In turn, this implies so = 0, so that (*) becomes rank
− A 0 B −C 0 D
< n+p
and this contradicts the second rank condition in (+). Conversely, supposing (++) is controllable, then
You're Reading a Preview 2 ...
B AB A B . . . = n+p D full CBaccess CAB Unlock with a free trial.
rank This implies
Download With n −1 Trial . . . Free B AB
rank
A
B
=n
in other words, the first rank condition in (+) holds. Now suppose < n+p rank C D Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title Then & The New York Times Useful Not useful s I − A 0 B A B
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rank
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o n
−C
so I p
D
< n +p
so = 0
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Linear System Theory, 2/E
Solutions
Solution 13.10 Since P −1 B
(P −1 AP )P −1 B . . . (P −1 AP )n −1 P −1 B
= P −1
B AB . . . A n −1 B
and controllability indices are defined by a left-to-right linear independence search, it is clear that controllab indices are unaffected by state variable changes. For the second part, let r k be the number of linearly dependent columns in A k B that arise in the left-tocolumn search of [ B AB . . . A n −1 B ]. Note r 0 = 0 since rank B = m. Then r k is the number of controllab indices that have value ≤ k . This is because for each of the r k columns of the form A k Bi that are dependen have ρi ≤ k , since for j > 0 the vector A k +j Bi also will be dependent on columns to its left. Th k = 1, . . . , m, r k −r k −1 gives the number of controllability indices with value k . Writing
BG ABG . . . A k BG
=
B AB . . . A k B
0 ... G ... . . . . . . . 0 0 ..
G 0 . . .
0 0 . . . G
and using the invertibility of G shows that the same sequence of r k ’s are generated by left-to-right column se in [ BG ABG . . . A n −1 BG ].
Solution 13.11
You're Reading a Preview For the time-invariant case, if Unlock full access with a free trial. p T A = p T λ , p T B = 0
implies p = 0, then
Download With Free Trial p T ( A +BK ) = p T λ , p T B = 0
obviously implies p = 0. Therefore controllability of the open-loop state equation implies controllability of closed-loop state equation. In the time-varying case, suppose the open-loop state equation is controllable on [ t o , t f ]. Thus g Read Free For 30 Days Sign up to vote on this title x (t o ) = xo there exists an input signal ua (t ) such that the corresponding solution xa (t ) satisfies xa (t f ) = 0. The closed-loop state equation Useful Not useful Cancel anytime. . Special offer for students: Only $4.99/month. z (t ) = [ A (t ) + B (t )K (t ) ] z (t ) + B (t )v (t )
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Linear System Theory, 2/E
Solutions
ˆ bˆ = B
0 0 . . . 0 1
Using × to denote various unimportant entries, set
ˆ bˆ = B Rbˆ = block diagonal B o
0 0 . . .
ρi × 1 , i = 1, . . . , m
1 0 . . .
.
0 1
× 1 . . .
0 0 0 0
... ... . . . ... ...
× × . . .
bˆ =
× 1
0 0 . . . 0 1
This gives a set of equations of the form 0 = bˆ 1 +
m
Σ × bˆ i
i
i =2
0 = bˆ 2 +
m
Σ × bˆ i
i
i =3
You're ..Reading a Preview
. Unlock full access with a free trial. 0 = bˆm −1 + × bˆm
Download 1 = bˆm With Free Trial
Clearly there is a solution for the entries of bˆ, regardless of the × ’s. Now it is easy to conclude controllabili the single-input state equation by calculation of the form of the controllability matrix. Then changing to original state variables gives the result since controllability is preserved. In the original variables, take K and b = bˆ. For an example to show that b alone does not suffice, take Exercise 13.11 with all ×’s zero.
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Supposing the rank of the controllability matrix is q, Theorem 13.1 gives an invertible P Special offer forSolution students:13.14 Only $4.99/month. that
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Linear System Theory, 2/E
Solutions
P −1 (P−a 1 APa )P =
A˜ 11
A˜ 21 A˜ 22 A˜ 23
0 CPa P =
0 A˜ 13 0 A˜ 33
˜ B 1
˜ , P −1 (P−a 1 B ) = B 2 0
˜ 0 C ˜ C 1 2
˜ is l × l, and in fact A˜ = Aˆ , C ˜ = C ˆ . It is easy to see that the state equation formed where A 11 33 22 2 2 ˜ ˜ ˜ C 1 , A 11 , B 1 is both controllable and observable. Also an easy calculation using block triangular structure sh that the impulse response of the state equation defined by (*) is A˜ 11 t ˜
˜ e C 1
B 1
It remains only to show that l = s. Using the effect of variable changes on the controllability and observab matrices and the special structure of (*) give C CA . . . CA n −1
˜ C 1
B AB . . . A
n −1
B
=
˜ A˜ C 1 11 . . . ˜ A˜ n −1 C 1 11
˜ A˜ B˜ . . . A˜ n −1 B˜ B 1 11 1 11 1
You're Reading a Preview
Thus
˜ C Unlock full access with a free trial. 1 ˜ ˜ C 1 A 11 ˜ . . . A˜ n −1 B˜ . B˜ 1 A˜ 11 B 1 11 1 . Download With Free Trial . ˜ A˜ n −1 C 1 11
rank
=s
But
Master your semester withC ˜ Scribd & The New York Timesrank C ˜ A.˜ = rank B˜
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1
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11
. .
˜ A˜ C
l −1
1
Useful −1 ˜ Not useful ˜ ˜ lCancel . . . A anytime. A˜ 11 B =l 1 11 B 1
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CHAPTER 14
Solution 14.2
For any t f > 0, t f
W=
∫ e − BB e − At
T
A T t
dt
0
is symmetric and positive definite by controllability, and t f You're Reading a− At Preview d ___ T − A T t T
∫
e BB e AW + WA = − dt 0 Unlock full access with a free trial. = − e
dt
− At f T − A T t f BB e + BB T
Download With Free Trial Letting K = − B T W −1 , we have ( A + BK )W + W ( A + BK )T = − ( e
− At f T − A T t f BB e + BB T )
Master your semester with Scribd Read For 30 Days Sign up to vote this ≠title Suppose λ is an eigenvalue of A +BK . Then λ is an eigenvalue of ( A+BK )Free , and weon let 0 be a correspon p eigenvector. ThenTimes & The New York Useful Not useful T
Special offer for students: Only $4.99/month. Also,
( A + BK ) p = λ p T
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Linear System Theory, 2/E
Solutions
Solution 14.5 (a) For any n × 1 vector x, x H ( A + A T ) x = x H A x + x H A T x ≥ − 2αm x H x
If λ is an eigenvalue of A, and x is a unity-norm eigenvector corresponding to λ, then _ H T A x = λ x , x A = λ x H and we conclude
_
λ + λ ≥ −2 αm Therefore any eigenvalue of A satisfies Re [λ] ≥ −αm , and this implies that for α > αm all eigenvalues of have positive real parts. Therefore all eigenvalues of −( A T +α I ) = (− A −α I )T have negative real parts. (b) Using Theorem 7.11, with α > αm , the unique solution of Q (− A − α I )T + (− A − α I )Q = − BB T
is ∞
Q=
∫ e −
T BB T e −( A +α I )t dt
( A +α I )t
0
Reading a Preview Clearly Q is positive semidefinite. If x QxYou're = 0, then T
( A +α I )t Unlock access free 0 , a t ≥ 0trial. x T e −full B =with
and the usual sequential differentiation and evaluation at t = 0 gives a contradiction to controllability. Thus Download With Free Trial positive definite. (c) Now consider the linear state equation . z (t ) = ( A+ α I − BB T Q −1 ) z (t )
Master your semester with Scribd Using (*) to write BB Q − gives Read Free Foron 30this Days Sign up to vote title . − z (t ) = −Q ( A+ α I ) Q z (t ) Useful Not useful & The New York Times T
1
T
1
Cancel anytime.
)T ]Q −1 has negative-real-part eigenvalues, which proves that (**) is exponentially stable. Q [ −( A + α I Special offer forBut students: Only $4.99/month. (d) Invoking Lemma 14.6 gives that
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Linear System Theory, 2/E
Solutions
A+BK +Bbk = A +B (K+bk )
has the specified characteristic polynomial. Thus for the original state equation, the feedback law u (t ) = (K+bk ) x (t )
yields a closed-loop state equation with specified characteristic polynomial.
Solution 14.8
Without loss of generality we can assume the change of variables in Theorem 13.1 has
performed so that A =
where A 11 is q
A 11 A 12 0 A 22
, B=
B 1 0
× q, and rank
λ I − A 11
B1 = q
for all complex values of λ . Then the eigenvalues of A comprise the eigenvalues of A 11 and the eigenvalu A 22 . Also, for any complex λ , rank
λ I − A
λ I − A 11 − A 12 B 1 0 λ I − A 22 0 = q + rank λ I − A 22 You're Reading a Preview
B = rank
Now suppose rank [λ I − A B ] = nUnlock for allfull nonnegative-real-part access with a free trial.eigenvalues of A . Then by (+) any eigenvalue must be an eigenvalue of A 11 , which implies that all eigenvalues of A 22 have negative real parts. we can compute an m × q matrix K 1 such that A 11 + B 1 K 1 has negative-real-part-eigenvalues. So se Download With Free Trial K = [ K 1 0 ] we have that A + BK =
A 11 +B 1 K 1 A 12 0 A 22
Master your semester with Scribd has negative-real-part eigenvalues. On the other hand, if there exists a K = [K K ] such that & The New York Times 1
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A + BK =
2
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A 11 +B 1 K 1 A 12 +B 1 K 2 0 A 22
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Linear System Theory, 2/E
Solutions
tr [ A+BLC ] = tr [ A ] + tr [ BLC ] = tr [ A ] + tr [CBL ] = tr [ A ] > 0
Thus at least one eigenvalue of A+BLC has positive real part, regardless of L.
Solution 14.12
Write the k th -row of G(s ) in terms of the k th -row C k of C as C k (sI − A )−1 B =
∞
Σ C A Bs − k
j
( j+1)
j =0
j The k th -relative degree κ k is such that, since L A [C k ](t ) B (t ) = C k A j B,
κ −2 C k B = . . . = C k A k B = 0
C k A
κ k −1 B ≠ 0
Thus in the k th -row of G(s ), the minimum You're difference betweenathe numerator and denominator polynomial deg Reading Preview among the entries Gk1 (s ), . . . , Gkm (s ) is κ k . Unlock full access with a free trial.
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CHAPTER 15
Solution 15.2
The closed-loop state equation can be written as . x (t ) = Ax (t ) + BMz (t ) +BNv (t ) = Ax (t ) + BMz (t ) +BNC [ Lz (t )+x (t )]
. z (t ) = Fz(t ) + GC [ Lz (t )+x (t )]
You're Reading a Preview Making the variable change w (t ) = x (t )+Lz (t ) gives the description . w (t ) = Ax (t )Unlock + BMzfull BNCw LFztrial. (t ) + (t )a+free (t ) + LGCw (t ) access with = Ax (t ) + [ BM+LF ] z (t ) + [ BN+LG ]Cw (t )
Download With Free Trial
= [ A − HC ]w (t )
. z (t ) = Fz(t ) +GCw (t ) Thus the closed-loop state equation in matrix form is Master your semester with Scribd . A − HC 0 w (t ) = . GC F z (t ) & The New York Times
Special offer forand students: Only the result is $4.99/month. clear.
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Linear System Theory, 2/E
Solutions
τ+δ
∫ τ B (σ
τ+δ
)2
d σ ≤ γ
2
∫ τ Φ(τ, σ) B (σ) B
(σ)ΦT (τ, σ ) d σ
T
≤ γ 2 n ε1 =∆ β1 Now for any τ, and t ∈ [τ+k δ, τ+(k +1)δ], k = 0, 1, . . . , τ+(k +1)δ
t
∫ τ B (σ
)2
∫ τ
d σ ≤
B (σ)2 d σ
k τ+( j +1)δ
≤Σ
j =0
∫
B (σ)2 d σ
τ+j δ
≤ (k +1) β1 ≤ [1 + (t −τ) / δ ] β1 This bound is independent of k , so letting β 2 = β1 / δ we have t
∫ τ B (σ)
2
d σ ≤ β 1 + β2 (t −τ)
for all t , τ with t ≥ τ. (Of course this provides a simplification of the hypotheses of Theorem 15.5 bounded-A (t ) case.)
You're Reading a Preview
Unlock full access with a free trial. Write the given state equation in the partitioned form . za (t ) A 11 A 12 za (t ) B 1 = + . Download With Free Trial A A z t B 2 ( ) 21 22 b zb (t )
Solution 15.6
y (t ) =
I p
c
2
u (t )
za (t ) zb (t )
0
Master your semester with Scribd and the reduced-dimension observer and feedback in the form Read Free Foron 30this Days Sign up to vote title . & The New York Not − Huseful z (t Times + ( Useful A − H A A ] z (t ) ) = [ A − H A ] z (t ) + [ B − HB ]u (t ) + [ A ) H c
22
12
Special offer for students: Only (t ) = zc (t ) + Hza (t ) zˆb$4.99/month.
1
21
22Cancel anytime. 12
11
a
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Linear System Theory, 2/E
Solutions
Y(s ) = I p
1 0 (sI − A − BK )− BN R(s )
which is the same as if a static state feedback gain K is used.
Solution 15.9
Similar in style to Solution 14.8.
Solution 15.10 Since u = Hz + Jv = Hz + JC 2 x + JD 21 r +JD 22 u
∆ we assume that I − JD 22 is invertible, and let L = ( I − JD 22 )−1 to write u = LHz + LJC 2 x + L JD 21 r
Then, substituting for u, . x = ( A+BL JC 2 ) x + BLHz + BLJD 21 r . z = (GC 2 +GD 22 LJC 2 ) x + (F+GD 22 L H ) z + (GD 22 +GD 22 L JD 21 )r y = (C 1 +D 1 L JC 2 ) x + D 1 L Hz + D 1 L JD 21 r
This gives the closed-loop coefficients
You're Reading a Preview
ˆ= A
A+BL JC 2 Unlock full BLH BLJD 21 access with a free trial. , Bˆ = GC 2 +GD 22 L JC 2 F+GD 22 L H GD 22 +GD 22 L JD 21
ˆ= C
C 1 +D 1 L JC 2
Download With ˆ = DFree L JDTrial D LH , D 1
1
21
These expressions can be rewritten using L = ( I − JD 22 )−1 = I + J ( I − D 22 J )−1 D 22
Master your semester with Scribd which follows from Exercise 28.2 or is easily verified using the identity Exercise Read Free For28.1. 30this Days Signinup to vote on title & The New York Times Useful Not useful Special offer for students: Only $4.99/month.
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CHAPTER 16
Solution 16.4
By Theorem 16.16 there exist polynomial matrices X (s ), Y (s ), A (s ), and B (s ) such that N (s ) X (s ) + D (s )Y (s ) = I p N a (s ) A (s ) + Da (s ) B (s ) = I p
Since D −1 (s ) N (s ) = D −a1 (s ) N a (s ), N a (s ) = Da (s ) D −1 (s ) N (s ). Substituting this into (**) gives (s ) A (s ) +aDPreview Da (s )You're D −1 (s ) N a (s ) B (s ) = I p Reading that is,
Unlock full access with a free trial. N (s ) A (s ) + D (s ) B (s ) = D (s ) D −a1 (s )
Similarly, N (s ) = D (s ) D −a 1 (s ) N a (s ), and substituting into (*) gives Download With Free Trial N a (s ) X (s ) + Da (s )Y (s ) = Da (s ) D −1 (s )
Therefore D (s ) D −a1 (s ) and [ D (s ) D −a1 (s ) ]−1 both are polynomial matrices, and thus both are unimodular.
Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title Solution 16.5 From the given equality, & The New York Times Useful Not useful N (s ) D (s ) − D (s ) N (s ) = 0 Special offer for students: Only $4.99/month.
L
L
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and since N (s ) and D (s ) are right coprime there exist polynomial matrices X (s ) and Y (s ) such that
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Linear System Theory, 2/E
Solutions
I 0
−[ X (s ) B (s )+Y (s ) A (s )] I
gives D (s ) − D (s )[ X (s ) B (s )+Y (s ) A (s )]+B (s ) − N (s ) N (s )[ X (s ) B (s )+Y (s ) A (s )]+A (s )
X (s ) Y (s ) N L (s ) D L (s )
= I
That is X (s ) Y (s ) N L (s ) D L (s )
−1
=
D (s ) − D (s )( X (s ) B (s )+Y (s ) A (s ))+B (s ) − N (s ) N (s )( X (s ) B (s )+Y (s ) A (s ))+A (s )
which is another polynomial matrix. Thus X (s ) Y (s ) N L (s ) D L (s )
is unimodular.
Solution 16.7
The relationship −1 R Preview You're Reading (P ρ s+P 1 s+R 0 ρ−1 ) = a
holds if R 1 and R 0 are such that
Unlock full access with a free trial.
I = (P ρ s+P ρ−1 ) ( R 1 s+R 0 ) = P ρ R 1 s 2 + (P ρ R 0 +P ρ−1 R 1 )s + P ρ−1 R 0
Download With Free Taking R 0 = P −ρ−11 and R 1 = −P −ρ−11 P ρ P −ρ−11 , it remains to verify that PTrial ρ R 1 = 0. We have I = (P ρ s ρ + . . . + P 0 ) (Q η s η + . . . + Q 0 ) = P ρ Q η s η+ρ + (P ρ Q η−1 +P ρ−1 Q η )s η+ρ−1 + . . .
Master your semester Q − invertible.with with P − and ThereforeScribd Read Free Foron 30this Days Sign up to vote title 0 P Q = 0 , P Q − +P − Q = & The New York Times Useful Not useful ρ 1
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Linear System Theory, 2/E
Solutions
˜ (s )u (s ) + . . . + D˜ (s )u (s ) + . . . + D˜ (s )u (s ) , k = 1, . . . , m Dk (s ) = D 1 1,k J J , k m m,k Using a similar column notation for D hc and D l (s ) gives D hc k s
ck [ D ]
˜] hc c 1 [ D
˜ s + D lk (s ) = [ D 1
˜
˜ l (s )] u (s ) + . . . + [ D˜ hc s c J [ D ] +D˜ l (s )] u (s ) +D 1 1,k J J J , k ˜
˜ hc s cm [ D ] +D˜ l (s )] u (s ) , k = 1, . . . , m + . . . + [ D m m m,k We claim that ck [ D ] =
max
j = 1, . . . , m
{ c j [ D˜ ]+degree u j,k (s ) } hc
hc
˜ , . . . , D˜ as follows. Let This is shown by a an argument using linear independence of D 1 m c˜ =
and let µ j,k be the coefficient of s term on the right side is
˜] c˜ −c j [ D
max
j = 1, . . . , m
{ c j [ D˜ ]+degree u j,k (s ) }
in u j,k (s ). Then not all the µ j,k are zero, and the vector coefficient of th m
Σµ
˜ hc j,k D j
j =1
By linear independence this sum is nonzero, which implies ck [ D ] = c˜. Now, using the definition of J ,
You're Reading a Preview
˜ ] ≤ . . . ≤ c [ D˜ ] , k = 1, . . . , J −1 ck [ D ] < c J [ D m Unlock full access with a free trial. and this implies u J , k (s ) = . . . = um,k (s ) = 0. Thus U (s ) has the form U a (s ) U b (s ) Download With Free Trial U (s ) =
0(m− J+1) × J U c (s )
where U a (s ) is ( J −1) × J , from which rank U (s ) ≤ m −1 for all values of s . This contradicts unimodularity, ˜ ]. The proof is complete since the roles of D (s ) and D˜ (s ) can be reversed. c J [ D ] = c J [ D
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CHAPTER 17
Solution 17.1
If . x (t ) = A x (t ) + Bu (t ) y (t ) = C x (t )
is a realization of G T (s ), then .
T
T
z (t ) =Reading A x (t ) +aC Preview v (t ) You're w (t ) = B T z (t ) Unlock full access with a free trial.
is a realization for G (s ) since T T ) ] = [ C (sI With − A )−1Free G (s ) = [ G T (sDownload B ] = B T (sI − A T )−1 C T Trial
Furthermore, easy calculation of the controllability and observability matrices of the two realizations shows one is minimal if and only if the other is. Now, if N (s ) and D (s ) give a coprime left polynomial frac description for G (s ), then there exist polynomial matrices X (s ) and Y (s ) such that
Master your semester with Scribd N (s ) X (s ) + D (s )Y (s ) =Read I Free Foron 30this Days Sign up to vote title Therefore & The New York Times Useful Not useful Special offer for students: Only $4.99/month.
X T (s ) N T (s ) + Y T (s ) D T (s ) = I
Cancel anytime.
which implies that N T ( ) and D T ( ) are right coprime. Also, since D ( ) is row reduced, D T ( ) is column redu
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Linear System Theory, 2/E
Solutions
Therefore T T −1 SB T o [ sI − A o − Q VB o ]
−1
= D −1 (s )ΨT (s )
Using the definition of N (s ), T T −1 −1 −1 D −1 (s ) N (s ) = SBT o [ sI − ( A o + Q VB o ) ] Q B
−1 = CQ [ sI − Q −1 AQ ] Q −1 B
= C (sI − A )−1 B
Note that D (s ) is row reduced since Dlr = S −1 , which is invertible. Finally, if the state equation is controllab well as observable, hence minimal, then it is clear from the definition of D (s ) that the degree of the polyno fraction description equals the dimension of the minimal realization. Therefore D −1 (s ) N (s ) is a coprime polynomial fraction description.
Solution 17.5
Suppose there is a nonzero h with the property that for each uo there is an xo such that t
∫
hCe xo + hCe A (t −σ) Buo e At
so σ
d σ = 0 , t ≥ 0
0
Suppose G (s ) = N (s ) D −1 (s ) is a coprime right polynomial fraction description. Then taking Laplace transfo You're Reading a Preview gives −1 (sa)free access with (s ) D −so )−1 = 0 hC (sI −Unlock A )−1 xofull + hN uo (strial.
that is,
Download With Free Trial
(s −so )hC (sI − A )−1 xo + hN (s ) D −1 (s )uo = 0 If so is not a pole of G (s ), then D (so ) is invertible. Thus evaluating at s = so gives hN (so ) D −1 (so )uo = 0
Master your semester with Scribd and we have that if s is not a pole of G (s ), then for every u˜ & The New York Times hN (s )u˜ = 0 o
o
o
o
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Useful
Special offer forThus students: $4.99/month. hN (sOnly o ) = 0, that is rank N (so ) < p < m, which implies that so is a transmission zero.
Conversely, suppose s is a transmission zero that is not a pole of G (s ). Then for a right-
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Linear System Theory, 2/E
Solution 17.9
Solutions
Using a coprime right polynomial fraction description G (s ) = N (s ) D −1 (s ) =
(s ) adj D (s ) N ____________ det D (s )
suppose for some i, j and complex so we have
[ N (s ) adj D (s ) ]
o o ij ∞ = Gij (so ) = ___________________
det D (so )
Since the numerator is the magnitude of a polynomial, it is finite for every so , and this implies det D (so ) = is, so is a pole of G (s ). Now suppose s o is such that det D (so ) = 0. By coprimeness of the right polynomial fraction descrip N (s ) D −1 (s ), there exist polynomial matrices X (s ) and Y (s ) such that X (s ) N (s ) + Y (s ) D (s ) = I m
for all s. Therefore
[ X (s )G (s ) + Y (s ) ] D (s ) = I m for all s, and thus det [ X (s )G (s ) + Y (s ) ] det D (s ) = 1 for all s. This implies that at s = so we must have
You're Reading a Preview det [ X (so )G (so ) + Y (so ) ] = ∞ Unlock full access with a free trial.
Since the entries of the polynomial matrices X (so ) and Y (so ) are finite, some entry of G (so ) must have infi magnitude.
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CHAPTER 18
Solution 18.2
(a) If x ∈ A ( A −1 V ), then clearly x ∈ Im [ A ], and there exists y ∈ A −1 V such that x = Ay, which implies Therefore A ( A −1 V ) ⊂ V ∩ Im [ A ]. Conversely, suppose x ∈ V ∩ Im [ A ]. Then x ∈ Im [ A ] implies there exi such that x = Ay, and x ∈ V implies y ∈ A −1 V . Thus x ∈ A ( A −1 V ), that is, V ∩ Im [ A ] ⊂ A ( A −1 V ). (b) If x
∈ V +Ker [ A ], then we can write x = You're xa + xb ,Reading xa ∈ V , a xPreview b ∈ Ker [ A ]
−1 and Ax = Axa ∈ A V . Thus x ∈ A −1 ( A V ), which ] ⊂ A V +Ker Unlockgives full access with[ A a free trial.( A V ). Conversely, if x there exists y ∈ V such that Ax = Ay, that is, A ( x − y ) = 0. Thus writing
∈ A −1 ( A V
) ∈ V Free [ A ] x = y + ( x − yWith +Ker Trial Download gives A −1 ( A V ) ⊂ V +Ker [ A ]. (c) If A V ⊂ W , then using (b) gives A −1 ( A V ) = V + Ker [ A ] ⊂ A −1 W . Thus V ⊂ A −1 W . Conversely, V ⊂ A −1 W implies, using (a),
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Linear System Theory, 2/E
Solutions
Solution 18.9 Clearly C < A | B > = Y if and only if rank C B AB . . . A n −1 B
=p
and thus the proof involves showing that the rank condition is equivalent to positive definiteness of t f
∫ Ce
A (t f −t )
A T (t f −t )
BB T e
C T dt
0
This is carried out in Solution 9.11.
Solution 18.10
We show equivalence of the negations. First suppose 0 subspace. Then picking a friend F of V we have
≠ V ⊂ Ker [C ] is a controlled inva
( A + BF )V ⊂ V ⊂ Ker [C ] Selecting 0 ≠ xo
∈ V , this gives e ( A +BF )t xo ∈ V , t ≥ 0
and thus
You're Reading Ce ( A +BF )t xo = 0a, Preview t ≥ 0
Thus the closed-loop state equation is notUnlock observable, since the zero-input response to x o ≠ 0 is identical to full access with a free trial. zero-input response to the zero initial state. Conversely, suppose the closed-loop state equation is not observable for some F . Then Download With Free Trial n −1 k N = ∩ Ker [C ( A + BF ) ] ≠ 0 k =0
Thus 0≠ xo
∈ N implies, using the Cayley-Hamilton theorem,
Master your semester with = C ( A + BF ) x = C ( A + BF ) x = . . . 0 = Cx Scribd Read Free Foron 30this Days Sign up to vote title inva That is, ( A + BF ) x ∈ , which gives ( A + BF ) ⊂ . Clearly ⊂ Ker [C ], so is a nonzero controlled & The New York Times Useful Not useful subspace contained in Ker [C ]. o
o
N
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N
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N
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CHAPTER 19
Solution 19.1
First we show ( W + S ) ⊥ = W ⊥
∩ S ⊥
An n × 1 vector x satisfies x ∈ ( W + S ) ⊥ if and only if x T (w + s ) = 0 for all w ∈ W and s ∈ S . This is equiv to x T w + x T s = 0 for all w ∈ W and s ∈ S , and by taking first s = 0 and then w = 0 this is equivalent to x for all w ∈ W and x T s = 0 for all s ∈ S . These conditions hold if and only if x ∈ W ⊥ and x ∈ S ⊥ , tha x ∈ W ⊥ ∩ S ⊥ . You're Reading a Preview Next we show Unlock fullT access with a free trial. ( A S ) ⊥ = A −1 S ⊥
An n × 1 vector x satisfies x ∈ ( A T S ) ⊥ if and only if x T y = 0 for all y ∈ A T S , which holds if and on Trial ⊥ as ( Ax )T z =With 0 forFree all z ∈ x T A T z = 0 for all z ∈ S , which is the sameDownload S , which is equivalent to Ax ∈ S , whi equivalent to x ∈ A −1 S ⊥ . Finally we prove that ( S ⊥ ) ⊥ = S . It is easy to show that S ⊂ ( S ⊥ ) ⊥ since x ∈ S implies y T x = 0 fo ⊥ y ∈ S , that is, x T y = 0 for all y ∈ S ⊥ , which implies x ∈ ( S ⊥ ) ⊥ . To show ( S ⊥ ) ⊥ ⊂ S , suppose 0 ≠ x ∈ ( S ⊥ ) ⊥ . Then for all y ∈ S ⊥ we have x T y = 0. That is, if y T z T Free Foron 30this Days Sign to vote title y =up all z ∈ S , then x T y = 0. Equivalently, if z T y = 0 for all z ∈ S , then xRead 0. Thus
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Ker
z T = Ker
T
z , x T
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Linear System Theory, 2/E
Solutions
0 V k +1 V
= K = K ∩ A −1 (V k + B ) = V k ∩ A −1 (V k + B )
For k = 0 the claim becomes ( K ⊥ ) ⊥ = K , which is established in Exercise 19.1. So suppose for some noneg integer K we have (W K ) ⊥ = V K . Then, using Exercise 19.1, ⊥ (W K +1 ) ⊥ = W K + A T [ W K ∩ B ⊥ ] = (W K ) ⊥
⊥ A T (W K ∩ B ⊥ )
∩
A T [ (V K ) ⊥
= V K ∩
∩ B ⊥ ]
⊥
But further use of Exercise 19. 1 gives (V K ) ⊥
A T
⊥
∩ B ⊥
= A −1
(V K ) ⊥
∩ B ⊥
⊥
= A −1 (V K + B )
Thus (W K +1 ) ⊥ = V K ∩ A −1 (V K + B ) = V K +1 This completes the induction proof, and gives V * = V n = (W n ) ⊥ .
You're Reading a Preview
Solution 19.4
full access a freeof trial. We establish the Hint byUnlock induction, for F with a friend V *. For k = 1, k
∩ Free * =Trial * ∩ ( A .0 + ( ∩ * ) = Σ ( A + BF ) −Download With j 1
B
B
V
V
V
B )
j =1
= R 1
Assume now that for some positive integer K we have
Master your semester with Scribd ( ) − ( ∩ * ) = A + BF Σ & The New York Times Then K
j 1
B
V
j =1
Special offer for students: Only $4.99/month. K +1
Σ
( A + BF ) j −1 (B
K −1 ( Aup ) 30this B on ∩ ∩For Read Days R Free Sign to vote title Useful Not useful Cancel anytime.
K R = V *
K
V * )
B
V *
+ ( A + BF )
Σ ( A + BF ) − ( j 1
B
V * )
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Linear System Theory, 2/E
Solution 19.7
Solutions
The closed-loop state equation . x (t ) = ( A + BF ) x (t ) + ( E + BK )w (t ) +BGv (t ) y (t ) = Cx (t )
is disturbance decoupled if and only if C (sI − A − BF )−1 ( E + BK ) = 0
That is, if and only if
< A +BF Im [ E +BK ]> ⊂ Ker [C ] Thus we want to show that there exist F and K such that (*) holds if and only if Im [ E ] ⊂ V * + B , where V maximal controlled invariant subspace contained in Ker [C ] for the plant. First suppose F and K are such that (*) holds. Since < A +BF Im [ E +BK ]> is invariant under ( A + is a controlled invariant subspace contained in Ker [C ] for the plant. Then Im [ E +BK ] ⊂ < A +BF Im [ E +BK ]> ⊂ V *
∈ X there is a v ∈ V * such that ( E + BK ) x = v. Therefore Ex = v + B (−K x ) You're Reading a Preview which implies Im [ E ] ⊂ V * + B . Conversely, suppose Im [ E ] ⊂ V * + B , where V * is the maximal controlled invariant subspace conta Unlock access K with a free trial. Im [ E +BK ] ⊂ V *. Then we can pi in Ker [C ] for the plant. We first show how to full compute such that That is, for any x
friend F of V * and the proof will be finished since we will have
Download With Free Trial ]> ⊂ < A +BF V * ⊂ Ker [C ] Im [ E +BK If w 1 , . . . , wq is a basis for W , then there exist v 1 , . . . , vq
∈ V * and u 1 , . . . , uq ∈ U such that
Ew j = v j + Bu j , j = 1, . . . , q
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( E + BK )
= Ew
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Linear System Theory, 2/E
Since Im [ BG 1 ] ⊂ B ∩ R 1 *
Solutions
ˆ = C P = C 1 1
ˆ C 11 0 0
ˆ = C P = C 2 2
ˆ 0 C 11 0
⊂ R 1 * and BG 1 = PBˆ 1 we have Bˆ 1 =
Similarly, Im [ BG 2 ] ⊂ B ∩ R 2 *
ˆ B 11 0 ˆ B 13
⊂ R 2 * gives 0
ˆ Bˆ 2 = B 22 ˆ B 23
Finally, ( A + BF )R i *
⊂ R i *, i = 1, 2, and ( A + BF )P = PAˆ give Aˆ =
ˆ A 11
0
0
ˆ 0 A 0 22 ˆ ˆ ˆ A 31 A 32 A
33
You're Reading a Preview
That is, with z (t ) = P −1 x (t ), the closed-loop state equation takes the partitioned form . ˆ z (t Unlock ˆ full access with a free trial. za (t ) = A 11 a ) + B 11 r 1 (t ) . ˆ z (t ) + Bˆ r (t ) zb (t ) = A 22 b 22 2 Download With Free Trial . ˆ ˆ ˆ z (t ) + Bˆ r (t ) + Bˆ r (t ) zc (t ) = A 31 za (t ) + A 32 zb (t ) + A 33 c 13 1 23 2
ˆ z (t ) y 1 (t ) = C 11 a ˆ z (t ) y (t ) = C Master your semester with Scribd & The New York Times 2
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CHAPTER 20
Solution 20.1
A sketch shows that v (t ) is a sequence of unit-height rectangular pulses, occurring eve seconds, with the width of the k th pulse given by k/ 5, k = 0, . . . , 5. This is a piecewise-continuous (actu piecewise-constant) input, and the continuous-time solution formula gives
z (t ) = e
t
∫
F (t −t o )
z (t o ) + e F (t −σ) Gv (σ) d σ t o
Reading a Preview Evaluate this at t = (k +1)T and t o = kT to You're get (k +1)T
Unlock trial. FT full access with aF free (kT+T −σ)
z [(k +1)T ] = e z (kT ) +
∫
Gv (σ) d σ
e
kT
Let τ = kT+T −σ in the integral, to obtain
Download With Free Trial T
∫
z [(k +1)T ] = e z (kT ) + e F τ Gv (kT +T −τ) d τ FT
0
Then the special form of v (t ) gives Master your semester with Scribd & The New York Times z [(k +1)T ] = e z (kT ) + ∫ − T
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T u (k )T
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The integral term is not linear in the input sequence u (k ), so we approximate the integral when u (k ) is s
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Linear System Theory, 2/E
Solutions
u˜
x˜ =
u˜
2
,
y˜ = u˜
2
Easy calculation gives the linearized state equation
x δ (k +1) = y δ (k ) =
−1
0
0
−1
2u˜
x δ (k ) +
−1
2 u δ (k ) 4u˜
x δ (k ) + 2u˜ u δ (k )
Since A k = (−1)k I and CB = 0, the zero-state solution formula easily gives
y δ (k ) = 2u˜ u δ (k ) Thus the zero-state behavior of the linearized state equation is that of a pure gain.
Solution 20.10 Computing Φ ( j +q, j ) for the first few values of q ≥ 0 easily leads to the general formul
Φ(k , j ):
0 a 1 (k −1)a 2 (k −2)a 1 (k −3)a 2 (k −4) . . . a 1 ( j ) . . . a 2 (k −1)a 1 (k −2)a 2 (k −3)a 1 (k −4) a 2 ( j ) 0
,
k − j odd, ≥ 1
,
k − j even, ≥ 1
You're Reading a Preview a 1 (k −1)a 2 (k −2)a 1 (k −3)a 2 (k −4) . . . a 2 ( j ) 0 Unlock full access with a free trial. 0 a 2 (k −1)a 1 (k −2)a 2 (k −3)a 1 (k −4) . . . a 1 ( j )
Download With Free Trial Solution 20.11
By definition, for k ≥ j +1,
ΦF (k ,
j ) = F (k −1)F (k −2) . . . F ( j +1)F ( j )
Master your semester with (1−k ) A (2−k ) . . . A (−1− j ) A (− j ) , k ≥ j +1 = A Scribd Read Free Foron 30this Days Sign up to vote title Therefore, for k ≥Times j +1, & The New York Useful Not useful T
Φ T F (k , j ) = A (− j ) A (− j −1) 2, that is, k ≥ j 1,
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T
T
T
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. . . A (−k+ 2) A (−k+1)
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Linear System Theory, 2/E
Solutions
Φ(k , k o ) ≤
k −1 1 _____ Φ(k , j ) Φ( j, k ) , k ≥ k 1 +1 ≥ k o +1 k −k 1 j =k 1
Σ
Solution 20.16 Given A (k ) and F we want P (k ) to satisfy F = P −1 (k +1) A (k )P (k ) for all k . Assuming F is invertible and A (k ) is invertible for every k , it is easy to verify that
P (k ) = Φ A (k , 0)F −k
is the correct choice. Obviously if F = I , then the variable change is P (k ) = Φ A (k , 0). Using this in Exa 20.19, where 1 a (k ) 0 1
A (k ) = gives
k −1
P (k ) = Φ A (k , 0) =
1
Σ a (i )
i =0
0
, k ≥ 1
1
You're Reading a Preview and k −1 trial. Unlock full access with a free
P −1 (k +1) = Φ A (0, k +1) =
1
− Σ a (i ) i =0
Download With0Free Trial 1
, k ≥ 0
Then an easy multiplication verifies the property.
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CHAPTER 21
Solution 21.3 Using z-transforms, ( zI − A )−1 =
z −1 12 z+ 7
−1
=
1 _________ 2 z +7 z+ 12
z +7 1 −12 z
and
Y ( z ) = zc ( zI − A )−1 xo + c ( zI − A )−1 b U ( z )
You're Reading a Preview
z z −1 ____ z _________ 1 / 20 _________ − z −19 z −1 1 / 20 + 2 = 2 Unlock full access with a free trial. z +7 z+ 12 z −1 z +7 z+ 12 = 0
Download With Free Trial
Therefore the complete solution is y (k ) = 0, k ≥ 0.
Solution 21.4 First compute the corresponding discrete-time state equation Master your semester with Scribd (kT ) up x ([(k +1)T ] = Fx (kT ) + guRead Free Foron 30this Days Sign to vote title & The New York Times y (kT ) = hx (kT ) Useful Not useful Special offer forUsing students: 0, it$4.99/month. is easy to compute A 2 =Only
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Linear System Theory, 2/E
Solutions
x (k ) = (1+r/l )k xo +
k −1
Σ (1+r/l ) − − b k j 1
j =0
= (1+r/l )k xo + b (1+r/l )k −1
1−1 / (1+r/l )k ____________ 1−1 / (1+r/l )
= (1+r/l )k ( xo +bl/r ) − bl/r (b) In one year a deposit x o yields
x (l ) = (1+r/l )l xo so (1+r/l )l xo − xo _____________ effective interest rate = xo
× 100% = [(1+r/l )l − 1] × 100%
For r = 0.05, l = 2, the effective interest rate is 5.06%. For r = 0.05, l = 12, the effective interest rate is 5.12% (c) Set 50,000 (−50,000) _______ _________ + You're Reading a0.05 Preview 0.05
0 = x (19) = (1.05)19 xo +
and solve to obtain xo = $604,266. Of course this means you have actually won only $654,26 Unlock full access with a free trial. congratulations remain appropriate.
Download With Free Trial
Solution 21.9 With T = T d / l and v (t ) = v (kT ), kT ≤ t ≤ (k +1)T , evaluate the solution formula t
z (t ) = e
F (t −τ)
∫ τ
z (τ) + e F (t −σ) Gv (σ−T d ) d σ , t ≥ T
Master your semester with Scribd at t = (k +1)T , τ = kT to obtain & The New York Times Special offer for students: Only $4.99/month.
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∫
z [(k +1)T ] =e FT z (kT ) + e F τ 0
Useful Not useful d τ G v [(k −Cancel l )T ] anytime.
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Linear System Theory, 2/E
x (k +1) =
ˆ (k ) = y
Solutions
0 .. . 1 .. . . . . . . . 0 0 0 .. . 0 0 0 .. .
A B 0 0 . . . . . .
0 0 . . x (k ) + .
0 0 . . .
1 0
0 1
u (k ) ,
x (0) =
z (0) v (−lT ) . . . v (−2T ) v (−T )
C 0 . . . 0 x (k )
The dimension of the initial state is n +l. The transfer function of this state equation is the same as the tran function of
z (k +1) = Az (k ) + Bu (k −l ) y (k ) = Cz (k ) Taking the z -transform, using the right shift property, gives
Y ( z ) = C ( zI − A )−1 Bz −l U ( z )
Solution 21.12
Easy calculation shows that for 1 0 0 1 You're Reading , M ba =Preview M a = 0 0 0 0 Unlock full access with a free trial.
M a has a square root, with √ M a = M a , but M b does not.
Download With Free Trial
Solution 21.13
By Lemma 21.6, given any k o there is a K -periodic solution of the forced state equation if only if there is an x o satisfying k o +K −1 Φ(k o +K , j +1) f ( j ) [ I − Φ(k o +K , k o )] xo =
Master your semester with Scribd Σ Read Free Foron 30this Days Sign up to vote title Similarly there is Times a K -periodic solution of the unforced state equation ifUseful if there is a z satisfying & The New York Not useful and only j =k o
Special offer for students: Only $4.99/month. Since there is no
[ I − Φ(k o +K , k o )] zo = 0
≠ 0 satisfying (**), it follows that [ I −Φ(k +K
Cancel anytime.
o
k )] is invertible. This implies that for
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Linear System Theory, 2/E
Solutions
we have by linear algebra that there exits a nonzero, n × 1 vector p such that [ I − Φ(k o +K , k o )]T p = 0 and
p T
k o +K −1
Σ
Φ(k o +K ,
∆ j +1) f ( j ) = q ≠ 0
j =k o
Now pick any xo . Then it is easy to show that the corresponding solution satisfies p T x (k o +jK ) = p j = 1, 2, . . . . This shows that the solution is unbounded.
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CHAPTER 22
Solution 22.1
Similar to Solution 6.1.
Solution 22.4
If the state equation is uniformly exponentially stable, then there exist γ ≥ 1 and 0 ≤ λ <
that
Φ(k , j ) ≤ γ λk − j , k ≥ j
You're Reading a Preview
Equivalently, for every k ,
Unlock full access with j a free trial.
Φ(k +j , k ) ≤ γ λ , j ≥ 0
which implies
Download With Free Trial φ j = sup Φ(k +j , k ) ≤ γ λ j k
Then
φ = lim (γ Master your semester with lim Scribd → ∞ → ∞ ≤ λ & The New York Times j
Special offer for students: Only $4.99/month. Now suppose
1 /j j
1 /j
j
< 1
λ) = λ lim γ 1 /j j → ∞ Read Free Foron 30this Days Sign up to vote title
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Linear System Theory, 2/E
Solutions
Similarly, for j > J ,
Φ(k +j , k ) ≤ sup Φ(k +j , j ) k
= φ j < (1−ε) j = λ j
≤ γ λ j This implies uniform exponential stability.
Solution 22.6
For λ = 0 the problem is trivial, so suppose λ ≠ 0 and write k
k λk = k λk = k ( e lnλ ) , k ≥ 0 Let η = −lnλ, so that η > 0 since λ < 1. Then max k λk ≤ max t e −η t k ≥ 0 t ≥ 0 and a simple maximization argument (as in Exercise 6.10) gives 1 ___ You'remax Reading te −η t ≤ a Preview ηe t ≥ 0 Therefore
Unlock full access with a free trial.
________ ∆ = β , k ≥ 0 Download −e lnWith λ Free Trial
k λk ≤
1
To get a decaying exponential bound, write
k λk = k ( √λ )k ( √λ )k = 2 β ( √λ )k , k ≥ 0
Then Master your semester with Scribd ∞ Σ k λ & The New York Times
k
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2β ≤ _______ 1− √λ
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Linear System Theory, 2/E
Solutions
which is equivalent to
Φ A T (−k ) (k , j ) ≤ γ λk − j , k ≥ j which is equivalent to uniform exponential stability of A T (−k ). However for the case of A T (k ), consider the example where A (k ) is 3-periodic with
A (0) =
0 2 1 / 2 0
, A (1) =
0 1 / 2 1 / 2 0
, A (2) =
2 0 0 1 / 2
Then
Φ A (k ) (3, 0) =
1 / 2 0 0 1 / 2
and it is easy to conclude uniform exponential stability. However
Φ A
(k ) (3,
T
0) =
2 0 0 1 / 8
and it is easy to see that there will be unbounded solutions.
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CHAPTER 23
Solution 23.1 With Q = q I , where q > 0 we compute A T (k )QA (k )−Q to get the sufficient condition uniform exponential stability:
a 21 (k ), a 22 (k ) ≤ 1−
ν __ q
,
ν > 0
Thus the state equation is uniformly exponentially stable if there exists a constant α < 1 such that for all k
≤ α k ), a 2 (k a)Preview a 1 (Reading You're With
Unlock full access with a free trial.
Q=
q1 0 0 q2
Download With Free Trial where q 1 , q 2 > 0, the sufficient condition for uniform exponential stability becomes existence of a constant such that for all k ,
q 2 − ν
q 1 − ν
_____ _____ , a (k ) ≤ (k ) ≤ Master your semester with aScribd q Free For 30 Days q Read Sign up to vote on this title These conclusions show uniform exponential stability under weaker one bounded coeffic & The New York Times Useful where Not useful conditions, can be larger than unity if the other bounded coefficient is suitably small. For example, 2 1
1
2 2
2
Cancel anytime.
Special offer for students: Only $4.99/month. sup | a 2 (k ) = α < ∞. Then we can take q 1 = α2 +0.01, q 2 = 1, and ν = 0.01 to conclude uniform expone k
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Linear System Theory, 2/E
Solutions
holds if a 21 (k ), a 22 (k ) ≤ α < 1 for all k , but it also holds under weaker conditions. For example suppose th bound is violated only for k = 0, and
a 21 (0) > 1 , a 21 (0)a 22 (1) < α Then we can conclude uniform exponential stability. (More sophisticated analyses should be possible . . .
Solution 23.6
If the state equation is exponentially stable, then by Theorem 23.7 there is for any symmetr a unique symmetric Q such that
A T QA − Q = − M Write
M =
m1 m2 m2 m3
Q=
,
q1 q2 q2 q3
and write the discrete-time Lyapunov equation as the vector equation
−1 0 1 The condition
0
− 1− a 0 −2
a 20 a0 0
q1 q2 q3
−m 1 −m 2 −m 3
=
You're Reading a Preview
Unlock full −1access 0 witha 20a free trial.
0 −1−a 0 a 0 ≠ 0 −2 Free 1 With 0 Download Trial det
reduces to the condition a 0 ≠ 0, 1, −2. Assuming this condition we compute Q for M = I , and use the fact Q > 0 since M > 0. The expression
−1 0 q Master your semester with Scribd q = 0 −1−a −2 q 1 & The New York Times 1 2 3
0
a 20 a0 0
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2
−1
−1
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Linear System Theory, 2/E
Solutions
p H A T QAp − p H Qp = − p H Mp That is,
( λ2 −1 ) p H Qp = − p H Mp If p H Mp > 0, then λ2 −1 < 0, which gives λ < 1. But suppose p H Mp = 0. Then for k ≥ 0, _ k H 2 k 0 = λ p Mp = λ p H Mp λk = p H ( A T )k MA k p
= ( Re [ p ])T ( A T )k MA k ( Re [ p ]) + ( Im [ p ])T ( A T )k MA k ( Im [ p ]) Since M ≥ 0, this implies 0 = ( Re [ p ])T ( A T )k MA k ( Re [ p ]) = ( Im [ p ])T ( A T )k MA k ( Im [ p ]) By hypothesis this implies lim A k ( Re [ p ]) = lim A k ( Im [ p ]) = 0 k → ∞
k → ∞
Therefore lim A k p = lim λk p = 0 k → ∞
k → ∞
which implies λ < 1.
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CHAPTER 24
Solution 24.1 Since T
A (k ) A (k ) =
a 22 0 0 a 21
it is clear that / 2 (k ) = max [ a 1 (k ), a 2 (k ) ] λ 1max You're Reading a Preview
Thus Corollary 24.3 states that the state equation is uniformly stable if there exists a constant γ such that Unlock full access with a free trial. k
Π max [ a 1 (i ), a 2 (i ) ] ≤ γ
i =j
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for all k , j with k ≥ j. (Note that this condition holds if
max [ a 1 (k ), a 2 (k ) ] ≤ 1 for all but a finite number of values of k .) Of course the condition (#) is not necessary. Consider
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L in e a r S y s t e m T h e o r y , 2 / E
k
r (k +1)
Π
j =k o
S o lu t io n s
1 __________ 1+η( j ) ν( j )
k −1
≤ r (k ) Π
j =k o
1 __________ 1+η( j ) ν( j )
+ ν(k )ψ (k )
k
Π
j =k o
1 __________ 1+η( j ) ν( j )
≥ k o + , k ≥
Iterating this inequality gives
r (k ) ≤
k −1
Σ
k −1
ν( j )ψ ( j ) Π [ 1+η(i ) ν(i ) ] , k ≥ ≥ k o +1 i = j +1
j =k o
and substituting this into (*) yields the result.
Solution 24.7
By assumption Φ A (k , j ) ≤ γ for for k ≥ j. Treating f ( ( k , z (k )) )) as an input, the complete solu
formula is
z (k ) = Φ A (k , k o ) z (k o ) +
k −1
Σ Φ (k , j +1) f ( ( j, z ( j )) , A
≥ k o +1 k ≥
j =k o
This gives
z (k ) ≤ γ z (k o ) +
k −1
Σ γ f ( ( j, z ( j ))
j =k o
≤ γ z (k o ) +
k −1
Σ γ α z ( j ) ,
≥ k o +1 k ≥
j
j =k o
Applying Lemma 24.5,
z (k ) ≤ γ z (k o ) exp [ γ
k −1
Σα] j
j =k o
∞
≤ γ z (k o ) exp [ γ Σ α j ]
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CHAPTER 25
Solution 25.1
M (k o , k f ) is not invertible, then there exists a nonzero, n × 1 vector x a such that If M 0 = x T a M (k o , k f f ) xa k f −1 T T = x T a Φ ( j, k o )C ( j )C ( j )Φ( j, k o ) xa
Σ
j =k o k f −1
=
Σ C ( j )Φ( j, k ) x o
a
2
j =k o
This implies . , k f C ( j )Φ( j, k o ) xa = 0 , j = k o , . . . , f −1
which shows that the nonzero initial state x a yields the same output on the interval as does the zero initial s Therefore the state equation is not observable. On the other hand, for any initial state x o we can write, just as in the proof of Theorem 25.9,
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o
T
o
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If M M (k k ) is invertible, then the initial state is uniquely determined by
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L in e a r S y s t e m T h e o r y , 2 / E
S o lu t io n s
The claim is true if A (k ) is invertible at each k . Let k f = n so that n −1 Φ(n, j +1)b ( j )b T ( j )ΦT (n, j +1) W (0, n ) =
Σ
j =0
Since Φ(n, j +1) is invertible for j = 0, . . . , n −1, let . , n −1 0, . . . , b (k ) = Φ−1 (n, k +1)ek +1 , k = 0, . where e k is the k th -column of I I n . Then n −1
W (0, n ) =
Σe
T j +1 e j +1
= I n
j =0
and the state equation is reachable on [0, n ].
Solution 25.7 Suppose W O (k o , k f f ) is invertible. Given a p × 1 vector y f , let −1 (k , k ) y , k = k , . . . , . , k f u (k ) = B T (k )ΦT (k f f , k +1)C T (k f f )W O o f f f o f −1
and let u (k ) = 0 for other values of k . Then it is easy to show that the zero-state response to this input y y (k f ) = y f . Thus the state equation is output reachable on [ k o , k f ]. Conversely, suppose the state equation is output reachable on [ k o , k f W O (k o , k f f ) is not invertible, f ]. If W there exists a nonzero p × 1 vector y a such that 0 = y T a W O (k o , k f ) ya k f −1
=
Σ y C (k )Φ (k , j +1) B ( j ) B ( j )Φ (k , j +1)C (k ) y T a
T
f f
f f
T
T
f f
T
f f
a
j =k o k f −1
=
Σ y C (k )Φ(k , j +1) B ( j ) T a
f f
f f
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j =k o
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But by output reachability, with y f = ya , there exists an input u a (k ) such that
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Linear System Theory, 2/E
Solution 25.13
Solutions
We will prove that the state equation is reconstructible if and only if
C CA . . .
z = 0 implies A n z = 0
CA n −1
That is, if and only if the null space of the observability matrix is contained in the null space of A n . First, suppose the state equation is not reconstructible. Then there exist n × 1 vectors xa and xb such xa ≠ xb and
C . . .
xa =
CA n −1
C . . .
xb ,
A n xa ≠ A n xb
CA n −1
That is
C . . .
( xa − xb ) = 0 ,
A n ( xa − xb ) ≠ 0
CA n −1
You're Reading a Preview
Thus the condition (*) fails. Now suppose the condition (*) fails Unlock and z isfull such thatwith a free trial. access
C . . Download Free Trial ≠ 0 0 and z =With A n z . CA n −1 Obviously z ≠ 0. Then for x (0) = z the zero-input response is
Master your semester with Scribd y (k ) = 0 , k = 0, . . . , n −1 Read Free Foron 30this Days Sign up to vote title and x (n ) ≠ 0. But the same output sequence is produced by x (0) = 0, and for this initial state x (n ) = 0. Thu & The New York Times Useful Not useful cannot determine from the output (+) whether x (n ) = z or x (n ) = 0, which implies the state equation is Special offer forreconstructible. students: Only $4.99/month.
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CHAPTER 26
Solution 26.2
For the linear state equation
x (k +1) =
1 k x (k ) + 1 1
0 1
u (k )
easy computations give
R 2 (k ) = and
R 3 (k ) =
B (k )
B (k )
Φ(k +1, k ) B (k −1)
=
You're Reading a Preview
0 k 1 1
full access with a free trial. Φ(k +Unlock 1, k ) B (k −1) Φ(k +1, k −1) B (k −2)
=
0 k 2k −1 1 1 k
From the respective ranks the state equation is 3-step reachable, but Trial not 2-step reachable. Download With Free
Solution 26.4
The (n +1)-dimensional state equation
A 0 b Master your semester with z (Scribd ) k +1) = z (k ) + Read u (k Free Foron 30this Days to vote title c 0 d Sign up & The New York Times Useful Not useful (k ) y (k ) = 0 × 1 z (k ) − u
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1 n
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Linear System Theory, 2/E
Solutions
m
G ( z ) =
σl
ΣΣG
lr
l =1 r =1
z ______ ( z −λl )r
_
Here λ1 , . . . , λm are distinct complex numbers __ such that if λ L is complex, then λ M = λ L for some M . Furtherm the p × m complex matrices satisfy G Mr = G Lr for r = 1, . . . , σ L . From Table 1.10 the corresponding unit p response is m σl k 1−r λ k+ G (k ) = Glr l − l 1 l =1 r =1
ΣΣ
Thus we can state that a unit pulse response G (k ) is realizable if and only if (a) there exist positive integers m , σ1 , . . . , σm , distinct complex numbers λ 1 , . . . , λm , and σ 1 + . . . +σm com for all k ≥ 1, and p × m matrices G lr such that (#) holds _ (b) if λ L is complex, then λ M = λ L for some M . Furthermore the p × m complex matrices satisfy G Mr = r = 1, . . . , σ L .
Solution 26.8
Suppose the given state equation is minimal and of dimension n. We can write its (str proper, rational) transfer function as . adj( zI − A ) . b _c____________ G ( z ) = det ( zI − A )
You're Reading a Preview
where the polynomial det ( zI − A ) has degree n. If the numerator and denominator polynomials have a com root, then this root can be canceled without changing inverse z transform of G ( z ). Therefore, follo Unlock full accessthe with a free trial. Example 26.10, we can write by inspection a dimension-( n −1) realization of the unit pulse response of original state equation. This contradicts the assumed minimality, and the contradiction gives that the polynomials cannot have a common root. Download With Free Trial Now suppose the polynomials det ( zI − A ) and c . adj( zI − A ) . b have no common root, but that the given equation is not minimal. Then there is a minimal realization
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. adj ( zI − A ) . b h . adj ( zI −F ) . g ______________ _c____________ = det ( zI F ) det ( zI A )
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Linear System Theory, 2/E
Solutions
From the latter approach, setting
cb = 0, cAb = 1, cA 2 b = 1 / 2, cA 3 b = 1 / 2 easily yields c 1 = 0, c 0 = 1, a 0 = 1 / 4, a 1 = 1 / 2.
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CHAPTER 27
Solution 27.1
Similar to Solution 12.1.
Solution 27.4
Suppose the entry G ij ( z ) has one pole at z = 1, that is
Gij ( z ) =
N ij ( z ) __________ ( z −1) Dij ( z )
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where all roots of the polynomial D ij ( z ) have magnitude less than unity (so D ij (1) ≠ 0), and the polynomial satisfies N ij (1) ≠ 0. Suppose that the m ×Unlock 1 U ( zfull ) has all components zero except for U j ( z ) = z / ( z −1). The access with a free trial. i th -component of the output is given by ( z ) Trial z N ijFree ___________ Download With Y ( z ) = i
( z −1)2 Dij ( z )
By partial fraction expansion y i (k ) includes decaying exponential terms, possibly a constant term, and the term
(1) N ______ Master your semester with Scribd k , k ≥ 0 D (1) Read Free Foron 30this Days Sign up to vote title & The New York Since this term is Times unbounded, every realization of G ( z ) fails to be uniform bounded-input, bounded-output sta Useful Not useful ij
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Linear System Theory, 2/E
Solutions
µ = sup u (k ) , k ≥ 0
η =
∞
Σ G (k )
k =0
The first constant is finite for a well-defined sequence that goes to zero, and the second is finite by unif bounded-input, bounded-output stability. Then there is a positive integer K 1 such that ∞ ε ε ___ ___ , k ≥ K 1 , G (k ) ≤ u (k ) ≤ 2µ 2η k =K 1
Σ
Let K = 2K 1 . Then for k ≥ K we have
y (k ) ≤ µ
K 1 −1
Σ
j =0
≤ µ
k
Σ
k
ε ___ G (k − j ) G (k − j ) + 2 η k =K 1
q =k −K 1
Σ
ε k −K 1 ___ G (q ) G (q ) + 2 η q =0
Σ
ε 2η
ε ___ η = ε + ≤ µ ___ 2µ
Solution 27.8
Similar to Solution 12.12.
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CHAPTER 28
Solution 28.2
Lemma 16.18 gives that if V 11 and V are invertible, then
V −1 =
V 11 V 12 V 21 V 22
−1
V −111 +V −111 V 12 V −a1 V 21 V −111 −V −a1 V 21 V −111
=
−V −111 V 12 V −a1
−1
V −a1
where V a = V 22 −V 21 V −111 V 12 . From the expression V V −1 = I , written as
You're Reading a Preview
V 11 V 12 W 11 W 12 = I V 21 full V 22access W with 21 W Unlock a 22 free trial. we obtain
Download With Free Trial V 11 W 11 + V 12 W 21 = I V 21 W 11 + V 22 W 21 = 0
Under the assumption that V and V are invertible these imply Master your semester with Scribd − W = V − − V − V W , W =Read V − up V Free Foron 30this Days Sign toW vote title & The New Times SolvingYork for W gives Useful Not useful 11
22
11
1 11
1 11
12
21
21
11
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W 11 = (V 11 −V 12 V −221 V 21 )−1
1 22
21
11
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Linear System Theory, 2/E
Solutions
ˆ T ( A ˆ T )n K = − B
n
Σ
−1
ˆ k B ˆ B ˆ T ( A ˆ T )k A
ˆ A
n +1
k =0
That is,
K = −α B (α A )
T n
T
n
−1
Σ (α A ) (α B)(α B) (α A ) k
T
T k
(α A )n +1
k =0
= − B ( A ) T
T n
n
Σ
−1
α−2(n −k ) A k BB T ( A T )k
A n +1
k =0
Solution 28.4
Similar to Solution 13.11. However for the time-invariant case the reachability matrix rank can be used, rather than the eigenvector test, by writing
B ( A+BK ) B ( A+BK )2 B
.. .
=
B AB A 2 B
.. .
I 0 0 . . .
KB KAB+(KB )2 . . . .. . I KB .. . 0 I . . . . . . . . .
You're Reading a Preview Solution 28.6
Similar to Solution 2.8. Unlock full access with a free trial.
Solution 28.8
Download With Free Trial
Supposing that the linear state equation is reachable, there exists K such that all eigenvalu A+BK have magnitude less than unity. Therefore ( I − A − BK ) is invertible, and if we suppose
A − I B C 0
Master your semester with Scribd − Free Foron 30this Days Sign up to vote title is invertible, then C ( I − A − BK ) B is invertible from Exercise 28.6. Read Then given any diagonal, m × m matrix can choose & The New York Times Useful Not useful 1
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ˆ
N = [C ( I − A − BK )−1 B ]−1 Λ
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Linear System Theory 2 e Sol Uploaded by Shruti Mahadik
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This is the solution manual of Rugh W.J Linear system theory. 2 edition.
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Khalil - Nonlinear Systems Slides
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Table of Contents for Linear
Linear System Theory and
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CHAPTER 29
Solution 29.1
The error e b (k ) satisfies
eb (k +1) = z (k +1) − Pb (k +1) x (k +1)
˜ (k )C (k )−P (k +1) A (k )] x (k ) + [G ˜ (k )−P (k +1) B (k )]u (k ) = ˜ F (k ) z (k ) + [G b b a b = ˜ F (k ) z (k ) − ˜ F (k )Pb (k ) x (k ) = ˜ F (k )eb (k )
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full access with a free trial. ∞. Now Therefore e b (k ) → 0 exponentially as k →Unlock
∆ ˆ (k ) e (k ) = x (k ) − x
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= x (k )− H (k )C (k ) x (k )− J (k ) z (k )
= − J (k )eb (k ) + [ I − H (k )C (k )− J (k )Pb (k )] x (k ) (k )e (k ) = − J Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title k → Therefore if J (k ) is bounded, that is, J (k ) ≤ α < ∞ for all k , then e (k ) → 0 implies e (k ) → 0, as & The New York Times Useful Not useful xˆ (k ) is an asymptotic estimate of x (k ). b
b
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