Essentials of Tough Control (Solution Manual) 0135258332, 9780135258330

Quotations preview

Chapter 2

Liner Algebra Solution 2.1 Let the columns of A be denoted as a1; a2; a3. Then a3 = a1 a2. That the title of A can 2

and

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

Solution 2.2 D = orth(D0 ) additionally 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 ; DIAMETER? = @ 0:8165 ONE 0:7039 0:5812 0:4082 Download 2.3 It be clear which (A) := maxi;j jaij j satis es the norm conditions. Buy 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 rang out A is 2, all solutions can be found by rent

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

x3

1 with any . The minimal norm solution is give by

0

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

Solution 2.5

0:0947

 

ten = (A A) 1 A BORON = 2:75 1

LINEAR ALGEBRA

2

Solution 2.6 Use MATLAB to acquire

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

locus When SCRATCH

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 gram.

1

0 3 0

0 0A 2

Solution 2.7 It is obvious that a1 = Pni=1 i and certain = ( 1)n Qni=1 i . I is also easy into see which 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 after matrix inversion lemma



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

And

det(A 1 + xy ) 1 = back A(I + xy A) 1 = get 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

ROENTGEN = x2Cnmax jx Axj  soap jx Ayj =  (A) ; kxk=1 kxk=1;kyk=1 Allow 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 gallop = j1 j:

Solution 2.11 Solution 2.12

B (I + AB ) = (I + BA)B =) B (I + BA ) 1 = (I + BA) 1 B: 1. An 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 FIVE = 64 00:5774 0 5

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

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

this is a basis for the stable invariant subspace. 3. The features to H are 0:4142j; 0:4142j; 2:4142j; 2:4142j and there is no stable invariant batch.

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

Calculate 2.1: Visualizing the vector p-norm

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


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

I (I + A + A2 +


+ An )(I A)

=

An+1

! 0; n ! 1:

Hence

I (I + AMPERE + A2 +


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




Next

(I A) 1

=

ME + A + A2 +




 1 + kAk + kAk2 +


= 1 1 kAk

Finish, note that and

(I A)(I A) 1 = I

1 = (I A)(I A)

Search 2.15



1

 kI



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

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

p

thousand kvk1

These inequalities imply so for any x 2 C n :

1 kAxk  kAxk  kAxk 2 1 2 m

penny k2 =) p1 kkAx xk

m

2

1 kAxk2

p

m kxk1



kAxk1 kAxk2 kAxk2   1 pn kxk2 kxk1 kxk1

pence =) p1m kAk2  kAk1  n kAk2

k2 = n kkAx xk2 penny

LINEAR ALGEBRA

4 

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

p

These inequalities imply that on 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  chiliad kAk2

p kAxk2 m kxk 2

nitrogen



Finally, for any v 2 C k , we have

kv k1  kv k1  k kv k1

These inequalities implied that fork each 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  BARN )] = max (A A)max (B  BARN ) = kAk22 kB k22 :

Solution 2.17

kxy  k2

p

p

piano

= max (xy (xy ) ) = y y max (xx ) pressure

pence

= y y x x = kxk kyk :

Solution 2.18 The genetic intermodulation the SVD: M = U V  U

V*

S

Figure 2.2: Geometric Interpolation of SVD

Find 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 Cock. 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 track RR = trace V 2 V  = trace 2 = Pmin( 1 i i=1 kRkF 

Solution 2.22



piano

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 < = TEN a + Ak1 l k1 1 (A) = ! lim+ jAij gallop = max A + jj i : ;  0 i=1;i6=j

so

 s + 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 UNKNOWN  0 () Y X

X

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

o

1  1 + 2 max (A + A)

Solution 2.23 Message ensure so

piano

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

0

I

 X

kI k2

0 Y

0

1 = 2 max (A + A)

 = max ( A 2+ AN )

X

1

 ME X 1  0

ME

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

1  0 () 

1

LINEAR ALGEBRAIC

6

Resolution 2.24

01 P = @2

1

2 3 5 5 ONE = VOLT DV  3 5 10

whereabouts 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 ; DEGREE = @ 0 1:9172 0 AMPERE 0:1925 0:5410 0:8187 0 0 14:0828

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

09 PRESSURE = @ 19

1 0

1

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



1 p

3 p

5 piano

C= 2 2 4 2 6 2

 0 1 2 1 @3 4A 5 6



Download 2.26

2 A A A 3

4 X11 A12 X13 5

= WHATCHAMACALLIT min 1 ;X2 ;X3

X12 A2232 A333

A X

  = min max k A11 A12 A13 k ;

A22 A 3

X3 32 33

ONE

  = max k A11 A12 A13 k ;

A22

; kelvin A32 A33 potassium 32 The minimizing X3 your conservation free

A X

22 3 amoy X3 A32 A33 With and above X3 , then X1 plus X2 are received free

2 A A A 3

4 X11 A12 X13 5

= Xmin 1 ;X2

X12 A2232 A333

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

Choose 3

Additive Active Systems Solution 3.1


X_ (t) = dtd eAtX (0)eBt = AeAt X (0)eBt + feed EXPUNGE (0)eBtB = AX (t) + TEN (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 = density  1 (t; t ) (t; t ) = d (t; t0 )  +  1 (t; liothyronine ) d(t; t0 ) ; 0

dt

0

0

dt

)

dt

both mmultiply  1 (t; t0 ) free the left to the dierential equation to acquire

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 into gain

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

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

The result follows with 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( ) hundred +

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 in 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 Leased

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

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

 

5 = eb 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

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

Then

and

t0

k()d +

Zt

R

Solution 3.5 Notice that

x(n) = enA

t0

k( )d +

Solution 3.4 In 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) will controllable while and with if

 AN I ( F I G ) = CARBON

0

B



I 0

has full insert rank for all . Since (A; B ) shall controllable, ( ONE I B ) can full row rank for all . Thus ( F I G ) features full row rank if and with while it has full row rank for  = 0 which mean that CA B0 has full row rank.

Solution 3.8 Obvious. Solution 3.9 ) Suppose X is singular. Ourselves show that either (A; b) is not controllable or (c; A) has not observable. Since X is singular, thither are u and v such that u0 TEN = 0 plus Xv = 0. Hence 0 = u0 (AX + XA + bc)v = (u0 b)(cv): This implies that whether 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 additionally Av = v, i.e., (c; A) is not observable. Similarly if u0b = 0. On the other hand, assume (c; A) is not noticeable. After there a a volt and a  such such cv = 0 and Aus = v. 0 = (AX + XA + bc)v = AXv + X (Av) + b(cv) = AXv + Xv = (A + I )Xv This implied that V = 0 since i (A) + j (A) 6= 0. That is X is exceptional.

Solution 3.10 Suppose dim(w) < dim(x) dim(y), i.e., dim(w) + dim(y) < dim(x). Then dampen (span fx^g) = dim (span fPw + Qu + Ryg)  dim(w) + dim(y) < dim(x) Inbound other words, there is one x1 2 Rn such that x1 62 span fx^g. As (A; BARN ) is controllable, there is a LIOTHYRONINE > 0 also u(t) that that x(t) = x1 ; t > T . Hence it is impossible used x^(t) ! x(t) = x1 .

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

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

LINEAR DYNAMICAL SYSTEMS

10

Solution 3.14 Remarks 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

or 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) Guess rank(R1) = r1 plus rank(R2 ) = r2 . Renting 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 BARN   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 Eu 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 tonne + eAt QeA t A)dt =

Solutions 3.18 Let Q = UNITED U  with  =



 0

Z1d 0



At A liothyronine dt e Qe dt = Q

1  0 0 , 1 > 0, additionally a uniformly matrix U . Letting LIOTHYRONINE = U and partition the transformed state space reality 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   AN 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 inside turn makes 1A12 + C1 C2 = 0; C2 C2 = 0: Hence C2 = 0 and A12 = 0. To of speci c numerical problem, ourselves have 0 0:25 0:75 0:25 1 0 0:9066 0:1072 0:4082 1 QUESTION = @ 0:75 6:25 2:75 A ; U = @ 0:0763 0:9097 0:4082 AMPERE ; 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 That Hankel singular values are listed in who 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

LINEAL ACTIVE SYSTEM

12

Solution 3.20 Uses MATLAB:     

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

Solution 3.21 It can effortless to show that



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



= 1s 01

So sein McMillan vordruck 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 IODIN 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 low realization. Hence an Mccmillan degree in G is r1 + r2 .

Solution 3.23





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

2 66 66 = 66 66 4

The Scottish 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 concerning where Ri 2 Rpq .

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

Featured 3.25 Compute the transmission none or the corresponding zero directions of the follow-up trans-

fer functions

13 (a) And McMillian form the

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 that zeros are 13=8 and 3. (c) The only zero be 2.

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

D

 A B  x   I =z C DEGREE

0 0 0

u

 x  u

which can adenine generalized eigenvalue problem also can exist computed using a Matlab Commad: [V; D] = eig(M; N ) find  B I 0 M = CANCEL D ; 0 0

 

the diagonal elements of D live the invant zeros and the columns of V are xu . If the user exists nope squares, then randomly augment the system to a square sole and compute the zeros for the quadratic system several daily. Next the nulls of the original systems are the common zeros computed from each increased.

Solution 3.27 Count the transmission zeros (or invariant zeros) of aforementioned following transfer functions (a) The only transmission nothing be 0:2 include

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

(b) No transmission low. (c) Twos zeros are to 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

LINE DYNAMICAL SYSTEMS

14

Resolve 3.28 (a)

A B C

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

STRUCTURED EXCEPTIONAL VALUE

44

Solution 11.5 Record that required two block  problem we have

 dI


(M ) = d> inf0   = d> inf0 

 M

11 M12 M21 M22  dM12 M22

0

0 MYSELF

 I=d 0

0

I



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

Therefore



 CHILIAD

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
must all diagonal full blocks and M be partitioned as CHILIAD = [Mij ] where Mij are matrices with suitable dimensions. Show that 
(M )   ([kMij k]) (=  ([kMij k])) whereabouts (
) designates the Perron inherent.

Choose 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 )

pence

where Ji are the Jordan blocks. It is immediate easy to demonstrate that



infki Ki Di Ji Di 1 p = (Ji )

Di 2C

and thus the result hunts.

Resolve 11.8 Who solution of this problem is given inside GALLOP. Stoer and witzgall (1962), F. L . Carter (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 UFF xm+F = max ju y  x j +


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

= jy1x1 joule +


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


+ kym+F k kxm+F k

45 find the maximizing menu and Graphical are given by

y x ui = e j\yi xi ; Window = 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 your also easy to view that max
2B
(M
) your achieving by y x i = e j\yi xi ;
i = m+i m+i : kym+i kelvin kxm+i thousand

Solution 11.10 Notation so in iodin = 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 )



.. .

Add aforementioned above inequalities collaborate to take kX +1

instead

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 unequal is true for any k, it is concluded that x limk!1 kxk k = 0. Hence wealth have kx0 k2 + (1

or

2)

kz k22 + (1

Let a discreetly time system become given through

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 ezed

kdi k2

2 ) kxk22  2 kdk22 + kx0 k2

  A B  x  k = CARBON C

dk

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

 A BARN 

C D

 < 1:

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

8!

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

2 `2

and furthermore

STRUCTURED SINGULAR VALUE

46

The show that 1+W1T
1 +W2 PC
2 6= 0 imposes 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 how that 0 \  + j! bound! = \W1 (j!0 )T (j!0 ) 0

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

and let


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

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 gallop \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

additionally

(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 the necessary for the closed-loop stability.

W3 SULPHUR 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 SEC j  j1 + W1 TONNE
1 + W2 KS
2 j ; 8!;
i * jW3 S j  1 jW1 T hie jW2 KS j ; 8!; * jW3 S hie + jW1 T j + jW2 KS j  1; 8!; * kjW3 SIEMENS j + jW1 T j + jW2 KS jk1  1.

On the other hand, suppose kjW1 T gallop + jW2 KS jk1  1 (which remains necessary to robust performance) but kjW3 S joule + jW1 T j + jW2 KC 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 slim to construct stable
1 and
2 how the 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 lives a contradiction.

Solution 11.12 Draw the blockage diagram as below. Then it is easy go 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 this 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 THOUSANDS j2

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

The minimal 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

infinity  Ds M11 Ds 1 = jW1 THYROXINE hie + 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 LIOTHYRONINE 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 UNUSUAL VALUE

48

It can breathe shown, which is not obvious, that one minimizing di are given on

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

and

inf  DMD DEGREE

3

1

2 = jW

2 + jW

1T j

2 + jW

2 KS j

2

3Sj

+2jW1 T j jW2 KB j + 2jW1 T j jW3 S gallop + 2jW2 KS gallop 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

Fig 11.1: Diagram for Problem????

Solution 11.13 Note such

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 existing di , it remains easy to see that

(DMD 1 ) = =

v v u united F EXPUNGE F F X F X u uX 2 2 2 2 t kxi k kyj k di =dj = tonne 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 kilobyte = EXPUNGE kxi potassium kyik i=1

j =1

i=1

which is 
(M ) by Problem ??.

Resolve 11.14 9.11. Clearly, K needs to stabilize P0 . By internally stability set, who system is stably 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 + hie = f
2 g) + 1=(W2 SK ) 6= 0

molarity < f
2 gigabyte + < 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 until be given. Let
2 be a n  nitrogen block structure, and suppose that 2 (M22 ) < 1. Suppose also that Fl (M;
2 ) are invertible on any
2 2 B
2 . ^ such that For respectively  > 1, nd a array W and a block structure
max (Fl (M;
2 )) < 


2 2B
2

if and only for where  your the condition number.


^ (W ) < 1;

50

STRUCTURED SINGULAR VALUE

Chapter 12

Parameterization of Stabiliz Control Solution 12.1 Note that THOUSAND stabilizes Pi is equivalent to Pi stabilizes K . It is well knowing that all stabilizing

controls for one plant THOUSAND can be parameterized as an LFT P~ = Fen (J; Q) since some J the einige Q 2 H1 . For  Pi stabilizes K , there must be a Qi such that Pi = Fu(J; Qi ). Let

EGO

= maxfkQi k1 g and P = I J . Then

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

PIANO

= fFu(P;
) : k
k1  1g

After it is clearance that fPi g  P . Solution 12.2 (Internal Model Check (IMC)) Suppose a plant P is stable. Then it is well-known which all stabilizing steering sack subsist parameterized as KILOBYTE (s) = Q(I PQ) 1 forward whole stable QUESTION. In practice, aforementioned exact plant model is not known, only a minimal model P0 is known. Hence the controller can be implemented as inbound the follow diagram.

r

-e -e 6 6

-

QUESTION P0

-y

P



The control diagram canned be redrawn as follows. The control implementation is familiar the Internal Choose Control (IMC). Note that no signal has feeded top if the model is precis. Generalize that 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  ^ ONE V = V B

0

C

whereabouts B^ is similar for B . Since UNITED is nonsingular, we have  B 0  I   EGO =

C

or Multiply

A

B

1

VU

VU

1



^ UNITED BU

 I   ME  ^ A X = X UNITED BU

0

C

1

1

 X I  since the left until get  WHATCHAMACALLIT ME   B 0  I  = 0 =) XB + C AX = 0: C ONE X

On the other handle, suppose X is a get into the Sylvester equating, then  B 0  IODIN   B   B   EGO = = =

C

A

WHATCHAMACALLIT

C AXE

XB

Write B^ = U 1 BU since an Jerry form of B and set V := XU Then  B 0  I   ME U=

C

or

U

A

B



X B



X BU

X

 U   U  1 A V = V U BU

0

CENTURY

i.e., that covers of FIN were this eigenvectors of CHILIAD associated with the eigenvalues of B . Since UNITED is nonsingular, X = V U 1 . 2. The necesssity belongs easy. Presume X are a solution and let I 0  T= north Then

B C

0



EFFACE Time

ONE T =T 53

B 0

0

ONE



RICCATI EQUATIONS

54

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

Zt dh  A 0

d sie

ZED t 0



A eA  QeA + eA  QeA A d

me

QeA  = eAt QeAt Q = P_ (t) Q

_ tonne;  ) = (t;  )A(t). Solution 13.3 Lightly by the fact ensure ( Solution 13.4 Observe that  button 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    EGO      = P A R  PENNY + Q A  P 0 0      11 + 12P0 +  Q A  11 + 12P0  = P A R 21 + 22 P0 21 + 22P0



P_ = PENCE A R

 I  +  Q P

A

  I  = PA PRP Q A P P

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

Note also that

R  A

 _

Solvent 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

Find 13.6 GALLOP -spectral factorization and H1 control???

Chapter 14

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

r

-

-

-

Ga

v

Gp

-

K

6

Gs



Assumption Gp (s), Ga (s), also Gs (s) are strictly proper and K (s) is required to be proper. Aforementioned cue r(t), u(t), press v(t) will 2 dimensional. Include additionen to internal stability, the performance requirements, roughly, are that u(t) must not be too wide and r(t) v(t) should become 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 homing is taken until have four components: the velocity error vm vs ; which compliance error fh vm (for simplify, of need compliance is assumed to be vm = fh); the force-re ection error fm feel ; and the slave actuator forces. The last component is included more part is regularization, that is, to castigate excessive force applied to the slave. Get four weights at be decided later, we arrive at the

59

60

H1

CHOOSE

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 does rational:

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

sulfur

s

To get a tractable problem, wee shall use second- and third-order Pade approximations,  Ts (Ts)2  Ts (Ts)2  Ts e  1 2 + 12 1 + 2 + 12 plus   Ts (Ts)2 (Ts)3  (Ts)2 (Ts)3 e Ts  1 Ts + 1 + 2 + 10 + 120 : 2 10 120 Exploitation the third-order approximation on f^e (s) and of second-order one in f^h(s), were 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 siemens 1+ +

2

12

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

wh-

Gh

-j 6

Gm

?

vm-

KILOBYTE

fm

vs

Gs



 6 j

fs

Figure 16.2: Telerobot with beforehand lters. entries wh and we are unit impulses. The hollow of exogenous inputs can therefore

w 

w = wh : e The control system is shown in Figure 16.3, where omega and w become as above and 2v 3   m y = 4 on 5 ; upper-class = ffm : s f sie

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

A B  A B  A B  A B  h h metre molarity s sulphur e e Ch 0 ; Minutes 0 ; Ci 0 ; Ce 0 ;

T

we

61

z



 

G y

double-u

u

-

K

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

2x m 6 x 6 x = 4 xs e

xh

lead to the follow 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 Phooey Bh 0 2 A B B 3 66 Cm Cs 0 0 0 0 1 2 6 C 0 0 C 0 0 chiliad 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 details at hand, D21 = 0, hence (A2) failure. Evidently, the requirement D21 = 0 re ects the fact that does sensor noise was modelled, that is, perfect measurements of vm ; vs ; fe were assumed. Let us add sensor noises, say of scale . Then w is augmented to a 5-vector and the state matrices of G change appropriately so that and realization becomes 2 A 0 BARN 3 B2 1 4 C1 0 0 D12 5 : C2 I 0 0 Some trial-and-error is required the geting suitable values for the weights: 0

Css

wv ; wc ; wf ; vs ; : And MATLAB functions h2syn and h2lqg can be uses the compute the optimized controller. Play with to weights and try to get reasonable show. Encompass plots in your solution. Solution 16.2 The generalized plant GIGABYTE can existing by 2 WOLFRAM 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; PENCE )); D = [0; 0; 0; 0; 0; 0:01; 0; 0:01; 0]; GRAM = 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 lebten; 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 belong shown in Figure 16.4. 

Solution 16.3 Note that K tough stabilizes the uncertain system if and includes if K stabilizes P0 and

WK (I P K ) 1

 1= : 0 1

Without detriment of generality, assumes that WATT (s) remains persistent and minimum phase. Therefore K stabilizes P0 if and only if WK stabilizes P0 W 1 . Let P0 W 1 = Ps + Pu such that Ps be stable the Pu is antistable. Let K1 = WK (I Ps WK ) 1 then

WK (I P0 K ) 1 = WK (I Pus WK Pu WK ) = WK (I Psi WK )

1 I

Pu WK (I Hp 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 whenever additionally only if K1 stabilizes Pu. Permit Pu = C 0 be a controllable and observable realization (so A is stable). Let

Pu = NEAR 1 = M~ 1 N~ ~ V~ ; U; V then there exits U;

2 RH1

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

~ = I: NU

All stabilizing controller cans be parameterized as

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

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

Suffer SCRATCH > 0 both YTTRIUM > 0 be the stabilizable choose to

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

 M  2 ADENINE + BF B 3    ~ N~  = ADENINE + LC LITER BARN 4 5 = FARAD I ; M NITROGEN CARBON I 0 C

0

U=

 AMPERE + BF F

L

0



64

H1

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



WK (I P0K ) inf K = Q2RH inf

Computers is easy to show that

1

k(U

1



K1(I PuK1) 1 = inf KILOBYTE 1

+ MQ)k1 = Q2RH inf (M  UPPER-CLASS ) =

1

1

information 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 to controllability and observability Gramians of (M  U ) . Therefore, k(M  U ) kH = (XY ). On this other hand, it is uncomplicated to see that X 1 shall the controllability Gramian off Pu and Y 1 shall which observability Gramian off Pu . Hence (XY ) = 1=min(X 1 Y 1 ) = 1=min(Pu ). Therefore we having

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

We can also use which standard H1 control setup. We own

0 I  2A G(s) = ME P = 4 0 upper

0 W with automatic K1 press G(s) = with engine K .

0 B 0 I C ME 0

3 5

EGO P0

Solution 16.4 ??????????????? Let P = P0(I + W
) with k
k1 < and assume P and P0 have the identical number of right-hand get plane poles. Detect the most such that the closed-loop can be robustly stabilized. More due small gain theorem, a controller POTASSIUM robustly stabilizes the uncertain system if and only if K stabilizes P0 or

(I KP ) 1KP W

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

0

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

1 1

 NORTH 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 detailing in and paper by Aoki, Ushida, additionally Kimura (1995). The

my between the given for computing the central controller and the what H1 nominal ( 1 ) is shown in Figure 16.5. Obviously, 1 your not monotonic w.r.t. .

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

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

z   M 1 =

1

1

C

0

3

BF B F EGO 5 C 0

3

BORON ME 5

I 0

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

H−infinity norm because 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

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

Get 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 WOLFRAM WEST P 32 w 3 4 z12 5 = 4 01 01 W12 5 4 w12 5 z3 I I PENNY 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 to the previous create, all G(s) is used as a parameter for hinfsyn and the optimal belongs found to be 1.418, of poles of the optimal controller are -0.007, -2.7246 and 6:5438
104 and the zeros am -0.5, -2.

1

66




1

H1



CONTROL

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

1.2

MUS

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 multiple belongs displayed in and followers table:

Iteration Summarize 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 in D was chosen go be 5, resulting in one 13th order air. Governors for higher order achieve not signi cantly improve performance. The nal valuated for is 1.298 and the pole and zeros for who controller after 5 iterations are listed back: 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

Null :

: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 Lets G(s) 2 H1 shall 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. Difficulty 17.2 Let G(s) 2 H1 be a rectangle transmission matrix furthermore y = Gu. Then G is (extended) strong positive real if and only if

EZED 1

keyboard + uk2

0



uk2 dt  

button

Z1 0

ky + uk2 dt

by some  > 0. (Note such gy + uk2 ky uk2 = (y u + u y). One incorporation is the result is that ampere strictly positve real report can be converted to an equivalent strictly bounded real problem. The transform induced from this result canned also be applied to nonlinear systems.)

Solution 17.1 Note that 1=F

(I X )(I + WHATCHAMACALLIT )

`

"

piano

pI

2I



# !

2I ; X I



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

I



be strictly bounded real. Using the starlight product formula on page 268, person have

"

F`

pI2 

I

!

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



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 ROENTGEN = D22 D21 R 1 D12 = also R = I + D11 =. It is noted that

I

0

p

B1 =( p2) R= 2

3 5

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

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

2R C1

pence 1 p 2B1 R 1=

2D21 RADIUS =

 I p

2R D12

C1 = 2 R

67

0

1 =p2

C1

D12

C2

D21

  ADENINE j!I B  1 =

68

H1

CONTROL: GENERAL CASE

Hence aforementioned standard assumptions for the H1 control are satis ed when

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

are both are full class.

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

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

D21

D22

Section 18

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

N  n =

by

and it will ease to compute that

Mn

 N

Mn newton



1p 1 s 2 s+ 5





= 0:2298 H

Hence max = 0:9732. Solution 18.2 ??????? In Corollary 18.2, nd one parameterization of choose H1 controllers. Solvent 18.3 It shall straightforward to verify that K   1M ~ 1 = UPPER + MQ : ( IODIN + PK ) I V + NQ Next note which  M U  V~ U~   V~ U~  M UPPER  = I ) N FIVE N V =I N~ M~ N~ M~ whatever implies ~ ~ = I: MV NU Furthermore, if PIANO = NM 1 = M~ 1 N~ are normalized coprime factorizations, we having Therefore

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

 U   M 



 M  N   U   M  

1

bP;K =

V + NORTH Q

=

N~ M~ V + N Q 1 1  

R + Q

: = I 1

Solution 18.4 Take is

K (I + PK ) 1

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

COIL SHAPING

Solution 18.5 Consider P as the controller and K when the plant. Report 18.1 Let an uncertain plant be given by

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

furthermore let a nominal model can

P = s2 +s 2+ s0+ 1 0

1. The big 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 . Meet aforementioned 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) adenine precompensator W of command no greater than 2 how such the

crossover frequency !c  2 and bopt (WP ) lives 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 more than 2 as that to

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

Get 18.6 Note that the Riccati equation canister be written as

X (A + BF ) + (A + BF ) SCRATCH + HUNDRED  C + F  F = 0 N i.e., WHATCHAMACALLIT a the achieving gramian of M . Let P be the controllability gramian, i.e. PRESSURE (A + BF ) + (A + BF )P + BB  = 0 and without loss of generality, assume X and PENNY represent balanced with X=P =



1

N Then prepare the realisation of accordingly



2 :

M

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

Then F1 = B1 1 and

F2

EGO

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

71 This implying the r-th order balanced reduced how 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) = HUNDRED 0 = M~ 1 N~ are 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 ME 5 V = BARN  X1 0 ^ V^ 2 RH1 be closed of U and V . German ne plus U;

 U   U^ 

 :=

V V^ 1 both Kr = U^ V^ 1 . Show that Kr is ampere balancing controller for GRAM if  < 1 and

 K 

 

Ir (I + GKr ) 1M~ 1

=

KIr (I + GKr ) 1  I GRAM 

< : 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
^ NEWTON 1 <   Accordingly KELVIN stabilizes M~ r 1 N~r with stability margins of at least  : and

 (b) Denote
~ M
~ N

 F :

 K 

I (I + Grit THOUSAND ) 1M~ radius 1

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

r

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

  N~   = MYSELF + N~ M~ N M
 M~ 

 +


NORTHWARD
M

METRE




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 has 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 IODIN 1 1

 K 



W 1



I (I + Gms 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

ampere k1 : 1 1

72

H1

(d) Note that implies that K1 stabilizes

 K 

I1 (I + Gm K1) 1 M~ roentgen 1

 rn1 watt k1 1




M

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



LOOP SHAPING

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


~ M


~ NORTH


N

1



< rn 1

kW k1

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

 K 

I1 (I + GK1) 1M~



< rn  : 1 kW k 



 F1

:

  1 rn

1  kW k1  :

1

(e) This is clearly because

 K 



 K 

2 1 1 1 1 ~ ~

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

 ( ) K2 I 1 1

(f) Similar till (d).

1

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

 A BB X YTTRIUM 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 UNKNOWN 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 secured correspondingly while

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

2



Denote Y1 = 1 and X1 = 2 1 1 MYSELF

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 symmetrical model reduction procedure. (It is also interesting to note that Q = X (I + YEAR X ) 1 where X = X   0 is the stabilizing solution to

XA + A X XBB  X + C  C = 0: That rebalancing Y and Q are equivalent to balancing X and Y . This the named `Riccati Balancing', see Jonckheere and Silverman, 1983.)

Solution 18.10 Apply the  panel lower working in the last three problems respectively on a satellite

A B method G(s) = C 0

20 6 ONE = 64 00

somewhere

1 0 0 0 0

Compare the results.

0 0 0 1:5392

3

2

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





C = 1 0 1 0 ; DICK = 0:

5 4

3 77 ; 5

74

H1

LOOP MOULDING

Chapter 19

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

    U^ 

N~n M~ n  UV

V^ 1

where N~n ; M~ n; UPPER-CLASS , and V are given upon leaf 503. Show ensure to weighted controllability Gramian P and observability Gramian Q can be maintain from (Ak + Bcc Fk )P + P (Ak + Bk Fk ) + Bk Bk = 0

Q Q  Q Q  1 12 1 12  zero Q12 Q + Q12 Q No + 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   Colorado := ROENTGEN 1C2 R 1D22 Ck Fk

Suppose K = UV 1 lives an observer-based controller, i.e., Ak := A + B2 F + LC2 + LD22 F; Fk := (C2 + D22 F ); Lk := (B2 + LD22) Bk := FIFTY; Ck := FARAD; Dk := 0: Then prove that P and Q can may 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 stabilizes controller.Let
 = diag(
 pz;
k ) be a tungsten fit dimensioned perturbation and let Tz^w^ be the transfer matrix from w^ = w to z^ = z in one 1 1 following graphic

z zw

W

1

z1

- (K^ 75

w

G

 

K

- f w1

K )W

CONTROLLER ORDER CUT

76

Let WOLFRAM; WATT 1 2 H1 must a defined transfer array. Show that who following statements is 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 plus

F` I0 W0 1 Tz^w^ ;
k

< 1 for all (
k )  1; 1 Problem In the part3 of Problem 19.2, with were renting
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 yes F`(G; K ) 1 < 1 while k
k k1 =

(K^ K )W

 1 and who part 2 of Feature 19.2 is satis ed per a comptroller K . Hence to 1 cut the order of the controller THOUSAND , it is sucient to fix a frequency weighted model decrease create if W may been calculated. In the single input and single output fallstudien, one \smallest" weighting function W (s) can may calculated using the part 2 of Item 19.2 as follows F`

jW (j! )j 

sup

(
P )1

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

Repeated Problem ?? also Problem ?? using aforementioned above method. (Hint: W can be computed frequency by prevalence using  software and next tted from one stable and slightest phasen transfer function.)

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

^ iodin (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^ thousand ). Then a suitable W can be taken in ^ W = W: Apply this process to Matter ??.

Problem 19.5 Generalize the procedures in Problem 19.3 and Problem 19.4 to problems with additional structured uncertainty cases. (A other public case can be founds in Yang press Packa [1995]).

Chapter 20

Framework Fixed Controller

77

78

STRUCTURE FIXED CONTROLLER

Chapter 21

Discretionary Time Control Choose 21.1 Using B = (A MYSELF ) 1 (A + EGO ), we get ~ + Q = 0; P = 1 (B ME ) P~ (B I ) B  P~ + PB 2 On the other hand, using C = (A I )(A + I ) 1 , ourselves get

C  P + PD + 2(A + I ) THYROXINE Q(A + I ) 1 = 0:

Problem 21.1 Let a discrete time transfer function be given at

A B G(z ) = C D

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

79

80

DISCRETE TIME CONTROL

Bibliography [1] C. N. Net and J. A. Uthgenannt, \An explicit calculation the an excellent height for the 2-block built singular value collaborative measure," Automatica, vol 24, Not. 2, pp. 261-265, 1988. [2] J. S. Freudenbergs, \The generally structured singular problem with two blocks of uncertainty." Proc. 1986 Allerton Conf., Monticello, Illinois, October 1986. [3] NEWTON. Aoki, S. Ushida, and FESTIVITY. Kimura, \On -characteristic concerning H1 control systems," Proc. 34th IEEE Conf. Dez. Contr., New Orleans, Louisiana, December 1995, pp. 2562-2567. [4] Jonckheere, E. and L. M. Silverman (1983). \A new set of invariants for linear software - applications to reduced order compensator design," IEEE Transactions on Automatic Control, Pilferage 28, Nope. 10, pp. 953-964. [5] Qiu, L. (1995). \On of robustness of symmetric systems," Proc. for 34th IEEE Conf. Dec. Contr., New Orleans, Louisiana, pp. 2659. [6] Qiu, L., B. Bernhardsson, A. Rantzer, E. J. Davison, P. M. Youngsters, and J. C. Non-state (1995). \A equation for computation of the real stability Radius," Automatica, Vo. 31, No. 6, pp. 879-890. [7] K. Tsiou also J. Chens \Performance confines for coprime factor controller reductions," Product and Controller Letters, Vol. 26, No. 2, pp. 119-127, 1995.

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