Solutions Manual for SIGNALS AND SYSTEMS A MATLAB Integrated Approach 9781466598560

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SOLUTIONS MANUAL FOR SIGNALS AND SYSTEMS A MATLAB® Integrated Approach

by

Oktay Alkin

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2014 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20140611 International Standard Book Number-13: 978-1-4665-9856-0 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

2

Contents 1 Signal Representation and Modeling

5

2 Analyzing Continuous-Time Systems in the Time Domain

41

3 Analyzing Discrete-Time Systems in the Time-Domain

85

4 Fourier Analysis for Continuous-Time Signals and Systems

121

5 Fourier Analysis for Discrete-Time Signals and Systems

149

6 Sampling and Reconstruction

189

7 Laplace Transform for Continuous-Time Signals and Systems

207

8 z-Transform for Discrete-Time Signals and Systems

269

9 State-Space Analysis of Systems

331

10 Analysis and Design of Filters

367

11 Amplitude Modulation

401

3

4

CONTENTS

Chapter 1 Signal Representation and Modeling replacemen

1.1. The signal xa (t) 3

a.

Amplitude

2 1 0 −1 −1

1

0

2

3

4

5

t (sec)

The signal xb (t) 3

b.

Amplitude

2 1 0 −1 −2 −1

0

1

3

2

4

5

t (sec)

1.2. a.

b.

 0,       2t + 2 , −t + 2 , x a (t ) =    1,    −t + 3 ,

t < −1 or t > 3 −1 < t < 0 0> >> >>

ek = exp(− j * 6 * pi * k / 8 ) ; ck ek dk

CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS

182

Print the coefficients c˜k and d˜k and in the range k = −7, . . . , 7 and check symmetry properties using the following code:

5.38. 1 2 3 4 5

k = [ −7:7]; xn = [ 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 ] ; ck = s s _ d t f s ( xn , k ) gn = [ 5 , 6 , 7 , 0 , 1 , 2 , 3 , 4 ] ; dk = s s _ d t f s ( gn , k )

5.39.

The script “pr_05_39.m” listed below computes the DTFS coefficients of the periodic pulse

train: 1 2 3 4 5 6 7

k = [ 0 : N−1]; k ( 1 ) = eps ; % Avoid d i v i s i o n by zero ck = sin ( pi * k/N* ( 2 * L + 1 ) ) . / (N* sin ( pi * k/N) ) ; subplot (211) stem ( k , abs ( ck ) ) ; subplot (212) stem ( k , angle ( ck ) ) ;

% DTFS c o e f f i c i e n t s

The graphs can be obtained by using the script as shown below:

a. >> >> >>

N = 30; L = 5; pr_05_39 ;

>> >> >>

N = 30; L = 8; pr_05_39 ;

>> >> >>

N = 40; L = 10; pr_05_39 ;

>> >> >>

N = 40; L = 15; pr_05_39 ;

b.

c.

d.

183

5.40. The script “pr_05_40.m” listed below computes the DTFS coefficients of the periodic pulse train. It is a slightly modified version of the script used in Problem 5.39. 1 2 3 4 5

k = [−M:M]+ eps ; N = 2 *M+1; Omg = 2 * pi * k/N; ck = sin ( pi * k/N* ( 2 * L + 1 ) ) . / (N* sin ( pi * k/N) ) ; stem (Omg, ck ) ;

% DTFS c o e f f i c i e n t s

The graphs can be obtained by using the script as shown below:

a. >> M = 20; >> L = 3 ; >> pr_05_40 ;

b. >> M = 35; >> L = 3 ; >> pr_05_40 ;

c. >> M = 60; >> L = 3 ; >> pr_05_40 ;

5.41. a. 1 2 3 4 5 6 7 8 9 10

b.

k = [ −3:3]+ eps ; n = [ − 50: 50] ; ck = sin ( 7 * pi * k / 4 0 ) . / ( 4 0 * sin ( pi * k / 4 0 ) ) ; % DTFS c o e f f i c i e n t s sum = 0 ; for kk = 1 : 7 , sum = sum+ck ( kk ) * exp ( j * 2 * pi /40 * k ( kk ) * n ) ; end ; sum = real (sum ) ; % Clean up imaginary part that i s % due to roundoff e r r o r . stem (n ,sum ) ;

CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS

184 1 2 3 4 5 6 7 8 9 10

k = [ −3:3]+ eps ; n = [ − 50: 50] ; ck = sin (13 * pi * k / 4 0 ) . / ( 4 0 * sin ( pi * k / 4 0 ) ) ; % DTFS c o e f f i c i e n t s sum = 0 ; for kk = 1 : 7 , sum = sum+ck ( kk ) * exp ( j * 2 * pi /40 * k ( kk ) * n ) ; end ; sum = real (sum ) ; % Clean up imaginary part that i s % due to roundoff e r r o r . stem (n ,sum ) ;

5.42. a. The output of the length-4 moving average filter can be computed using the following script: >> >> >> >> >> >> >> >> >>

clear a l l n = [0:49]; inp = exp ( j * 0 . 2 * pi * n ) ; out = [ ] ; for nn = 1: 50 , xn = inp (nn ) ; yn = ss_movavg4 ( xn ) ; out = [ out , yn ] ; end ;

% Clear any p e r s i s t e n t v a r i a b le s % Input stream % Output stream

The steady-state response was found in Problem 5.24 and is computed using the following statement: >>

yss = 0.7694 * exp ( j * ( 0 . 2 * pi * n− 0. 9425) ) ;

% Steady s t a t e response

The two responses may be printed on the screen for comparison using >> >>

out yss

b. 1 2 3 4 5 6 7 8 9 10 11 12

clear a l l % Clear any p e r s i s t e n t v a r i a b le s n = [0:49]; inp = cos ( 0 . 2 * pi * n ) ; % Input stream out = [ ] ; % Output stream for nn = 1: 50 , xn = inp (nn ) ; yn = ss_movavg4 ( xn ) ; out = [ out , yn ] ; end ; yss = 0.7694 * cos ( 0 . 2 * pi * n−0.9425); % Compare the r e s u l t s [ n ’ , out ’ , yss ’ ]

185

c. 1 2 3 4 5 6 7 8 9 10 11 12

clear a l l % Clear any p e r s i s t e n t v a r i a b le s n = [0:49]; inp = 2 * sin ( 0 . 3 * pi * n ) ; % Input stream out = [ ] ; % Output stream for nn = 1: 50 , xn = inp (nn ) ; yn = ss_movavg4 ( xn ) ; out = [ out , yn ] ; end ; yss = 1.0474 * sin ( 0 . 3 * pi * n−1.4137); % Compare the r e s u l t s [ n ’ , out ’ , yss ’ ]

d. 1 2 3 4 5 6 7 8 9 10 11 12

clear a l l % Clear any p e r s i s t e n t v a r i a b le s n = [0:49]; inp = 3 * cos ( 0 . 1 * pi * n)−5 * sin ( 0 . 2 * pi * n ) ; % Input stream out = [ ] ; % Output stream for nn = 1: 50 , xn = inp (nn ) ; yn = ss_movavg4 ( xn ) ; out = [ out , yn ] ; end ; yss = 2.8179 * cos ( 0 . 1 * pi * n−0.9425) −3.8470 * sin ( 0 . 2 * pi * n−0.9425); % Compare the r e s u l t s [ n ’ , out ’ , yss ’ ]

5.43. a. The script “pr_5_43a.m” listed below computes the output by iterating through the difference equation for n = 0, . . . , 49. 1 2 3 4 5 6 7 8 9 10 11 12 13 14

n = [0:49]; inp = ss_per ( xper , n ) ; out = [ ] ; ynm1 = 0 ; ynm2 = 0 ; xnm1 = 0 ; for nn=1:50 , xn = inp (nn ) ; yn = xn+2 *xnm1−ynm1−0.89 * ynm2; xnm1 = xn ; ynm2 = ynm1; ynm1 = yn ; out = [ out , yn ] ; end ;

CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS

186

The variable “xper” should hold samples of one period prior to using this script. The variable “out” holds the response of the system after the script completes.

b.

Compute the output of the system for the input signal in part (a) of Problem 5.3 using the following statements: >> >> >>

xper = [ 4 , 3 , 2 , 1 , 0 , 1 , 2 , 3 ] ; pr_5_43a ; out1 = out ( 2 6 : 5 0 ) ;

% One period o f the s i g n a l . % Use the s c r i p t from part ( a ) . % Discard the f i r s t 25 samples .

c.

Compute the steady-state output using DTFS coefficients. Us an anonymous function to represent H (Ω) and then evaluate it at the frequencies Ωk = 2πk/8. Compute the DTFS coefficients of the input and the output signals and construct the output signal. >> >> >> >> >> >> >>

H = @(Omg) (1+2 * exp(− j *Omg) ) . / ( 1 + exp(− j *Omg) + 0. 89 * exp(− j * 2 *Omg) ) ; k = [0:7]; Hk = H( 2 * pi * k / 8 ) ; n = [25:49]; ck = s s _ d t f s ( xper , k ) ; dk = Hk. * ck ; out2 = real ( s s _ i n v d t f s ( dk , n ) ) ; % Clean imaginary part that i s % due to roundoff e r r o r

The output computed from the difference equation and the correct steady-state output computed through DTFS may be graphed simultaneously using the following statements: >> >> >> >>

stem (n , out1 ) ; hold on ; plot (n , out2 , ’ ro ’ ) ; hold o f f ;

5.44. a. 1 2 3 4 5 6 7

b.

The function ss_dftmat(..) is listed below. function W = ss_dftmat (N) W = zeros (N,N) ; for k =0:N−1, for n=0:N−1, W( k+1 ,n+1) = exp ( j * 2 * pi /N* n * k ) ; end ; end ;

Using ss_dftmat(..) the 10-point DFT of x[n] is computed as >> W = ss_dftmat ( 1 0 ) ; >> xn = [ 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 ] ’ ; >> Xk=W* xn ;

187

c.

Using ss_dftmat(..) the 20-point DFT of x[n] is computed as >> W = ss_dftmat ( 2 0 ) ; >> xn = [ ones ( 1 , 1 0 ) , zeros ( 1 , 1 0 ) ] ’ ; >> Xk=W* xn ;

5.45. a.

One period of the signal x˜ (t ) is x˜ (t ) =

½

sin (2πt ) , 0 ≤ t < 0.5 0, 0.5 ≤ t < 1

The following script computes estimated EFS coefficients using the function ss_efsapprox(..): 1 2 3 4 5

b.

t = [0:99]/100; x = sin ( 2 * pi * t ) ; x = x . * ( x >=0); k = [ − 15: 15] ; ck_est = ss_efsapprox ( x , k )

Actual EFS coefficients were determined in Example 4-10 of Chapter 4 as   0, k odd and k 6= ∓1      − j /4 , k =1  ck = j /4 , k = −1    −1    ¢ , k even  ¡ 2 π k −1

They can be computed with the following script: >> >>

kk = k+eps ; % Avoid d i v i s i o n by zero . ck_act = −0.25 * j * ( k ==1)+0.25 * j * ( k== −1) −1./( pi * ( kk . * kk − 1 ) ) . * (mod( k , 2) = = 0)

c.

The statement below allows the actual and estimated coefficients to be printed side by side for comparison: >>

conj ( [ k ’ , ck_act ’ , ck_est ’ ] )

Note: In the last MATLAB statement, vectors “ k_a t” and “ k_est” are transposed so that they can be tabulated on the screen in column format. However, it must be remembered that they are complex vectors. The transpose operation in MATLAB has an additional effect of conjugating complex vectors. The conj(..) function is used for counteracting that.

5.46. scriupt listed below defined an anonymous function to return the DTFT of the signal a. The ¡ ¢ x[n] = n u[n] − u[n − 12] .

CHAPTER 5. FOURIER ANALYSIS FOR DISCRETE-TIME SIGNALS AND SYSTEMS

188 1 2 3

Q = @(n ,Omg) n * exp(− j * n *Omg) ; X = @(Omg) Q( 1 ,Omg)+Q( 2 ,Omg)+Q( 3 ,Omg)+Q( 4 ,Omg)+Q( 5 ,Omg)+Q( 6 ,Omg) + . . . Q( 7 ,Omg)+Q( 8 ,Omg)+Q( 9 ,Omg)+Q( 10 ,Omg)+Q( 11 ,Omg) ;

b.

The transform X (Ω) may be evaluated at 10 equally spaced frequencies and the corresponding inverse transform may be computed with the following code. >> >> >>

k = [0:9]; Sk = X( 2 * pi * k / 1 0 ) ; sn = i f f t ( Sk )

c.

The transform X (Ω) may be evaluated at 8 equally spaced frequencies and the corresponding inverse transform may be computed with the following code. >> >> >>

k = [0:7]; Sk = X( 2 * pi * k / 8 ) ; sn = i f f t ( Sk )

Chapter 6 Sampling and Reconstruction 6.1. a. xs (t) 1

−1

b.

t

1

Using A = 1 and τ = 1 the transform of the original signal x a (t ) is ¡ ¢ X a () = sinc2 f

The transform of the impulse sampled signal x s (t ) is X s () =

1 Ts

=5

c.

∞ X

k=−∞ ∞ X

k=−∞

¡ ¢ Xa f − k fs

¡ ¢ sinc2 f − 5k

Xs (f ) 6 5 4 3 2 1 0 −1 −10

−5

0

f (Hz)

189

5

10

CHAPTER 6. SAMPLING AND RECONSTRUCTION

190

6.2. X s () =

∞ X

1 Ts

k=−∞ ∞ X

= 100

¢ ¡ Xa f − k fs

k=−∞

¡ ¢ X a f − 100k

Xs (f ) 100 ...

... −135 −105 −65 −95

6.3.

−35

−5 5

35

65

105

f (Hz)

135

95

Start with X a (ω) shown below. Xa (ω) 1

−70π −10π 10π

ω (rad/s)

70π

The spectrum of the discrete-time signal x[n] is 1 X (Ω) = Ts

∞ X

Xa

k=−∞

= 100

∞ X

µ

Xa

k=−∞

Ω − 2πk Ts

µ

¶

Ω − 2πk 0.01

¶

X (Ω) 100 ...

... −2.7π

−2π

−1.3π −0.7π

0.7π

1.3π

2π

6.4.

2.7π

Ω (rad)

If aliasing is to be avoided, the minimum sampling rate required is f s,mi n = 70 Hz. Let the sampling rate be 90 percent of the required minimum, that is, f s = 63 Hz. µ ¶ ∞ Ω − 2πk 1 X Xa X (Ω) = T s k=−∞ Ts = 63

∞ X

k=−∞

Xa

µ

Ω − 2πk 1/63

¶

191 Xa (Ω/Ts ) 1

−

10π 9

10π 9

Ω (rad)

X (Ω) 63 ...

... −2π

2π −

10π 9

−

8π 9

8π 9

Ω (rad)

10π 9

X (Ω)

63 ...

... −2π

2π −

10π 9

−

8π 9

8π 9

Ω (rad)

10π 9

6.5. a.

The continuous-time signal is

Using A = 1 and τ = 1 sec, we get

  1 − A |t | , τ x a (t ) =  0,

x a (t ) =

(

−τ < t < τ otherwise

1 − |t | ,

−τ < t < τ

0,

otherwise

The sampling interval is T s = 1/ f s = 1/12 sec. The sampled signal x[n] = x a (nT s ) is obtained as   1 − |n| , 12 x[n] =  0,

−12 ≤ n ≤ 12 otherwise

CHAPTER 6. SAMPLING AND RECONSTRUCTION

192

The signal x[n]

Amplitude

1 0.8 0.6 0.4 0.2 0 −0.2

−15

−10

−5

5

0

10

15

Sample index n

b.

Let us use Eqn. (6.25). The spectrum of the continuous-time signal x a (t ) is ³ ωτ ´ ³ω´ X a (ω) = Aτ sinc2 = sinc2 2π 2π

and X (Ω) is computed using Eqn. (6.25) as 1 X (Ω) = Ts

∞ X

Xa

k=−∞

µ

µ

¶ Ω − 2πk 1 = Ts Ts

µ

¶ ¶ µ ∞ X Ω − 2πk 2 6Ω = 12 − 12k sinc 2π (1/12) π k=−∞

∞ X

sinc

2

k=−∞

Ω − 2πk 2π T s

¶

Using T s = 1/12 s, we get X (Ω) = 12

∞ X

sinc

k=−∞

2

|X (Ω)|

Magnitude

15

10

5

0 −π

−3π/4

−π/2

0

−π/4

π/2

π/4

Ω (rad)

6.6. The discrete-time signal is

or, equivalently

 ´ ³  sin πn , 15 x[n] =  0,

x[n] = sin Let

³ πn ´ ¡

15

n = 0, . . . , 14 otherwise

u[n] − u[n − 15]

p[n] = u[n] − u[n − 15]

¢

3π/4

π

193 so that P (Ω) =

14 X

n=0

Using the modulation property, X (Ω) is X (Ω) = − j

e − j Ωn =

sin (15 Ω/2) − j 7Ω e sin (Ω/2)

1 1 P (Ω − π/15) + j P (Ω + π/15) 2 2 |X (Ω)|

12

Magnitude

10 8 6 4 2 0 −π

−3π/4

−π/2

0

−π/4

π/4

π/2

π

3π/4

Ω (rad)

6.7. a.

Not bandlimited. Cannot be sampled without loss of information.

b.

Not bandlimited. Cannot be sampled without loss of information.

c.

Bandlimited.

f max = 75 Hz,

d.

Bandlimited.

f max = 175 Hz,

e.

Not bandlimited. Cannot be sampled without loss of information.

f s ≥ 150 Hz. f s ≥ 350 Hz.

6.8. a, b. xa (t) and x[n] 1 0.5 0 −0.5 −1 0

1

2

3

t (sec)

4

5

CHAPTER 6. SAMPLING AND RECONSTRUCTION

194

c.

¡ ¢¯¯ x[n] = sin 2π f a t ¯

t =n/ f s

= sin

µ

¶ µ ¶ 2πn 2πn = sin + 2πr n 2.4 2.4

r, n : Integer

The normalized frequency is

F= For r = 1 :

For r = 2 :

For r = 3 :

F = 1.4167

F = 2.4167

F = 3.4167

⇒ ⇒ ⇒

fa +r fs

f a = 3400 Hz

f a = 5800 Hz

f a = 8200 Hz

replacemen

6.9. a.

Xa (f ) for fa = 1 kHz j 0.5

−j 0.5

−1

replacemen

−8

−6

−4

−2

1 0

2

4

6

8

6

8

f (kHz)

Xa (f ) for fa = 3.4 kHz j 0.5

−j 0.5

−3.4

replacemen

−8

−6

−4

3.4 −2

0

2

4

f (kHz)

Xa (f ) for fa = 5.8 kHz j 0.5

−j 0.5

−5.8 −8

−6

5.8 −4

−2

0

f (kHz)

2

4

6

8

replacemen

195 Xa (f ) for fa = 8.2 kHz j 0.5

−j 0.5

−8.2 −8

8.2 −6

−4

0

−2

2

4

6

8

f (kHz)

b.

Since all four continuous-time signals lead to the same discrete-time signal when sampled; they all have the same DTFT spectrum. µ ¶ 2πn x[n] = sin = sin (0.8333πn) 2.4 X (Ω) =

π j

∞ £ X ¤ δ (Ω − 0.8333π − 2πm) − δ (Ω + 0.8333π − 2πm)

m=−∞

X (Ω) jπ

−jπ

−0.8333π −2π

−π

0.8333π 0

π

2π

Ω (rad)

6.10. a.

¯ ¯ x[n] = x a (t )¯

t =n/ f s

µ

¶ 500πn = sin = sin (0.4πn + 2πr n) , fs

The normalized frequency is

F=

f a 250 = = 0.2 + r fs fs

Using r = 0 for proper sampling, the sampling rate must be fs =

b. For r = 1 :

250 = 1.2 fs

⇒

f s = 208.33 Hz

250 = 1250 Hz. 0.2

r, n : Integer

CHAPTER 6. SAMPLING AND RECONSTRUCTION

196 For r = 2 :

c.

250 = 2.2 fs

⇒

f s = 113.64 Hz

Let the new signal frequency be f¯a .

For r = 1 : For r = 2 :

f¯a = 1.2 1250 f¯a = 2.2 1250

f¯a = 0.2 + r fs

⇒

f¯a = 1500 Hz

⇒

f¯a = 2750 Hz

6.11. a. x[n] = 3 cos (πn) + 5 sin (2.5πn) = 3 cos (πn) + 5 sin (0.5πn) ∞ X © ª F 3 cos (πn) = 6π δ (Ω − π − 2πm) m=−∞

Let

© ª 5π F 5 sin (πn) = j

∞ £ X

m=−∞

δ (Ω − π/2 − 2πm) − δ (Ω + π/2 − 2πm)

¤

X˜ (Ω) = 3π δ (Ω − π) + 3π δ (Ω + π) − j 5π δ (Ω − π/2) + j 5π δ (Ω + π/2) so that X (Ω) can be written as X (Ω) =

∞ X

m=−∞

X˜ (Ω − 2πm)

X (Ω) −j5π

6π

j5π

−j5π

−π/2

−3π/2 −π

6π

π/2 0

Ω (rad)

b.

j5π

The recovered signal would be xˆ (t ) = 3 cos (100πt ) + 5 sin (50πt )

3π/2 π

197

6.12. a.

¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ X a (ω) = π δ f − 100π + π δ f + 100π + π δ f − 120π + π δ f + 120π Xa (ω) π

π

π

−120π −100π

π

100π 120π

ω (rad/s)

b.

The period is T s = 1/ f s = 1/400 = 0.0025 s, or equivalently T s = 2.5 ms. This corresponds to a duty cycle of d = τ/T s = 0.2. X¯ s (ω) = d

∞ X

k=−∞

= 0.2

c.

∞ X

sinc (kd ) X a (ω − kωs )

k=−∞

sinc (0.2k) X a (ω − 800πk)

¯ s (ω) X 0.2π

0.2 sinc (0.2) π

−800π

0.2 sinc (0.2) π

0

800π

ω (rad/s)

6.13.

The spectrum of the zero-order hold sampled signal is X¯ s (ω) = d sinc

µ

¶ ∞ ωd T s − j ωdTs /2 X e X a (ω − kωs ) 2π k=−∞ µ

¶ ∞ 0.2ω − j 0.1ω/800 X = 0.2 sinc e X a (ω − 800πk) 800π k=−∞

CHAPTER 6. SAMPLING AND RECONSTRUCTION

198

¯ s (ω) X

−800π

0

800π

ω (rad/s)

6.14. xr (t) 1 0.5 0 −0.5 −1 0

0.05

0.1

0.15

0.2

0.25

0.15

0.2

0.25

t (sec)

6.15. xr (t) 1 0.5 0 −0.5 −1 0

0.05

0.1

t (sec)

6.16. Using the geometric series formula 1 1 − e j 2πm D 1 − e j 2πm/D For m 6= nD then the numerator is equal to zero, and the denominator is nonzero, resulting in w [m] = 0. w [m] =

If m = nD then both the numerator and the denominator are equal to zero, requiring the use of

199 L’Hospital’s rule: w [nD] =

6.17.

¯ ¯ 1 j 2π e j 2πm ¯ =1 ¯ j 2πm/D D j (2π/D) e m=nD

Using the inverse DTFT relationship 1 h r [n] = 2π

ˆ

π/L

(L) e

j Ωn

−π/L

The result can be simplified to h r [n] =

¯π/L L e j Ωn ¯¯ dΩ = 2π j n ¯−π/L

³ πn ´ L = sinc (n/L) sin πn L

6.18. a. Not bandlimited. Cannot be downsampled without loss of information. b.

Not bandlimited. Cannot be downsampled without loss of information.

c.

Ωmax = π/3. May be downsampled with D = 3.

d.

Ωmax = 2π/7. May be downsampled with D = 3.

6.19. a. H zoh (Ω) =

b.

2 X

n=0

Using Euler’s formula

£ ¤ e − j Ωn = 1 + e − j Ω + e − j 2Ω = 1 + 2 cos (Ω) e − j Ω

x[n] = The transform of the input signal is replacemen

X (Ω) =

π j

∞ £ X

m=−∞

i 1 h j 0.1πn e − e − j 0.1πn 2j

δ (Ω − 0.1π − 2πm) − δ (Ω + 0.1π − 2πm)

X (Ω) jπ

−jπ

jπ

−jπ

−0.1π −2π

−π

jπ

−jπ

0.1π 0

Ω (rad)

π

2π

¤

CHAPTER 6. SAMPLING AND RECONSTRUCTION

200

¶ µ 2 1 X Ω − 2πk X1 = X 3 k=0 3 X1 (Ω)

j

π 3

−j

π 3

j

π 3

−j

−0.3π −2π

π 3

j

π 3

−j

0.3π

−π

0

π

2π

π

2π

Ω (rad)

X 2 (Ω) = X 1 (3 Ω) X2 (Ω)

j

π 3

−j

−0.1π −2π

−π

π 3

0.1π 0

Ω (rad)

X 3 (Ω) = H zoh (Ω) X 2 (Ω) X3 (Ω)

0.1π

−0.1π −2π

−π

0

Ω (rad)

π

2π

π 3

201

6.20. The impulse response of the first-order hold interpolation filter is h f oh [n] = { 1/3, 2/3, 1, 2/3, 1/3 } ↑ n=−2

and its transform is H f oh (Ω) =

1 j 2Ω 2 j Ω 2 1 e + e + 1 + e − j Ω + e − j 2Ω 3 3 3 3

= 1+

2 4 cos (Ω) + cos (2Ω) 3 3

The spectra X (Ω), X 1 (Ω) and X 2 (Ω) are as shown in the solution to Problem 6.19. X 3 (Ω) = H f oh (Ω) X 2 (Ω) X3 (Ω)

0.1π

−0.1π −2π

−π

0

π

Ω (rad)

6.21. a. 1 2 3 4 5 6 7 8 9 10 11 12 13

Xa = @( f ) 2. / ( 1+ 4 * pi * pi * f . * f ) ; f = [ −7:0.01:7]; fs = 3; Ts = 1/ f s ; Xs = zeros ( s i z e ( f ) ) ; for k= −5:5 , Xs = Xs+ f s * Xa ( f−k * f s ) ; end ; plot ( f , Xs ) ; axis ( [ − 7 , 7 , − 1 , 10] ) ; t i t l e ( ’ X_ { s } ( f ) ’ ) ; xlabel ( ’ f (Hz) ’ ) ; grid ;

% Original spectrum

2π

CHAPTER 6. SAMPLING AND RECONSTRUCTION

202

b. 1 2 3 4 5 6 7 8 9 10 11 12 13

c.

Xa = @( f ) 2. / ( 1+ 4 * pi * pi * f . * f ) ; f = [ −7:0.01:7]; fs = 4; Ts = 1/ f s ; Xs = zeros ( s i z e ( f ) ) ; for k= −5:5 , Xs = Xs+ f s * Xa ( f−k * f s ) ; end ; plot ( f , Xs ) ; axis ( [ − 7 , 7 , − 1 , 10] ) ; t i t l e ( ’ X_ { s } ( f ) ’ ) ; xlabel ( ’ f (Hz) ’ ) ; grid ;

% Original spectrum

Aliasing effect is present in both cases, but is less pronounced for f s = 4 Hz.

6.22. ¡ ¢ Compute and graph the spectrum X a f : 1 2 3 4 5 6 7

Xa = @( f , A , tau ) A * tau * sinc ( f * tau ) . * sinc ( f * tau ) ; f = [ −10:0.02:10]; plot ( f , Xa ( f , 1 , 1 ) ) ; axis ( [ − 10 , 10 , − 0. 25 , 1. 25] ) ; xlabel ( ’ f (Hz) ’ ) ; t i t l e ( ’ X_ { a } ( f ) ’ ) ; grid ;

Compute and graph the spectrum X s (s) for the sampling rate f s = 5 Hz. 1 2 3 4 5 6 7 8 9 10 11

fs = 5; % Edit t h i s f o r other sampling r a t e s . Ts = 1/ f s ; Xs = zeros ( s i z e ( f ) ) ; for k= −15:15 , Xs = Xs+1/Ts * Xa ( f−k * fs , 1 , 1 ) ; end ; plot ( f , Xs ) ; axis ([ −10 ,10 , −1 ,6]); xlabel ( ’ f (Hz) ’ ) ; t i t l e ( ’ X_ { s } ( f ) ’ ) ; grid ;

6.23. 1 2 3

A = 1; tau = 1 ; Xa = @(omg) A * tau * ( sinc (omg* tau / ( 2 * pi ) ) ) . ^ 2 ;

203 4 5 6 7 8 9 10 11 12 13 14 15 16 17

f s = 12; Omg = [ − 1 : 0 . 0 0 1 : 1 ] * pi ; % Use Eqn . ( 6 . 2 5 ) X = zeros ( s i z e (Omg) ) ; for k= −10:10 , X = X+12 * Xa ( (Omg−2* pi * k ) * f s ) ; end ; p1 = plot (Omg, X ) ; axis ([ − pi , pi , − 2 , 1 5 ] ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; t i t l e ( ’ | X( \Omega ) | $ ’ ) ; axis ([ − pi , pi , 0 , 1 5 ] ) ; grid ;

6.24. 1 2 3 4 5 6 7 8 9 10

% Compute the transform o f p [ n ] . Omg = [ − 1 : 0 . 0 0 2 : 1 ] * pi+eps ; P = @(Omg) sin ( 7 . 5 *Omg) . / sin ( 0 . 5 *Omg) . * exp(− j * 7 *Omg) ; % Use the modulation property o f the DTFT . X = −0.5 * j * P (Omg−pi / 15) + 0. 5 * j * P (Omg+pi / 1 5 ) ; plot (Omg, abs (X ) ) ; axis ([ − pi , pi , 0 , 1 2 ] ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; t i t l e ( ’ | X( \Omega ) | ’ ) ; grid ;

6.25. a. 1 2 3 4 5 6 7 8 9 10 11

xa = @( t ) sin ( 2 * pi * 1000 * t ) ; t = [ 0 : 5 e −6:5e −3]; f s = 2400; Ts = 1/ f s ; n = [0:14]; plot (1000 * t , xa ( t ) ) ; axis ( [ 0 , 5 , − 1 . 2 , 1 . 2 ] ) ; hold on ; stem(1000 * n * Ts , xa (n * Ts ) , ’ r ’ ) ; hold o f f ; grid ;

b. 1 2 3

xa = @( t ) sin ( 2 * pi * 1000 * t ) ; xb = @( t ) sin ( 2 * pi * 3400 * t ) ; xc = @( t ) sin ( 2 * pi * 5800 * t ) ;

CHAPTER 6. SAMPLING AND RECONSTRUCTION

204 4 5 6 7 8 9 10 11 12 13 14

xd = @( t ) sin ( 2 * pi * 8200 * t ) ; t = [ 0 : 5 e −6:1e −3]; f s = 2400; Ts = 1/ f s ; n = [0:14]; plot (1000 * t , xa ( t ) , ’b ’ ,1000 * t , xb ( t ) , ’b−− ’ ,1000 * t , xc ( t ) , ’ g−− ’ ,1000 * t , xd ( t ) , ’ k−− ’ ) ; axis ( [ 0 , 1 , − 1 . 2 , 1 . 2 ] ) ; hold on ; stem(1000 * n * Ts , xa (n * Ts ) , ’ r ’ ) ; hold o f f ; grid ;

6.26. a. Lines 2 and 3 create a naturally sampled version of the signal rather then a zero-order hold version. The vector “t1” contains 0s and 1s depending on whether the corresponding time instant in vector “t” is within an active pulse of the signal p˜ (t ) or not (refer to Fig. 6.18). The loop between lines 5 and 15 searches through this naturally sampled signal. When the left edge of a pulse is encountered, its value is saved in the variable “value”, and the amplitude is adjusted to the saved level for the duration of the pulse.

b.

A modified version of the function ss_zohsamp(..) is listed below as ss_zohsamp2(..). This function is only for the case d = 1. The function ss_zohsamp2(..) is listed below: 1 2 3 4 5 6

function xzoh = ss_zohsamp2( xa , Ts , t ) d e l t = t (2) − t ( 1 ) ; % Time increment used in v e c t o r " t " L = f l o o r ( Ts/ d e l t ) ; % Number o f samples per pulse ( must be i n t e g e r ! ) pulse = ones ( 1 , L ) ; xzoh = downsample ( xa ( t ) , L ) ; xzoh = kron ( xzoh , pulse ) ;

The script below can be used for testing. 1 2 3 4 5

x = @( t ) exp(−abs ( t ) ) ; t = [ −4:0.001:3.999]; xzoh = ss_zohsamp2( x , 0 . 2 , t ) ; plot ( t , xzoh ) ; axis ( [ − 4 , 4 , − 0 . 2 , 1 . 2 ] ) ;

6.27. a. 1 2 3 4

xa = @( t ) exp(−abs ( t ) ) ; t = [ −4:0.001:4]; xzoh = ss_zohsamp( xa , 0 . 2 , 0 . 9 0 , t ) ; plot ( t , xzoh ) ;

205

b. 1 2 3 4

1 2 3 4

1 2 3 4

% System with a=3 sys = t f ( [ 3 ] , [ 1 , 3 ] ) ; y = lsim ( sys , xzoh , t ) ; plot ( t , xzoh , t , y ) ;

% System with a=2 sys = t f ( [ 2 ] , [ 1 , 2 ] ) ; y = lsim ( sys , xzoh , t ) ; plot ( t , xzoh , t , y ) ;

% System with a=1 sys = t f ( [ 1 ] , [ 1 , 1 ] ) ; y = lsim ( sys , xzoh , t ) ; plot ( t , xzoh , t , y ) ;

6.28. The system can be simulated with the following code: 1 2 3 4 5 6

n = [0:199]; xn = sin ( 0 . 1 * pi * n ) ; % Signal x [ n ] . x1 = downsample ( xn , 3 ) ; x2 = upsample( x1 , 3 ) ; hzoh = ones ( 1 , 3 ) ; % Impulse response o f i n t e r p o la t i o n f i l t e r . x3 = conv ( x2 , hzoh ) ;

Individual signals can be graphed using following code segments: 1 2 3 4

1 2 3 4

1 2 3 4

stem (n , xn ) ; axis ( [ − 0 . 5 , 1 9 9 . 5 , − 1 . 2 , 1 . 2 ] ) ; xlabel ( ’n ’ ) ; t i t l e ( ’ x [n] ’ ) ;

stem ( [ 0 : 6 6 ] , x1 ) ; axis ( [ − 0 . 5 , 6 6 . 5 , − 1 . 2 , 1 . 2 ] ) ; xlabel ( ’n ’ ) ; t i t l e ( ’ x_ { 1 } [ n ] ’ ) ;

stem ( [ 0 : 2 0 0 ] , x2 ) ; axis ( [ − 0 . 5 , 2 0 0 . 5 , − 1 . 2 , 1 . 2 ] ) ; xlabel ( ’n ’ ) ; t i t l e ( ’ x_ { 2 } [ n ] ’ ) ;

CHAPTER 6. SAMPLING AND RECONSTRUCTION

206 1 2 3 4

stem ( [ 0 : 2 0 0 ] , x3 ( 1 : 2 0 1 ) ) ; axis ( [ − 0 . 5 , 2 0 0 . 5 , − 1 . 2 , 1 . 2 ] ) ; xlabel ( ’n ’ ) ; t i t l e ( ’ x_ { 3 } [ n ] ’ ) ;

6.29. The system can be simulated with the following code: 1 2 3 4 5 6

n = [0:199]; xn = sin ( 0 . 1 * pi * n ) ; % Signal x [ n ] . x1 = downsample ( xn , 3 ) ; x2 = upsample( x1 , 3 ) ; hfoh = [ 1 / 3 , 2 / 3 , 1 , 2 / 3 , 1 / 3 ] ; x3 = conv ( x2 , hfoh ) ;

Individual signals can be graphed using following code segments: 1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

stem (n , xn ) ; axis ( [ − 0 . 5 , 1 9 9 . 5 , − 1 . 2 , 1 . 2 ] ) ; xlabel ( ’n ’ ) ; t i t l e ( ’ x [n] ’ ) ;

stem ( [ 0 : 6 6 ] , x1 ) ; axis ( [ − 0 . 5 , 6 6 . 5 , − 1 . 2 , 1 . 2 ] ) ; xlabel ( ’n ’ ) ; t i t l e ( ’ x_ { 1 } [ n ] ’ ) ;

stem ( [ 0 : 2 0 0 ] , x2 ) ; axis ( [ − 0 . 5 , 2 0 0 . 5 , − 1 . 2 , 1 . 2 ] ) ; xlabel ( ’n ’ ) ; t i t l e ( ’ x_ { 2 } [ n ] ’ ) ;

stem ( [ 0 : 2 0 0 ] , x3 ( 1 : 2 0 1 ) ) ; axis ( [ − 0 . 5 , 2 0 0 . 5 , − 1 . 2 , 1 . 2 ] ) ; xlabel ( ’n ’ ) ; t i t l e ( ’ x_ { 3 } [ n ] ’ ) ;

Chapter 7 Laplace Transform for Continuous-Time Signals and Systems 7.1. a. jω

Using Laplace transform definition ∞

X (s) =

ˆ

=

ˆ

=

1 s +2

e −2t u (t ) e −st d t

−∞ ∞

e

−2t −st

e

0

¯∞ e −(s+2)t ¯¯ dt = − (s + 2) ¯0

σ

−2

For convergence of the integral, we need Re {s + 2} > 0

⇒

|

Re {s} > −2

{z

}

Re {s} > −2

b. jω

X (s) =

ˆ

=

ˆ

∞

e −2t u (t ) e −st d t

−∞ ∞

e −2t e −st d t

1

σ

−2

¯∞ e −2 e −s e −(s+2)t ¯¯ = = − (s + 2) ¯1 s +2

For convergence of the integral, we need Re {s + 2} > 0

⇒

|

Re {s} > −2

{z Re {s} > −2

207

}

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

208

c. jω

X (s) =

ˆ

=

ˆ

∞

e 2t u (−t ) e −st d t

−∞ 0 −∞

e 2t e −st d t =

¯0 e (2−s)t ¯¯ 1 = ¯ (2 − s) −∞ 2 − s

2

σ

For convergence of the integral, we need Re {2 − s} > 0

⇒

Re {s} < 2

|

{z

}

Re {s} < 2

d. jω

X (s) =

ˆ

=

ˆ

∞ −∞ 1

e 2t u (−t + 1) e −st d t 2t −st

e e −∞

¯1 e 2 e −s e (2−s)t ¯¯ = dt = 2−s (2 − s) ¯−∞

2

For convergence of the integral, we need Re {2 − s} > 0

⇒

Re {s} < 2

|

{z

}

Re {s} < 2

e. X (s) = =

ˆ

1 0

(1) e −st d t +

−st ¯¯1

ˆ

2

(−1) e −st d t

1

−st ¯¯2

¤ £ e e ¯ ¯ = 1 1 − 2 e −s + e −2s − ¯ ¯ −s 0 −s 1 s

The ROC is the entire s-plane with the exception of Re {s} → −∞.

σ

209

f. jω

X (s) = =

ˆ

1

(1) e

0 −st ¯¯1

−st

dt +

ˆ

∞

(−1) e −st d t

1

−st ¯¯∞

¤ £ e ¯ e ¯ = a 1 − 2 e −s + ¯ ¯ −s 0 s 1 s

σ

For convergence of the second integral, we need Re {s} > 0

|

{z

} |

Re {s} < 0

{z

}

Re {s} > 0

7.2. Using the definition of the Laplace transform given by Eqn. (7.1) ˆ ∞·∞ ¸ X −anT X (s) = e δ (t − nT ) e −st d t −∞

= =

∞ X

e

n=0

−anT

n=0

∞

−∞

n=0 ∞ X

ˆ

δ (t − nT ) e −st d t

e −anT e −snT =

∞ X

n=0

e −(s+a)T n =

The closed form expression is valid only if ¯ −(s+a)T ¯ © ª ¯e ¯ − Re a

which establishes the ROC. The poles of X (s) are found by solving

jω j 4π/T

e −(s+a)T = 1 which, using the identity e j 2πk = 1, can also be written as e −(s+a)T = e j 2πk

j 2π/T σ

−a

The poles are at s k = −a − j

−j 2π/T

2πk , T

−j 4π/T

all integer k

as shown in the pole-zero diagram. (We are assuming that a is real-valued.)

|

{z Re {s} > −a

}

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

210

7.3. For the Fourier transform to exist, the ROC needs to include the j ω-axis.

a.

ROC:

b.

ROC:

c.

ROC:

d.

ROC:

© ª − 1 < Re s < 1 © ª Re s > −1

© ª Re s < 1

© ª Re s > −1

7.4. For the signal to be causal the ROC needs to be to the right of a vertical line in the s-plane and include s → ∞.

a.

ROC:

b.

ROC:

c.

ROC:

d.

ROC:

© ª Re s > 1

© ª Re s > −1

© ª Re s > 2

© ª Re s > −1

7.5. The ROC of the transform an anti-causal signal is to the left of a vertical line in the s-plane.

a.

ROC:

b.

ROC:

c.

ROC:

d.

ROC:

7.6. a.

© ª Re s < −1

© ª Re s < −2

© ª Re s < 1

© ª Re s < −1

¯ © ª ¯ F x (t ) = X (s)¯

s= j ω

Since s = j ω is included in the ROC the Fourier transform exists, and is found as © ª F x (t ) = ¡

b.

1 − ω jω−1 ¢ =¡ ¢2 2 + ω2 + j ω jω − jω−2

The desired Fourier transform is related to X (s) by ¯ ª © ¯ F x (t ) e t = X (s)¯

s=−1+ j ω

211 Since the trajectory s = −1 + j ω is not included in the ROC, the Fourier transform of x (t ) e t does not exist.

c.

The desired Fourier transform is related to X (s) by ¯ © ª ¯ F x (t ) e −3t = X (s)¯

s=3+ j ω

Since the trajectory s = 3 + j ω is not included in the ROC, the Fourier transform of x (t ) e −3t does not exist.

7.7. a.

¯ © ª ¯ F x (t ) = X (s)¯

b.

c.

¯ ª © ¯ F x (t ) e −t = X (s)¯

s= j ω

=¡

1 1 ¢ =¡ ¢2 2 5 − ω + j 2ω j ω + j 2ω + 5

1 1 ¢ =¡ =¡ ¢2 ¡ ¢ s=1+ j ω 8 − ω2 + j 4ω 1+ jω +2 1+ jω +5

The desired Fourier transform is related to X (s) by

¯ ª © ¯ F x (t ) e t = X (s)¯

s=−1+ j ω

The trajectory s = −1 + j ω is not included in the ROC. Therefore the Fourier transform of x (t ) e t does not exist.

d.

The desired Fourier transform is related to X (s) by ¯ ª © ¯ F x (t ) e 3t = X (s)¯

s=−3+ j ω

The trajectory s = −3 + j ω is not included in the ROC. Therefore the Fourier transform of x (t ) e 3t does not exist.

7.8. jω

a.

The transform has a zero at s = 2 and poles at s = −1, −2. Since x (t ) is causal, the ROC is © ª Re s > −1

σ −2 −1

2

The Fourier transform X (ω) is ¯ ¯ X (ω) = X (s)¯

s= j ω

=¡

jω−2 jω−2 ¢ =¡ ¢2 2 − ω2 + j 3ω j ω + j 3ω + 2

|

{z Re {s} > −1

}

212

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS jω

b.

The transform has a zero at s = 0 and poles at s = ∓1. Since x (t ) is causal, the ROC is σ

© ª Re s > 1

−1

1

The ROC does not include the j ω-axis of the s-plane. Therefore the Fourier transform does not exist. {z

|

}

Re {s} > 1

jω

c.

The transform has a zero at s = −1 and poles at s = 1, 3. Since x (t ) is specified to be anti-causal, the ROC is © ª Re s < 1

σ −1

1

3

The ROC includes the j ω-axis. The Fourier transform X (ω) is ¯ ¯ X (ω) = X (s)¯

s= j ω

=¡

jω+1 jω+1 ¢ =¡ ¢2 3 − ω2 − j 4ω j ω − j 4ω + 3

|

{z

}

Re {s} < 1

jω

d. The transform has zeros at s = 0, 1 and poles at s = −2, 3. Since x (t ) is specified to be anti-causal, the ROC must be σ

© ª Re s < −2

−2

The ROC does not include the j ω-axis. Consequently, the Fourier transform X (ω) does not exist. | {z } Re {s} < −2

1

3

213 jω

e. The transform has zeros at s = 0, 1 and poles at s = −3, −5. Since the Fourier transform of x (t ) e 4t exists, the ROC must include the trajectory s = −4 + j ω. Therefore the ROC is © ª − 5 < Re s < −3

σ −5

−3

The ROC does not include the j ω-axis. Consequently, the Fourier transform X (ω) does not exist.

|

{z

}

−5 < Re {s} < −3

7.9. a.

1 , s +1 © ª 1 , L e −3t u (t ) = s +3 Using the linearity of the Laplace transform we obtain © ª L e −t u (t ) =

X (s) =

© ª Re s > −1

© ª Re s > −3

3 5 −2 (s − 2) , − = s + 1 s + 3 (s + 1) (s + 3)

© ª Re s > −1

b.

1 , s +1 © ª −1 , L e 3t u (−t ) = s −3 Using the linearity of the Laplace transform we obtain © ª L e −t u (t ) =

X (s) =

c.

© ª Re s < 3

3 2 s − 11 − = , s + 1 s − 3 (s + 1) (s − 3) © ª L δ (t ) = 1 ,

1 , s +1 Using the linearity of the Laplace transform we obtain © ª L e −t u (t ) =

X (s) = 1 +

d.

© ª Re s > −1

2 s +3 = , s +1 s +1

© ª 1 L u (t ) = , s

© ª −1 < Re s < 3

all s © ª Re s > −1 © ª Re s > −1 © ª Re s > 0

1

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

214

1 , s +1 Using the linearity of the Laplace transform we obtain © ª L e −t u (t ) =

X (s) =

e.

© ª Re s > −1

1 1 1 , − = s s + 1 s (s + 1)

© ª L cos (2t ) u (t ) =

s s2 + 4

,

© ª Re s > 0 © ª Re s > 0

3 , s2 + 9 Using the linearity of the Laplace transform we obtain

© ª Re s > 0

s 3 + 6 s 2 + 9 s + 24 ¢¡ ¢ , X (s) = ¡ 2 s + 4 s2 + 9

© ª Re s > 0

© ª L sin (3t ) u (t ) =

f.

The signal x (t ) can be written as x (t ) =e −2t

·

¸ 1 j 3t 1 − j 3t e + e u (t ) 2 2

1 1 = e (−2+ j 3)t u (t ) + e (−2− j 3)t u (t ) 2 2 n o © ª 1 , Re s > −2 L e (−2+ j 3)t u (t ) = s +2− j3 n o © ª 1 L e (−2+ j 3)t u (t ) = , Re s > −2 s +2+ j3

Using the linearity of the Laplace transform we obtain X (s) =

1/2 1/2 s +2 + = , s + 2 − j 3 s + 2 + j 3 s 2 + 4 s + 13

© ª Re s > −2

7.10. a.

© ª 1 , Re s > −2 s +2 Using the time shifting property of the Laplace transform we get © ª L e −2t u (t ) =

© ª e −s X (s) = L e −2(t −1) u (t − 1) = , s +2

b.

© ª Re s > −2

Scaling the transform pair found in part (a) we obtain © ª © ª e −(s+2) , L e −2t u (t − 1) = e −2 L e −2(t −1) u (t − 1) = s +2

© ª Re s > −2

215

c.

Starting with the transform pair © ª L e −2t u (t ) =

1 , s +2

© ª Re s > −2

and using the time scaling property of the Laplace transform we get © ª L e 2t u (−t ) =

−1 1 − , −s + 2 s − 2

© ª Re s < 2

Finally, using the time shifting property on this result leads to © ª −e s X (s) = L e 2(t +1) u (−t − 1) = , s −2

d.

Using the transform pair found in part (c) with the linearity property of the Laplace transform we

obtain

e.

© ª Re s < 2

© ª © ª −e (s−2) , X (s) = L e 2t u (−t − 1) = e −2 L e 2(t +1) u (−t − 1) = s −2

From earlier parts of the problem we have © ª −1 L e 2t u (−t ) = , s −2

© ª Re s < 2

Applying the time shifting property of the Laplace transform we get ª −e −s © , L e 2(t −1) u (−t + 1) = s −2

and therefore

© ª Re s < 2

© ª −e −s e −2 L e 2t u (−t + 1) = , s −2

© ª Re s < 2

© ª −e −(s−2) , X (s) = L e 2t u (−t + 1) = s −2

© ª Re s < 2

7.11. a. Using the appropriate trigonometric identity, x (t ) is x (t ) = cos (3t ) cos (π/6) u (t ) − sin (3t ) sin (π/6) u (t ) Transforms of individual terms are s

© ª L cos (3t ) u (t ) =

s2 + 9

© ª L sin (3t ) u (t ) =

s2 + 9

Using the linearity of the Laplace transform we get X (s) =

3

,

© ª Re s > 0

,

© ª Re s > 0

cos (π/6) s 3 sin (π/6) 0.866 s − 1.5 − = s2 + 9 s2 + 9 s2 + 9

© ª Re s < 2

216

b.

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS Using Euler’s formula, x (t ) is x (t ) =

1 1 j π/6 j 3t e e u (t ) + e − j π/6 e − j 3t u (t ) 2 2

Transforms of individual terms are n o L e j 3t u (t ) =

1 , s − j3

n o L e − j 3t u (t ) =

1 , s + j3

Using the linearity of the Laplace transform we get

© ª Re s > 0 © ª Re s > 0

1 j π/6 11 e − j π/6 e X (s) = 2 +2 s − j3 s + j3 ¡ ¢ ¡ ¢ 1 j π/6 s + j 3 + 12 e − j π/6 s − j 3 2e = s2 + 9

=

0.866 s − 1.5 , s2 + 9

© ª Re s > 0

© ª 1 L u (t ) = , s

© ª Re s > 0

=

7.12. a. Using the relationships and

cos (π/6) s − 3 sin (π/6) s2 + 9

© ª © ª e −s , Re s > 0 L u (t − 1) = s the transform is found through the use of the linearity property as

X (s) =

1 − e −s , s

© ª Re s > −∞

© ª The ROC is the entire s-plane with the exception of Re s → −∞ since the pole at s = 0 is canceled when ¡ ¢ the two terms are added. This is evident from the fact that X 0 + j 0 = 1 and that x (t ) is a finite length signal. x (t) 1 t

1

b.

Using the relationships © ª 1 L u (t ) = , s

© ª Re s > 0

217 © ª e −s , L u (t − 1) = s

and

© ª Re s > 0

© ª e −2s © ª L u (t − 2) = , Re s > 0 s the transform is found through the use of the linearity property as © ª 1 − 2 e −s + e −2s , Re s > −∞ s © ª The ROC is the entire s-plane with the exception of Re s → −∞ since the pole at s = 0 is canceled when ¡ ¢ 1 the two terms are added. This is evident from the fact that X 0 + j 0 = 0 and that x (t ) is a finite length signal.

X (s) =

x (t) 1 2 1

c. and

t

Using the relationships © ª 1 L u (t ) = , s © ª e −s L u (t − 1) = , s

© ª Re s > 0 © ª Re s > 0

the transform is found through the use of the linearity property as X (s) =

1 − 2 e −s , s

© ª Re s > 0

The ROC is the same as the ROCs of the two individual transforms that were added. Note that the pole at s = 0 is not canceled in this case (the numerator of X (s) is nonzero for s = 0), and the signal x (t ) is of infinite length. x (t) 1

1

t

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

218

7.13. a. The signal x (t ) can be written as x (t ) = u (t ) − 2 u (t − 1) + u (t − 2) which leads to the transform X (s) =

b.

¢ 1¡ 1 − 2 e −s + e −2s , s

© ª Re s > −∞

An expression for x (t ) using unit step functions is x (t ) = u (t ) − 1.5 u (t − 1) + u (t − 2) − 0.5 u (t − 3)

It leads to the transform X (s) =

¢ 1¡ 1 − 1.5 e −s + e −2s − 0.5 e −3s , s

© ª Re s > −∞

7.14. The Laplace transform of the unit ramp function r (t ) = t u (t ) was found in Example 7-18 as © ª 1 R (s) = L r (t ) = 2 , s

© ª Re s > 0

© ª 1 U (s) = L u (t ) = , s

© ª Re s > 0

The transform of the unit step signal is

Writing a signal using these two functions allows computation of the Laplace transform in terms of the corresponding transforms.

a. Signal: Transform:

x (t ) = r (t ) − r (t − 1) − r (t − 2) + r (t − 3) X (s) =

¢ 1 ¡ 1 − e −s − e −2s + e −3s 2 s

b. Signal: Transform:

x (t ) = r (t ) − u (t − 1) − 2 r (t − 1) + u (t − 2) + r (t − 2) X (s) =

¢ ¢ 1¡ 1 ¡ − e −s + e −2s 1 − 2 e −s + e −2s + 2 s s

c. Signal: Transform:

x (t ) = r (t ) − 2 r (t − 1) + r (t − 2) X (s) =

¢ 1 ¡ 1 − 2 e −s + e −2s 2 s

219

7.15. a. Using Euler’s formula the signal x (t ) can be written as £ ¤ x (t ) = sin (πt ) u (t ) − u (t − 2) ´£ ¤ 1 ³ j πt e − e − j πt u (t ) − u (t − 2) = 2j

=

1 j πt 1 − j πt 1 j πt 1 − j πt e u (t ) − e u (t − 2) − e u (t ) e u (t − 2) 2j 2j 2j 2j

The transforms needed are

n o L e j πt u (t ) =

1 s − jπ

n o n o e −2s L e j πt u (t − 2) = e j 2π L e j π(t −2) u (t − 2) = s − jπ n o 1 L e − j πt u (t ) = s + jπ n o n o e −2s L e − j πt u (t − 2) = e − j 2π L e − j π(t −2) u (t − 2) = s + jπ

In the relationships above we have recognized that e ± j 2π = 0. The transform X (s) is ¡ ¢ · ¸ π 1 − e −2s 1 1 − e −2s 1 − e −2s X (s) = = − 2j s − jπ s + jπ s 2 + π2

b.

Let us write x (t ) as x (t ) = sin (πt ) u (t ) − sin (π (t − 2)) u (t − 2)

The transforms needed are

and

© ª L sin (πt ) u (t ) =

π s 2 + π2

© ª e −2s π L sin (π (t − 2)) u (t − 2) = 2 s + π2 Using the linearity of the Laplace transform we obtain ¢ ¡ π 1 − e −2s π e −2s π X (s) = 2 − = s + π2 s 2 + π2 s 2 + π2

7.16. a. Given that

1 s −a through the use of the time differentiation property we get © ª L e at u (t ) =

© ª L t e at u (t ) =

1 (s − a)2

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

220

Setting a = −2 yields L

b.

−1

½

1 (s + 2)2

¾

= t e −2t u (t )

From the transform table in Appendix B.3 © ª L e at cos (ω0 t ) u (t ) =

Setting a = −3 and ω0 =

p 2 we have L

c.

−1

½

s +3

(s + 3)2 + 2

¾

s −a

(s − a)2 + ω20

= e −3t cos

³p ´ 2t u (t )

From the transform table in Appendix B.3 © ª L e at sin (ω0 t ) u (t ) =

Setting a = −1 and ω0 =

p 6 we have

n

L e and L −1

½

−t

ω0 (s − a)2 + ω20

³p ´ o sin 6t u (t ) =

p

6

(s + 1)2 + 6

¾

³p ´ 1 −t = e sin 6t u (t ) p (s + 1)2 + 6 6

1

7.17. a. The transform X (s) has the partial fraction expansion X (s) = with residues

and

The partial fraction expansion for X (s) is

k2 k1 + s +1 s +2

¯ 1 ¯¯ =1 k1 = s + 2 ¯s=−1

¯ 1 ¯¯ k2 = = −1 s + 1 ¯s=−2

X (s) =

1 1 − s +1 s +2

Since the ROC is Re {s} > −1 both terms in the partial fraction expansion correspond to causal signal components. Therefore x (t ) = e −t u (t ) − e −2t u (t )

221

b.

The partial fraction expansion is the same as in part (a). Since the ROC is Re {s} < −2 both terms in the partial fraction expansion correspond to anti-causal signal components. Therefore x (t ) = −e −t u (−t ) + e −2t u (−t )

c.

The partial fraction expansion is the same as in part (a). Since the ROC is −2 < Re {s} < −1 the first term in the partial fraction expansion corresponds to an anti-causal signal component, and teh second term corresponds to a causal signal component, that is X (s) =

1 1 − s + 1 s + | {z } | {z2}

anti-causal

Therefore

causal

x (t ) = −e −t u (−t ) − e −2t u (t )

d.

The transform X (s) has the partial fraction expansion X (s) =

with residues

and

k1 k2 k3 + + s +1 s +2 s +3

¯ (s − 1) (s − 2) ¯¯ =3 k1 = (s + 2) (s + 3) ¯s=−1 ¯ (s − 1) (s − 2) ¯¯ k2 = = −12 (s + 1) (s + 3) ¯s=−2 ¯ (s − 1) (s − 2) ¯¯ = 10 k3 = (s + 1) (s + 2) ¯s=−3

The partial fraction expansion for X (s) is

X (s) =

3 12 10 − + s +1 s +2 s +3

Using the causal signal component for each term we have x (t ) = 3 e −t u (t ) − 12 e −2t u (t ) + 10 e −3t u (t )

e.

The transform X (s) has the partial fraction expansion X (s) =

with residues

k2 k3 k1 + + s −1 s +2 s +3

¯ 1 (s + 1) (s − 2) ¯¯ =− ¯ 6 (s + 2) (s + 3) s=1 ¯ 4 (s + 1) (s − 2) ¯¯ k2 = =− ¯ 3 (s − 1) (s + 3)

k1 =

s=−2

and

¯ 5 (s + 1) (s − 2) ¯¯ = k3 = ¯ (s − 1) (s + 2) s=−3 2

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

222

The partial fraction expansion for X (s) is 1/6 4/3 5/2 X (s) = − − + | s{z− 1} | s + 2{z s + 3} anti-causal

causal

Since the Fourier transform of x (t ) exists, the ROC of X (s) must include the j ω-axis of the s-plane. Therefore, the term with the pole at s = 1 must correspond to an anti-causal signal component, and the other two terms must correspond to causal signal components. x (t ) =

1 t 4 5 e u (−t ) − e −2t u (t ) + e −3t u (t ) 6 3 2

7.18. a. Since the numerator order is the same as the denominator order, an constant term must be extracted from X (s) to write it in the form X (s) = 2 −

1 s +2

The ROC indicates a causal signal. Therefore x (t ) = 2 δ (t ) − e −2t u (t )

b.

The transform X (s) has the partial fraction expansion X (s) =

with residues

and

k2 k1 + s +2 s +3

¯ s (s + 1) ¯¯ =2 k1 = s + 3 ¯s=−2

¯ s (s + 1) ¯¯ k2 = = −6 s + 2 ¯s=−3

The partial fraction expansion for X (s) is

X (s) =

6 2 − s +2 s +3

The ROC for the transform is Re {s} < −3, and it indicates an anti-causal signal. Therefore x (t ) = −2 e −2t u (−t ) + 6 e −3t u (−t )

c.

The transform X (s) has the partial fraction expansion X (s) =

with residues k1 =

k2 k1 + s + j2 s − j2

¯ 5 1 s + 5 ¯¯ = +j s − j 2 ¯s=− j 2 2 4

223 and k 2 = k 1∗ =

1 5 −j 2 4

The partial fraction expansion for X (s) is ¶ ¶ µ µ 1 1 5 5 1 1 +j + −j X (s) = 2 4 s + j2 2 4 s − j2 The ROC indicates a causal signal. Therefore ¶ µ ¶ µ 1 5 − j 2t 5 j 2t 1 e u (t ) + − j e u (t ) x (t ) = + j 2 4 2 4 ´ ´ 1³ 5 ³ − j 2t = e − j 2t + e j 2t u (t ) + j e − e j 2t u (t ) 2 4 = cos (2t ) u (t ) +

d.

Factored form of X (s) is

and leads to the partial fraction form

X (s) = ¡ X (s) =

The residues are

and

5 sin (2t ) u (t ) 2

s +6 ¢¡ ¢ s +1+ j2 s +1− j2

k1 k2 + s +1+ j2 s +1− j2

¯ s + 6 ¯¯ 5 1 k1 = = +j ¯ s + 1 − j 2 s=−1− j 2 2 4

k 2 = k 1∗ =

5 1 −j 2 4

The ROC indicates a causal signal. Therefore µ ¶ ¶ µ 1 1 5 (−1− j 2)t 5 (−1+ j 2)t x (t ) = + j e e u (t ) + − j u (t ) 2 4 2 4 ³ ´ ³ ´ 5 1 = e −t e − j 2t + e j 2t u (t ) + j e −t e − j 2t − e j 2t u (t ) 2 4 =e −t cos (2t ) u (t ) +

e.

5 −t e sin (2t ) u (t ) 2

Partial fraction form of X (s) is X (s) =

k 1,1 s +1

The residues are k 1,2 =

+

k 1,2 2

(s + 1)

+

k2 s +2

¯ s (s − 1) ¯¯ =2 s + 2 ¯s=−1

224

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS ¯ s (s − 1) ¯¯ =6 k2 = ¯ (s + 1)2 s=−2

and

¯ · ¸¯ d s (s − 1) ¯¯ s 2 + 4s − 2 ¯¯ k 1,1 = = = −5 ¯ ds s + 2 ¯s=−1 (s + 2)2 s=−1

The partial fraction expansion for X (s) is

X (s) = −

2 6 5 + + 2 s + 1 (s + 1) s +2

The ROC indicates a causal signal. Therefore x (t ) = −5 e −t u (t ) + 2 t e −t u (t ) + 6 e −2t u (t )

7.19. a. Pole-zero diagram is shown below: jω

−1

1

|

{z

σ

}

−1 < Re {s} < 1

b.

Possible choices for the ROC are: 1. 2. 3. 4.

c.

© ª Re s < −2 © ª −2 < Re s < −1 © ª −1 < Re s < 1 © ª Re s > 1

The partial fraction form of X (s) is X (s) =

The residues are

k1 k2 k3 + + s −1 s +1 s +2

¯ 1 s (s − 3) ¯¯ =− k1 = 3 (s + 1) (s + 2) ¯s=1

225 k2 = and k3 = The partial fraction expansion for X (s) is

¯ s (s − 3) ¯¯ = −2 (s − 1) (s + 2) ¯s=−1

¯ 10 s (s − 3) ¯¯ = ¯ 3 (s − 1) (s + 1) s=−2

X (s) = −

1/3 2 10/3 − + s −1 s +1 s +2

The inverse transform is shown below for each choice of the ROC: © ª 1 10 −2t Re s < −2 : x (t ) = e t u (−t ) + 2 e −t u (−t ) − e u (−t ) 3 3

© ª 10 −2t 1 e u (t ) −2 < Re s < −1 : x (t ) = e t u (−t ) + 2 e −t u (−t ) + 3 3 © ª 10 −2t 1 e u (t ) −1 < Re s < 1 : x (t ) = e t u (−t ) − 2 e −t u (t ) + 3 3 © ª 1 10 −2t Re s > 1 : x (t ) = − e t u (t ) − 2 e −t u (t ) + e u (t ) 3 3

The inverse transform x (t ) is square integrable if the ROC includes the j ω-axis. The only choice for ROC that satisfies this condition is −1 < Re {s} < 1.

7.20. a. The transform can be written as X (s) = −

2 1 + s +2 s +3

Possible choices for the ROC and the corresponding signals x (t ) are as follows: © ª Re s < −3 : x (t ) = e −2t u (−t ) − 2 e −3t u (−t )

© ª −3 < Re s < −2 : x (t ) = e −2t u (−t ) + 2 e −3t u (t )

© ª Re s > −2 : x (t ) = − e −2t u (t ) + 2 e −3t u (t )

b.

The transform can be written as X (s) =

1 2 2 − − 2 s + 2 (s + 2) s +3

Possible choices for the ROC and the corresponding signals x (t ) are as follows: © ª Re s < −3 : x (t ) = −2 e −2t u (−t ) + t e −2t u (−t ) + 2 e −3t u (−t )

© ª −3 < Re s < −2 : x (t ) = −2 e −2t u (−t ) + t e −2t u (−t ) − 2 e −3t u (t ) © ª Re s > −2 : x (t ) = 2 e −2t u (t ) − t e −2t u (t ) − 2 e −3t u (t )

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

226

c.

The transform can be written as X (s) = 1 −

12 7 + s + 3 (s + 3)2

Possible choices for the ROC and the corresponding signals x (t ) are as follows: © ª Re s < −3 : x (t ) = δ (t ) + 7 e −3t u (−t ) − 12 t e −3t u (−t ) © ª Re s > −3 : x (t ) = δ (t ) − 7 e −3t u (t ) + 12 t e −3t u (t )

d.

The transform can be written as X (s) =

1 2

− j 13

s +2+ j3

+

1 2

+ j 31

s +2− j3

Possible choices for the ROC and the corresponding signals x (t ) are as follows: ¶ ¶ µ µ © ª 1 1 1 (−2− j 3)t 1 (−2+ j 3)t e e Re s < −2 : x (t ) = − − j u (−t ) − + j u (−t ) 2 3 2 3 2 = −e −2t cos (3t ) u (−t ) + e −2t sin (3t ) u (−t ) 3 µ ¶ ¶ µ © ª 1 1 (−2− j 3)t 1 (−2+ j 3)t 1 e e u (t ) + + j u (t ) −j Re s > −2 : x (t ) = 2 3 2 3 2 = e −2t cos (3t ) u (t ) − e −2t sin (3t ) u (t ) 3

7.21. a. Let the transform X 1 (s) be defined as X 1 (s) =

1 s +1

so that x 1 (t ) = e −t u (t ) Since we conclude that

¡ ¢ X (s) = 1 − e −s X 1 (s)

x (t ) =x 1 (t ) − x 1 (t − 1) =e −t u (t ) − e −(t −1) u (t − 1)

b.

Let the transform X 1 (s) be defined as X 1 (s) =

s 1 = 1− s +1 s +1

so that x 1 (t ) = δ (t ) − e −t u (t )

227 Since we conclude that

¡ ¢ X (s) = 1 − e −s X 1 (s)

x (t ) =x 1 (t ) − x 1 (t − 1) =δ (t ) − e −t u (t ) − δ (t − 1) + e −(t −1) u (t − 1)

c.

Let the transform X 1 (s) be defined as X 1 (s) =

1 1 1 = − s (s + 1) s s + 1

so that x 1 (t ) = u (t ) − e −t u (t ) Since we conclude that

¡ ¢ X (s) = 1 − e −s X 1 (s)

x (t ) =x 1 (t ) − x 1 (t − 1) =u (t ) − e −t u (t ) − u (t − 1) + e −(t −1) u (t − 1)

c.

Let the transform X 1 (s) be defined as X 1 (s) =

1 s

so that x 1 (t ) = u (t ) Since we conclude that

¢ ¡ X (s) = 1 − e −s + e −2s − e −3s X 1 (s)

x (t ) = u (t ) − u (t − 1) + u (t − 2) − u (t − 3)

7.22.

Taking the Laplace transform of each side yields the following:

a. 3s Y (s) + 2 Y (s) = 7 X (s)

⇒

H (s) =

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

b. s 2 Y (s) + 4s Y (s) + 3 Y (s) = s X (s) + X (s)

⇒

H (s) =

s +1 Y (s) = 2 X (s) s + 4s + 3

c. s 2 Y (s) + 4 Y (s) = s 2 X (s) + s X (s) + 3 X (s)

⇒

H (s) =

Y (s) s 2 + s + 3 = X (s) s2 + 4

228

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

7.23. a. H (s) =

Y (s) 1 = s + 4 X (s)

⇒

s Y (s) + 4 Y (s) = X (s)

d y (t ) + 4 y (t ) = x (t ) dt

b. H (s) =

s Y (s) = s + 4 X (s)

⇒

s Y (s) + 4 Y (s) = X (s)

d y (t ) + 4 y (t ) = x (t ) dt

c. H (s) =

s +1

s2 + 5 s + 6

=

Y (s) X (s)

s 2 Y (s) + 5s Y (s) + 6 Y (s) = s X (s) + X (s) d 2 y (t ) d x (t ) d y (t ) + 6 y (t ) = + x (t ) +5 2 dt dt dt

d. H (s) =

s +1

s2 + 5 s + 6

=

Y (s) X (s)

s 2 Y (s) + 5s Y (s) + 6 Y (s) = s X (s) + X (s) d 2 y (t ) d x (t ) d y (t ) + 6 y (t ) = + x (t ) +5 2 dt dt dt

e. H (s) =

Y (s) s2 − s = s 3 + 5 s 2 + 8 s + 6 X (s)

s 3 Y (s) + 5s 2 Y (s) + 8s Y (s) + 6 Y (s) = s 2 X (s) − s X (s) d 3 y (t ) d 2 y (t ) d y (t ) d 2 x (t ) d x (t ) + 5 + 8 + 6 y = − (t ) dt3 dt2 dt dt2 dt

f. H (s) =

s2 + s − 2 Y (s) = 3 2 s + 7 s + 19 s + 13 X (s)

s 3 Y (s) + 7s 2 Y (s) + 19s Y (s) + 13 Y (s) = s 2 X (s) + s X (s) − 2 X (s) d 2 y (t ) d y (t ) d 2 x (t ) d x (t ) d 3 y (t ) + 7 + 19 + 13 y = + − 2 x (t ) (t ) dt3 dt2 dt dt2 dt

229

7.24. Let the system function be H (s). For a unit-step input signal we have X (s) = 1/s, and the Laplace transform of the unit-step response of the system is Yu (s) =

H (s) s

The solution method is based on determining H (s) from this relationship, and then finding a differential equation from the knowledge of H (s).

a. Yu (s) =

1 s +1

⇒

H (s) = s Yu (s) =

s s +1

(s + 1) Y (s) = s X (s) d x (t ) d y (t ) + y (t ) = dt dt

b. Yu (s) =

1 1 1 − = s s + 1 s (s + 1)

H (s) = s Yu (s) =

⇒

1 s +1

(s + 1) Y (s) = X (s) d y (t ) + y (t ) = x (t ) dt

c. Yu (s) =

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

H (s) = s Yu (s) =

⇒

s (s + 1) (s + 2)

¢ s 2 + 3 s + 2 Y (s) = s X (s)

d 2 y (t ) d y (t ) d x (t ) +3 + 2 y (t ) = 2 dt dt dt

d. Yu (s) =

1 1 2 2s 2 + 3s + 2 − + = 3 s s + 1 s + 2 s + 3s 2 + 2s ¡

⇒

H (s) = s Yu (s) =

2s 2 + 3s + 2 s 2 + 3s + 2

¢ ¡ ¢ s 2 + 3s + 2 Y (s) = 2s 2 + 3s + 2 X (s)

d y (t ) d 2 x (t ) d x (t ) d 2 y (t ) + 3 + 2 y = 2 +3 + 2 x (t ) (t ) dt2 dt dt2 dt

e. Yu (s) =

1 0.3 (s + 1) 0.7 s 2 + 1.7 s + 5 ¡ ¢ − = s (s + 1)2 + 4 s s2 + 2 s + 5

⇒

H (s) = s Yu (s) =

0.7 s 2 + 1.7 s + 5 s2 + 2 s + 5

230

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

¡

¢ ¡ ¢ s 2 + 2 s + 5 Y (s) = 0.7 s 2 + 1.7 s + 5 X (s)

d y (t ) d 2 x (t ) d x (t ) d 2 y (t ) + 2 + 5 y = 0.7 + 1.7 + 5 x (t ) (t ) 2 2 dt dt dt dt

7.25. Let the system function be H (s). For a unit-ramp input signal we have X (s) = 1/s 2 , and the Laplace transform of the unit-step response of the system is Yu (s) =

H (s) s2

The solution method is based on determining H (s) from this relationship, and then finding a differential equation from the knowledge of H (s).

a. Yr (s) =

1 1 1 − = s s + 1 s (s + 1)

⇒

H (s) = s 2 Yr (s) =

s Y (s) = s + 1 X (s)

(s + 1) Y (s) = s X (s) d x (t ) d y (t ) + y (t ) = dt dt

b. Yr (s) =

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

H (s) = s 2 Yr (s) = ¡

−s 4 + s 3 + 4 s 2 Y (s) = 3 2 s + 6 s + 11 s + 6 X (s)

¢ ¢ ¡ s 3 + 6 s 2 + 11 s + 6 Y (s) = −s 4 + s 3 + 4 s 2 X (s)

d 3 y (t ) d 4 x (t ) d 3 x (t ) d 2 y (t ) d y (t ) d 2 x (t ) + 6 y = − + 6 + 11 + + 4 (t ) dt3 dt2 dt dt4 dt3 dt2

c. Yr (s) =

0.7 s 2 + 1.7 s + 5 1 0.3 (s + 1) ¡ ¢ − = s (s + 1)2 + 4 s s2 + 2 s + 5

H (s) = s 2 Yr (s) = ¡

0.7 s 3 + 1.7 s 2 + 5 s Y (s) = s2 + 2 s + 5 X (s)

¢ ¡ ¢ s 2 + 2 s + 5 Y (s) = 0.7 s 3 + 1.7 s 2 + 5 s X (s)

231 d y (t ) d 3 x (t ) d 2 x (t ) d x (t ) d 2 y (t ) + 2 + 5 y = 0.7 + 1.7 +5 (t ) dt2 dt dt3 dt2 dt

7.26. The response of the CTLTI system with system function H (s) to the signal x (t ) = e s0 t is © ª Sys e s0 t = e s0 t H (s 0 )

a. s 0 = −0.5 ,

H (−0.5) =3.3333 + j 0 =3.3333 e j 0

y (t ) = 3.3333 e −0.5t

b. ¡ ¢ H −0.5 + j 2 = −0.0047 − j 0.6212

s 0 = −0.5 + j 2 ,

=0.6212 e − j 1.5783

y (t ) = 0.6212 e −0.5t − j 1.5783

c. s0 = j 3 ,

¡ ¢ H j 3 =0.0462 − j 0.3692

=0.3721 e − j 1.4464

y (t ) = 0.3721 e j (3t −1.4464)

d.

The signal x (t ) is in the form x (t ) = e s0 t + e s1 t

with s 0 = − j 3 and s 1 = j 3. s0 = − j 3 ,

¡ ¢ H − j 3 =0.0462 + j 0.3692

=0.3721 e j 1.4464

s1 = j 3 ,

¡ ¢ H j 3 =0.0462 − j 0.3692

=0.3721 e − j 1.4464

y (t ) =0.3721 e j (−3t +1.4464) + 0.3721 e j (3t −1.4464) =0.7442 cos (3t − 1.4464)

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

232

e.

The signal x (t ) is in the form x (t ) = e s0 t + e s1 t

with s 0 = j 2 and s 1 = j 3. ¡ ¢ H j 2 =0.1500 − j 0.5500

s0 = j 2 ,

=0.5701 e − j 1.3045

¡ ¢ H j 3 =0.0462 − j 0.3692

s1 = j 3 ,

=0.3721 e − j 1.4464

y (t ) = 0.5701 e j (2t −1.3045) + 0.3721 e j (3t −1.4464)

7.27. Using the Laplace transform definition, H (s 0 ) is found as H (s 0 ) =

ˆ

∞

h (t ) e −s0 t d t

−∞

Conjugating both sides yields H ∗ (s 0 ) = =

·ˆ

∞

h (t ) e −s0 t d t

−∞

ˆ

∞

¸∗

∗

h ∗ (t ) e −s0 t d t

−∞

Since h (t ) is real-valued we have h ∗ (t ) = h (t ) and H ∗ (s 0 ) = Therefore

ˆ

∞ −∞

¡ ¢ ∗ h (t ) e −s0 t d t = H s 0∗

¡ ¢ H s 0∗ = H ∗ (s 0 ) = H0 e − j Θ0

7.28. a. s0 = j 2 ,

¡ ¢ H j 2 = − 0.1379 + 0.3448

=0.3714 e j 1.9513

y (t ) = 0.3714 cos (2t + 1.9513)

b. s 0 = −0.5 + j 2 ,

¡ ¢ H −0.5 + j 2 = − 0.3479 + 0.2879

=0.4516 e j 2.4503

233 y (t ) = 0.4516 e −0.5t cos (2t + 2.4503)

c. ¡ ¢ H −1 + j 2 = − 0.6154 + j 0.0769

s 0 = −1 + j 2 ,

=0.6202 e j 3.0172

y (t ) = 0.6202 e −t sin (2t + 3.0172)

7.29. The system function is

a.

© ª H (s) = L h (t ) =

1 s +1

Laplace transform of the input signal is

© ª X (s) = L cos (2t ) u (t ) =

s s2 + 4

Laplace transform of the output signal is found as Y (s) = H (s) X (s) =

(s + 1)

Partial fraction expansion of Y (s) is Y (s) = with the residues

s ¡

¢=

s2 + 4

s ¡

¢¡ ¢ (s + 1) s + j 2 s − j 2

k2 k3 k1 + + s +1 s + j2 s − j2

s ¯¯ = −0.2 ¯ s 2 + 4 s=−1 ¯ ¯ s ¡ ¢ ¯¯ = 0.1 + j 0.2 + 1) s − j 2

k1 = k2 = and

(s

s=− j 2

k 3 = k 2∗ = 0.1 − j 0.2

The output signal is found as ¡ ¢ ¡ ¢ y (t ) = − 0.2 e −t u (t ) + 0.1 + j 0.2 e − j 2t u (t ) + 0.1 − j 0.2 e j 2t u (t ) ³ ´ ³ ´ = − 0.2 e −t u (t ) + 0.1 e − j 2t + e j 2t u (t ) + j 0.2 e − j 2t − e j 2t u (t )

= −0.2 e −t u (t ) + 0.2 cos (2t ) u (t ) + 0.4 sin (2t ) u (t ) {z } | {z } | transient

steady-state

The transient component of the response is

y t (t ) = −0.2 e −t u (t )

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

234

and the steady-state component is y ss (t ) = 0.2 cos (2t ) + 0.4 sin (2t )

b.

Laplace transform of the input signal is © ª X (s) = L sin (3t ) u (t ) =

3 s2 + 9

Laplace transform of the output signal is found as Y (s) = H (s) X (s) = Partial fraction expansion of Y (s) is

3 3 ¡ ¢= ¡ ¢¡ ¢ 2 (s + 1) s + 9 (s + 1) s + j 3 s − j 3

Y (s) = with the residues

k2 k3 k1 + + s +1 s + j3 s − j3

¯ ¯ ¢ ¯¯ = 0.3 s 2 + 9 s=−1 ¯ ¯ 3 ¡ ¢ ¯¯ k2 = = −0.15 + j 0.05 (s + 1) s − j 3 s=− j 3

k1 = ¡

and

3

k 3 = k 2∗ = −0.15 − j 0.05

The output signal is found as ¡ ¢ ¡ ¢ y (t ) =0.3 e −t u (t ) + −0.15 + j 0.05 e − j 3t u (t ) + −0.15 − j 0.05 e j 3t u (t ) ³ ´ ³ ´ =0.3 e −t u (t ) − 0.15 e − j 3t + e + j 3t u (t ) + j 0.05 e − j 3t − e j 3t u (t )

= 0.3 e −t u (t ) − 0.3 cos (3t ) u (t ) + 0.1 sin (3t ) u (t ) {z } | {z } | transient

steady-state

The transient component of the response is

y t (t ) = 0.3 e −t u (t ) and the steady-state component is y ss (t ) = −0.3 cos (3t ) + 0.1 sin (3t )

7.30. a. H (s) = 1 −

2 s +3

⇒

h (t ) = δ (t ) − 2 e −3t u (t )

235

b. H (s) = 1 + Let

−3s − 1

s 2 + 3s + 2

= 1+

−3s − 1 (s + 1) (s + 2)

H1 (s) =

−3s − 1 (s + 1) (s + 2)

H1 (s) =

k2 k1 + s +1 s +2

so that

The residues are k1 = and k2 = The system function is

¯ −3s − 1 ¯¯ =2 s + 2 ¯s=−1

¯ −3s − 1 ¯¯ = −5 s + 1 ¯s=−2

H (s) = 1 + H1 (s) = 1 + and the impulse response is

5 2 − s +1 s +2

h (t ) = δ (t ) + 2 e −t u (t ) − 5 e −2t u (t )

c. H (s) = = The residues are

s2 − 1 ¢¡ ¢ (s + 2) s + 1 + j s + 1 − j ¡

k2 k3 k1 + + s +2 s +1+ j s +1− j

¯ s 2 − 1 ¯¯ k1 = 2 = 1.5 s + s + 2 ¯s=−2 ¯ ¯ s2 − 1 ¡ ¢ ¯¯ k2 = = −0.25 − j 0.75 (s + 2) s + 1 − j s=−1− j

and

k 3 = k 2∗ = −0.25 + j 0.75

The impulse response is ¡ ¢ ¡ ¢ h (t ) =1.5 e −2t u (t ) + −0.25 − j 0.75 e (−1− j )t u (t ) + −0.25 + j 0.75 e (−1+ j )t u (t ) ´ ³ ´ ³ =1.5 e −2t u (t ) − 0.25 e −t e − j t + e j t u (t ) − j 0.75 e −t e − j t − e j t u (t )

=1.5 e −2t u (t ) − 0.5 e −t cos (t ) u (t ) − 1.5 e −t sin (t ) u (t )

d. H (s) =

−4

2

(s + 1)

+

1 s +1

236

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

The corresponding impulse response is h (t ) = −4t e −t u (t ) + e −t u (t ) = (−4t + 1) e −t u (t )

7.31. The system function is H (s) = The input signal can be written as

1 s +1

1 j 2t 1 − j 2t e + e 2 2 Let s 0 = j 2. The system function evaluated at s = s 0 = j 2 is x (t ) =

H (s 0 ) =

1 = 0.2 − j 0.4 j2+1

The system function evaluated at s = s 0∗ = − j 2 is ¡ ¢ H s 0∗ =

The response of the system is

1 = 0.2 + j 0.4 −j2+1

© ª y (t ) = Sys cos (2t ) n o 1 n o 1 = Sys e j 2t + Sys e − j 2t 2 2 ¢ j 2t 1 ¡ ¢ 1¡ = 0.2 − j 0.4 e + 0.2 + j 0.4 e − j 2t 2 2 ³ ´ ´ ¢³ 1 1¡ = (0.2) e j 2t + e − j 2t − j 0.4 e j 2t − e − j 2t 2 2

=0.2 cos (2t ) + 0.4 sin (2t )

The result found matches the steady-state component of the result found in part (a) of Problem 7.29.

7.32. The system function is found as © ª H (s) = L e −t u (t ) − e −t u (t − 2) © ª © ª = L e −t u (t ) − L e −t u (t − 2) ª © ª © = L e −t u (t ) − e −2 L e −(t −2) u (t − 2) £ ¤ = 1 − e −2 (s+1)

1 s +1

Laplace transform of the input signal is

X (s) =

s + 0.5 2

(s + 0.5) + 4

=

s + 0.5

s 2 + s + 4.25

237 and the Laplace transform of the output signal is Y (s) =H (s) X (s) ¤ £ = 1 − e −2 (s+1)

Let

Y1 (s) = so that

s + 0.5 ¡ ¢ (s + 1) s 2 + s + 4.25

s + 0.5 ¡ ¢ (s + 1) s 2 + s + 4.25

£ ¤ Y (s) = 1 − e −2 (s+1) Y1 (s)

The transform Y1 (s) can be expanded into partial fractions as Y1 (s) = with the residues

k2 k3 k1 + + s + 1 s + 0.5 + j 2 s + 0.5 − j 2

¯ s + 0.5 ¯¯ k1 = 2 = −0.1176 s + s + 4.25 ¯s=−1 ¯ ¯ s + 0.5 ¡ ¢ ¯¯ = 0.0588 + j 0.2353 k2 = (s + 1) s + 0.5 − j 2 s=−0.5− j 2

and

k 3 = k 2∗ = 0.0588 − j 0.2353

Using the residues, y 1 (t ) is found as ¡ ¢ ¡ ¢ y 1 (t ) = − 0.1176 e −t u (t ) + 0.0588 + j 0.2353 e (−0.5− j 2)t u (t ) + 0.0588 − j 0.2353 e (−0.5+ j 2)t u (t ) ³ ´ ³ ´ = − 0.1176 e −t u (t ) + 0.0588 e −0.5t e − j 2t + e j 2t u (t ) + j 0.2353 e −0.5t e − j 2t − e j 2t u (t ) = − 0.1176 e −t u (t ) + 0.1176 e −0.5t cos (2t ) u (t ) + 0.4706 e −0.5t sin (2t ) u (t )

The signal y (t ) is y (t ) =y 1 (t ) − e −2 y 1 (t − 2) £ ¤ = −0.1176 e −t + 0.1176 e −0.5t cos (2t ) + 0.4706 e −0.5t sin (2t ) u (t ) £ ¤ + 0.1176 e −t − 0.1176 e −0.5t −1 cos (2t − 4) − 0.4706 e −0.5t −1 sin (2t − 4) u (t − 2)

7.33. a. jω

The system function has a zero at s = −1 and poles at s = −2, −3.

σ −3

−2

−1

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

238

b. jω

H (s) =

¡

s−j

¢¡

s+j

¢

j1

(s + 2) (s + 3)

σ

The system function has zeros at s = ∓ j 1 and poles at s = −2, −3.

−3

−2 −j1

c. jω j3

H (s) =

(s − 1) (s + 1) ¢¡ ¢ (s + 2) s + 2 + j 3 s + 2 − j 3 ¡

σ

The system function has zeros at s = ∓1 and poles at s = −2 and s = −2 ± j 3.

−2

−1

1

−j3

7.34. a. H (s) = K

¡

s+j

¢¡

s−j

(s + 1) (s + 2)

µ ¶ 1 =1 H (0) = K 2

H (s) =

b. H (s) = K ¡

¢

=K

s2 + 1 s2 + 3 s + 2

⇒

K =2

¡ ¢ 2 s2 + 1

s2 + 3 s + 2

s 2 − 2.25 (s − 1.5) (s + 1.5) ¢¡ ¢ =K 2 s + 2 s + 3.25 s + 1 − j 1.5 s + 1 + j 1.5 µ

¶ −2.25 13 =1 ⇒ K =− 3.25 9 µ ¶ s 2 − 2.25 13 H (s) = − 9 s 2 + 2 s + 3.25

H (0) = K

239

c. H (s) = K

¡

s−j

¢¡

H (s) =

H (s) = K

¢

(s − 1) (s + 1)

H (0) = −K = 1

d.

s+j

=K

s2 + 1 s2 − 1

⇒ K = −1 ¡ ¢ − s2 + 1 s2 − 1

s − 1.5 s − 1.5 ¢¡ ¢ =K 3 2 s + 3.5 s + 6.25 s + 4.875 (s + 1.5) s + 1 − j 1.5 s + 1 + j 1.5 ¶ µ −1.5 =1 ⇒ K = −3.25 H (0) = K 4.875 −3.25 (s − 1.5) H (s) = 3 s + 3.5 s 2 + 6.25 s + 4.875 ¡

7.35. a.

replacemen

H (ω) =

jω−1 jω+2 H (ω)

|H (ω)| 6

π

Phase (rad)

Magnitude

1 0.8 0.6 0.4

π 2 0 −

0.2 0 −10

−5

0

5

π 2

−π −10

10

−5

ω (rad/s)

b.

0

5

10

5

10

ω (rad/s)

¡

replacemen

|H (ω)|

¢¡ ¢ jω− j1 jω+ j1 ¢¡ ¢ H (ω) = ¡ jω+2 jω+3 6

H (ω)

π

Phase (rad)

Magnitude

1 0.8 0.6 0.4 0.2 0 −10

−5

0

ω (rad/s)

5

10

π 2 0 −

π 2

−π −10

−5

0

ω (rad/s)

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

240

c.

¡

¢¡ ¢ jω−1 jω+1 ¢¡ ¢ H (ω) = ¡ jω+1+ j2 jω+1− j2 |H (ω)| 6

Phase (rad)

Magnitude

1.5

1

0.5

0 −10

−5

0

10

5

π 2 0 −

π 2

−π −10

−5

ω (rad/s)

3 2

jω

1 2 poles

0

System is stable. −1 −2 −3 −3

−2

−1

0

σ

1

0

ω (rad/s)

7.36. a.

−4

H (ω)

π

2

3

5

10

241

b. 3 2

jω

1 0

System is unstable. −1 −2 −3 −4

−3

−2

−1

0

1

2

3

σ

c. 3 2

jω

1 0

System is stable. −1 −2 −3 −4

−3

−2

−1

0

σ

1

2

3

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

242

d. 3 2

jω

1 0

System is unstable. −1 −2 −3 −4

−3

−2

−1

0

1

2

3

σ

7.37. For a system function to represent a system that is both causal and stable, the following conditions must be met: • The order of the numerator of the system function must not be greater than the order of the denominator. • All poles of the system function must be in the left half s-plane. We need Re {s} < 0 for all poles.

a.

The system function can be written as H (s) =

s −1 (s + 1) (s + 2)

Poles are at s = −1 and s = −2. This could be the system function of a causal and stable system.

b.

The system function can be written as H (s) = ¡

s (s + 1) ¢¡ ¢ s +1− j2 s +1+ j2

Poles of the system function are at s = −1 ± j 2. This could be the system function of a causal and stable system.

c.

The system function can be written as H (s) =

s2 + 1 s (s + 2) (s + 3)

Poles of the system function are at s = 0, s = −2 and s = −3. This could not be the system function of a causal and stable system since there is a pole on the j ω-axis.

243

d.

The system function can be written as H (s) = ¡

s +3 ¢¡ ¢ s −3− j1 s −3− j1

Poles of the system function are at s = 3 + j 1 and s = 3 − j 1. This could not be the system function of a causal and stable system since the poles are in the right half of the s-plane.

7.38. a. Taking the Laplace transform of both sides of the differential equation yields s 2 Y (s) = −2s Y (s) − a Y (s) + X (s) which can be rearranged as

¡

and the system function is

¢ s 2 + 2 s + a Y (s) = X (s)

H (s) =

1 Y (s) = 2 X (s) s + 2 s + a

b.

For the system to be causal and stable, the poles of the system function must be in the left half of the s-plane. The poles of H (s) are found as s2 + 2 s + a = 0

p s 1,2 = −1 ± 1 − a

⇒

If a > 1, the poles are complex-valued, and they are at s 1,2 = −1 ± j

p

a −1

If a < 1, both poles are real-valued. The system cannot be stable if at least one is in the right half s plane or on the j ω-axis. p −1+ 1−a ≥ 0 ⇒ a≤0 Therefore, for stability we need a > 0.

7.39. a.

Evaluating H (s) for s = j ω yields

The phase of the system function is

¯ ¯ H (ω) = H (s)¯

s= j ω

=

jω−2 jω+2

∡ H (ω) = tan(−ω/2) − tan(ω/2) = −2 tan(ω/2)

244

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS ∡H (ω) π

Phase (rad)

π 2 0

−

π 2

−π −10

−5

0

10

5

ω (rad/s)

b.

The transform of the input signal is X (s) =

1 s +1

and the transform of the corresponding output signal is Y (s) = H (s) X (s) =

s −2 (s + 1) (s + 2)

Using partial fraction expansion on Y (s) we get Y (s) = The residues are

k2 k1 + s +1 s +2

¯ s − 2 ¯¯ = −3 k1 = s + 2 ¯s=−1

and

and the response of the system is

¯ s − 2 ¯¯ k2 = =4 s + 1 ¯s=−2

y (t ) = −3 e −t u (t ) + 4 e −2t u (t )

c.

For the input signal we obtain 1 X (ω) = 1+ jω

⇒

and for the output signal we obtain −2 + j ω Y (ω) = 2 − ω2 + j 3ω

⇒

¯ ¯ ¯ X (ω)¯ =

r

1 1 + ω2

v u ¯ ¯ u 4 + ω2 ¯Y (ω)¯ = t ¡ ¢2 2 − ω2 + 9ω2

245 |X (ω)|

|Y (ω)|

1

1

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0 −10

0

−5

5

0 −10

10

0

−5

ω (rad/s)

5

ω (rad/s)

7.40. a. The frequency response of the system is computed by evaluating H (s) for s = j ω: ¯ −ω2 + 5 − j 2ω H (ω) = H (s)¯s= j ω = −ω2 + 5 + j 2ω

The phase characteristic is

∡ H (ω) = tan−1

µ

¶ µ ¶ µ ¶ −2ω 2ω 2ω −1 −1 − tan = −2 tan 5 − ω2 5 − ω2 5 − ω2 ∡H (ω)

π

Phase (rad)

π 2 0

−

π 2

−π −10

−5

0

5

10

ω (rad/s)

b.

The transform of the input signal is X (s) =

1 s +1

and the transform of the output signal is Y (s) =

s2 − 2 s + 5 s2 − 2 s + 5 ¡ ¢= 3 s + 3 s2 + 7 s + 5 (s + 1) s 2 + 2 s + 5

10

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

246

which can be expressed in partial fraction form as Y (s) = The residues are

k2 k3 k1 + + s +1 s +1+ j2 s +1− j2

¯ s 2 − 2 s + 5 ¯¯ =2 s 2 + 2 s + 5 ¯s=−1 ¯ ¯ 1 s2 − 2 s + 5 ¡ ¢ ¯¯ = − − j1 k2 = 2 (s + 1) s + 1 − j 2 s=−1− j 2

k1 =

and

1 k 3 = k 2∗ = − + j 1 2

The output signal is µ µ ¶ ¡ ¶ ¡ ¢ ¢ 1 1 1 1 y (t ) =2 e −t u (t ) + − − j 1 e − 2 + j 1 t u (t ) + − + j 1 e − 2 − j 1 t u (t ) 2 2 1

1

=2 e −t u (t ) − e − 2 t cos (t ) u (t ) − 2 e − 2 t sin (t ) u (t )

c.

For the input signal 1 X (ω) = 1+ jω

⇒

and for the output signal 5 − ω2 − j 2ω ¡ ¢ Y (ω) = 5 − 3ω2 + j ω 7 − ω2

⇒

¯ ¯ ¯ X (ω)¯ =

r

1 1 + ω2

v ¡ ¢2 u ¯ ¯ u 5 − ω2 + 4ω2 ¯Y (ω)¯ = t ¡ ¢2 ¡ ¢2 5 − 3ω2 + ω2 7 − ω2

|X (ω)|

|Y (ω)|

1

1

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0 −10

−5

0

5

10

0 −10

ω (rad/s)

−5

0

ω (rad/s)

7.41. a. The inverse of the system is H −1 (s) =

(s + 3)(s + 4) (s + 1)(s − 2)

5

10

247 The inverse system is causal but not stable since there is a pole at s = 2 in the right half s-plane.

b.

Let us express H (s) as

¶ µ (s + 1)(s − 2) s + 2 H (s) = (s + 3)(s + 4) s + 2

Let

H1 (s) = and

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

s −2 s +2 so that H (s) = H1 (s) H2 (s) and the subsystem H2 (s) is an all pass system. H2 (s) =

c.

The inverse system H1−1 (s) is H1−1 (s) =

It is causal and stable.

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

d. H (s) H1−1 (s) =

s −2 = H2 (s) s +2

7.42. a. Taking the Laplace transform of each side we obtain (s + 5) Y (s) = (2s + 3) X (s) which leads to the system function µ ¶ 1 (s + 5) Y (s) = H (s) = X (s) 2 s + 3/2

The inverse system has the system function Hi (s) = H −1 (s) =

2 (s + 3/2) s +5

The inverse system is both causal and stable. Its differential equation is d x (t ) d y (t ) + 5 y (t ) = 2 + 3 x (t ) dt dt

b.

Taking the Laplace transform of each side we obtain

which leads to the system function

¡

¢ ¡ ¢ s 2 + 7s + 12 Y (s) = s 2 + s X (s)

H (s) =

s2 + s s (s + 1) Y (s) = 2 = X (s) s + 7s + 12 (s + 3) (s + 4)

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

248

The system function for the inverse system is Hi (s) = H −1 (s) =

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

Since there is a pole at s = 0 the inverse system cannot be stable.

c.

Taking the Laplace transform of each side we obtain

which leads to the system function

¡

¢ s 2 + 3 Y (s) = (s + 5) X (s)

H (s) = The system function for the inverse system is

Y (s) s +5 = 2 X (s) s + 3

Hi (s) = H −1 (s) =

s2 + 3 s +5

The numerator order for the inverse system is 2, and the denominator order is 1. Therefore the inverse system cannot be causal.

d.

Taking the Laplace transform of each side we obtain

which leads to the system function

¡ ¢ (s − 3) Y (s) = s 2 + 2s + 1 X (s)

H (s) =

Y (s) s 2 + 2s + 1 = X (s) s −3

The numerator order is 2, and the denominator order is 1. Furthermore, there is a pole in the right half of the s-plane. The original system cannot be causal and stable. The system function for the inverse system is s −3 s −3 = Hi (s) = H −1 (s) = 2 s + 2s + 1 (s + 1)2 The inverse system is causal and stable. Its differential equation is

d x (t ) d y (t ) d 2 y (t ) + y (t ) = − 3 x (t ) +2 2 dt dt dt

7.43.

Asymptotically we have 20 log10 |H (ω)| = 20 log10 |H1 (ω)| + 20 log10 |H2 (ω)| + 20 log10 |H3 (ω)| + 20 log10 |H4 (ω)|

At ω = 5 rad/s

20 log10 |H1 (5)| = 20 log10 (5) = 13.98 dB

and 20 log10 |H2 (ω)| = 20 log10 |H3 (ω)| = 20 log10 |H4 (ω)| = 0 resulting in 20 log10 |H (5)| = 13.98 dB

249 At ω = 300 rad/s we have 20 log10 |H (300)| = 20 log10 (300/1) − 20 log10 (300/5) − 20 log10 (300/40) + 20 log10 (300/300) = −3.52 dB

7.44. a. 1 + 3 s −1 1 + 2 s −1 ¢ ¢ ¡ ¡ 1 + 2 s −1 Y (s) = 1 + 3 s −1 X (s) H (s) =

1 X(s)

1 W (s) = H1 (s) = X (s) 1 + 2 s −1 H2 (s) =

1/s b

b

3

Y (s)

−2

Y (s) = 1 + 3 s −1 W (s)

b. H (s) =

s −1 + s −3 1 + 9 s −1 + 26 s −2 + 24 s −3 1

X (s)

1/s b

1/s

1/s b

1 b

Y (s)

−9 −26 −24

c. H (s) =

0.5 s + 2.5 0.5 s −2 + 2.5 s −3 = s 3 + 2 s + 3.5 1 + 2 s −2 + 3.5 s −3 0.5

X (s)

1/s

1/s b

1/s

−2 −3.5

b

2.5

Y (s)

250

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

7.45. a. H (s) = Let H1 (s) = so that

s +1 s +3

(s + 1) (s + 2) ¡ ¢ (s + 3) s 2 + 2s + 2 and

H2 (s) =

s +2

s 2 + 2s + 2

H (s) = H1 (s) H2 (s) 1 X(s)

1

1/s b

1 b

1/s

−3

1/s b

b

2

Y (s)

−2 −2

b. H (s) = Let H1 (s) = so that

1 s +2

and

s2 + 1 (s + 2)(s + 3) (s + 4) H2 (s) =

s2 + 1 s2 + 1 = 2 (s + 3) (s + 4) s + 7s + 12

H (s) = H1 (s) H2 (s) 1

X(s)

1/s b

1

−2

b

1/s

1/s b

b

1

−7 −12

c. H (s) = Let H1 (s) = so that

s −1 s +1

(s − 1) (s − 2) (s + 1) (s + 2) and

H2 (s) =

H (s) = H1 (s) H2 (s)

s −2 s +2

Y (s)

251 1 X(s) b

1

1/s

−1 b

1/s b

−1

b

−2

Y (s)

−2

7.46. a.

The system function can be written as 0.4 0.6 s + 0.4 + 2 s + 3 s + 2s + 2

H (s) =

0.6 1/s

1/s b

b

0.4

−2 −2 X(s)

Y (s)

1/s b

0.4

−3

b.

The system function can be written as H (s) =

10 8.5 2.5 − + s +2 s +3 s +4 1/s b

2.5

−2

X(s)

−10

1/s b

−3

1/s −4

b

8.5

Y (s)

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

252

c.

The system function can be written as H (s) = 1 +

12 6 − s +1 s +2

1

X(s)

6

1/s b

Y (s)

−1

1/s

−12 b

−2

7.47. a. G u (s) =

ˆ

∞ 0

x (t − 1) e −st d t

Using the variable change t − 1 = λ we obtain G u (s) =

∞

ˆ

x (λ) e

−s(λ+1)

−1

If x (t ) = 0 for −1 < t < 0, then

dλ = e

−s

ˆ

∞ −1

x (λ) e −sλ d λ =

x (t − 1) u (t ) = x (t − 1) u (t − 1)

and G u (s) = e

−s

ˆ

∞ 0

b. G u (s) =

ˆ

x (λ) e −sλ d λ = e −s X u (s)

∞ 0

x (t + 2) e −st d t

Using the variable change t + 2 = λ we obtain G u (s) = If x (t ) = 0 for 0 < t < 2, then

ˆ

2

∞

x (λ) e

−s(λ−2)

dλ = e

2s

ˆ

∞ 2

x (λ) e −sλ d λ =

x (t + 2) u (t ) = x (t + 2) u (t + 2)

253 and G u (s) = e 2s

ˆ

∞ 0

c. G u (s) =

x (λ) e −sλ d λ = e 2s X u (s)

ˆ

∞

x (2t ) e −st d t

0

Using the variable change 2t = λ we obtain G u (s) =

ˆ

d. G u (s) =

∞

x (λ) e

−sλ/2

0

ˆ

∞

e

−2t

x (t ) e

−st

0

dt =

e. X u (s) = d X u (s) = ds

ˆ

ˆ

1 1 dλ = 2 2

∞ 0

ˆ

2

ˆ

∞

∞ 0

∞

x (λ) e −sλ/2 d λ =

x (t ) e −(s+2)t d t = X u (s + 2)

x (t ) e −st d t

0

−t x (t ) e −st d t = −G u (s)

Therefore G u (s) = −

d X u (s) ds

7.48. a. Writing KVL around the loop we obtain x (t ) = R i (t ) + L Recall that i (t ) = C

d i (t ) + y (t ) dt

d y (t ) dt

and

d 2 y (t ) d i (t ) =C dt dt2 Substituting these two relationships into the differential equation yields x (t ) = RC

³s´ 1 Xu 2 2

d y (t ) d 2 y (t ) + LC + y (t ) dt dt2

Rearranging terms we obtain ¶ µ ¶ µ µ ¶ 1 1 R d y (t ) d 2 y (t ) y = x (t ) + + (t ) dt2 L dt LC LC

Finally, substituting numerical values yields d 2 y (t ) d y (t ) +2 + 8 y (t ) = 8 x (t ) 2 dt dt

254

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

b.

Taking the Laplace transform of both sides of the differential equation using zero initial conditions (a requirement for the system function) we get ¡

and

¢ s 2 + 2 s + 8 Y (s) = 8 X (s)

H (s) =

8 Y (s) = X (s) s 2 + 2 s + 8

c.

We can not use the system function found in part (b) to determine the transform of the output signal as Y (s) = H (s) X (s) since initial conditions are specified. Recall that the system function is only meaningful when the system is CTLTI which requires all initial conditions to be zero. In this case we need to use the unilateral Laplace transform. ¯ ¯ d y (t ) ¯¯ i (0) 0.5 d y (t ) ¯¯ = =4 ⇒ = i (0) = C d t ¯t =0 d t ¯t =0 C 1/8 L

½

¯ ¾ d y (t ) ¯¯ d 2 y (t ) 2 =s Y (s) − s y (0) − dt2 d t ¯t =0

=s 2 Y (s) − 2s − 4

d y (t ) = s Y (s) − y (0) = s Y (s) − 2 dt Computing the unilateral Laplace transform of both sides of the differential equation leads to £ ¤ s 2 Y (s) − 2s − 4 + 2 s Y (s) − 2 + 8 Y (s) = 8 X (s)

and

¡

¢ s 2 + 2 s + 8 Y (s) =8 X (s) + 2s + 8

8 = + 2s + 8 s

The transform Y (s) is

2 s2 + 8 s + 8 ¢ Y (s) = ¡ 2 s s +2s +8

Its partial fraction expansion is Y (s) =

k1 k2 k3 + p + p s s +1+ j 7 s +1− j 7

with residues k1 = 1 ,

k 2 = 0.5 + j 0.945 ,

k 3 = 0.5 − j 0.945

The output signal is p ¢ p ¢ ¡ ¢ ¡ ¡ ¢ ¡ y (t ) =u (t ) + 0.5 + j 0.945 e −1− j 7 t u (t ) + 0.5 − j 0.945 e −1+ j 7 t u (t ) ³p ´ ³p ´ =u (t ) + e −t cos 7t u (t ) + 1.89 e −t sin 7t u (t )

255

d.

Taking the unilateral Laplace transform of the differential equation leads to ¡

¢ s 2 + 2 s + 8 Y (s) =8 X (s) + 2s + 8

=

The transform Y (s) is Y (s) = Its partial fraction expansion is Y (s) =

8 + 2s + 8 s +2

2 s 2 + 12 s + 24 ¡ ¢ (s + 2) s 2 + 2 s + 8

k1 k2 k3 + p + p s +2 s +1+ j 7 s +1− j 7

with residues k1 = 1 ,

k 2 = 0.5 + j 1.323 ,

k 3 = 0.5 − j 1.323

The output signal is p ¢ p ¢ ¡ ¢ ¡ ¡ ¢ ¡ y (t ) =u (t ) + 0.5 + j 1.323 e −1− j 7 t u (t ) + 0.5 − j 1.323 e −1+ j 7 t u (t ) ³p ´ ³p ´ =e −2t u (t ) + e −t cos 7t u (t ) + 2.646 e −t sin 7t u (t )

7.49. a. The output voltage is y (t ) = R 2 i 2 (t ) = R 2C 2 which leads to the relationship

¤ d £ v 0 (t ) − y (t ) dt

d v 0 (t ) 1 d y (t ) = y (t ) + dt R 2C 2 dt

(P.7.49.1)

Additionally, writing the KCL at the center node we get

which can be simplified to

¤ v 0 (t ) − x (t ) d v0 d £ +C 1 +C 2 v 0 (t ) − y (t ) = 0 R1 dt dt

v 0 (t ) + R 1 (C 1 +C 2 )

d v 0 (t ) d y (t ) = x (t ) + R 1C 2 dt dt

(P.7.49.2)

Substituting Eqn. (P.7.49.1) into Eqn. (P.7.49.2) and rearranging terms yields v 0 (t ) = −R 1C 1

d y (t ) R 1 (C 1 +C 2 ) − y (t ) + x (t ) dt R 2C 2

(P.7.49.3)

Differentiating both sides of Eqn. (P.7.49.3) d 2 y (t ) R 1 (C 1 +C 2 ) d y (t ) d x (t ) d v 0 (t ) = −R 1C 1 − + dt dt2 R 2C 2 dt dt

(P.7.49.4)

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

256

Equating the right sides of Eqns. (P.7.49.1) and (P.7.49.4) leads to the desired differential equation: µ ¶ 1 d x (t ) R 1C 1 + R 1C 2 + R 2C 2 d y (t ) d 2 y (t ) + + y (t ) = R 1C 1 2 dt R 2C 2 dt R 2C 2 dt

Substituting numerical values into the differential equation we obtain d 2 y (t ) d y (t ) d x (t ) +3 + y (t ) = 2 dt dt dt

b.

Taking the Laplace transform of both sides of the differential equation using zero initial conditions (a requirement for the system function) we get ¡

and

¢ s 2 + 3 s + 1 Y (s) = s X (s)

H (s) =

s Y (s) = 2 X (s) s + 3 s + 1

c. y (0) = R 2 i 2 (0) = 2 Since i 2 (t ) = C 2 it follows that

The current of the capacitor C 1 is

⇒

i 2 (0) =

y (0) =2 R2

¤ d £ v 0 (t ) − y (t ) dt

¯ ¯ d y (t ) ¯¯ d v 0 (t ) ¯¯ −C 2 i 2 (0) = C 2 d t ¯t =0 d t ¯t =0 ¯ ¯ d y (t ) ¯¯ d v 0 (t ) ¯¯ − 2= ¯ d t t =0 d t ¯t =0

C1

d v 0 (t ) x (t ) − v 0 (t ) = − i 2 (t ) dt R1

At time t = 0 we have

¯ 1−3 x (0) − v 0 (0) d v 0 (t ) ¯¯ − 2 = −4 − i 2 (0) = = C1 d t ¯t =0 R1 1

and we obtain

⇒

¯ d y (t ) ¯¯ = −6 d t ¯t =0

We are now ready to solve the differential equation. L

½

¯ ¾ d 2 y (t ) d y (t ) ¯¯ 2 =s Y − s y − (s) (0) dt2 d t ¯t =0

=s 2 Y (s) − 2s + 6

d y (t ) = s Y (s) − y (0) = s Y (s) − 2 dt

¯ d v 0 (t ) ¯¯ = −4 d t ¯t =0

257 Computing the unilateral Laplace transform of both sides of the differential equation leads to

and

£

¤ £ ¤ s 2 Y (s) − 2s + 6 + 3 s Y (s) − 2 + Y (s) = s X (s) ¡

¢ s 2 + 3 s + 1 Y (s) =s X (s) + 2s

=1 + 2s

The transform Y (s) is Y (s) = Its partial fraction expansion is Y (s) = + with residues

2s +1

s2 + 3 s + 1

k2 k1 + s + 0.3820 s + 2.6180

k 1 = 0.1056 ,

k 2 = 1.8944

The output signal is y (t ) =0.1056 e −0.3820 t u (t ) + 1.8944 e −2.6180 t u (t ) d. Taking the unilateral Laplace transform of the differential equation leads to ¡

¢ s 2 + 3 s + 1 Y (s) =s X (s) + 2s

=

The transform Y (s) is Y (s) = Its partial fraction expansion is Y (s) = +

s 2 s2 + 5 s + 2s = s +2 s +2

2 s2 + 5 s ¡ ¢ (s + 2) s 2 + 3 s + 1

k1 k2 k1 + + s + 2 s + 0.3820 s + 2.6180

with residues k1 = 2 ,

k 2 = −0.4472 ,

k 3 = 0.4472

The output signal is y (t ) =e −2t u (t ) − 0.4472 e −0.3820 t u (t ) + 0.4472 e −2.6180 t u (t )

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

258

7.50. a. 3 2

jω

1 0 −1 −2 −3 −4

−3

−2

−1

0

1

2

3

σ

b.

MATLAB script to evaluate the magnitude of X (s) at a grid of complex points in the s-plane: [ sr , s i ] = meshgrid ( [ − 5 : 0 . 1 : 5 ] , [ − 5 : 0 . 1 : 5 ] ) ; s = sr+ j * s i ; Xs = @( s ) s . * ( s + 2 ) . / ( ( s + 1) . ^ 2+ 9) ; XsMag = abs ( Xs ( s ) ) ; XsMag = XsMag . * ( XsMag< = 2) + 2. * (XsMag> 2 ) ;

c. 1 2 3 4 5 6 7 8 9 10 11 12

¯ ¯ Script to produce a three dimensional mesh plot of ¯ X (s)¯:

shading interp ; % Shading method : I n t e r p o la t e d colormap copper ; % S p e c i f y the c o lo r map used . m1 = mesh( sr , s i , XsMag ) ; axis ( [ − 5 , 5 , − 5 , 5] ) ; % Adjust transparency o f s u r f a c e l i n e s . set (m1, ’ EdgeAlpha ’ , 0 . 6 ’ , ’ FaceAlpha ’ , 0 . 6 ) ; % S p e c i f y x , y , z a xi s l a b e l s . xlabel ( ’ \sigma ’ ) ; ylabel ( ’ j \omega ’ ) ; zlabel ( ’ | X( s ) | ’ ) ; % S p e c i f y viewing angles . view ( gca , [ 2 3 . 5 , 3 8 ] ) ;

d. Modify the script in part (c) to also evaluate the Laplace transform for s = j ω and draw it over the three dimensional mesh plot: 1

% Define the t r a j e c t o r y s= j * omega

259 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

omega = [ − 5 : 0 . 0 1 : 5 ] ; t r = j * omega ; % Produce a mesh p lo t and hold i t . shading interp ; colormap copper ; m1 = mesh( sr , s i , XsMag ) ; hold on ; % Superimpose a p lo t o f X( s ) magnitude values evaluated on the % t r a j e c t o r y using ’ plot3 ’ function . m2 = plot3 ( real ( t r ) , imag ( t r ) , abs ( Xs ( t r ) ) , ’b− ’ , ’ LineWidth ’ , 1 . 5 ) ; hold o f f ; axis ( [ − 5 , 5 , − 5 , 5] ) ; % Adjust transparency o f s u r f a c e l i n e s . set (m1, ’ EdgeAlpha ’ , 0 . 6 ’ , ’ FaceAlpha ’ , 0 . 6 ) ; % S p e c i f y x , y , z a xi s l a b e l s . xlabel ( ’ \sigma ’ ) ; ylabel ( ’ j \omega ’ ) ; zlabel ( ’ | X( s ) | ’ ) ; % S p e c i f y viewing angles . view ( gca , [ 2 3 . 5 , 3 8 ] ) ;

7.51. a. 1 2 3 4 5 6 7 8 9 10

num = [ 1 , − 2] ; den = [ 1 , 3 , 2 ] ; pl s = roots ( den ) z r s = roots (num) plot ( [ − 1 0 , 1 0 ] , [ 0 , 0 ] , ’ k− ’ , [ 0 , 0 ] , [ − 1 0 , 1 0 ] , ’ k ’ , . . . real ( z r s ) , imag ( z r s ) , ’bo ’ , real ( pl s ) , imag ( pl s ) , ’ bx ’ ) ; axis ( [ − 4 , 3 , − 3 . 5 , 3 . 5 ] ) ; xlabel ( ’ \sigma ’ ) ; ylabel ( ’ j \omega ’ ) ; grid ;

b. 1 2 3 4 5 6 7 8 9 10

c.

num = [ 1 , 0 ] ; den = [ 1 , 0 , − 1] ; pl s = roots ( den ) z r s = roots (num) plot ( [ − 1 0 , 1 0 ] , [ 0 , 0 ] , ’ k− ’ , [ 0 , 0 ] , [ − 1 0 , 1 0 ] , ’ k ’ , . . . real ( z r s ) , imag ( z r s ) , ’bo ’ , real ( pl s ) , imag ( pl s ) , ’ bx ’ ) ; axis ( [ − 4 , 3 , − 3 . 5 , 3 . 5 ] ) ; xlabel ( ’ \sigma ’ ) ; ylabel ( ’ j \omega ’ ) ; grid ;

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

260 1 2 3 4 5 6 7 8 9 10

num = [ 1 , 1 ] ; den = [ 1 , − 4 , 3] ; pl s = roots ( den ) z r s = roots (num) plot ( [ − 1 0 , 1 0 ] , [ 0 , 0 ] , ’ k− ’ , [ 0 , 0 ] , [ − 1 0 , 1 0 ] , ’ k ’ , . . . real ( z r s ) , imag ( z r s ) , ’bo ’ , real ( pl s ) , imag ( pl s ) , ’ bx ’ ) ; axis ( [ − 4 , 3 , − 3 . 5 , 3 . 5 ] ) ; xlabel ( ’ \sigma ’ ) ; ylabel ( ’ j \omega ’ ) ; grid ;

d. 1 2 3 4 5 6 7 8 9 10

num = [ 1 , − 1 , 0] ; den = [1 , −1 , −6]; pl s = roots ( den ) z r s = roots (num) plot ( [ − 1 0 , 1 0 ] , [ 0 , 0 ] , ’ k− ’ , [ 0 , 0 ] , [ − 1 0 , 1 0 ] , ’ k ’ , . . . real ( z r s ) , imag ( z r s ) , ’bo ’ , real ( pl s ) , imag ( pl s ) , ’ bx ’ ) ; axis ( [ − 3 , 4 , − 3 . 5 , 3 . 5 ] ) ; xlabel ( ’ \sigma ’ ) ; ylabel ( ’ j \omega ’ ) ; grid ;

e. 1 2 3 4 5 6 7 8 9 10

num = [ 1 , − 1 , 0] ; den = [ 1 , 8 , 1 5 ] ; pl s = roots ( den ) z r s = roots (num) plot ( [ − 1 0 , 1 0 ] , [ 0 , 0 ] , ’ k− ’ , [ 0 , 0 ] , [ − 1 0 , 1 0 ] , ’ k ’ , . . . real ( z r s ) , imag ( z r s ) , ’bo ’ , real ( pl s ) , imag ( pl s ) , ’ bx ’ ) ; axis ( [ − 6 , 1 , − 3 . 5 , 3 . 5 ] ) ; xlabel ( ’ \sigma ’ ) ; ylabel ( ’ j \omega ’ ) ; grid ;

7.52. a. 1 2 3 4 5 6 7

%% X = @( s ) ( s − 2) . / ( s .^2+3 * s + 2 ) ; omg = [ − 1 0 : 0 . 0 2 : 1 0 ] ; Xomg = X( j *omg ) ; plot (omg, abs (Xomg ) ) ; axis ( [ − 1 0 , 1 0 , 0 , 1 . 2 ] ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ;

261 8 9 10 11 12 13 14 15 16 17

ylabel ( ’ Magnitude ’ ) ; t i t l e ( ’ |H( \omega ) | ’ ) ; grid ; %% plot (omg, angle (Xomg ) ) ; axis ([ −10 ,10 , − pi , pi ] ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Phase ( rad ) ’ ) ; t i t l e ( ’ \ angle H( \omega) ’ ) ; grid ;

c. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

%% X = @( s ) ( s + 1 ) . / ( s .^2−4 * s + 3 ) ; omg = [ − 1 0 : 0 . 0 2 : 1 0 ] ; Xomg = X( j *omg ) ; plot (omg, abs (Xomg ) ) ; axis ( [ − 1 0 , 1 0 , 0 , 0 . 4 ] ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; t i t l e ( ’ |H( \omega ) | ’ ) ; grid ; %% plot (omg, angle (Xomg ) ) ; axis ([ −10 ,10 , − pi , pi ] ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Phase ( rad ) ’ ) ; t i t l e ( ’ \ angle H( \omega) ’ ) ; grid ;

7.53. a. 1 2 3

syms s t Xs = 1/( s ^2+3 * s + 2 ) ; x t = i l a p l a c e ( Xs )

b. 1 2 3

syms s t Xs = ( s −1) * ( s −2)/(( s + 1) * ( s + 2) * ( s + 3 ) ) ; x t = i l a p l a c e ( Xs )

c. 1 2 3

syms s t Xs = s * ( s −1)/(( s +1)^2 * ( s + 2) ) x t = i l a p l a c e ( Xs )

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

262

7.54. a. 1

sys = t f ( [ 1 , 1 , 0 ] , [ 1 , 4 , 1 3 ] )

b. 1 2 3 4 5 6 7

t = [ −1:0.01:10]; x = exp( −0.2 * t ) . * ( t >=0); plot ( t , x ) ; axis ( [ − 1 , 10 , − 0. 25 , 1. 25] ) ; xlabel ( ’ t ( sec ) ’ ) ; title ( ’x( t ) ’ ) ; grid ;

c. 1 2 3 4 5 6

y = lsim ( sys , x , t ) ; plot ( t , y ) ; axis ( [ − 1 , 10 , − 0. 25 , 1. 25] ) ; xlabel ( ’ t ( sec ) ’ ) ; title ( ’y( t ) ’ ) ; grid ;

7.55. The code 1 2 3 4 5

% Enter numerator and denominator polynomials num = [1 , −9 ,30 , −42 ,20]; den = [ 1 , 12 , 59 , 152 , 200 , 96] ; % Compute the p o le s and the r e s i d u e s [ r , p , k ] = residue (num, den )

results in the response r = 62.5000 −74.0000 4.5500 4.5500 3.4000

+ + − + +

0.0000 i 0.0000 i 7.1500 i 7.1500 i 0.0000 i

p = −4.0000 + 0.0000 i −3.0000 + 0.0000 i −2.0000 + 2.0000 i

263 −2.0000 − 2.0000 i −1.0000 + 0.0000 i k = []

Correspondingly, the partial fraction expansion is X (s) =

3.4 4.55 + j 7.15 4.55 − j 7.15 −74 62.5 + + + + s +1 s +2+ j2 s +2− j2 s +3 s +4

7.56. a. 1 2 3 4 5

num1 = [ 1 , 1 ] ; den1 = [ 1 , 5 , 6 ] ; sys1 = t f (num1, den1 ) ; bode( sys1 ) ; grid ;

b. 1 2 3 4 5

num2 = [ 1 , − 1 , 0] ; den2 = [ 1 , 5 , 8 , 6 ] ; sys2 = t f (num2, den2 ) ; bode( sys2 ) ; grid ;

c. 1 2 3 4 5

num3 = conv ( [ 1 , − 1 ] , [ 1 , 2 ] ) ; % Use conv ( ) f o r polynomial m u lt i p li c a t i o n . den3 = conv ( [ 1 , 1 ] , [ 1 , 6 , 1 3 ] ) ; sys3 = t f (num3, den3 ) ; bode( sys3 ) ; grid ;

7.57. 1 2

Use the following script to define H (s) as a function of parameters ζ and ω0 :

numH = @( zeta , omg0) [omg0 * omg0 ] ; denH = @( zeta , omg0) [ 1 , 2 * zeta * omg0, omg0 * omg0 ] ;

Try with ζ = 0.01 and ω0 = 5 rad/s: 1 2 3

sys1 = t f (numH( 0 . 0 1 , 5 ) ,denH ( 0 . 0 1 , 5 ) ) ; bode( sys1 , { 1 , 1 0 } ) ; grid ;

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

264

Try with ζ = 0.1 and ω0 = 5 rad/s: 1 2 3

sys2 = t f (numH( 0 . 1 , 5 ) ,denH ( 0 . 1 , 5 ) ) ; bode( sys2 , { 1 , 1 0 } ) ; grid ;

Try with ζ = 0.5 and ω0 = 5 rad/s: 1 2 3

sys3 = t f (numH( 0 . 5 , 5 ) ,denH ( 0 . 5 , 5 ) ) ; bode( sys3 , { 1 , 1 0 } ) ; grid ;

Try with ζ = 1 and ω0 = 5 rad/s: 1 2 3

sys4 = t f (numH( 1 , 5 ) ,denH ( 1 , 5 ) ) ; bode( sys4 , { 1 , 1 0 } ) ; grid ;

Try with ζ = 2 and ω0 = 5 rad/s: 1 2 3

sys5 = t f (numH( 2 , 5 ) ,denH ( 2 , 5 ) ) ; bode( sys5 , { 1 , 1 0 } ) ; grid ;

Graph all five diagrams together for comparison. 1 2 3 4 5 6 7 8 9 10

bode( sys1 , { 1 , 1 0 } ) ; hold on ; bode( sys2 , { 1 , 1 0 } ) ; bode( sys3 , { 1 , 1 0 } ) ; bode( sys4 , { 1 , 1 0 } ) ; bode( sys5 , { 1 , 1 0 } ) ; hold o f f ; legend ( ’ \ zeta =0.01 ’ , ’ \ zeta =0.1 ’ , ’ \ zeta =0.5 ’ , ’ \ zeta=1 ’ , ’ \ zeta=2 ’ , . . . ’ Location ’ , ’ SouthWest ’ ) grid ;

7.58. a. 1 2 3

b.

Define H (s), H1 (s), and H1−1 (s). H = zpk([ −1 ,2] ,[ −3 , −4] ,1) H1 = zpk([ −1 , −2] ,[ −3 , −4] ,1) H1inv = zpk([ −3 , −4] ,[ −1 , −2] ,1)

Graph |H (s)| and ∠ H (s).

265 1 2 3 4 5 6 7 8

omg = [ − 5 : 0 . 0 1 : 5 ] ; Homg = freqresp (H,omg ) ; subplot ( 1 , 2 , 1 ) ; plot (omg, abs (Homg( 1 , : ) ) ) ; grid ; subplot ( 1 , 2 , 2 ) ; plot (omg, angle (Homg( 1 , : ) ) ) ; grid ;

Graph |H1 (s)| and ∠ H1 (s). 1 2 3 4 5 6 7

H1omg = freqresp (H1,omg ) ; subplot ( 1 , 2 , 1 ) ; plot (omg, abs (H1omg ( 1 , : ) ) ) ; grid ; subplot ( 1 , 2 , 2 ) ; plot (omg, angle (H1omg ( 1 , : ) ) ) ; grid ;

¯ ¯ Graph ¯ H1−1 (s)¯ and ∠ H1−1 (s). 1 2 3 4 5 6 7

c. 1 2 3

H1invomg = freqresp ( H1inv ,omg ) ; subplot ( 1 , 2 , 1 ) ; plot (omg, abs (H1invomg ( 1 , : ) ) ) ; grid ; subplot ( 1 , 2 , 2 ) ; plot (omg, angle (H1invomg ( 1 , : ) ) ) ; grid ;

Graph unit step response of H (s). clf ; step (H, [ 0 : 0 . 0 1 : 5 ] ) grid ;

Graph unit step response of H (s) H1−1 (s). 1 2 3

clf ; step (H* H1inv , [ 0 : 0 . 0 1 : 5 ] ) grid ;

7.59. a. Recall that the differential equation is d 2 y (t ) d y (t ) + 8 y (t ) = 8 x (t ) +2 2 dt dt

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

266

Using the initial conditions y (0) = 2

and

¯ d y (t ) ¯¯ =4 d t ¯t =0

with unilateral Laplace transform leads to £ 2 ¤ £ ¤ s Y (s) − 2s − 4 + 2 s Y (s) − 2 + 8 Y (s) = 8 X (s)

Solve for the output transform Y (s). >> >> >> >> >> >>

syms xt = Xs = Y2 = Y1 = Ys =

s t Ys heaviside ( t ) ; % x ( t )=u( t ) laplace ( x t ) ; % Laplace transform o f s * s * Ys−2* s −4; % Laplace transform o f s * Ys −2; % Laplace transform o f solve ( Y2+2 * Y1+8 * Ys−8* Xs , Ys ) % Solve f o r

x( t ) d2y / dt2 dy / dt Y( s )

Find y (t ) through inverse Laplace transform. >>

y t = i l a p l a c e ( Ys )

% I n v e r s e Laplace transform o f Y ( s )

Graph the output signal. >> >>

b.

ezplot ( yt , [ 0 , 5 ] ) ; grid ; axis ( [ 0 , 5 , − 1 , 3 ] ) ;

Solve for the output transform Y (s). >> >> >> >> >> >>

syms xt = Xs = Y2 = Y1 = Ys =

s t Ys exp(−2 * t ) * heaviside ( t ) ; % x ( t )= exp(−2 * t ) * u( t ) laplace ( x t ) ; % Laplace transform o f x ( t ) s * s * Ys−2* s −4; % Laplace transform o f d2y / dt2 s * Ys −2; % Laplace transform o f dy / dt solve ( Y2+2 * Y1+8 * Ys−8* Xs , Ys ) % Solve f o r Y ( s )

Find y (t ) through inverse Laplace transform. >>

y t = i l a p l a c e ( Ys )

% I n v e r s e Laplace transform o f Y ( s )

Graph the output signal. >> >>

ezplot ( yt , [ 0 , 5 ] ) ; grid ; axis ( [ 0 , 5 , − 1 , 3 ] ) ;

7.60. a. Recall that the differential equation is d 2 y (t ) d y (t ) d x (t ) +3 + y (t ) = 2 dt dt dt

267 Using the initial conditions y (0) = 2 with unilateral Laplace transform leads to £

and

¯ d y (t ) ¯¯ = −6 d t ¯t =0

¤ £ ¤ s 2 Y (s) − 2s + 6 + 3 s Y (s) − 2 + Y (s) = s X (s)

Solve for the output transform Y (s). >> >> >> >> >> >>

syms xt = Xs = Y2 = Y1 = Ys =

s t Ys heaviside ( t ) ; % x ( t )=u( t ) laplace ( x t ) ; % Laplace transform o f x ( t ) s * s * Ys−2* s +6; % Laplace transform o f d2y / dt2 s * Ys −2; % Laplace transform o f dy / dt solve ( Y2+3 * Y1+Ys−s * Xs , Ys ) % Solve f o r Y ( s )

Find y (t ) through inverse Laplace transform. >>

y t = i l a p l a c e ( Ys )

% I n v e r s e Laplace transform o f Y ( s )

Graph the output signal. >> >>

b.

ezplot ( yt , [ 0 , 5 ] ) ; grid ; axis ( [ 0 , 5 , − 1 , 3 ] ) ;

Solve for the output transform Y (s). >> >> >> >> >> >>

syms xt = Xs = Y2 = Y1 = Ys =

s t Ys exp(−2 * t ) * heaviside ( t ) ; % x ( t )= exp(−2 * t ) * u( t ) laplace ( x t ) ; % Laplace transform o f x ( t ) s * s * Ys−2* s +6; % Laplace transform o f d2y / dt2 s * Ys −2; % Laplace transform o f dy / dt solve ( Y2+3 * Y1+Ys−s * Xs , Ys ) % Solve f o r Y ( s )

Find y (t ) through inverse Laplace transform. >>

y t = i l a p l a c e ( Ys )

Graph the output signal. >> >>

ezplot ( yt , [ 0 , 5 ] ) ; grid ; axis ( [ 0 , 5 , − 1 , 3 ] ) ;

% I n v e r s e Laplace transform o f Y ( s )

268

CHAPTER 7. LAPLACE TRANSFORM FOR CONTINUOUS-TIME SIGNALS AND SYSTEMS

Chapter 8 z-Transform for Discrete-Time Signals and Systems 8.1. a.

Applying the z-transform definition: X (z) = 1 + z −1 + z −2 =

Zeros: Poles:

z 1 = −0.5 + j 0.866 ,

p1 = p2 = 0

z2 + z + 1 z2

z 2 = −0.5 − j 0.866

The transform converges at every point except the origin of the z-plane. ROC:

b.

|z| > 0 Applying the z-transform definition: X (z) = 1 + z −1 + z −2 + z −3 + z −4 =

Zeros: Poles:

z 1 = 0.309 + j 0.9511 ,

p1 = p2 = p3 = p4 = 0

z 2 = 0.309 − j 0.9511 ,

z4 + z3 + z2 + z + 1 z4

z 3 = −0.809 + j 0.5878 ,

z 4 = −0.809 − j 0.5878

The transform converges at every point except the origin of the z-plane. ROC:

c.

|z| > 0 Applying the z-transform definition: X (z) = z 2 + z + 1 + z −1 + z −2 =

Zeros: Poles:

z 1 = 0.309 + j 0.9511 ,

p1 = p2 = 0

z 2 = 0.309 − j 0.9511 ,

z4 + z3 + z2 + z + 1 z2

z 3 = −0.809 + j 0.5878 ,

z 4 = −0.809 − j 0.5878

In addition to the two poles at the origin of the z-plane, the transform also has two poles at infinity (due to the z 2 term). Therefore the ROC must exclude both the origin of the z-plane and infinity. ROC:

d.

0 < |z| < ∞ Applying the z-transform definition: X (z) = z 4 + z 3 + z 2 + z + 1

Zeros:

z 1 = 0.309 + j 0.9511 ,

z 2 = 0.309 − j 0.9511 , 269

z 3 = −0.809 + j 0.5878 ,

z 4 = −0.809 − j 0.5878

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

270

The transform X (z) has no finite poles. There are four poles at infinity. Therefore X (z) converges at all points except where |z| → ∞. ROC:

|z| < ∞

8.2. a.

No, since the ROC does not include the unit circle.

b.

Yes, since the ROC includes the unit circle. ¯ ¯ X (Ω) = X (z)¯

c.

¡ ¢ e jΩ e jΩ +2 ¢¡ ¢ = ¡ jΩ z=e j Ω e + 1/2 e j Ω + 3/2

Yes, since the ROC includes the unit circle. ¯ ¯ X (Ω) = X (z)¯

d.

e j 2Ω + 5 e j Ω + 6

Yes, since the ROC includes the unit circle. ¯ ¯ X (Ω) = X (z)¯

8.3. a. b.

z=e j Ω

e j 2Ω

=

ROC:

¡

¢¡ ¢ e jΩ +1 e jΩ −1 ¢¡ ¢¡ ¢ = ¡ jΩ z=e j Ω e +2 e jΩ −3 e jΩ −4

|z| > 0.75

1 Two complex poles are at 0.5 ∓ j 0.5 = p e ∓ j π/4 2 1 ROC: p < |z| < 1.25 2

c.

ROC:

|z| > 0.5

d.

ROC:

1 < |z| < 1.25 3

8.4. a.

Using z-transform definition X (z) =

9 X

n z −n

n=0

Let A (z) be defined as A (z) =

9 X

z −n

n=0

Using the closed form formula for the finite-length geometric series A (z) =

1 − z −10 , 1 − z −1

¯ ¯ ¯z ¯ > 0

271 Differentiating A (z) with respect to z yields 9 X ¤ d £ n z −n−1 A (z) = − dz n=0

and therefore −z

9 ¤ X d £ A (z) = n z −n = X (z) dz n=0

The derivative of the closed form expression for A (z) is

and X (z) is found as

¤ −z −2 + 10 z −11 − 9 z −12 d £ A (z) = , ¡ ¢2 dz 1 − z −1

X (z) = −z

b.

¯ ¯ ¯z ¯ > 0

¤ z −1 − 10 z −10 + 9 z −11 d £ A (z) = , ¡ ¢2 dz 1 − z −1

¯ ¯ ¯z ¯ > 0

The transform X (z) can be written as X (z) = X 1 (z) + X 2 (z)

with X 1 (z) = and X 2 (z) = From part (a) we have X 1 (z) =

9 X

n z −n

n=0 ∞ X

10 z −n

n=10

z −1 − 10 z −10 + 9 z −11 , ¡ ¢2 1 − z −1

X 2 (z) is found by using the variable change m = n − 10: X 2 (z) = 10

∞ X

m=0

z −(m+10) =

10 z −10 , 1 − z −1

and X (z) =

¯ ¯ ¯z ¯ > 1

z −1 − 10 z −10 + 9 z −11 10 z −10 + ¡ ¢2 1 − z −1 1 − z −1

z −1 − z −11 = ¡ ¢2 , 1 − z −1

c.

¯ ¯ ¯z ¯ > 0

¯ ¯ ¯z ¯ > 1

The transform X (z) can be written as X (z) = X 1 (z) + X 2 (z)

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

272 with

X 1 (z) = and X 2 (z) = From part (a) we have X 1 (z) =

19 X

n=10

9 X

n z −n

n=0

(−n + 20) z −n

z −1 − 10 z −10 + 9 z −11 , ¡ ¢2 1 − z −1

X 2 (z) is found by using the variable change m = n − 10: X 2 (z) =

9 X

m=0

(−m + 10) z −(m+10) 9 X

= −z −10

m=0

= −z −10

"

= The transform X (z) is

m z −m + 10 z −10

9 X

z −m

m=0

# · −10 ¸ z −1 − 10 z −10 + 9 z −11 −10 1 − z + 10 z ¡ ¢2 1 − z −1 1 − z −1

10 z −10 − 11 z −11 + z −21 , ¡ ¢2 1 − z −1

X (z) = X 1 (z) + X 2 (z) =

8.5.

¯ ¯ ¯z ¯ > 0

¯ ¯ ¯z ¯ > 0

z −1 − 2 z −11 + z −21 , ¡ ¢2 1 − z −1

¯ ¯ ¯z ¯ > 0

Let X (z) be written as X (z) = X 1 (z) + X 2 (z)

with X 1 (z) = and X 2 (z) =

∞ X

z −n

n=0 n even

∞ X

(2/3)n z −n

n=1 n odd

Using n = 2m for the indices of the summation for X 1 (z) yields X 1 (z) =

∞ X

m=0

z −2m =

1 , 1 − z −2

¯ −2 ¯ ¯z ¯ < 1

⇒

¯ ¯ ¯z ¯ > 1

273 Similarly, using n = 2m + 1 for the indices of the summation for X 2 (z) yields X 2 (z) = = =

∞ X

(2/3)2m+1 z −(2m+1)

m=0

µ

¶ ∞ µ ¶ 2 −1 X 4 −2 m z z 3 m=0 9 2 3

1−

z −1 4 9

¯ ¯ ¯ 4 −2 ¯ ¯ z ¯ 3

1 6 z −1 9 + 6 z −1 − 4 z −2 − 6 z −3 + = , 1 − z −2 9 − 4 z −2 9 − 13 z −2 + 4 z −4

or, using non-negative powers of z X (z) =

8.6. a.

9 z4 + 6 z3 − 4 z2 − 6 z , 9 z 4 − 13 z 2 + 4

¯ ¯ ¯z ¯ > 1

¯ ¯ ¯z ¯ > 1

Using Euler’s formula g [n] =

The transform is

h i 1 (0.9)n e j 0.3n + e − j 0.3n u[n] 2

´n 1 X ´n ∞ ³ ∞ ³ 1 X 0.9 e j 0.3 z −1 0.9 e − j 0.3 z −1 2 n=0 2 n=0 µ ¶ µ ¶ 1 1 1 1 = + 2 1 − 0.9 e j 0.3 z −1 2 1 − 0.9 e − j 0.3 z −1

G (z) =

= =

1 − 0.9 cos (0.3) z −1 1 − 1.8 cos (0.3) z −1 + 0.81 z −2 £ ¤ z z − 0.9 cos (0.3)

z 2 − 1.8 cos (0.3) z + 0.81 ¡ ¢ ¯ ¯ z z − 0.8598 ¯z ¯ > 0.9 = 2 , z − 1.7196 z + 0.81

b.

Since x[n] = g [2n], the corresponding transform is 1 ¡p ¢ 1 ¡ p ¢ G z + G − z 2 2 ¢ ¢ µ ¶ p ¡p µ ¶ p ¡p z z − 0.8598 z z + 0.8598 1 1 = + p p 2 z − 1.7196 z + 0.81 2 z + 1.7196 z + 0.81

X (z) =

=

z (z − 0.6685)

z 2 − 1.3370 z + 0.6561

,

¯ ¯ ¯z ¯ > 0.81

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

274

c.

The signal x[n] is x[n] = g [2n] = (0.9)2n cos ((0.3) 2n) u[2n] = (0.81)n cos (0.6n) u[n]

and the corresponding transform is X (z) =

£ ¤ z z − 0.81 cos (0.6)

z 2 − 1.62 cos (0.6) z + 0.6561 ¯ ¯ z (z − 0.6685) ¯z ¯ > 0.81 = 2 , z − 1.3370 z + 0.6561

8.7. a. b. c.

Using z-transform definition

¯ ¯ ¯z ¯ < ∞

X (z) = z −3 ,

¯ ¯ ¯z ¯ > 0

Using z-transform definition

Using z-transform definition with the closed form formula for the infinite-length geometric series X (z) =

d.

X (z) = z 2 ,

1 1 2

1 − z −1

=

z 1 2

z−

¯ ¯ 1 ¯z ¯ > 2

,

From part (c) © ª Z (1/2)n u[n] = © ª Z (1/3)n u[n] =

z z−

1 2

z z−

1 3

¯ ¯ 1 ¯z ¯ > 2

,

¯ ¯ 1 ¯z ¯ > 3

,

Using linearity of the z-transform X (z) =

z

+

z

z− z − 31 ¢ ¡ z 2 z − 56 = , z 2 − 56 z + 61

e.

1 2

From part (c) ª © Z (1/2)n u[n] =

z z − 21

¯ ¯ 1 ¯z ¯ > 2 ¯ ¯ 1 ¯z ¯ > 2

,

Using the time shifting property of the z-transform

¢ © ª ¡ X (z) = Z (1/2)n−1 u[n − 1] = z −1

z z−

1 2

=

1 z−

1 2

,

¯ ¯ 1 ¯z ¯ > 2

275

f.

From part (c)

Using linearity of the z-transform

© ª Z (1/2)n u[n] =

z z − 21

,

¯ ¯ 1 ¯z ¯ > 2

ª © ª © 2z Z (1/2)n−1 u[n] = (1/2)−1 Z (1/2)n u[n] = , z − 21

g.

From part (c)

Using linearity of the z-transform

ª © Z (1/2)n u[n] =

z z−

1 2

,

¯ ¯ 1 ¯z ¯ > 2

© ª © ª Z (1/2)n+1 u[n] = (1/2) Z (1/2)n u[n] =

h.

¯ ¯ 1 ¯z ¯ > 2

From part (d) ª © Z (1/3)n u[n] =

z z − 31

Applying the time reversal property of the z-transform ª ª © © Z (1/3)−n u[−n] = Z (3)n u[−n] =

,

1 2

z

z−

1 2

¯ ¯ 1 ¯z ¯ > 2

,

¯ ¯ 1 ¯z ¯ > 3

z −1 z −1 − 13

=

3 , 3−z

¯ ¯ ¯z ¯ < 3

Replacing n with n + 1 and using time shifting property of the z-transform

and using linearity

i.

ª © Z (3)n+1 u[−n − 1] = (z)

3 3z = , 3−z 3−z

µ ¶ © ª 1 3z z X (z) = Z (3)n u[−n − 1] = = , 3 3−z 3−z

¯ ¯ ¯z ¯ < 3 ¯ ¯ ¯z ¯ < 3

From part (h)

¯ ¯ 3 ¯z ¯ < 3 , 3−z Replacing n with n − 1 and using time shifting property of the z-transform © ª Z (3)n u[−n] =

¢ © ª ¡ Z (3)n−1 u[−n + 1] = z −1

3 3 z −1 3 , = = 3 − z 3 − z z (3 − z)

¯ ¯ 0 < ¯z ¯ < 3

Note that the ROC for the new transform must also exclude the origin of the z-plane due to the pole introduced at z = 0. Another way to explain the exclusion of the origin from the ROC is to realize that the rightmost sample of x[n] with nonzero amplitude is now at n = 1 due to the right shift. Using linearity we obtain µ ¶ ¯ ¯ © ª 3 1 1 = , 0 < ¯z ¯ < 3 X (z) = Z (3)n u[−n + 1] = 3 z (3 − z) z (3 − z)

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

276

replacemen

8.8. a.

xL [n] 1

Let

n

(1/2) , n ≥ 0 (1/2)n , n < 0

Amplitude

x[n] =

½

replacemen µ ¶−n 1 x L [n] = u[−n − 1] = (2)n u[−n − 1] 2 µ ¶n 1 u[n] x R [n] = 2

0.5

0 −8

−6

−4

−2

0

Amplitude

x[n] = x L [n] + x R [n] The two signal components are shown on the right.

−6

−4

¯ ¯ © ª z ¯z ¯ < 2 X L (z) = Z (2)n u[−n − 1] = − , z −2 ¾ ½ µ ¶n ¯ ¯ 1 z 1 ¯z ¯ > u[n] = , X R (z) = Z 2 2 z − 12

Using linearity of the z-transform

X (z) = X L (z) + X R (z)

=

8.9. a.

We know that

Using the time reversal property

z z + z − 2 z − 12 − 23 z 5 2

z2 − z + 1

© ª Z u[n] =

© ª X (z) = Z u[−n] =

8

2

4

6

8

0 −2

0

Index n

=−

6

0.5

−8

c.

4

1

so that

b.

2

Index n xR [n]

,

z , z −1

1 ¯¯ ¯¯ < z 1

z1 1 = , z −1 − 1 1 − z

¯ ¯ ¯z ¯ < 1

277

b.

Applying the time reversal property to the transform pair © ª Z u[n − 1] =

yields the transform pair

© ª X (z) = Z u[−n − 1] =

c.

From part (a) © ª Z u[−n] =

In addition we have ©

ª

Z u[n − 5] = Applying the time reversal property

¡

1 , z −1

¯ ¯ ¯z ¯ > 1

1 z = , z −1 − 1 1 − z 1 , 1−z

¢ z −5 z

z −1

¯ ¯ ¯z ¯ < 1

z −4 , z −1

=

and, applying the linearity property

X (z) =

d.

¯ ¯ ¯z ¯ > 1

z5 z4 = , z −1 − 1 1 − z

© ª Z u[−n − 5] =

¯ ¯ ¯z ¯ < 1

1 z5 1 + z5 + = , 1−z 1−z 1−z

¯ ¯ ¯z ¯ < 1 ¯ ¯ ¯z ¯ < 1

The z transform of the unit-ramp function was found in Example 8-21 to be © ª Z n u[n] =

z (z − 1)

2

¯ ¯ ¯z ¯ > 1

,

Applying the time reversal property leads to © ª Z − n u[−n] = ¡

z −1

z z = , ¢2 = 2 −1 (1 − z) (z − 1)2 z −1

¯ ¯ ¯z ¯ < 1

and, through the use of the linearity property, we get

© ª X (z) = Z n u[−n] =

e.

−z

(z − 1)

2

,

¯ ¯ ¯z ¯ < 1

The transform of a cosine signal was found in Example 8-15 to be © ª Z cos (Ω0 n) u[n] =

z [z − cos (Ω0 )] , z 2 − 2 cos (Ω0 ) z + 1

Applying the time reversal property yields ©

ª

Z cos (−Ω0 n) u[−n] = =

¯ ¯ ¯z ¯ > 1

£ ¤ z −1 z −1 − cos (Ω0 )

z −2 − 2 cos (Ω0 ) z −1 + 1

1 − cos (Ω0 ) z , 1 − 2 cos (Ω0 ) z + z 2

¯ ¯ ¯z ¯ < 1

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

278

Since the cosine function is even, that is, cos (−Ω0 n) = cos (Ω0 n), we get © ª X (z) = Z cos (Ω0 n) u[−n] =

1 − cos (Ω0 ) z , 1 − 2 cos (Ω0 ) z + z 2

¯ ¯ ¯z ¯ < 1

8.10. a.

Starting with

¯ ¯ z ¯z ¯ > 1 , z −1 the following relationships can be obtained through the use of the differentiation property of the z-transform: ¯ ¯ © ª z d h z i ¯z ¯ > 1 , = Z n u[n] = −z 2 dz z −1 (z − 1) · ¸ ¯ ¯ © 2 ª d z z (z + 1) ¯z ¯ > 1 Z n u[n] = −z = , 2 3 d z (z − 1) (z − 1) © ª Z u[n] =

Using the relationships found, X (z) is obtained as X (z) =

b.

z (z + 1) (z − 1)

3

+3

z (z − 1)

2

+5

z 5 z3 − 6 z2 + 3 z , = z −1 (z − 1)3

¯ ¯ ¯z ¯ > 1

Starting with

¯ ¯ © ª −z ¯z ¯ < 1 , Z u[−n − 1] = z −1 the following relationships can be obtained through the use of the differentiation property of the z-transform: ¯ ¯ © ª −z d h −z i ¯z ¯ < 1 = , Z n u[−n − 1] = −z dz z −1 (z − 1)2 ¸ · ¯ ¯ © ª −z (z + 1) −z d ¯z ¯ < 1 , = Z n 2 u[−n − 1] = −z 2 3 d z (z − 1) (z − 1)

Using the relationships found, X (z) is obtained as X (z) =

c.

−z (z + 1) (z − 1)

3

−5

5 z 3 − 11 z 2 + 4 z −z = z −1 (z − 1)3

¯ ¯ ¯z ¯ < 1

The transform of a cosine signal was found in Example 8-15 to be © ª Z cos (Ω0 n) u[n] =

z [z − cos (Ω0 )] , z 2 − 2 cos (Ω0 ) z + 1

Using the differentiation property of the z-transform

¯ ¯ ¯z ¯ > 1

¸ · © ª d z [z − cos (Ω0 )] X (z) = Z n cos (Ω0 n) u[n] = −z , d z z 2 − 2 cos (Ω0 ) z + 1

It can be shown that

¸ · − cos (Ω0 ) z 2 + 2 z − cos (Ω0 ) z [z − cos (Ω0 )] d = ¢2 ¡ d z z 2 − 2 cos (Ω0 ) z + 1 z 2 − 2 cos (Ω0 ) z + 1

¯ ¯ ¯z ¯ > 1

279 and the transform X (z) is ¤ £ z cos (Ω0 ) z 2 − 2 z + cos (Ω0 ) , X (z) = ¢2 ¡ z 2 − 2 cos (Ω0 ) z + 1

d.

¯ ¯ ¯z ¯ > 1

The transform of a sine signal was found in Example 8-16 to be © ª Z sin (Ω0 n) u[n] =

sin (Ω0 ) z , z 2 − 2 cos (Ω0 ) z + 1

¯ ¯ ¯z ¯ > 1

Using the differentiation property of the z-transform along with the linearity property © ª © ª X (z) = Z n sin (Ω0 n) u[n] + Z sin (Ω0 n) u[n] ¸ · d sin (Ω0 ) z sin (Ω0 ) z = −z + 2 , d z z 2 − 2 cos (Ω0 ) z + 1 z − 2 cos (Ω0 ) z + 1

It can be shown that

and the transform X (z) is

¡ ¢ ¸ · − sin (Ω0 ) z 2 − 1 d sin (Ω0 ) z =¡ ¢2 d z z 2 − 2 cos (Ω0 ) z + 1 z 2 − 2 cos (Ω0 ) z + 1

X (z) = ¡ =

¡ ¢ sin (Ω0 ) z z 2 − 1

sin (Ω0 ) z ¢2 + 2 z − 2 cos (Ω0 ) z + 1 z 2 − 2 cos (Ω0 ) z + 1

2 sin (Ω0 ) z 2 [z − cos (Ω0 )] ¢2 , ¡ z 2 − 2 cos (Ω0 ) z + 1

8.11. a. £

¤ x[0] = lim X (z) = lim z→∞

z→∞

b. £ ¤ x[0] = lim X (z) = lim z→∞

c.

z→∞

¯ ¯ ¯z ¯ > 1

·

¸ z2 =1 z2

·

¸ z2 =0 z3

The first step is to write X (z) using non-negative powers of z: X (z) =

z2 + z − 1 z 3 + 0.7 z 2 + 1.2 z − 1.5

Applying the initial value property £

¤ x[0] = lim X (z) = lim z→∞

z→∞

·

¸ z2 =0 z3

¯ ¯ ¯z ¯ > 1

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

280

8.12. a. Using the correlation property, the transform of the autocorrelation function is ¡ ¢ R xx (z) = X (z) X z −1

The transform of x[n] is

X (z) = 1 + z −1 + z −2 Therefore

¡ ¢¡ ¢ R xx (z) = 1 + z −1 + z −2 1 + z + z 2 = z + 2 + z −1

and

r xx [m] = δ[m + 1] + 2 δ[m] + δ[m − 1]

b.

The transform of x[n] is X (z) =

Replacing z with z −1 leads to

¡ ¢ 1 1 − z −N , −1 1−z

¢ ¡ X z −1 =

Therefore

1 (1 − z) , 1−z ¡

R xx (z) = X (z) X z =

−1

¢

R xx (z) can be written as R xx (z) = which leads to the autocorrelation function

¯ ¯ ¯z ¯ < ∞

µ

1 − z −N = 1 − z −1

z N +1 − 2 z + z −N +1 (z − 1)

¯ ¯ ¯z ¯ > 0

2

z (z − 1)2

£

¶µ

1 − zN 1−z

¶

¯ ¯ 0 < ¯z ¯ < ∞

,

z N − 2 + z −N

¤

r xx [m] = (m + N ) u[m + N ] − 2m u[m] + (m − N ) u[m − N ]

8.13. The transforms of the two signals are X (z) = 1 + z −1 + z −2 and Y (z) = 1 + z −1 + z −2 + z −3 + z −4 The transform of the cross correlation function is ¢ ¢¡ ¢ ¡ ¡ R X Y (z) = X (z) Y z −1 = 1 + z −1 + z −2 1 + z + z 2 + z 3 + z 4 = z 4 + 2 z 3 + 3 z 2 + 3 z + 3 + 2 z −1 + z −2

and the cross correlation function is found as r X Y [m] = { 1, 2, 3, 3, 3 , 2, 1 } ↑

m=0

281

8.14. a. The convolution of x[m] and y[m] is x[m] ∗ y[m] =

n=−∞

Replacing y[m] with y[−m] gives x[m] ∗ y[−m] =

b.

∞ X

n=−∞

x[n] y[m − n]

x[n] y[n − m] = r x y [m]

Using correlation property of the z-transform ¡ ¢ R x y (z) = X (z) Y z −1

Since it follows that

© ª Z −1 X (z) = x[m]

and

© ¡ ¢ª Z −1 Y z −1 = y[−m]

© ¡ ¢ª r x y [m] = Z −1 X (z) Y z −1 = x[m] ∗ y[−m]

8.15. a. The transform of X (z) is X (z) = Using the summation property W (z) =

b.

∞ X

z , z −a

¯ ¯ ¯ ¯ ¯ z ¯ > ¯a ¯

z z2 , X (z) = z −1 (z − 1) (z − a)

¯ ¯ ¡ ¯ ¯¢ ¯z ¯ > max 1, ¯a ¯

Partial fraction form of W (z) is W (z) =

with the residues k1 = Inverse transform is

1 , 1−a

k1 z k2 z + z −1 z −a and

k2 =

a a −1

1 a u[n] + a n u[n] 1−a a −1 ¢ 1 ¡ 1 − a n+1 u[n] = 1−a

w [n] =

c.

Using the closed form formula for finite-length geometric series w [n] =

for n ≥ 0.

n X

k=−∞

a k u[k] =

n X

k=0

ak =

1 − a n+1 1−a

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

282

8.16.

Let x[n] = n u[n]

so that w [n] = The transform of x[n] is

n X

k=−∞

x[k] =

X (z) =

n X

k=−∞

k u[k] =

n X

k

k=0

z (z − 1)2

Using the differentiation property of the z-transform W (z) =

z2 z X (z) = z −1 (z − 1)3

Following transform relationships were derived earlier (see Example 8-22) Z

n u[n] ←→ Z

n 2 u[n] ←→

z (z − 1)2 z2 + z

=

z2 − z

(z − 1)3

(z − 1)3

Combining these transform pairs using linearity we get ¡ 2 ¢ Z n + n u[n] ←→

Therefore w [n] =

8.17. a.

2 z2 (z − 1)3

n (n + 1) u[n] 2

1 X (z) k1 k2 = = + z (z + 1) (z + 2) (z + 1) (z + 2)

Next step is to determine the residues:

Partial fraction expansion for X (z) is

¯ 1 ¯¯ =1 k1 = (z + 2) ¯z=−1 ¯ 1 ¯¯ k2 = = −1 (z + 1) ¯z=−2

z z − z +1 z +2 Based on the ROC specified, x[n] is an anti-causal signal. Therefore X (z) =

x[n] = − (−1)n u[−n − 1] + (−2)n u[−n − 1]

283

b. The residues are

k1 z +1 k2 X (z) k3 = = + + z z (z + 1/2) (z + 2/3) z (z + 1/2) (z + 2/3) ¯ ¯ z +1 ¯ =3 k1 = (z + 1/2) (z + 2/3) ¯z=0 ¯ z + 1 ¯¯ k2 = = −6 z (z + 2/3) ¯z=−1/2

and

k3 = Partial fraction expansion for X (z) is

¯ z + 1 ¯¯ =3 z (z + 1/2) ¯z=−2/3

X (z) = 3 −

6z 3z + z + 1/2 z + 2/3

The specified ROC indicates a causal signal, therefore x[n] = 3 δ[n] − 6 (−1/2)n u[n] + 3 (−2/3)n u[n]

c. The residues are

and

z +1 X (z) k1 k2 = = + z (z − 0.4) (z + 0.7) (z − 0.4) (z + 0.7) ¯ ¯ 14 z +1 ¯ = k1 = ¯ (z − 0.4) (z + 0.7) z=0.4 11

k2 = Partial fraction expansion for X (z) is

¯ ¯ 3 z +1 ¯ =− ¯ 11 (z − 0.4) (z + 0.7) z=−0.7

µ ¶ ¶ 3 14 z z − X (z) = 11 (z − 0.4) 11 (z + 0.7) µ

The specified ROC indicates a causal signal, therefore µ ¶ µ ¶ 14 3 x[n] = (0.4)n u[n] − (−0.7)n u[n] 11 11

d. The residues are

z +1 X (z) k1 k2 k3 = = + + z (z + 3/4) (z − 1/2) (z − 3/2) (z + 3/4) (z − 1/2) (z − 3/2) ¯ ¯ z +1 4 ¯ k1 = = (z − 1/2) (z − 3/2) ¯z=−3/4 45 ¯ ¯ 6 z +1 ¯ =− k2 = ¯ + 3/4) − 3/2) 5 (z (z z=1/2

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

284 and

k3 = Partial fraction expansion for X (z) is

¯ ¯ z +1 10 = =¯¯ 9 (z + 3/4) (z − 1/2) z=3/2

X (z) =

4 45

z

(z + 3/4)

−

6 5

z

(z − 1/2)

+

10 9

z

(z − 3/2)

Based on the ROC specified, the terms with poles at z = 1/2 and z = −3/4 correspond to causal signal components whereas the term with pole at z = 3/2 corresponds to an anti-causal signal component: X (z) =

10 6 5z 9 z + − (z + 3/4) (z − 1/2) (z − 3/2) {z } | {z } | 4 45

z

causal

The signal x[n] is x[n] =

µ

anti-causal

µ ¶ µ ¶ ¶ 6 10 4 (−3/4)n u[n] − (1/2)n u[n] − (3/2)n u[−n − 1] 45 5 9

e.

Partial fraction expansion for X (z) is as found in part (d). However, since the ROC is different than that of part (d), the terms with poles at z = −3/4 and z = 3/2 correspond to anti-causal signal components whereas the term with pole at z = 1/2 corresponds to a causal signal component: X (z) =

4 45

z

−

x[n] = −

µ

z

10 9

+

z

(z + 3/4) (z − 1/2) (z − 3/2) | {z } | {z } | {z } anti-causal

The signal x[n] is

6 5

causal

anti-causal

µ ¶ µ ¶ ¶ 6 10 4 (−3/4)n u[−n − 1] − (1/2)n u[n] − (3/2)n u[−n − 1] 45 5 9

8.18. a. Im {z} Unit circle

−2

−1

−1/2

1

2

Re {z}

285

b.

k1 k2 X (z) k3 k4 (z + 1) (z − 2) = = + + + z z (z + 1/2) (z − 1) (z + 2) z (z + 1/2) (z − 1) (z + 2)

Next step is to determine the residues:

¯ ¯ (z + 1) (z − 2) ¯ =2 k1 = (z + 1/2) (z − 1) (z + 2) ¯z=0 ¯ 10 (z + 1) (z − 2) ¯¯ k2 = =− z (z − 1) (z + 2) ¯z=−1/2 9 ¯ 4 (z + 1) (z − 2) ¯¯ k3 = =− z (z + 1/2) (z + 2) ¯z=1 9 ¯ 4 (z + 1) (z − 2) ¯¯ k4 = =− ¯ z (z + 1/2) (z − 1) z=−2 9

Partial fraction expansion for X (z) is

X (z) = 2 −

c.

10 9

z

(z + 1/2)

−

4 9

z

(z − 1)

−

4 9

z

(z + 2)

There are four possible choices for the ROC:

Case 1:

|z| < 1/2 x[n] = 2 δ[n] +

Case 2:

µ

µ ¶ ¶ µ ¶ 10 4 4 u[−n − 1] + (−1/2)n u[−n − 1] + (−2)n u[−n − 1] 9 9 9

1/2 < |z| < 1 x[n] = 2 δ[n] −

Case 3:

1 < |z| < 2

µ

µ ¶ ¶ µ ¶ 4 4 10 u[−n − 1] + (−1/2)n u[n] + (−2)n u[−n − 1] 9 9 9

µ ¶ ¶ µ ¶ 4 10 4 n x[n] = 2 δ[n] − u[n] + (−1/2) u[n] − (−2)n u[−n − 1] 9 9 9

Case 4:

|z| > 2

µ

µ

µ ¶ ¶ µ ¶ 10 4 4 n x[n] = 2 δ[n] − u[n] − (−1/2) u[n] − (−2)n u[n] 9 9 9

8.19. a. Factored form of X (z) is

and

X (z) = ¡

z (z + 3) ¢¡ ¢ z − 0.7 − j 0.6 z − 0.7 + j 0.6

X (z) k1 k2 (z + 3) ¢¡ ¢= =¡ + z z − 0.7 − j 0.6 z − 0.7 + j 0.6 z − 0.7 − j 0.6 z − 0.7 + j 0.6

Residues are found as

¯ ¯ ¢ ¯¯ = 0.5 − j 3.0833 k1 = ¡ z − 0.7 + j 0.6 z=0.7+ j 0.6

(z + 3)

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

286 and

k 2 = k 1∗ = 0.5 + j 3.0833 Partial fraction expansion of X (z) is X (z) =

¡ ¢ 0.5 − j 3.0833 z

z − 0.7 − j 0.6

+

¡ ¢ 0.5 + j 3.0833 z

z − 0.7 + j 0.6

and the inverse transform is ¡ ¢¡ ¢n ¡ ¢¡ ¢n x[n] = 0.5 − j 3.0833 0.7 + j 0.6 u[n] + 0.5 + j 3.0833 0.7 − j 0.6 u[n]

Since

0.5 − j 3.0833 = 3.1236 e − j 1.41

0.7 + j 0.6 = 0.922 e j 0.7086

and

the inverse transform can be written as ´n ´³ ´n ³ ´³ ³ x[n] = 3.1236 e − j 1.41 0.922 e j 0.7086 u[n] + 3.1236 e j 1.41 0.922 e − j 0.7086 u[n] = 6.2472 (0.922)n cos (0.7086n − 1.41) u[n]

b.

and

Factored form of X (z) is X (z) = ¡

z2 ¢¡ ¢ z − 0.8 − j 0.6 z − 0.8 + j 0.6

X (z) k1 z k2 ¢¡ ¢= =¡ + z z − 0.8 − j 0.6 z − 0.8 + j 0.6 z − 0.8 − j 0.6 z − 0.8 + j 0.6

Residues are found as

k1 = ¡

and

¯ ¯ 2 1 ¢ ¯¯ = −j 3 z − 0.8 + j 0.6 z=0.8+ j 0.6 2

z

k 2 = k 1∗ =

1 2 +j 2 3

Partial fraction expansion of X (z) is X (z) =

¡1

2

−j

2 3

¢

z

z − 0.8 − j 0.6

+

¡1 2

+j

2 3

¢

z

z − 0.8 + j 0.6

and the inverse transform is ¶ µ ¶ µ ¢n ¢n 1 2 ¡ 2 ¡ 1 0.8 + j 0.6 u[n] + + j 0.8 − j 0.6 u[n] −j x[n] = 2 3 2 3 Since

2 5 1 − j = e − j 0.9273 and 0.8 + j 0.6 = e j 0.6436 2 3 6 the inverse transform can be written as µ ¶ µ ¶ 5 − j 0.9273 j 0.7086n 5 x[n] = e e u[n] + e j 0.9273 e − j 0.7086n u[n] 6 6 =

5 cos (0.9273n − 0.7086) u[n] 3

287

c.

Factored form of X (z) is X (z) = ³

and

Residues are found as

z−

p 3 2

z (z + 3) ´³ ´ p − 21 z − 23 + j 12

X (z) z +3 k2 k1 ´³ ´= p =³ + p p p 3 3 3 1 1 1 z z− 2 −2 z− 2 +j 2 z − 2 − 2 z − 23 + j ¯ ¯ ¯ ´ ¯¯ k1 = ³ p z − 23 + j 12 ¯

z +3

and

p z= 23 + j 12

1 2

= 0.5 − j 3.866

k 2 = k 1∗ = 0.5 + j 3.866 Partial fraction expansion of X (z) is X (z) =

¡

¢ 0.5 − j 3.866 z

z−

p

3 2

−j

1 2

+

¡ ¢ 0.5 + j 3.866 z

z−

p

3 2

+j

1 2

and the inverse transform is ¡

¢ x[n] = 0.5 − j 3.866

Ãp !n !n Ãp ¡ ¢ 1 1 3 3 +j −j u[n] + 0.5 + j 3.866 u[n] 2 2 2 2

Since 0.5 − j 3.866 = 3.8982 e

− j 1.4422

and

p

1 3 + j = e j 0.5236 2 2

the inverse transform can be written as ´ ³ ´ ³ x[n] = 3.8982 e − j 1.4422 e j 0.5236n u[n] + 3.8982 e j 1.4422 e − j 0.5236n u[n] = 7.7964 cos (0.5236n − 1.4422) u[n]

8.20. a. Factored form of X (z) is X (z) = and

Residues are found as

(z + 1) (z + 2) (z − 1)2

k 2,2 X (z) (z + 1) (z + 2) k 1 k 2,1 = + + = 2 z z z − 1 (z − 1)2 z (z − 1) ¯ (z + 1) (z + 2) ¯¯ k1 = =2 ¯ (z − 1)2 z=0 ¯ (z + 1) (z + 2) ¯¯ k 2,2 = =6 ¯ z z=1

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

288 and

¯ · ¸¯ z 2 − 2 ¯¯ d (z + 1) (z + 2) ¯¯ = −1 = k 2,1 = ¯ dz z z 2 ¯z=1 z=1

Partial fraction expansion of X (z) is

X (z) = 2 −

b.

z 6z + z − 1 (z − 1)2

Dividing X (z) by z yields k 1,2 k 1,1 z +1 k2 X (z) = + = + 2 2 z z + 0.9 z −1 (z + 0.9) (z − 1) (z + 0.9)

Residues are found as ¯ z + 1 ¯¯ = −0.0526 z − 1 ¯z=−0.9 ¯ · ¸¯ d z + 1 ¯¯ −2 ¯¯ k 1,1 = = = −0.5540 d z z − 1 ¯z=−0.9 (z − 1)2 ¯z=−0.9

k 1,2 =

and

k2 = Partial fraction expansion of X (z) is

X (z) = −

c.

(z

¯ ¯ ¯ 2¯ + 0.9)

z +1

z=1

= 0.5540

0.0526 z 0.5540 z 0.5540 z − + 2 z + 0.9 z −1 (z + 0.9)

Dividing X (z) by z yields k 2,1 k 2,2 k 3,2 k 3,1 X (z) k1 z2 + 4 z − 7 = = + + + + 2 2 2 z z z + 0.9 (z + 0.9) z − 1.2 (z − 1.2)2 z (z + 0.9) (z − 1.2)

Residues are found as k1 =

¯ ¯ ¯ 2¯ − 1.2)

z2 + 4 z − 7

(z + 0.9)2 (z

z=0

= −6.0014

¯ z 2 + 4 z − 7 ¯¯ = 2.4666 z (z − 1.2)2 ¯z=−0.9 ¯ z 2 + 4 z − 7 ¯¯ = −0.1436 k 3,2 = z (z + 0.9)2 ¯z=1.2

k 2,2 =

¯ · ¸¯ d z 2 + 4 z − 7 ¯¯ −z 4 − 8 z 3 + 32.04 z 2 − 33.6 z + 10.08 ¯¯ k 2,1 = = 4.5355 = ¯ d z z (z − 1.2)2 ¯z=−0.9 z 2 (z − 1.2)4 z=−0.9 ¯ · ¸¯ −z 4 − 8 z 3 + 14.61 z 2 + 25.2 z + 5.67 ¯¯ d z 2 + 4 z − 7 ¯¯ = 1.4658 = k 3,1 = ¯ d z z (z + 0.9)2 ¯z=1.2 z 2 (z + 0.9)4 z=1.2

Partial fraction expansion of X (z) is

289 X (z) = −6.0014 +

2.4666 z 0.1436 z 4.5355 z 1.4658 z + − + 2 z + 0.9 z − 1.2 (z + 0.9) (z − 1.2)2

8.21. a. Taking the z-transform of the difference equation Y (z) = (1 + c) z −1 Y (z) − X (z) and H (z) =

b. X (z) = −A + B z −1

c.

−z z − (1 + c)

³ z ´ B −A z + A + B = −A + = z −1 z −1 z −1 µ

d.

¶ A +B Az z− A Y (z) = H (z) X (z) = (z − 1 − c) (z − 1)

Partial fraction expansion for Y (z) is found through µ

¶ A +B A z− k1 k2 Y (z) A = = + z (z − 1 − c) (z − 1) z − 1 − c z − 1

The residues are k1 = A − The solution is

B , c

and

k2 =

µ

B c

¶ B B y[n] = A − (1 + c)n u[n] + u[n] c c

Setting y[N ] = 0 we obtain B=

8.22. a. X (z) =

A c (1 + c)N

(1 + c)N − 1

z2 + 3 z z 2 − 1.4 z + 0.85

290

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

Long division is carried out as follows: 1 + 4.4z −1 + 5.31z −2 + 3.694z −3 + 0.6581z −4 z 2 −1.4z +0.85 | z 2 2

+3z

z −1.4z +0.85 4.4z −0.85

4.4z −6.16 +3.74z −1 5.31 −3.74z −1 5.31 −7.434z −1

+4.5135z −2

3.694z −1

−4.5135z −2

3.694z −1

−5.1716z −2 +3.1399z −3 0.6581z −2 −3.1399z −3

The inverse transform x[n] is x[n] = { 1 , 4.4, 5.31, 3.694, 0.6581, −2.2186; −3.6654, . . . } ↑

n=0

b. X (z) = Long division is carried out as follows:

z2 z 2 − 1.6 z + 1

1 + 1.6z −1 + 1.56z −2 + 0.896z −3 − 0.1264z −4 z 2 −1.6z +1 | z 2

z 2 −1.6z

+1

1.6z

−1

1.6z −2.56

+1.6z −1

1.56

−1.6z −1

1.56 −2.496z −1

+1.56z −2

0.896z −1

−1.56z −2

0.896z −1

−1.4336z −2 +0.896z −3 −0.1264z −2 −0.896z −3

The inverse transform x[n] is x[n] = { 1 , 1.6, 1.56, 0.896, −0.1264, −1.0982; −1.6308, . . . } ↑

n=0

c. X (z) =

z2 + 3 z z 2 − 1.7321 z + 1

291 Long division is carried out as follows: 1 + 4.7321z −1 + 7.1962z −2 + 7.7324z −3 + 6.1971z −4 z 2 −1.7321z +1 | z 2

+3z

z 2 −1.7321z

+1

4.7321z

−1

4.7321z −8.1962 +4.7321z −1 7.1962 −4.7321z −1

7.1962 −12.4645z −1

+7.1962z −2

7.7324z −1

−7.1962z −2

7.7324z −1 −13.3933z −2 +7.7324z −3 6.1971z −2 −7.7324z −3 The inverse transform x[n] is x[n] = { 1 , 4.7321, 7.1962, 7.7324, 6.1971, 3, −1, . . . } ↑

n=0

8.23.

There are four possible choices for the ROC:

Case 1:

|z| < 1/2

z2 − z − 2 −2 − z + z 2 = z 3 + 1.5 z 2 − 1.5 z − 1 −1 − 1.5 z + 1.5 z 2 + z 3 Long division is carried out as follows: X (z) =

2 − 2z − 5z 2 − 8.5z 3 + 18.25z 4 +z 2

− 1 −1.5z +1.5z 2 + z 3 | −2 −z

−2 −3z +3z 2

+2z 3

2z −2z 2

−2z 3

−3z 3

−2z 4

−5z 2

+z 3

+2z 4

2z +3z 2

−5z 2 −7.5z 3

+7.5z 4

5z 5

8.5z 3

−5.5z 4

−5z 5

8.5z 3 +12.75z 4 −12.75z 5 −8.5z 6 −18.25z 4 +7.75z 5 +8.5z 6 The inverse transform x[n] is x[n] = { . . . , −35.125, 18.25, −8.5, 5, −2, 2 , 0, 0, . . . } ↑

n=0

Case 2:

1/2 < |z| < 1 X (z) = 2 −

10 9

z

z+

1 2

−

4 9

z

z −1

−

4 9

z

z +2

= X L (z) + X R (z)

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

292

10 9

X R (z) = 2 − X L (z) = −

4 9

z

−

z

z+

4 9

z

z −1 z +2 For X R (z), long division is carried out as follows:

1 2

=

=

0.8889 z + 1 z + 0.5

−0.4444 z − 0.8889 z 2 −2 + z + z 2

0.8889 + 0.5556z −1 − 0.2778z −2 + 0.1389z −3 − 0.0694z −4 z +0.5 | 0.8889z

+1

0.8889z +0.4444

0.5556 0.5556 +0.2778z −1 −0.2778z −1

−0.2778z −1 −0.1389z −2 0.1389z −2 0.1389z −2 +0.0694z −3 −0.0694z −3

For X L (z), long division is carried out as follows: 0.2222z + 0.5556z 2 + 0.3889z 3 + 0.4722z 4 + 0.4306z 5 − 2 +z +z 2 | −0.4444z −0.8889z 2

−0.4444z +0.2222z 2 +0.2222z 3 −1.3333z 2 −0.2222z 3

−1.3333z 2 +0.5556z 3 +0.5556z 4 −0.7778z 3 −0.5556z 4

−0.7778z 3 +0.3889z 4 +0.3889z 5 −0.9445z 4 −0.3889z 5

−0.9445z 4 +0.4722z 5 +0.4722z 6 −0.8611z 5 −0.4722z 6

The inverse transform x[n] is x[n] = { . . . , 0.4306, 0.4722, 0.3889, 0.5556, 0.2222, 0.8889, 0.5556, −0.2778, 0.1389, −0.0694, . . . } ↑

n=0

Case 3:

1 < |z| < 2 X (z) = 2 − X R (z) = 2 −

10 9

z

z+ 10 9

1 2

z

z+

1 2

− −

4 9

z

z −1 4 9

z

z −1

− =

4 9

z

z +2

= X L (z) + X R (z)

0.4444 z 2 − 0.1111 z − 1 z 2 − 0.5 z − 0.5

293

X L (z) = −

4 9

z

z +2 For X R (z), long division is carried out as follows:

=

−0.4444 z 2+z

0.4444 + 0.1111z −1 − 0.7222z −2 − 0.3056z −3 − 0.5139z −4 z 2 −0.5z −0.5 | 0.4444z 2 −0.1111z

0.4444z 2 −0.2222z −0.2222 0.1111z −0.7778

0.1111z −0.0556 −0.0556z −1 −0.7222 +0.0556z −1

−0.7222 +0.3611z −1

+0.3611z −2

−0.3056z −1

−0.3611z −2

−0.3056z −1

+0.1528z −2 +0.1528z −3

−0.5139z −2 −0.1528z −3 For X L (z), long division is carried out as follows: −0.2222z + 0.1111z 2 − 0.0556z 3 + 0.0278z 4 2 +z | −0.4444z

−0.4444z −0.2222z 2 0.2222z 2 0.2222z 2 +0.1111z 3 −0.1111z 3

−0.1111z 3 −0.0556z 4 0.0556z 4 0.0556z 4 +0.0278z 5 −0.0278z 5 The inverse transform x[n] is x[n] = { . . . , 0.0278, −0.0556, 0.1111, −0.2222, 0.4444, 0.1111, −0.7222, −0.3056, −0.5139 . . . } ↑

n=0

Case 4:

|z| > 2 X (z) =

z2 − z − 2 z 3 + 1.5 z 2 − 1.5 z − 1

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

294

Long division is carried out as follows: z −1 − 2.5z −2 + 3.25z −3 − 7.625z −4 + 13.8125z −5 z 3 +1.5z 2 −1.5z − 1 | z 2

−z

2

−2

z +1.5z

−1.5

−z −1

−2.5z

−0.5

+z −1

−2.5z −3.75 +3.75z −1

+2.5z −2

3.25 −2.75z −1

−2.5z −2

3.25 +4.875z −1

−4.875z −2

−3.25z −3

−7.625z −1

+2.375z −2

+3.25z −3

−7.625z −1 −11.4375z −2 +11.4375z −3 +7.625z −4 13.8125z −2 −8.1875z −3 −7.625z −4 The inverse transform x[n] is x[n] = { 0 , 1, −2.5, 3.25, −7.625, 13.8125, −28.9063, . . . } ↑

n=0

8.24. a. X (z) = Long division:

z z2 + 3 z + 2

z −1 − 3z −2 + 7z −3 − 15z −4 + 31z −5 z 2 +3z +2 | z

z +3 +2z −1 −3 −2z −1

−3 −9z −1 −6z −2 7z −1 +6z −2

7z −1 +21z −2 +14z −3 15z −2 −14z −3

15z −2 −45z −3 −30z −4 31z −3 +30z −4 The inverse transform x[n] is x[n] = { 0 , 1, −3, 7, −15, 31, −63, . . . } ↑

n=0

b. X (z) =

z +1

z 2 + 1.1667 z + 0.3333

295 Long division: z −1 − 0.1667z −2 − 0.1388z −3 + 0.2175z −4 − 0.2075z −5 z 2 +1.1667z +0.3333 | z

+1

z +1.1667 +0.3333z −1 −0.1667 −0.3333z −1

−0.1667 −0.1945z −1 −0.0556z −2 −0.1388z −1 +0.0556z −2

−0.1388z −1 −0.1619z −2 −0.0463z −3 0.2175z −2 +0.0463z −3

0.2175z −2 +0.2538z −3 +0.0725z −4 0.2075z −3 −0.0725z −4

The inverse transform x[n] is x[n] = { 0 , 1, −0.1667, −0.1388, 0.2175, −0.2075, 0.1696, . . . } ↑

n=0

c. X (z) = Long division:

z2 + z z 2 + 0.3 z − 0.28

1 + 0.7z −1 + 0.07z −2 + 0.175z −3 − 0.0329z −4 z 2 +0.3z −0.28 | z 2 2

+z

z +0.3z −0.28 0.7z +0.28

0.7z +0.21 −0.196z −1 0.07 +0.196z −1

0.07 +0.021z −1 −0.0196z −2 0.175z −1 +0.0196z −2

0.175z −1 +0.0525z −2 −0.049z −3 −0.0329z −2 +0.049z −3 The inverse transform x[n] is x[n] = { 1 , 0.7, 0.07, 0.175, −0.0329, 0.0589, −0.0269, . . . } ↑

n=0

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296

8.25. The partial fraction for X (z) is X (z) =

− 65 z

1 2

z−

+

4 45

z

z+

3 4

+

10 9

z

z − 32

a.

For the specified ROC, the poles at z = 1/2 and z = −3/4 correspond to causal terms, and the pole at z = 3/2 corresponds to an anti-causal term. The causal part of the transform is X R (z) =

− 56 z

z − 21

+

4 45

z

z + 43

=

−1.1111 z 2 − 0.9444 z z 2 + 0.25 z − 0.375

Long division: −1.1111 − 0.6667z −1 − 0.25z −2 − 0.1875z −3 − 0.0329z −4 z 2 +0.25z −0.375 | −0.1111z 2 −0.9444z

−0.1111z 2 −0.2778z +0.4167 −0.6667z +0.4167 −0.6667z +0.1667

+0.25z −1

−0.25

−0.25z −1

−0.25 −0.0625z −1

0.0938z −2

−0.1875z −1 −0.0938z −2

−0.1875z −1 −0.0469z −2 +0.0703z −3 −0.0329z −2 +0.049z −3 The anti-causal part of the transform is X L (z) =

10 9

z

z−

3 2

=

1.1111 z −1.5 + z

Long division: −0.7407z − 0.4938z 2 − 0.3292z 3 − 0.2195z 4 −1.5 +z | 0.1111z

0.1111z −0.7407z 2 0.7407z 2 0.7407z 2 −0.4938z 3 0.4938z 3 0.4938z 3 −0.3292z 4 0.3292z 4 0.3292z 4 −0.2195z 5 0.2195z 5

297 The inverse transform x[n] is x[n] = { . . . , −0.1463, −0.2195, −0.3292, −0.4938, −0.7407, −1.1111, −0.6667, −0.25, −0.1875, −0.0469, −0.0586, . . . } ↑

n=0

b.

For the specified ROC, the pole at z = 1/2 corresponds to a causal term. The poles at and z = −3/4 and z = 3/2 correspond to anti-causal terms. The causal part of the transform is X R (z) =

− 56 z

z−

1 2

=

−1.2 z −0.5 + z

Long division: −1.2 − 0.6z −1 − 0.3z −2 − 0.15z −3 − 0.075z −4 z −0.5 | −1.2z

−1.2z +0.6 −0.6

−0.6 +0.3z −1 −0.3z −1

−0.3z −1 +0.15z −2 −0.15z −2

−0.15z −2 +0.075z −3 −0.075z −3

The anti-causal part of the transform is X L (z) =

4 45

z

z + 34

+

10 9

z

z − 32

=

1.2 z 2 + 0.7 z z 2 − 0.75 z − 1.125

Long division: −0.6222z − 0.6519z 2 − 0.1185z 3 − 0.5004z 4 + 0.2283z 5 − 1.125 −0.75z +z 2 | 0.7z

+1.2z 2

0.7z +0.4667z 2 −0.6222z 3 0.7333z 2 +0.6222z 3

0.7333z 2 +0.4889z 3 −0.6519z 4 0.1333z 3 +0.6519z 4

0.1333z 3 +0.0889z 4 −0.1185z 5 0.5630z 4 +0.1185z 5

0.5630z 4 +0.3753z 5 −0.5004z 6 −0.2568z 5 +0.5004z 6

The inverse transform x[n] is x[n] = { . . . , 0.2283, −0.5004, −0.1185, −0.6519, −0.6222, −1.2, −0.6, −0.3, −0.15, −0.075, −0.0375, . . . } ↑

n=0

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298

8.26. a. Take the z-transform of both sides of the difference equation: Y (z) = 0.9 z −1 Y (z) + X (z) + z −1 X (z) The system function is H (z) = and its partial fraction form is

The corresponding impulse response is

Y (z) 1 − z −1 z −1 = = X (z) 1 − 0.9 z −1 z − 0.9 1 10 9z H (z) = − 9 z − 0.9

h[n] =

b.

10 1 δ[n] − (0.9)n u[n] 9 9

Take the z-transform of both sides of the difference equation: Y (z) = 1.7 z −1 Y (z) − 0.72 z −2 Y (z) + X (z) − 2 z −1 X (z)

The system function is H (z) =

1 − 2 z −1 z (z − 2) Y (z) = = −1 −2 X (z) 1 − 1.7 z + 0.72 z (z − 0.9) (z − 0.8)

H (z) z −2 k2 k1 = + = z (z − 0.9) (z − 0.8) z − 0.9 z − 0.8

The residues are

¯ z − 2 ¯¯ = −11 k1 = z − 0.8 ¯z=0.9

and

¯ z − 2 ¯¯ k2 = = 12 z − 0.9 ¯z=0.8

and the partial fraction form of the system function is H (z) = −

12 z 11 z + z − 0.9 z − 0.8

and the impulse response is found as h[n] = −11 (0.9)n u[n] + 12 (0.8)n u[n]

c.

Take the z-transform of both sides of the difference equation: Y (z) = 1.7 z −1 Y (z) − 0.72 z −2 Y (z) + X (z) + z −1 X (z) + z −2 X (z)

The system function is H (z) =

1 + z −1 + z −2 z2 + z + 1 Y (z) = = X (z) 1 − 1.7 z −1 + 0.72 z −2 (z − 0.9) (z − 0.8)

299 Division of H (z) by z yields H (z) k1 z2 + z + 1 k2 k3 = = + + z z (z − 0.9) (z − 0.8) z z − 0.9 z − 0.8 The residues are

¯ ¯ z2 + z + 1 ¯ = 1.3889 k1 = (z − 0.9) (z − 0.8) ¯z=0 ¯ z 2 + z + 1 ¯¯ = 30.111 k2 = z (z − 0.8) ¯z=0.9

and

¯ z 2 + z + 1 ¯¯ k3 = = −30.5 z (z − 0.9) ¯z=0.8

Partial fraction form of the system function is

H (z) = 1.3889 +

30.111 z 30.5 z − z − 0.9 z − 0.8

The impulse response is found as h[n] = 1.3889 δ[n] + 30.1111 (0.9)n u[n] − 30.5 (0.8)n u[n]

d.

Take the z-transform of both sides of the difference equation: Y (z) = z −1 Y (z) − 0.11 z −2 Y (z) − 0.07 z −3 Y (z) + z −1 X (z)

The system function is H (z) =

z −1 z2 Y (z) = = X (z) 1 − z −1 + 0.11 z −2 + 0.07 z −3 z 3 − z 2 + 0.11 z + 0.07

In factored form the system function is H (z) =

z2 (z − 0.7) (z − 0.5) (z + 0.2)

Dividing H (z) by z we get

The residues are

and

H (z) z = z (z − 0.7) (z − 0.5) (z + 0.2) ¯ ¯ z ¯ k1 = = 2.7222 (z − 0.5) (z + 0.2) ¯z=0.7 ¯ ¯ z ¯ = −1.7857 k2 = (z − 0.7) (z + 0.2) ¯z=0.5 ¯ ¯ z ¯ k3 = = 0.0635 (z − 0.7) (z − 0.5) ¯z=−0.2

Partial fraction form of the system function is H (z) =

2.7222 z 1.7857 z 0.0635 z − + z − 0.7 z − 0.5 z + 0.2

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

300 The impulse response is

h[n] = 2.7222 (0.7)n u[n] − 1.7857 (0.5)n u[n] + 0.0635 (−0.2)n u[n]

8.27. a. Begin by writing H (z) using negative powers of z: H (z) =

z −1 + z −2 Y (z) = X (z) 1 + 5 z −1 + 6 z −2

The relationship between X (z) and Y (z) is ¤ ¤ £ £ Y (z) 1 + 5 z −1 + 6 z −2 = X (z) z −1 + z −2

or, equivalently

Y (z) + 5 z −1 Y (z) + 6 z −2 Y (z) = z −1 X (z) + z −2 X (z) Taking inverse transforms of both sides leads to y[n] + 5 y[n − 1] + 6 y[n − 2] = x[n − 1] + x[n − 2]

b.

Multiply out numerator and denominator factors, and write the result using negative powers of z to obtain 1 − 2 z −1 + z −2 Y (z) = H (z) = 11 −1 3 −2 4 −3 X (z) z − z − z 1+ 30 5 15 Using the same steps as in part (a) ¸ · £ ¤ 11 −1 3 −2 4 −3 Y (z) 1 + = X (z) 1 − 2 z −1 + z −2 z − z − z 30 5 15 Y (z) +

3 4 −3 11 −1 z Y (z) − z −2 Y (z) − z Y (z) = X (z) − 2 z −1 X (z) + z −2 X (z) 30 5 15

and y[n] +

c.

11 3 4 y[n − 1] − y[n − 2] − y[n − 3] = x[n] − 2 x[n − 1] + x[n − 2] 30 5 15

The system function is H (z) =

which leads to

z −1 + z −3 Y (z) = X (z) 1 + 1.2 z −1 − 1.8 z −3

£ ¤ £ ¤ Y (z) 1 + 1.2 z −1 − 1.8 z −3 = X (z) z −1 + z −3

and to the corresponding difference equation

y[n] + 1.2 y[n − 1] − 1.8 y[n − 3] = x[n − 1] + x[n − 3]

301

8.28. a. Take the z-transform of both sides of the difference equation: Y (z) = −0.1 z −1 Y (z) + 0.56 z −2 Y (z) + X (z) − 2 z −1 X (z) The system function is H (z) =

b.

Y (z) 1 − 2 z −1 z (z − 2) = = X (z) 1 + 0.1 z −1 − 0.56 z −2 (z + 0.8) (z − 0.7)

Division of H (z) by z yields H (z) k2 k1 (z − 2) = + = z (z + 0.8) (z − 0.7) z + 0.8 z − 0.7

The residues are found as

¯ z − 2 ¯¯ = 1.8667 k1 = z − 0.7 ¯z=−0.8

and

k2 =

¯ z − 2 ¯¯ = −0.8667 z + 0.8 ¯z=0.7

Partial fraction form of the system function is H (z) =

1.8667 z 0.8667 z − , z + 0.8 z − 0.7

ROC:

and the impulse response is found as

¯ ¯ ¯z ¯ > 0.8

h[n] = 1.8667 (−0.8)n u[n] − 0.8667 (0.7)n u[n]

c.

If x[n] = u[n] then

X (z) =

and the transform of the output signal is Y (z) = H (z) X (z) =

z , z −1

ROC:

¯ ¯ ¯z ¯ > 1

z 2 (z − 2) (z + 0.8) (z − 0.7) (z − 1)

ROC:

Partial fraction form of Y (z) is Y (z) =

¯ ¯ ¯z ¯ > 1

0.8296 z 2.0222 z 1.8519 z + − z + 0.8 z − 0.7 z −1

which leads to the output signal £ ¤ y[n] = 0.8296 (−0.8)n + 2.0222 (0.7)n − 1.8519 u[n]

d.

If x[n] = u[−n] then

X (z) =

−1 , z −1

ROC:

¯ ¯ ¯z ¯ < 1

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

302

and the transform of the output signal is Y (z) = H (z) X (z) =

−z (z − 2) (z + 0.8) (z − 0.7) (z − 1)

¯ ¯ ROC: 0.8 < ¯z ¯ < 1

Partial fraction form of Y (z) is Y (z) =

1.0370 z 2.8889 z 1.8519 z − + z + 0.8 z − 0.7 z −1

which leads to the output signal y[n] = 1.0370 (−0.8)n u[n] − 2.8889 (0.7)n u[n] − 1.8519 u[−n − 1]

8.29. a. Take the z-transform of both sides of the difference equation: 1 1 1 5 Y (z) = − z Y (z) − z 2 Y (z) + z X (z) + z 2 X (z) 6 6 6 6 The system function is H (z) =

b.

Y (z) z (z + 1) , = X (z) (z + 2) (z + 3)

ROC:

¯ ¯ ¯z ¯ < 2

Division of H (z) by z yields k2 k1 H (z) (z + 1) = + = z (z + 2) (z + 3) z + 2 z + 3

The residues are found as

¯ z + 1 ¯¯ k1 = = −1 z + 3 ¯z=−2

and

¯ z + 1 ¯¯ k2 = =2 z + 2 ¯z=−3

Partial fraction form of the system function is H (z) =

2z −z + , z +2 z +3

ROC:

and the impulse response is found as

¯ ¯ ¯z ¯ < 2

h[n] = (−2)n u[−n − 1] − 2 (−3)n u[−n − 1]

c.

If x[n] = u[n] then

X (z) =

and the transform of the output signal is Y (z) = H (z) X (z) =

z , z −1

ROC:

z 2 (z + 1) (z − 1) (z + 2) (z + 3)

¯ ¯ ¯z ¯ > 1 ¯ ¯ ROC: 1 < ¯z ¯ < 2

303 Partial fraction form of Y (z) is Y (z) = which leads to the output signal y[n] =

d.

If x[n] = u[−n] then

z

z −1

+

− 23 z

z +2

+

3 2

z

z +3

2 3 1 u[n] + (−2)n u[−n − 1] − (−3)n u[−n − 1] 6 3 2

X (z) =

and the transform of the output signal is Y (z) = H (z) X (z) =

−1 , z −1

ROC:

¯ ¯ ¯z ¯ < 1

−z (z + 1) (z − 1) (z + 2) (z + 3)

Partial fraction form of Y (z) is Y (z) = which leads to the output signal y[n] =

1 6

− 16 z

z −1

+

− 13 z

z +2

+

1 2

ROC:

¯ ¯ ¯z ¯ < 1

z

z +3

1 1 1 u[−n − 1] + (−2)n u[−n − 1] − (−3)n u[−n − 1] 6 3 2

8.30. a.

The transform of the causal sinusoidal signal is X (z) =

sin (0.01) z sin (0.01) z ¢¡ ¢ =¡ j z 2 − 2 cos (0.01) z + 1 z − e 0.01 z − e − j 0.01

and the transform of the output signal is Y (z) = H (z) X (z) = =

0.04 sin (0.01) z 2 ¡ ¢¡ ¢ (z − 0.96) z − e j 0.01 z − e − j 0.01

0.2264 z 0.4856 e − j 1.8061 z 0.4856 e j 1.8061 z + + z − 0.96 z − e j 0.01 z − e − j 0.01

The output signal is y[n] =0.2264 (0.96)n u[n] + 0.4856 e − j 1.8061 e j 0.01n + 0.4856 e j 1.8061 e − j 0.01n =0.2264 (0.96)n u[n] + 0.9713 cos (0.01n − 1.8061) u[n]

b.

The system function evaluated at ω0 = 0.01 rad/s is ¯ ¯ H (0.01) = H (z)¯ = 0.9445 − j 0.2264 = 0.9713 e − j 0.2353 z=e j 0.01

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

304

and the steady-state response of the system is y ss [n] =0.9713 sin (0.01n − 0.2353) =0.9713 cos (0.01n − 0.2353 − π/2) =0.9713 cos (0.01n − 1.8061)

c.

The difference between the two responses is the transient term y t [n] = 0.2264 (0.96)n u[n]

It can be shown that y t [140] ≈ 0.0007 which is less than 0.1 percent of the amplitude of the steady-state response.

8.31. a. Starting with © ª Z cos (0.2 πn) u[n] =

=

z [z − cos (0.2 π)] z +1

z 2 − 2 cos (0.2 π) z (z − 0.8090)

z 2 − 1.6180 z + 1

and using the multiplication by an exponential property of the z-transform, the system function is obtained as ³ z ´ h³ z ´ i − 0.8090 ª © Z (0.8)n cos (0.2 πn) u[n] = ³ 0.8´2 0.8 ³ z ´ z +1 − 1.6180 0.8 0.8 z (z − 0.6472) = 2 z − 1.2944 z + 0.64

b. Im {z}

The zeros of the system function are at

Unit circle

z 1 = 0 , z 2 = 0.6472 and its poles are at Re {z}

p 1,2 = 0.6472 ± 0.4702 Pole zero plot is shown. Poles are inside the unit circle. Since h[n] is causal, it is also stable.

305

c.

The difference equation for the system is y[n] − 1.2944 y[n − 1] + 0.64 y[n − 2] = x[n] − 0.5178 x[n − 1]

d.

The z-transform of the output signal is Y (z) =H (z) X (z) =

z 2 (z − 0.6472)

z 3 − 2.2944 z 2 + 1.9344 z − 0.64

¡ ¢ ¡ ¢ −0.0104 − j 0.6804 z −0.0104 + j 0.6804 z 1.0209 z + + = z −1 z − 0.6472 − j 0.4702 z − 0.6472 + j 0.4702

=

1.0209 z 0.6804 e − j 1.5861 z 0.6804 e j 1.5861 z + + z −1 z − 0.8 e j 0.6283 z − 0.8 e − j 0.6283

The output signal is ´n ´n ³ ³ y[n] =1.0209 u[n] + 0.6804 e − j 1.5861 0.8 e j 0.6283 u[n] + 0.6804 e j 1.5861 0.8 e j 0.6283 u[n]

=1.0209 u[n] + 1.3608 (0.8)n cos (0.6283n − 1.5861)

8.32. a. X (z) =

Y (z) =

The system function is

z z−

3z

+

,

ROC:

|z| >

1 2

2z

z − 34 ¡ ¢ 5 z z − 13 20 =¡ ¢¡ ¢, z − 12 z − 43 z−

1 2

1 2

ROC:

¡ ¢ 13 Y (z) 5 z − 20 = H (z) = , X (z) z − 34

|z| >

ROC:

3 4

|z| >

3 4

The system is both causal and stable.

b. X (z) =

z z−

1 2

,

ROC:

|z| >

1 2

The output signal may be written as y[n] =

µ ¶ µ ¶n+2 µ ¶ µ ¶n+1 1 1 1 1 u[n + 2] + u[n + 1] 2 2 2 2

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CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

and leads to the transform ! µ ¶ Ã ! µ ¶ Ã z 1 z 1 2 z z + Y (z) = 2 2 z − 21 z − 21 ¡1¢ 2 z (z + 1) 1 = 2 , ROC: < |z| < ∞ 1 2 z−2

The system function is

µ ¶ 1 Y (z) z (z + 1) , = H (z) = X (z) 2

ROC:

|z| < ∞

and the impulse response is h[n] = { 0.5 , 0.5 } ↑

n=−1

The system is stable but not causal.

c.

z , ROC: z −1 · ¸ z 1 −1 Y (z) = z , = 2 (z − 1) (z − 1)2 X (z) =

The system function is

H (z) = The system is causal but not stable.

d.

Y (z) 1 = , X (z) z (z − 1)

|z| > 1 ROC:

ROC:

z −1 0.25 z = , z − 0.8 z − 0.8 z Y (z) = , ROC: z − 0.8

X (z) = 1.25 −

The system function is

H (z) = The system is causal but not stable.

Y (z) z = , X (z) z − 1

|z| > 1

ROC:

ROC:

|z| > 1

Y (z) = H1 (z) E (z) E (z) = X (z) + H2 (z) Y (z) Combining the two relationships we get Y (z) = H1 (z) [X (z) + H2 (z) Y (z)] H (z) =

|z| > 0.8

|z| > 0.8

8.33. a.

Therefore

|z| > 1

H1 (z) Y (z) = X (z) 1 − H1 (z) H2 (z)

307

b. ⇒

H1 (z) =

h 2 [n] = K δ[n − 1]

⇒

H2 (z) = K z −1 =

H1 (z) = H (z) = 1 − H1 (z) H2 (z)

c.

z z −1

h 1 [n] = u[n]

K z

z z z −1 ³ z ´ µK ¶ = z −1−K 1− z −1 z

For a causal system, the ROC of the system function found in part (b) is ¯ ¯ ¯ ¯ ¯ z ¯ > ¯1 + K ¯

ROC:

For the system to be stable, the ROC must include the unit circle of the z-plane. This requires

d.

¯ ¯ ¯1 + K ¯ < 1

−1 < 1 + K < 1

⇒

⇒

−2 < K < 0

With K = 3/2 the system function is z

H (z) = The transform of the unit step response is à Y (z) =

z

z − 52

Partial fraction form of Y (z) is Y (z) = and the unit-step response is

!

z − 25

³ z ´ z2 =¡ ¢ z −1 z − 25 (z − 1)

1.6667 z z−

5 2

−

0.6667 z z −1

·

µ ¶n ¸ 5 y[n] = − 0.6667 + 1.6667 u[n] 2

8.34. Using the z-transform definition we have H (z 0 ) = and

∞ X

n=−∞

h[n] z 0−n

∞ X ¡ ¢ ¡ ¢−n H z 0∗ = h[n] z 0∗ n=−∞

Conjugating both sides of the transform relationship for H (z 0 ) yields £

¤∗ H (z 0 ) =

·

∞ X

n=−∞

h[n] z 0−n

¸∗

=

∞ X

n=−∞

¡ ¢−n (h[n])∗ z 0∗

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

308

Since h[n] is specified to be real, (h[n])∗ = h[n], and therefore £

H (z 0 )

¤∗

=

∞ X

n=−∞

¡ ¢ ¡ ¢−n = H z 0∗ h[n] z 0∗

which implies that ¯ ¯ ¯ ¡ ¢¯ ¯H (z 0 )¯ = ¯ H z ∗ ¯ = H0

and

0

¡

¢

∡ H z 0∗ = −∡ H (z 0 ) = −Θ0

8.35. a.

The input signal can be expressed as x[n] = z 0n with z 0 = 0.8 e j 0.4π

The steady-state response of the system for a complex exponential signal is y[n] = H (z 0 ) x[n] = H (z 0 ) z 0n The system function can be evaluated at z = z 0 to yield

Therefore

¡ ¢2 ¡ ¢ ³ ´ 0.8 e j 0.4π + 3 0.8 e j 0.4π j 0.4π =¡ H (z 0 ) = H 0.8 e = 3.8717 e j 3.0777 ¢2 ¡ ¢ 0.8 e j 0.4π − 1.4 0.8 e j 0.4π + 0.85

y[n] = 3.8717 (0.8)n e j (0.4 πn+3.0777)

b.

Let z 1 = (0.9)n e 0.3π . The input signal can be expressed as x[n] =

1 n 1 ¡ ∗ ¢n z z + 2 1 2 1

The system function can be evaluated at z = z 1 and at z = z 1∗ to yield

and

¡ ¢2 ¡ ¢ ³ ´ 0.9 e j 0.3π + 3 0.9 e j 0.3π j 0.3π = 11.3347 e − j 3.0515 =¡ H (z 1 ) = H 0.9 e ¢2 ¡ ¢ 0.9 e j 0.3π − 1.4 0.9 e j 0.3π + 0.85 ´ ³ ´ ³ ¡ ¢ H z 1∗ = H 0.9 e − j 0.3π = H ∗ 0.9 e j 0.3π = 11.3347 e j 3.0515

¡ ¢ In the last step we recognized that H (z) has only real coefficients and therefore H z 1∗ = H ∗ (z 1 ) (see Problem 8.34). The output signal is µ ¶ ³ ´n µ 1 ¶ ³ ´n 1 − j 3.0515 j 0.3π y[n] = 11.3347 e 0.9 e + 11.3347 e j 3.0515 0.9 e − j 0.3π 2 2

= 11.3347 (0.9)n cos (0.3πn − 3.0515)

309

8.36. a. Im {z} Unit circle

The zero of the system function is at z 1 = 0.4 and its poles are at p1 = 0 ,

Re {z}

p 2,3 = 0.7 ± 0.6

Pole zero plot is shown.

b. Im {z}

¯ ¯ ¯ A1¯ = ¯ ¯ ¯B 1 ¯ = ¯ ¯ ¯B 2 ¯ = ¯ ¯ ¯B 3 ¯ =

−→

B2

p2 −→

B1

π/6

−→

A1

p1

Re {z}

z1

0.6835 ,

∡ A 1 = 0.8206 rad

1,

∡ B 1 = 0.5236 rad

0.1938 ,

∡ B 2 = −0.5421 rad

1.1125 ,

∡ B 3 = 1.4210 rad

The magnitude and the phase of the system function are

−→

B3 p3

and

¯ ¯ ¯ H (π/6)¯ =

0.6835 = 3.1701 (1) (0.1938) (1.1125)

∡ H (π/6) = 0.8206−0.5236+0.5421−1.4210 = −0.5819 rad

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

310

c. Im {z}

¯ ¯ ¯ A1¯ = ¯ ¯ ¯B 1 ¯ = ¯ ¯ ¯B 2 ¯ = ¯ ¯ ¯B 3 ¯ =

−→

B2

p2

−→

B1

π/4

−→

A1

p1

Re {z}

z1

0.7709 ,

∡ A 1 = 1.1611 rad

1,

∡ B 1 = 0.7854 rad

0.1073 ,

∡ B 2 = 1.5045 rad

1.3071 ,

∡ B 3 = 1.5654 rad

The magnitude and the phase of the system function are

−→

B3 p3

and

¯ ¯ ¯ H (π/4)¯ =

0.7709 = 5.4944 (1) (0.1073) (1.3071)

∡ H (π/4) = 1.1611−0.7854−1.5045−1.5654 = −2.6942 rad

8.37. a. The system has zeros at z = ±2 and a pole at z = −1/2. The system function has a second-order

numerator and a first-order denominator, and therefore does not converge at |z| → ∞. The ROC may be one of the following: ¯ ¯ 1 ¯z ¯ < : Neither stable nor causal 2 1 ¯¯ ¯¯ < z < ∞ : Stable, but not causal 2

In the first case the ROC includes neither the unit circle nor infinity. In the second case it includes the unit circle but excludes infinity. Consequently, this cannot be the system function of a system that is both stable and causal.

b. The system has two zeros at z = −1 and poles at z = −1/3 and z = 1/2. The ROC may be one of the following: ¯ ¯ 1 ¯z ¯ < 3 1 ¯¯ ¯¯ 1 < z < 3 2 ¯ ¯ 1 ¯z ¯ > 2

: Neither stable nor causal : Neither stable nor causal : Stable and causal

This system function could correspond to a system that is both stable and causal.

c.

The system has zeros at z = 0 and z = −1. Its poles are at z = −1/2 and z = −3/2. The ROC may be

311 one of the following: ¯ ¯ 1 ¯z ¯ < 2 1 ¯¯ ¯¯ 3 < z < 2 2 ¯ ¯ 3 ¯z ¯ > 2

: Neither stable nor causal : Stable, but not causal : Causal, but not stable

This cannot be the system function of a system that is both stable and causal.

8.38. a. Taking the z-transform of the difference equation leads to Y (z) = 1.5 z −1 Y (z) − 0.54 z −2 Y (z) + X (z) + 3 z −1 X (z) from which the system function can be obtained as H (z) =

Y (z) 1 + 3 z −1 = X (z) 1 − 1.5 z −1 + 0.54 z −2

The factored form of the system function is H (z) =

z (z + 3) (z − 0.6) (z − 0.9)

and the pole-zero diagram is shown below. Im {z} Unit circle

Re {z} −3

0

0.6 0.9

Since the system is specified to be causal, the ROC of the system function is ¯ ¯ ¯z ¯ > 0.9

The ROC includes the unit circle, therefore the system is stable.

b.

Taking the z-transform of the difference equation leads to Y (z) = −0.64 z −2 Y (z) + 2 X (z)

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CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

from which the system function can be obtained as H (z) =

2 2 z2 Y (z) ¡ ¢¡ ¢ = = X (z) 1 + 0.64 z −2 z − j 0.8 z + j 0.8

The pole-zero diagram is shown below.

Im {z} Unit circle j0.8

Two zeros Re {z}

0 −j0.8

Since the system is specified to be causal, the ROC of the system function is ¯ ¯ ¯z ¯ > 0.8 The ROC includes the unit circle, therefore the system is stable.

c.

Taking the z-transform of the difference equation leads to Y (z) = 0.25 z −1 Y (z) − 0.125 z −2 Y (z) − 0.5 z −3 Y (z) + X (z)

from which the system function can be obtained as H (z) = or, using non-negative powers of z

1 Y (z) = −1 X (z) 1 − 0.25 z + 0.125 z −2 + 0.5 z −3

H (z) =

z3 z 3 − 0.25 z 2 + 0.125 z + 0.5

There are three zeros at z = 0. Poles are at z = −0.6718 and z = 0.4609 ± 0.7293. The pole-zero diagram is shown below. Im {z} Unit circle

Three zeros Re {z} −0.67 0

313 Since the system is specified to be causal, the ROC of the system function is ¯ ¯ ¯z ¯ > 0.8627

The ROC includes the unit circle, therefore the system is stable.

d.

Taking the z-transform of the difference equation leads to Y (z) = 0.25 z −1 Y (z) − 0.5 z −2 Y (z) − 0.75 z −3 Y (z) + X (z) + z −1 X (z)

from which the system function can be obtained as H (z) =

1 + z −1 Y (z) = X (z) 1 − 0.25 z −1 + 0.5 z −2 + 0.75 z −3

or, using non-negative powers of z H (z) =

z 2 (z + 1) z 3 − 0.25 z 2 + 0.5 z + 0.75

The factored form of the system function is H (z) =

z 2 (z + 1)

¡

¢¡ ¢ (z + 0.6709) z − 0.4605 − j 0.9518 z − 0.4605 + j 0.9518

and the pole-zero diagram is shown below.

Im {z} Unit circle Two zeros Re {z} −1 −0.67 0

Since the system is specified to be causal, the ROC of the system function is ¯ ¯ ¯z ¯ > 1.0573

The ROC does not include the unit circle, therefore the system is not stable.

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

314

8.39. a. Im {z}

The system function is H (z) =

Unit circle

−z z − (1 + c)

It has a zero at z 1 = 0 and a pole at p 1 = 1 + c. The ROC of the system function is |z| > 1 + c which excludes the unit circle for any positive interest rate c. Therefore, the system is unstable for any positive interest rate c.

b.

1+c

Re {z}

Consider the input signal x[n] = −A δ[n]

which corresponds to borrowing the amount A initially and never making a payment. The transform of the input signal is X (z) = −A, and the transform of the output signal is Y (z) = H (z) X (z) =

Az z − (1 + c)

which corresponds to the output signal y[n] = A (1 + c)n u[n] Alternatively, let x[n] = −A δ[n] + D u[n − 1] The transform of this input signal is

The output transform is

µ ¶ A +D −A z − A X (z) = z −1

¶ A +D Az z− A Y (z) = H (z) X (z) = (z − 1) (z − 1 − c) µ

which can be expressed in partial fraction form as Y (z) = The output signal is

µ

(D/c) z (A − D/c) z + z −1 z −1−c

¶ µ ¶ D D u[n] + A − (1 + c)n u[n] c c

If A − D/c 6= 0, the output signal would grow unbounded even though the input signal is bounded.

315

c.

For the second input signal in part (b), let A−

D =0 c

D = Ac

⇒

This results in the output signal x[n] = A u[n]. From a practical perspective this corresponds to paying exactly the amount of interest each month, and yields a constant monthly balance. The result found is intuitively satisfying.

8.40. Frequency response of the allpass filter is ¯ ¯ H (Ω) = H (z)¯

z=e j Ω

=

e j Ω − r e j Ω0

e j Ω − (1/r ) e j Ω0

Multiply both the numerator and the denominator with the complex conjugate of the denominator: ¯ ¯ H (z)¯

z=e j Ω

=

µ

e j Ω − r e j Ω0

e j Ω − (1/r ) e j Ω0

¶µ

e − j Ω − (1/r ) e − j Ω0

e − j Ω − (1/r ) e − j Ω0

¶

2 − (1/r ) cos (Ω − Ω0 ) − r cos (Ω − Ω0 ) − j (1/r ) sin (Ω − Ω0 ) + j r sin (Ω − Ω0 ) 1 + 1/r 2 − (2/r ) cos (Ω − Ω0 ) ¡ 2 ¢ ¡ 2 ¢ r − 1 sin (Ω − Ω0 ) 2r − r + 1 cos (Ω − Ω0 ) ¢ ¢ =¡ 2 +j ¡ 2 r + 1 /r − 2 cos (Ω − Ω0 ) r + 1 /r − 2 cos (Ω − Ω0 )

=

The phase characteristic is

∡ H (Ω) = tan−1

·

" ¡ # ¢ ¸ 2 r − 1 sin − Ω ) (Ω Im {H (Ω)} 0 ¡ ¢ = tan−1 Re {H (Ω)} 2r − r 2 + 1 cos (Ω − Ω0 )

8.41. a. W (z) 1

X(z)

b

−1

1+2z H (z) = 1 − (1/2) z −1

1/2

z −1 2 b

Y (z)

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

316

b. W (z)

X(z)

−0.8

H (z) =

−1

Y (z)

z −1 1 b

−3

z +z 1 + 0.8 z −1 − 2.2 z −2 + 0.6 z −3

z −1 2.2 b

−0.6

z −1 1 b

c. W (z) 0.5

X(z)

H (z) =

−1

b

−0.4

−2

1+2z −3z +4 2 + 0.8 z −1 + 1.8 z −2 + 3.2 z −3

z −1 1 b

0.5 + z −1 − 1.5 z −2 + 2 = 1 + 0.4 z −1 + 0.9 z −2 + 1.6 z −3

−0.9

z −1

−1.5

b

−1.6

z −1 2 b

8.42. a. H (z) = H1 (z) H2 (z) H1 (z) = H2 (z) =

z +1 1 + z −1 = z + 1/2 1 + (1/2) z −1 z −1 1 = z + 2/3 1 + (2/3) z −1

1

X (z)

Y (z)

b

−1/2

z −1

−2/3

1

z −1 1 b

b

H1 (z)

H2 (z)

Y (z)

317

b. H (z) = H1 (z) H2 (z) H1 (z) = H2 (z) =

z 1 = z − 0.4 1 − 0.4 z −1 z +1 1 + z −1 = z + 0.7 1 + 0.7 z −1

1

X (z)

1

Y (z)

b b

z −1

−0.7

0.4

z −1 1 b

H1 (z)

H2 (z)

c. H (z) = H1 (z) H2 (z) H1 (z) =

z 1 = z + 0.6 1 + 0.6 z −1

z −1 + z −2 z +1 = z 2 − 1.4, z + 0.85 1 − 1.4, z −1 + 0.85 z −2

H2 (z) =

1

X (z)

Y (z)

b

−0.6

z −1

z −1 1.4

1 b

H1 (z)

−0.85

z −1 1 b

H2 (z)

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

318

8.43. a. b

The system function can be written as

−1/2

z −1 3 b

H (z) = H¯ 1 (z) + H¯ 2 (z) =

X(z)

¯ 1 (z) H

3 −2 + z + 1/2 z + 2/3

Y (z)

b

3 z −1 −2 z −1 = + 1 + (1/2) z −1 1 + (2/3) z −1

−2/3

z −1

−2

b

¯ 2 (z) H

b. 1.2727 b

z −1

The system function can be written as H (z) = H¯ 1 (z) + H¯ 2 (z)

0.4 X(z)

Y (z) ¯ 1 (z) H

1.2727 z −0.2727 z + z − 0.4 z + 0.7 1.2727 −0.2727 = + −1 1 − 0.4 z 1 + 0.7 z −1 =

−0.2727 b

−0.7

z −1

¯ 2 (z) H

c.

Write H (z) as H (z) = H¯ 1 (z) + H¯ 2 (z) ¡ ¢ ¡ ¢ −0.0976 − j 1.0447 z −0.0976 + j 1.0447 z 0.1951 z + + = z + 0.6 z − 0.7 − j 0.6 z − 0.7 + j 0.6 = =

0.1951 z −0.1951 z 2 + 1.3902 z + z + 0.6 z 2 − 1.4 z + 0.85

−0.1951 + 1.3902 z −1 0.1951 + 1 + 0.6 z −1 1 − 1.4 z −1 + 0.85 z −2

319 0.1951 b

−0.6

z −1

X(z)

¯ 1 (z) H

Y (z)

−0.1951 b

z −1 1.4

1.3902 b

−0.85

z −1

¯ 2 (z) H

8.44. a.

Using the definition of the unilateral z-transform ∞ © ª X Z u x[n + 1] = x[n + 1] z −n n=0

Let a new variable m be defined as m = n + 1: ∞ © ª X Z u x[n + 1] = x[m] z −m+1 m=1

It is possible to start the summation at m = 0 by adding and subtracting the missing term: ∞ © ª X Z u x[n + 1] = x[m] z −m+1 − z x[0] m=0

=z

∞ X

m=0

x[m] z −m − z x[0]

=z X u (z) − z x[0]

b.

∞ © ª X Z u x[n + 2] = x[n + 2] z −n n=0

Let a new variable m be defined as m = n + 2:

∞ © ª X Z u x[n + 2] = x[m] z −m+2 m=2

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

320

It is possible to start the summation at m = 0 by adding and subtracting the two missing terms: ∞ © ª X x[m] z −m+2 − z 2 x[0] − z x[1] Z u x[n + 2] = m=0

=z 2

∞ X

m=0

x[m] z −m − z 2 x[0] − z x[1]

=z X u (z) − z 2 x[0] − z x[1] 2

c.

∞ © ª X x[n + k] z −n Z u x[n + k] = n=0

Let a new variable m be defined as m = n + k:

∞ © ª X x[m] z −m+k Z u x[n + k] = m=k

Since ∞ X

m=0

x[m] z −m+k = =

k−1 X

x[m] z −m+k +

k−1 X

x[m] z k−m + z k X u (z)

m=0

m=0

∞ X

x[m] z −m+k

m=k

∞ k−1 X © ª X Z u x[n + k] = x[m] z −m+k = z k X u (z) − x[n] z k−n m=k

8.45. a.

n=0

¢ ¢ ¡ ¡ Yu (z) − 1.4 z −1 Yu (z) + y[−1] + 0.85 z −2 Yu (z) + z −1 y[−1] + y[−2] = 0

Using initial values we obtain

Yu (z) =

1.05 − 4.25 z −1 z (1.05 z − 4.25) = 2 −1 −2 1 − 1.4 z + 0.85 z z − 1.4 z + 0.85

The partial fraction form of Yu (z) is ¡ ¢ ¡ ¢ 0.5250 − j 2.9292 z 0.5250 + j 2.9292 z + Yu (z) = z − 0.7 − j 0.6 z − 0.7 + j 0.6 ´n ´n ¡ ¢³ ¡ ¢³ y[n] = 0.5250 + j 2.9292 0.9220 e j 0.7086 u[n] + 0.5250 − j 2.9292 0.9220 e − j 0.7086 u[n]

=1.05 (0.9220)n cos (0.7086n) u[n] + 5.8583 (0.9220) n sin (0.7086n) u[n]

b.

¢ ¢ ¡ ¡ Yu (z) − 1.6 z −1 Yu (z) + y[−1] + 0.64 z −2 Yu (z) + z −1 y[−1] + y[−2] = 0

321 Using initial values we obtain Yu (z) =

5.12 − 1.28 z −1 z (5.12 z − 1.28) z (5.12 z − 1.28) = 2 = −1 −2 1 − 1.6 z + 0.64 z z − 1.6 z + 0.64 (z − 0.8)2

The partial fraction form of Yu (z) is Yu (z) =

2.8160 z (z − 0.8)

2

+

5.12 z z − 0.8

µ

¶ 2.8160 y[n] = n (0.8)n u[n] + 5.12 (0.8)n u[n] 0.8

8.46. a. The transform of the input signal is X u (z) =

z (z − cos (0.2π)) z (z − 0.8090) = 2 z + 1 z − 1.6180 z + 1

z 2 − 2 cos (0.2π)

Using the unilateral z-transform we have ¤ £ Yu (z) − 2 z −1 Yu (z) + 5 = X u (z)

Solving for Yu (z) yields

¡ ¢ 11 z z 2 − 1.5445 z + 0.9091 ¡ ¢ Yu (z) = (z − 2) z 2 − 1.6180 z + 1

The partial fraction form of Yu (z) is Yu (z) =

11.3503 z 0.3765 e − j 2.0548 0.3765 e j 2.0548 + + z −2 z − e j 0.6283 z − e − j 0.6283

and the output signal is y[n] = 11.3503 (2)n u[n] + 0.3765 e − j 2.0548 e j 0.6283 n + 0.3765 e j 2.0548 e − j 0.6283 n = 11.3503 (2)n u[n] + 0.7530 cos (0.6283 n − 2.0548) u[n]

b.

The transform of the input signal is X u (z) =

z z −1

Using the unilateral z-transform we have

Solving for Yu (z) yields

¤ £ Yu (z) + 0.6 z −1 Yu (z) − 3 = X u (z) + z −1 X u (z)

Yu (z) =

z +1 1.8 z 2.8 z (z − 0.2857) X u (z) + = z + 0.6 z + 0.6 (z − 1) (z + 0.6)

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

322

The partial fraction form of Yu (z) is Yu (z) = and the output signal is

c.

1.25 z 1.55 z + z − 1 z + 0.6

¤ £ y[n] = 1.25 + 1.55 (−0.6)n u[n]

The transform of the input signal is X u (z) =

z z −1

Using the unilateral z-transform we have ¤ ¤ £ £ Yu (z) − 0.2 z −1 Yu (z) − 1 − 0.48 z −2 Yu (z) + 2 − z −1 = X u (z)

Solving for Yu (z) yields

z2 0.76 z 2 − 0.48 z + X (z) u z 2 − 0.2 z − 0.48 z 2 − 0.2 z − 0.48 ¡ ¢ z 1.76 z 2 − 1.24 z + 0.48 ¡ ¢ = (z − 1) z 2 − 0.2 z − 0.48

Yu (z) =

The partial fraction form of Yu (z) is

Yu (z) =

3.125 z 2.1943 z 0.8293 z − + z −1 z − 0.8 z + 0.6

and the output signal is ¤ £ y[n] = 3.125 − 2.1943 (0.8)n + 0.8293 (−0.6)n u[n]

8.47. a. y[n] = y[n − 1] + y[n − 2]

b.

and

y[n] − y[n − 1] − y[n − 2] = 0 ,

⇒

y[1] = 1 ,

Using unilateral z-transform yields ¢ ¢ ¡ ¡ Yu (z) − z −1 Yu (z) − y[−1] − z −2 Yu (z) − z −1 y[−1] − y[−2] = 0

1 + z −1 z (z + 1) = 1 − z −1 − z −2 z 2 − z − 1 The transform can be put into partial fraction form as follows: Yu (z) =

Yu (z) =

0.17082 z 1.17802 z − z − 1.618034 z + 0.618034

The sequence can be found as the inverse transform. y[n] = 1.17082 (1.618034) n − 0.17082 (−0.618034) n

y[−2] = 0

323

c.

Using the result found in part (b), y[n + 1] is y[n + 1] = 1.17082 (1.618034) n+1 − 0.17082 (−0.618034) n+1

and the ratio of two consecutive numbers in the sequence is y[n + 1] 1.17082 (1.618034) n+1 − 0.17082 (−0.618034) n+1 = y[n] 1.17082 (1.618034) n − 0.17082 (−0.618034) n For large n, the terms (−0.618034) n+1 and (−0.618034) n approach zero. Therefore we have · ¸ y[n + 1] 1.17082 (1.618034) n+1 ϕ = lim = lim = 1.618034 n→∞ y[n] n→∞ 1.17082 (1.618034) n

8.48. a. Begin with © ª Z u[n] − u[n − 8] =

z8 − 1 z 7 (z − 1)

Using the multiplication by an exponential signal property of the z-transform ³ z ´8 −1 © ª X (z) = Z (0.8)n (u[n] − u[n − 8]) = ³ 0.8 ´ z ´7 ³ z −1 0.8 0.8

The zeros of the transform X (z) are found by solving ³ z ´8 = 1 e j 2πk , 0.8

k = 0, . . . , 7

The zeros are at z k = 1 e j 2πk/8; ,

k = 1, . . . , 7

The zero for k = 1 is canceled by the denominator factor (z − 1) so that there is neither a zero nor a pole at z = 1. The transform has seven poles at z = 0.

b. 1 2 3 4 5 6 7 8 9

c.

% Anonymous function f o r the transform X = @( z ) ( ( z / 0 . 8 ) . ^ 8 − 1 ) . / ( ( ( z / 0 . 8 ) . ^ 7 ) . * ( ( z / 0 . 8 ) − 1 ) ) ; % Create a grid o f z values [ zr , z i ] = meshgrid ( [ − 1 . 5 : 0 . 0 5 : 1 . 5 ] , [ − 1 . 5 : 0 . 0 5 : 1 . 5 ] ) ; z = ( zr+eps)+ j * ( z i +eps ) ; % Avoid d i v i s i o n by 0 % Evaluate the magnitude | X( z ) | Xmag = abs (X( z ) ) ; % Clip the peak Xmag = Xmag. * ( Xmag 10) ;

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

324 1 2 3 4 5 6 7

clf ; shading interp ; colormap copper ; m1 = mesh( zr , zi ,Xmag ) ; xlabel ( ’ Real ( z ) ’ ) ; ylabel ( ’ Imag( z ) ’ ) ; zlabel ( ’ | X( z ) | ’ ) ;

d. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

r = 1; omg = [ 0 : 0 . 0 0 5 : 1 ] * 2 * pi+eps ; t r = r * exp ( j *omg ) ; % Circular t r a j e c t o r y shading interp ; colormap copper ; m1 = mesh( zr , zi ,Xmag ) ; hold on ; p2 = plot3 ( . . . real ( t r ) , imag ( t r ) , zeros ( s i z e ( t r ) ) , . . . % Draw unit c i r c l e real ( t r ) , imag ( t r ) , abs (X( t r ) ) , . . . % Draw | X( z ) | evaluated on the unit c i r c l e [ −1.5 ,1.5] ,[0 ,0] ,[0 ,0] ,... [0 ,0] ,[ −1.5 ,1.5] ,[0 ,0]); set (m1, ’ FaceAlpha ’ , 0 . 4 , ’ EdgeAlpha ’ , 0 . 6 ) ; set ( p2 ( 1 ) , ’ Color ’ , [ 0 . 8 , 0 , 0 ] , ’ Linewidth ’ , 1 . 5 ) ; set ( p2 ( 2 ) , ’ Color ’ , [ 0 , 0 , 1 ] , ’ Linewidth ’ , 1 . 5 ) ; set ( p2 ( 3 ) , ’ Color ’ , [ 0 , 0 , 0 ] ) ; set ( p2 ( 4 ) , ’ Color ’ , [ 0 , 0 , 0 ] ) ; hold o f f ; xlabel ( ’ Real ( z ) ’ ) ; ylabel ( ’ Imag( z ) ’ ) ; zlabel ( ’ | X( z ) | ’ ) ;

8.49. a. In the script in part (b), modify line 2 as follows: 2

b. 2

X = @( z ) ( ( z / 0 . 6 ) . ^ 8 − 1 ) . / ( ( ( z / 0 . 6 ) . ^ 7 ) . * ( ( z / 0 . 6 ) − 1 ) ) ;

In the script in part (b), modify line 2 as follows: X = @( z ) ( ( z / 0 . 4 ) . ^ 8 − 1 ) . / ( ( ( z / 0 . 4 ) . ^ 7 ) . * ( ( z / 0 . 4 ) − 1 ) ) ;

8.50. a. A system object can be obtained as follows: >> s f = zpk ( [ 0 ] , [ 0 . 9 6 ] , 0 . 0 4 , 1 ) Zero / pole / gain :

325 0.04 z −−−−−−−− ( z −0.96) Sampling time : 1

b.

The output signal is computed and graphed with the following statements: >> >> >> >>

n = [0:49]; x = sin ( 0 . 0 1 * n ) ; y = lsim ( s f , x , n ) ; stem (n , y )

8.51. a. The script below computes and displays the impulse response by iterating through the difference equation: % S e t i n i t i a l conditions to zero ynm1 = 0 ; % y [ −1] ynm2 = 0 ; % y [ −2] xn = 1 ; % x [ 0 ] = 1 s i n c e x [ n] i s a unit impulse xnm1 = 0 ; % x[ −1] = 0 s i n c e x [n ] i s a unit impulse out = [ ] ; % Empty v e c t o r to s t a r t for n=0:10 , yn = 1.2944 * ynm1−0.64 * ynm2+xn−0.6472 *xnm1 ; out = [ out , yn ] ; ynm2 = ynm1; ynm1 = yn ; xnm1 = xn ; xn = 0 ; % x [ n ] = 0 f o r n > 0 end ; % Display output s i g n a l n = [0:10]; [ n ’ , out ’ ]

For comparison, the specified impulse response can be computed from its analytical expression and displayed for n = 0, . . . , 10 using the following code: % A n a ly t i c a l r e s u l t f o r the impulse response y = ( 0 . 8 ) . ^ n . * cos ( 0 . 2 * pi * n ) ; [n’ , y ’ ]

b.

The script below computes and displays the unit-step response by iterating through the difference equation: % S e t i n i t i a l conditions to zero ynm1 = 0 ; % y [ −1] ynm2 = 0 ; % y [ −2]

326

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS xn = 1 ; % x [ 0 ] = 1 s i n c e x [ n] i s a unit s t e p xnm1 = 0 ; % x[ −1] = 0 s i n c e x [n ] i s a unit s t e p out = [ ] ; % Empty v e c t o r to s t a r t for n=0:10 , yn = 1.2944 * ynm1−0.64 * ynm2+xn−0.6472 *xnm1 ; out = [ out , yn ] ; ynm2 = ynm1; ynm1 = yn ; xnm1 = xn ; end ; % Display output s i g n a l n = [0:10]; [ n ’ , out ’ ]

For comparison, the unit-step response can be computed from the analytical result found in Problem 8.31 and displayed for n = 0, . . . , 10 using the following code: % Compute and d i s p la y the a n a l y t i c a l s o lu t i o n y = 1. 0209+ 1. 36 08 * ( 0 . 8 ) . ^ n . * cos ( 0. 6283 * n−1.5861); [n’ , y ’ ]

c.

The script below creates a system object and uses it to compute and display the impulse response: sys = t f ( [ 1 , − 0. 6472 , 0] , [ 1 , − 1. 2944 , 0. 64] , − 1) n = [0:10]; h = impulse ( sys , n ) ; % Display the impulse response [n’ ,h]

For the unit-step response, use the following: y = step ( sys , n ) ; % Display the unit−s t e p response [n’ , y ]

8.52. a. The code listed below can be used for computing and graphing the frequency response H (Ω): % Anonymous function f o r H( z ) H = @( z ) ( z . * z+3 * z ) . / ( z . * z −1.4 * z + 0 . 8 5 ) ; % Evaluate H( z ) at z=exp ( j * Omega) Omg = [ − 1 : 0 . 0 0 1 : 1 ] * pi ; s f = H( exp ( j *Omg) ) ; % Graph the magnitude sfmag = abs ( s f ) ; plot (Omg, sfmag ) ; t i t l e ( ’ Magnitude of the system function ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ;

327 ylabel ( ’ Magnitude ’ ) ; grid ; % Graph the phase sfmag = angle ( s f ) ; plot (Omg, sfmag ) ; t i t l e ( ’ Phase of the system function ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Phase ( rad ) ’ ) ; grid ;

The code below graphs the magnitude and the phase of H (z) evaluated for z = 0.8 e j Ω . Critical points |H (z 0 )| and ∡ H (z 0 ) are marked on magnitude and phase graphs. % Evaluate H( z ) at z =0.8 * exp ( j *Omega) s f1 = H( 0 . 8 * exp ( j *Omg) ) ; % Graph the magnitude and the phase sf1mag = abs ( s f1 ) ; sf1phs = angle ( s f1 ) ; z0 = 0 . 8 * exp ( j * 0 . 4 * pi ) ; clf ; subplot ( 2 , 1 , 1 ) ; plot (Omg, sf1mag , 0 . 4 * pi , abs (H( z0 ) ) , ’ r * ’ ) ; t i t l e ( ’H( z ) evaluated at z =0.8 * exp ( j *Omega) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ; subplot ( 2 , 1 , 2 ) ; plot (Omg, sf1phs , 0 . 4 * pi , angle (H( z0 ) ) , ’ r * ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Phase ( rad ) ’ ) ; grid ;

The code¯ below graphs the magnitude and the phase of H (z) evaluated for z = 0.9 e j Ω . Critical points ¡ ¢ ¡ ¢¯ |H (z 1 )|, ¯ H z 1∗ ¯, ∡ H (z 1 ) ∡ H z 1∗ are marked on magnitude and phase graphs. % Evaluate H( z ) at z =0.9 * exp ( j *Omega) s f1 = H( 0 . 9 * exp ( j *Omg) ) ; % Graph the magnitude and the phase sf1mag = abs ( s f1 ) ; sf1phs = angle ( s f1 ) ; z1 = 0 . 9 * exp ( j * 0 . 3 * pi ) ; clf ; subplot ( 2 , 1 , 1 ) ; plot (Omg, sf1mag , [ − 0. 3 * pi , 0 . 3 * pi ] , [ abs (H( conj ( z1 ) ) ) , abs (H( z1 ) ) ] , ’ r * ’ ) ; t i t l e ( ’H( z ) evaluated at z =0.9 * exp ( j *Omega) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ; subplot ( 2 , 1 , 2 ) ; plot (Omg, sf1phs , [ − 0. 3 * pi , 0 . 3 * pi ] , [ angle (H( conj ( z1 ) ) ) , angle (H( z1 ) ) ] , ’ r * ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Phase ( rad ) ’ ) ; grid ;

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS

328

b.

The steady state response for the input signal x[n] = (0.8)n e j 0.4πn is computed and graphed as follows: % Steady−s t a t e response found in part ( a ) o f Problem 8 . 3 5 : n = [0:19]; y = H( z0 ) * ( 0 . 8 ) . ^ n . * exp ( j * 0 . 4 * pi * n ) ; clf ; subplot ( 2 , 1 , 1 ) ; stem (n , real ( y ) ) ; t i t l e ( ’ y [n] ’ ) ; ylabel ( ’ Real part ’ ) ; grid ; subplot ( 2 , 1 , 2 ) ; stem (n , imag ( y ) ) ; xlabel ( ’ Sample index n ’ ) ; ylabel ( ’ Imag . part ’ ) ; grid ;

The steady state response for the input signal x[n] = (0.9)n cos (0.3πn) is computed and graphed as follows: % Steady−s t a t e response found in part ( b ) o f Problem 8 . 3 5 : n = [0:19]; y = 0 . 5 *H( z1 ) * ( 0 . 9 * exp ( j * 0 . 3 * pi ) ) . ^ n + 0 . 5 *H( conj ( z1 ) ) * ( 0 . 9 * exp(− j * 0 . 3 * pi ) ) . ^ n ; clf ; stem (n , real ( y ) ) ; % Take r e a l part as a precaution t i t l e ( ’ y [n] ’ ) ; xlabel ( ’ Sample index n ’ ) ; grid ;

8.53. % Anonymous function f o r a l l p a s s H( z ) with s p e c i f i e d r and theta H = @( r , theta , z ) ( z−r * exp ( j * theta ) ) . / ( z−1/ r * exp ( j * theta ) ) ; % Create v e c t o r z=exp ( j *Omega) Omg = [ − 1 : 0 . 0 0 2 : 1 ] * pi ; z = exp ( j *Omg) ; % Evaluate system function f o r r = 0 . 2 , 0 . 4 , 0 . 6 , 0 . 8 H1 = H( 0 . 2 , 0 , z ) ; H2 = H( 0 . 4 , 0 , z ) ; H3 = H( 0 . 6 , 0 , z ) ; H4 = H( 0 . 8 , 0 , z ) ; plot (Omg, angle (H1) ,Omg, angle (H2) ,Omg, angle (H3) ,Omg, angle (H4 ) ) ; legend ( ’ r =0.2 ’ , ’ r =0.4 ’ , ’ r =0.6 ’ , ’ r =0.8 ’ ) ; t i t l e ( ’ Phase of f i r s t −order a l l p a s s system ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Phase ( rad ) ’ ) ; grid ;

329

8.54. a. % Anonymous function f o r a l l p a s s H( z ) with s p e c i f i e d r and theta H = @( r , theta , z ) ( z−r * exp ( j * theta ) ) . / ( z−1/ r * exp ( j * theta ) ) ; % Create v e c t o r z=exp ( j *Omega) Omg = [ − 1 : 0 . 0 0 2 : 1 ] * pi ; z = exp ( j *Omg) ; % Evaluate system function H1 = H( 0 . 4 , pi /6 , z ) . * H(0.4 , − pi /6 , z ) ; H2 = H( 0 . 6 , pi /6 , z ) . * H(0.6 , − pi /6 , z ) ; H3 = H( 0 . 8 , pi /6 , z ) . * H(0.8 , − pi /6 , z ) ; plot (Omg, angle (H1) ,Omg, angle (H2) ,Omg, angle (H3 ) ) ; legend ( ’ r =0.4 ’ , ’ r =0.6 ’ , ’ r =0.8 ’ ) ; t i t l e ( ’ Phase of second−order a l l p a s s system , \ theta =\ pi /6 ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Phase ( rad ) ’ ) ; grid ;

b. % Evaluate system function H1 = H( 0 . 8 , pi /6 , z ) . * H(0.8 , − pi /6 , z ) ; H2 = H( 0 . 8 , pi /4 , z ) . * H(0.8 , − pi /4 , z ) ; H3 = H( 0 . 8 , pi /3 , z ) . * H(0.8 , − pi /3 , z ) ; plot (Omg, angle (H1) ,Omg, angle (H2) ,Omg, angle (H3 ) ) ; legend ( ’ \ theta =\ pi /6 ’ , ’ \ theta =pi /4 ’ , ’ \ theta=pi /3 ’ ) ; t i t l e ( ’ Phase of second−order a l l p a s s system , r =0.8 ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Phase ( rad ) ’ ) ; grid ;

8.55. a. The first difference equation can be solved iteratively with the following: ynm1 = 5 ; % Placeholder f o r y [ n−1] , i n i t i a l l y s e t to y [ −1] ynm2 = 7 ; % Placeholder f o r y [ n−2] , i n i t i a l l y s e t to y [ −2] y = []; % Empty array ( output stream ) for n=0:10 , yn = 1 . 4 * ynm1−0.85 * ynm2; % y [ n ] = 1. 4 y [ n−1]−0.85 y [ n−2] y = [ y , yn ] ; % Append to output stream ynm2 = ynm1; % Update y [ n−2] f o r next index ynm1 = yn ; % Update y [ n−1] f o r next index end ; [[0:10] ’ , y ’ ] % Tabulate s o lu t i o n

For the second difference equation, use

330

CHAPTER 8. Z-TRANSFORM FOR DISCRETE-TIME SIGNALS AND SYSTEMS ynm1 = 2 ; % Placeholder f o r y [ n−1] , i n i t i a l l y s e t to y [ −1] ynm2 = −3; % Placeholder f o r y [ n−2] , i n i t i a l l y s e t to y [ −2] y = []; % Empty array ( output stream ) for n=0:10 , yn = 1 . 6 * ynm1−0.64 * ynm2; % y [ n ] = 1. 6 y [ n−1]−0.64 y [ n−2] y = [ y , yn ] ; % Append to output stream ynm2 = ynm1; % Update y [ n−2] f o r next index ynm1 = yn ; % Update y [ n−1] f o r next index end ; [[0:10] ’ , y ’ ] % Tabulate s o lu t i o n

b.

The first difference equation can be solved symbolically and then evaluated numerically with the following: syms z n Yz Y1 = 5+z ^( −1) * Yz ; % z−transform o f y [ n−1] Y2 = 7+5 * z^(−1)+z ^( −2) * Yz ; % z−transform o f y [ n−2] Yz = solve ( Yz−1.4 * Y1+0.85 * Y2 , Yz ) % Solve f o r Y ( z ) yn = iztrans ( Yz ) % I n v e r s e z−transform o f Y ( z ) y = eval ( subs ( yn , ’n ’ , [ 0 : 1 0 ] ) ) % Evaluate y [ n ] f o r n = 0 , . . , 1 0

For the second difference equation, use syms z n Yz Y1 = 2+z ^( −1) * Yz ; % z−transform o f y [ n−1] Y2 = −3+2* z^(−1)+z ^( −2) * Yz ; % z−transform o f y [ n−2] Yz = solve ( Yz−1.6 * Y1+0.64 * Y2 , Yz ) % Solve f o r Y ( z ) yn = iztrans ( Yz ) % I n v e r s e z−transform o f Y ( z ) y = eval ( subs ( yn , ’n ’ , [ 0 : 1 0 ] ) ) % Evaluate y [ n ] f o r n = 0 , . . , 1 0

Chapter 9 State-Space Analysis of Systems 9.1. 

and

  −1 0   2 x (t ) + 0 −2 1

2 2  x˙ (t ) = −3 0 1 1 y (t ) =

·

1 0 0 1

1 1

¸

x (t ) +

·

 1 −1  r (t ) 1 ¸

1 0 1 0

r (t )

9.2. a. z (t ) = x (t ) − F r (t )

x (t ) = z (t ) + F r (t )

⇒

Differentiating both sides yields x˙ (t ) = z˙ (t ) + F

r (t ) dt

Substituting this result into the state equation we obtain z˙ (t ) + F

£ ¤ r (t ) d r (t ) = A z (t ) + F r (t ) + B r (t ) + E dt dt

¡ ¢ ¡ ¢ d r (t ) z˙ (t ) = A z (t ) + B + A F r (t ) + E − F dt

Choose F = E to eliminate the term d r (t )/d t so that

b.

¡ ¢ z˙ (t ) = A z (t ) + B + A E r (t )

B+AE =

·

0 1

¸ · −1 + −2

=

·

0 1

¸ · +

The state-space model is x˙ (t ) =

·

−1 3 −2 −2

¸

3 −2

¸·

¸

=

·

x (t ) +

·

1 −5

1 −6

331

1 −5 ¸

2 1

¸

¸

r (t )

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

332

9.3. R1 = 1 Ω

R2 = 1 Ω N1

N2

b

− +

r(t)

+ vC1 (t) −

+

+

1 C1 = F 2

1 C2 = F 3

vC2 (t) −

y(t) −

b

Begin by writing KCL at node “N1”: v C 1 (t ) − r (t ) v C 1 (t ) − v C 2 (t ) d v C 1 (t ) =0 + +C 1 R1 R2 dt Rearranging terms we obtain d v C 1 (t ) R1 + R2 1 1 =− v C 1 (t ) + v C 2 (t ) + r (t ) dt R 1 R 2C 1 R 2C 1 R 2C 1 Similarly, writing the KCL at node “N2” yields d v C 2 (t ) v C 2 (t ) − v C 1 (t ) +C 2 =0 R2 dt which can be rearranged to become 1 1 d v C 2 (t ) = v C 1 (t ) − v C 2 (t ) dt R 2C 2 R 2C 2 Defining the two state variables as x 1 (t ) = v C 1 (t ) and x 2 (t ) = v C 2 (t ) and substituting numerical values of circuit components we get d x 1 (t ) = − 4 x 1 (t ) + 2 x 2 (t ) + 2 r (t ) dt d x 2 (t ) =3 x 1 (t ) − 3 x 2 (t ) dt In matrix form the state equation is x˙ (t ) =

·

−4 2 3 −3

¸

x (t ) +

·

2 0

¸

r (t )

Since y (t ) = x 2 (t ) = v C 2 (t ) the output equation is y (t ) =

£

0 1

¤

x (t )

9.4. The inductor and the capacitor have the same voltage, therefore L

d i L (t ) = v C (t ) dt

⇒

d i L (t ) 1 = v C (t ) dt L

333 Next step is to write the KCL to obtain r (t ) − v C (t ) d v C (t ) = i L (t ) +C R dt which can be rearranged to yield 1 1 1 d v C (t ) = − i L (t ) − v C (t ) + r (t ) dt C RC RC Defining the two state variables as x 1 (t ) = i L (t ) and x 2 (t ) = v C (t ) and substituting numerical values of circuit components we get d x 1 (t ) =1.2 x 2 (t ) dt d x 2 (t ) = − 5 x 1 (t ) − 5 x 2 (t ) + 5 r (t ) dt In matrix form the state equation is x˙ (t ) =

·

0 1.2 −5 −5

¸

x (t ) +

·

0 5

¸

r (t )

Since y (t ) = x 2 (t ) = v C (t ) the output equation is y (t ) =

9.5. a.

£

0 1

¤

x (t )

State variables can be defined as x 1 (t ) =y (t ) x 2 (t ) =

d y (t ) dt

d x 1 (t ) =x 2 (t ) dt

=⇒

Recognizing that d 2 y (t ) d x 2 (t ) = dt2 dt the differential equation can be written as d x 2 (t ) = −3 x 2 (t ) − 2 x 1 (t ) + 2 r (t ) dt In matrix form, the state-space model is x˙ (t ) =

·

0 1 −2 −3 y (t ) =

£

¸

x (t ) +

1 0

¤

·

0 2

x (t )

¸

r (t )

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

334

b.

State variables can be defined as x 1 (t ) =y (t ) x 2 (t ) =

d y (t ) dt

d x 1 (t ) =x 2 (t ) dt

=⇒

Recognizing that d 2 y (t ) d x 2 (t ) = dt2 dt the differential equation can be written as d x 2 (t ) = −4 x 2 (t ) − 3 x 1 (t ) + 3 r (t ) dt In matrix form, the state-space model is x˙ (t ) =

·

0 1 −3 −4 y (t ) =

c.

£

¸

x (t ) +

1 0

¤

·

0 3

¸

r (t )

x (t )

State variables can be defined as x 1 (t ) =y (t ) x 2 (t ) =

d y (t ) dt

=⇒

d x 1 (t ) =x 2 (t ) dt

Recognizing that d 2 y (t ) d x 2 (t ) = dt2 dt the differential equation can be written as d x 2 (t ) = −x 1 (t ) + r (t ) dt In matrix form, the state-space model is x˙ (t ) =

·

0 1 −1 0

y (t ) =

d.

£

¸

x (t ) +

1 0

¤

·

0 1

¸

r (t )

x (t )

State variables can be defined as x 1 (t ) =y (t ) x 2 (t ) =

d y (t ) dt

=⇒

d x 1 (t ) =x 2 (t ) dt

x 3 (t ) =

d 2 y (t ) dt2

=⇒

d x 2 (t ) =x 3 (t ) dt

335 Recognizing that d 3 y (t ) d x 3 (t ) = dt3 dt the differential equation can be written as d x 3 (t ) = −6 x 3 (t ) − 11 x 2 (t ) − 6 x 1 (t ) + r (t ) dt In matrix form, the state-space model is 

0  0 x˙ (t ) = −6

y (t ) =

e.

   0 0 1  x (t ) +  0  r (t ) −6 1

1 0 −11 £

1 0

0

¤

x (t )

Begin by writing the differential equation as d 3 y (t ) d r (t ) d 2 y (t ) d y (t ) − 2 = −6 − 11 − 6 y (t ) + r (t ) 3 2 dt dt dt dt

Let x 1 (t ) =y (t ) x 2 (t ) =

d y (t ) dt

x 3 (t ) =

d 2 y (t ) − 2 r (t ) =⇒ dt2

=⇒

d x 1 (t ) =x 2 (t ) dt d x 2 (t ) =x 3 (t ) + 2 r (t ) dt

It follows that

£ ¤ d x 3 (t ) = −6 x 3 (t ) + 2 r (t ) − 11 x 2 (t ) − 6 x 1 (t ) + r (t ) dt In matrix form, the state-space model is 

   0 1 0 0 0 1  x (t ) +  2  r (t ) x˙ (t ) =  0 −6 −11 −6 −11

y (t ) =

9.6. a. G (s) = Using residue formulas

and

£

1 0

0

¤

x (t )

k2 k1 Y (s) X 1 (s) X 2 (s) + = + = R (s) R (s) R (s) s +1 s +2 ¯ 2 ¯¯ k1 = =2 s + 2 ¯s=−1

¯ 2 ¯¯ = −2 k2 = s + 1 ¯s=−2

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

336 Therefore

X 1 (s) 2 = R (s) s +1

and

−2 X 2 (s) = R (s) s +2

In matrix form

x˙ =

·

d x 1 (t ) = −x 1 (t ) + 2 r (t ) dt

⇒

d x 2 (t ) = −2 x 2 (t ) − 2 r (t ) dt

⇒ −1 0

0 −2

y (t ) =

b. G (s) = Using residue formulas

x˙ =

c. G (s) =

and

Therefore

¸

r (t )

x (t )

·

⇒

d x 1 (t ) = −2 x 1 (t ) − 9.5 r (t ) dt

⇒

d x 2 (t ) = −4 x 2 (t ) − 19.5 r (t ) dt

−2 0 0 −4

¸

y (t ) =

Using residue formulas

¤

2 −2

¯ 10 s + 1 ¯¯ k2 = = −19.5 s + 2 ¯s=−4

X 2 (s) −19.5 = R (s) s +4

In matrix form

1 1

·

¯ 10 s + 1 ¯¯ = −9.5 k1 = s + 4 ¯s=−2

X 1 (s) −9.5 = R (s) s +2

and

£

x (t ) +

Y (s) X 1 (s) X 2 (s) k2 k1 + = + = R (s) R (s) R (s) s +2 s +4

and

Therefore

¸

£

x (t ) + 1 1

· ¤

−9.5 −19.5

¸

r (t )

x (t )

k2 k3 k1 Y (s) X 1 (s) X 2 (s) X 3 (s) + + = + + = R (s) R (s) R (s) R (s) s +1 s +2 s +3 ¯ (s − 1) (s + 4) ¯¯ k1 = = −3 (s + 2) (s + 3) ¯s=−1 ¯ (s − 1) (s + 4) ¯¯ k2 = =6 (s + 1) (s + 3) ¯s=−2 ¯ (s − 1) (s + 4) ¯¯ = −2 k3 = (s + 1) (s + 2) ¯s=−3

X 1 (s) −3 = R (s) s +1

⇒

d x 1 (t ) = −x 1 (t ) − 3 r (t ) dt

337

and

In matrix form

9.7. a.

X 2 (s) 6 = R (s) s +2

⇒

d x 2 (t ) = −2 x 2 (t ) + 6 r (t ) dt

−2 X 3 (s) = R (s) s +3

⇒

d x 3 (t ) = −3 x 3 (t ) − 2 r (t ) dt



   −1 0 0 −3 0  x (t ) +  6  r (t ) x˙ =  0 −2 0 0 −3 −2 £ ¤ y (t ) = 1 1 1 x (t )

Taking the Laplace transform of both sides of the differential equation yields ¡

¢ s 3 + 5 s 2 + 11 s + 15 Y (s) = (3 + 7 s) X (s)

which leads to the system function

G (s) =

Y (s) 3+7s = 3 R (s) s + 5 s 2 + 11 s + 15

Note: We assume zero initial conditions in taking the Laplace transform of the differential equation, consistent with the facts that the system is CTLTI.

b.

A simulation diagram can be constructed for G (s) as shown. 7

r(t)

dx3 (t) dt R

x3 (t)

dt

b

−5

dx2 (t) dt

R

dt

x2 (t) b

dx1 (t) dt

R

dt

x1 (t)

−11 −15

d x 1 (t ) =x 2 (t ) dt d x 2 (t ) =x 3 (t ) dt d x 3 (t ) = − 15 x 1 (t ) − 11 x 2 (t ) − 5 x 3 (t ) + r (t ) dt y (t ) =3 x 1 (t ) + 7 x 2 (t )

b

3

y(t)

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

338 In matrix form we have

and

   0 1 0 0 0 0 1  x (t ) +  0  r (t ) x˙ (t) =  1 −15 −11 −5 

y (t ) =

£

3

¤

7 0

x (t )

9.8. a.

In order to find the transformation matrix P we need to determine the eigenvalues of the state matrix A. Eigenvalues are the solutions of ¯ ¯ ¯λ I − A¯ = 0 · ¸ · ¸ · ¸ 1 0 −4 3 λ + 4 −3 λI−A = λ − = 0 1 −1 0 1 λ ¯ ¯ ¯λ I − A¯ =λ (λ + 4) − (−3) (1)

=λ2 + 4 λ + 3 = 0

Eigenvalues are λ1 = −1 and λ2 = −3. Let the corresponding eigenvectors be v1 =

·

1 b

¸

and ·

−4 3 −1 0

− 4 + 3 b = −1

⇒

A v1 = λ1 v1

⇒

A v2 = λ2 v2

·

⇒

·

a 1

¸·

1 b

¸

¸

=−

·

1 b

¸

·

a 1

¸

b =1 ¸·

3 0

−4 −1

− 4 a + 3 = −3 a

v2 =

a 1

¸

= −3

a=3

⇒

The eigenvectors of the state matrix A are v1 =

·

1 1

¸

and the transformation matrix is P=

b.

and

·

1 3 1 1

v2 =

·

3 1

¸

¸

The similarity transformation z (t ) = P x (t ) transforms the state-space model from x˙ (t ) = A x (t ) + B r (t ) y (t ) = C x (t ) + d r (t )

to ˜ z (t ) + B ˜ r (t ) z˙ (t ) = A

339 ˜ z (t ) + d˜ r (t ) y (t ) = C The coefficient matrices are −1

˜ =P A

AP =

·

¸·

1.5 −0.5

−0.5 0.5

˜ = P−1 B = B

·

−0.5 0.5

˜ = CP = C

£

3

1.5 −0.5 ¤

1

¸·

−4 3 −1 0

·

1 3 1 1

¸

0.5 0.5

¸·

2 1

¸

¸

=

£

·

0.5 0.5

1 3 1 1

=

·

·

=

−1 0 0 −3

¸

¸

¤

4 10

d˜ = d = 2 The equivalent state-space model is z˙ (t ) =

·

¸

−1 0 0 −3

y (t ) =

£

z (t ) + ¤

4 10

¸

r (t )

z (t ) + (2) r (t )

9.9. Let x (t ) = P z (t ). P=

·

0 1 1 0

¸

˜ = P−1 A P = A

·

0 1

¸·

1 0

˜ = P−1 B = B ˜ = CP = C

£

−5

−1

⇒

·

−6 1

−5 0

0 1 1 0

−7

¤

·

P

=

¸·

0 1

¸·

1 0

0 1 1 0

¸ ¸

·

1 0

¸

·

0 1

=

=

£

¸

0 1 1 0 =

−7

·

0 1 −5 −6

¸

¸

−5

¤

d˜ = d = 1

9.10. a. Equations in open form are d x 1 (t ) = −2 x 1 (t ) − 2 x 2 (t ) + r (t ) dt d x 2 (t ) = x 1 (t ) − 5 x 2 (t ) dt y (t ) = 5 x 2 (t )

(P.9.10.1) (P.9.10.2) (P.9.10.3)

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

340

Differentiating Eqn. (P.9.10.3) and using it in conjunction with Eqn. (P.9.10.2) yields d y (t ) d x2 =5 dt dt (P.9.10.4)

=5 x 1 (t ) − 25 x 2 (t ) Differentiating Eqn. (P.9.10.4) one more time we obtain d x 1 (t ) d x 2 (t ) d 2 y (t ) =5 − 25 2 dt dt dt ¤ ¤ £ £ =5 − 2 x 1 (t ) − 2 x 2 (t ) + r (t ) − 25 x 1 (t ) − 5 x 2 (t ) = − 35 x 1 (t ) + 115 x 2 (t ) + 5 r (t )

b.

Solving for x 2 (t ) from Eqn. (P.9.10.3) we get µ ¶ 1 y (t ) x 2 (t ) = 5

Substituting this result into Eqn. (P.9.10.4) leads to µ ¶ 1 d y (t ) x 1 (t ) = + y (t ) 5 dt These two results can be used in Eqn. (P.9.10.5) to produce the differential equation d 2 y (t ) d y (t ) = −7 − 12 y (t ) + 5 r (t ) 2 dt dt or in scaled form

c.

7 d y (t ) 12 1 d 2 y (t ) =− − y (t ) + r (t ) 2 5 dt 5 dt 5

New state variables can be defined as µ ¶ 1 y (t ) z 1 (t ) = 5 µ ¶ 1 d y (t ) z 2 (t ) = 5 dt

Recognizing that

=⇒ z 2 (t ) =

d z 1 (t ) dt

µ ¶ 2 1 d y (t ) d z 2 (t ) = 5 dt2 dt

the differential equation can be written as d z 2 (t ) = −7 z 2 (t ) − 12 z 1 (t ) + r (t ) dt In matrix form, the state-space model is z˙ (t ) =

·

0 1 −12 −7

¸

z (t ) +

·

0 1

¸

r (t )

(P.9.10.5)

341 y (t ) =

£

· −2 1

¸ · ¸ −2 1 x (t ) + r (t ) −5 0

5 0

¤

x (t )

9.11. The original state-space model is x˙ (t ) =

£ ¤ y (t ) = 0 5 x (t )

and the state-space model in phase-variable canonical form is ·

0 z˙ (t ) = −12

¸ · ¸ 1 0 z (t ) + r (t ) −7 1

£ y (t ) = 5

¤ 0 z (t )

Let the new state vector z be defined through the transformation z (t ) = P x (t ) where P is a transformation matrix in the form P=

·

p 11 p 21

p 12 p 22

¸

The first state variable z 1 (t ) can be written as z 1 (t ) = p 11 x 1 (t ) + p 12 x 2 (t ) Let P1 and P2 be the row vectors constructed using the rows of matrix P, that is, P1 = so that

£

p 11

p 12

¤

and

P2 =

£

p 21

p 22

¤

z 1 (t ) = P1 x (t )

(P.9.11.1)

z 2 (t ) = P2 x (t )

(P.9.11.2)

and Differentiating Eqn. (P.9.11.1) we get d z 1 (t ) =P1 x˙ (t ) dt £ ¤ =P1 A x (t ) + B r (t )

=P1 A x (t ) + P1 B r (t )

In the phase-variable canonical form d z 1 (t )/d t should not have a r (t ) term. Therefore we require that P1 B = 0, and d z 1 (t ) (P.9.11.3) = z 2 (t ) = P1 A x (t ) dt

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

342 Similarly, differentiating Eqn. (P.9.11.2) yields

d z 2 (t ) =P2 x˙ (t ) dt £ ¤ =P2 A x (t ) + B r (t )

=P2 A x (t ) + P2 B r (t )

(P.9.11.4)

which leads to the requirement P2 B = 1. It should be noted that d z 2 (t ) /d t can also be obtained by differentiating Eqn. (P.9.11.3) which results in d z 2 (t ) =P1 A x˙ (t ) dt £ ¤ =P1 A A x (t ) + B r (t )

=P1 A2 x (t ) + P1 A B r (t )

(P.9.11.5)

Comparing Eqns. (P.9.11.4) and (P.9.11.5) we conclude that P2 = P1 A and P1 A B = 0. Therefore, the transformation matrix P is in the form " # P1 P= P1 A Furthermore, we have the condition P1 which leads to

£

P1 =

B AB £

0

B=

1 0

¸

and

P1 = and the vector P2 is P2 = The transformation matrix is

£

£

0

0

1

1

¤

¤

P=

0 1

¤ ¤−1

·

−2 −2 1 −5

¸·

1 0

¸

1 −2 0 1

¸

=

£

0

1

¤

−2 −2 1 −5

¸

=

£

1

−5

·

¸

·

·

£

B AB

AB =

The vector P1 is found as

=

¤£

1

Using the original state-space model, we have ·

¤

0 1 1 −5

=

·

−2 1

¸

¤

It can easily be verified that the matrix P converts the original state-space model to phase-variable canonical form: · ¸ · ¸ £ ¤ 0 1 ˜ = C P−1 = 5 0 ˜ = P A P−1 = ˜ = PB = 0 A B C −12 −7 1

343

9.12. a. State equations in open form are d x 1 (t ) = −2 x 1 (t ) − 2 x 2 (t ) + R (s) dt d x 2 (t ) = x 1 (t ) − 5 x 2 (t ) dt and the output equation is y (t ) = 5 x 2 (t ) Taking the Laplace transform of each state equation yields s X 1 (s) = −2 X 1 (s) − 2 X 2 (s) + R (s)

(P.9.12.1)

s X 2 (s) = X 1 (s) − 5 X 2 (s)

(P.9.12.2)

The Laplace transform of the output equation is Y (s) = 5 X 2 (s)

b.

(P.9.12.3)

Solving for X 1 (s) from Eqn. (P.9.12.2) we obtain X 1 (s) = (s + 5) X 2 (s)

Substituting this result into Eqn. (P.9.12.1) leads to s (s + 5) X 2 (s) = −2 (s + 5) X 2 − 2 X 2 + R (s) and therefore

R (s) s 2 + 7 s + 12 Substituting this result into the s-domain output equation given by Eqn. (P.9.12.3) yields the system function 5 Y (s) = 2 G (s) = R (s) s + 7 s + 12 X 2 (s) =

c.

From the system function we get ¡

¢ s 2 + 7 s + 12 Y (s) = 5 R (s)

and the corresponding differential equation is

d y (t ) d 2 y (t ) +7 + 12 y (t ) = 5 r (t ) dt2 dt

9.13. Choosing the state variables as x 1 (t ) = y 1 (t )

and

x 2 (t ) = y 2 (t ) − r 1 (t )

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

344 the state equations become

¤ £ d x 1 (t ) = − 2 x 1 (t ) − x 2 (t ) + r 1 (t ) + r 1 (t ) + r 2 (t ) dt

= − 2 x 1 (t ) − x 2 (t ) + r 2 (t )

and

d x 2 (t ) = x 1 (t ) − 3 x 2 (t ) − 3 r 1 (t ) dt

Output equations are y 1 (t ) = x 1 (t ) y 2 (t ) = x 2 (t ) + r 1 (t ) In matrix form we have x˙ (t) =

·

−2 1

and y (t ) =

·

−1 −3

1 0 0 1

¸ ¸

x (t ) +

x (t ) +

·

·

0 −3 0 1

0 0

1 0

¸

¸

r (t )

r (t )

9.14. a. In order to find the transformation matrix P we need to determine the eigenvalues of the state matrix A. Eigenvalues are the solutions of

¯ ¯ ¯λ I − A¯ = 0 · ¸ · ¸ · 1 0 −1 2 λ+1 λI−A = λ − = 0 1 −3 −6 3

−2 λ+6

¸

¯ ¯ ¯λ I − A¯ = (λ + 1) (λ + 6) − (3) (−2)

=λ2 + 7 λ + 12 = 0

Eigenvalues are λ1 = −3 and λ2 = −4. Let the corresponding eigenvectors be · ¸ · ¸ v 11 v 21 v1 = and v2 = v 12 v 22 A v1 = λ1 v1

⇒

·

−1 −3

− v 11 + 2 v 12 = −3 v 11 A v2 = λ2 v2

⇒

·

− v 21 + 2 v 12 = −4 v 21

−1 −3

2 −6

¸·

⇒

¸

= −3

·

v 11 v 12

¸

v 21 v 22

¸

v 11 = −v 12

⇒ 2 −6

v 11 v 12

¸·

v 21 v 22

¸

= −4

·

v 21 = − (2/3) v 22

Arbitrarily choosing v 12 = 1 and v 22 = 3 the other elements of vectors v1 and v2 can be determined from these relationships, and lead to eigenvectors · ¸ · ¸ −1 −2 v1 = and v2 = 1 3

345 and the transformation matrix is P=

·

−1 −2 1 3

¸

The similarity transformation x (t ) = P z (t ) transforms the state-space model from x˙ (t ) = A x (t ) + B r (t ) y (t ) = C x (t ) + D r (t ) to ˜ z (t ) + B ˜ r (t ) z˙ (t ) = A ˜ z (t ) + D ˜ r (t ) y (t ) = C The coefficient matrices are ˜ = P−1 A P = A

·

−3 −2 1 1

˜ = P−1 B = B ˜ = CP = C

·

·

¸·

¸·

−3 −2 1 1

−1 −3 2 0

2 −6

−1 −3

¸·

˜ =D= D

¸·

1 2 0 −1

¸

−3 −2 1 1

¸

·

0 4

1 −1

¸

−1 −2 1 3

=

·

−3 0

=

·

−3 −4 1 1

¸

=

·

−2 −2

¸

−7 −4

0 −4

¸

¸

The equivalent state-space model is z˙ (t ) =

·

y (t ) =

b.

·

−3 0 −2 −2

0 −4 −7 −4

¸

z (t ) +

¸

·

z (t ) +

·

−3 −4 1 1 0 1 4 −1

¸ ¸

r (t )

r (t )

Using the state equation d z 1 (t ) = −3 z 1 (t ) − 3 r 1 (t ) − 4 r 2 (t ) dt

⇒

s Z1 (s) = − 3 Z1 (s) − 3 R 1 (s) − 4 R 2 (s) Z1 (s) = −

d z 2 (t ) = −4 z 2 (t ) + r 1 (t ) + r 2 (t ) dt

⇒

3 4 R 1 (s) − R 2 (s) s +3 s +3

s Z2 (s) = − 4 Z2 (s) + R 1 (s) + R 2 (s) Z2 (s) =

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

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

346 Using the output equation y 1 (t ) = − 2 z 1 (t ) − 7 z 2 (t ) + r 2 (t )

Y1 (s) = − 2 Z1 (s) − 7 Z2 (s) + R 2 (s) ¸ · ¸ · 4 1 1 3 R 1 (s) − R 2 (s) − 7 R 1 (s) + R 2 (s) + R 2 (s) =−2 − s +3 s +3 s +4 s +4 · · 2 ¸ ¸ s + 11 s + 6 s + 15 = 2 R 1 (s) + 2 R 2 (s) s + 7 s + 12 s + 7 s + 12 y 2 (t ) = − 2 z 1 (t ) − 4 z 2 (t ) + 4 r 1 (t ) − r 2 (t ) Y2 (s) = − 2 Z1 (s) − 4 Z2 (s) + 4 R 1 (s) − R 2 (s) · ¸ · ¸ 4 1 3 1 R 1 (s) − R 2 (s) − 4 R 1 (s) + R 2 (s) + 4 R 1 (s) − R 2 (s) =−2 − s +3 s +3 s +4 s +4 · 2 ¸ ¸ · 2 −s − 3 s + 8 4 s + 30 s + 60 + R R 2 (s) = (s) 1 s 2 + 7 s + 12 s 2 + 7 s + 12

c.



s 2 + 6 s + 15 s 2 + 7 s + 12

  Y (s) =   4 s 2 + 30 s + 60 s 2 + 7 s + 12

 s + 11 s 2 + 7 s + 12    R (s) −s 2 − 3 s + 8 

s 2 + 7 s + 12

9.15. The state variables were found as µ

¶ 7 5 −t 11 −2t 11 −3t x 1 (t ) = + e − e + e u (t ) 6 2 2 6 µ ¶ 7 11 −2t −t x 2 (t ) = + 5 e − e u (t ) 2 2 ¢ ¡ x 3 (t ) = 1 + e −t u (t )

Evaluating each state variable at t = 0 we get

7 5 11 11 + =0 x 1 (0) = + − 6 2 2 6 7 11 x 2 (0) = + 5 − =3 2 2 x 3 (0) =1 + 1 = 2

347 matching the desired initial state vector. Derivatives of state variables are ¶ µ d x 1 (t ) 5 −t 11 −3t −2t u (t ) = − e + 11 e − e dt 2 2 ¢ d x 2 (t ) ¡ = −5 e −t + 11 e −2t u (t ) dt

d x 3 (t ) = − e −t u (t ) dt

It can be shown that d x 1 (t ) = − 3 x 1 (t ) + x 2 (t ) dt d x 2 (t ) = − 2 x 1 (t ) + 5 x 2 (t ) + 2 u (t ) dt d x 3 (t ) = − x 3 (t ) + u (t ) dt

9.16. The state equations in open form are d x 1 (t ) = − 3 x 1 (t ) + 2 x 2 (t ) dt d x 2 (t ) = − x 2 (t ) + r (t ) dt The second equation can be solved with r (t ) = u (t ) and x 2 (0) = 3 to yield x 2 (t ) =e

−t

=3 e

x 2 (0) +

−t

+e

−t

ˆ

ˆ

t

e −(t −τ) u (τ) d τ

0 t

eτ dτ

0

¢ ¡ = 1 + 2 e −t u (t )

Now the first state equation can be written as

¡ ¢ d x 1 (t ) = − 3 x 1 (t ) + 2 1 + 2 e −t u (t ) dt ¢ ¡ = − 3 x 1 (t ) + 2 + 4 e −t u (t )

It can be solved with the initial condition x 1 (0) = −2 to yield x 1 (t ) =e

−3t

x 1 (0) +

= −2e

−3t

+e

ˆ

t 0

−3t

¡ ¢ e −3 (t −τ) 2 + 4 e −τ u (τ) d τ

ˆ

0

t

¡

¢ 2 e 3τ + 4 e 2τ d τ

¶ 14 −3t 2 −t e u (t ) = +2e − 3 3 µ

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

348 The output signal is

y (t ) =x 1 (t ) + x 2 (t ) + r (t ) ¶ µ 14 −3t 8 −t e u (t ) = +4e − 3 3

9.17. The eigenvalues of the state matrix A are found as follows:

λI−A = λ

·

¯ ¯ ¯λ I − A¯ = 0

¸ · ¸ · 1 0 −3 2 λ+3 − = 0 1 0 −1 0 ¯ ¯ ¯λ I − A¯ = (λ + 3) (λ + 1) = 0

−2 λ+1

¸

Eigenvalues are λ1 = −3 and λ2 = −1. Let the corresponding eigenvectors be v1 = A v1 = λ1 v1 A v2 = λ2 v2

⇒

1 a

·

−3 2 0 −1

·

⇒

¸

·

and ¸·

2 −1

−3 0

Thus the transformation matrix is P=

·

v2 = 1 a

¸·

¸

1 1 0 1

= −3

¸

1 b

·

=−

¸

1 b ·

·

¸

1 a 1 b

a=0

⇒

¸

⇒

b=1

¸

The similarity transformation x (t ) = P z (t ) transforms the state-space model from x˙ (t ) = A x (t ) + B r (t ) y (t ) = C x (t ) + d r (t ) to ˜ z (t ) + B ˜ r (t ) z˙ (t ) = A

˜ z (t ) + d˜ r (t ) y (t ) = C The coefficient matrices are ˜ = P−1 A P = A

·

1 0

−1 1

˜ = P−1 B = B ˜ = CP = C

£

¸· ·

1

2 −1

−3 0

1 0 1

¸·

1 0

1 1

¸

0 1

¸

·

−1 1

¸

1 2

¤

−1 1

¸·

·

1 1 0 1

¤

¸

=

=

£

=

·

−3 0 0 −1

¸

349

d˜ = d = 1 The equivalent state-space model is z˙ (t ) =

·

0 −1

−3 0

y (t ) =

£

¸

1

z (t ) + ¤

2

·

−−1 1

¸

r (t )

z (t ) + r (t )

The initial conditions must also be translated to the new state-space model. −1

z (0) = P

x (0) =

·

¸·

1 −1 0 1

−2 3

¸

=

·

−5 3

¸

The two state equations are now decoupled, and can be solved independently of each other. With r (t ) = u (t ) the first state equation is d z 1 (t ) = −3 z 1 (t ) − u (t ) , dt

z 1 (0) = −5

It can be solved for z 1 (t ) as z 1 (t ) =e −3t z 1 (0) − = −5e

−3t

−e

ˆ

t

e −3 (t −τ) u (τ) d τ

0

−3t

ˆ

t

e 3τ d τ

0

µ ¶ 1 14 −3t = − − u (t ) e 3 3

The second state equation is d z 2 (t ) = −z 2 (t ) + u (t ) , dt

z 2 (0) = 3

and can be solved to yield z 2 (0) +

ˆ

=3 e −t + e −t

ˆ

z 2 (t ) =e

The output signal is

−t

t

e −(t −τ) u (τ) d τ

0 t

eτ dτ

0

¡ ¢ = 1 + 2 e −t u (t )

y (t ) =z 1 (t ) + 2 z 2 (t ) + u (t ) ¶ µ 14 −3t 8 −t u (t ) e = +4e − 3 3

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

350

9.18. a. s I−A = s

·

1 0 0 1

¸ · −3 − 0

Resolvent matrix:

State transition matrix:

£ ¤−1 Φ (s) = s I − A =

¸

=

·

1 (s + 1) (s + 3)

·

s +1 0

2 −1

s +3 0

−2 s +1 2 s +3

¸

¸

φ (t ) = L −1 {Φ (s)} The elements of the state transition matrix are: φ11 (t ) = L φ12 (t ) = L −1

½

−1

½

¾ 1 = e −3t u (t ) s +3

½ ¾ ¾ ¢ ¡ 1 2 1 = L −1 = e −t − e −3t u (t ) − s +1 s +3 (s + 1) (s + 3)

φ21 (t ) = L −1 {0} = 0 φ11 (t ) = L

−1

½

Thus, the state transition matrix is φ (t ) =

"

e −3t u (t )

"

e −6

b. x (2) = φ (2) x (0) =

c. x (5) = φ (5) x (0) =

"

¾ 1 = e −t u (t ) s +1

e −t u (t )

0

0

e −15 0

# ¢ ¡ −t e − e −3t u (t )

¡ −2 ¢ #· ¸ ¸ · e − e −6 0.3936 −2 = 0.4060 3 e −2 ¡ −5 ¢ #· ¸ · ¸ e − e −15 −2 0.0202 = 3 0.0202 e −5

9.19. a.

Let the vector X (s) be defined as X (s) =

·

X 1 (s) X 2 (s)

¸

Taking unilateral Laplace transform of both sides of the state equation yields s X (s) − x (0) = A X (s) + B R (s)

(P.9.19.1)

351

b.

Rearranging the terms in Eqn. (P.9.19.1) we get £ ¤ s I − A X (s) = x (0) + B R (s)

Multiplying both sides of this result by [s I − A]−1 leads to £ ¤−1 £ ¤−1 X (s) = s I − A x (0) + s I − A B R (s)

Recall that

L −1 Therefore L −1 and L −1

n£

s I−A

¤−1

n£

n£

s I−A

s I−A

¤−1 o

= φ (t ) = e A t

o x (0) = e A t x (0)

¤−1

(P.9.19.2)

n o n£ o ¤−1 o ∗ L −1 B R (s) B R (s) = L −1 s I − A

=e A t ∗ B r (t ) =

ˆ

t

e A (t −τ) B r (τ) d τ

0

Combining Eqns. (P.9.19.2) and (P.9.19.3) we get x (t ) = e A t x (0) +

ˆ

t

e A (t −τ) B r (τ) d τ

0

c. Y (s) =C X (s) + d R (s) £ ¤−1 £ ¤−1 =C s I − A x (0) + C s I − A B R (s)

d. y (t ) = C e

At

x (0) +

ˆ

t

C e A (t −τ) B r (τ) d τ

0

9.20. The state variables were found as ¶ 5 28 −3t 27 −4t u (t ) + e − e x 1 (t ) = 12 3 4 ¶ µ 1 14 −3t 27 −4t u (t ) + e − e x 2 (t ) = 12 3 4 µ

Evaluating each state variable at t = 0 we get 5 28 27 + − =3 12 3 4 1 14 27 − = −2 x 2 (0) = + 12 3 4 x 1 (0) =

(P.9.19.3)

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

352

matching the desired initial state vector. Derivatives of state variables are ¢ d x 1 (t ) ¡ = −28 e −3t + 27 e −4t u (t ) dt ¢ d x 2 (t ) ¡ = −14 e −3t + 27 e −4t u (t ) dt

It can be shown that

d x 1 (t ) = − 2 x 1 (t ) − 2 x 2 (t ) + u (t ) dt d x 2 (t ) = x 1 (t ) − 5 x 2 (t ) dt

9.21. G (s) = s I−A = s £

The system function is G (s) = =

·

s I−A

£ ¤−1 Y (s) = C s I−A B+d R (s)

1 0 0 1

¸ · −2 − 26

¤−1

1 2 s + 2 s + 26

=

£ ¤ 1 −2 0 2 s + 2 s + 26

−1 0

¸

·

s 26

·

s 26

=

·

s +2 1 −26 s

−1 s +2

−1 s +2

9.22. a. Expressing y[n] as a function of the other terms leads to y[n] = 0.9 y[n − 1] + r [n] Define the state variable x 1 [n] as x 1 [n] = y[n − 1] so that the difference equation can be written as x 1 [n + 1] = 0.9 x[n] + r [n] In matrix form, the state-space model is x[n + 1] =

£

y[n] =

0.9 £

¤

0.9

x[n] + ¤

£

1

¤

x[n] + r [n]

¸

¸·

s 2 + 26 s 2 + 2 s + 26

r [n]

¸

1 0

¸

+1

353

b.

Expressing y[n] as a function of the other terms leads to y[n] = 1.7 y[n − 1] − 0.72 y[n − 2] + 3 r [n]

State variables can be defined as x 1 [n] =y[n − 2] x 2 [n] =y[n − 1] =⇒ x 1 [n + 1] =x 2 [n] Recognizing that y[n] = x 2 [n + 1] the difference equation can be written as x 2 [n + 1] = 1.7 x 2 [n] − 0.72 x 1 [n] + 3 r [n] In matrix form, the state-space model is x[n + 1] =

·

y[n] =

c.

£

0 −0.72

1 1.7

−0.72

¸

1.7

x[n] +

¤

·

0 3

¸

r [n]

x[n] + 3 r [n]

Expressing y[n] as a function of the other terms leads to y[n] = y[n − 1] − 0.11 y[n − 2] − 0.07 y[n − 3] + r [n]

State variables can be defined as x 1 [n] =y[n − 3] x 2 [n] =y[n − 2] =⇒ x 1 [n + 1] =x 2 [n] x 3 [n] =y[n − 1] =⇒ x 2 [n + 1] =x 3 [n] Recognizing that y[n] = x 3 [n + 1] the difference equation can be written as x 3 [n + 1] = x 3 [n] − 0.11 x 2 [n] − 0.07 x 1 [n] + r [n] In matrix form, the state-space model is 

0  0 x[n + 1] = −0.07 y[n] =

£

−0.07

1 0 −0.11 −0.11

   0 0   1 x[n] + 0  r [n] 1 1

1

¤

x[n] + r [n]

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

354

9.23. a. Partial fraction expansion of the system function is H (z) =

Y (z) X 1 (z) X 2 (z) k1 k2 = + = + R (z) R (z) R (z) z + 1/2 z + 1/3

The residues are

¯ z + 1 ¯¯ =3 k1 = z + 2/3 ¯z=−1/2

and

Therefore

and

X 1 (z) 3 = R (z) z + 1/2

¯ z + 1 ¯¯ k2 = −2 z + 1/2 ¯z=−2/3

2 X 2 (z) =− R (z) z + 2/3

In matrix form x[n + 1] =

·

1 x 1 [n + 1] = − x 1 [n] + 3 r [n] 2

⇒

⇒

2 x 2 [n + 1] = − x 2 [n] − 2 r [n] 3

¸ · ¸ 0 3 x[n] + r [n] −2/3 −2 £ ¤ y[n] = 1 1 x[n]

−1/2 0

b.

In this case the numerator order is the same as the denominator order and we will therefore use a slightly different form of partial fraction expansion compared to part (a): H (z) =

Y (z) X 1 (z) X 2 (z) k1 z k2 z = + = + R (z) R (z) R (z) z + 1/2 z + 1/3

The residues are

¯ z + 1 ¯¯ = 1.2727 k1 = z + 0.7 ¯z=0.4

and

k2 = Therefore

and

¯ z + 1 ¯¯ = −0.2727 z − 0.4 ¯z=−0.7

W1 (z) 1.2727 z = R (z) z − 0.4

⇒

w 1 [n + 1] = 0.4 w 1 [n] + 1.2727 r [n + 1]

0.2727 z W2 (z) =− R (z) z + 2/3

⇒

w 2 [n + 1] = −0.7 w 2 [n] − 0.2727 r [n + 1]

The terms w 1 [n] and w 2 [n] can not be selected as the state variables of the system since we do not want the r [n + 1] terms to appear in state equations. Let us substitute n → n − 1 and write the two equations as w 1 [n] =0.4 w 1 [n − 1] + 1.2727 r [n] w 2 [n] = − 0.7 w 2 [n − 1] − 0.2727 r [n]

355 and define x 1 [n] and x 2 [n] as follows: x 1 [n + 1] = w 1[n]

x 2 [n + 1] = w 2 [n]

It follows that x 1 [n + 1] =0.4 x 1 [n] + 1.2727 r [n] x 2 [n + 1] = − 0.7 x 2 [n] − 0.2727 r [n] The output signal is y[n] =w 1 [n] + w 2[n] =x 1 [n + 1] + x 2 [n + 1] =0.4 x 1 [n] + 1.2727 r [n] − 0.7 x 2 [n] − 0.2727 r [n] =0.4 x 1 [n] − 0.7 x 2 [n] + r [n] In matrix form x[n + 1] =

·

0.4 0

y[n] =

c.

£

0 −0.7 0.4

¸

x[n] +

−0.7

¤

·

1.2727 −0.2727

¸

r [n]

x[n] + r [n]

Partial fraction expansion of the system function is H (z) =

The residues are

and

Y (z) X 1 (z) X 2 (z) X 3 (z) k2 k3 k1 + + = + + = R (z) R (z) R (z) R (z) z + 3/4 z − 1/2 z − 3/2 ¯ ¯ z (z + 1) ¯ = −1/15 k1 = (z − 1/2) (z − 3/2) ¯z=−3/4 ¯ ¯ z (z + 1) ¯ k2 = = −3/5 (z + 3/4) (z − 3/2) ¯z=1/2

k3 = Therefore

In matrix form

X 1 (z) −1/15 = R (z) z + 3/4

¯ ¯ z (z + 1) ¯ = 5/3 (z + 3/4) (z − 1/2) ¯z=3/2

⇒

3 1 x 1 [n + 1] = − x 1 [n] − r [n] 4 15

X 2 (z) −3/5 = R (z) z − 1/2

⇒

x 2 [n + 1] =

1 3 x 2 [n] − r [n] 2 5

5/3 X 3 (z) = R (z) z − 3/2

⇒

x 3 [n + 1] =

3 5 x 3 [n] + r [n] 2 3



−3/4 x[n + 1] =  0 0

0 1/2 0

   0 −1/15 0  x[n] +  −3/5  r [n] 3/2 1/3

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

356 y[n] =

£

1 1

1

¤

x[n]

9.24. a. The system function is r[n]

z +1 H (z) = 2 z + (7/6) z + 1/3

y[n] x2 [n + 1] D x2 [n]

−7/6

1

b

x 1 [n + 1] =x 2 [n] x 2 [n + 1] = −

x1 [n + 1]

1 7 x 1 [n] − x 2 [n] + r [n] 3 6

D x1 [n]

−1/3

1

b

y[n] =x 1 [n] + x 2 [n] In matrix form x[n + 1] =

·

b.

¸ · ¸ 1 0 x[n] + r [n] −7/6 1 £ ¤ y[n] = 1 1 x[n]

0 −1/3

The system function is H (z) =

z2 + z z 2 + 0.3 z − 0.28

1

r[n]

b

x2 [n + 1] D

x 1 [n + 1] =x 2 [n]

b

x 2 [n + 1] =0.28 x 1 [n] − 0.3 x 2 [n] + r [n]

x1 [n + 1] D

y[n] =x 2 [n + 1] + x 2 [n] =0.28 x 1 [n] − 0.3 x 2 [n] + r [n] + x 2[n]

0.28

=0.28 x 1 [n] + 0.7 x 2 [n] + r [n] In matrix form

x2 [n]

−0.3

·

¸ · ¸ 0 1 0 x[n + 1] = x[n] + r [n] 0.28 −0.3 1 £ ¤ y[n] = 0.28 0.7 x[n] + r [n]

x1 [n]

1

y[n]

357

c. r[n]

The system function is H (z) =

y[n] x3 [n + 1]

z2 + z z 3 − (5/4) z 2 − (3/4) z + 9/16

D x3 [n]

5/4

1

b

x2 [n + 1]

x 1 [n + 1] =x 2 [n]

D

x 3 [n + 1] = −

x2 [n]

3/4

x 2 [n + 1] =x 3 [n]

1

b

3 5 9 x 1 [n] + x 2 [n] + x 3 [n] + r [n] 16 4 4

x1 [n + 1] D x1 [n]

−9/16

y[n] =x 2 [n] + x 3[n] In matrix form



   1 0 0 0 1  x[n] +  0  r [n] 1 3/4 5/4 £ ¤ y[n] = 0 1 1 x[n]

0 0 x[n + 1] =  −9/16

9.25. a. The system function is H (z) =

z +1 1 + z −1 = −1 1 − 0.9 z z − 0.9

1

r[n]

b

x1 [n + 1]

x 1 [n + 1] =0.9 x 1 [n] + r [n]

D b

=x 1 [n] + 0.9 x 1 [n] + r [n] =1.9 x 1 [n] + r [n] In matrix form

x1 [n]

0.9

y[n] =x 1 [n] + x 1 [n + 1]

¤ £ ¤ 0.9 x[n] + 1 r [n] £ ¤ y[n] = 1.9 x[n] + r [n]

x[n + 1] =

£

1

y[n]

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

358

b. The system function is H (z) =

3 + 2 z −2 3 z2 + 2 = 1 − 1.7 z −1 + 0.72 z −2 z 2 − 1.7 z + 0.72

3

r[n]

b

y[n]

x2 [n + 1] D x2 [n]

x 1 [n + 1] =x 2 [n]

1.7

x 2 [n + 1] = − 0.72 x 1 [n] + 1.7 x 2 [n] + r [n]

x1 [n + 1]

b

D

y[n] =2 x 1 [n] + 3 x 2[n + 1] ¡ ¢ =2 x 1 [n] + 3 − 0.72 x 1 [n] + 1.7 x 2 [n] + r [n]

x1 [n]

−0.72

2

b

= − 0.16 x 1 [n] + 5.1 x 2 [n] + 3 r [n]

In matrix form x[n + 1] =

·

y[n] =

c.

£

0 −0.72 −0.16

1 1.7 5.1

¸

x[n] +

¤

·

0 1

¸

r [n]

x[n] + 3 r [n]

The system function is H (z) =

−1

−2

1−z +z −1 1 − z + 0.11 z −2 + 0.07 z −3

z3 − z2 + z = 3 z − z 2 + 0.11 z + 0.07

1

r[n]

b

x3 [n + 1] D b

x2 [n + 1] D

x 1 [n + 1] =x 2 [n]

x2 [n]

−0.11

x 2 [n + 1] =x 3 [n]

b

x 3 [n + 1] = − 0.07 x 1 [n] − 0.11 x 2 [n] + x 3 [n] + r [n] y[n] =x 2 [n] − x 3 [n] + x 3[n + 1]

x1 [n + 1] D −0.07

= − 0.07 x 1 [n] + 0.89 x 2 [n] + r [n] In matrix form

x3 [n] −1

1



   0 1 0 0    0 0 1 x[n] + 0  r [n] x[n + 1] = −0.07 −0.11 1 1 £ ¤ y[n] = −0.07 0.89 0 x[n] + r [n]

x1 [n]

1

y[n]

359

9.26. x[1] = A x[0] + B r [0] =

·

0 0.25

y[0] = C x[0] = ·

−0.5 0.75

¸·

£

1

3

¤

2 0 ·

¸ · · ¸ ¸ 2 2 + (1) = 1 1.5 ¸ 2 =6 0

¸·

¸ · · ¸ ¸ 2 2 1.25 x[2] = A x[1] + B r [1] = + (1) = 1.5 1 2.625 · ¸ £ ¤ 2 y[1] = C x[1] = 3 1 = 7.5 1.5

0 0.25

·

0 0.25

·

0 0.25

−0.5 0.75

¸·

¸ · · ¸ ¸ 1.25 2 0.6875 x[3] = A x[2] + B r [2] = + (1) = 2.625 1 3.2813 ¸ · £ ¤ 1.25 = 6.375 y[2] = C x[2] = 3 1 2.625

−0.5 0.75

¸·

¸ · · ¸ ¸ 2 0.3594 + (1) = 1 3.6328 · ¸ £ ¤ 0.6875 y[3] = C x[3] = 3 1 = 5.3438 3.2813

x[4] = A x[3] + B r [3] =

9.27. a. Resolvent matrix is found as £ ¤ The first step is to find the matrix z I − A .

z I−A = z

The resolvent matrix is

£

·

z I−A

1 0

0 1

¤−1

0.6875 3.2813

£ ¤−1 Φ (z) = z z I − A

¸ · −0.1 − −0.8

−0.7 0

1 = (z − 0.7) (z + 0.8)

£ ¤−1 Φ (z) = z z I − A =

b.

−0.5 0.75

·

1 (z − 0.7) (z + 0.8)

¸

=

·

z −0.8 ·

z + 0.1 0.8 −0.7 z + 0.1

z2 −0.8 z

0.7 z ¸

−0.7 z z (z + 0.1)

The state transition matrix is φ[n] = [Φ (z)]

Φ11 (z) =

−1

=

·

φ11 [n] φ12 [n] φ21 [n] φ22 [n]

¸

¸

z2 (7/15) z (8/15) z = + z − 0.7 z + 0.8 (z − 0.7) (z + 0.8)

¸

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

360

φ11 [n] =

µ

µ ¶ ¶ 7 8 (0.7)n u[n] + (−0.8)n u[n] 15 15

−0.7 z (7/15) z (7/15) z + =− z − 0.7 z + 0.8 (z − 0.7) (z + 0.8)

Φ12 (z) =

µ

µ ¶ ¶ 7 7 n φ12 [n] = − (0.7) u[n] + (−0.8)n u[n] 15 15

−0.8 z (8/15) z (8/15) z + =− z − 0.7 z + 0.8 (z − 0.7) (z + 0.8)

Φ21 (z) =

µ ¶ ¶ 8 8 n φ21 [n] = − (0.7) u[n] + (−0.8)n u[n] 15 15 µ

Φ22 (z) =

z (z + 1) (8/15) z (7/15) z + = z − 0.7 z + 0.8 (z − 0.7) (z + 0.8) µ

µ ¶ ¶ 8 7 n φ22 [n] = (0.7) u[n] + (−0.8)n u[n] 15 15

c.

Using Eqn. (9.177) we have X (z) = Φ (z) x (0) + z −1 Φ (z) B R (z)

The resolvent matrix was found in part (a). Φ (z) x (0) =

1 (z − 0.7) (z + 0.8)

·

z2 −0.8 z

−0.7 z z (z + 0.1)

¸·

¸

2 0

=

1 (z − 0.7) (z + 0.8)

·

2 z2 −1.6 z

The second term in the expression for X (z) is z

−1

Φ (z) B R (z) = z =

−1

1 − 0.7) (z (z + 0.8)

·

1 (z − 1) (z − 0.7) (z + 0.8)

z2 −0.8 z ·

Thus X (z) is found as 1 X (z) = − 1) − 0.7) (z (z (z + 0.8) and the output transform is Y (z) =

−0.7 z z (z + 0.1)

3 z 2 − 0.7 z z 2 − 2.3 z ·

¸·

3 1

¸

2 z 3 + z 2 − 0.7 z −0.6 z 2 − 0.7 z

¸

¡ ¢ z 4 z 2 + 2.6 z − 0.7

(z − 1) (z − 0.7) (z + 0.8)

The output signal can be found using partial fraction expansion:

y[n] = 10.9259 − 0.0815 (−0.8)n − 6.8444 (0.7)n ,

n≥0

¸³

z ´ z −1

¸

361

9.28. The system function is found as

£ ¤−1 H (z) = C z I − A B £ ¤ The first step is to find the matrix z I − A and its inverse:

z I−A = z £

H (z) =

=

£

·

z I−A

2

1 0

¤−1

0 1

¸ · −0.1 − −0.8

1 = (z − 0.7) (z + 0.8)



z ¤  (z − 0.7) (z + 0.8) −1   −0.8 (z − 0.7) (z + 0.8)

5 z + 0.9 (z − 0.7) (z + 0.8)

¸

−0.7 0

·

=

·

z −0.8

z + 0.1 0.8 −0.7 z + 0.1

0.7 z

¸

¸

 −0.7 · ¸ (z − 0.7) (z + 0.8)   3  1 z + 0.1 (z − 0.7) (z + 0.8)

9.29. The system function must first be written using powers of z −1 : H (z) =

0 + z −1 + 0 z −2 − 7 z −3 + 6 z −4 1 − 0.2 z −1 − 0.93 z −2 + 0.198 z −3 − 0.1296 z −4

It is important to ensure that numerator and denominator polynomials have the same length for the use of the function tf2ss (..) . We also account for missing terms using zero-valued coefficients. The state-space model for the system is found with the statements >> >> >>

num = [ 0 , 1 , 0 , − 7 , 6] ; den = [ 1 , − 0. 2 , − 0. 93 , 0. 198 , 0. 1296] ; [ A , B , C, d ] = t f 2 s s (num, den )

A = 0.2000 1.0000 0 0 B = 1 0 0 0 C =

0.9300 0 1.0000 0

−0.1980 0 0 1.0000

−0.1296 0 0 0

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

362 1

0

−7

6

d = 0

The transformation matrix P to convert this state-space model to an alternative one with a diagonal state matrix is found using the following: >>

[ P , lambda ] = eig (A)

P = 0.4211 −0.4678 0.5198 −0.5776

0.3367 0.4209 0.5262 0.6577

0.1743 0.2904 0.4841 0.8068

−0.0258 0.0859 −0.2862 0.9540

0 0.8000 0 0

0 0 0.6000 0

0 0 0 −0.3000

lambda = −0.9000 0 0 0

˜ for the alternative state-space model are determined using the ˜ B ˜ and C Coefficient matrices A, following: >>

A _ t i l d e = inv (P ) * A * P

A_tilde = −0.9000 0.0000 −0.0000 0.0000 >>

−0.0000 0.8000 0.0000 −0.0000

−0.0000 −0.0000 0.6000 −0.0000

−0.0000 −0.0000 −0.0000 −0.3000

1.6265

7.7014

B _ t i l d e = inv (P ) * B

B_tilde = 1.1316 4.0654 −4.5906 1.7647 >>

C _ t i l d e = C* P

C_tilde = −6.6833

0.5998

363

9.30. Code for the script is listed below: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

% S e t up the s t a t e space model A = [0 , −0.5;0.25 ,0.75]; B = [2;1]; C = [3 ,1]; xn = [ 2 ; 0 ] ; % "xn " r e p r e s e n t s the current s t a t e v e c t o r x [ n ] % Solve i t e r a t i v e l y n = [0:99]; out = [ ] ; % Placeholder f o r the output s i g n a l f o r n = 0 , . . . , 9 9 rn = 1 ; % Unit−s t e p input ; always 1 f o r n>=0 for nn=0:99 , xnp1 = A * xn+B * rn ; % "xnp1" r e p r e s e n t s v e c t o r x [n+1] yn = C* xn ; out = [ out , yn ] ; % Append y [ n ] to the output stream xn = xnp1 ; % Get ready f o r next i t e r a t i o n end ; % Graph the output s i g n a l stem (n , out ) ;

9.31. Set up the state space model: 1 2 3

A = [ − 0. 1 , − 0. 7; − 0. 8 , 0] ; B = [3;1]; C = [ 2 , − 1] ;

Compute the resolvent matrix and the state transition matrix. 1 2 3 4

z = tmp rsm stm

sym( ’ z ’ ) ; = z * eye(2) −A ; = z * inv (tmp) = iztrans (rsm)

% Resolvent matrix % S t a t e t r a n s i t i o n matrix

Find X (z), the z-transform of the state vector. 1 2

Xz = rsm * [ 2 ; 0 ] + 1 / ( z −1) * rsm * B xn = iztrans ( Xz )

Find Y (z), the output transform. Afterwards compute and graph the output signal y[n]. 1 2 3 4 5 6 7 8

Yz = C* Xz yn = iztrans ( Yz ) n = [0:60]; y = subs ( yn , n ) ; stem (n , y ) ; axis ( [ − 0 . 5 , 6 0 . 5 , 0 , 1 2 ] ) ; xlabel ( ’n ’ ) ; ylabel ( ’ y [ n ] ’ ) ;

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

364

9.32. Code for the script is listed below: 1 2 3 4 5 6 7 8 9

%% S e t up the s t a t e −space model A = [ − 0. 1 , − 0. 7; − 0. 8 , 0] ; B = [3;1]; C = [ 2 , − 1] ; % Find H( z ) using symbolic math functions z = sym( ’ z ’ ) ; tmp = z * eye(2) −A ; H = C * inv (tmp) * B ; s i mpl i fy (H)

The answer produced by the script is 31/ ( 3 * ( 5 * z + 4 ) ) + 88/ ( 3 * ( 10 * z − 7 ) )

and is equivalent to H (z) =

31/3 88/3 5 z + 0.9 + = 5 z + 4 10 z − 7 (z − 0.7) (z + 0.8)

9.33.

Set up the continuous-time state space model and convert to a discrete-time model using Euler’s method with T = 0.02 seconds: 1 2 3 4 5 6 7 8 9

A = [ −2 , −2;1 , −5]; B = [1;0]; C = [0 ,5]; d = 0; Ts = 0 . 0 2 ; A_bar = eye ( 2) +A * Ts B_bar = B * Ts C_bar = C d_bar = d

Solve the approximate discrete-time model iteratively: 1 2 3 4 5 6 7 8

xn = [ 3 ; − 2 ] ; % I n i t i a l value o f s t a t e v e c t o r n = [0:150]; % Vector o f i n d i c e s yn = [ ] ; % Empty v e c t o r to s t a r t for nn=0:150 , xnp1 = A_bar * xn+B_bar ; % ’ xnp1 ’ r e p r e s e n t s x [ n+1] yn = [ yn , C* xn ] ; % Append to v e c t o r ’ yn ’ xn = xnp1 ; % New becomes old f o r next i t e r a t i o n end ;

Graph actual and approximate solutions together:

365 1 2 3 4 5

t = [0:0.01:3]; y = 5/12+70/3 * exp(−3 * t ) −135/4 * exp(−4 * t ) ; plot (n * Ts , yn , ’ r . ’ , t , y , ’b− ’ ) ; grid ; ht = t i t l e ( ’ y_ { a } ( t ) and y [ n ] ’ ) ; hx = xlabel ( ’ t ( sec ) ’ ) ;

% From Example 9.13

9.34. a.

Set up the continuous-time state space model and convert to a discrete-time model with T = 0.1 seconds: 1 2 3 4 5 6 7 8 9

b. 1 2 3 4 5 6 7 8

A = [0 ,1 ,0;0 ,0 ,1; −15 , −11 , −5] B = [0;0;3] C = [1 ,0 ,0] d = 0 Ts = 0 . 1 ; A_bar = expm(A * Ts ) B_bar = inv (A ) * ( A_bar−eye ( 3 ) ) * B C_bar = C d_bar = d

Solve the approximate discrete-time model iteratively: xn = [ 0 ; 0 ; 0 ] ; % I n i t i a l value o f s t a t e v e c t o r n = [0:60]; % Vector o f i n d i c e s yn = [ ] ; % Empty v e c t o r to s t a r t for nn=0:60 , xnp1 = A_bar * xn+B_bar ; % ’ xnp1 ’ r e p r e s e n t s x [ n+1] yn = [ yn , C* xn ] ; % Append to v e c t o r ’ yn ’ xn = xnp1 ; % New becomes old f o r next i t e r a t i o n end ;

The system function for the continuous-time system is H (s) =

3 s 3 + 5 s 2 + 11 s + 15

Create a system object “sys” and use it to compute the unit-step response of the system through the use of the function lsim(..) Afterwards graph the output signal along with that obtained through the iterative solution of the discretized state space model: 1 2 3 4 5 6 7

sys = t f ( [ 3 ] , [ 1 , 5 , 1 1 , 1 5 ] ) ; t = [0:0.01:6]; xa = ones ( s i z e ( t ) ) ; ya = lsim ( sys , xa , t ) ; plot (n * Ts , yn , ’ r . ’ , t , ya , ’b− ’ ) ; grid ; ht = t i t l e ( ’ Actual and approximate s o l u t i o n s ’ ) ; hx = xlabel ( ’ t ( sec ) ’ ) ;

CHAPTER 9. STATE-SPACE ANALYSIS OF SYSTEMS

366

9.35. a. Set up the continuous-time state space model and convert to a discrete-time model using Euler’s method with T = 0.1 seconds: 1 2 3 4 5 6 7 8 9

b. 1 2 3 4 5 6 7 8

A = [0 ,1 ,0;0 ,0 ,1; −15 , −11 , −5]; B = [0;7; −32]; C = [1 ,0 ,0]; d = 0; Ts = 0 . 1 ; A_bar = eye ( 3) +A * Ts B_bar = B * Ts C_bar = C d_bar = d

Solve the approximate discrete-time model iteratively: xn = [ 0 ; 0 ; 0 ] ; % I n i t i a l value o f s t a t e v e c t o r n = [0:60]; % Vector o f i n d i c e s yn = [ ] ; % Empty v e c t o r to s t a r t for nn=0:60 , xnp1 = A_bar * xn+B_bar ; % ’ xnp1 ’ r e p r e s e n t s x [ n+1] yn = [ yn , C* xn ] ; % Append to v e c t o r ’ yn ’ xn = xnp1 ; % New becomes old f o r next i t e r a t i o n end ;

Create a system object “sys” and use it to compute the unit-step response of the system through the use of the function lsim(..) Afterwards graph the output signal along with that obtained through the iterative solution of the discretized state space model: 1 2 3 4 5 6 7

sys = t f ( [ 7 , 3 ] , [ 1 , 5 , 1 1 , 1 5 ] ) ; t = [0:0.01:6]; xa = ones ( s i z e ( t ) ) ; ya = lsim ( sys , xa , t ) ; plot (n * Ts , yn , ’ r . ’ , t , ya , ’b− ’ ) ; grid ; ht = t i t l e ( ’ Actual and approximate s o l u t i o n s ’ ) ; hx = xlabel ( ’ t ( sec ) ’ ) ;

Chapter 10 Analysis and Design of Filters 10.1. a. ωc =

1 1 = = 1000 rad/s RC 1000 × 10−6

The peak magnitude occurs at ω = 0 and its value is |H (0)| = 1. For at most 1 percent deviation from the peak magnitude s ¶ µ 100 2 − 1 = 142.5 rad/s ω0 = 1000 100 − 1 Therefore we require |ω| ≤ 142.5 rad/s.

b.

The slope of the phase characteristic at ω = 0 is ¯ d Θ ¯¯ 1 1 =− =− ¯ d ω ω=0 ωc 1000

If the phase characteristic were perfectly linear, the phase at ω0 = 142.5 rad/s would be ˆ (142.5) = − Θ

142.5 = −0.1425 radians 1000

The actual phase at ω0 = 142.5 rad/s is Θ (142.5) = − tan−1

µ

¶ 142.5 == −0.1415 radians 1000

and the percent deviation from linear phase is µ ¶ −0.1415 + 0.1425 100 = −0.67 percent −0.1425

c.

If the phase characteristic were linear, the time delay for the frequency ω0 would be tˆd (ω0 ) = −

ˆ (ω0 ) 0.1425 Θ = = 0.001 sec ω0 142.5

The actual time delay for the frequency ω0 is t d (ω0 ) = −

Θ (ω0 ) 0.1415 = 0.000993 sec = ω0 142.5

367

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

368

d.

For at most 2 percent deviation from the peak magnitude s ¶ µ 100 2 − 1 = 203.1 rad/s ω0 = 1000 100 − 2

Therefore we require |ω| ≤ 203.1 rad/s. The slope of the phase characteristic at ω = 0 is ¯ d Θ ¯¯ 1 1 =− =− ¯ d ω ω=0 ωc 1000

If the phase characteristic were perfectly linear, the phase at ω0 = 203.1 rad/s would be ˆ (203.1) = − Θ

203.1 = −0.2031 radians 1000

The actual phase at ω0 = 203.1 rad/s is −1

Θ (203.1) = − tan

µ

¶ 203.1 == −0.2003 radians 1000

and the percent deviation from linear phase is µ ¶ −0.2003 + 0.2031 100 = −1.34 percent −0.2031 If the phase characteristic were linear, the time delay for the frequency ω0 would be tˆd (ω0 ) = −

ˆ (ω0 ) 0.2031 Θ = 0.001 sec = ω0 203.1

The actual time delay for the frequency ω0 is t d (ω0 ) = −

Θ (ω0 ) 0.2003 = = 0.000987 sec ω0 203.1

10.2. a. H (ω) =

¯ ¯ 1 1 ¯ =¡ p p ¢ ¯ s 2 + 2 s + 1 s= j ω 1 − ω2 + j 2ω

¯ ¯ ¯ H (ω)¯ = q ¡

1 1 =p ¢2 1 + ω4 1 − ω2 + 2 ω2

For the magnitude response to stay within p percent of the of the peak we need 1

100 − p ≤ p p 1 + ω4 For p = 1 ω0 =

½µ

100 99

⇒ ¶2

ω0 =

½µ

100 100 − p

¶2

¾1/4 −1 = 0.3775 rad/s

¾1/4 −1

369

b. Θ (ω) = ∡ H (ω) = − tan It can be shown that

−1

Ãp

2ω 1 − ω2

!

p ¡ ¢ d Θ (ω) − 2 1 + ω2 = dω 1 + ω4

At ω = 0 we get

¯ p d Θ (ω) ¯¯ =− 2 ¯ d ω ω=0

If the phase characteristic were perfectly linear, the phase at ω0 = 0.3775 rad/s would be p ˆ (0.3775) = − 2 (0.3775) = −0.5338 radians Θ The actual phase at ω0 = 0.3775 rad/s is Θ (0.3775) = − tan−1

Ãp

! 2 (0.3775) = −0.5568 radians 1 − 0.37752

and the percent deviation from linear phase is µ ¶ −0.5568 + 0.5338 100 = 4.31 percent −0.5338

c.

If the phase characteristic were linear, the time delay for the frequency ω0 would be ˆ (ω0 ) 0.5338 Θ = 1.4142 sec = ω0 0.3775

tˆd (ω0 ) = −

The actual time delay for the frequency ω0 is Θ (ω0 ) 0.5568 = 1.4751 sec = ω0 0.3775

t d (ω0 ) = −

d.

For at most 2 percent deviation from the peak magnitude ω0 =

½µ

100 98

¶2

¾1/4 −1 = 0.4506 rad/s

Therefore we require |ω| ≤ 0.4506 rad/s. The slope of the phase characteristic at ω = 0 is ¯ p d Θ (ω) ¯¯ =− 2 ¯ d ω ω=0

If the phase characteristic were perfectly linear, the phase at ω0 = 0.4506 rad/s would be p ˆ (0.4506) = − 2 (0.4506) = −0.6373 radians Θ

The actual phase at ω0 = 0.4506 rad/s is −1

Θ (0.4506) = − tan

Ãp

! 2 (0.4506) = −0.6745 radians 1 − 0.45062

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

370

and the percent deviation from linear phase is ¶ µ −0.6745 + 0.6373 = 5.85 percent 100 −0.6373 If the phase characteristic were linear, the time delay for the frequency ω0 would be tˆd (ω0 ) = −

ˆ (ω0 ) 0.6373 Θ = = 1.4142 sec ω0 0.4506

The actual time delay for the frequency ω0 is t d (ω0 ) = −

Θ (ω0 ) 0.6745 = 1.4969 sec = ω0 0.4506

10.3. a.

a. Using Eqn. (10.29) HLP (ω) = Π

or in terms of f ¡ ¢ HLP f = Π

b.

³ ω ´ e − j 0.1ω 400 π µ

¶ f e − j 0.2πω 200

Using Eqn. (10.31) µ ¶ ¡ ¢ 200 π 200 π h LP (t ) = sinc (t − 0.1) = 200 sinc 200 (t − 0.1) π π

c. hLP (t) 250

Amplitude

200 150 100 50 0 −50 −0.1

−0.05

0

0.05

0.1

0.15

t (sec)

10.4. a.

Using Eqn. (10.33) with ωb = 450 π rad/s and 2 ω0 = 500 pi rad/s and incorporating the time delay of t d = 0.3 seconds we obtain · µ ¶ µ ¶¸ ω − 450 π ω + 450 π HB P (ω) = Π +Π e − j 0.3ω 500 π 500 π

371 or in terms of f

b.

· µ ¶ µ ¶¸ ¡ ¢ f − 225 f + 225 HB P f = Π +Π e − j 0.6π f 250 250

Using Eqn. (10.37) with f b = 225 Hz and 2 f 0 = 250 Hz we obtain ¡ ¢ ¡ ¢ h B P (t ) = 500 sinc 250 (t − 0.3) cos 450 π (t − 0.3)

c. hBP (t)

Amplitude

400 200 0 −200 0.28

0.29

0.3

0.31

0.32

t (sec)

10.5. Using parameters ωc = 2 rad/s and N = 4, the Butterworth squared magnitude function is ¯ ¯ ¯ H (ω)¯2 =

Using Eqn. (10.49)

¯ ¯2 ¯¯ H (s) H (−s) = ¯ H (ω)¯ ¯

1 ³ ω ´8 1+ 2

ω2 =−s 2

1

= 1+

µ

256 = ¶ 4 256 + s 8 −s 2 4

The poles of H (s) H (−s) are the solutions of s 8 = −256 which can be written in the equivalent form s 8 = 256 e j π e j 2πk = 256 e j (2k+1)π The poles of H (s) H (−s) are at p k = 4 e j (2k+1)π/8 ,

k = 0, . . . , 7

The poles in the left half s-plane are associated with H (s). H (s) = ¡

s − p2

¢¡

s − p3

K ¢¡

s − p4

¢¡

s − p5

¢

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

372 Poles p 2 and p 5 form a conjugate pair: p 2 =e j 5π/8

p 5 =e j 11π/8 = e − j 5π/8 ¡

s − p2

¢¡

´ ´³ ¢ ³ s − p 5 = s − 2 e j 5π/8 s − 2 e − j 5π/8

=s 2 + 1.5307 s + 4

Similarly, poles p 3 and p 4 form a conjugate pair: p 3 =e j 7π/8 p 4 =e j 9π/8 = e − j 7π/8 ¡

s − p3

¢¡

´ ´³ ¢ ³ s − p 4 = s − 2 e j 7π/8 s − 2 e − j 7π/8

=s 2 + 3.6955 s + 4

The system function is H (s) = ¡ =

K ¢¡ ¢ s 2 + 1.5307 s + 4 s 2 + 3.6955 s + 4

16 s 4 + 5.2263 s 3 + 13.6569 s 2 + 20.9050 s + 16.0000

where the gain factor K is adjusted to achieve |H (0)| = 1.

10.6. Use Eqn. (10.65) to determine N :

N≥

log10

q¡ ¢ ¡ ¢ 103 − 1 / 100.1 − 1

log10 (4.5/3)

= 10.18

Filter order must be chosen as N = 11. If the excess tolerance that results from rounding up N to 11 is to be used for improving the stopband response, we need to obtain ωc from Eqn. (10.62): µ

3 ωc

¶22

= 101/10 − 1

⇒

ωc = 3.19 rad/s

Alternatively, the excess tolerance can be used for improving the response in the passband by solving for ωc from Eqn. (10.63): µ

4.5 ωc

¶22

= 1030/10 − 1

⇒

ωc = 3.2875 rad/s

373

10.7. ¡ ¢ In order to evenly distribute the excess tolerance, we need a passband ripple of R p − ∆ dB and a stopband attenuation of (A s + ∆) dB where ∆ is a positive quantity. Using Eqns. (10.62) and (10.63) we get µ ¶ ω1 2N = 10(R p −∆)/10 − 1 (P.10.7.1) ωc

and

µ

ω2 ωc

¶2N

= 10(A s +∆)/10 − 1

(P.10.7.2)

Combining the two requirements yields µ

Let K=

µ

ω1 ω2

¶2N

,

ω1 ω2

¶2N

=

10R p /10 10−∆/10 − 1 10 A s /10 10∆/10 − 1

A 1 = 10R p /10 ,

A 2 = 10 A s /10

and

D = 10∆/10

The condition for equal distribution of the excess tolerance between the passband and the stopband is K=

A 1 /D − 1 A2 D − 1

or, equivalently A 2 K D 2 + (1 − K ) D − A 1 = 0 Solving for D we obtain D=

(K − 1) ±

p

(1 − K )2 + 4A 1 A 2 K 2A 2 K

Use the positive solution since ∆ = 10 log10 (D). Once ∆ is found, solve for ωc from either Eqn. (P.10.7.1) or Eqn. (P.10.7.2).

10.8. Parameter αk is found as αk =

(2k + 1) π , 8

k = 0, . . . , 7

and the parameter βk is

sinh−1 (1/0.3) = 0.4797 4 Poles of H (s) H (−s) are found through the use of Eqn. (10.86) as · µ ¶ µ ¶ ¸ (2k + 1) π (2k + 1) π p k = 2 cos cosh (0.4797) − j sin sinh (0.4797) , 8 8 βk =

k = 0, . . . , 7

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

374 Locations of the poles of H (s) H (−s) are as follows:

p 0 =0.3814 + j 2.0645 p 1 =0.9208 + j 0.8551 p 2 =0.9208 − j 0.8551 p 3 =0.3814 − j 2.0645 p 4 = − 0.3814 − j 2.0645 p 5 = − 0.9208 − j 0.8551 p 6 = − 0.9208 + j 0.8551 p 7 = − 0.3814 + j 2.0645 Poles p 4 , p 5 , p 6 and p 7 are in the left half s-plane, and are therefore associated with H (s). System function is constructed as H (s) = ¡ =

A ¢¡ ¢¡ ¢¡ ¢ s + 0.3814 + j 2.0645 s + 0.9208 + j 0.8551 s + 0.9208 − j 0.8551 s + 0.3814 − j 2.0645 A

s 4 + 2.6044 s 3 + 7.3915 s 2 + 9.3217 s + 6.9602

The gain factor A is adjusted to achieve |H (ω)|max = 1. With N = 4, this requires that (see Fig. 10.21) ¯ ¯ 1 ¯ H (0)¯ = p 1 = 0.9578 =p 1 + ε2 1 + (0.3)2

and therefore

A = 0.9578 6.9602

A = 6.6667

⇒

The system function is H (s) =

6.6667 s 4 + 2.6044 s 3 + 7.3915 s 2 + 9.3217 s + 6.9602

10.9. Parameter αk is found as αk =

(2k + 1) π , 8

k = 0, . . . , 7

and the parameter βk is

sinh−1 (1/0.3) = 0.4797 4 Poles of H (s) H (−s) are found through the use of Eqn. (10.109) as βk =

pk =

j2

µ ¶ , (2k + 1) π (2k + 1) π cosh (0.4797) − j sin sinh (0.4797) cos 8 8 µ

¶

k = 0, . . . , 7

375 Locations of the poles of H (s) H (−s) are as follows: p 0 = − 0.3461 − j 1.8736 p 1 = − 2.3324 − j 2.1661 p 2 = − 2.3324 + j 2.1661 p 3 = − 0.3461 + j 1.8736 p 4 =0.3461 + j 1.8736 p 5 =2.3324 + j 2.1661 p 6 =2.3324 − j 2.1661 p 7 =0.3461 − j 1.8736 Poles p 0 , p 1 , p 2 and p 3 are in the left half s-plane, and are therefore associated with H (s). Zeros of the system function H (s) are found using Eqns. (10.111) and (10.112): zk =

± j ω2 ¶ , (2k − 1) π cos 2N µ

k = 1, 2

Locations of the zeros of H (s) are as follows: z 1 =0 ± j 2.1648 z 2 =0 ± j 5.2263 System function is constructed as ¡ ¢¡ ¢¡ ¢¡ ¢ A s + j 2.1648 s − j 2.1648 s + j 5.2263 s − j 5.2263 ¢¡ ¢¡ ¢¡ ¢ H (s) = ¡ s + 0.3461 + j 1.8736 s + 2.3324 + j 2.1661 s + 2.3324 − j 2.1661 s + 0.3461 − j 1.8736 =

0.2873 s 4 + 9.1951 s 2 + 36.7805 s 4 + 5.3571 s 3 + 16.9916 s 2 + 23.9481 s + 36.7805

The peak of the system function occurs at ω = 0. The gain factor A is adjusted to achieve |H (0)| = 1.

10.10. a.

The minimum filter order is found using Eqn. (10.65): q¡ ¢ ¡ ¢ 1020/10 − 1 / 101/10 − 1 log10 = 5.3129 N≥ log10 (3.5/2)

Filter order must be chosen as N = 6. If the excess tolerance that results from rounding up N to 6 is to be used for improving the passband response, we need to obtain ωc from Eqn. (10.63): ¶ µ 3.5 12 = 1020/10 − 1 ⇒ ωc = 2.3865 rad/s ωc The poles of H (s) H (s) are at p k = 2.3865 e j (2k+1)π/12 ,

k = 0, . . . , 11

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

376

Using the poles in the left half s-plane, H (s) is constructed as H (s) =

b.

184.75 s 6 + 9.2208 s 5 + 42.5116 s 4 + 124.256 s 3 + 242.124 s 2 + 299.109 s + 184.75

Minimum filter order is found using Eqn. (10.94) and Eqn. (10.101): F= N≥

s

1020/10 − 1 101/10 − 1

= 19.5538

cosh−1 (19.5538) cosh−1 (3.5/2)

= 3.1633

which must be rounded up to N = 4. The parameter ε is found using Eqn. (10.89): ε=

p

101/10 − 1 = 0.5088

The poles of the product H (s) H (−s) can now be found using Eqns. (10.82), (10.84) and (10.85). Parameter αk is found as (2k + 1) π , k = 0, . . . , 7 αk = 8 and the parameter βk is sinh−1 (1/0.5088) = 0.3570 βk = 4 Poles of H (s) H (−s) are found through the use of Eqn. (10.86) as · µ ¶ µ ¶ ¸ (2k + 1) π (2k + 1) π p k = 2 cos cosh (0.3570) − j sin sinh (0.3570) , k = 0, . . . , 7 8 8 Locations of the poles of H (s) H (−s) are as follows: p 0 =0.2791 − j 1.9668 p 1 =0.6737 − j 0.8147 p 2 =0.6737 + j 0.8147 p 3 =0.2791 + j 1.9668 p 4 = − 0.2791 + j 1.9668 p 5 = − 0.6737 + j 0.8147 p 6 = − 0.6737 − j 0.8147 p 7 = − 0.2791 − j 1.9668 Poles p 4 , p 5 , p 6 and p 7 are in the left half s-plane, and are therefore associated with H (s). System function is constructed as H (s) = ¡ =

A ¢¡ ¢¡ ¢¡ ¢ s + 0.2791 − j 1.9668 s + 0.6737 − j 0.8147 s + 0.6737 + j 0.8147 s + 0.2791 + j 1.9668 A

s 4 + 1.9056 s 3 + 5.8157 s 2 + 5.9410 s + 4.4100

377 The gain factor A is adjusted to achieve |H (ω)|max = 1. With N = 4, this requires that (see Fig. 10.21) ¯ ¯ 1 ¯ H (0)¯ = p 1 = 0.8913 =p 1 + ε2 1 + (0.5088)2

and therefore

A = 0.8913 4.4100

A = 3.9305

⇒

The system function is H (s) =

c.

3.9305 s 4 + 1.9056 s 3 + 5.8157 s 2 + 5.9410 s + 4.4100

Minimum filter order is found using Eqn. (10.117) and Eqn. (10.118): F= N≥

s

1020/10 − 1 101/10 − 1

= 19.5538

cosh−1 (19.5538) cosh−1 (3.5/2)

= 3.1633

which must be rounded up to N = 4. The parameter ε is found using Eqn. (10.114): ε= p Parameter αk is found as αk =

1 1020/10 − 1

(2k + 1) π , 8

= 0.1005

k = 0, . . . , 7

and the parameter βk is

sinh−1 (1/0.1005) = 0.7483 4 Poles of H (s) H (−s) are found through the use of Eqn. (10.109) as βk =

pk =

j2

µ ¶ , (2k + 1) π (2k + 1) π cosh (0.7483) − j sin sinh (0.0.7483) cos 8 8 µ

¶

Locations of the poles of H (s) H (−s) are as follows: p 0 = − 0.7198 + j 2.7402 p 1 = − 3.2378 + j 2.1149 p 2 = − 3.2378 − j 2.1149 p 3 = − 0.7198 − j 2.7402 p 4 =0.7198 − j 2.7402 p 5 =3.2378 − j 2.1149 p 6 =3.2378 + j 2.1149 p 7 =0.7198 + j 2.7402

k = 0, . . . , 7

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

378

Poles p 0 , p 1 , p 2 and p 3 are in the left half s-plane, and are therefore associated with H (s). Zeros of the system function H (s) are found using Eqns. (10.111) and (10.112): zk =

± j ω2 ¶ , (2k − 1) π cos 2N

k = 1, 2

µ

Locations of the zeros of H (s) are as follows: z 1 =0 ± j 3.7884 z 2 =0 ± j 9.1459 System function is constructed as ¡ ¢¡ ¢¡ ¢¡ ¢ A s + j 3.7884 s − j 3.7884 s + j 9.1459 s − j 9.1459 ¢¡ ¢¡ ¢¡ ¢ H (s) = ¡ s + 0.7198 − j 2.7402 s + 3.2378 − j 2.1149 s + 3.2378 + j 2.1149 s + 0.7198 + j 2.7402 =

0.1 s 4 + 9.8 s 2 + 120.05 s 4 + 7.9152 s 3 + 32.3049 s 2 + 73.5079 s + 120.05

The peak of the system function occurs at ω = 0. The gain factor A is adjusted to achieve |H (0)| = 1.

10.11. Let the system functions for lowpass and bandpass filters be HLP (s) and HB P (λ) respectively. Furthermore, let ω and p be the radian frequency variables in s and λ domains so that s = j ω and λ = j p. The lowpass to bandpass frequency transformation is s=

λ2 + ω20 Bλ

jω=

⇒

−p 2 + ω20

⇒

jBp

−ω =

−p 2 + ω20 Bp

p (rad/s)

Lowpass to bandpass frequency transformation

ωL1 −ωL1

−ωB3

−ωB2

ωB2

ωB3

ω (rad/s)

The goal is to map the band of frequencies −ωL1 < ω < ωL1 for the lowpass filter to the band of frequencies ωB 2 < p < ωB 3 for the bandpass filter. At the two critical frequencies ω = ±ωL1 we have − ωL1 =

−ω2B 3 + ω20 B ωB 3

and

ωL1 =

−ω2B 2 + ω20 B ωB 2

379 which can be solved together to yield ω2B 3 − ω20 B ωB 3

=

−ω2B 2 + ω20 B ωB 2

and B=

ω0 =

and

p ωB 2 ωB 3

ωB 3 − ωB 2 ωL1

10.12. Let the system functions for lowpass and band-reject filters be HLP (s) and HB R (λ) respectively. Furthermore, let ω and p be the radian frequency variables in s and λ domains so that s = j ω and λ = j p. The lowpass to bandpass frequency transformation is s=

Bλ

⇒

λ2 + ω20

jω=

jBp

⇒

−p 2 + ω20

ω=

−B p

p 2 − ω20

p (rad/s)

Lowpass to band-reject frequency transformation

ωL1 −ωL1

−ωS4 −ω0 −ωS1

ωS1 ω0

ωS4

ω (rad/s)

The goal is to map the band of frequencies −ωL1 < ω < ωL1 for the lowpass filter to the three bands of frequencies, namely Band 1:

− ∞ < p < ωS4

Band 2:

ωS1 < p < ωS1

Band 3:

ωS4 < p < ∞

for the band-reject filter. At the two critical frequencies ω = ±ωL1 we have − ωL1 =

−B ωS4

ω2S4 − ω20

and

ωL1 =

−B ωS1

ω2S1 − ω20

which can be solved together to yield B ωS4 ω2S4 − ω20

=

−B ωS1

ω2S1 − ω20

and

ω0 =

and B = (ωS4 − ωS1 ) ωL1

p ωS1 ωS4

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

380

10.13. a. ω0 = ωL1 ωH2 = (2) (5) = 10 The transformation is s=

b.

Let s = j ω and λ = j p so that jω =

10 jp

10 λ

⇒

ω=−

10 p

If ω1 = 6 rad/s, the corresponding frequency of the highpass filter is p1 = −

10 = −1.6667 rad/s 6

c. H (λ) = µ At λ = j 5

Similarly, for G (s) at s = j 2 we have

λ 2 ¶ = 10 λ+5 +2 λ

¯ ¯ ¯ j5 ¯ ¯ = p1 = ¯¯ j5+5¯ λ= j 5 2

¯ ¯¯ ¯ H (λ)¯¯¯ ¯ ¯¯ ¯G (s)¯¯¯

¯ ¯ ¯ j2 ¯ ¯ ¯ = p1 =¯ j2+2¯ s= j 2 2

10.14. a.

Let the passband edge frequency for the lowpass filter be ωL1 = 1 rad/s. For the lowpass to highpass transformation we have ω0 = ωL1 ωH2 = (1) (5) = 5 Therefore, the transformation is

5 λ The frequency ωL1 for the lowpass filter corresponds to the frequency −ωH2 for the highpass filter. Similarly, the frequency ωL2 for the lowpass filter should correspond to the frequency −ωH1 for the highpass filter, therefore 5 = 2.5 rad/s ωL2 = ωH1 s=

The specifications for the lowpass prototype are ωL1 = 2 rad/s

ωL2 = 2.5 rad/s

R p = 1 dB

A s = 30 dB

381

b.

The minimum filter order is found using Eqn. (10.65):

N≥

log10

q¡

¢ ¡ ¢ 1030/10 − 1 / 101/10 − 1

log10 (2.5/1)

= 4.5062

Filter order must be chosen as N = 5. If the excess tolerance that results from rounding up N to 5 is to be used for improving the passband response, we need to obtain ωc from Eqn. (10.63): µ

c.

2.5 ωc

¶10

= 1030/10 − 1

⇒

ωc = 1.2531 rad/s

The poles of G (s) G (s) are at p k = 1.2531 e j kπ/5 ,

k = 0, . . . , 9

Using the poles in the left half s-plane, G (s) is constructed as G (s) =

3.0897 s 5 + 4.0551 s 4 + 8.2219 s 3 + 10.3028 s 2 + 7.9791 s + 3.0897

d. ¯ ¯ H (λ) = G (s)¯

3.0897 = µ ¶5 µ ¶4 µ ¶3 µ ¶2 µ ¶ s=5/λ 5 5 5 5 5 + 4.0551 + 8.2219 + 10.3028 + 7.9791 + 3.0897 λ λ λ λ λ =

λ5 λ5 + 12.91 λ4 + 83.36 λ3 + 332.63 λ2 + 820.28 λ + 1011.4

e. ¯ ¯ 20 log10 ¯H (2)¯ = − 30 dB ¯ ¯ 20 log10 ¯H (5)¯ = − 0.43 dB

10.15. a.

The partial fraction form of G (s) is G (s) =

2 2 − s +1 s +2

and the corresponding impulse response is g (t ) = 2 e −t u (t ) − 2 e −2t u (t )

b. h[n] = 0.5 g (0.5 n) = e −0.5 n u[n] − e −n u[n]

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382

c.

z z 0.2387 z − = 2 −0.5 −1 z −e z −e z − 0.9744 z + 0.2231

H (z) =

10.16. s= The system function is H (z) = µ = =

2 1 − z −1 4 (z − 1) = T 1 + z −1 z +1

2 ¶µ ¶ 4 (z − 1) 4 (z − 1) +1 +2 z +1 z +1

2 (z + 1)2 (5 z − 3) (6 z − 2) 2 z2 + 4 z + 2 30 z 2 − 28 z + 6

10.17. a.

The system function is H (Ω) =

NX −1

h[n] e − j Ωn

n=0

Since N is even, the summation can be split into two halves: H (Ω) =

NX /2−1 n=0

h[n] e − j Ωn +

NX −1

h[n] e − j Ωn

n=N /2

Using the variable change n = N − 1 − m, the second summation can be written as NX −1

n=N /2

h[n] e − j Ωn =

0 X

m=N /2−1

h[N − 1 − m] e − j Ω (N −1−m)

Recognizing that h[N − 1 − m] = h[m] and that the summation limits can be swapped, we have NX −1

n=N /2

h[n] e − j Ωn = e − j (N −1)

NX /2−1

h[m] e j Ω m

m=0

and H (Ω) is H (Ω) = =

NX /2−1 n=0

h[n] e − j Ωn + e − j (N −1)

NX /2−1 n n=0

NX /2−1

h[n] e j Ωn

n=0

h[n] e − j Ωn + e − j (N −1) h[n] e j Ωn

o

383 Factoring out e − j (N −1)/2 the result can be written as H (Ω) =e − j (N −1)/2 =e

− j (N −1)/2

NX /2−1 n n=0

NX /2−1 n=0

|

=e − j (N −1)/2 A (Ω)

b.

h[n] e − j Ωn e j Ω (N −1)/2 + h[n] e j Ωn e − j Ω (N −1)/2

o

¶¸ · µ N −1 2 h[n] cos Ω n − 2 {z } Purely real

The system function is H (Ω) =

NX −1

h[n] e − j Ωn

n=0

Since N is even, the summation can be split into two halves: H (Ω) =

NX /2−1 n=0

h[n] e − j Ωn +

NX −1

h[n] e − j Ωn

n=N /2

Using the variable change n = N − 1 − m, the second summation can be written as NX −1

n=N /2

h[n] e − j Ωn =

0 X

m=N /2−1

h[N − 1 − m] e − j Ω (N −1−m)

Recognizing that h[N − 1 − m] = −h[m] and that the summation limits can be swapped, we have NX −1

n=N /2

h[n] e − j Ωn = −e − j (N −1)

NX /2−1

h[m] e j Ω m

m=0

and H (Ω) is H (Ω) = =

NX /2−1 n=0

h[n] e − j Ωn − e − j (N −1)

NX /2−1 n n=0

NX /2−1

h[n] e j Ωn

n=0

h[n] e − j Ωn − e − j (N −1) h[n] e j Ωn

o

Factoring out e − j (N −1)/2 the result can be written as H (Ω) =e − j (N −1)/2

NX /2−1 n n=0

h[n] e − j Ωn e j Ω (N −1)/2 − h[n] e j Ωn e − j Ω (N −1)/2

¶¸ · µ N −1 − j 2 h[n] sin Ω n − 2 n=0 · µ ¶¸ NX /2−1 N −1 −2 h[n] sin Ω n − =e − j (N −1)/2 e j π/2 2 n=0 | {z }

=e − j (N −1)/2

NX /2−1

Purely real

=e − j (N −1)/2 e j π/2 B (Ω)

o

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

384

10.18. a.

The system function is H (Ω) =

NX −1

h[n] e − j Ωn

n=0

Since N is odd, the summation must be split as follows: H (Ω) =

−3)/2 (N X n=0

h[n] e − j Ωn + h[(N − 1) /2] e − j ω(N −1)/2 +

NX −1

h[n] e − j Ωn

n=(N +1)/2

Using the variable change n = N − 1 − m, the second summation can be written as NX −1

n=(N +1)/2

h[n] e − j Ωn =

0 X

m=(N −3)/2

h[N − 1 − m] e − j Ω (N −1−m)

Recognizing that h[N − 1 − m] = h[m] and that the summation limits can be swapped, we have NX −1

n=(N +1)/2

h[n] e − j Ωn = e − j (N −1)

−3)/2 (N X

h[m] e j Ω m

m=0

and H (Ω) is H (Ω) = =

−3)/2 (N X n=0

h[n] e − j Ωn + h[(N − 1) /2] e − j ω(N −1)/2 + e − j (N −1)

−3)/2 n (N X

−3)/2 (N X

h[n] e j Ωn

n=0

o h[n] e − j Ωn + e − j (N −1) h[n] e j Ωn + h[(N − 1) /2] e − j ω(N −1)/2

n=0

Factoring out e − j (N −1)/2 the result can be written as # " o −3)/2 n (N X − j Ωn j Ω (N −1)/2 j Ωn − j Ω (N −1)/2 − j (N −1)/2 + h[(N − 1) /2] h[n] e e + h[n] e e H (Ω) =e n=0

=e

− j (N −1)/2

"

|

−3)/2 (N X n=0

Purely real

=e − j (N −1)/2 A (Ω)

b.

# ¶¸ · µ N −1 + h[(N − 1) /2] 2 h[n] cos Ω n − 2 {z }

The system function is H (Ω) =

NX −1

h[n] e − j Ωn

n=0

It should be noted that, for the odd symmetry to work in this case, the center sample must be zero, that is, h[(N − 1) /2] = 0 as no other value would be equal to its own negative. The summation can be split into two halves as follows: H (Ω) =

−3)/2 (N X n=0

h[n] e − j Ωn +

NX −1

n=(N +1)/2

h[n] e − j Ωn

385 Using the variable change n = N − 1 − m, the second summation can be written as NX −1

n=(N +1)/2

h[n] e − j Ωn =

0 X

m=(N −3)/2

h[N − 1 − m] e − j Ω (N −1−m)

Recognizing that h[N − 1 − m] = −h[m] and that the summation limits can be swapped, we have NX −1

n=(N +1)/2

h[n] e − j Ωn = −e − j (N −1)

−3)/2 (N X

h[m] e j Ω m

m=0

and H (Ω) is H (Ω) = =

−3)/2 (N X n=0

h[n] e − j Ωn − e − j (N −1)

−3)/2 n (N X n=0

−3)/2 (N X

h[n] e j Ωn

n=0

h[n] e − j Ωn − e − j (N −1) h[n] e j Ωn

o

Factoring out e − j (N −1)/2 the result can be written as H (Ω) =e − j (N −1)/2

−3)/2 n (N X n=0

h[n] e − j Ωn e j Ω (N −1)/2 − h[n] e j Ωn e − j Ω (N −1)/2

o

¶¸ · µ N −1 − j 2 h[n] sin Ω n − =e 2 n=0 · µ ¶¸ −3)/2 (N X N −1 − j (N −1)/2 j π/2 −2 h[n] sin Ω n − =e e 2 n=0 | {z } − j (N −1)/2

−3)/2 (N X

Purely real

=e − j (N −1)/2 e j π/2 B (Ω)

10.19. Spectrum of an ideal bandpass filter can be expressed as the difference of the spectra of two lowpass filters: ¶ ¶ µ µ Ω2 Ω2 Ω2 n Ω2 n HB P (Ω) = − sinc sinc π π π π =0.7 sinc (0.7 n) − 0.3 sinc (0.3 n) Since N = 2M + 1 = 19 we have M = 9. Hamming window with M = 9 is ¶ µ π (n + M ) w [n] = 0.54 − 0.46 cos M Coefficients are listed in the table on the left. The impulse response of the designed filter is h[n] = h T [n − 9] w [n − 9] and is listed in the table on the right.

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

386

n −9 −8 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 8 9

h T [n] 0.0000 0.0757 0.0000 −0.0624 0.0000 −0.0935 0.0000 0.3027 0.0000 −0.4000 0.0000 0.3027 0.0000 −0.0935 0.0000 −0.0624 0.0000 0.0757 0.0000

w [n] 0.0800 0.1077 0.1876 0.3100 0.4601 0.6199 0.7700 0.8924 0.9723 1.0000 0.9723 0.8924 0.7700 0.6199 0.4601 0.3100 0.1876 0.1077 0.0800

h T [n] w [n] 0.0000 0.0082 0.0000 −0.0193 0.0000 −0.0580 0.0000 0.2702 0.0000 −0.4000 0.0000 0.2702 0.0000 −0.0580 0.0000 −0.0193 0.0000 0.0082 0.0000

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

h[n] 0.0000 0.0082 0.0000 −0.0193 0.0000 −0.0580 0.0000 0.2702 0.0000 −0.4000 0.0000 0.2702 0.0000 −0.0580 0.0000 −0.0193 0.0000 0.0082 0.0000

10.20. Using the inverse DTFT with the desired spectrum Hd (Ω) yields 1 h d [n] = 2π =

ˆ

ˆ

π

π

Hd (Ω) e j Ωn d Ω

−π

(1) e − j Ω (N −1)/2 e j Ωn d Ω

−π

µ ¶¸ · N −1 Ωc Ωc n− sinc = π π 2

If normalized frequency F c is used, then µ · ¶¸ N −1 h d [n] = 2F c sinc 2F c n − 2

Truncating the ideal impulse response h d [n] to keep only the samples for n = 0, . . . , N − 1 yields h T [n] =

½

h d [n] , 0,

n = 0, . . . , N − 1 otherwise

387 Check for linear phase: ¶¸ µ · N −1 h d [N − 1 − n] =2F c sinc 2F c N − 1 − n − 2 µ · ¶¸ N −1 =2F c sinc 2F c −n + 2

=h d [n] ,

n = 0, . . . , N − 1

10.21. a.

Triangular window:

¯ ¯ w [n] = 1 − α ¯n + β¯

The center of the window function needs to be at n (N − 1) /2. Therefore β=−

N −1 2

We also need w [0] = w [N − 1] = 0. Using the value at either end results in α=

2 N −1

Using the values of α and β found, the triangular window function is ¯ ¶¯ µ ¯ N − 1 ¯¯ 2 ¯ n− w [n] = 1 − N −1 ¯ 2 ¯

b.

Hanning window

¡ ¢ w [n] = 0.5 − 0.5 cos α n + β

The center of the window function needs to be at n (N − 1) /2. Therefore ¶ µ N −1 N −1 + β = −1 ⇒ α +β = π cos α 2 2 We also need w [0] = w [N − 1] = 0. Using the value at n = 0 results in ¡ ¢ cos β = 1

⇒

β=0

⇒

α=

2π N −1

Using the values of α and β found, the Hamming window function is µ ¶ 2πn w [n] = 0.5 − 0.5 cos N −1

c.

Hamming window

¡ ¢ w [n] = 0.54 − 0.46 cos α n + β

The center of the window function needs to be at n (N − 1) /2. Therefore µ ¶ N −1 N −1 cos α + β = −1 ⇒ α +β = π 2 2

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

388

We also need w [0] and w [N − 1] to be minimum, that is, w [0] = w [N − 1] = 0.08. Using the value at n = 0 results in ¡ ¢ 2π cos β = 1 ⇒ β=0 ⇒ α= N −1 Using the values of α and β found, the Hamming window function is ¶ µ 2πn w [n] = 0.54 − 0.46 cos N −1

d.

Blackman window

¡ ¢ ¡ ¢ w [n] = 0.42 − 0.5 cos α n + β + 0.08 cos α n + β

Parameters α and β are the same as in parts (b) and (c): α=

2π , N −1

0.42 − 0.5 cos

µ

β=0

¶ µ ¶ 4πn 2πn + 0.08 cos N −1 N −1

10.22. a. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

p = 1; % Determine the frequency omg0 f o r p percent magnitude v a r i a t i o n omg0 = 1000 * sqrt ((100/(100 − p))^2 −1); % Define anonymous function f o r H(omg) H = @(omg) 1. / ( 1+ j *omg/ 1000) ; % Graph the magnitude in the range −omg0, +omg0 omg = [−omg0 : 0 . 1 : omg0 ] ; subplot ( 3 , 1 , 1 ) ; plot (omg, abs (H(omg ) ) ) ; axis ([ −omg0, omg0, 0 . 9 9 , 1 ] ) ; t i t l e ( ’ |H( \omega ) | ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ; % Graph the phase in the range −omg0, +omg0 subplot ( 3 , 1 , 2 ) ; plot (omg, angle (H(omg ) ) ) ; axis ([ −omg0, omg0,− pi /18 , pi / 1 8 ] ) ; t i t l e ( ’ Phase of H( \omega) ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Phase ( rad ) ’ ) ; grid ; % Graph the time delay in the range −omg0, omg0 subplot ( 3 , 1 , 3 ) ; omg = omg+eps ; % Avoid d i v i s i o n by 0 tdelay = −angle (H(omg ) ) . / omg; plot (omg, tdelay ) ; axis ([ −omg0, omg0, 0 . 0 0 0 9 , 0 . 0 0 1 1 ] ) ;

389 29 30 31 32

b. 1

t i t l e ( ’Time delay fo r H( \omega) ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Delay ( sec ) ’ ) ; grid ;

Modify line 1 as follows and repeat. p = 2;

10.23. a. p = 1; % Determine the frequency omg0 f o r 1 percent magnitude v a r i a t i o n tmp = (100/(100 −p))^2 −1; omg0 = sqrt ( sqrt (tmp ) ) ; % Define anonymous function f o r H(omg) H = @(omg) 1./((1 −omg. * omg)+ j * sqrt ( 2 ) * omg ) ; % Graph the magnitude in the range −omg0, +omg0 omg = [−omg0 : 0 . 0 1 :omg0 ] ; subplot ( 3 , 1 , 1 ) ; plot (omg, abs (H(omg ) ) ) ; axis ([ −omg0, omg0, 0 . 9 9 , 1 ] ) ; t i t l e ( ’ |H( \omega ) | ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ; % Graph the phase in the range −omg0, +omg0 subplot ( 3 , 1 , 2 ) ; plot (omg, angle (H(omg ) ) ) ; axis ([ −omg0, omg0, − 0 . 7 5 , 0 . 7 5 ] ) ; t i t l e ( ’ Phase of H( \omega) ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Phase ( rad ) ’ ) ; grid ; % Graph the time delay in the range −omg0, omg0 subplot ( 3 , 1 , 3 ) ; omg = omg+eps ; % Avoid d i v i s i o n by 0 tdelay = −angle (H(omg ) ) . / omg; plot (omg, tdelay ) ; axis ([ −omg0, omg0, 1 . 2 , 1 . 5 ] ) ; t i t l e ( ’Time delay fo r H( \omega) ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Delay ( sec ) ’ ) ; grid ;

b.

Modify line 1 as follows and repeat.

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

390 1

p = 2;

10.24. a. function h = s s _ i l p (omg0, td , t ) f0 = omg0/ ( 2 * pi ) ; h = 2 * f0 * sinc ( 2 * f0 * ( t−td ) ) ; end

b.

% Eqn . ( 1 0 . 3 2 )

Duplicate the result of Problem 10.3. t = [ −0.1:0.0001:0.2]; hLP = s s _ i l p (200 * pi , 0 . 1 , t ) ; plot ( t , hLP ) ; grid ;

10.25. a. function h = ss_ibp (omg1, omg2, td , t ) h = s s _ i l p (omg2, td , t )− s s _ i l p (omg1, td , t ) ; end

b.

Duplicate the result of Problem 10.4. t = [0.275:0.00005:0.325]; hBP = ss_ibp (200 * pi , 700 * pi , 0 . 3 , t ) ; plot ( t , hBP ) ; axis ( [ 0. 275 , 0. 325 , − 350 , 550] ) ; xlabel ( ’ t ( sec ) ’ ) ; ylabel ( ’ Amplitude ’ ) ; t i t l e ( ’ h_ {BP } ( t ) ’ ) ; grid ;

10.26. The filter can be designed using the statement [num, den ] = butter ( 4 , 2 , ’ s ’ )

which results in numerator and denominator polynomials

391 num = 0

0

0

0

16.0000

den = 1.0000

5.2263

13.6569

20.9050

16.0000

The system function can be evaluated at a set of radian frequencies using the following code: omg = [ 0 : 0 . 0 1 : 5 ] ; H = freqs (num, den ,omg ) ;

Graph the magnitude response: plot (omg, abs (H) ) ; axis ( [ 0 , 5 , 0 , 1 . 2 ] ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; t i t l e ( ’ |H( \omega ) | ’ ) ; grid ;

Graph the phase response: plot (omg, angle (H) ) ; axis ([0 ,5 , − pi , pi ] ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Phase ( rad ) ’ ) ; t i t l e ( ’ Phase of H( \omega) ’ ) ; grid ;

10.27. The filter is designed using the following statements: N = 11; omgc = 3. 2875; [num, den ] = butter (N, omgc, ’ s ’ )

Evaluate the system function for ω = 0, . . . , 6 rad/s. omg = [ 0 : 0 . 0 1 : 6 ] ; H = freqs (num, den ,omg ) ;

Graph the dB magnitude of the system function. plot (omg, 2 0 * log10 ( abs (H) ) ) ; axis ( [ 0 , 6 , − 6 0 , 1 0 ] ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude (dB) ’ ) ; t i t l e ( ’ |H( \omega ) | _ {dB} ’ ) ; grid ;

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

392

10.28. a.

The system function for the filter is found using the following statements: omg1 = 2 ; epsilon = 0 . 3 ; % Maximum passband r i p p l e Rp = 10 * log10 (1+ epsilon ^2) [num, den ] = cheby1 ( 4 ,Rp, omg1, ’ s ’ )

b.

Evaluate the system function in the interval ω = 0, . . . , 5 rad/s. omg = [ 0 : 0 . 0 1 : 5 ] ; H = freqs (num, den ,omg ) ;

Graph |H (ω)|: plot (omg, abs (H) ) ; axis ( [ 0 , 5 , 0 , 1 . 2 ] ) ; t i t l e ( ’ |H( s ) | ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ;

Graph ∡ H (ω): plot (omg, angle (H) ) ; axis ([0 ,5 , − pi , pi ] ) ; t i t l e ( ’ Phase of H( s ) ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Phase ( rad ) ’ ) ; grid ;

c.

Pole-zero plot is obtained through the following: p = roots ( den ) z = roots (num) plot ( real (p ) , imag (p ) , ’ rx ’ , real ( z ) , imag ( z ) , ’ ro ’ ) ; axis ( [ − 2 , 2 , − 3 , 3] ) ; xlabel ( ’ Real ’ ) ; ylabel ( ’ Imag ’ ) ; grid ;

393

10.29. a.

The system function for the filter is found using the following statements: omg2 = 2 ; epsilon = 0 . 3 ; % Minimum stopband attenuation As = 10 * log10 ( ( 1+ epsilon ^2)/ epsilon ^ 2 ) ; % Compute system function using MATLAB function cheby2 ( ) [num, den ] = cheby2 ( 4 , As , omg2, ’ s ’ )

b.

Evaluate the system function in the interval ω = 0, . . . , 5 rad/s. omg = [ 0 : 0 . 0 1 : 5 ] ; H = freqs (num, den ,omg ) ;

Graph |H (ω)|: plot (omg, abs (H) ) ; axis ( [ 0 , 5 , 0 , 1 . 2 ] ) ; t i t l e ( ’ |H( s ) | ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ;

Graph ∡ H (ω): plot (omg, angle (H) ) ; axis ([0 ,5 , − pi , pi ] ) ; t i t l e ( ’ Phase of H( s ) ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Phase ( rad ) ’ ) ; grid ;

c.

Pole-zero plot is obtained through the following: p = roots ( den ) z = roots (num) plot ( real (p ) , imag (p ) , ’ rx ’ , real ( z ) , imag ( z ) , ’ ro ’ ) ; axis ( [ − 3 , 3 , − 6 , 6] ) ; xlabel ( ’ Real ’ ) ; ylabel ( ’ Imag ’ ) ; grid ;

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

394

10.30. a. omg1 = 2 ; omg2 = 3 . 5 ; Rp = 1 ; As = 20; % Butterworth design [N1, omgc] = buttord (omg1, omg2, Rp , As , ’ s ’ ) [num1, den1 ] = butter (N1, omgc, ’ s ’ ) % Chebyshev type−I design N2 = cheb1ord (omg1, omg2, Rp , As , ’ s ’ ) [num2, den2 ] = cheby1 (N2, Rp , omg1, ’ s ’ ) % Chebyshev type−I I design N3 = cheb2ord (omg1, omg2, Rp , As , ’ s ’ ) [num3, den3 ] = cheby2 (N3, As , omg2, ’ s ’ )

b.

The following code evaluates the system function for each design. omg = [ 0 : 0 . 0 1 : 6 ] ; H1 = freqs (num1, den1 ,omg ) ; H2 = freqs (num2, den2 ,omg ) ; H3 = freqs (num3, den3 ,omg ) ;

Graph |H (s)| for the Butterworth design. plot (omg, abs (H1 ) ) ; axis ( [ 0 , 6 , 0 , 1 . 2 ] ) ; t i t l e ( ’ |H( s ) | fo r the Butterworth design ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ;

Graph |H (s)| for the Chebyshev type-I design. plot (omg, abs (H2 ) ) ; axis ( [ 0 , 6 , 0 , 1 . 2 ] ) ; t i t l e ( ’ |H( s ) | fo r the Chebyshev type−I design ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ;

Graph |H (s)| for the Chebyshev type-II design. plot (omg, abs (H3 ) ) ; axis ( [ 0 , 6 , 0 , 1 . 2 ] ) ; t i t l e ( ’ |H( s ) | fo r the Chebyshev type−I I design ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ;

395

c.

Plot the dB magnitude of all three designs simultaneously. plot (omg, 2 0 * log10 ( abs (H1) ) ,omg, 2 0 * log10 ( abs (H2) ) , omg, 2 0 * log10 ( abs (H3 ) ) ) ; axis ( [ 0 , 6 , − 5 0 , 1 0 ] ) ; t i t l e ( ’ |H( s ) | ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’dB Magnitude ’ ) ; legend ( ’ Butterworth ’ , ’ Chebyshev type−I ’ , ’ Chebyshev type−I I ’ ) ; grid ;

10.31. Design the filter: % S e t up f i l t e r s p e c i f i c a t i o n s omg1 = 1 ; omg2 = 2 . 5 ; Rp = 1 ; As = 30; % Design the lowpass prototype [N, omgc] = buttord (omg1, omg2, Rp , As , ’ s ’ ) [numL, denL ] = butter (N, omgc, ’ s ’ ) % Lowpass to highpass transformation [numH, denH] = lp2hp (num, den , 5 )

Evaluate and graph the dB magnitude: omg = [ 1 : 0 . 0 1 : 1 0 ] ; H = freqs (numH, denH,omg ) ; plot (omg, 2 0 * log10 ( abs (H) ) ) ; axis ( [ 0 , 10 , − 70 , 10] ) ; t i t l e ( ’ |H( \omega ) | _ {dB} ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude (dB) ’ ) ; grid ;

10.32. a.

The discrete-time filter can be designed using the following statements: num = [ 2 ] ; den = [ 1 , 3 , 2 ] ; [numz, denz ] = impinvar (num, den , 2 )

MATLAB response is

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

396 numz = 0 denz = 1.0000

0.2387

−0.9744

0.2231

and corresponds to H (z) =

0.2387 z 0 + 0.2387 z −1 = 1 − 0.9744 z −1 + 0.2231 z −2 z 2 − 0.9744 z + 0.2231

b. omg = [ 0 : 0 . 0 1 : 2 * pi ] ; G = freqs (num, den ,omg ) ; plot (omg, abs (G ) ) ; axis ( [ 0 , 2 * pi , 0 , 1 . 2 ] ) ; t i t l e ( ’ |G( \omega ) | ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ;

c. Omg = [ 0 : 0 . 0 1 : 1 ] * pi ; H = freqz (numz, denz ,Omg) ; plot (Omg, abs (H) , ’ r ’ ) ; axis ( [ 0 , pi , 0 , 1 . 2 ] ) ; t i t l e ( ’ |H( \Omega ) | ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ;

d.

Use ω = Ω/T s = 2 Ω. Graph |G (ω)| and |Y a (ω) /X a (ω)| up to f max = f s /2 = 1 Hz, or equivalently, up to ωmax = ωs /2 = 2π rad/s. plot (omg, abs (G) , ’b ’ , 2 *Omg, abs (H) , ’ r ’ ) ; axis ( [ 0 , 2 * pi , 0 , 1 . 2 ] ) ; t i t l e ( ’ |G( \omega ) | and | Y_ { a } ( \ omega) / X_ { a } ( \ omega ) | ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; legend ( ’ |G( \omega ) | ’ , ’ | Y_ { a } ( \ omega) / X_ { a } ( \ omega ) | ’ ) ; grid ;

10.33. a.

The discrete-time filter can be designed using the following statements:

397 num = [ 2 ] ; den = [ 1 , 3 , 2 ] ; [numz, denz ] = impinvar (num, den , 4 )

MATLAB response is numz = 0 denz = 1.0000

0.0861

−1.3853

0.4724

and corresponds to H (z) =

0 + 0.0861 z −1 0.0861 z = 2 −1 −2 1 − 1.3853 z + 0.4724 z z − 1.3853 z + 0.4724

b. omg = [ 0 : 0 . 0 1 : 4 * pi ] ; G = freqs (num, den ,omg ) ; plot (omg, abs (G ) ) ; axis ( [ 0 , 4 * pi , 0 , 1 . 2 ] ) ; t i t l e ( ’ |G( \omega ) | ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ;

c. Omg = [ 0 : 0 . 0 1 : 1 ] * pi ; H = freqz (numz, denz ,Omg) ; plot (Omg, abs (H) , ’ r ’ ) ; axis ( [ 0 , pi , 0 , 1 . 2 ] ) ; t i t l e ( ’ |H( \Omega ) | ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ;

d.

Use ω = Ω/T s = 4 Ω. Graph |G (ω)| and |Y a (ω) /X a (ω)| up to f max = f s /2 = 2 Hz, or equivalently, up to ωmax = ωs /2 = 4π rad/s. plot (omg, abs (G) , ’b ’ , 4 *Omg, abs (H) , ’ r ’ ) ; axis ( [ 0 , 4 * pi , 0 , 1 . 2 ] ) ; t i t l e ( ’ |G( \omega ) | and | Y_ { a } ( \ omega) / X_ { a } ( \ omega ) | ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; legend ( ’ |G( \omega ) | ’ , ’ | Y_ { a } ( \ omega) / X_ { a } ( \ omega ) | ’ ) ; grid ;

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

398

This design has a lower degree of aliasing compared to the one in Problem 10.32 due to the use of a higher sampling rate. It should also be noted, however, that the bandwidth of the equivalent analog system is approximately halved in the process.

10.34. a.

The discrete-time filter can be designed using the following statements: num = [ 2 ] ; den = [ 1 , 3 , 2 ] ; [numz, denz ] = b i l i n e a r (num, den , 2 )

MATLAB response is numz = 0.0667 denz = 1.0000

0.1333

0.0667

−0.9333

0.2000

and corresponds to H (z) =

0.0667 + 0.1333 z −1 + 0.0667 0.0667 z 2 + 0.1333 z + 0.0667 = 1 − 0.9333 z −1 + 0.2 z −2 z 2 − 0.9333 z + 0.2

b. omg = [ 0 : 0 . 0 1 : 2 * pi ] ; G = freqs (num, den ,omg ) ; plot (omg, abs (G ) ) ; axis ( [ 0 , 2 * pi , 0 , 1 . 2 ] ) ; t i t l e ( ’ |G( \omega ) | ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ;

c. Omg = [ 0 : 0 . 0 1 : 1 ] * pi ; H = freqz (numz, denz ,Omg) ; plot (Omg, abs (H) , ’ r ’ ) ; axis ( [ 0 , pi , 0 , 1 . 2 ] ) ; t i t l e ( ’ |H( \Omega ) | ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ;

d.

Use ω = Ω/T s = 2 Ω. Graph |G (ω)| and |Y a (ω) /X a (ω)| up to f max = f s /2 = 1 Hz, or equivalently, up to ωmax = ωs /2 = 2π rad/s.

399 plot (omg, abs (G) , ’b ’ , 2 *Omg, abs (H) , ’ r ’ ) ; axis ( [ 0 , 2 * pi , 0 , 1 . 2 ] ) ; t i t l e ( ’ |G( \omega ) | and | Y_ { a } ( \ omega) / X_ { a } ( \ omega ) | ’ ) ; xlabel ( ’ \omega ( rad / s ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; legend ( ’ |G( \omega ) | ’ , ’ | Y_ { a } ( \ omega) / X_ { a } ( \ omega ) | ’ ) ; grid ;

10.35. a. 1 2 3 4 5 6 7 8 9 10 11

h = [0.0000 ,0.0082 ,0.0000 , −0.0193 ,0.0000 , −0.0580 ,0.0000 ,0.2702 , . . . 0.0000 , −0.4000 ,0.0000 ,0.2702 ,0.0000 , −0.0580 ,0.0000 , −0.0193 ,... 0. 0000 , 0. 0082 , 0. 0000] ; Omg = [ −256:255]/256 * pi ; % Compute DFT f r e q u e n c i e s Hmag = abs ( f f t s h i f t ( f f t ( h , 5 1 2 ) ) ) ; % 512−point DFT plot (Omg,Hmag) ; axis ([ − pi , pi , 0 , 1 . 2 ] ) ; t i t l e ( ’ |H( \Omega ) | ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ;

b. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

c.

% I d e a l lowpass f i l t e r impulse response hLP = @(n ,Omgc) Omgc/ pi * sinc (Omgc* n/ pi ) ; % Eqn . ( 10. 172) % I d e a l bandpass f i l t e r impulse response hBP = @(n ,Omg1,Omg2) hLP (n ,Omg2)−hdLP(n ,Omg1) ; % Bandpass % Anonymous function f o r Hamming window wHamm = @(n ,M) 0.54 −0.46 * cos ( pi * ( n+M) /M) ; % Eqn . ( 10. 181) % Truncated impulse response f o r bandpass f i l t e r n = [ − 22: 22] ’ ; % N=45=2M+1 t h e r e f o r e M=22 hT = hBP(n , 0 . 7 * pi , 0 . 3 * pi ) ; % Let h1 be the impulse response o f the bandpas f i l t e r designed using % the Hamming window h1 = hT . *wHamm(n , 2 2 ) ; % Compute and graph the magnitude c h a r a c t e r i s t i c H1mag = abs ( f f t s h i f t ( f f t ( h1 , 5 1 2 ) ) ) ; plot (Omg, 1 0 * log10 (Hmag) ,Omg, 1 0 * log10 (H1mag ) ) ; axis ([ − pi , pi , − 50 , 20] ) ; t i t l e ( ’ |H( \Omega ) | _ {dB} and | H_ { 1 } ( \Omega ) | _ {dB} ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; legend ( ’ |H( \Omega ) | _ {dB} ’ , ’ | H_{ 1 } ( \Omega ) | _ {dB} ’ , ’ Location ’ , ’ North ’ ) ; grid ;

CHAPTER 10. ANALYSIS AND DESIGN OF FILTERS

400 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

% Anonymous function f o r Blackman window wBlk = @(n ,M) 0.42 −0.5 * cos ( pi * ( n+M) /M) + . . . 0. 08 * cos ( 2 * pi * ( n+M) /M) ; % Eqn . ( 10. 183) % Let h2 be the impulse response o f the bandpas f i l t e r designed using % the Blackman window h2 = hT . * wBlk (n , 2 2 ) ; % Compute and graph the magnitude c h a r a c t e r i s t i c H2mag = abs ( f f t s h i f t ( f f t ( h2 , 5 1 2 ) ) ) ; plot (Omg, 1 0 * log10 (H1mag) ,Omg, 1 0 * log10 (H2mag ) ) ; axis ([ − pi , pi , − 50 , 20] ) ; t i t l e ( ’ | H_{ 1 } ( \Omega ) | _ {dB} and | H_{ 2 } ( \Omega ) | _ {dB} ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; legend ( ’ | H_{ 1 } ( \Omega ) | _ {dB} ’ , ’ | H_ { 2 } ( \Omega ) | _ {dB} ’ , ’ Location ’ , ’ North ’ ) ; grid ;

10.36. a. function hd = s s _ f i r 1 (Omgc,N) n = [ 0 : N−1]; tdelay = (N−1)/2; hd = Omgc/ pi * sinc (Omgc * ( n−tdelay ) / pi ) ; end

b.

Compute and graph the impulse response h d [n]: n = [0:35]; hd = s s _ f i r 1 ( 0 . 4 * pi , 3 6 ) ; stem (n , hd ) ; t i t l e ( ’ Impulse response ’ ) ; xlabel ( ’n ’ ) ; ylabel ( ’h[ n ] ’ ) ;

Compute and graph the magnitude |Hd ()|: Omg = [ −256:255]/256 * pi ; Hmag = abs ( f f t s h i f t ( f f t (hd , 5 1 2 ) ) ) ; plot (Omg,Hmag) ; axis ([ − pi , pi , 0 , 1 . 2 ] ) ; t i t l e ( ’ |H( \Omega ) | ’ ) ; xlabel ( ’ \Omega ( rad ) ’ ) ; ylabel ( ’ Magnitude ’ ) ; grid ;

Chapter 11 Amplitude Modulation 11.1. a.

µ=

Am 2 = = 0.7 Ac Ac

b.

µ=

Am 8 = = 0.85 Ac Ac

c.

µ=

Am Am = 0.9 = Ac 2

d.

µ=

A m 2.5 = = 0.83 Ac 3

⇒ ⇒ ⇒

Ac =

2 = 2.86 0.7

Ac =

3 = 3.53 0.85

A m = 1.8

11.2. a. 5

0

−5 0

0.5

1

1.5

2

1.5

2

t (ms)

b. 5

0

−5 0

0.5

1

t (ms)

401

CHAPTER 11. AMPLITUDE MODULATION

402

c. 10 5 0 −5 −10

0

0.2

0.4

0.6

0.8

1

t (ms)

11.3. a. m (t) 5

m min = −2.5746 0

Ac =

|m min | 2.5746 = = 3.68 µ 0.7 −5

0

0.5

1

1.5

2

2.5

3

3.5

4

2.5

3

3.5

4

t (sec)

b. m (t) 5

m min = −4 0

Ac =

|m min | 4 = = 5.71 µ 0.7 −5

0

0.5

1

1.5

2

t (sec)

403

c. m (t) 5

m min = −4.971 0

Ac =

|m min | 4.971 = = 7.10 µ 0.7 −5

0

0.5

1

1.5

2

2.5

3

3.5

4

t (sec)

11.4. a.

The signal x AM (t ) is £ ¡ ¢¤ ¡ ¢ x AM (t ) = A c 1 + µ cos 2π f m t cos 2π f c t £ ¤ = 3 1 + 0.6 cos (4πt ) cos (20πt )

= 3 cos (20πt ) + 0.9 cos (16πt ) + 0.9 cos (24πt )

The carrier signal can be written as n o n o x car (t ) = Re X car e j 2π f c t = Re 3 e j 20πt

Similarly, lower and upper sideband signals are n o n o x lsb (t ) = Re X lsb e j 2π( f c − f m )t = Re 0.9 e j 16πt

and

n o n o x usb (t ) = Re X usb e j 2π( f c + f m )t = Re 0.9 e j 24πt

At the time instant t = 0.35 seconds we get o n © ª x car (0.35) = Re 3 e j 21.9911 = Re −3 + j 0 n o © ª x lsb (0.35) = Re 0.9 e j 17.5929 = Re 0.2781 − j 0.8560 n o © ª x usb (0.35) = Re 0.9 e j 26.3894 = Re 0.2781 + j 0.8560 im

˜ usb X fm

X car

−fm

fc ˜ lsb X

re

CHAPTER 11. AMPLITUDE MODULATION

404

b.

The signal x AM (t ) is £ ¡ ¢¤ ¡ ¢ x AM (t ) = A c 1 + µ cos 2π f m t cos 2π f c t £ ¤ = 3 1 + 1.3 cos (4πt ) cos (20πt )

= 3 cos (20πt ) + 1.95 cos (16πt ) + 1.95 cos (24πt )

The carrier signal can be written as o o n n x car (t ) = Re X car e j 2π f c t = Re 3 e j 20πt

Similarly, lower and upper sideband signals are o o n n x lsb (t ) = Re X lsb e j 2π( f c − f m )t = Re 1.95 e j 16πt

o o n n x usb (t ) = Re X usb e j 2π( f c + f m )t = Re 1.95 e j 24πt

and

At the time instant t = 0.6 seconds we get n o © ª x car (0.6) = Re 3 e j 37.6991 = Re 3 + j 0 n o © ª x lsb (0.6) = Re 1.95 e j 30.1593 = Re 0.6026 − j 1.8546 n o © ª x usb (0.35) = Re 1.95 e j 45.2389 = Re 0.6026 + j 1.8546 im ˜ usb X fm

fc X car

re −fm ˜ lsb X

c.

The modulation index is µ=

Am 4 = = 0.8 Ac 5

and the signal x AM (t ) is £ ¡ ¢¤ ¡ ¢ x AM (t ) = A c 1 + µ cos 2π f m t cos 2π f c t £ ¤ = 5 1 + 0.8 cos (16πt ) cos (50πt )

= 5 cos (50πt ) + 2 cos (34πt ) + 2 cos (66πt )

The carrier signal can be written as n o n o x car (t ) = Re X car e j 2π f c t = Re 5 e j 50πt

405 Similarly, lower and upper sideband signals are o o n n x lsb (t ) = Re X lsb e j 2π( f c − f m )t = Re 2 e j 34πt

o o n n x usb (t ) = Re X usb e j 2π( f c + f m )t = Re 2 e j 66πt

and

im

At the time instant t = 150 milliseconds we get o n © ª x car (0.6) = Re 5 e j 23.5619 = Re − j 5 o n © ª x lsb (0.6) = Re 2 e j 16.0221 = Re −1.9021 − j 0.6180 o n © ª x usb (0.35) = Re 2 e j 31.1018 = Re 1.9021 − j 0.6180

re X car fm

fc −fm

˜ usb X

˜ lsb X

11.5. The AM signal is in the form ¤ ¢ µA c ¤ ¢ ¡ £ ¡ £ ¡ ¢ µA c cos 2π f c − f m t + cos 2π f c + f m t x AM (t ) = A c cos 2π f c t + 2 2 and the frequency spectrum is in the form ¡ ¢ Ac ¡ ¢ Ac ¡ ¢ µA c ¡ ¢ µA c ¡ ¢ X AM f = δ f − fc + δ f + fc + δ f − fc − fm + δ f − fc + fm 2 2 4 4

+

a.

¢ µA c ¡ ¢ µA c ¡ δ f + fc + fm + δ f + fc − fm 4 4

¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ X AM f = 1.4286 δ f − f c + 1.4286 δ f + f c + 0.5 δ f − f c − f m + 0.5 δ f − f c + f m ¢ ¢ ¡ ¡ + 0.5 δ f + f c + f m + 0.5 δ f + f c − f m XAM (f )

1.4286

1.4286

−fc−fm −fc −fc+fm

fc−fm fc fc+fm f

CHAPTER 11. AMPLITUDE MODULATION

406

b. ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ X AM f = 1.7647 δ f − f c + 1.7647 δ f + f c + 0.75 δ f − f c − f m + 0.75 δ f − f c + f m ¢ ¢ ¡ ¡ + 0.75 δ f + f c + f m + 0.75 δ f + f c − f m XAM (f ) 1.7647

1.7647

−fc−fm −fc −fc+fm

fc−fm fc fc+fm f

c. ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ X AM f = δ f − f c + δ f + f c + 0.45 δ f − f c − f m + 0.45 δ f − f c + f m ¢ ¢ ¡ ¡ + 0.45 δ f + f c + f m + 0.45 δ f + f c − f m XAM (f )

1

1

−fc−fm −fc −fc+fm

fc−fm fc fc+fm f

d. ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ X AM f = 1.5 δ f − f c + 1.5 δ f + f c + 0.625 δ f − f c − f m + 0.625 δ f − f c + f m ¢ ¢ ¡ ¡ + 0.625 δ f + f c + f m + 0.625 δ f + f c − f m XAM (f ) 1.5

1.5

−fc−fm −fc −fc+fm

fc−fm fc fc+fm f

407

11.6. a. Using A c = 3.68 we obtain x AM (t ) = 3.68 cos (20πt ) + cos (4πt ) cos (20πt ) + 2 cos (6πt ) cos (20πt ) = 3.68 cos (20πt ) +

1 1 cos (16πt ) + cos (24πt ) + cos (14πt ) + cos (26πt ) 2 2

¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ X AM f = 1.84 δ f − 10 + 1.84 δ f + 10 + 0.25 δ f − 8 + 0.25 δ f + 8 + 0.25 δ f − 12 + 0.25 δ f + 12 ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ + 0.5 δ f − 7 + 0.5 δ f + 7 + 0.5 δ f − 13 + 0.5 δ f + 13 |XAM (f )|

1.84

−20

−10

1.84

0

10

20

f (Hz)

b.

Let us write m (t ) as m (t ) = cos (4πt ) + 3 cos (6πt − π/2)

Using A c = 5.71 we obtain x AM (t ) = 5.71 cos (20πt ) + cos (4πt ) cos (20πt ) + 3 cos (6πt − π/2) cos (20πt ) = 5.71 cos (20πt ) +

1 3 3 1 cos (16πt ) + cos (24πt ) + cos (14πt + π/2) + cos (26πt − π/2) 2 2 2 2

¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ X AM f = 2.855 δ f − 10 + 2.855 δ f + 10 + 0.25 δ f − 8 + 0.25 δ f + 8 + 0.25 δ f − 12 + 0.25 δ f + 12 ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ + j 0.75 δ f − 7 − j 0.75 δ f + 7 − j 0.75 δ f − 13 + j 0.75 δ f + 13 |XAM (f )|

2.885

−20

−10

2.885

0

f (Hz)

10

20

CHAPTER 11. AMPLITUDE MODULATION

408 6

XAM (f )

3

Phase (rad)

2 1 0 −1 −2 −3 −20

−15

−10

−5

0

5

10

15

20

f (Hz)

c.

Let us write m (t ) as m (t ) = 3 cos (4πt − π/2) + 2 cos (6πt − π/3)

Using A c = 7.10 we obtain x AM (t ) = 7.10 cos (20πt ) + 3 cos (4πt − π/2) cos (20πt ) + 2 cos (6πt − π/3) cos (20πt ) = 7.10 cos (20πt ) +

3 3 cos (16πt + π/2) + cos (24πt − π/2) + cos (14πt + π/3) + cos (26πt − π/3) 2 2

¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ X AM f = 3.55 δ f − 10 + 3.55 δ f + 10 + j 0.75 δ f − 8 − j 0.75 δ f + 8 − j 0.75 δ f − 12 + j 0.75 δ f + 12 ¡ ¢ ¡ ¢ ¡ ¢ ¡ ¢ + 0.5 e j π/3 δ f − 7 + 0.5 e − j π/3 δ f + 7 + 0.5 e − j π/3 δ f − 13 + 0.5 e j π/3 δ f + 13 |XAM (f )| 3.55

−20

3.55

−10

0

10

20

f (Hz)

6

XAM (f )

3

Phase (rad)

2 1 0 −1 −2 −3 −20

−15

−10

−5

0

f (Hz)

5

10

15

20

409

11.7. ¡ ¢ ¡ ¢ x AM (t ) = A c cos 2π f c t + m (t ) cos 2π f c t ¢ Ac ¡ ¢ 1 ¡ ¢ 1 ¡ ¢ ¡ ¢ Ac ¡ δ f − fc + δ f + fc + M f − fc + M f + fc X AM f = 2 2 2 2 XAM (f ) 0.75

0.75 0.5

−13

−10

−7

7

10

13

f (kHz)

11.8. a.

The terms in the spectrum of the AM signal are listed in the table below: Index 1 2 3 4

Term ¤ 3 1 + 0.8 m (t ) cos (20πt ) 2 m (t ) 1.8 cos2 (20πt ) 0.8 m 2 (t ) £

Frequencies 7 ≤ f ≤ 13 kHz −3 ≤ f ≤ 3 kHz f = 0, and f = 20 kHz −6 ≤ f ≤ 6 kHz

M (f )

(a)

−3

f (kHz)

3 Vout (f )

(b) 1 3

3

2

1 3

4

−10

f (kHz) 10

BT = 6 kHz

CHAPTER 11. AMPLITUDE MODULATION

410

b.

To select the desirable terms (row 1 of the table), the passband of the filter should span the frequency range 7 ≤ f ≤ 13 kHz

11.9. a.

The carrier completes 15 full cycles in a duration of 1 ms. Therefore the carrier frequency is f c = 15 kHz.

b.

Observing the envelope, we see that it completes one full cycle in about 0.84 ms. The message frequency is approximately f m = 1.2 kHz.

c.

For a single-tone modulated AM carrier the maximum value of the positive envelope is ¯ ¯ ¯x AM (t )¯

max

≈ Ac + Am

and the minimum value of the positive envelope is ¯ ¯ ¯x AM (t )¯

From the graph we observe

Ac + Am ≈ 7

min ≈ |A c

− Am |

and

Ac − Am ≈ 7

and

Am ≈ 3

yielding carrier and message amplitudes Ac ≈ 4 The modulation index is µ=

d.

Am ≈ 0.75 Ac

Approximate efficiency is η≈

(0.75)2 2 + (0.75)2

= 0.2195

11.10. a. A 2c = (3.68)2 = 13.5424

® 1 4 m 2 (t ) = + = 2.5 2 2 ­ 2 ® m (t ) 2.5 η= 2 ­ 2 ® = = 0.1558 13.5424 + 2.5 A c + m (t ) ­

b.

A 2c = (5.71)2 = 32.6041 ­

® 1 9 m 2 (t ) = + = 5 2 2

411 ­ 2 ® m (t ) 5 = 0.1330 η= 2 ­ 2 ® = 32.6041 + 5 A c + m (t )

c.

A 2c = (7.10)2 = 50.41

® 9 4 m 2 (t ) = + = 6.5 2 2 ­ 2 ® m (t ) 6.5 = 0.1142 η= 2 ­ 2 ® = 50.41 + 6.5 A c + m (t ) ­

11.11. a. vin (t) 10 5 0 −5 −10

0

5

10

15

20

25

30

20

25

30

t (ms)

b. vout (t) 10 5 0 −5 −10

0

5

10

15

t (ms)

c.

The message bandwidth is W = 100 Hz. The carrier frequency is f c = 800 Hz. The significant frequency components of the modulated carrier are at 700, 800 and 900 Hz. A reasonable choice for a bandpass filter to isolate the AM signal may be one with a passband for 600 < f < 1000 Hz

CHAPTER 11. AMPLITUDE MODULATION

412

11.12. The output of the first AM modulator: £ ¤ ¡ ¢ w 1 (t ) = A c + m (t ) cos 2π f c t

The output of the second AM modulator:

£ ¤ ¡ ¢ w 2 (t ) = A c − m (t ) cos 2π f c t

Adding the two signals yields

¡ ¢ x DSB (t ) = w 1 (t ) + w 2 (t ) = 2 A c cos 2π f c t

11.13. a.

¡ ¢ x DSB (t ) = A c m (t ) cos 2π f c t

¢ Ac ¡ ¢ ¡ ¢ Ac ¡ M f − fc + M f + fc X DSB f = 2 2 XDSB (f ) 0.75

−13

−10

−7

7

10

13

10

13

f (kHz)

b. XSSB,U (f ) 0.75

−13

−10

f (kHz)

c. XSSB,L (f ) 0.75

−10

−7

7

10

f (kHz)

413

11.14. The modulation index is

3.2 = 0.8 4 Using Eqn. (11.64) with µ = 0.8 and f m = 100 Hz, we obtain µ=

1 τ≤ 200π

Ãp

! 1 − (0.8)2 = 0.0012 0.8

Let C = 1 µF. Required resistor value is R = 1200 Ω.

11.15. The signal r (t ) is

£ ¤ ¡ ¢ ¡ ¢ r (t ) = 2 A c + m (t ) cos 2π f c t cos 2π f c t + θ

Using the appropriate trigonometric identity

¡ ¢ ¡ ¢ r (t ) = A c cos 4π f c t + θ + A c cos (θ) + m (t ) cos 4π f c t + θ + m (t ) cos (θ)

The bandpass filter removes the first three terms, resulting in

ˆ (t ) = m (t ) cos (θ) m The effect of the phase error is multiplication of the message signal by the factor cos (θ) which is a constant as long as the phase error θ is constant. For values of θ close to π/2 radians, the factor cos (θ) is small, potentially degrading the performance of the demodulator in the presence of random noise.

11.16. The signal r (t ) is

£ ¤ ¡ ¢ ¡ £ ¤ ¢ r (t ) = 2 A c + m (t ) cos 2π f c t cos 2π f c + ∆ f t

Using the appropriate trigonometric identity

¡ £ ¤ ¢ ¡ ¢ ¡ £ ¤ ¢ ¡ ¢ r (t ) = A c cos 2π 2 f c + ∆ f t + A c cos 2π ∆ f t + m (t ) cos 2π 2 f c + ∆ f t + m (t ) cos 2π ∆ f t

The bandpass filter removes the first three terms, resulting in

¡ ¢ ˆ (t ) = m (t ) cos 2π ∆ f t m

¡ ¢ The effect of the frequency error is multiplication of the message signal by the factor cos 2π ∆ f t . The ˆ (t ) is still a modulated term, with the carrier frequency ∆ f . output signal m

11.17. Let us begin by defining an anonymous function for the AM signal: 1

xAM = @( t , Ac ,mu, fc , fm) Ac * (1+mu* cos ( 2 * pi * fm * t ) ) . * cos ( 2 * pi * fc * t ) ;

CHAPTER 11. AMPLITUDE MODULATION

414

Afterwards, each AM signal in question can be produced using this anonymous function.

a. 1 2 3 4

t = [ 0 : 2 e −6:2e −3]; plot (1000 * t ,xAM( t , 3 , 0 . 6 , 1 0 0 0 0 , 2 0 0 0 ) ) ; axis ( [ 0 , 2 , − 8 , 8 ] ) ; xlabel ( ’ t (ms) ’ ) ;

b. 1 2 3 4

t = [ 0 : 2 e −6:2e −3]; plot (1000 * t ,xAM( t , 3 , 1 . 3 , 1 0 0 0 0 , 2 0 0 0 ) ) ; axis ( [ 0 , 2 , − 8 , 8 ] ) ; xlabel ( ’ t (ms) ’ ) ;

c. 1 2 3 4 5

mu = 4/ 5; t = [ 0 : 1 e −6:1e −3]; plot (1000 * t ,xAM( t , 3 ,mu, 25000 , 8000) ) ; axis ( [ 0 , 1 , − 1 0 , 1 0 ] ) ; xlabel ( ’ t (ms) ’ ) ;

11.18. a. 1 2 3 4 5 6 7 8

t = [0:0.005:4]; Ac = 3 . 6 8 ; m = cos ( 4 * pi * t )+2 * cos ( 6 * pi * t ) ; x_am = ( Ac+m) . * cos (20 * pi * t ) ; plot ( t , x_am ) ; xlabel ( ’ t ( sec ) ’ ) ; t i t l e ( ’ x_ {AM} ( t ) ’ ) ; grid ;

b. 1 2 3 4 5 6 7

Ac = 5 . 7 1 ; m = cos ( 4 * pi * t )+3 * sin ( 6 * pi * t ) ; x_am = ( Ac+m) . * cos (20 * pi * t ) ; plot ( t , x_am ) ; xlabel ( ’ t ( sec ) ’ ) ; t i t l e ( ’ x_ {AM} ( t ) ’ ) ; grid ;

415

c. 1 2 3 4 5 6 7

Ac = 7 . 1 0 ; m = 3 * sin ( 4 * pi * t )+2 * cos ( 6 * pi * t−pi / 3 ) ; x_am = ( Ac+m) . * cos (20 * pi * t ) ; plot ( t , x_am ) ; xlabel ( ’ t ( sec ) ’ ) ; t i t l e ( ’ x_ {AM} ( t ) ’ ) ; grid ;

11.19. a.

Using A c = 3.68 we obtain x AM (t ) = 3.68 cos (20πt ) + cos (4πt ) cos (20πt ) + 2 cos (6πt ) cos (20πt ) = 3.68 cos (20πt ) +

1 1 cos (16πt ) + cos (24πt ) + cos (14πt ) + cos (26πt ) 2 2

The frequency components in x AM (t ) are f 1 = 10 Hz ,

f 2 = 8 Hz ,

f 3 = 12 Hz ,

f 4 = 7 Hz ,

f 5 = 13 Hz

Fundamental frequency f 0 is found through (see Section 1.3.4) m1 m2 m3 m4 m5 1 = = = = = f0 10 8 12 7 13 Using the integers m 1 = 10 ,

m2 = 8 ,

m 3 = 12 ,

m4 = 7 ,

m 5 = 13

the fundamental frequency is found as f 0 = 1 Hz corresponding to a fundamental period of T0 = 1 s. In order to generate 1024 samples in one period, samples need to be taken 1/1024 seconds apart. 1 2 3 4 5 6 7 8

b.

t = [ 0: 1023] / 102 4 ; x_am = 3. 68 * cos (20 * pi * t )+ cos ( 4 * pi * t ) . * cos (20 * pi * t )+2 * cos ( 6 * pi * t ) . * cos (20 * pi * t ) ; k =[ −20:20]; c = ss_efsapprox (x_am , k ) ; stem ( k , c ) axis ( [ − 2 0 . 5 , 2 0 . 5 , 0 , 3 ] ) ; xlabel ( ’ k ’ ) ; t i t l e ( ’ | c_ { k } | ’ ) ;

Writing m (t ) as m (t ) = cos (4πt ) + 3 cos (6πt − π/2)

and using A c = 5.71 we obtain x AM (t ) = 5.71 cos (20πt ) + cos (4πt ) cos (20πt ) + 3 cos (6πt − π/2) cos (20πt ) = 5.71 cos (20πt ) +

1 3 3 1 cos (16πt ) + cos (24πt ) + cos (14πt + π/2) + cos (26πt − π/2) 2 2 2 2

As in part (a) of the problem, the fundamental frequency and the fundamental period are f 0 = 1 Hz and T0 = 1 s respectively.

CHAPTER 11. AMPLITUDE MODULATION

416 1 2 3 4 5 6 7 8

c.

t = [ 0: 1023] / 102 4 ; x_am = 5. 71 * cos (20 * pi * t )+ cos ( 4 * pi * t ) . * cos (20 * pi * t )+3 * cos ( 6 * pi * t−pi / 2 ) . * cos (20 * pi * t ) ; k =[ −20:20]; c = ss_efsapprox (x_am , k ) ; stem ( k , abs ( c ) ) ; axis ( [ − 2 0 . 5 , 2 0 . 5 , 0 , 3 ] ) ; xlabel ( ’ k ’ ) ; t i t l e ( ’ | c_ { k } | ’ ) ;

Writing m (t ) as m (t ) = 3 cos (4πt − π/2) + 2 cos (6πt − π/3)

and using A c = 7.10 we obtain x AM (t ) = 7.10 cos (20πt ) + 3 cos (4πt − π/2) cos (20πt ) + 2 cos (6πt − π/3) cos (20πt ) 3 3 3 3 cos (16πt + π/2) + cos (24πt − π/2) + cos (14πt + π/3) + cos (26πt − π/3) 2 2 2 2 As in parts (a) and (b) of the problem, the fundamental frequency and the fundamental period are f 0 = 1 Hz and T0 = 1 s respectively. = 7.10 cos (20πt ) +

1 2 3 4 5 6 7 8

t = [ 0: 1023] / 102 4 ; x_am = 7. 10 * cos (20 * pi * t )+3 * cos ( 4 * pi * t−pi / 2 ) . * cos (20 * pi * t )+2 * cos ( 6 * pi * t−pi / 3 ) . * cos (20 * pi * t ) ; k =[ −20:20]; c = ss_efsapprox (x_am , k ) stem ( k , abs ( c ) ) ; axis ( [ − 2 0 . 5 , 2 0 . 5 , 0 , 4 ] ) ; xlabel ( ’ k ’ ) ; t i t l e ( ’ | c_ { k } | ’ ) ;

11.20. 1 2 3 4 5 6 7 8 9 10 11

Compute and graph the signal v i n (t ).

t = [0:0.05:30]/1000; Bc = 5 ; fc = 800; c a r r i e r = Bc * cos ( 2 * pi * fc * t ) ; message = 3 * cos (200 * pi * t ) ; vin = c a r r i e r +message ; plot (1000 * t , vin ) ; axis ( [ 0 , 30 , − 10 , 10] ) ; xlabel ( ’ t (ms) ’ ) ; t i t l e ( ’ v_ { in } ( t ) ’ ) ; grid ;

Compute and graph the signal v out (t ). 1 2 3 4 5 6

vout = vin . * ( vin >=0); plot (1000 * t , vout ) ; axis ( [ 0 , 30 , − 10 , 10] ) ; xlabel ( ’ t (ms) ’ ) ; t i t l e ( ’ v_ { out } ( t ) ’ ) ; grid ;

417 Compute and graph the signal x AM (t ) using the function ss_switchmod(..) 1 2 3 4 5 6

x_am = ss_switchmod ( message , Bc , fc , 0 . 0 5 e −3 ,600 ,1000); plot (1000 * t , x_am ) ; axis ( [ 0 , 3 0 , − 5 , 5 ] ) ; xlabel ( ’ t (ms) ’ ) ; t i t l e ( ’ x_ {AM} ( t ) ’ ) ; grid ;

11.21. 1 2 3 4 5 6 7 8 9 10 11 12

Function ss_switchmod2(..).

function x_am = ss_switchmod2 (msg, Bc , fc , Ts , f1 , f2 ) nSamp = length (msg ) ; % Number o f samples in "msg " . t = [ 0 : nSamp−1] * Ts ; % Vector o f time i n s t a n t s . c a r r i e r = Bc * cos ( 2 * pi * fc * t ) ; % Compute input to the diode switch . v_in = c a r r i e r +msg ; % Eqn . ( 1 1 . 3 6 ) % Simulate the diode switch . v_out = ( v_in − 0 . 6 ) . * ( v_in > = 0 . 6 ) ; % Design the bandpass f i l t e r . [numz, denz ] = butter ( 5 , [ 2 * f1 * Ts , 2 * f2 * Ts ] , ’ bandpass ’ ) ; % P r o c e s s switch output through bandpass f i l t e r . x_am = f i l t e r (numz, denz , v_out ) ;

Compute and graph the signal x AM (t ) using the functions ss_switchmod(..) and ss_switchmod2(..) 1 2 3 4 5 6

x_am1 = ss_switchmod ( message , Bc , fc , 0 . 0 5 e −3 ,600 ,1000); x_am2 = ss_switchmod2 ( message , Bc , fc , 0 . 0 5 e −3 ,600 ,1000); plot (1000 * t , x_am1 , ’b ’ ,1000 * t , x_am2 , ’ r ’ ) ; axis ( [ 0 , 3 0 , − 5 , 5 ] ) ; xlabel ( ’ t (ms) ’ ) ; grid ;

11.22. Script to simulate envelope detector using the value τ = 0.0012 seconds found in Problem 11.14: 1 2 3 4 5 6 7

t = [ 0 : 2 0 e−6:20e −3]; x_am = ( 4+ 3. 2 * cos (200 * pi * t ) ) . * cos (2000 * pi * t ) ; Ts = 20e−6; tau = 0. 0012; x_env = ss_envdet (x_am , Ts , tau ) ; plot ( t , x_am , ’b ’ , t , x_env , ’ r ’ ) ; grid ;

Repeat using half the time constant value found:

CHAPTER 11. AMPLITUDE MODULATION

418 1 2 3

x_env = ss_envdet (x_am , Ts , 0 . 5 * tau ) ; plot ( t , x_am , ’b ’ , t , x_env , ’ r ’ ) ; grid ;

Repeat using twice the time constant value found: 1 2 3

x_env = ss_envdet (x_am , Ts , 2 * tau ) ; plot ( t , x_am , ’b ’ , t , x_env , ’ r ’ ) ; grid ;

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