-loop transfer function

. is

which is a second-order system. The closed-loop characteristic equation is Denoting the natural frequency and the damping ratio of the closed-loop system as

and

, we have

486

As a requirement, the closed-loop poles are located at

. This implies that rad/s

where c.

, 2. Substituting these values into the previous equation set yields

.

The steady-state value of the unit step response of the plant is

Thus, the steady-state error is

and d.

and

. For the closed-loop system, we have

.

Below is the MATLAB session. sys = tf([2],[1 5]); kp = 6.5; ki = 112.5; sysc = tf([kp ki],[1 0]); clp = feedback(sysc*sys,1); step(sys,clp); legend('Plant','Closed-loop system');

Figure PS10-4 No5 6.

Consider the feedback control system shown in Figure 10.45. a. Find the values for kp and TI such that the maximum overshoot in the response to a unit-step reference input is less than 15% and the peak time is less than 0.4 s. b.

Verify the results in Part (a) using MATLAB by plotting the unit-step response of the closed-loop system. If the maximum overshoot and the peak time exceed the requirements, fine-tune and reduce them to be approximately the specified values or less.

Figure 10.45 Problem 6.

487

Solution a.

The closed-loop transfer function

is

which is a second-order system. The closed-loop characteristic equation is Denoting the natural frequency and the damping ratio of the closed-loop system as

and

, we have

and

The two requirements for the transient response of the closed-loop system are

s. This

implies that

rad/s Choose

and

rad/s. The undamped natural frequency is rad/s

Substituting the values of b.

and

into the previous equation set yields

Below is the MATLAB session. sys = tf([1],[1 2]); kp = 16.135; Ti = 0.1104; sysc = tf([kp*Ti kp],[Ti 0]); clp = feedback(sysc*sys,1); step(clp);

Figure PS10-4 No6 7.

The unit-step response of a plant is shown in Figure 10.46.

and

.

488

a.

The lag time L and the reaction rate R can be determined from the figure. Find the P, PI, and PID controller parameters using the Ziegler–Nichols reaction curve method.

b.

Assume that the transfer function of the plant is 3/(10s2+8s+1). Use MATLAB to plot the unit-step response of the closed-loop system with P, PI, or PID control.

Figure 10.46 Problem 7.

Solution a.

Estimate the lag time L and the reaction rate R:

The proportional, PI, and PID controller parameters are listed in the table. Table PS10-4 No7 Type of controller Optimum gains P PI or PID or or b.

Below is the MATLAB session. sys = tf([3],[10 8 1]); R = 0.2857; L = 1.5; % P Control kp = 1/(R*L); clp1 = feedback(kp*sys,1); % PI Control kp = 0.9/(R*L); Ti = L/0.3; ki = kp/Ti; s = tf([1 0],[1]); sysc2 = kp+ki/s; clp2 = feedback(sysc2*sys,1);

489

% PID Control kp = 1.2/(R*L); Ti = 2*L; Td = 0.5*L; ki = kp/Ti; kd = kp*Td; sysc3 = kp+ki/s+kd*s; clp3 = feedback(sysc3*sys,1); % Step responses step(clp1,clp2,clp3); legend('P','PI','PID'); 1.4 P PI

1.2

PID

1

Amplitude

0.8

0.6

0.4

0.2

0 0

5

10

15

Time (seconds)

Figure PS10-4 No7 8.

Consider a proportional feedback control system. As shown in Figure 10.47, the closed-loop system becomes marginally stable when the proportional gain is 0.75. Find the P, PI, and PID controller parameters using the Ziegler–Nichols ultimate sensitivity method.

Figure 10.47 Problem 8.

Solution The proportional, PI, and PID controller parameters are listed in the table, where Pu ≈ 18 and K u is given as 0.75.

490

Table PS10-4 No8 Optimum gains = k p 0.5 = K u 0.375

Type of controller P

= k p 0.45 = K u 0.3375

PI

= k I k= 0.0225 = TI P= 15 or p TI u 1.2 = k p 0.6 = K u 0.45

PID

= k I k= 0.05 = TI P= 9 or p TI u 2 k D k= 1.0125 = TD P= 2.25 or= pTD u 8

Problem Set 10.5 1.

Roughly sketch the root locus with respect to K for the equation of 1 + KL(s) = 0 and the following choices for L(s). Make sure to give the asymptotes, arrival or departure angles, and points crossing the imaginary axis. Verify your results using MATLAB. 1 1 b. a. L( s ) = L( s ) = ( s + 1)( s + 3) ( s + 1)( s + 3)( s + 7) c.

L( s ) =

s+5 ( s + 1)( s + 3)( s + 7)

L( s ) =

d.

s ( s + 5) ( s + 1)( s + 3)( s + 7)

Solution a.

poles: −1 , −3 zeros: none Asymptotes for large gain values: (1) the number of asymptotes: n − m = 2 − 0 = 0 (2 poles and no zeros) −1 − 3 = −2 (2) the centroid of asymptotes: α = 2 (3) the angle of asymptotes: φ1 = 90 , φ2 = −90 Root Locus

2

-1

)

1.5 1

Imaginary Axis (seconds

0.5 0 -0.5 -1

-1.5 -2 -3.5

-3

-2.5

-2

-1.5

Real Axis (seconds

b.

-1

-0.5 -1

)

Figure PS10-5 No1a zeros: none poles: −1 , −3 , −7 Asymptotes for large gain values: (1) the number of asymptotes: n − m = 3 − 0 = 0 (3 poles and no zeros) −1 − 3 − 7 11 (2) the centroid of asymptotes: α = = − 3 3 (3) the angle of asymptotes: φ1 = 60 , φ2 = 180 , φ3 = −60

0

0.5

491

Crossing jω-axis points:

K 1 + KL ( jω ) = 1+ 0 = ( jω + 1)( jω + 3)( jω + 7 ) − jω3 − 11ω2 + 31jω + 21 + K = 0 which is equivalent to

−11ω2 + 21 + K = 0  3 0  −ω + 31ω = Solving for ω gives ω = ± 31 and K = 320 (Note that the solution, ω = 0 and K = −21 , is not valid). Root Locus

15

-1

)

10

Imaginary Axis (seconds

5

0

-5

-10

-15 -25

-20

-15

-10

-5

Real Axis (seconds

c.

0 -1

5

10

)

Figure PS10-5 No1b zeros: −5 poles: −1 , −3 , −7 Asymptotes for large gain values: (1) the number of asymptotes: n − m = 3 − 1 = 2 (3 poles and 1 zero) ( −1 − 3 − 7 ) − ( −5) (2) the centroid of asymptotes: α = = −4 3 −1 (3) the angle of asymptotes: φ1 = 90 , φ2 = −90 Root Locus

20

-1

)

15 10

Imaginary Axis (seconds

5 0 -5

-10 -15 -20 -8

-6

-4 Real Axis (seconds

d.

-2 -1

)

Figure PS10-5 No1c zeros: 0, −5 poles: −1 , −3 , −7 (Note that all the branches are on the real axis.)

0

492

Root Locus

1.5

-1

)

1

Imaginary Axis (seconds

0.5

0

-0.5

-1

-1.5 -25

-20

-15

-10

Real Axis (seconds

-5 -1

5

0

)

Figure PS10-5 No1d 2.

Repeat Problem 1 for the following choices for L(s). 1 b. a. L( s ) = 2 s + 2 s + 10 c.

L( s ) =

s + 4s + 8

L( s ) =

2

L( s ) =

d.

( s + 4)( s 2 + 2 s + 10)

1 ( s + 4)( s + 2 s + 10) 2

( s + 1)( s 2 + 4 s + 8) ( s + 4)( s 2 + 2 s + 10)

Solution a.

poles: −1 ± 3j zeros: none Asymptotes for large gain values: (1) the number of asymptotes: n − m = 2 − 0 = 0 (2 poles and no zeros) −1 − 1 (2) the centroid of asymptotes: α = = −1 2 (3) the angle of asymptotes: φ1 = 90 , φ2 = −90 Departure angle from the complex pole at s =−1 + 3j :

ϕdep = Σψ i − Σϕi − 180 = −90 − 180 = −270 (equivalent to 90 ) Root Locus

15

-1

)

10

Imaginary Axis (seconds

5

0

-5

-10

-15 -2.5

-2

-1.5

-1

Real Axis (seconds

b.

-0.5 -1

0

)

Figure PS10-5 No2a zeros: none poles: −4 , −1 ± 3j Asymptotes for large gain values: (1) the number of asymptotes: n − m = 3 − 0 = 0 (3 poles and no zeros)

0.5

493

−4 − 1 − 1 = −2 3 (3) the angle of asymptotes: φ1 = 60 , φ2 = 180 , φ3 = −60 Departure angle from the complex pole at s =−1 + 3 j :

(2) the centroid of asymptotes: α =

ϕdep = Σψ i − Σϕi − 180 = − 90 + 45  − 180 = −315 (or 45 ) Crossing jω-axis points:

K = 1 + KL ( jω ) = 1+ 0 2 ( jω + 4 ) ( jω ) + 2 jω + 10 − jω3 − 6ω2 + 18jω + 40 + K = 0 which is equivalent to

−6ω2 + 40 + K = 0  3 0  −ω + 18ω = Solving for ω gives ω = ± 18 and K = 68 (Note that the solution, ω = 0 and K = −20 , is not valid). Root Locus

15

-1

)

10

Imaginary Axis (seconds

5

0

-5

-10

-15 -20

-15

-10

-5

Real Axis (seconds

c.

0 -1

5

10

)

Figure PS10-5 No2b zeros: −2 ± 2 j poles: −4 , −1 ± 3j Asymptotes for large gain values: (1) the number of asymptotes: n − m = 3 − 2 = 1 (3 poles and 2 zeros) − 4 − 1 − 1 − ( −2 − 2 ) (2) the centroid of asymptotes: α = = −2 1 (3) the angle of asymptotes: φ1 = 180 Departure angle from the complex pole at s =−1 + 3j :

 45 + tan −1 ( 15 )  − 90 + 45  − 180 = ϕdep = Σψ i − Σϕi − 180aaaaaa = −191 (or 169 ) Arrival angle to the complex zero at s =−2 + 2 j :

 −135 + 180 − tan −1 ( 15 ) + 45  − 90 + 180 = Σϕi − Σψ i + 180aaaaaaa = ϕarr = 101

494

Root Locus

4

-1

)

3 2

Imaginary Axis (seconds

1 0 -1 -2 -3 -4 -8

-6

-4

-2

Real Axis (seconds

d.

-1

0

)

Figure PS10-5 No2c zeros: −1, −2 ± 2 j poles: −4 , −1 ± 3j Note that all the branches start at poles and end at zeros. Departure angle from the complex pole at s =−1 + 3j :

90 + 45 + tan −1 ( 15 )  − 90 + 45  − 180 = ϕdep = Σψ i − Σϕi − 180aaaaaaa = −101 Arrival angle to the complex zero at s =−2 + 2 j :

 −135 + 180 − tan −1 ( 15 ) + 45  − 180 − tan −1 ( 12 ) + 90  + 180 = ϕarr = Σϕi − Σψ i + 180aaaaaaaa = −15 Root Locus

4

-1

)

3 2

Imaginary Axis (seconds

1 0 -1 -2 -3 -4 -4

-3

-2 Real Axis (seconds

-1 -1

0

)

Figure PS10-5 No2d 3.

A control system is represented using the block diagram shown in Figure 10.59. Sketch the root locus with respect to the proportional control gain K. Determine all the values of K for which the closed-loop system is stable.

Figure 10.59 Problem 3.

495

Solution The closed-loop characteristic equation is

Poles of : 0, , , Asymptotes for large gain values: (1) the number of asymptotes:

zero of

:

(5 poles and 1 zero)

(2) the centroid of asymptotes: , (3) the angle of asymptotes: Departure angle from the complex pole at

,

, : (or

)

Crossing jω-axis points:

and : , Solving for closed-loop system is stable when

and

, 6.98 (Note that the solution, . The root locus is shown in the figure below.

is not valid). The

Figure PS10-5 No3 4.

A control system is represented using the block diagram shown in Figure 10.60, in which the parameter a is subjected to variations. Sketch the root locus with respect to the parameter a. Determine all the values of a for which the closed-loop system is stable.

Figure 10.60 Problem 4.

496

Solution The closed-loop characteristic equation is

which is equivalent to We can write the equation in the form of

The root locus with respect to the parameter >> L = tf([4 8 0],[4 8 3 9]); >> rlocus(L);

, where

can be plotted based on

. Below is the MATLAB session.

Figure PS10-5 No4 The stability range of a can be determined by finding the crossing jω-axis points.

which is equivalent to

Solving for ω and solution, 5.

: and a = 0.15. The closed-loop system is stable when is not valid).

(Note that the

Figure 10.61 shows the root locus of a unity negative feedback control system, where K is the proportional control gain. a. Determine the transfer function of the plant. Use MATLAB to plot the root locus based on your choice of the plant, compare it with the root locus shown in Figure 10.61, and check the accuracy of your plant transfer function. b. Give comments on the stability of the closed-loop system when K varies from 0 to ∞. c. Give comments on the transient performance of the closed-loop system when K = 0.5 and 5. Use MATLAB to plot the corresponding unit-step responses and verify your comments.

497

Figure 10.61 Problem 5.

Solution a.

Note that there are three branches. Two of them end at zeros and the other ends at infinity. This implies that the plant is a third-order system and has two zeros. When K = 0 , which corresponds to having no control, three poles are located at −2 , 0.5 ± 2.5j and two zeros are at 0.5, −1. Thus, a possible transfer function of the plant is

= L( s )

10

4

K = 0.5

25

K=5

0.8 0.6

2 0.4 0.2

0 Amplitude

c.

which is proved to be correct by comparing its root locus with the one shown in Figure 10.59. When 0 ≤ K < 1.37 , two complex poles are in the right half s-plane, and thus the closed-loop system is unstable. When K = 1.37 , two complex poles appear on the imaginary axis, and thus the closed-loop system becomes marginally stable. When 1.37 < K < 26 , all three branches are in the left half s-plane, and thus the closed-loop system is stable. For K > 26 , one of the branches crosses the imaginary axis and enter the right half s-plane. The closed-loop system becomes unstable. Therefore, the closed-loop system is stable when 1.37 < K < 26 . When K = 0.5, two complex poles are in the right half s-plane and the closed-loop system is unstable. Thus, the response is divergent. When K = 5, two complex poles are closer to the origin than the real pole, and also the damping ratio is between 0 and 1. Thus, the closed-loop system exhibits decaying oscillations.

Amplitude

b.

( s + 0.5)( s + 1) ( s + 0.5)( s + 1) = ( s + 2 )( s − 0.5 − 2.5j)( s − 0.5 + 2.5j) s ( s 2 − s + 6.5)

-2

0 -0.2 -0.4

-4 -0.6 -6

-0.8 0

50

100

150

200

0

Time (seconds)

2

4

6 Time (seconds)

Figure PS10-5 No5

8

10

12

14

498

6.

Figure 10.62 shows the root locus of a unity negative feedback control system, where K is the proportional control gain. a. Determine the transfer function of the plant. Use MATLAB to plot the root locus based on your choice of the plant, compare it with the root locus shown in Figure 10.62, and check the accuracy of your plant transfer function. b. Find the range of values of K for which the system has damped oscillatory response. What is the value of K when pure harmonic oscillations occur? Also, what is the frequency of pure harmonic oscillations? Use MATLAB to plot the corresponding unit-step response and verify the accuracy of your computed frequency.

Figure 10.62 Problem 6.

Solution a.

b.

Note that there are three branches, all of which start at poles and end at infinity by approaching to asymptotes. , which corresponds to having no This implies that the plant is a third-order system and has no zeros. When , and . Thus, a possible transfer function of the plant is control, three poles are located at 0.5,

which is proved to be correct by comparing its root locus with the one shown in Figure 10.62. Note that the branch starting at 0.5 crosses the imaginary axis into the left half s-plane when K = 12. This branch break away from the real axis when and the poles split into a together with the one starting at . These two complex branches cross the imaginary axis into the right half complex conjugate pair when . Thus, the closed-loop system has damped oscillatory response when . s-plane when when pure harmonic oscillations occur is about 192.5. The frequency of pure harmonic The value of oscillations can be determined by solving for the crossing jω-axis points. We have

Solving either the real part

or the imaginary part

gives

. Therefore,

the closed-loop system oscillates harmonically at a frequency of rad/s (or 4.36 rad/s) when the proportional control gain is about 192.5. As shown in the figure below, it takes 8.646 seconds to complete 6 full-cycle oscillations. Thus, the period of oscillation is around 1.44 s, which corresponds to a frequency of 4.36 rad/s.

499

Figure PS10-5 No6 7. a. b. c.

Consider the feedback system shown in Figure 10.63. Find the locus of the closed-loop poles with respect to K. Find a value of K such that the maximum overshoot in the response to a unit-step reference input is less than 5%. What is the corresponding steady-state error of the closed-loop system? Plot the unit-step response of the closed-loop system to verify the result for Part (b).

Figure 10.63 Problem 7.

Solution a.

The characteristic equation of the closed-loop system is

which is equivalent to Dividing by

b.

gives

which is in the form of . We can plot the root locus based on sys = tf([1],[0.1 1.8 1]); rlocus(sys); grid on; axis([-20 5 -15 15]); The requirement for the transient response of the closed-loop system is

.

. This implies that

As shown in the root locus, the dashed diagonal lines indicate pole locations with a damping ratio of 0.69. Only . The steady-state error the part of the root locus in between the two diagonal lines is acceptable. Choose of the closed-loop system can be found using the final-value theorem.

500

Figure PS10-5 No7a c.

The following is the MATLAB session. K = 12; sysc = tf([1 K],[1]); sys = tf([1],[0.1 0.8 1]); clp = feedback(sysc*sys,1); step(clp,1);

Figure PS10-5 No7b 8. a. b. c.

Consider the feedback system shown in Figure 10.64. Find the locus of the closed-loop poles with respect to K. Find a value of K such that the maximum overshoot in the response to a unit-step reference input is less than 20% and the 2% settling time is less than 1.1 s. Plot the unit-step response of the closed-loop system to verify the result in Part (b).

Figure 10.64 Problem 8.

Solution a.

The characteristic equation of the closed-loop system is

which is in the form of

. The root locus is shown in Plot (a).

501

Root Locus

50 0.72

0.58

0.44

0.32

0.22

0.1

-1

)

0.86

0.96

Imaginary Axis (seconds

40

50

0

20

30

10

0.96

0.86

0.72

0.58

0.44

0.32

0.1

0.22

-50 -40

-50

-20

-30

-10

b.

0

-1

Real Axis (seconds

)

Figure PS10-5 No8a The requirements for the transient response of the closed-loop system are overshoot M p < 20% and 2% settling time ts < 1.1 s. This implies that

ln M p

ζ=

π 2 + ( ln M p )

2

ln 0.20

>

π 2 + ( ln 0.20 )

2

=0.46

For 2% settling time ts < 1.1 s and the smallest possible damping ζ = 0.46, we have

ζωn = −

(

ln ∆ 1 − ζ 2

) > − ln (0.02 1 − 0.46 ) = 3.67 2

ts 1.1 As shown in Plot (a), the dashed diagonal lines indicate pole locations with a damping ratio of 0.46, and the dashed vertical line indicates pole locations with a real part of −3.67 . Only the part of the root locus between the two diagonal lines and to the left of the vertical line is acceptable. Zoom in the desired region. Left-clicking on the root locus, holding down the left mouse button and moving the mouse along the root locus, we can see the values of the pole and the gain varying correspondingly. Choose K = 150 . Root Locus

10 10

0.29

0.4

0.52

0.13

0.2

0.06

-1

)

8 0.7

6

0.9 System: olp

2

5

4

Imaginary Axis (seconds

Gain: 150 0Pole: -3.88 + 4.23i Damping: 0.676 2

Overshoot (%): 5.6 0.9 Frequency (rad/s): 5.73 -5

4 6

0.7

8 0.29

0.4

0.52

-10

0.06

0.13

0.2

10 -7

-6

-5

-4

-3

Real Axis (seconds

c.

-2

-1

0

1

-1

)

Figure PS10-5 No8b The following is the MATLAB session. The closed-loop step response is shown in Plot (c). K = 150; sysc = tf(K*[1 5.5],[1 40]); sys = tf([10],conv([1 5 0],[1 10])); clp = feedback(sysc*sys,1); step(clp,2);

502

System: clp 1.4 Peak amplitude: 1.07 Overshoot (%): 6.66 1.2

System: clp

At time (seconds): 0.736 Settling time (seconds): 1.07 1

Amplitude

0.8

0.6

0.4

0.2

0 0

1

0.5

1.5

2

Time (seconds)

Figure PS10-5 No8c

Problem Set 10.6 1.

Sketch the asymptotes of the Bode plot magnitude and phase for the following open-loop transfer functions. Make sure to give the corner frequencies, slopes of the magnitude plot, and phase angles. Verify the results using MATLAB. s +1 s +1 b. a. G ( s ) = G (s) = ( s + 10)( s + 1000) s + 10 c.

G (s) =

s +1 s ( s + 10)( s + 1000)

G (s) =

d.

( s + 1) 2 s ( s + 10)( s + 1000)

Solution The frequency response function is

= G ( jω)

( jω + 1) jω + 1 = jω jω + 10 10 10 +1

(

)

The basic terms include: 1. One constant term 1/10 (or 20log10 101 = −20 dB) 2. Two first-order terms: jω + 1 in the numerator with 1 τ = 1 rad/s and jω 10 + 1 in the denominator with 1 τ = 10 rad/s Bode Diagram

0

Magnitude (dB)

-4 -8 -12 20 dB/dec

-16 -20 90 75

Phase (deg)

a.

60 45 30 15 0 10

-2

10

-1

10

0

10

1

Frequency (rad/s)

Figure PS10-6 No1a

10

2

10

3

503

b.

The frequency response function is

= G ( jω)

( jω + 1) jω + 1 = jω ( jω + 10 )( jω + 1000 ) 10000 10jω + 1 1000 +1

)

)(

(

The basic terms include: 1 1. One constant term 1/10000 ( 20 log10 10000 = −80 dB) 2.

Three first-order terms: jω + 1 in the numerator with 1 τ = 1 rad/s, jω 10 + 1 in the denominator with 1 τ = 10 rad/s, and jω 1000 + 1 in the denominator with 1 τ = 1000 rad/s Bode Diagram

-60

Magnitude (dB)

-70 -80 -20 dB/dec -90

-20 dB/dec

-100 -110 90

Phase (deg)

60 30 0 -30 -60 -90 10

-2

10

0

10

2

10

4

Frequency (rad/s)

Figure PS10-6 No1b The frequency response function is

( jω + 1) jω + 1 = jω jω ( jω + 10 )( jω + 1000 ) 10000 jω 10 +1

= G ( jω)

(

)(

jω 1000

)

+1

The basic terms include: 1 1. One constant term 1/10000 ( 20 log10 10000 = −80 dB) One integrator term Three first-order terms: jω + 1 in the numerator with 1 τ = 1 rad/s, jω 10 + 1 in the denominator with 1 τ = 10 rad/s, and jω 1000 + 1 in the denominator with 1 τ = 1000 rad/s Bode Diagram

0 -40

Magnitude (dB)

2. 3.

-80 -120 -20 dB/dec

-160

-20 dB/dec

-200 -40 dB/dec -240 0 -30

Phase (deg)

c.

-60 -90 -120 -150 -180 10

-2

10

0

10

2

Frequency (rad/s)

Figure PS10-6 No1c

10

4

504

d.

The frequency response function is

( jω + 1) ( jω + 1) = jω jω ( jω + 10 )( jω + 1000 ) 10000 jω 10 +1 2

= G ( jω)

(

2

)(

jω 1000

)

+1

The basic terms include: 1 1. One constant term 1/10000 ( 20 log10 10000 = −80 dB) 2.

One integrator term

3.

Four first-order terms: ( jω + 1) in the numerator with 1 τ = 1 rad/s, jω 10 + 1 in the denominator with 2

1 τ = 10 rad/s, and jω 1000 + 1 in the denominator with 1 τ = 1000 rad/s Bode Diagram

-30 -40

Magnitude (dB)

-50 -60 -70 -80 -20 dB/dec

-90

-20 dB/dec

20 dB/dec

-100 -110 90

Phase (deg)

60 30 0 -30 -60 -90 10

-2

10

0

10

2

10

4

Frequency (rad/s)

Figure PS10-6 No1d 2.

Repeat Problem 1 for the following open-loop transfer functions. 1 a. G ( s ) = 2 s + 4 s + 100 s + 0.5 b. G ( s ) = 2 s + s + 25 c.

G( s) =

s 2 + 0.1s + 25 s 2 + 0.24 s + 144

d.

G( s) =

100( s 2 + 7 s + 49) ( s + 1)( s + 500)

Solution a.

The frequency response function is

= G ( jω)

1 = 2 ( jω ) + 4 jω + 100

0.01

( )

jω 2 10

jω + 2 ( 0.2 ) ( 10 ) +1

The basic terms include: 1. One constant term 0.01 ( 20log10 0.01 = −40 dB) 2. One second-order term in the denominator with ω n = 10 rad/s and ζ = 0.2

505

Bode Diagram

-30

Magnitude (dB)

-40 -50 -60 -70

-40 dB/dec

-80 0

Phase (deg)

-30 -60 -90 -120 -150 -180 10

0

10

1

10

2

Frequency (rad/s)

Figure PS10-6 No2a b.

The frequency response function is

= G ( jω)

jω + 0.5 = 2 ( jω ) + jω + 25

( )

jω + 1) 0.02 ( 0.5

jω 2 5

+ 2 ( 0.1) ( jω5 ) + 1

The basic terms include: 1. One constant term 0.02 ( 20log10 0.02 = −34 dB) 2. One first-order term in the denominator with 1τ = 0.5 rad/s 3.

One second-order term in the denominator with ω n = 5 rad/s and ζ = 0.1 Bode Diagram

0

Magnitude (dB)

-10 -20 -30 20 dB/dec

-40

-20 dB/dec -50 90

Phase (deg)

60 30 0 -30 -60 -90 10

-2

10

-1

10

0

10

1

10

2

Frequency (rad/s)

Figure PS10-6 No2b c.

The frequency response function is

( jω ) + 0.1jω + 25 = 2 ( jω ) + 0.24 jω + 144 2

= G ( jω)

2 25 ( jω5 ) + 2 ( 0.01) ( jω5 ) + 1   jω 2 jω  144  ( 12 ) + 2 ( 0.01) ( 12 ) + 1  

The basic terms include: 25 25 1. One constant term 144 ( 20 log10 144 = −15 dB) 2. Two second-order terms: one in the numerator with ω n = 5 rad/s and ζ = 0.01 and one in the denominator with ω n = 12 rad/s and ζ = 0.01

506

Figure PS10-6 No2c d.

The frequency response function is

The basic terms include: dB) 1. One constant term 49/5 ( 2. Two first-order terms in the denominator with the corner frequencies at 1 and 500 rad/s rad/s and 3. One second-order term in the numerator with

Figure PS10-6 No2d 3.

For each of the following open-loop transfer functions, construct a Bode plot for K = 1 using the MATLAB command bode. Estimate the GM, PM, and their associated crossover frequencies from the plot. Verify the results using the MATLAB command margin. Determine the stability of the corresponding closed-loop system. a. b.

507

c.

KG ( s ) = K

d.

KG ( s ) = K

s + 0.1 s ( s + 5) 2 1 ( s + 5)( s + 0.5) 2

Solution a.

KG ( s ) = >> >> >> >>

K

s ( s + 2 s 2 + 64 ) 2

sys = tf([1],[1 2 64 0]); w = logspace(-2,3,100); bode(sys,w); grid on;

From the plot, we have GM ≈ 43 dB, ωcg ≈ 8 rad/s, PM ≈ 90 , and ωcp ≈ 0.015 rad/s. Using the command >> [gm,pm,wcg,wcp] = margin(sys) returns GM = 128 (or 42.14 dB), ωcg = 8 rad/s, PM = 89.97 , and ωcp = 0.0156 rad/s. Because GM > 0 dB and PM > 0 , the closed-loop system with proportional controller K = 1 is stable. System: sys1

Bode Diagram

Frequency (rad/s): 0.015

50

System: sys1

Magnitude (dB): 0.335

Frequency (rad/s): 8

Magnitude (dB)

0

Magnitude (dB): -43

-50 -100 -150 -200 -90 System: sys1

Phase (deg)

System: sys1

Frequency (rad/s): 8

Frequency (rad/s): 0.015

Phase (deg): -180

Phase (deg): -90

-180

-270 10

-2

10

-1

10

0

10

1

10

2

10

3

Frequency (rad/s)

Figure PS10-6 No3a b.

KG ( s ) = K

1000

( s + 8)( s 2 + 2 s + 4) >> sys = tf([1000],conv([1 8],[1 2 4])); >> w = logspace(-1,2,100); >> bode(sys,w); >> grid on; From the plot, we have GM ≈ −15.4 dB, ωcg ≈ 4.5 rad/s, PM ≈ −36 , and ωcp ≈ 9.16 rad/s. Using the command margin returns GM = 0.1682 (or −15.48 dB), ωcg = 4.47 rad/s, PM = −36 , and ωcp = 9.17 rad/s. Because

GM < 0 dB and PM < 0 , the closed-loop system with proportional controller K = 1 is unstable.

508

Bode Diagram System: sys2 50

Frequency (rad/s): 9.16 Magnitude (dB): 0.019

Magnitude (dB)

0 System: sys2 -50

Frequency (rad/s): 4.5 Magnitude (dB): 15.4

-100

-150 0

Phase (deg)

-90

System: sys2 Frequency (rad/s): 9.16 Phase (deg): -216

-180 System: sys2 -270 Frequency (rad/s): 4.5 -1 0 10Phase (deg): -18010

10

1

10

2

10

3

Frequency (rad/s)

Figure PS10-6 No3b c.

KG ( s ) = K

s + 0.1

s ( s + 5) 2 >> sys = tf([1 0.1],conv([1 0],[1 10 25])); >> w = logspace(-3,3,100); >> bode(sys,w); >> grid on; From the plot, we have PM ≈ 92.2 and ωcp ≈ 0.004 rad/s. Because the phase plot never crosses the line of −180 , the gain margin is ∞ . Using the command margin returns GM = ∞ , PM = 92.20 , and ωcp = 0.004 rad/s. Because GM > 0 dB and PM > 0 , the closed-loop system with proportional controller K = 1 is stable. Bode Diagram

System: sys3 50

Frequency (rad/s): 0.004 Magnitude (dB): 0.00638

Magnitude (dB)

0

-50

-100

-150 0

Phase (deg)

-45 -90 System: sys3 -135 -180

Frequency (rad/s): 0.00402 Phase (deg): -87.8 10

-2

10

0

Frequency (rad/s)

Figure PS10-6 No3c d.

KG ( s ) = K >> >> >> >>

1

( s + 5)( s + 0.5)

2

sys = tf([1],conv([1 5],[1 1 0.25])); w = logspace(-2,2,100); bode(sys,w); grid on;

10

2

509

From the plot, we have GM ≈ 29.8 dB and ωcg ≈ 2.31 rad/s. Note that the magnitude plot never crosses the line of 0 dB and thus the phase margin is ∞ . Using the command margin returns GM = 30.25 (or 29.61 dB), ωcg = 2.29 rad/s, and PM = ∞ . Because GM > 0 dB and PM > 0 , the closed-loop system with proportional controller K = 1 is stable. Bode Diagram

Magnitude (dB)

0

-50

System: sys4 Frequency (rad/s): 2.31 Magnitude (dB): -29.8

-100

-150 0

Phase (deg)

System: sys4 -90

Frequency (rad/s): 2.31 Phase (deg): -180

-180

-270 10

-2

10

-1

10

0

10

1

10

2

Frequency (rad/s)

Figure PS10-6 No3d 4.

The Bode plot of a dynamic system is shown in Figure 10.78, in which the asymptotes are also given. Following the rules of sketching Bode plots, find the transfer function of the system.

Figure 10.78 Problem 4.

Solution

As shown in Figure 10.78, the Bode magnitude plot starts from −20 dB, the slope increases from 0 dB/decade to 20 dB/decade at 1 rad/s and then decreases to −20 dB/decade at 100 rad/s, and there is a lightly-damped resonant peak at 100 rad/s. This implies that the frequency response function of the system consists of one constant term, one first-order term in the numerator, and one second-order term in the denominator, K ( jω / τ + 1) G ( jω) = . ( jω / ωn ) 2 + 2ζ ( jω / ωn ) + 1 −20

20 The initial magnitude −20 dB = 20 log10K, which gives= K 10 0.1 . The slope increases by 20 dB/decade at 1 = rad/s, indicating the corner frequency of the first-order term is 1/τ = 1 rad/s. Similarly, the slope decreases by −40

510

dB/decade at 100 rad/s, indicating the corner frequency, also the natural frequency, of the second-order term is ωn = 100 rad/s. The value of the peak is approximately 55 dB, or 35 dB above the asymptotes, indicating 1 20 log10 = 35 dB 2ζ which gives 1 = ζ = 0.01 35 2 10 20

( )

Thus, the frequency response function of the system is

0.1( jω + 1) G ( jω) = ( jω / 100) 2 + 2(0.01)( jω / 100) + 1 and replacing jω with s gives the transfer function of the system as 1000( s + 1) G ( jω) = 2 s + 2 s + 10000 5.

Figure 10.79 shows the Bode plot for an open-loop transfer function KG(s) with K = 500. a. Determine the stability of the closed-loop system with K = 500. b. Determine the value of K that would yield a PM of 45°. 100

Magnitude (dB)

39 dB 19 dB 0

-100 10

-1

10

0

10

1

10

2

10

3

10

3

Phase (deg)

-100 -150 -180

-135

o

o

-200 -214

o

-250 10

-1

10

0

10

1

10

2

Frequency (rad/s)

Figure 10.79 Problem 5.

Solution a.

b.

6.

As shown in the figure, the magnitude is 19 dB when the phase curve crosses −180 . This implies that GM = −19 dB. The phase is −214 when the magnitude curve crosses 0 dB. This implies that PM = −34 . Because GM < 0 dB and PM < 0 , the closed-loop system with a proportional controller K = 500 is unstable. To yield a phase margin of 45 , the phase when the magnitude plot crosses 0 dB should be −135 . It is observed from the figure that the magnitude is 39 dB when the phase is −135 . To make the magnitude be 0 dB, the magnitude plot should slide downward by 39 dB. This is the effect of dividing by a constant term of 1039 20 = 89.13 and the new value of the proportional control gain K is 500 89.13 = 5.61 . The Bode plot for an open-loop transfer function KG(s) is shown in Figure 10.80. a. Determine the stability of the closed-loop system. b. Assume that the proportional control gain K is increased by a factor of 10 and 100, respectively. Will the closed-loop system still be stable with the new values of K?

511

Figure 10.80 Problem 6.

Solution a.

b.

As shown in the figure, the magnitude is about dB when the phase curve crosses . This implies that dB. The phase is about when the magnitude curve crosses 0 dB. This implies that . dB and , the closed-loop system is stable. Because is increased by a factor of 10, which corresponds to 20 dB, then the magnitude If the proportional control gain plot will slide upward by 20 dB. The magnitude will be still less than 0 dB when the phase curve crosses . The new gain margin is approximately dB. Thus the new closed-loop system is still stable. is increased by a factor of 100, which corresponds to 40 dB, then the If the proportional control gain magnitude plot will slide upward by 40 dB. The magnitude will become greater than 0 dB when the phase curve crosses . The new gain margin is approximately dB. Thus the new closed-loop system is unstable.

7. a. b. c. d.

Consider the unity negative feedback system shown in Figure 10.81. Use MATLAB to obtain the Bode plot KG(s) for K = 5. Determine the stability of the closed-loop system when K = 5 using the stability margins. Determine the value of K that would yield a PM of 40°. Verify the result obtained in Part (c) by using the MATLAB command margin.

Figure 10.81 Problem 7.

Solution a.

b.

The Bode plot of when is shown in the figure below. The following is the MATLAB session. >> num = 100*[1 20]; >> den = conv([1 5 0],[1 10]); >> G = tf(num,den); >> K = 5; >> bode(K*G); >> grid on; Using either the MATLAB command >> [gm,pm,wcg,wcp] = margin(K*G); dB (or or the pull-down menu in the Bode diagram, we can find the stability margins, which are . Because dB and , the closed-loop system is unstable. 0.30) and

512

c.

Figure PS10-6 No7 To yield a phase margin of , the phase when the magnitude plot crosses 0 dB should be . It is observed with that the frequency corresponding to is 4.08 rad/s, at which the from the Bode plot of magnitude is 31.1 dB. To make the magnitude be 0 dB, the magnitude plot should slide downward by 31.1 dB. This is the effect of dividing by a constant term of and the new value of the proportional control gain

is

.

b.

To verify the result obtained in Part (c), we use the MATLAB command margin which returns >> [gm,pm,wcg,wcp] = margin(0.14*G);

8.

Reconsider the feedback system in Figure 10.64. Using the Bode plot technique, find a value of K such that the maximum overshoot in the response to a unit-step reference input is less than 20% and the 2% settling time is less than 1.1 s. Plot the unit-step response of the closed-loop system to verify the result.

.

Solution The open-loop transfer function is

The Bode plot of

with

is shown in Plot (a).

Figure PS10-6 No8a: K = 1 and 2% settling time Note that the requirements are given as overshoot

s. These conditions

513

correspond to ζ > 0.46 and ζω n > 3.67 rad/s. The relationship between the damping ratio and phase margin, i.e.,

PM ≈ 100ζ , yields the requirement PM > 46 . As shown in the Plot (a), PM = 89.8 , which meets the requirement. However, the frequency ωc when the Bode magnitude plot of KG ( s ) crosses −3 dB is only about 0.04 rad/s, which is too slow. To meet both requirements, we must adjust the value of the proportional control gain K . Let us pick PM = 55 . This implies that the phase when the magnitude plot crosses 0 dB should be −125 . It is observed from the figure above that the frequency corresponding to −125 is 4.8 rad/s, at which the magnitude is −46.2 dB. To make the magnitude be 0 dB, the magnitude plot should slide upward by 46.2 dB. This is the effect of multiplying a constant term of 1046.2 20 = 204.17 which is the value of the proportional control gain K . Let us set K to be 204. The Bode plot of the open-loop transfer function KG ( s ) with the new value of K is shown in Plot (b). The phase margin is 54.8 and the crossover frequency ωc is about 6 rad/s. Bode Diagram

50

System: untitled1 Gain Margin (dB): 19.3 At frequency (rad/s): 19.4

Magnitude (dB)

0

Closed loop stable? Yes System: untitled1

-50

Frequency (rad/s): 6.26 Magnitude (dB): -2.99

-100

System: untitled1 Phase Margin (deg): 54.8 Delay Margin (sec): 0.199

-150 -90

At frequency (rad/s): 4.81

Phase (deg)

Closed loop stable? Yes

-180

-270 10

-1

10

0

10

1

10

2

10

3

Frequency (rad/s)

Figure PS10-6 No8(b): K = 204 The unit-step response of the closed-loop system is shown in Plot (c). The overshoot is 13% and the rise time is 0.884 s. Both requirements are met. System: clp Peak amplitude: 1.13

Step Response

Overshoot (%): 13.1

1.2At time (seconds): 0.568 System: clp Settling time (seconds): 0.884

1

Amplitude

0.8

0.6

0.4

0.2

0 0

0.5

1

1.5

Time (seconds)

Figure PS10-6 No8(c)

Problem Set 10.7 1.

For the system shown in Figure 10.83, derive the state-space equations using the state variables indicated. Make sure to give the A, B, C, and D matrices. Also determine the poles of the system.

514

Figure 10.83 Problem 1.

Solution

Note that the input to the subsystem 2 ( s + 3) is u − x2 and the input to the subsystem 1 ( s + 5 ) is x1 . The outputs of these two subsystems are defined as the states x1 and x2 . Thus, in the Laplace domain, we have

X 1 (s) 2  U ( s ) − X ( s ) = s + 3  2  X s ( ) 1 2  =  X 1 (s) s + 5 or

3) 2 (U ( s ) − X 2 ( s ) )  X 1 ( s ) ( s +=  X ( s ) ( s + 5) = X 1 (s)  2 Taking the inverse Laplace transform gives −3 x1 − 2 x2 + 2u  x1 =  x1 + 3 x1 =2u − 2 x2 or     = x x1 − 5 x2 x + x = x 5   2 2 1 2 The output of the system is y = x1 + u − x2 Writing the state equations and the output equation in matrix form results in  x1   −3 −2   x1   2  =     +  u  x2   1 −5   x2   0  y=

[1

x  −1]  1  + 1 ⋅ u  x2 

The characteristic equation of the system is

sI − A =

2 s+3 = s 2 + 8s + 17 = 0 −1 s + 5

which yields the poles s1,2 =−4 ± j . 2.

Repeat Problem 1 for the system shown in Figure 10.80.

Figure 10.84 Problem 2.

Solution In the Laplace domain, we have

515

 X 1 (s) −8 =  − +4 U s X s s ( ) ( ) 3   X 2 (s) 1 =  X 1 (s) s + 2   X 3 (s) 2 =  X 2 (s) s + 6  or

−8U ( s ) + 8 X 3 ( s )  X 1 (s) ( s + 4 ) =  X ( s ) s + X 1 (s) ( 2) =  2  X 3 ( s) ( s + 6 ) = 2 X 2 ( s) 

Taking the inverse Laplace transform gives

−4 x1 + 8 x3 − 8u  x1 =  x1 − 2 x2  x= 2  =  x3 2 x2 − 6 x3 The output of the system is y= x1 + x2 . Writing the state equations and the output equation in matrix form results in  x1   −4 0 8   x1   −8  x1            , y [1 1 0]  x2  + 0 ⋅ u  x2  =  1 −2 0   x2  +  0  u=  x   0 2 −6   x   0  x   3  3   3   The characteristic equation of the system is −8 s+4 0 sI − A = −1 s + 2 0 =( s + 4 ) ( s 2 + 8s + 12 ) − 8(2) =s 3 + 12 s 2 + 44 s + 32 =0 −2 s + 6 0 which yields the poles s1 = −0.96 and s2,3 = −5.52 ± 1.72 j . 3.

Determine the controllability and observability for each of the following systems.  x1   −5 −3 0   x1   2   x1            a. =  x2   2 0 0   x2  +  0= u , y [ 0 1 6]  x2   x   0 1 0   x   0  x   3    3   3  x1   −1 −1 −2   x1   2   x1             b.= 0 u , y [ 2 2 1]  x2   x2   4 0 0   x2  + =  x   0 0 2   x   0     3    3   x3  c.

d.

 x1  1 0 0   x1  1  x1       x  + 1u , y =  x = 0 − 2 0 1 1 0 [ ]  x2   2   2    x  0 7 −6   x  1 x   3    3   3  x1  1 0 0   x1   0   x1         x2  0 1 0   x2  + 0 u= , y [1 1 1]  x2  =  x  0 4 3   x  1  x   3    3   3

Solution a.

The controllability matrix P is

 2 −10 38  = P  B AB = A B   0 4 −20   4   0 0 2

516

which is nonsingular because det( P= ) 32 ≠ 0 . Thus, the system is controllable. The observability matrix Q is

 C  = Q = CA    CA 2 

b.

 0 1 6  2 6 0    2 −6 0 which is nonsingular because det(Q) = −144 ≠ 0 . Thus, the system is observable. The controllability matrix P is  2 −2 −6  2 = P  B AB= A B   0 8 −8    0 0 0  which is singular. Thus, the system is uncontrollable. The observability matrix Q is

 C   Q =  CA=   2 CA  c.

2 1   2  6 −2 −2     −14 −6 −16

which is nonsingular because det(= Q) 224 ≠ 0 . Thus, the system is observable. The controllability matrix P is 1  1 1 2  = P  B AB A = B  1 −2 4    1 1 −20 which is nonsingular because det( P= ) 63 ≠ 0 . Thus, the system is controllable. The observability matrix Q is

d.

 C   = Q  CA =   2 CA  which is singular. Thus, the system is unobservable. The controllability matrix P is

1 1 0 1 −2 0   1 4 0

0 0 0 P = B AB A B  0 0 0  =   1 3 9  which is singular. Thus, the system is uncontrollable. The observability matrix Q is  C  1 1 1  = Q = CA  1 5 3     2 CA  1 17 9  2

which is singular because det(Q) = 0 . Thus, the system is unobservable. 4.

Consider the two-degree-of-freedom mass–spring system as shown in Figure 10.81, in which two masses are to be controlled by two equal and opposite forces f. The equations of motion of the system are given by mx1 + 2kx1 − kx2 = f, mx2 − kx1 + 2kx2 = −f. Show that the system is uncontrollable. Using the concept of mode discussed in Section 9.4, associate a physical meaning with the controllable and uncontrollable modes.

Figure 10.85 Problem 4.

517

Solution Choosing the state vector x = [ x1

x2

x1

x 2 ] and writing the equations of motion in the state-space form T

 x1   x   2 =   x1      x2  The controllability matrix is

 0  0   − 2mk  k  m

0 0 k m 2k m

−

1 0 0 0

0  x1   0  1   x2   0   +  f 0  x1   m1     0  x 2   − m1 

1 0 − m3k2   0 m   3k 1 0  0 −m m2  2 3 = P = B AB A B A B   1 0 − m3k2 0   m  3k  − m1 0 0  m2  which is not full rank because the first two rows are dependent with each other and it is same for the last two rows. Thus, the system is uncontrollable. Using the concept of mode discussed in Section 9.4, we are able to identify the controllable and uncontrollable modes. Rewrite the equations of motion in the second-order form x1   2k −k   x1   f   m 0    +    0 m      =    x2   − k 2 k   x2   − f  Substituting the mass and stiffness matrices into the frequency equation defined by Eq.(9.86), we have 2k − ω 2 m −k K − ω2 M = = m 2 ω 4 − 4kmω 2 + 3k 2 = 0 −k 2k − ω 2 m The two natural frequencies are

k 3k and ω 2 = m m 2 Inserting ω r ( r = 1, 2 ) into Eq. (9.84) to solve for the modal vectors, which are ω1 =

1 1 X 1 =   and X 2 =   1    −1 Clearly, mode 1, the masses moving together in the same direction, is uncontrollable because they are controlled by two equal and opposite forces. Mode 2, the masses moving in the opposite direction, is controllable. 5.

Reconsider Example 10.24. Using the approach in Section 4.4.1, find the controllable canonical form for the plant transfer function and then design a full-state feedback controller that places the closed-loop poles at the same locations as in the example.

Solution The transfer function of the plant is given by

Y ( s) 3.778 = 2 U ( s ) s + 16.883s where the output y is the position of the cart and the input u is the voltage applied to the DC motor. Convert the transfer function to the controller canonical form 1  0 0 = x  x +  u , y = [3.778 0] x  0 −16.883 1  ( s) G =

For a second-order system, the gain matrix K is a 1 × 2 matrix and K = [ k1 k2 ] . The theoretical characteristic equation of the closed-loop system is s −1 sI − A + BK = = s 2 + (16.883 + k2 ) s + k1 = 0 k1 s + 16.883 + k2 We still choose the desired closed-loop poles located at −8.775 ± 10.26j. The desired characteristic equation is

518

Equating the two characteristic equations, we have

. So the full-state feedback controller is

which gives

6.

A regulation system has a plant with the transfer function

a. b.

. Transform the plant transfer function into the state-space form with the state vector Determine the state-feedback gain matrix K such that the closed-loop poles are located at p1,2 = −3 ± 4j and p3 = −8.

c.

Verify the result in Part (b) by using the MATLAB command place.

Solution a.

Choose the state vector as

. The transfer function can be converted to the state-space form ,

b.

For a third-order single-input-single-output system, the gain matrix

is a

matrix and

The theoretical characteristic equation of the closed-loop system is

The desired closed-loop poles are located at

and

, which gives the desired characteristic

equation Equating the two characteristic equations, we have

which gives c.

7.

. So the full-state feedback controller is

The following is the MATLAB session used to compute a full-state feedback control gain matrix pole placement design. >> A = [0 1 0; 0 0 1; -6 -4 -3]; >> B = [0 0 5]'; >> p = [-3+4j -3-4j -8]; >> K = place(A,B,p); Consider the system

using

519

a.

Design a state-feedback controller so that the closed-loop poles have a damping ratio ζ = 0.7 and a natural frequency ωn = 3 rad/s.

b.

Verify the result in Part (a) by using the MATLAB command place.

Solution a.

For a second-order single-input-single-output system, the gain matrix

is a

matrix and

The theoretical characteristic equation of the closed-loop system is

The closed-loop poles have a damping ratio ζ = 0.7 and a natural frequency ωn = 5 rad/s. Thus, the desired closed-loop poles are located at which gives the desired characteristic equation Equating the two characteristic equations, we have

which gives

. So the full-state feedback controller is

b.

The following is the MATLAB session used to compute a full-state feedback control gain matrix pole placement design. >> A = [0 1; 0 -1]; >> B = [0 10]'; >> zeta = 0.7; >> wn = 3; >> p = [-zeta*wn+j*wn*sqrt(1-zeta^2) -zeta*wn-j*wn*sqrt(1-zeta^2)]; >> K = place(A,B,p);

8.

Consider the system

a.

using

Design a state-feedback controller so that the closed-loop response has an overshoot of less than 5% and a rise time under 0.5 s. Set one of the closed-loop poles at −10.

b.

Verify the result in Part (a) by using the MATLAB command place.

Solution a.

For the plant

using the state vector

, we convert the transfer function to the state-space form

For a third-order single-input-single-output system, the gain matrix The theoretical characteristic equation of the closed-loop system is

is a

matrix and

.

520

The time-domain specifications, and

and

s, indicate that

and

rad/s. Choose

rad/s. Two of the closed-loop poles are complex conjugate and located at

And the third one is given,

. Thus, the desired characteristic equation of the closed-loop system is

Equating the two characteristic equations, we have

which gives b.

. So the full-state feedback controller is

The following is the MATLAB session. A = [0 1 0; 0 0 1; -6 -11 -6]; B = [0 0 1]'; zeta = 0.75; wn = 5; p = [-zeta*wn+j*wn*sqrt(1-zeta^2) -zeta*wn-j*wn*sqrt(1-zeta^2) -10]; K = place(A,B,p);

Problem Set 10.8 1.

Consider the control system shown in Figure 10.42 (Problem Set 10.4, Problem 3). Using the results obtained in Part (a), build a Simulink block diagram to simulate the feedback control system and find the unit-step response of the closed-loop system.

Solution

Figure PS10-8 No1 2.

Repeat Problem 1 for the control system shown in Figure 10.43 (Problem Set 10.4, Problem 4).

Solution

Figure PS10-8 No2

521

3.

Repeat Problem 1 for the control system shown in Figure 10.44 (Problem Set 10.4, Problem 5).

Solution

Figure PS10-8 No3 4.

Repeat Problem 1 for the control system shown in Figure 10.45 (Problem Set 10.4, Problem 6).

Solution

Figure PS10-8 No4 5.

Consider the control system shown in Figure 10.63 (Problem Set 10.5, Problem 7). Using the results obtained in Part (b) of the cited problem, build a Simulink block diagram to simulate the feedback control system and find the unit-step response of the closed-loop system.

Solution

Figure PS10-8 No5 6.

Repeat Problem 5 for the control system shown in Figure 10.64 (Problem Set 10.5, Problem 8).

Solution

Figure PS10-8 No6

522

7.

Consider Problem 6 in Problem Set 10.7. Using the state-space model obtained in Part (a) and the full-state feedback controller obtained in Part (b), build a Simulink block diagram to simulate the resulting feedback control system. Find the closed-loop response if the initial conditions are x1(0) = 0.1 and x2(0) = 0.

Solution

Figure PS10-8 No7a

Figure PS10-8 No7b 8.

Consider Problem 7 in Problem Set 10.7. Using the full-state feedback controller obtained in Part (b), build a Simulink block diagram to simulate the resulting feedback control system. Find the closed-loop response if the initial conditions are x1(0) = 0.1 and x2(0) = 0.

Solution

Figure PS10-8 No8a

523

Figure PS10-8 No8b 9.

, is

Consider the rotational mass–spring–damper system in Example 5.22. A PD controller,

designed to adjust the input torque τ so that the rotational disk can quickly return to the equilibrium position regardless of disturbances applied to the system. The performance requirements of the closed-loop system are overshoot Mp < 5% and rise time tr < 0.004 s. a. Design a PD controller to meet the performance requirements. b.

Build a block diagram of the feedback control system, in which the plant is constructed using Simscape blocks and the controller is constructed using Simulink blocks. Find the closed-loop response if the disk is initially 0.1 rad away from the equilibrium position.

Solution a.

The differential equation of motion of the rotational mass–spring–damper system is or For a feedback control system with a PD controller

, the differential equation of motion of the

closed-loop system is or The characteristic equation is Denoting the natural frequency and the damping ratio of the closed-loop system as

The two requirements for the transient response of the closed-loop system are implies that

and

, we have

and

s. This

524

rad/s Choose

and and

b.

rad/s. Substituting the values of

and

into the above equation set yields

.

The block diagram is given in Plot (a), and Plot (b) shows the resulting closed-loop response.

Figure PS10-8 No9a

Figure PS10-8 No9b 10.

Consider the mass–spring–damper system shown in Figure 5.107 (Example 5.21). Assume that f is a control force to maintain the system at equilibrium regardless of disturbances applied to the system. a. Design a full-state feedback controller such that the closed-loop poles are located at −10±10j, −15, and −16. . Assume the state vector b. Build a Simulink block diagram of the feedback control system. Find the closed-loop response if mass 1 is initially 0.1 m from the equilibrium position.

Solution a.

The differential equations of motion of the system is

525

Assume the state vector x = [ x1

x2

x1

x2 ]T . Converting to the state-space form yields

0 1 0  0 0  0 0 1   x1     x1   0  x1  0  x         x  k1 + k2 k2  1 0 0 0   x2   0   2 0 0   2 = = +  1 u y    −  +  u m1 m1 0 1 0 0  x3  0  x3     x3   m   x4   k2  x   1   x4  k − 2 0 0   4   0   m2  m2  The following is the MATLAB session, which returns the control gain matrix K= [ −3349 −2057 −306 −1699] % system parameters m1 = 6; m2 = 1.65; k1 = 6000; k2 = 800; % state-space matrices A = [0 0 1 0; 0 0 0 1; -(k1+k2)/m1 k2/m1 0 0; k2/m2 -k2/m2 0 0]; B = [0 0 1/m1 0]'; C = [1 0 0 0; 0 1 0 0]; D = [0 0]'; % closed-loop poles pc = [-10+10j -10-10j -15 -16]; % controller K = place(A,B,pc); b.

The block diagram is given in Plot (a), and Plot (b) shows the resulting closed-loop response.

x' = Ax+Bu y = Cx+Du State-Space xdot = Ax + Bu y=x

x1 C* u C

x2

-K* u -K

Figure PS10-8 No10a

526

2

0.15

1.5

1

0.1

x

x

1

2

(m)

(m)

0.5

0

0.05 -0.5

-1

0 -1.5

0

0.5

1

1.5

2

0

0.2

0.4

0.6

Time (s)

0.8

1

1.2

1.4

1.6

1.8

2

Time (s)

Figure PS10-8 No10b

Review Problems 1.

Consider the feedback control system as shown in Figure 10.91. Determine the range of K for closed-loop stability.

Figure 10.91 Problem 1.

Solution The closed-loop transfer function Y ( s ) R ( s ) is

s+2 ( s + 10 ) ( s 2 − 1) ( s + 5) Y (s) Ks + 2 K = = 4 3 2 s+2 R( s) 1 + K s + 15s + 49 s + ( K − 15 ) s + 2 K − 50 ( s + 10 ) ( s 2 − 1) ( s + 5) K

The closed-loop characteristic equation is s 4 + 15s 3 + 49 s 2 + ( K − 15 ) s + 2 K − 50 = 0 , for which we can construct the Routh array

s4 : 1 3

s : 15

49

2 K − 50

K − 15

0

750 − K 2 K − 50 15 − K 2 + 315 K s1 : 0 750 − K s 0 : 2 K − 50 To make the closed-loop system stable, all elements in the first column of Routh array must be positive. Therefore,  750 − K >0   15 750 − K > 0  K < 750  − K 2 + 315K  2  > 0  − K + 315 K > 0  0 < K < 315  750 − K  2 K − 50 > 0  K > 25    2 K − 50 > 0  which leads to 25 < K < 315 . s2 :

527

2.

Consider the feedback control system shown in Figure 10.92. a. Design a PD controller such that the closed-loop poles are at b. c.

.

Estimate the rise time, overshoot, peak time, and 1% settling time for the unit-step response of the closed-loop system. Use MATLAB to plot the unit-step response of the closed-loop system. Find the values of the rise time tr, overshoot Mp, peak time tp, and 1% settling time ts.

Figure 10.92 Problem 2.

Solution a.

The closed-loop transfer function

is

which is a second-order system. The closed-loop characteristic equation is Denoting the closed-loop natural frequency and damping ratio as

As a requirement, the closed-loop poles are located at

and

, we have

. This implies that rad/s

Substituting the values of b.

and

into the previous equation set yields

and

.

The rise time, overshoot, peak time, and 1% settling time are s

s

s c. >> >> >> >> >> >>

The unit step response of the closed-loop system is shown in the figure below. plant = zpk([],[0 -2],1); kp = 8; kd = 2; pd = tf([kd kp],[1]); clp = feedback(pd*plant,1); step(clp)

528

Figure Review10 No2 3.

Consider the feedback control system as shown in Figure 10.93. a. Assuming C(s) = kp, determine the value of the proportional gain that makes the closed-loop system marginally stable. Find the frequency of the sustained oscillation. b. Using the gain and the frequency obtained in Part (a), apply the ultimate sensitivity method of Ziegler–Nichols tuning rules to design a PID controller. Plot the unit-step response of the resulting closed-loop system. Find the values of the rise time tr, c. overshoot Mp, peak time tp, and 1% settling time ts.

Figure 10.93 Problem 3.

Solution a.

The closed-loop characteristic equation is

The value of the proportional gain that makes the closed-loop system marginally stable and the frequency of the pure harmonic oscillation can be determined by solving for the crossing jω-axis points. We have

which is equivalent to

Solving for b.

and

gives

,

and

, 12.

The period of oscillation is s and the ultimate gain is 12. Using the ultimate sensitivity method of Ziegler-Nichols tuning rules, we can determine the PID control gains, which are

529

c.

The unit step response of the resulting closed-loop system is shown in the figure below.

Figure Review10 No3 4. a. b. c.

Consider the feedback control system as shown in Figure 10.94. Determine the value of the gain K such that the undamped natural frequency ωn and the damping ratio ζ of the dominant closed-loop poles are roughly 2 rad/s and 0.5, respectively. Determine the values of all closed-loop poles. Plot the unit-step response of the resulting closed-loop system. Find the values of the rise time tr, overshoot Mp, peak time tp, and 2% settling time ts.

Figure 10.94 Problem 4.

Solution a.

The closed-loop characteristic equation is

Plots (a) and (b) show the root locus and the zoom-in of the designed region, where the dashed diagonal lines indicate poles with a damping ratio of about 0.5 and the dashed semicircle indicates poles with a natural frequency of 2 rad/s. The intersections of the diagonals and semicircle are the desired dominant pole locations. The corresponding gain value is about 190.

530

Root Locus 30

20

Imaginary Axis

10

0

-10

-20

-30 -60

-50

-40

-30

-20

-10

0

10

Real Axis

Figure Review10 No4a Root Locus 3

K = 190 2

Imaginary Axis

1

0

ω = 2 rad/s

-1

ζ = 0.5

-2

-3 -5

-4.5

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

Real Axis

Figure Review10 No4b b.

The two dominant poles are about

p1,2 = −ζω n ± jω n 1 − ζ 2 ≈ −0.5(2) ± j2 1 − 0.52 = −1 ± 1.732 j Using MATLAB rlocfind, we can determine the values of the closed-loop poles. >> sysc = zpk([-3 -3],[-20 -20],1); >> sys = zpk([],[0 0],1); >> sysh = zpk([],[-10],10); >> p = -1+j*1.732; >> [K,poles] = rlocfind(sysc*sys*sysh,p) K = 190.6313 poles = -28.9837 -9.5045 + 7.6418i -9.5045 - 7.6418i -1.0037 + 1.7241i -1.0037 - 1.7241i

531

c.

The unit-step response of the closed-loop system with

is shown in Plot (c).

Figure Review10 No4c 5.

Consider a unity negative feedback system with the open-loop transfer function

a. b. c.

Use MATLAB to draw the Bode plots for K = 1. Determine the range of K for which the closed-loop system will be stable. Determine the range of K for closed-loop stability by sketching the root locus. Using Routh’s criterion, determine the range of K for closed-loop stability.

Solution a.

The Bode plot shows that the phase curve crosses at frequency 4.24 rad/s, and the corresponding dB. Thus, the gain margin is about 44.2 dB or 162.18. This indicates that the magnitude is about . closed-loop system is stable for

Figure Review10 No5a b.

, zeros: none poles: 0, Asymptotes for large gain values: (1) the number of asymptotes:

(3 poles and no zeros)

532

0−3−6 = −3 3 (3) the angle of asymptotes: φ1 = 60 , φ2 = 180 , φ3 = −60 Crossing jω-axis points:

(2) the centroid of asymptotes: α =

K 1 + KL ( jω ) = 1+ 0 = jω ( jω + 3)( jω + 6 ) − jω3 − 9ω2 + 18jω + K = 0 which is equivalent to

 −9ω2 + K = 0  3 0 −ω + 18ω = Solving for ω gives = ω 0, ± 3 2 and K = 0, 162 . As shown in Plot (b), the closed-loop system is stable for 0 < K < 162 . Root Locus

15 System: G Gain: 162 Pole: 0.00113 + 4.24i

10 Damping: -0.000266

-1

)

Overshoot (%): 100 Frequency (rad/s): 4.24

Imaginary Axis (seconds

5

0

-5

-10

-15 -25

-20

-15

-10 Real Axis (seconds

-5

5

0

10

-1

)

Figure Review10 No4b c.

The closed-loop characteristic equation is s ( s + 3)( s + 6 ) + K = s 3 + 9 s 2 + 18s + K = 0 for which we can construct the Routh array 18 s3 : 1

s2 : 9 s1 : 18 −

K K 9(18) − K = 9 9

0

s0 : K To make the closed-loop system stable, all elements in the first column of Routh array must be positive. Therefore, K  18 − > 0 9   K > 0 which leads to 0 < K < 162 . 6.

The Bode plot of a dynamic system is shown in Figure 10.95, in which the asymptotes are also given. Following the rules of sketching Bode plots, find the transfer function of the system.

533

Figure 10.95 Problem 6.

Solution Note the following observations: (1) The initial magnitude is 20dB, and the initial phase is 0 degree, indicating a gain of K = 10. (2) There is a lightly-damped resonance at ωn = 1 rad/sec, indicating a second-order term. The value of the peak is approximately 35 dB, or 15 dB above the asymptotes, indicating

which gives

(3) There is a first-order term at ω = 10 rad/sec, indicating τ = 0.1. Combining these observation, the FRF is

or

7.

Consider the system

a. b.

Design a state-feedback controller so that the closed-loop unit-step response has an overshoot of less than 10% and a peak time under 0.5 s. Verify the results of Part (a) in MATLAB.

Solution a.

For the given system, , matrix

is a

matrix and

. The theoretical characteristic equation of the closed-loop system is

534

The time-domain specifications, rad/s and

and

s, indicate that

and

rad/s. Choose

. Two of the closed-loop poles are complex conjugate and located at

and the desired characteristic equation of the closed-loop system is Equating the two characteristic equations, we have

which gives

. So the full-state feedback controller is

b.

The following MATLAB command session returns . A = [0 1; 0 -10]; B = [0; 210]; wn = 8; zeta = 0.7; p = [-zeta*wn+j*wn*sqrt(1-zeta^2) -zeta*wn-j*wn*sqrt(1-zeta^2)]; K = place(A,B,p);

8.

Consider the two-degree-of-freedom quarter-car model shown in Figure 5.34, in which the force f, applied between the car body and the wheel–tire–axle assembly, is controlled by feedback and represents the active . Build a Simulink components of the suspension system. Assume that block diagram of the feedback control system. Find the displacement responses x1(t) and x2(t) if initially x1 = −0.05 m and x2 = −0.05m. Ignore the displacement input z(t). What are the system responses x1(t) and x2(t) without control?

Solution The Simulink block diagram is shown below, where the quarter-car model is represented in the state-space form,

and the full-state feedback controller is

Figure Review10 No8a The displacement responses x1(t) and x2(t) for initial values of x1 = −0.05 m and x2 = −0.05m are shown below.

535

Figure Review10 No8b Setting the control gain matrix K to be zero (a 1 by 4 row vector) and running the simulation yields the system responses x1(t) and x2(t) without control.

Figure Review10 No8c 9.

Consider the DC motor–driven wheeled mobile robot shown in Figure 6.82, in which the voltage applied to , where xr is a the DC motor is computed by a controller. Assume that reference trajectory that the cart should follow. Build a block diagram of the feedback control system, where the mobile robot is constructed using Simscape blocks and the controller is constructed using Simulink blocks. Find the displacement response x(t) of the mobile robot if a unit-step reference command signal is sent to the system.

Solution The Simulink block diagram is shown below, where the controller is

536

Figure Review10 No9a For a unit-step reference input, the displacement response x(t) of the mobile robot is shown below.

Figure Review10 No9b 10. Case Study Consider the cart–inverted-pendulum system shown in Figure 5.79. Assume that the mass of the cart is 0.8 kg, the mass of the pendulum is 0.2 kg, and the length of the pendulum is 0.6 m. a. Determine the poles of the linearized system. Is it stable or unstable? b. Design a full-state feedback controller for the linearized system using the pole placement method. Assume that two of the closed-loop poles are complex conjugate, with a natural frequency of 3.6 rad/s and a damping ratio of 0.6. They dominate the effect of the other two poles, which are assumed at −10 and −20. c. Assume that the initial angle of the inverted pendulum is 5° away from the vertical reference line. Using the state feedback gain matrix K obtained in Part (b), examine the responses of the nonlinear and linearized closed-loop systems using Simulink.

Solution a.

The state-space model of the cart-inverted-pendulum system in Figure 5.79 is

537

0 1 0 0 0        x x  1 0  x1  0 0 1  1   0   x   x     1 0 0 0   x2   0   2   2   3Mg 4 + u y = 0 − 0 0  =   0 1 0 0  x  + 0 u.   x   M + 4m   3    x3     3   M + 4m   x4    x4    x4  6( M + m ) g 6    0 L( M + 4 m ) 0 0   − L( M + 4 m )      The following MATLAB session returns the open-loop poles, which are 0, 0, and ±5.37 . Thus, the system is unstable. % m M L g % A

system parameters = 0.8; % mass of the cart = 0.2; % mass of the pendulum = 0.6; % length of the pendulum = 9.81; state space matrices = [0 0 1 0; 0 0 0 1; 0 -3*M*g/(M+4*m) 0 0; 0 6*(M+m)*g/L/(M+4*m) 0 0]; B = [0; 0; 4/(M+4*m); -6/L/(M+4*m)]; C = [1 0 0 0; 0 1 0 0]; D = zeros(2,1); % open-loop poles po = eig(A); b.

Using the following MATLAB commands, wn = 3.6; zeta = 0.6; pc(1:2) = [-zeta*wn+j*wn*sqrt(1-zeta^2) –zeta*wn-j*wn*sqrt(1-zeta^2)]; pc(3:4) = [-10 -20]; K = place(A,B,pc); we can determine the gain matrix of the full-state feedback controller, K= [ −89.83 −162.21 −43.42 −29.04]

c.

The Simulink block diagram for the linear simulation is shown below. 1 s

B* u

Integrator

B

A* u A

C* u x C 180/pi r2d

theta

-K* u -K

Figure Review10 No10a To construct the Simulink block diagram for nonlinear simulation, substituting the parameter values into the nonlinear model,

538

1 1 θ cos θ − MLθ 2 sin θ = f ML 2 2 1 1 2 1 ML θ + MLx cos θ − MgL sin θ = 0 3 2 2

( m + M )  x+

yields

 x + 0.06 θ cos θ − 0.06θ 2 sin θ = f 0.024 θ + 0.06  x cos θ − 0.06 g sin θ = 0

Applying Cramer’s rule to solve for x and  θ gives

240 f + 14.4θ 2 sin θ 240 − 36cos2 θ 600 g sin θ − 600 f cos θ − 36θ 2 sin θ cos θ  θ= 240 − 36cos2 θ where f = −Kx. Below is the Simulink block diagram for the nonlinear simulation, in which the function block Fcn is defined using the equation of x and Fcn1 is defined using the equation of  θ.  x=

Figure Review10 No10b 5

0.07

Linear

Linear Nonlinear

Nonlinear

4

0.05

3

0.04

2

Angle of pendulum (deg)

Position of cart (m)

0.06

0.03

0.02

0.01

1

0

-1

-2

0

-3

-0.01 0

1

2

3

4

5

0

1

2

3

Time (s)

Time (s)

Figure Review10 No10c

4

5