Essentials of Robust Control (Solution Manual) 0135258332, 9780135258330

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Chapter 2

Linear Algebra Solution 2.1 Let the columns of A be denoted as a1; a2; a3. Then a3 = a1 a2. Hence the rank of A is 2

and

011 Im(A) = spanfa1 ; a2 g; Ker(A) = spanf@ 1 Ag;

Solution 2.2 D = orth(D0 ) and D? = null(D0)

0 0:4287 D = @ 0:5663

1

011 011 Im(A ) = spanf@ 1 A ; @ 0 Ag

1

0

0

1

1

0:8060 0:4082 0:1124 A ; D? = @ 0:8165 A 0:7039 0:5812 0:4082 Solution 2.3 It is clear that (A) := maxi;j jaij j satis es the norm conditions. Now let   A = 13 24 ; 4 = (A) < (A) = 5:37 and 1 2 1 1 A = 3 4 ; B = 1 1 ; 7 = (AB ) > (A)(B ) = 4:

Solution 2.4 Since the rank of A is 2, all solutions can be found by letting

 1 2  x   3   0  1 4 1 x2 = 1 x3 + 1

and so

 x   1 2  1  3  1 x2

= 4 1

   0:7143   0:2857  = 1:8571 x3 + 0:1429 0 x 1 0 0:7143 1 0 0:2857 1 @ x12 A = @ 1:8571 A  + @ 0:1429 A 0 1 x3 + 1

x3

1 for any . The minimal norm solution is given by

0

0 0:2181 1 x = A (AA ) 1 B = @ 0:0329 A

Solution 2.5

0:0947

 

x = (A A) 1 A B = 2:75 1

LINEAR ALGEBRA

2

Solution 2.6 Use MATLAB to get

0 0:7379 0:4243 V = @ 0:4216 0:5657

where Then X

A = V DV

1

1

0

0:8944 1 0:4472 A ; D = @ 0 0:5270 0:7071 0:0000 0 (A) = spanfv2 ; v3 g and X+ (A) = spanfv1 g.

1

0 3 0

0 0A 2

Solution 2.7 It is obvious that a1 = Pni=1 i and an = ( 1)n Qni=1 i . It is also easy to see that det(I

A) = n (a11 + a22 + : : : + ann )n 1 +


+ an . Hence a1 = trace(A). Let  = 0 to get an = det( A) = ( 1)n det(A)

Solution 2.8 It follows easily from matrix inversion lemma



A : (A 1 + xy ) 1 = A Ax(1 + y Ax) 1 y A = A 1Axy  + y Ax

Furthermore

det(A 1 + xy ) 1 = det A(I + xy A) 1 = det A det(I + xy A) 1 = det A (det(I + xy A)) 1 = det A (det(I + y Ax)) 1 = 1 +detyAAx : Solution 2.9 Obvious.

Solution 2.10

R = x2Cnmax jx Axj  max jx Ayj =  (A) ; kxk=1 kxk=1;kyk=1 Let x1 and 1 be such that Ax1 =  1x1 , kx1 k = 1, and (A) = j1 . Then If A = A , then (A) = (A).

R  jx1 Ax1 j = j1 x1 x1 j = j1 j:

Solution 2.11 Solution 2.12

B (I + AB ) = (I + BA)B =) B (I + AB ) 1 = (I + BA) 1 B: 1. The eigenvalues of H are f2; 3; 2; 3g and the eigenvectors corresponding to 2 and

3 are the columns of

2 0 3 0:707 6 :5774 0:7071 77 V = 64 00:5774 0 5

0:5774 0 which form a basis for the stable invariant subspace. 2. The eigenvalues of H are f 1  j; 1; 1g and the eigenvector corresponding to 1 is

2 0:2887 3 6 2887 77 V = 64 00::2887 5 0:8860

which is a basis for the stable invariant subspace. 3. The eigenvalues of H are 0:4142j; 0:4142j; 2:4142j; 2:4142j and there is no stable invariant subspace.

3 p=1,1.5,2,5,10, infinity 1.5

1

0.5

0

−0.5

−1

−1.5 −1.5

−1

−0.5

0

0.5

1

1.5

Figure 2.1: Visualizing the vector p-norm

Solution 2.13 The vector p-norm for p = 1; 1:5; 2; 5; 10; 1 are shown. Solution 2.14 It is esay to verify that I (I + A + A2 +


+ An )(I A) = An+1 Since kAk < 1, we have

I (I + A + A2 +


+ An )(I A)

=

An+1

! 0; n ! 1:

Hence

I (I + A + A2 +


)(I A) = 0 or (I A) 1 = I + A + A2 +




Next

(I A) 1

=

I + A + A2 +




 1 + kAk + kAk2 +


= 1 1 kAk

Finally, note that and

(I A)(I A) 1 = I

1 = (I A)(I A)

Solution 2.15



1

 kI



Ak (I A) 1  (1 + kAk) (I A) 1

Note that for any v 2 C k , we have kv k1  kv k2 

p

k kvk1

These inequalities imply that for any x 2 C n :

1 kAxk  kAxk  kAxk 2 1 2 m

p k2 =) p1 kkAx xk

m

2

1 kAxk2

p

m kxk1



kAxk1 kAxk2 kAxk2   1 pn kxk2 kxk1 kxk1

p =) p1m kAk2  kAk1  n kAk2

k2 = n kkAx xk2 p

LINEAR ALGEBRA

4 

Similarily, for any v 2 C k , we have kv k2  kv k1 

p

These inequalities imply that for any x 2 C n : kAxk2  kAxk1 

=)

k kvk2

p

m kAxk2 1 kAxk2  kAxk2  kAxk1  pm kAxk2 p kxk1 kxk1 kxk1 n kxk2



p =) p1 kAk2  kAk1  m kAk2

p kAxk2 m kxk 2

n



Finally, for any v 2 C k , we have

kv k1  kv k1  k kv k1

These inequalities imply that for any x 2 C n :

kAxk1  kAxk1  m kAxk1 k1 kAxk1 kAxk1 kAxk1 =) n1 kkAx   m xk1 kxk1 kxk1 kxk1 =) n1 kAk1  kAk1  m kAk1

Solution 2.16

kA B k22

m

kAxk1 kxk1

= max [(A B ) (A B )] = max [(A A) (B  B )] = max (A A)max (B  B ) = kAk22 kB k22 :

Solution 2.17

kxy  k2

p

p

p

= max (xy (xy ) ) = y y max (xx ) p

p

= y y x x = kxk kyk :

Solution 2.18 The geometric interpolation of SVD: M = U V  U

V*

S

Figure 2.2: Geometric Interpolation of SVD

Solution 2.19

A 12



0:5574 + 0:5574j = 0:55740:5574 0:5574j 1:1547





1



and B = 1 j .

5

Solution 2.20

See Horn and Johnson. Let P = P  =

P

i (P11 ); 8 1  i  k.

P12 P22

11 P12





0 with P11 2 C kk . Show i (P ) 

p;m)  2  min(p; m) 2 . Solution 2.21 Let R = U V . Then trace RR = trace V 2 V  = trace 2 = Pmin( 1 i i=1 kRkF 

Solution 2.22



p

min(p; m)  max (R):

Note that for a suciently small  > 0

8 9 < = X  jAij j kI + Ak1 kI k1 = max 1 + Aii + i : ; 1 j =1;j 6=i 8 9 < = X =  max A + jAij j ii i : ; j =1;j 6=i 8 9 < = X kI + Ak1 kI k1 = max A + jAij j 1 (A) = ! lim+ ii i : ;  0 j =1;j 6=i

so



8 9 < = X kI + Ak1 kI k1 = max 1 + Ajj +  jAij j j : ; 1 i=1;i6=j 8 9 < = X =  max A + jAij j jj j : ; i=1;i6=j 8 9 < = X kI + Ak1 kI k1 1 (A) = ! lim+ jAij j = max A + jj i : ;  0 i=1;i6=j

so

 kI + Ak2 

p

kI k2

p

1 + max (A + A)

so

n

2 (A) = ! lim+ kI + Ak2 0

X I   I = 1 I Y

X I  I Y  0 () Y X

X

1 = max (I + (A + A) + 2 A A)

o

1  1 + 2 max (A + A)

Solution 2.23 Note that so

p

= max (I + A) (I + A)

0

I

 X

kI k2

0 Y

0

1 = 2 max (A + A)

 = max ( A 2+ A )

X

1

 I X 1  0

I

min (X 1=2 Y X 1=2 )  1 () min (XY )  1:

1  0 () 

1

LINEAR ALGEBRA

6

Solution 2.24

01 P = @2

1

2 3 5 5 A = V DV  3 5 10

where Hence

Solution 2.25

So

0 0:9623 V = @ 0:1925

1

0

1

0:0596 0:2656 0 0 0 0:8389 0:5092 A ; D = @ 0 1:9172 0 A 0:1925 0:5410 0:8187 0 0 14:0828

0 0:2696 0:5766 0:7713 1 P 1=2 = V D1=2 V  = @ 0:5766 1:9473 0:9358 A 0:7713 0:9358 2:9205

09 P = @ 19

1 0

1

19 29 1 2 1 0 A @ 41 63 = 3 4 A 0 2 29 63 97 5 6



1 p

3 p

5 p

C= 2 2 4 2 6 2

 0 1 2 1 @3 4A 5 6



Solution 2.26

2 A A A 3

4 X11 A12 X13 5

= X min 1 ;X2 ;X3

X12 A2232 A333

A X

  = min max k A11 A12 A13 k ;

A22 A 3

X3 32 33

A

  = max k A11 A12 A13 k ;

A22

; k A32 A33 k 32 The minimizing X3 is obtained from

A X

22 3 min X3 A32 A33 With the above X3 , then X1 and X2 are obtained from

2 A A A 3

4 X11 A12 X13 5

= Xmin 1 ;X2

X12 A2232 A333

p Solution 2.27 0 = jaj2 + jbj2 and x = jbbajc2 .

Chapter 3

Linear Dynamical Systems Solution 3.1


X_ (t) = dtd eAtX (0)eBt = AeAt X (0)eBt + eAt X (0)eBtB = AX (t) + X (t)B:

Solution 3.2 ???? where

_ t; t0 ) = A(t)(t; t0 ); (t0 ; t0 ) = I: ( (Show also that  1 (t;  ) = (; t) and (t; )(;  ) = (t;  ).) Note that 1
0 = d  1 (t; t ) (t; t ) = d (t; t0 )  +  1 (t; t ) d(t; t0 ) ; 0

dt

0

0

dt

)

dt

and mmultiply  1 (t; t0 ) from the left to the dierential equation to get

d 1 (t; t0 ) =  1 A dt

 1 (t; t0 ) (x_ A(t)x(t)) =  1 (t; t0 )B (t)u(t) i.e.,

d  1 (t; t )x(t)
=  1 (t; t )B (t)u(t) 0 0 dt

Integrate both sides to get

 1 (t; t0 )x(t)  1 (t0 ; t0 )x(t0 ) = i.e.

x(t) = (t; t0 )x(t0 ) +

The result follows by noting

x(t)  c +

c+c

t0

Zt

 1 (; t0 )B ( )u( )d

 1 (t; t0 ) 1 (; t0 )B ( )u( )d

t0

k( )x( )d  c +

=c+c

k( )d +

t0

t0

 1 (t; t0 ) 1 (; t0 ) = (t;  ):

Solution 3.3

Zt

Zt

Zt

Zt t0

Zt t0

k( )d

k( )d +

Z t0

"

Zt

Zt t0

t0

k( )d

k() c + c 7

 Z

k( ) c +

Z t0

Z t0

t0



k()x()d d

k()x()d

k( )d +

Z t0

k( )

Z t0

#

k( )x( )d d

LINEAR DYNAMICAL SYSTEMS

8

" Zt

=c 1+

t0

Zt

k( )d + +

t0

1+

+

Zt t0

k( )d

k( )d

" Zt

c

Keep substituting to get

Zt

t0

t0

Z t0

k( )d

Z

k()d

Z t

k( )

t0

k( )d

t0

k()d

Z

Z t0

+

Z t0

k()

t0

t0

k( )d

#

k( )x( )d

2 Z t

k( )

t0

k( )d

Z

t0

k( )d

3 #

k( )x( )d

h(t) = b0h 1 + b1 h_ 1 + b2 h1 : x(t) = eA(t t0 ) x(t0 )





=

  

 0:7071



4 2



 2

 0:7071

 1   0:7071 =

1 1 2

0 0 1

0:4472 0:7071 0:8944

0:4472 0:7071 0:8944

 ln 2 0

Solution 3.6 Let

0 0

1

=

Then we have 1 (t) 2 (t)



3 1 2 0

 2n

0 0 1

1



1

 0:7071

0:4472 0:7071 0:8944

Zt



0:4472 0:7071 0:8944

 0:7071

xi (t) := x_ 1 (t) = x_ 2 (t) =

 

5 = eA 1 2 2

 1

5 2

0:4472 0:7071 0:8944

0



4 = eA 1 ; 2 1

eA =

 0:7071 A=

Z

t0

Z

t x(t)  ce t0 k( )d :

Then

and

t0

k()d +

Zt

R

Solution 3.5 Note that

x(n) = enA

t0

k( )d +

Solution 3.4 By linearity,

Hence

Z

0:4472 0:7071 0:8944



1

 1:3862 =

i ( )d

.. . x_ n (t) = n (t) y(t) = 1 (t)x1 + 2 (t)x2 +


+ n (t)xn

 1 1  0

0:6931 1:3862 0:6931



9

Solution 3.7 Use PBH test: (F; G) is controllable if and only if

 A I ( F I G ) = C

0

B



I 0

has full row rank for all . Since (A; B ) is controllable, ( A I B ) is full row rank for all . Thus ( F I G ) has full row rank if and only if it has full row rank for  = 0 which implies that CA B0 has full row rank.

Solution 3.8 Obvious. Solution 3.9 ) Suppose X is singular. We show that either (A; b) is not controllable or (c; A) is not observable. Since X is singular, there are u and v such that u0 X = 0 and Xv = 0. Hence 0 = u0 (AX + XA + bc)v = (u0 b)(cv): This implies that either u0b = 0 or cv = 0. Suppose cv = 0. Then

0 = (AX + XA + bc)v = XAv i.e., Ker(X ) is a A-invariant subspace. Therefore there is a v 2 Ker(X ) such that cv = 0 and Av = v, i.e., (c; A) is not observable. Similarly if u0b = 0. On the other hand, suppose (c; A) is not observable. Then there is a v and a  such that cv = 0 and Av = v. 0 = (AX + XA + bc)v = AXv + X (Av) + b(cv) = AXv + Xv = (A + I )Xv This implies that Xv = 0 since i (A) + j (A) 6= 0. That is X is singular.

Solution 3.10 Suppose dim(w) < dim(x) dim(y), i.e., dim(w) + dim(y) < dim(x). Then dim (span fx^g) = dim (span fPw + Qu + Ryg)  dim(w) + dim(y) < dim(x) In other words, there is a x1 2 Rn such that x1 62 span fx^g. Since (A; B ) is controllable, there is a T > 0 and u(t) such that x(t) = x1 ; t > T . Hence it is impossible for x^(t) ! x(t) = x1 .

Solution 3.11 See page 115. Solution 3.12 See page 66. Solution 3.13 Note that x_ = Ax + B1 u + B2 u_ y = C1 x + D1 u De ne  = x B2 u. Then

_ = A + (B1 + AB2 )u y = C1  + (D1 + C1 B2 )u It is easy to see that (A; AB2 ) is controllable if A is nonsingular.

LINEAR DYNAMICAL SYSTEMS

10

Solution 3.14 Note that y=

Zn

a1 y( ) a~1 y(  ) + b1 u( ) + ~b1 u(  )

9 > Zh = i + a2y(t) a~2 y(t  ) + b2 u(t) + ~b2 u(t  ) dt> d | {z }; x2

and de ne x1 = y. Then

x_ 1 (t) = x_ 2 (t) =

a1y(t) a~1 y(t  ) + b1 u(t) + ~b1 u(t  ) + x2 (t) a2y(t) a~2 y(t  ) + b2 u(t) + ~b2 u(t  )

or

x_ 1 (t) = a1x1 (t) a~1 x1 (t  ) + x2 (t) + b1 u(t) + ~b1 u(t  ) x_ 2 (t) = a2x1 (t) a~2 x1 (t  ) + b2 u(t) + ~b2 u(t  ) y(t) = x1 (t)

Solution 3.15 (a) Suppose rank(R1) = r1 and rank(R2 ) = r2 . Let R1 = C1B1 and R2 = C2 B2 with C1 2 Rpr1 ; B1 2 Rr1 q ; C2 2 Rpr2 ; B2 2 Rr2 q ; Then

G(s) = C1 (sIr1 + Ir1 ) 1 B1 + C2 (sIr2 + Ir2 ) 1 (sIr2 + Ir2 ) 1 B2    I B   r2 2 = CIr1 B01 + C Ir2 I0r2 Ir2 0 1 2 2 I 3 0 0 B1 r1 6 Ir2 Ir2 0 77 = 64 00 0 Ir2 B2 5 C1 C2 0 0

(b)

G(s) =

 1 

1 0 s+1 +

2 =4



1 1 1 1 0 1

1 1

3 5

2 66 6 = 66 64

Solution 3.16 Note that y(t) =

Z

[b1 ()u() a1 ()y()] +



1 1 1 0

1 1 0 1 0 0 0 0 1 1 0 1

Z



 

1  1 0 + 1 0 1 + 0 1 0 0 s+1 s + 2 3 10

2  0 + 66 4 0

2 0 0 2 0 1 3 10 3 0 0 0 0 0 0 1 0 77 2 0 1 0 77 0 2 0 1 77 0 1 0 15 3 10 0 0

1 0 0 0

0 1 1 0



3 77 5

 

[b2 ( ) b_ 1 ( )]u( ) + [_a1 ( ) a2 ( )]y( ) d d

11 De ne x1 = y and

x2 (t) =

Then

or

Z



[b2 ( ) b_ 1 ( )]u( ) + [_a1 ( ) a2 ( )]y( ) d

x_ 1 (t) = b1 (t)u(t) a1 (t)y(t) + x2 (t) x_ 2 (t) = (b2 (t) b_ 1 (t))u(t) + [_a1 (t) a2 (t)]y(t)









b1 (t) a1 (t) 1 a_ 1 (t) a2 (t) 0 x(t) + b2 (t) b_ 1 (t) u(t)   y(t) = 1 0 x x_ =

Solution 3.17 Use the following relation AP + PA =

Z1 0





(AeAt QeA t + eAt QeA t A)dt =

Solution 3.18 Let Q = U U  with  =



 0

Z1d 0



At A t dt e Qe dt = Q

1  0 0 , 1 > 0, and a unitary matrix U . Let T = U and partition the transformed state space realization accordingly as

 TAT CT

1 1

2

3

 11 A12 B1 TB = 4 A A 21 A22 B2 5 D C1 C2 D

Then the Lyapunov equation gives   0  A A   A A    0   C C C C  1 11 12 11 12 1 1 1 1 2 0 0 + C2 C1 C2 C2 = 0 0 0 A21 A22 + A21 A22 which in turn gives 1A12 + C1 C2 = 0; C2 C2 = 0: Hence C2 = 0 and A12 = 0. For the speci c numerical problem, we have 0 0:25 0:75 0:25 1 0 0:9066 0:1072 0:4082 1 Q = @ 0:75 6:25 2:75 A ; U = @ 0:0763 0:9097 0:4082 A ; 0:25 2:75 1:25 0:4151 0:4012 0:8165 0 0:1986 0 0 1 =@ 0 7:5514 0 A 0 0 0 and 2 2:6050 7:2139 0 0:9066 0:9066 3  TAT 1 TB  66 0:1346 0:3950 0 0:1072 0:1072 77 6 2:4908 1 0:4082 2:8577 77 CT 1 D = 64 02::4075 2625 2:2206 0 0 0 5 0:9829 1:0169 0 0 0 Solution 3.19 The Hankel singular values are listed in the following table:

1 2  = 2 1.572 0.641  = 4 0.992 0.689  = 20 0.589 0.549  = 100 0.517 0.510 Hence i ! 0:5 as  ! 1.

3

0.209 0.418 0.497 0.500

4

0.061 0.245 0.449 0.500

5

0.017 0.156 0.416 0.483

LINEAR DYNAMICAL SYSTEMS

12

Solution 3.20 Use MATLAB:     

set T =?? g = nd2sys([ 0:05; 1]; [0:05; 1]); g5 = mmult(g; g; g; g; g); g5 = sysbal(g5); % to avoid numerical problem. g10 = mmult(g5; g5); gt = nd2sys(1; [T; 1]); sys = mmult(g10; gt); [sysb; sig] = sysbal(sys);

Solution 3.21 It is easy to show that



G(s) = (s + 1)(1 s + 2) 1s 1s



= 1s 01

So its McMillan form is

 

1 (s+1)(s+2)

0 0

1 (s+1)(s+2)

0 0

0

0

and its McMillan degree is 2.



 1 1 



0 1 :

Solution 3.22 Let rank R1 = r1 and rank R2 = r2 . Then there are C1 Rpr2 ; B2 2 Rr2 q such that

R1 = C1 B1 ; R2 = C2 B2

2 I r1 G(s) = 4 0

and

2 Rpr1 ; B1 2 Rr1 q ; C2 2

3

0 B1 2Ir2 B2 5 C1 C2 0 is a minimal realization. Hence the McMillan degree of G is r1 + r2 .

Solution 3.23





G(s) = s +1 1 00 13 + s +1 2

2 66 66 = 66 66 4

The McMillan degree is 5.

1 0 0 0 0 1 3



0 2 0 0 0 1 0





0 0 0 0 0 0 3 0 0 4 2 0 1 2

0 1 0 1 1 0 0

0 + 1 2 s+3

1 0 0 0 2 0 0 0 2





2 0 + 1 0 0 1 0 s+4 2 0 1 0 0 0 0 0 0



3 77 77 77 77 5

Solution 3.24 ??????? Find a formula for the McMillan degree of where Ri 2 Rpq .

G(s) = 1s R1 + s12 R2 ;

Solution 3.25 Compute the transmission zeros and the corresponding zero directions of the following trans-

fer functions

13 (a) The McMillian form is

G(s) =

"

#

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

s2 +16s+14 s 2

p

so the zeros are 8  5 2. (b) The McMillian form is

G(s) =

"

#

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

8s2 +37s+39 s+2

so the zeros are 13=8 and 3. (c) The only zero is 2.

 A zI B  Solution 3.26 Note that det = 0 is equivalent to the existence of x and u such that C

D

 A B  x   I =z C D

0 0 0

u

 x  u

which is a generalized eigenvalue problem and can be computed using a Matlab Commad: [V; D] = eig(M; N ) where  B I 0 M = CA D ; 0 0

 

the diagonal elements of D are the invariant zeros and the columns of V are xu . If the system is not square, then randomly augment the system to a square one and compute the zeros for the squared system several times. Then the zeros of the original systems are the common zeros computed from each augmentation.

Solution 3.27 Compute the transmission zeros (or invariant zeros) of the following transfer functions (a) The only transmission zero is 0:2 with

2 0:0466 3  y  66 0:0466 77 6 77 v = 64 00::1866 9702 5 0:1399

(b) No transmission zero. (c) Two zeros are at 2:1861 and 0:6861 with

2 :3015 3 2 0:8739 3  x  6 00:8098 77 66 0:1627 77 6 u = 4 0:1507 5 ; 4 0:4370 5 0:4803

0:1371

0:1920

0:3181

2 :0878 3 2 0:4636 3  y  6 00:4719 77 66 0:1726 77 6 v = 4 0:8560 5 ; 4 0:8088 5

LINEAR DYNAMICAL SYSTEMS

14

Solution 3.28 (a)

A B C

2 inf  inf0 fd (M12 ); (M21 )=dg 0 M21 =d = d> p0 = (M12 ) (M21 ): 43

STRUCTURED SINGULAR VALUE

44

Solution 11.5 Note that for two block  problem we have

 dI


(M ) = d> inf0   = d> inf0 

 M

11 M12 M21 M22  dM12 M22

0

0 I

 I=d 0

0

I



M11 M21 =d     = d> inf0  M011 M0 + M 0 =d dM0 12 22 21

Hence



 M

0

11





0

dM12



 
(M ) M22 M21 =d 0  M 0   0 dM  11  inf0  M =d 0 12 0 M22 + d> 21

0

inf  d>0

Solution 11.6 Let
be all diagonal full blocks and M be partitioned as M = [Mij ] where Mij are matrices with suitable dimensions. Show that 
(M )   ([kMij k]) (=  ([kMij k])) where (
) denotes the Perron eigenvalue.

Solution 11.7

infnn DMD

D 2C

1

p = D2inf C nn

0 J1 0

B 0 J2

D B

@ 0 0 0 0

0 0 1 0 0 C CD ... 0 A 0 Jm



1

= (M )

p

where Ji are the Jordan blocks. It is now easy to show that



infki Ki Di Ji Di 1 p = (Ji )

Di 2C

and thus the result follows.

Solution 11.8 The solution of this problem is given in J. Stoer and witzgall (1962), F. L . Bauer (1963), Safonov (1982), and Safonov and Doyle (1984).

Solution 11.9 Let x and y be partitioned compatibly with the
:

0 y1 1 0 x1 1 B y2 C B x2 C x=B A @ ... C A; y = B @ ... C xm+F

Then



ym+F


(M ) = max (MU ) = max (y Ux) U 2U U 2U



u yx +


+ umym xm + ym +1U1xm+1 +


+ ym +F UF xm+F = max U 2U 1 1 1







 xm j + y U1 xm+1 +


+ y UF xm+F = max ju y  x j +


+ jum ym m+1 m+F U 2U 1 1 1

= jy1x1 j +


+ jym xm j + kym+1 k kxm+1 k +


+ kym+F k kxm+F k

45 where the maximizing ui and Ui are given by

y x ui = e j\yi xi ; Ui = ky m+ki kmx +i k + U~i m+i m+i

such that U~i xm+i = 0, ym +i U~i = 0, and Ui is unitary. It is also easy to show that max
2B
(M
) is achieved by y x i = e j\yi xi ;
i = m+i m+i : kym+i k kxm+i k

Solution 11.10 Note that for i = 0; 1; : : :; k kx1 k2 + kz0 k2

kx2 k2 + kz1 k2

kxk+1 k2 + kzk k2



2 (kx0 k2 + kd0 k2 ) 2 (kx1 k2 + kd1 k2 )



2 (kxk k2 + kdk k2 )



.. .

Adding the above inequalities together to get kX +1

or

i=1 kxk+1 k2

kxi k2 +

k X i=0

k X

kzi k2  2 (

kx0 k2 + (1

2)

k X i=0

i=0

kxi k2 +

kxi k2 +

k X i=0

k X i=0

kzi k2  2

Since the above inequality is true for any k, it is concluded that x limk!1 kxk k = 0. Hence we have kx0 k2 + (1

or

2)

kz k22 + (1

Let a discrete time system be given by

x

k+1 zk

1 X i=0

kxi k2 +

1 X i=0

kdi k2 )

kzi k2  2

2 `2

1 X i=0

k X i=0

kdi k2

and z

kdi k2

2 ) kxk22  2 kdk22 + kx0 k2

  A B  x  k = C C

dk

The interpretation of the above is that the state space realization satis es

 A B 

C D

 < 1:

Solution 11.11 Clearly, K needs to stabilize P0. (a) By internally stability condition, the system is stable if and only if for all
i: 1 + P (j!)K (j!) 6= 0; m

8!

1 + W1 T
1 + W2 SK
2 6= 0 * k jW1 T j + jW2 KS j k1  1

2 `2

and furthermore

STRUCTURED SINGULAR VALUE

46

To show that 1+W1T
1 +W2 SK
2 6= 0 implies k jW1 T j + jW2 KS j k1  1. Suppose k jW1 T j + jW2 KS j k1 > 1, i.e.,

:= jW1 (j!0 )T (j!0)j + jW2 (j!0 )K (j!0 )S (j!0 )j > 1 for some frequency !0 . Let   0 and  0 be such that 0 \  + j! j! = \W1 (j!0 )T (j!0 ) 0

0 \ + j! j! = \W2 (j!0 )K (j!0 )S (j!0 ) 0

and let


1 = 1  + ss ;
2 = 1 + ss :

Then k
i k1 < 1 and

j \W1 (j!0 )T (j!0 )


1 (j!0 ) = jW (j! )T (j!e )j + jW (j! )K (j! )S (j! )j 1 0 0 2 0 0 0 j \W2 (j!0 )K (j!0 )S (j!0 )


2 (j!0 ) = jW (j! )T (ej! )j + jW (j! )K (j! )S (j! )j 1 0 0 2 0 0 0

and

(b)

1 + W1 (j!0 )T (j!0)
1 (j!0 ) + W2 (j!0 )K (j!0)S (j!0 )
2 (j!) = 0 i.e., j!0 is a closed-loop pole. Therefore, k jW1 T j + jW2 KS j k1  1 is necessary for the closed-loop stability.

W3 S 3 = Tzd = 1 +WPK 1 + W1 T
1 + W2 KS
2 : kTzd k1  1 * W S 3 1+W1 T
1+W2 KS
2  1; 8!;
i

* jW3 S j  j1 + W1 T
1 + W2 KS
2 j ; 8!;
i * jW3 S j  1 jW1 T j jW2 KS j ; 8!; * jW3 S j + jW1 T j + jW2 KS j  1; 8!; * kjW3 S j + jW1 T j + jW2 KS jk1  1.

On the other hand, suppose kjW1 T j + jW2 KS jk1  1 (which is necessary for robust performance) but kjW3 S j + jW1 T j + jW2 KS jk1 > 1 i.e., there is an !0 such that

0 := jW3 (j!0 )S (j!0 )j + jW1 (j!0 )T (j!0 )j + jW2 (j!0 )K (j!0 )S (j!0 )j > 1: Then there exists a > 1 such that 1 1 jW (j! )S (j! )j + jW (j! )T (j! )j + jW (j! )K (j! )S (j! )j > 1: 3

0

0

1

0

0

2

0

0

0

It is also easy to construct stable
1 and
2 such that k
i k1 = 1= < 1 and
1 (j!0 ) = e \W1 (j!0 )T (j!0 ) = ;
2 (j!0 ) = e \W2 (j!0 )K (j!0 )S(j!0 ) = : Thus

47 1

jW3 (j!0 )S (j!0 )j > 1

jW1 (j!0 )T (j!0 )j

1

jW2 (j!0 )K (j!0 )S (j!0 )j

+ jW3 (j!0 )S (j!0 )j > j1 + W1 (j!0 )T (j!0 )
1 (j!0 ) + W2 (j!0 )K (j!0 )S (j!0 )
2 (j!0 )j + jW3 (j!0 )S (j!0 )j >1 j1 + W1 (j!0 )T (j!0 )
1 (j!0 ) + W2 (j!0 )K (j!0 )S (j!0 )
2 (j!0 )j + kTzw k1 > 1.

This is a contradiction.

Solution 11.12 Draw the block diagram as below. Then it is easy to show that

0z 1 0 W T @ z12 A = @ W21KS z

with

10 1

W1 T W2 KS S

S

 WT 1

 



W1 T W1 T W2 KS W2 KS = W2 KS ( 1 Hence by problem ??, 
s (M11 ) = jW1 T j + jW2 KS j. Next note that M11 =




s

 Ds M11 Ds 1 = 

= max

 dW T 

W2 KS ( 1=d

s s

p

= 1 + 1=d2 max

p

q

1 ) ( 1=d

1)



1)

1 )

 dW T  1

W2 KS

 dW T  dW T  1

W2 KS

1

W2 KS

 dW T   dW T  1

W2 KS

q

1

W2 KS

= 1 + 1=d2 jdW1 T j2 + jW2 KS j2

jdW1 T j2 + jW2 KS j2 + jW1 T j2 + jW2 KS=dj2

The minimum is achieved by and

W2 KS ( 1=d

1

= 1 + 1=d2 max

=

1

 dW T  p

0 1

W1 TW3 d1 d1 W2 KSW3 A @ d2 A = M @ d2 A SW3 d d

2 KS j d2 = jW jW T j




1

inf  Ds M11 Ds 1 = jW1 T j + jW2 KS j D s

Similarly, it is easy to see that

0 WT 1 1 M = @ W2 KS A ( 1 S

and





1

W3 )

 DMD 1 2 = jW1 T j2 + jW2 KS j2 + jW3 S j2 + jW2 Td1 =d2 j2 +jW1 KSd2=d1 j2 + jW3 Td1j2 + jW2 S=d2 j2 + jW3 KSd2 j2 + jW1 S=d1 j2

STRUCTURED SINGULAR VALUE

48

It can be shown, which is not obvious, that the minimizing di are given by

W1 S j ; d2 = jW2 S j d21 = jjW T j 2 jW KS j 3

and

inf  DMD D

3

1

2 = jW

2 + jW

1T j

2 + jW

2 KS j

2

3Sj

+2jW1 T j jW2 KS j + 2jW1 T j jW3 S j + 2jW2 KS j jW3 S j = (jW1 T j + jW2 KS j + jW3 S j)2

- W2 z2-
2

d

d2

?

W3

-e 6

-

K

P0

- W1 z1-
1 - ?ed2- ?e - ?e -z

Figure 11.1: Diagram for Problem????

Solution 11.13 Note that

v u F X F uX (DMD 1 ) =  ([xi yj di =dj ]) = t kxi k2 kyj k2 d2i =d2j i=1 j =1

With the given di , it is easy to see that

(DMD 1 ) = =

v v u u F X F F X F X u uX 2 2 2 2 t kxi k kyj k di =dj = t kxi k kyj k kxj k kyi k i=1 j =1 i=1 j =1 v u F F F u tX kxi k kyi k X kxj k kyj k = X kxi k kyik i=1

j =1

i=1

which is 
(M ) by Problem ??.

Solution 11.14 9.11. Clearly, K needs to stabilize P0 . By internally stability condition, the system is stable if and only if for all
i :

1 + P (j!)K (j!) 6= 0; m

8!

1 + W2 SK
2 6= 0 m

(< f
2 g + j = f
2 g) + 1=(W2 SK ) 6= 0

m < f
2 g + < f(1=(W2 SK ))g + j (= f
2 g + = f(1=(W2 SK ))g) 6= 0 m < f(1=(W2 SK ))g > ; = f(1=(W2 SK ))g >

49

Solution 11.15 Consider M 2 C 2n2n to be given. Let
2 be a n  n block structure, and suppose that 2 (M22 ) < 1. Suppose also that Fl (M;
2 ) is invertible for all
2 2 B
2 . ^ such that For each  > 1, nd a matrix W and a block structure
max (Fl (M;
2 )) < 


2 2B
2

if and only if where  is the condition number.


^ (W ) < 1;

50

STRUCTURED SINGULAR VALUE

Chapter 12

Parameterization of Stabilizing Controllers Solution 12.1 Note that K stabilizes Pi is equivalent to Pi stabilizes K . It is well known that all stabilizing

controllers for a plant K can be parameterized as an LFT P~ = Fu (J; Q) for some J and some Q 2 H1 . Since  Pi stabilizes K , there must be a Qi such that Pi = Fu(J; Qi ). Let

I

= maxfkQi k1 g and P = I J . Then

P~ = Fu (J; Q) = Fu (P; Q= ): De ne

P

= fFu(P;
) : k
k1  1g

Then it is clear that fPi g  P . Solution 12.2 (Internal Model Control (IMC)) Suppose a plant P is stable. Then it is known that all stabilizing controllers can be parameterized as K (s) = Q(I PQ) 1 for all stable Q. In practice, the exact plant model is not known, only a nominal model P0 is known. Hence the controller can be implemented as in the following diagram.

r

-e -e 6 6

-

Q P0

-y

P



The control diagram can be redrawn as follows. This control implementation is known as Internal Model Control (IMC). Note that no signal is feeded back if the model is exact. Generalize the IMC to unstable systems.

r

-e 6

-

Q

51

-

P

-

P0

-y - e?

52

PARAMETERIZATION OF STABILIZING CONTROLLERS

Chapter 13

Riccati Equations Solution 13.1

B

1. Note that

 U   U  ^ A V = V B

0

C

where B^ is similar to B . Since U is nonsingular, we have  B 0  I   I =

C

or Multiply

A

B

1

VU

VU

1



^ U BU

 I   I  ^ A X = X U BU

0

C

1

1

 X I  from the left to get  X I   B 0  I  = 0 =) XB + C AX = 0: C A X

On the other hand, suppose X is a solution to the Sylvester equation, then  B 0  I   B   B   I = = =

C

A

X

C AX

XB

Write B^ = U 1 BU for the Jordan form of B and set V := XU Then  B 0  I   I U=

C

or

U

A

B



X B



X BU

X

 U   U  1 A V = V U BU

0

C

i.e., the columns of V are the eigenvectors of M associated with the eigenvalues of B . Since U is nonsingular, X = V U 1 . 2. The necesssity is easy. Suppose X is a solution and let I 0  T= n Then

B C

0



X Im

A T =T 53

B 0

0

A



RICCATI EQUATIONS

54

Solution 13.2 A P (t) + P (t)A = =

Zt dh  A 0

d e

Z t 0



A eA  QeA + eA  QeA A d

i

QeA  = eAt QeAt Q = P_ (t) Q

_ t;  ) = (t;  )A(t). Solution 13.3 Easy by the fact that ( Solution 13.4 Note that  or and Then

So

 _

11

   _ 12 = A R ;

Hence i.e.,

  _ 22 = Q

21



A 

P (11 + 12 P0 ) = 21 + 22 P0 P_ (11 + 12 P0 ) = P (_ 11 + _ 12 P0 ) + _ 21 + _ 22 P0  I    I      = P A R  P + Q A  P 0 0      11 + 12P0 +  Q A  11 + 12P0  = P A R 21 + 22 P0 21 + 22P0



P_ = P A R

 I  +  Q P

A

  I  = PA PRP Q A P P

_ t) = HeH (t T ) = (T ) = I; (

Note also that

R  A

 _

Solution 13.5 Note that or



A Q

_ t) = HeHt = (

 A

Q



R (t) A

_ 11 = A11 + R12 ; _ 21 = Q11 A 21

P 11 = 21 P_ 11 = _ 21 + P _ 11 = Q11 + A 21 + P (A11 + R12 ) P_ = Q + A P + PA + PRP

Solution 13.6 J -spectral factorization and H1 control???

Chapter 14

H2 Optimal Control Problem 14.1 The following block diagram shows a plant Gp , actuators Ga , sensors Gs , and controller K : u

r

-

-

-

Ga

v

Gp

-

K

6

Gs



Assume Gp (s), Ga (s), and Gs (s) are strictly proper and K (s) is required to be proper. The signals r(t), u(t), and v(t) are 2 dimensional. In addition to internal stability, the performance requirements, roughly, are that u(t) should not be too large and r(t) v(t) should be small. Each component of r(t) is given by

8 > < ri (t) = > :

10t; 0t 2;

(16.2)

to mimic a ramp-up, ramp-down command. The generalized error vector is taken to have four components: the velocity error vm vs ; the compliance error fh vm (for simplicity, the desired compliance is assumed to be vm = fh); the force-re ection error fm fe ; and the slave actuator force. The last component is included as part of regularization, that is, to penalize excessive force applied to the slave. Introducing four weights to be decided later, we arrive at the

59

60

H1

CONTROL

2 w (v v ) 3 v m s 6 7 w c 6 z = 4 w ((ffh vfm )) 75 : f m e

generalized error vector

ws fs

The Laplace transforms of fe and fh are not rational:

f^e (s) = 10 1 e 0:2s ; f^h(s) = 22 1 e s 2 :

s

s

To get a tractable problem, we shall use second- and third-order Pade approximations,  Ts (Ts)2  Ts (Ts)2  Ts e  1 2 + 12 1 + 2 + 12 and   Ts (Ts)2 (Ts)3  (Ts)2 (Ts)3 e Ts  1 Ts + 1 + 2 + 10 + 120 : 2 10 120 Using the third-order approximation for f^e (s) and the second-order one for f^h(s), we get   0:2s (0:2s)2 (0:2s)3  23s2 f^e (s)  20 02:2 + 0:120 1 + 2 + 10 + 120 =: g^e (s)

f^h(s)



2

,

2 2 s s 1+ +

2

12

=: g^h(s): Incorporating these two pre lters into the preceding block diagram leads to Figure 16.2. The two exogenous

wh-

Gh

-j 6

Gm

?

vm-

K

fm

vs

Gs



 6 j

fs

Figure 16.2: Telerobot with pre lters. inputs wh and we are unit impulses. The vector of exogenous inputs is therefore

w 

w = wh : e The control system is shown in Figure 16.3, where z and w are as above and 2v 3   m y = 4 vs 5 ; u = ffm : s f e

Beginning with state models for Gh ; Gm ; Gs ; Ge , namely,

A B  A B  A B  A B  h h m m s s e e Ch 0 ; Cm 0 ; Cs 0 ; Ce 0 ;

Ge

we

61

z



 

G y

w

u

-

K

Figure 16.3: Telerobot con gured in the standard form. with corresponding states xh ; xm ; xs ; xe , using the interconnections in Figure 16.2, and de ning the state

2x m 6 x 6 x = 4 xs e

xh

lead to the following state model for G:

3 77 5

2 Am 0 0 BmCh 0 0 66 0 As Bs Ce 0 0 0 66 0 0 Ae 0 0 Be 66 0 0 0 Ah Bh 0 2 A B B 3 66 Cm Cs 0 0 0 0 1 2 6 C 0 0 C 0 0 m h 4 C1 0 D12 5 := 66 0 0 C 0 0 0 e 66 0 0 0 C2 0 0 0 0 0 66 66 0 0 0 64 0 0 Ce

3 Bs 77 0 77 0 77 0 77 0 77 0 77 : I 77 77 0 77 5

Bm

0

0 0 0 0 0

I

0

(16.3)

0 0 0 0 0 0 0 Cm 0 0 0 0 0 0 0 For the data at hand, D21 = 0, so (A2) fails. Evidently, the condition D21 = 0 re ects the fact that no sensor noise was modelled, that is, perfect measurements of vm ; vs ; fe were assumed. Let us add sensor noises, say of magnitude . Then w is augmented to a 5-vector and the state matrices of G change appropriately so that the realization becomes 2 A 0 B 3 B2 1 4 C1 0 0 D12 5 : C2 I 0 0 Some trial-and-error is required to get suitable values for the weights: 0

Cs

wv ; wc ; wf ; ws ; : The MATLAB functions h2syn and h2lqg can be used to compute the optimal controller. Play with the weights and try to get reasonable performance. Include plots in your solution. Solution 16.2 The generalized plant G is given by 2 W 0 WP 3 2 W 3 20 0 0   4 0 0 1 5 = 4 0 5 1 0 P + 4 0 0 1 F 2 FP F 0 2 0

3 5

62

H1              

CONTROL

W = nd2sys([1]; [2:5=pi; 1]); W = mmult(W; W ); sys1 = abv(W; 0; F ); sys2 = sbs(1; 0; mmult( 1; P )); D = [0; 0; 0; 0; 0; 0:01; 0; 0:01; 0]; G = madd(mmult(sys1; sys2); D); [K; Tzw; gamma] = hinfsyn(G; 1; 1; 0; 10; 0:0001) w = logspace( 1; 3; 400); Tzwf = frsp(Tzw; w); vplot(0 liv; lm0; Tzwf ) Tf = vnorm(Tzwf ); vplot(0 liv; lm0; Tf ) axis([0:1; 1000; 0:00001; 1]) grid

2 66 66 6 G(s) = 66 66 64

1:2566 1:2566 0 1:2566 0 0 0 0 0 0 1:1210 0 0 0 0 0

0 0 0 1:1193 6:2832 2:5029 0 0:01 0 0:0282 0 0 0 0 2:5066 0

0 1:5841 3:5423 0:0282 20 0 0 0

0 0 1:1210 0 2:5066 0 0 0 0 0 0 0 0 0 0 0:01

0 0 0 0:9985 1:4132 0 0:01 0

3 77 77 77 77 77 5

We get opt = 0:08,

:4598  106 (s + 0:01)(s + 20)(s + 6:2832)(s + 3:9612) K (s) = (s +810578)( s + 619:0752)(s + 35:381)(s + 1:3224)(s + 1:1862) 799:7664(s + 0:01)(s + 20)(s + 6:2832)(s + 3:9612) (s + 619:0752)(s + 35:381)(s + 1:3224)(s + 1:1862) The closed loop frequency responses are shown in Figure 16.4. 

Solution 16.3 Note that K robustly stabilizes the uncertain system if and only if K stabilizes P0 and

WK (I P K ) 1

 1= : 0 1

Without loss of generality, assume that W (s) is stable and minimum phase. Then K stabilizes P0 if and only if WK stabilizes P0 W 1 . Let P0 W 1 = Ps + Pu such that Ps is stable and Pu is antistable. Let K1 = WK (I Ps WK ) 1 then

WK (I P0 K ) 1 = WK (I Ps WK Pu WK ) = WK (I Ps WK )

1 I

Pu WK (I Ps WK )

1

1 1

63 0

10

−1

10

the largest singular value of Tzw

T21 −2

10

−3

10

−4

10

T22

T12

T11

−5

10

−1

0

10

1

10

2

10

10

3

10

Figure 16.4: Frequency responses of the closed-loop = K1 (I Pu K1 ) 1 : Clearly, K stabilizes  A BP0 if and only if K1 stabilizes Pu. Let Pu = C 0 be a controllable and observable realization (so A is stable). Let

Pu = NM 1 = M~ 1 N~ ~ V~ ; U; V then there exist U;

2 RH1

such that ~ = I; MV ~ V~ M UN

~ = I: NU

All stabilizing controller can be parameterized as

K1 = (U + MQ) 1 (U + NQ) 1 ; Q 2 RH1 and

K1 (I Pu K1 ) 1 = (U + MQ)M~

Let X > 0 and Y > 0 be the stabilizing solutions to

XA + A X XBB  X = 0; Y A + AY Y C  CY = 0 and let F = B X and L = Y C  . Then

 M  2 A + BF B 3    ~ N~  = A + LC L B 4 5 = F I ; M N C I 0 C

0

U=

 A + BF F

L

0



64

H1

and M  M = I; M~ M~  = I . Hence



WK (I P0K ) inf K = Q2RH inf

It is easy to show that

1

k(U

1



K1(I PuK1) 1 = inf K 1

+ MQ)k1 = Q2RH inf (M  U ) =

1

1

inf 1 = Q2RH

k(M  U

 A 



(U + MQ)M~

1 1

+ Q)k1 = k(M  U ) kH

XB

L

CONTROL



0 Moreover, X and Y are the controllability and observability Gramians of (M  U ) . Therefore, k(M  U ) kH = (XY ). On the other hand, it is easy to see that X 1 is the controllability Gramian of Pu and Y 1 is the observability Gramian of Pu . Hence (XY ) = 1=min(X 1 Y 1 ) = 1=min(Pu ). Therefore we have

 min (Pu ) = min ([P0 W 1 ]+ ) = 1=(XY ):

We can also use the standard H1 control setup. We have

0 I  2A G(s) = I P = 4 0 u

0 W with controller K1 or G(s) = with controller K .

0 B 0 I C I 0

3 5

I P0

Solution 16.4 ??????????????? Let P = P0(I + W
) with k
k1 < and assume P and P0 have the same number of right half plane poles. Find the largest such that the closed-loop can be robustly stabilized. Again by small gain theorem, a controller K robustly stabilizes the uncertain system if and only if K stabilizes P0 and

(I KP ) 1KP W

 1= 0 0 1 We have  0 I  G(s) = P W P 0

0

Solution 16.5 Let G 2 RH1 and let G = NM 1 be a coprime factorization with M inner, i.e., M M = I . Then

1 1

 N I  

= Q2RH inf kN QM k1 = Q2RH inf

F` M 0 ; Q

1 1 1 kG

inf Q2RH

Qk1 = Q2RH inf NM

Q 1

1

Solution 16.6 This problem was discussed in detail in the paper by Aoki, Ushida, and Kimura (1995). The

relationship between the given for computing the central controller and the actually H1 norm ( 1 ) is shown in Figure 16.5. Obviously, 1 is not monotonic w.r.t. .

Solution 16.7 The problem can be set in a LFT from:

2w 3  2A  1 0 M M 4 w 2 5; N = 4 z2 P I P w3  M 1 0 M 1 2 A B 0 4 F I 0 G(s) = P I P =

z   M 1 =

1

1

C

0

3

BF B F I 5 C 0

3

B I 5

I 0

65 H−infinity Norm of the Central Controller Is NOT Monotonic w.r.t. gamma 1.23

H−infinity norm with central controller

1.22 1.21 1.2 1.19 1.18 1.17 1.16 1.15 1.14 1

1.2

1.4

1.6 gamma

1.8

2

2.2

Figure 16.5: A counter example to monotonicity properties of H1 central controllers. The Matlab function lqr is used to determine that   F = 1 339:7965 :0001 :0087 : The function pck is then used to create G(s) and this system is entered as a parameter for the function [k; t; r] = hinfsyn(G(s); 1; 1; 0; 5; :000001) : This function determines that the optimal is 3.5913 and that the optmal controller has poles at :004613  j 1:5385, :01436 and 1:6371
107, and zeros at -.0012, :00462  j 1:539.

Solution 16.8

 z   W K (I + PK ) 1 1 = 2 1 W1 (I + PK )

z2

W2 K (I + PK ) W1 (I + PK ) 1

1

 w  1

w2

2 z 3 2 W W W P 32 w 3 4 z12 5 = 4 01 01 W12 5 4 w12 5 z3 I I P w3 2 s=1:245+1 s=1:245+1  s=1:245+1  0:5(1 s)  3 50 s=0:007+1 50 s=0:007+1 50 s=0:007+1 (s+2)(s+0:5) 7 6 6 75 G(s) = 4 0 0 0:1256 s=s=0:502+1 2+1

0:5(1 s) 1 (s+2)(s+0:5) 2 :5 3 0 0 0 0 1 66 0 2 0 0 0 1 77 6 0 :007 :348 :348 :178 77 (Gilbert Realization) = 66 :0212 :1065 1 :281 :281 0 77 64 0 :7496 0 0 0 :5004 5 0:5 1 0 1 1 0 As in the previous problem, this G(s) is used as a parameter for hinfsyn and the optimal is found to be 1.418, the poles of the optimal controller are -0.007, -2.7246 and 6:5438
104 and the zeros are -0.5, -2.

1

66




1

H1



CONTROL

For the second situation where
=
2 , the Matlab function dkit is used to nd the optimal controller. The values of  with respect to frequency are shown in the the gure below. CLOSED−LOOP MU: CONTROLLER #5 1.4

1.2

MU

1

0.8

0.6

0.4

0.2 −1 10

0

10

1

10

2

3

10 10 FREQUENCY (rad/s)

4

10

5

10

6

10

Figure 16.6: Frequency response of the closed-loop The data after ve iterations is displayed in the following table:

Iteration Summary Iteration # 1 2 3 4 5 Controller Order 3 13 13 13 13 Total D Scale Order 0 10 10 10 10

Achieved 1:418 1:298 1:298 1::298 1:298 Peak  V alue 1:416 1:298 1:297 1:298 1:298

The order of D was chosen to be 5, resulting in a 13th order controller. Controllers of higher order do not signi cantly improve performance. The nal value for is 1.298 and the poles and zeros for the controller after 5 iterations are listed below: 0:007 3:4503 :494 483:2229 1:3321 609:2501 1:8709 923:0608 Poles : 2:8353 1949:7079 2684:8153 9913:8806 10230:871

Zeros :

:4939 564:343 :5 1087:346 1:336 1910:324 1:932 6439:409 2 9898:695 3:02 538787:6

Chapter 17

H1 Control: General Case Problem 17.1 Let G(s) 2 H1 be a square transfer

matrix and  > 0. Show that G is (extended) strictly positive real if and only if (I G)(I + G) 1 1 < 1. Problem 17.2 Let G(s) 2 H1 be a square transfer matrix and y = Gu. Then G is (extended) strictly positive real if and only if

Z 1

ky + uk2

0



uk2 dt  

ky

Z1 0

ky + uk2 dt

for some  > 0. (Note that ky + uk2 ky uk2 = (y u + u y). One implication of the result is that a strictly positve real problem can be converted to an equivalent strictly bounded real problem. The transformation induced from this result can also be applied to nonlinear systems.)

Solution 17.1 Note that 1=F

(I X )(I + X )

`

"

p

pI

2I



# !

2I ; X I



Then nding a controller K such that F`(G; K ) is (extended) strictly postive real is equivalent to nding a controller such that " ! p # I 2 I p2 F` ; F`(G; K ) I 

I



is strictly bounded real. Using the star product formula on page 268, we have

"

F`

pI2 

I

!

2I ; F (G; K ) = F (G; ` ` ~ K) I



2 A B R 1C = p1 1 1 G~ (s) = 4 2R C1

where

#

p

p

2B1 R 1 = B2 pB1 R 1 D12 = (Ip D11 =)R 1 2R 1 D12 1 1 C2 D21 R C1 = 2D21 R = D22 D21 R 1 D12 = and R = I + D11 =. It is noted that

I

0

p

B1 =( p2) R= 2

3 5

 A B R 1C = j!I B B R 1D =   A j!I B  1p 1 2 p1 12 2 = 1 1

 A B R 1C = j!I 1 1 C2 D21 R 1 C1 =

2R C1

p 1 p 2B1 R 1=

2D21 R =

 I p

2R D12

C1 = 2 R

67

0

1 =p2

C1

D12

C2

D21

  A j!I B  1 =

68

H1

CONTROL: GENERAL CASE

Hence the standard assumptions for the H1 control are satis ed if

 A j!I B   A j!I B  2 1 C1 D12 ; C2 D21

are both of full rank.

Solution 17.2 Assume D11 = 0 and D12 = 0 in the realization of G. Then ~ K) Q + (I + sN )F` (G; K ) = F`(G; with

2 3 B1 B2 A G~ = 4 C1 + NC1 A Q + NC1 B1 NC1 B2 5 C2

D21

D22

Chapter 18

H1 Loop Shaping Solution 18.1 See problem ???? in Chapter 9. Note that a normalized coprime factorization for G is given

N  n =

by

and it is easy to compute that

Mn

 N

Mn n



1p 1 s 2 s+ 5





= 0:2298 H

Hence max = 0:9732. Solution 18.2 ??????? In Corollary 18.2, nd the parameterization of all H1 controllers. Solution 18.3 It is straightforward to verify that K   1M ~ 1 = U + MQ : ( I + PK ) I V + NQ Next note that  M U  V~ U~   V~ U~  M U  = I ) N V N V =I N~ M~ N~ M~ which implies ~ ~ = I: MV NU Furthermore, if P = NM 1 = M~ 1 N~ are normalized coprime factorizations, we have Hence

 M  N   M  N    M  N    M  N   = N~ M~ N~ M~ N~ M~ N~ M~ = I:

 U   M 



 M  N   U   M  

1

bP;K =

V + N Q

=

N~ M~ V + N Q 1 1  

R + Q

: = I 1

Solution 18.4 Note that

K (I + PK ) 1

 1=b , K stabilizes P~ = P +
; k
k < b P;K a a P;K

PK (I + PK ) 1

1 1=b , K stabilizes P~ = (I +
)P; k
k < b

KP (I + KP ) 1

1  1=bP;K , K stabilizes P~ = P (I +
m ); k
m k < bP;K

(I + KP ) 1

1 1=b P;K, K stabilizes P~ = P (I +
) m1; k
mk < b P;K

(I + PK ) 1

1  1=bP;K , K stabilizes P~ = (I +
)f 1P; k
f k < b P;K P;K f f P;K 1 69

70

H1

LOOP SHAPING

Solution 18.5 Think P as the controller and K as the plant. Problem 18.1 Let an uncertain plant be given by

P = s2 +s 2+s+ 1 ;  2 [1; 3];  2 [0:2; 0:4]

and let a nominal model be

P = s2 +s 2+ s0+ 1 0

1. The largest possible k
add k1 ; = 4:2686 and k
mul k1 = 1:1415

  2. The largest possible
n
m 1 = 0:4552. 3. in (2), let

 part  (N ; M ) be a normalized coprime factorization of P . Find the largest possible
n
m 1 . P = N=M with s+2 s2 + 0:6s + 1 N = s2 + 1:9576 ; M = s + 2:2361 s2 + 1:9576s + 2:2361 is a normalized coprime factorization.

  4. Find the optimal nominal 0 and 0 such that the largest possible k
addk1 , k
mul k1 , and
n
m 1 are minimized respectively.

Problem 18.2 Let P = s(s101) . Design (a) a precompensator W of order no greater than 2 such that the

crossover frequency !c  2 and bopt (WP ) is as large as possible; (b) nd the optimal loop shaping controller K = K1 W .

s) Problem 18.3 Let P = 100(1 s(s+10) . Design (a) a precompensator W of order no greater than 2 such that the

crossover frequency !c  1 and bopt (WP ) is as large as possible; (b) nd the optimal loop shaping controller K = K1 W .

Solution 18.6 Note that the Riccati equation can be written as

X (A + BF ) + (A + BF ) X + C  C + F  F = 0 N i.e., X is the observability gramian of M . Let P be the controllability gramian, i.e. P (A + BF ) + (A + BF )P + BB  = 0 and without loss of generality, assume X and P are balanced with X=P =



1

N Then partition the realization of accordingly



2 :

M

2 A + BF B 3 2 A11 + B1F1 A12 + B1F2 B1 3 4 C 0 5 = 664 A21 +C B2F1 A22 +C B2F2 B02 775 1 2 F I F1

Then F1 = B1 1 and

F2

I

1 (A11 + B1 F1 ) + (A11 + B1 F1 ) 1 + C1 C1 + F1 F1 = 0

71 This implies the r-th order balanced reduced order model  N^  2 A11 + B1 F1 B1 4 C1 0 M^ =

F1

I

3 5

is also a normalized coprime factorization. Solution 18.7 (Reduced Order Controllers Model Reduction, McFarlane and Glover  A byBController  (1990), Zhou and Chen (1995)) Let G(s) = C 0 = M~ 1 N~ be a normalized left coprime factorization and let K (s) be a suboptimal controller given in Corollary 18.2 (with performance ). Let K = UV 1 be a right coprime factorization 2 3

U

A BB  X1 Y C  4 C I 5 V = B  X1 0 ^ V^ 2 RH1 be approximations of U and V . De ne and U;

 U   U^ 

 :=

V V^ 1 and Kr = U^ V^ 1 . Show that Kr is a stabilizing controller for G if  < 1 and

 K 

 

Ir (I + GKr ) 1M~ 1

=

KIr (I + GKr ) 1  I G 

< : 1 1 1  Solution 18.8 (a) Note that K stabilizes the family: n ~ ~ 1 ~ ~

 ~ ~ 

o F := (M +
M ) (N +
N ) :
M
N 1 < 

o

n

  (M~ r +
^ M ) 1 (N~r +
^ N ) :
^ M
^ N 1 <   Hence K stabilizes M~ r 1 N~r with stability margin of at least  : and

 (b) Denote
~ M
~ N

 F :

 K 

I (I + Gr K ) 1M~ r 1

 ( ) 1 : 1  :=  M~ M~ N~ N~ . Then r

r

W 1 W  = (N~ +
~ N )(N~ +
~ N ) + (M~ +
~ M )(M~ +
~ M )   
N  + 

  N~   = I + N~ M~ N M
 M~ 

 +


N
M

M




N
M



and W 1 1  1 +  and kW k1  1 1  . (c) This follows from the fact that Gr = (W M~ r ) 1 (W N~r ) 1 is a normalized left coprime factorization and

and

 K 



 K 

1 1 1 1 1

~

~ inf (I + Gr K1) (W Mr )  I (I + Gr K ) (W Mr )

K1 I 1 1

 K 



W 1



I (I + Gr K ) 1M~ r 1

1 W 1 1    1

 K 

 

I1 (I + Gr K1) 1M~ r 1



KI1 (I + Gr K1) 1(W M~ r ) 1

kW k1 : 1 1

72

H1

(d) Note that implies that K1 stabilizes

 K 

I1 (I + Gr K1) 1 M~ r 1

 rn1 kW k1 1




M

~ r +
M ) 1 (N~r +
N ) : F1 := (M Then the result follows from the fact that



LOOP SHAPING

(M~ +
~ M ) 1 (N~ +
~ N ) :


~ M


~ N


N

1



< rn 1

kW k1

i.e., K1 stabilizes M~ 1 N~ with stability margin of at least kWrnk1

 K 

I1 (I + GK1) 1M~



< rn  : 1 kW k 



 F1

:

  1 rn

1  kW k1  :

1

(e) This is obvious because

 K 



 K 

2 1 1 1 1 ~ ~

inf (I + Gr K2 ) Mr  I (I + Gr K ) Mr

 ( ) K2 I 1 1

(f) Similar to (d).

1

A B Solution 18.9 Let G(s) = C 0 = M~ 1N~ be a normalized left coprime factorization and let K (s) be a suboptimal controller given in Corollary 18.2 (with performance ):

 A BB X Y C C 1 K (s) = 

Y C

B X1

where



2

0



2

X1 = 2 1 Q I 2 1 Y Q

and



1

AY + Y A Y C  CY + BB  = 0 Q(A Y C  C ) + (A Y C  C ) Q + C  C = 0:

Suppose Y and Q are balanced,i.e.,

Y = Q = diag(1 ; : : : ; r ; r+1 ; : : : ; n ) = diag(1 ; 2 ) and let G(s) be partitioned accordingly as

2A A B 3 11 12 1 G(s) = 4 A21 A22 B2 5 :

2



Denote Y1 = 1 and X1 = 2 1 1 I

Kr (s) =

2 2

2 1 1

A

11



C1 1

C2

0

. Show that

B1 B1 X1 Y1 C1 C1 B1X1

Y1 C1 0



73 is exactly the reduced order controller obtained from the last problem with balanced model reduction procedure. (It is also interesting to note that Q = X (I + Y X ) 1 where X = X   0 is the stabilizing solution to

XA + A X XBB  X + C  C = 0: Hence balancing Y and Q is equivalent to balancing X and Y . This is called `Riccati Balancing', see Jonckheere and Silverman, 1983.)

Solution 18.10 Apply the  controller reduction methods in the last three problems respectively to a satellite

A B system G(s) = C 0

20 6 A = 64 00

where

1 0 0 0 0

Compare the results.

0 0 0 1:5392

3

2

0 0 77 ; B = 66 1:7319  10 0 5 4 1 0 2  0:003  1:539 3:7859  10





C = 1 0 1 0 ; D = 0:

5 4

3 77 ; 5

74

H1

LOOP SHAPING

Chapter 19

Controller Order Reduction Problem 19.1 Apply the frequency weighted balanced model reduction procedure to

    U^ 

N~n M~ n  UV

V^ 1

where N~n ; M~ n; U , and V are given on page 503. Show that the weighted controllability Gramian P and observability Gramian Q can be obtained from (Ak + Bk Fk )P + P (Ak + Bk Fk ) + Bk Bk = 0

Q Q  Q Q  1 12 1 12  o Q12 Q + Q12 Q Ao + C0 Co = 0  A + B D R 1C  B R~ 1 C

A where

Ao :=

2 k

2

2

k

Bk R 1 C2 Ak + Bk R 1D22 Ck   Co := R 1C2 R 1D22 Ck Fk

Suppose K = UV 1 is an observer-based controller, i.e., Ak := A + B2 F + LC2 + LD22 F; Fk := (C2 + D22 F ); Lk := (B2 + LD22) Bk := L; Ck := F; Dk := 0: Then show that P and Q can be obtained from (A + B2 F )P + P (A + B2 F ) + LL = 0 (A + LC2 ) Q + Q(A + LC2 ) + C2 C2 = 0 Problem 19.2 Let G be a generalized plant and K be a stabilizing controller.Let
 = diag(
 pz;
k ) be a w suitably dimensioned perturbation and let Tz^w^ be the transfer matrix from w^ = w to z^ = z in the 1 1 following diagram

z zw

W

1

z1

- (K^ 75

w

G

 

K

- f w1

K )W

CONTROLLER ORDER REDUCTION

76

Let W; W 1 2 H1 be a given transfer matrix. Show that the following statements are equivalent  I 0   1. 
0 W 1 Tz^w^ < 1; 2.

kF` (G; K )k1

1 Fu (Tz^w^ ;
p )

< 1 and W

1 < 1 for all  (
p )  1;

  

3. W 1 Tz1w1 1 < 1 and

F` I0 W0 1 Tz^w^ ;
k

< 1 for all (
k )  1; 1 Problem In the part3 of Problem 19.2, if we let
k = (K^ K )W , then Tz1 w1 = G22 (I KG22 ) 1 and  I 19.3  0

^ ^ ^ 0 W 1 Tz^w^ ;
k = F` (G; K ). Thus K will stabilizes the system and satis es F`(G; K ) 1 < 1 if k
k k1 =

(K^ K )W

 1 and the part 2 of Problem 19.2 is satis ed by a controller K . Hence to 1 reduce the order of the controller K , it is sucient to solve a frequency weighted model reduction problem if W can be calculated. In the single input and single output case, a \smallest" weighting function W (s) can be calculated using the part 2 of Problem 19.2 as follows F`

jW (j! )j 

sup

(
P )1

jFu (Tz^w^ (j! );
p )j:

Repeated Problem ?? and Problem ?? using the above method. (Hint: W can be computed frequency by frequency using  software and then tted by a stable and minimum phase transfer function.)

Problem 19.4 One way to generalize the method in Problem 19.3 to MIMO case is to take a diagonal W W = diag(W1 ; W2 ; : : : ; Wm ) and let W^ i be computed from

^ i (j!)j  jW

sup jeTi Fu (Tz^w^ (j!);
p )j

(
P )1

where ei is the i-th unit vector. Next let (s) be computed from ^ 1 Fu (Tz^w^ (j!);
p )j j(j! )j  sup jW (
P )1

where W^ = diag(W^ 1 ; W^ 2 ; : : : ; W^ m ). Then a suitable W can be taken as ^ W = W: Apply this method to Problem ??.

Problem 19.5 Generalize the procedures in Problem 19.3 and Problem 19.4 to problems with additional structured uncertainty cases. (A more general case can be found in Yang and Packard [1995]).

Chapter 20

Structure Fixed Controller

77

78

STRUCTURE FIXED CONTROLLER

Chapter 21

Discrete Time Control Solution 21.1 Using B = (A I ) 1 (A + I ), we get ~ + Q = 0; P = 1 (B I ) P~ (B I ) B  P~ + PB 2 On the other hand, using C = (A I )(A + I ) 1 , we get

C  P + PB + 2(A + I ) T Q(A + I ) 1 = 0:

Problem 21.1 Let a discrete time transfer function be given by

A B G(z ) = C D

and assume D = 0. Derive a computational procedure to compute the H1 norm of the transfer matrix G(z ) without the assumption that A is nonsingular.

79

80

DISCRETE TIME CONTROL

Bibliography [1] C. N. Nett and J. A. Uthgenannt, \An explicit formula and an optimal weight for the 2-block structured singular value interaction measure," Automatica, vol 24, No. 2, pp. 261-265, 1988. [2] J. S. Freudenberg, \The general structured singular problem with two blocks of uncertainty." Proc. 1986 Allerton Conf., Monticello, Illinois, October 1986. [3] N. Aoki, S. Ushida, and H. Kimura, \On -characteristic of H1 control systems," Proc. 34th IEEE Conf. Dec. Contr., New Orleans, Louisiana, December 1995, pp. 2562-2567. [4] Jonckheere, E. and L. M. Silverman (1983). \A new set of invariants for linear systems - applications to reduced order compensator design," IEEE Transactions on Automatic Control, Vol 28, No. 10, pp. 953-964. [5] Qiu, L. (1995). \On the robustness of symmetric systems," Proc. of 34th IEEE Conf. Dec. Contr., New Orleans, Louisiana, pp. 2659. [6] Qiu, L., B. Bernhardsson, A. Rantzer, E. J. Davison, P. M. Young, and J. C. Doyle (1995). \A formula for computation of the real stability Radius," Automatica, Vo. 31, No. 6, pp. 879-890. [7] K. Zhou and J. Chen \Performance bounds for coprime factor controller reductions," Systems and Control Letters, Vol. 26, No. 2, pp. 119-127, 1995.

81