Linear System Theory 2 e Sol

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This is the solution manual of Rugh W.J Linear system theory. 2 edition.

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Linear System Theory and

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Solutions Manual

LINEAR LINEAR SYSTEM SYSTEM THEO THEORY RY,, 2/E

Wilson Wilson J. Rugh Department of Electrical and Computer Engineering Johns Hopkins University

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PREFACE

With some lingering ambivalence about the merits of the undertaking, but with a bit more dedication Theory . Rou the first time around, I prepared this Solutions Manual for the second edition of Linear of Linear System Theory. 40% of the exercises are addressed, including all exercises in Chapter 1 and all others used in developments in text. This coverage complements the 60% of those in an unscientific unscientific survey who wanted a solutions manual perhaps does not overly overly upset the 40% who voted no. (The main contention between the two groups involved inevitable appearance of pirated student copies and the view that an available solution spoils the exercise.) I expect that a number of my solutions could be improved, and that some could be improved using techniques from the text. Also the press of time and my flagging enthusiasm for text processing impeded crafting of economical solutions—some solutions may contain too many steps or too many words. Howev hope that the error rate in these pages is low and that the value of this manual is greater than the price paid. Please send comments and corrections to the author at rugh@jhu.edu or ECE Department, Johns Hop University, Baltimore, MD 21218 USA.



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PREFACE

With some lingering ambivalence about the merits of the undertaking, but with a bit more dedication Theory . Rou the first time around, I prepared this Solutions Manual for the second edition of Linear of Linear System Theory. 40% of the exercises are addressed, including all exercises in Chapter 1 and all others used in developments in text. This coverage complements the 60% of those in an unscientific unscientific survey who wanted a solutions manual perhaps does not overly overly upset the 40% who voted no. (The main contention between the two groups involved inevitable appearance of pirated student copies and the view that an available solution spoils the exercise.) I expect that a number of my solutions could be improved, and that some could be improved using techniques from the text. Also the press of time and my flagging enthusiasm for text processing impeded crafting of economical solutions—some solutions may contain too many steps or too many words. Howev hope that the error rate in these pages is low and that the value of this manual is greater than the price paid. Please send comments and corrections to the author at rugh@jhu.edu or ECE Department, Johns Hop University, Baltimore, MD 21218 USA.



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CHAPTER 1

Solution 1.1 k = 2, ( A A + B )2 = A 2 + AB + BA + B 2 . If AB = BA, the A + B )2 = A 2 + 2 AB + B 2 . In ge (a) For k = then ( A k -fold product ( A + B ) can be written as a sum of terms of the form A j B k − j , j = 0, . . . , AB = BA, then the k -fold k number of terms that can be written as A j B k − j is given by the binomial coefficient . Therefore AB j implies

( A + B )k =

k

Σ

j =0

k j k − j A B j

(b) Write (b) Write det [λ I − − A (t )] = λn + an −1 (t )λn −1 + . . . + a 1 (t )λ + a 0 (t ) A (t ) implies a 0 (t ) ≠ 0. The Cayley-Hamilton theorem implies where invertibility of A A n (t ) + an −1 (t ) A n −1 (t ) + . . . + a 0 (t ) I = 0

Master your semester with Scribd − vote Free 30this Days − . . . − a − (t ) A −Read a (t ) I − A (t ) −up (t For ) on Sign to title _− ________________________________ A − (t ) = a (t )  Useful  Not useful & The New York Times for all t . Multiplying through by A −1 (t ) yields 1

1

n 1 0

n 2

n 1

Cancel anytime.

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Special offer forfor students: Only a$4.99/month. all t . Since )], a 0 (t ) = det A (t ). Assume ε > 0 is such that  det A (t ) ≥ ε for all t 0 (t ) = det [ − A (t )], for all t . Then, for all t  A (t ) ≤ α we have aij (t ) ≤ α, and thus there exists a γ such such that a j (t ) ≤ γ for

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Linear System Theory, 2/E

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Solutions

__________ _ _ H det (λ I − A ) = det (λ I − A ) = det (λ I − A ) H

(e) α A has eigenvalues α λ1 , . . . , αλn since Ap = λ p implies ( α A ) p = (αλ) p. (f) Eigenvalues of A T A are not nicely related to eigenvalues of A. Consider the example A =

0 α 0 0

,

A T A =

0 0 0 α

where the eigenvalues of A are both zero, and the eigenvalues of A T A are 0, α. (If A is symmetric, then applies.)

Solution 1.3

(a) If the eigenvalues of A are all zero, then det ( λ I − A ) = λn and the Cayley-Hamilton theorem shows that nilpotent. On the other hand if one eigenvalue, say λ1 is nonzero, let p be a corresponding eigenvector. T A k p = λ k _ 1 p ≠ 0 for all k ≥ 0, and A cannot be nilpotent. (b) Suppose Q is real and symmetric, and λ is an eigenvalue of Q. Then λ also _ _is_ an eigenvalue. From H H Qp = p p Qp = p p Qp = λ p, and eigenvalue/eigenvector equation we get . Also λ λ _ _ _ transposing H H H p Qp = λ p p. Subtracting the two results gives ( λ − λ) p p = 0. Since p ≠ 0, this gives λ = λ, that is, λ is real (c) If A is upper triangular, then λ I − A is upper triangular. Recursive Laplace expansion of the determinant a the first column gives det (λ I − A ) = (λ − a 11 ) . . . (λ − ann )

You're Reading a Preview

which implies the eigenvalues of A are the diagonal entries a 11 , . . . , ann .

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Solution 1.4

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(a) A =

0 0 1 0

implies A T A =

(b) Master your semester with Scribd 3 1 A = implies 1 3 & The New York Times

1 0 0 0

implies  A  = 1

10 6Free For 30 Days Read A T A = Sign up to vote on this title 6 10

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Then

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det (λ I − A T A ) = (λ − 16)(λ − 4)

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This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides

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Solution 1.6

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Solutions

By definition of the spectral norm, for any α ≠ 0 we can write

 A  = max  A x  = max  x  = 1

= max

α x  = 1

 x  = 1

 A x  _______  x 

α A x   A α x  __________ ________ = max  x  = 1 / α α x  α x 

Since this holds for any α ≠ 0,

 A x   A x  _______ _______ = max x ≠ 0  x   x  ≠ 0  x 

 A  = max Therefore

 A  ≥

 A x  _______  x 

for any x ≠ 0, which gives

 A x  ≤  A  x 

Solution 1.7

By definition of the spectral norm,

 AB  = You're max Reading ( AB ) x  =a max  A ( Bx ) Preview  x  = 1

 x  = 1

max { by trial. Exercise 1.6 ≤ Unlock  Aaccess  Bxwith } a, free full  x  = 1

=  A  max  Bx  =  A  B   x  = 1 Download With Free Trial

If A is invertible, then A A −1 = I and the obvious  I  = 1 give

1 =  A A −1  ≤  A  A −1  Therefore Master your semester with Scribd 1 _____  A −  ≥  A  & The New York Times 1



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

Solutions

A =

0 A 12 0 0

then partitioning the vector x similarly we see that max  A x  = max  A 12 x 2  =  A 12 

 x  = 1

Solution 1.9

 x 2  = 1

By the Cauchy-Schwarz inequality, and  x T  =  x ,

 x T A x  ≤  x T A  x  =  A T x  x 

≤  A T  x 2 =  A  x 2 This immediately gives x T A x ≥ −  A  x 2

If λ is an eigenvalue of A and x is a corresponding unity-norm eigenvector, then

λ = λ x  = λ x  =  A x  ≤  A  x  =  A 


Solution 1.10 Since Q = Q T , Q T Q = Q 2 , and the eigenvalues of Q 2 are λ 21 , . . . , λ 2n . Therefore full access2 with a free trial. λmax (Q ) = max λi  Q  = √ Unlock 1 ≤ i ≤ n

For the other equality Cauchy-Schwarz gives Download With Free Trial

 x T Qx |

≤  x T Q  x  = Qx  x  ≤ Q  x 2 = [ max λi  ] x T x 1 ≤ i ≤ n

Master your Scribd x . Choosing x as a unity-norm eigenvector of Q correspondi Thereforesemester | x Qx | ≤ Q  forwith all unity-norm  Read Free Foron 30this Days Sign up to vote title the eigenvalue that yields max λ  gives  ≤ ≤ & The New York Times  Useful  Not useful T

a

1

i

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

T

Qx 

Q 

n

i

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T  x T a Qxa  = x a [ max λi  ] xa = max λi  1 ≤ i ≤ n

1 ≤ i ≤ n

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

Solutions

max x T A T A x = λmax ( A T A )

 x  = 1

so we have  A  =

√λmax ( A T A ) .

Solution 1.12 Since A T A > 0 we have λi ( A T A ) > 0, i = 1, . . . , n, and ( A T A )−1 > 0. Then by Exercise 1.11  A −1 2 = λmax (( A T A )−1 ) = n

1 _________ λmin ( A T A )

Π λi ( A T A )

=

i =1 __________________ λmin ( A T A ) . det ( A T A )

=

 A 2(n −1) _________ (det A )2

[λmax ( A T A )]n −1

≤ _____________ 2 (det A )

Therefore

 A −1  ≤

 A n −1 ________ det A 

You're Reading a Preview Solution 1.13 Assume A ≠ 0, for the zero case is trivial. For any unity-norm x and y, Unlock full access with a free trial.  y T A x  ≤  y T  A x 

≤  y   A Free  x  =Trial  A  Download With Therefore max

 y T A x  ≤  A 

Master your semester x be suchwith Now let unity-norm that  A xScribd  =  A , and let Ax _____ & The New York Times y =  x ,  y  = 1

a

a

a

Special offer for students: Only $4.99/month. Then  ya  = 1 and

a

 A 


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

Solutions

example A (t ) =

t 0 0 t 2

,

 A (t ) =

t , 0 ≤ t ≤ 1 t 2 , 1 < t < ∞

Clearly the time derivative of  A (t ) is discontinuous at t = 1. (This overlaps Exercise 1.18 a bit.) Also the eigenvalues of continuously-differentiable A (t ) are not necessarily continuously differentiable, consi A (t ) =

0

1

−1 −t

An easy computation gives the eigenvalues

√t 2 − 4 t λ(t ) = __ ± _______ 2

2

Thus . t 1 ________ λ(t ) = __ ± 2 2 √t 2 − 4 and this function is not continuous at t = 2.

Solution 1.15 Clearly Q is positive definite, and by Rayleigh-Ritz if x ≠ 0,

You're Reading a Preview 0 < λmin (Q ) x T x ≤ x T Q x ≤ λ max (Q ) x T x Unlock full access with a free trial.

Choosing x as an eigenvector corresponding to λ min (Q ) (respectively, λ max (Q )) shows that these inequalitie tight. Thus

Download With Free Trial ε1 ≤ λmin (Q ) , λmax (Q ) ≤ ε2

Therefore 1 1 λmin (Q −1 ) = _______ ≥ ___ λmax (Q ) ε2 Read Free Foron 30this Days Sign up to vote title 1 1 ___ _______ 1 − ≤ ε Useful  Not useful λmax (Q ) = λmin (Q )  1 Cancel anytime.


Thus Rayleigh-Ritz for the positive definite matrix Q −1 gives

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Solution 1.17



Solutions

Using the product rule to differentiate A (t ) A −1 (t ) = I yields . d ___ A (t ) A −1 (t ) + A (t ) A −1 (t ) = 0 dt

which gives . d −1 ___ A (t ) = − A −1 (t ) A (t ) A −1 (t ) dt

Solution 1.18

Assuming differentiability of both x (t ) and  x (t ), and using the chain rule for s

functions,



d d ___ ___  x (t )  x (t )2  = 2 x (t ) dt dt = 2 x (t )

d ___  x (t ) dt

Also we can write, using the product rule and the Cauchy-Schwarz inequality,



. T . . d d ___ ___  x (t )2  =  x T (t ) x (t ) =  x (t ) x (t ) + x T (t ) x (t ) = 2 x T (t ) x (t ) dt dt . a Preview ≤ 2 x (t You're ) x (t )Reading 

Unlock full access For t such that x (t ) ≠ 0, comparing these expressions gives with a free trial.

d ___

.

 x (t ) ≤  x (t )  Download With Free Trial dt

If x (t ) = 0 on a closed interval, then on that interval the result is trivial. If x (t ) = 0 at an isolated point, continuity arguments show that the result is valid. Note that for the differentiable function x (t ) = t ,  x (t ) is not differentiable at t = 0. Thus we must make the assumption that  x (t ) is differentiable. (While inequality is not explicitly used in the book, the added differentiability hypothesis explains why  we T Read Free For 30 Days 2 Sign up to vote on this title differentiate  x (t ) = x (t ) x (t ) instead of  x (t ).) 

Master your semester with Scribd & The New York Times



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Cancel anytime.

Special offer for students: Only $4.99/month. Solution 1.19 To prove the contrapositive claim, suppose for each i, j there is a constant βij such that

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Solution 1.20



Solutions

If λ(t ), p (t ) are a pointwise-in- t eigenvalue/eigenvector pair for A −1 (t ), then

 A −1 (t ) p (t ) = λ(t ) p (t ) = λ(t ) p (t ) Therefore, for every t ,

λ(t ) =

 A −1 (t ) p (t ) _____________  p (t )

−1

 A (t ) p (t ) ≤ α ≤ _______________  p (t )

Since this holds for any eigenvalue/eigenvector pair,

det A (t ) =

1 1 _________________ ___________ = . 1 − λ1 (t ) . . λn (t ) det A (t )

1 > 0 ≥ ___ n

α

for all t .

Solution 1.21

Using Exercise 1.10 and the assumptions Q (t ) ≥ 0, t b t b

t b

t a

t a

∫ Q (σ) d σ = ∫ λ

max [Q (

≥ t a ,

t b

t b

t a

t a

σ)] d σ ≤ ∫ tr [Q (σ)] d σ = tr ∫ Q (σ) d σ

Note that t b

You're Reading a 0Preview Q (σ) d σ ≥

∫

t a


since for every x

t b

t b

Download With Free Trial x ∫ Q (σ) d σ x = ∫ x T Q (σ) x d σ ≥ 0 T

t a

t a

Thus, using a property of the trace on page 8 of Chapter 1, we have

Master your semester with ) d σ ≤ tr ∫ Q (σ) d σ ≤ n Read ) d Foron σFree ∫ Q (σScribd ∫ Q (up 30this Days Sign toσvote title & The New York Times  Useful  Not useful Finally, Special offer for students: Only $4.99/month.

t b

t b

t b

t a

t a

t a

Cancel anytime.

t b

∫

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

CHAPTER 2

Solution 2.3

. The nominal solution for u˜ (t ) = sin (3t ) is y˜ (t ) = sin t . Let x 1 (t ) = y (t ), x 2 (t ) = y (t ) to writ

state equation . x (t ) =

−

x 2 (t ) 3 (4 / 3) x 1 (t ) (1 / 3)u (t )

−

Computing the Jacobians and evaluating gives the linearized state equation


0 0 1 x δ (t ) + u δ (t ) 2 −4 sin −1trial. / 3 0 Unlock full t access with a free

. x δ (t ) = y δ (t ) =

where x δ (t ) = x (t ) −

sin t cos t

,

1

0 x δ (t )


u δ (t ) = u (t ) − sin (3t ) ,

y δ (t ) = y (t ) − sin t , x δ (0) = x (0) −

Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title & The New York Times Solution 2.5 For u˜ = 0 constant nominal solutions are solutions  of Useful  Not useful Special offer for students: Only $4.99/month.

Cancel anytime.

0 = x˜ 2 − 2 x˜ 1 x˜ 2 = x˜ 2 (1−2 x˜ 1 )

0 1

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

rank A = rank [ A b ]. Also, x˜ is a constant nominal with c x˜ = 0 if and only if

0 = A x˜ + bu˜ 0 = c x˜ that is, if and only if

−bu˜

A ˜ x = c

0

As above, this holds if and only if rank

A c

= rank

A b c 0

Finally, x˜ is a constant nominal with c x˜ = u˜ if and only if 0 = A x˜ + bu˜ = ( A + bc ) x˜ and this holds if and only if x˜ ∈ Ker [ A + bc ] (If A is invertible, we can be more explicit. For any u˜ the unique constant nominal is x˜ = − A −1 bu˜ . Then y˜ Reading −1.) u˜ ≠ 0 if and only if c A −1 b = 0 , and y˜ = u˜ ifYou're and only if c A −1 ba=Preview Unlock full access with a free trial.

Solution 2.8 (a) Since

Download With Free Trial A B C 0

is invertible, for any K

Master your semester with Scribd A + BK B = 0 C & The New York is invertible. Let Times Special offer for students: Only $4.99/month.

A + BK B

A B C 0

R 1 R 2

I 0

Read Free Foron 30this Days Sign to vote title K I up

 =


Useful

I 0

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Solution 2.10



Solutions

For u (t ) = u˜ , x˜ is a constant nominal if and only if 0 = ( A + Du˜ ) x˜ + bu˜

This holds if and only if bu˜

∈ Im [ A + Du˜ ], that is, if and only if rank ( A + Du˜ ) = rank A+Du˜

bu˜

If A +Du˜ is invertible, then x˜ = −( A + Du˜ )−1 bu˜

If A is invertible, then by continuity of the determinant det ( A + Du˜ ) ≠ 0 for all u˜ such that u˜  is suffici small, and (+) defines a corresponding constant nominal. The corresponding linearized state equation is . x δ (t ) = ( A + Du˜ ) x δ (t ) + [ b − D ( A + Du˜ )−1 bu˜ ] u δ (t ) y δ (t ) = C x δ (t )

Solution 2.12

For the given nominal input, nominal output, and nominal initial state, the nominal solu

satisfies 1

.

0

x˜ (t ) =You're − x˜ 3 (t ) a, Preview x˜ 1 (t )Reading x˜ (0) = −3 −2 x˜ 2 (t ) − 2 x˜ 3 (t ) Unlock full access with a free trial. 1 = x˜ 2 (t ) − 2 x˜ 3 (t )

Integrating for x˜ 1 (t ) and then x˜ 3 (t ) easilyDownload gives the nominal solution With Free Trialx˜ 1 (t ) = t , x˜ 2 (t ) = 2 t − 3, and x˜ 3 (t ) The corresponding linearized state equation is specified by A =

0 0 0 1 0 −1 0 1 −2

, B (t )=

0 t , C = 0

0 1

−2

Master your semester with Scribd  Free Foron 30this Days It is unusual that the nominal input and nominal output are constants,Read but up the linearization istitle time varying. Sign to vote  & The New York Times  Useful  Not useful Special offer forSolution students: 2.14 Only $4.99/month. Compute

Cancel anytime.

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

CHAPTER 3

Solution 3.2

∂ ___ ∂τ

Differentiating term k +1 of the Peano-Baker series using Leibniz rule gives σ1

t

σ2

∫ τ A (σ ) ∫ τ A (σ ) ∫ τ 1

2

σk

...

σ2

t

∫ τ A (σ

σ

k +1 ) d k +1

. . . d σ

1

σk

τ

τ

d _ . . d σ Reading a__ Preview t − A (τ) A (σ2 ) . . . d σk +1 . . . d σ2 A (t ) A (σ2 ) . . . A (σk +1You're ) d σk +1 . 2 d τ τ τ τ τ τ

∫

=

∫

∫

∫

∫

Unlock full access with a free trial. t

∫ τ

+ A (σ1 )

t

∂ ___

k +1

2

σ2

σ1

∂ τ ∫ semesterτ

∫ τ

= A (σ1 )

σk

. A (σ ) d σ . . . d σ ∫ τ A (σ ) ∫ τ . .Download ∫ τ With Free Trial

∂τ

∂ ___

σ2

σ1

A (σ2 )

t

∂τ

∫ τ . . . ∫ τ A (σ

k +1 )

1

2

2

d σk +1 . . . d σ1




σk

∫ A (σ ) ∫ A (σ ) ∫ . . . ∫ A (σ 1

1

σk

Master your with Scribd Repeating this process k times gives & The New York Times σ σ Special offer for students: Only $4.99/month. ∂ ___

k +1

σ

k +1 ) d k +1


Useful

. . . d σ

1

 

_ d

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Solution 3.6



Solutions

Writing the state equation as a pair of scalar equations, the first one is . −t x (t ) ______ x 1 (t ) = 1 1 + t 2

and an easy computation gives x 1 (t ) =

x 1o _________ (1 + t 2 )1 / 2

Then the second scalar equation then becomes x 1o . −4t x (t ) + _________ ______ x 2 (t ) = 2 (1 + t 2 )1 / 2 1 + t 2

The complete solution formula gives, with some help from Mathematica, . 1 ________ x 2o + x 2 (t ) = (1 + t 2 )2

t

(1 + σ2 )3 / 2 _________ d σ x 1o (1 + t 2 )2

∫ 0

1+t 2 (t 3 / 4+5t/ 8)+(3 / 8) sinh−1 (t ) 1 _√ ____________________________ ________ x 1o x + = 2o (1 + t 2 )2 (1 + t 2 )2 If x 1o = 1, then as t →∞, x 2 (t ) → 1 / 4, not zero.

You're Reading a Preview Solution 3.7

From the hint, letting

Unlock full access with a free trial. t

r (t ) = v (σ)φ(σ) d σ Downloadt ∫ With Free Trial o

. we have r (t ) = v (t )φ(t ), and

φ(t ) ≤ ψ (t ) + r (t )

Master your semester Scribd v (t ) gives Multiplying (*) through by thewith nonnegative Read Free Foron 30this Days Sign up to vote title v (t )φ(t ) ≤ v (t )ψ (t ) + v (t )r (t Useful & The New York Times )  Not useful Cancel anytime.

Special offer foror students: Only $4.99/month.

.

()

( ) ( ) ≤ ( ) ( )

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

Solutions

t

∫ v (τ) d τ

t

eo

gives t

∫ v (τ) d τ

t

r (t ) ≤

∫

v (σ)ψ (σ)e σ

d σ

t o

and using (*) yields the desired inequality.

Solution 3.10

Multiply the state equation by 2 z T (t ) to obtain . d ___  z (t )2 2 z T (t ) z (t ) = dt n

n

Σ Σ 2 z (t )a (t ) z (t )

=

i

ij

j

i =1 j =1 n

≤Σ

n

Σ 2a (t ) z (t ) z (t ) , ij

i

j

t ≥ t o

i =1 j =1

At each t ≥ t o let


a (t ) = 2n 2 max aij (t ) 1 ≤ i, j ≤ n Unlock full access with a free trial. Note a (t ) is a continuous function of t , as a quick sample sketch indicates. Then, since  zi (t ) ≤  z (t ), d Download ___ t )2 ,Trial t ≤ t o  z (t )2 ≤ aWith (t ) z (Free dt

Multiplying through by the positive quantity

Master your semester with Scribde gives York Times & The New

t

− ∫ a (σ) d σ t o

t


d ___ dt

− ∫ a (σ) d σ

e

t o

 z (t )2





Useful

≤ 0 , t ≤ t o

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Solutions

t

t t

∫

z (t ) = A (σ) z (σ) d σ + t o

t

=

t

∫

∫ σ

A (σ) + E (τ, σ ) d τ

t o

∆

∫ σ∫ E (τ, σ) d τ z(σ) d σ

t o

z (σ) d σ

t

∫

ˆ (t , = A

σ) z (σ) d σ

t o

Thus t

t

∫

∫

 z (t ) =  Aˆ (t , σ) z (σ) d σ ≤  Aˆ (t , σ) z (σ) d σ t o

t o

By continuity, given T > 0 there exists a finite constant α such that  Aˆ (t ,

σ) ≤ α for t o ≤ σ ≤ t ≤ t o + T . Thus

t

∫

 z (t ) ≤ α  z (t ) d σ , t ∈ [t o , t o +T ] t o

and the Gronwall-Bellman inequality gives  z (t ) = 0 for t ∈ [t o , t o +T ], implying that there can be no m than one solution.


From the Peano-Baker series, Unlock full access with a free trial. σ1 σk −1 t t . . . . . . . . . d σ Φ(t , τ) − I + A (σ1 ) d + With A (σ1Free ) σ1 + 1 Download Trial A (σk ) d σk

∫ τ

∞

=

∫ τ

σ1

t

∫ τ

∫ τ

σ j −1

Σ ∫ τ A (σ ) ∫ τ . . . ∫ τ A (σ ) d σ

. . . d σ 1

σ j−1

t

Master your semester with Scribd  Read Free Foron 30this Days Sign up to vote title  For any fixed T > 0 there is a finite constant α such that  A (t ) ≤ α for t ∈ [−T , T ], by continuity. Therefore & The New York Times Useful  Not useful  σ σ j

1

j

j =k +1

∞

t

1

Special offer for students: $4.99/month. ...  Only A (σ )

Σ ∫ τ

j =k +1

1

∫ τ

∫ τ A (σ ) d σ j

j

. . . d σ  ≤ 1

∞

1

Cancelσanytime. j − 1

Σ ∫ τ A (σ ) ∫ τ . . . ∫ τ A (σ ) d σ 1

j =k +1

j

j

. . . d σ  1

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Solutions

∞

α2T ) j _(_____ < ε j !

Σ

j =K +1

Using the hint, ∞

Σ

j =k +1

∞ (α2T )k +1+i α2T ) j __________ _(_____ = j ! i =0 (k +1+i )!

Σ

∞ (α2T )k +1 (α2T )i . ______ ≤ Σ ________ i i =0

(k +1)!

k

If k > α2T , then ∞

Σ

j =k +1

α2T ) j _(_____ j !

(α2T ) . ≤ ________ k +1

(k +1)!

(α2T ) 1 __________________ _________ = (k −1)!(k +1)(k −α2T ) 1 − α2T/k k +1

Because of the factorial in the denominator, given ε > 0 there exists a K > α2T such that (*) holds.

Solution 3.15

Writing the complete solution of the state equation at t f , we need to satisfy t f

H o xo + H f

Φ(t f , t o ) xo + ∫ Φ(t f , σ ) f ( σ) d σ

=h

t o

Thus there exists a solution that satisfies the boundary conditions if and only if t f


t o


h − H f ∫ Φ(t f , σ ) f ( σ) d σ ∈ Im[ H o + H f Φ (t f , t o ) ]

There exists a unique solution that satisfies the boundary conditions if H o + H f Φ(t f , t o ) is invertible. To com a solution x (t ) satisfying the boundary conditions: Download With Free Trial (1) Compute Φ(t , t o ) for t ∈ [t o , t f ] (2) Compute H o + H f Φ(t f , t o )

Master your semester with Scribd (3) Compute ∫ Φ(t , σ) f ( σ) d σ & The New York Times (4) Solve (+) for x t f

f

t o

o Special offer for students: Only $4.99/month. t

∫


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

CHAPTER 4

Solution 4.1

. An easy way to compute A (t ) is to use A (t ) = Φ(t , 0)Φ(0, t ). This gives A (t ) =

−2t −1 1 −2t

This A (t ) commutes with its integral, so we can write Φ(t , τ) as the matrix exponential t

−(t −τ) A (σ) d σ = exp ∫ (t −τ) You're Reading a Preview τ

Φ(t , τ) = exp

2

−(t −τ) −(t −τ)2


Solution 4.4

A linear state equation corresponding to the n th -order differential equation is


. x (t ) =

... ... . . . ... 0 0 −a 0 (t ) −a 1 (t ) . . . 0 0 . . .

Master your semester with Scribd & The New York Times The corresponding adjoint state equation is Special offer for students: Only $4.99/month.

0

1 0 . . .

...

0

0 0 . . . 1

x (t )

−anRead Free Foron 30this Days Sign to vote title −1 (t ) up  Useful  Not useful Cancel anytime.

a 0 (t )

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

Solutions

. zn −2 (t ) = − zn −3 (t ) + an −3 (t ) zn (t ) gives . d 2 d d 3 ____ ___ ____ [ an −1 (t ) zn (t ) ] a t z t + [ ( ) ( ) ] − z t = z t ( ) ( ) n −2 n n n −2 dt dt 2 dt 3 = − zn −3 (t ) + an −3 (t ) zn (t ) −

d 2 d ____ ___ [ an −2 (t ) zn (t ) ] + 2 [ an −1 (t ) zn (t ) ] dt dt

Continuing gives the n th -order differential equation d n −2 d n −1 d n _____ _____ ____ [ an −1 (t ) zn (t ) ] − n −2 [ an −2 (t ) zn (t ) ] zn (t ) = dt dt n −1 dt n d ___ [ a 1 (t ) zn (t ) ] + (−1)n +1 a 0 (t ) zn (t ) + . . . + (−1)n dt

Solution 4.6

For the first matrix differential equation, write the transpose of the equation as (transpose differentiation commute) . T X (t ) = A T (t ) X T (t ) , X T (t o ) = X T o This has the unique solution X T (t ) = Φ A T (t )You're (t , t o ) X T , so that a Preview oReading T X (full t ) =access X o Φ AT with (t ) (t o ) trial. Unlock a, t free

In the second matrix differential equation, let Φ k (t , τ) be the transition matrix for Ak (t ), k = 1, 2. Then it is to verify (Leibniz rule) that a solution is

Download With Free Trial t

X (t ) = Φ1 (t , t o ) X o Φ

T 2 (t , t o ) +

∫ Φ (t , σ)F (σ)Φ (t , σ) d σ 1

T 2

t o

Or, one can generate this expression by using the obvious integrating factors on the left and right sides of Master your semester with Scribd  Z on t )this differential. equation. (To show this is a unique solution, show that Read the (30 between Free For Days Signdifference up to vote title any two solu  satisfies Z (t ) = A (t ) Z (t ) + Z (t ) A (t ), with Z (t ) = 0. Integrate both sides and apply the Bellman-Gr & The New York Times Useful  Not useful  inequality to show Z (t ) is identically zero.) 1


T 2

o

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

Solutions

=

cos τ 0 0 cos τ

=

−sin τ

0

+

−sin τ

sin τ 0

cos τ sin τ cos τ

Replacing τ as noted above gives Φ(t , 0).

Solution 4.10

For sufficiency, suppose differentiable. Let z (t ) = T −1 (t ) x (t ) so that

Φ z (t ,

Φ x (t ,

0) = T (t )e Rt .

Then T (0) = I and T (t ) is continu

0) = T −1 (t )Φ x (t , 0)T (0) = T −1 (t )T (t )e Rt = e Rt

. Thus z (t ) = R z (t ). For necessity, suppose P (t ) is a variable change that gives . z (t ) = Ra z (t ) Then

Φ z (t ,

Ra t

0) = e

= P −1 (t )Φ x (t , 0)P (0)

that is,

Φ x (t ,

P −1 (0)

Ra t

0) = P (t )e


Let T (t ) = P (t )P −1 (0) and R = P (0) Ra P −1 (0). Then

Φ x (t ,

Unlock full access−1with free −trial. (0) RPa(0)t P 1 (0)

0) = T (t )P (0) e P

= T (t )P (0) [ P −1 (0)e Rt P (0) ] P −1 (0)

Download With Free Trial = T (t )e Rt

Solution semester 4.11 Suppose Master your with Scribd Φ(t , 0) = e & The New York Times

A 1 t A 2 t

e

Then


.

Φ(t

0)

d ___

A 1 t A 2 t

e

e

A 1 t

=e





Useful

A 2 t

( A +A ) e

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Solutions

d ___ dt

we have that Φ(t , 0) = e



A 1 t A 2 t

e

e

A 1 t A 2 t

= A (t )e

e

A 1 0 A 2 0

, e

e

= I

A 1 t A 2 t

e

.

Solution 4.13 Writing

∂ Φ (t , τ) = A (t )Φ (t , τ ) , ___ A ∂t A

Φ(τ, τ) = I

in partitioned form shows that

∂ Φ (t , τ) = A (t )Φ (t , τ ) , ___ 22 21 ∂t 21

Φ21 (τ, τ) = 0

Thus Φ21 (t , τ) is identically zero. But then

∂ Φ (t , τ) = A (t )Φ (t , τ ) , ___ ii ii ∂t ii

Φii (τ, τ) = I

for i = 1, 2, and

∂ Φ (t , τ) = A (t )Φ (t , τ ) + A (t )Φ (t , τ) , ___ 11 12 12 22 ∂t 12

Φ12 (τ, τ) = 0

Using Exercise 4.6 with F (t ) = A 12 (t ) Φ22 You're (t , τ) gives Reading a Preview t

∫ τ

with(σ a)free (t , σ) A (σ, τ) d σ τ) = full Φ12 (t ,Unlock Φ11access Φ22trial. 12

Download With Free Trial Solution 4.17

We need to compute a continuously-differentiable, invertible P (t ) such that t 1 = P −1 (t ) 1 t

. 0 1 P (t ) − P −1 (t )P (t ) 2 2−t 2 t

Master your semester with Scribd  Read Free Foron 30this Days Sign up to vote title  Multiplying on the left by P (t ), the result can be written as a dimension-4 linear state equation. Choos & The New York Times Useful  Not useful  initial condition corresponding to P (0) = I , some clever guessing gives Cancel anytime.


P (t ) =

1 0 t 1

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Solutions

∂ Φ T (−τ, −t ) = A T (−t )Φ T (−τ, −t ) ___ A ∂t A Since Φ(−τ,

−τ) = I , we have F (t ) = A T (−t ).

Or we can use the result of Exercise 3.2 to compute:

∂ ∂ Φ (−τ, −t ) =− _____ ___ A ∂(−t ) ∂t

Φ A (−τ, −t ) = Φ A (−τ, −t ) A (−t )

This implies

∂ Φ T (−τ, −t ) = A T (−t )Φ (−τ, −t ) ___ A ∂t A Since Φ(−τ,

−τ) = I , we have F (t ) = A T (−t ).

Solution 4.25

We can write t+ σ

Φ(t + σ, σ ) = I +

∞

τ1

t +σ

τk −1

∫ σ A (τ) d τ + Σ ∫ σ A (τ ) ∫ σ A (τ ) . . . ∫ σ A (τ ) d τ 1

k

2

k

. . . d τ 1

k =2

and _ k _ ∞a Preview You're Reading 1 ___ At (σ)t k e = I + At (σ)t + _ At (σ)t

Σ

! k =2 ak free Unlock full access with trial.

Then

_

 R (t , σ) = Φ(t + σ, σ)Download − e At (σ)t  With Free Trial ∞

= 

t +σ

Σ ∫ σ

k =2

τ1

τk −1

∫ σ

A (τ1 ) A (τ2 ) . . .

_ k 1 ___ At (σ)t k  A (τk ) d τk . . . d τ1 − k ! σ

∫

Master your with Scribd t ) ≤ α and the triangle From  A (semester inequality, Read Free Foron 30this Days Sign up to vote title ∞ 2 ∞ ___ − _t __ − α t  Not useful & The New York Times  R (t , σ) ≤ 2 Σ α k ! = α t Σk Useful ! k

Special offer for students: Only $4.99/month. Using

k =2

k

k 2

2 2

k =2

k 2

Cancel anytime.

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

CHAPTER 5

Solution 5.3

Using the series definition, which involves talent in series recognition, A 2k +1 =

0 1 1 0

, A 2k =

1 0 0 1

, k = 0, 1, . . .

gives At

e

=I+

=

Using the Laplace transform method,

0 t 3 t 2 0 0 t 1 1 ___ ___ + + t 3 0 0 t 2 a Preview t You're 0 3! 2! Reading

+ ...

−t ) / 2 (e t −e −t ) / 2 (e t +e cosh t sinh t Unlock full access with=a free trial. t t t t − − (e −e ) / 2 (e +e ) / 2 sinh t cosh t


(sI − A )−1 =

s −1 −1 = −1 s

1 _____ 2 s −1 s _____ s 2 −1

s _____ 2 s −1 1 _____ 2 s −1

Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title which gives again & The New York Times  Useful  Not useful cosh t sinh t Special offer for students: Only $4.99/month.

e At =

Cancel anytime.

sinh t cosh t

Using the diagonalization method, computing eigenvectors for A and letting

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

Solutions

t

Φ(t ,

∫ A (σ) d σ

0) = e 0

t 2 / 2 t t t 2 / 2

= exp

And since t 2 / 2 0 0 t 2 / 2

0 t t 0

,

commute,

Φ(t ,

1 0 0 1

0) = exp

t 2 / 2

0 1 1 0

. exp

t

Using Exercise 5.3 gives

Φ(t ,

Solution 5.7

2

0) =

cosh t sinh t = sinh t cosh t

e t / 2 0 2 0 e t / 2

2

2

e t / 2 cosh t e t / 2 sinh t 2 2 e t / 2 sinh t e t / 2 cosh t

To verify that t

You're Preview A Reading e A σ d σ = ea At − I

∫ 0


note that the two sides agree at t = 0, and the derivatives of the two sides with respect to t are identical. If A is invertible and all its eigenvalues have negative real parts, then lim t → ∞ e At = 0. This gives Download ∞ With Free Trial A e A σ d σ = − I

∫ 0

that is,

Master your semester with Scribd ∞ − A = − ∫ e & The New York Times

A σ

1

0


0

d σ =

Free Foron 30this Days Sign to vote title e A σRead d σ up

∫

∞




Useful

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Solutions

t

. x (t ) = bu (t ) + A



 Search document

t

∫ e

A (t −σ)

D

e

∫ σ u (τ) d τ

t

t

∫

D

bu (σ) d σ + Du (t ) e A (t −σ) e

0

∫ σ u (τ) d τ

bu (σ) d σ

0

= A x (t ) + Dx (t )u (t ) + bu (t )

Solution 5.12

We will show how to define β0 (t ), . . . , βn −1 (t ) such that n −1 . n −1 n −1 βk (t )Pk = βk (t ) APk , βk (0)Pk = I

Σ

Σ

k =0

Σ

k =0

k =0

which then gives the desired expression by Property 5.1. From the definitions, P 1 = AP 0 − λ1 I , P 2 = AP 1 − λ2 P 1 , . . . , Pn −1 = APn −2 − λn −1 Pn −2

Also Pn = ( A −λn I )Pn −1 = 0 by the Cayley-Hamilton theorem, so APn −1 = λn Pn −1 . Now we equate coefficien like Pk ’s in (*), rewritten as n −1 . n −1 βk (t )Pk = βk (t )[Pk+1 + λk +1 Pk ]

Σ

k =0

to get equations for the desired β k (t )’s:

Σ

k =0


. λ1 β0 (with P 0Unlock t ) =access t ) a free trial. : β0 (full . P 1 : β1 (t ) = β0 (t ) + λ2 β1 (t ) . Download With Free Trial . . . Pn −1 : βn −1 (t ) = βn −2 (t ) + λn βn −1 (t )

that is,

Master your semester with Scribd . λ 0 β. (t ) & The New York Times β .(t ) = 1.. λ.. 1

0

2

1


.

. .

.

.

0

0

... ... . . . ...

0 0 . . .

λ

1

0Read Foron 30this Days Sign up to vote title βFree 0 (t ) 0 Useful β1 (t )  Not useful  . Cancel anytime. . . . . . 0

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Solutions

Φ(t , t o ) = P −1 (t ) e P (t )SP o

= Q (t , t o )e

Solution 5.19



−1 (t ) (t −t ) o

o

P (t o ) = P −1 (t )P (t o ) e

S (t −t o )

P −1 (t o )P (t o )

S (t −t o )

From the Floquet decomposition and Property 4.9, T

det Φ(T , 0) = det e

RT

∫ tr [ A (σ)] d σ

= e0

Because the integral in the exponent is positive, the product of eigenvalues of Φ(T , 0) is greater than unity, w implies that at least one eigenvalue of Φ (T , 0) has magnitude greater than unity.Thus by the argument follo Example 5.12 there exist unbounded solutions.

Solution 5.20

Following the hint, define a real matrix S by e S 2T = Φ2 (T , 0)

and set Q (t ) = Φ(t , 0)e −St

Clearly Q (t ) is real and continuous, and


+2T ) Q (t +2T ) = Φ(t +2T ,Unlock = Φ(with t +2aT ,free T )Φ 0)e −S (tfull (T , 0)e −S 2T e −St access trial.

= Φ(t +T , 0)Φ(T , 0)e −S 2T e −St = Φ(t +T , T )Φ2 (T , 0)e −S 2T e −St

Download = Φ(t +T , T )e −St = Φ(t , With 0)e −St Free Trial = Q (t )

That is, Q (t ) is 2T -periodic. (For a proof of the hint, see Chapter 8 of D.L. Lukes, Differential Equat Classical to Controlled , Academic Press, 1982.)

Master your semester with Scribd  Read Free Foron 30this Days Sign up to vote title  & The New York Times  Useful  Not useful Solution 5.22 The solution will be T -periodic for initial state x if and only if x satisfies (see text equ Special offer for(32)) students: Only $4.99/month.

o

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

Solutions

t o +T

0 =

∫

t o +T

Φ(t o , σ ) f (σ) d σ =

z T o

t o

∫ z

(σ) f ( σ) d σ

T

t o

completes the proof.

Solution 5.24 Note A = − A T , and from Example 5.9, e At =

cos t sin t −sin t cos t

Therefore all solutions of the adjoint equation are periodic, with period of the form k 2π, where k is a pos integer. The forcing term has period T = 2π / ω, where we assume ω > 0. The rest of the analysis breaks d into 3 cases. Case 1: If ω ≠ 1, 1 / 2, 1 / 3, . . . then the adjoint equation has no T -periodic solution, so the condition (Exe 5.22) T

∫ z

(σ) f ( σ) d σ = 0

T

0

holds vacuously. Thus there will exist corresponding periodic solutions. Case 2: If ω = 1, then T You're Reading a Preview T A σ z f d = z ( ) ( ) σ σ σ ∫ ∫ o e f ( σ) d σ

T

T

0 full access with a free trial. Unlock

0

T

T

= − zo 1 ∫ sin (σ) d σ + zo 2 ∫ cos σ sin σ d σ Download With Free Trial 0 0 2

≠ 0

Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title & The New York Times ∫ cos σ sin (σ /k ) d σ = ∫ sin σ sin  (σ /k Useful ) d σ = 0  Not useful so there is no periodic solution. Case 3: If ω = 1 /k , k = 2, 3, . . . , then since


T

T

0

0

the condition (+) will hold, and there exist periodic solutions.

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

CHAPTER 6

If the state equation is uniformly stable, then there exists a positive γ such that for any t o the corresponding solution satisfies

Solution 6.1

 x (t ) ≤ γ  xo  , t ≥ t o Given a positive ε, take δ = ε / γ . Then, regardless of t o ,  xo  ≤ δ implies

 x (t ) ≤ γ δ = ε , t ≥ t o You're Reading a Preview Conversely, given a positive ε suppose positive δ is such that, regardless of t o ,  xo  ≤ δ implies  x (

t ≥ t o . For any t a

≥ t o let xa be such that


 xa  = 1 , Φ(t a , t o ) xa  = Φ(t a , t o )

With FreeatTrial Then xo = δ xa satisfies  xo  = δ, and the Download corresponding solution t = t a satisfies  x (t a ) = Φ(t a , t o ) xo  = δΦ(t a , t o ) ≤ ε Therefore

Master your semester with Scribd Φ(t , t ) ≤ ε / δ Read Free Foron 30this Days Sign up to vote title x can beTimes Such anYork selected for any t , t such that t ≥ t . Therefore Useful  Not useful & The New a

a

a

o


for all t and t with t ≥ t , and we can take

a

o

o

Φ(t , t o ) ≤ ε / δ

/ δ to obtain

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

Solutions

t

σ1

t

∫ τ

∫ τ

∫ τ

Φ(t , τ) =  I + A (σ) d σ + A (σ1 ) A (σ2 ) d σ2 d σ1 + . . .  t

σ1

t

∫ τ

∫ τ

∫ τ A (σ ) d σ d σ  + . . .

=  I  +  A (σ) d σ +  A (σ1 ) t

2

1

σ1

t

∫ τ

2

∫ τ

= 1 +   A (σ) d σ +   A (σ1 ) 

∫ τ  A (σ ) d σ d σ  + . . . 2

2

1

(Be careful of t < τ.) Since  A (t ) ≤ α for all t , t

σ1

t

∫ τ

∫ τ ∫ τ 1 d σ d σ  + . . .

Φ(t , τ) ≤ 1 + α 1 d σ + α   2

= 1 + αt −τ + α2

2

1

t −τ2 _|_____ + ... 2!

For | t −τ ≤ δ,

Φ(t , τ) ≤ 1 + α δ +

2 2 δ _α ____ + ... 2!

αδ

=e You're Reading a Preview Unlock full access with a free trial.

Solution 6.8

See the proof of Theorem 15.2.

Download With Free Trial Solution 6.10 Write Re [λ] = −η, where η > 0 by assumption, so that t e λt  = t e −ηt ,

t ≥ 0

A simple maximization argument (settingScribd the derivative to zero) gives Master your semester with Read Free Foron 30this Days Sign up to vote title 1 ∆ − ≤ ___ β ≥ = t , 0 t e & The New York Times  Useful  Not useful ηe ηt

Special offer forso students: Only $4.99/month. that

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Solutions

∞

∞ j +( j −1)+ . . . +1 j _2___________ e −(η / 2 )t dt t e  dt ≤ j (η e ) 0 . . . j +( j −1)+ +1 2 j . ___ ≤ _2___________ η (η e ) j . . . 22 j +( j −1)+ +1 _____________ = j e  Re [λ] j +1

∫ 0

j

∫

λt

Solution 6.12 By Theorem 6.4 uniform stability is equivalent to existence of a finite constant γ such e At  ≤ γ for all t ≥ 0. Writing e At =

m

σk

ΣΣ

W kj

k =1 j =1

where λ 1 , . . . ,

t j −1 λ t ______ e k ( j −1)!

λm are the distinct eigenvalues of A, suppose Re[λk ] ≤ 0 , k = 1, . . . , m Re[λk ] = 0 implies σ k = 1

λ t

λ t

Since t j −1 e k  is bounded if Re[ λk ] < 0 (for any j), and e k  = 1 if Re [λk ] = 0, it is clear that  You're Reading a Preview bounded for t ≥ 0. Thus (*) is a sufficient condition for uniform stability. A necessary condition for uniform stability is Unlock full access with a free trial. Re[λk ] ≤ 0 , k = 1, . . . , m

For if Re[λk ] > 0 for some k , the proof of Theorem 6.2 shows that Trial e At  grows without bound as t → ∞. The Download With Free between this necessary condition and the sufficient condition is illustrated by the two cases A =

0 0 0 0

,

A=

0 1 0 0

Both satisfy the necessary condition, neither satisfy the sufficient condition, and the first case is uniformly st Master your semester with Scribd  while the second case is not (unbounded solutions exist, as shown by easy computation oftitle the transition mat Read Free Foron 30this Days Sign up to vote  (It can be shown that a necessary and sufficient condition for uniform stability is that each eigenvalue of A & The New York Times  Useful  Not useful nonpositive real part and any eigenvalue of A with zero real part has algebraic multiplicity equal to its geom

Special offer formultiplicity.) students: Only $4.99/month.

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Solutions

 x (t a ) = Φ(t a , t o ) xa  = Φ(t a , t o ) ≤ γ e

−λ(t a −t o )

Since such an xa can be selected for any t o and t a > t o , we have

Φ(t , τ) ≤ γ e −λ(t −τ) for all t , τ such that t ≥ τ, and the proof is complete. . The variable change z (t ) = P −1 (t ) x (t ) yields z (t ) = 0 if and only if . P −1 (t ) A (t )P (t ) − P −1 (t )P (t ) = 0 . for all t . This clearly is equivalent to P (t ) = A (t )P (t ), which is equivalent to Φ A (t , τ) = P (t )P −1 (τ). Now is a Lyapunov transformation, that is P (t ) ≤ ρ < ∞ and det P (t ) ≥ η > 0 for all t , then P (τ)n −1 __________ Φ A (t , τ) ≤ P (t )P −1 (τ) ≤ P (t ) det P (τ)

Solution 6.18

≤ ρn / η =∆ γ for all t and τ. Conversely, suppose Φ A (t , τ) ≤ γ for all t and τ. Let P (t ) = Φ A (t , 0). Then P (t ) ≤ γ and P −1 (t )n −1 ___________ 1 a −Preview = P (t )n −1 det P (t ) P (t ) ≤ You're−Reading 1 det P (t ) Unlock full access with a free trial. 1 for all t . Using P (t ) ≥ 1 / P − (t ) gives 1 __________

det P (t ) ≥ Download With −1 (t )Trial n PFree and since P −1 (t ) = Φ A (0, t ) ≤ γ ,

det P (t ) ≥

1 ___

γ Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title Thus P (t ) is a Lyapunov transformation, and clearly . & The New York Times Useful  Not useful P − (t ) A (t )P (t ) − P − (t )P ( t ) = 0 1

Special offer for students: Only $4.99/month. for all t .

n

1

Cancel anytime.

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

CHAPTER 7

Solution 7.3

Let Aˆ = FA, and take Q = F −1 , which is positive definite since F is positive definite. Then sin

is symmetric,

ˆ T Q +QAˆ = A T FF −1 + F −1 FA = A T + A < 0 A This gives exponential stability by Theorem 7.4.

You're a Preview a (t )Reading By our default assumptions, is continuous. Since Q is constant, symmetric, and pos definite, the first condition of Theorem 7.2 holds. Checking the second condition, Solution 7.5

Unlock full access with a free trial. −a (t ) −a (t ) / 2 A (t )Q +QA (t ) = −a (t ) / 2 −1 T

≤ 0


gives the requirements

a (t ) ≥ 0 , 4a (t ) ≥ a 2 (t )

Thus the state equation is uniformly stable if a (t ) is a continuous function satisfying 0

Master your semester with Scribd Solution 7.6 With & The New York Times a (t ) 0 Q (t ) = Special offer for students: Only $4.99/month.

0

1

≤ a (t ) ≤ 4 for all t .

Read Free Foron 30this Days Sign up to vote title . Not useful  .Useful Cancelaanytime. (t ) 0

, A T (t )Q (t ) + Q (t ) A (t ) + Q (t ) =

0

−4

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Solutions

η ≤ a (t ) ≤ 1 / (2η) for all t . Then 2a (t ) + 1 − η ≥ η + 1 > 1 (t )+1 _a ______ a (t )

1 = 1 + η > 1 − η ≥ 1 + ______ 1 / (2η)

and Q (t )−η I ≥ 0, for all t , follows easily. Similarly, with ρ = (2η+1) / η we can show ρ I −Q (t ) ≥ 0 using 1 2η+1 ρ − 2a (t ) − 1 ≥ _____ − 2 ___ − 1 = 1 η 2η 1 2η+1 a (t )+1 ____ ≥ 1 ρ − _______ ≥ _____ − 1 − η a (t ) a (t ) Next consider . A T (t )Q (t ) + Q (t ) A (t ) + Q (t ) =

. 2a (t )−2a (t ) 0

0

. a (t ) _____ −2a (t )− 2 a (t )

≤ − ν I

This gives that for uniform exponential stability we also need existence of a small, positive constant ν such th . ν a 2 (t ) − 2a 3 (t ) ≤ a (t ) ≤ a (t )− ν / 2


for all t . For example, a (t ) = 1 satisfies these conditions.


Solution 7.11

Suppose that for every symmetric, positive-definite M there exits a unique, sym Download With Free Trial positive-definite Q such that A T Q + QA + 2µQ = − M

that is,

Master your semester with Scribd  ( A + µ I ) Q + Q ( A + µ I ) = − M Read Free Foron 30this Days Sign up to vote title  A + Then byYork the argument above Theorem 7.11 we conclude that all eigenvalues have negative real p µ I useful & The New Times Not  Useful of  That is, if T


Cancel anytime.

0 = det [ λ I − ( A +µ I ) ] = det [ ( λ − µ) I − A ]

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Solutions

∞

∞

∫

A σ σ ≤ ∫ x T Me A σ xa d σ ae

A T σ x T Me A σ xa d ae t

T

0

= x T a Qxa ≤ λ max (Q ) = Q 

Also, using a change of integration variable from σ to τ = σ − t , ∞ ∞ T A T σ A σ A T (t + τ) x a e Me xa d σ = x T Me A(t + τ) xa d τ ae

∫

∫

t

0

A t = x T Qe At xa ≥ λmin (Q )e At xa 2 = ae T

e At 2 _______ Q −1 

Therefore

e At 2 _______ ≤ Q  Q −1  Since t was arbitrary, this gives max e At  ≤ √Q Q −1  t ≥ 0

Solution 7.17

You're Let F = A + (µ−ε) I . Then F  ≤Reading  A +µ−aε,Preview all eigenvalues of F have real parts less tha

and

Unlock full access with a free trial. e Ft = e At e (µ−ε)t


Thus

e At  = e −(µ − ε)t e Ft  By Theorem 7.11 the unique solution of F T Q +QF = − I is ∞ T Q = e F σ e F σ d σ Read Free Foron 30this Days Sign up to vote title

Master your semester with Scribd∫ & The New York Times For any n × 1 vector x , 0

Special offer for students: Only $4.99/month. d ___

x T e F σ e F σ x = x T e F σ [ F T + F ] e F σ x T

T




Useful

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Solutions

x T e F t e Ft x ≤ 2 ( A +µ−ε) x T Qx , t ≥ 0 T

which gives

e Ft  ≤ √2 ( A +µ−ε)Q  , t ≥ 0 Thus the desired inequality follows from (*).

Solution 7.19

To show uniform exponential stability of A (t ), write the 1,2-entry of A (t ) as a (t ), an Q (t ) = q (t ) I , where 2+e −2t , t ≥ 1 / 2 q (t ) = q ⁄ (t ) , −1 / 2 < t < 1 / 2 1

2

3 , t ≤ − 1 / 2

Here q ⁄ (t ) is a continuously-differentiable ‘patch’ satisfying 2 ≤ q ⁄ (t ) ≤ 3 for −1 / 2 < t < 1 / 2, and an condition to be specified below. Then we have 2 I ≤ Q (t ) ≤ 3 I for all t . Next consider . . −2q (t )+q (t ) a (t )q (t . ) ≤ − ν I A T (t )Q (t ) + Q (t ) A (t ) + Q (t ) = a (t )q (t ) −6q (t )+q (t ) 1

1

2

We choose ν = 1 and show that

2

You're Reading a Preview .

−2q (t )+q (t )+1

a (t )q (t ) Unlock full access with a .free trial. ≤ 0 −6q (t )+q (t )+1 a (t )q (t )

. for all t . With t < −1 / 2 or t > 1 / 2 it is easy to show that q (t )−q (t )−1 ≥ 0, and a patch function can be sket Download With Free Trial such that this inequality is satisfied for −1 / 2 < t < 1 / 2. Then, for all t , . . −2q (t )+q (t )+1 ≤ −q (t ) ≤ 0 , −6q (t )+q (t )+1 ≤ −5q (t ) ≤ 0 . . [−2q (t )+q (t )+1][−6q (t )+q (t )+1] − a 2 (t )q 2 (t ) ≥ [5−a 2 (t )]q 2 (t ) ≥ 4q 2 (t ) ≥ 0

Master your semester with Scribd  Thus we have proven uniform exponential stability. Read Free Foron 30this Days Sign up to vote title  To show A (t ) is not uniformly exponentially stable, write the state equation as two scalar equation & The New York Times Useful  Not useful  compute T

Cancel anytime.


Φ

T

(t 0) =

e −t

0

t ≥ 0

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

CHAPTER 8

Solution 8.3

No. The matrix A =

−2 0

√8

−1

has negative eigenvalues, but 8 −4a √Preview You're A + Reading A T =

√8

has an eigenvalue at zero.

−2



Solution 8.6 Viewing F (t ) x (t ) as a forcing term, for any t o , xo , and t ≥ t o we can write t

x (t ) = Φ A +F (t , t o ) xo = Φ A (t , t o ) xo +

Master your semester with Scribd which gives, for suitable constants γ , λ > 0, & The New York Times − −

∫ Φ (t , σ)F (σ) x (σ) d σ A

t o


Useful  Not useful anytime. λ(t t o ) −λ(t −σ) F (σ)Cancel ≤ x t e x + e   ( )        x (σ) d σ γ γ o Special offer for students: Only $4.99/month. t

∫

t o

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

Solutions

t

 x (t ) ≤ γ e

−λ(t −t o )

∫ γ F (σ) d σ

t

 xo 

eo ∞

≤ γ e −λ(t −t ) e o

∫ γ F (σ) d σ

t o

 xo 

≤ γ e −λ(t −t ) e γ β  xo  o

and we conclude the desired uniform exponential stability.

Solution 8.8

We can follow the proof of Theorem 8.7 (first and last portions) to show that the solution ∞ T Q (t ) = e A (t )σ e A (t )σ d σ

∫ 0

of A T (t )Q (t ) + Q (t ) A (t ) = − I

is continuously-differentiable and satisfies, for all t , I ≤ Q (t ) ≤ ρ I You'reηReading a Preview

where η and ρ are positive constants. Then with

Unlock full access with a free . trial. F (t ) = A (t ) − 1⁄ 2Q −1 (t )Q (t )

an easy calculation shows


. F T (t )Q (t ) + Q (t )F (t ) + Q (t ) = A T (t )Q (t ) + Q (t ) A (t ) = − I Thus

. x (t ) = F (t ) x (t ) Master your semester with Scribd is uniformly exponentially stable by Theorem 7.4. & The New York Times

Special offer for students: Only $4.99/month. Solution 8.9 As in Exercise 8.8 we have, for all t ,





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

Solutions

Then the complete solution formula gives t

x (t ) = ΦF (t , t o ) xo +

.

∫ Φ (t , σ) ⁄ Q − (σ)Q (σ) x (σ) d σ 1

F

1

2

t o

and the result of Exercise 8.8 implies that there exists positive constants γ , λ such that, for any t o and t ≥ t o

 x (t ) ≤ γ e

−λ(t −t o )

t

∫

 xo  + γ e −λ(t −σ) t o

βρ2 ____ η

 x (σ) d σ

Therefore t

e  x (t ) ≤ γ e λt

λt o

 xo  +

____ λσ e  x (σ) d σ ∫ _γβρ η 2

t o

and the Gronwall-Bellman inequality (Lemma 3.2) implies t

∫ γβρ / η d σ 2

e  x (t ) ≤ γ e λt

λt o

 xo e

t o

Thus

 x (t ) ≤ γ e

−(λ−γβρ 2 / η)(t −t o )

 xo 


Now, writing the left side as Φ A (t , t o ) xo  and for any t o and t ≥ t o choosing the appropriate unity-norm x −(λ−γβρ 2 / η)(t −t o ) Unlock t o )access Φ A (t ,full  ≤ γ ewith a free trial.

For β sufficiently small this gives the desired uniform exponential stability. (Note that Theorem 8.6 also ca . Free used to conclude that uniform exponential Download stability of x With t ) implies uniform exponential stability o (t ) = F (t ) x (Trial . . x (t ) = [ F (t ) + 1⁄ 2Q −1 (t )Q (t ) ] x (t ) = A (t ) x (t ) for β sufficiently small.)

Master your semester with Scribd  Read Free 30 Days Sign up to vote title . Foron . this  t ) = A With F (t ) = A (t ) + (µ / 2) I we have that F (t ) ≤ α Useful ), and the eigenvalu Solution 8.10 Times & The New York Not(t useful  + µ / 2, F ( F (t ) satisfy Re [λ (t )] ≤ −µ / 2. The unique solution of Cancel anytime.

F


F T (t )Q (t ) + Q (t )F (t ) = − I

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Solutions

∞

− z

T F T (t )τ

e

e

F (t )τ

z =

d ∫ τ ___ d σ

z T e F (t )σ e F (t )σ z T

d σ

∞

≥ −(2α + µ) ∫ z T e F (t )σ e F (t )σ z d σ T

τ

∞

≥ −(2α + µ) ∫ z T e F (t )σ e F (t )σ z d σ T

0

≥ −(2α + µ) z T Q (t ) z Thus e F (t )τ e F (t )τ ≤ (2α + µ) Q (t ) , T

τ ≥ 0

and using e F (t )τ = e A(t )τ e (µ / 2) τ ,

τ ≥ 0

gives

e A(t )τ  ≤ √(2α+µ)ρ e (−µ / 2) τ , τ ≥ 0


Unlock full u access a free trial. Write (the chain rule is valid since (t ) iswith a scalar)

. q (t ) = − A −1 (u (t ))

. . dA _db __ ___ (u (t ))u (t ) (u (t ))u (t ) A −1 (u (t ))b (u (t )) − A −1 (u (t )) Download With Free Trial du du

∆ ˆ . = − B (t )u (t ) Then Master your semester with Scribd . Read Free Foron 30this Days Sign up to vote title x (t ) = A (u (t )) x (t ) + b (u (t )) & The New York Times = A (u (t )) [ x (t ) − q (t ) ] + A (u (t ))q Useful t )) Not useful (t ) + b (u (

Special offer for students: Only $4.99/month. gives

= A (u (t )) [ x (t ) − q (t ) ]

Cancel anytime.

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CHAPTER 9

Solution 9.7 Write B ( A −β I ) B ( A −β I )2 B . . .

=

B

AB −β B

A 2 B −2β AB+ β2 B . . .

I m −β I m 0 I m 2 You're=Reading a Preview . . . 0 0 B AB A B 0 0 Unlock full access with a free trial. . . . . . .

β2 I m . . . −2β I m . . . I m 0 . . .

... ... . . .

Download solution is even easier using rank tests Clearly the two controllability matrices have the sameWith rank.Free ( TheTrial Chapter 13.) eigenvalues, Solution semester 9.8 Since A has negative-real-part Master your with Scribd ∞ Q = ∫ e BB e & The New York Times At


is well defined, symmetric, and

0

T A T t

Read Free Foron 30this Days Sign up to vote title dt  Useful


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

Solutions

0 =

d j ___ dt j

x T e At B

= x T A j B t =0

for j = 0, 1, 2, . . . . But this implies x T B AB . . . A n −1 B

= 0

which contradicts the controllability hypothesis. Thus Q is positive definite.

Solution 9.9 Suppose λ is an eigenvalue of A, and p is a corresponding left eigenvector. Then p ≠ 0, and p T A = λ p T

This implies both

Now suppose Q is as claimed. Then

_ p H A = λ p H ,

A T p = λ p

_ p H AQp + p H QA T p = λ p H Qp + λ p H Qp = − p H BB T p

that is,

You're Reading a Preview 2 Re [λ] p H Q p = − p H BB T p Unlock full access with a free trial.

This gives Re [λ] ≤ 0 since Q is positive definite. Now suppose Re [λ] = 0. Then (*) gives p H B = 0. Also j = 1, 2, . . . , _ Download With Free _Trial j p H A j B = λ p H A j −1 B = . . . = λ p H B = 0 Thus p B AB . . . A − B = 0 Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title which contradicts the controllability assumption. Therefore Re [λ] < 0. & The New York Times  Useful  Not useful H

Special offer for students: Only $4.99/month. Solution 9.10 Let

n 1

Cancel anytime.

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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

Now suppose the state equation is output controllable on [ t o , t f ], but that W y (t o , t f ) is not invert invertible. ible. there exists a p × 1 vector ya ≠ 0 such that y aT W y (t o , t f ) ya = 0. Using by now familiar familiar arguments, this gives

∈ [t o , t f ] y aT C (t f )Φ(t f , t ) B (t ) = 0 , t ∈ Consider the initial state xo = Φ(t o , t f )C T (t f )[ C (t f )C T (t f ) ]−1 ya rank C (t f ) = p. There exists an input ua (t ) such that which is well defined and nonzero since rank C t f

0 = C (t f )Φ(t f , t o ) xo +

∫ C (t )Φ(t , σ) B (σ)u (σ) d σ f

f

a

t o t f

= ya +

∫ C (t )Φ(t , σ) B (σ)u (σ) d σ f

f

a

t o

Premultiplying by y aT gives 0= y aT ya This contradicts y a ≠ 0, and thus W y (t o , t f ) is invertible. The rank assumption on C (t f ) is needed in the necessity proof to guarantee that x o is well well define m = p = 1, invertibility of W W y (t o , t f ) is equivalent to existence of a t a ∈ (t o , t f ) such that C (t f )Φ(t f , t a ) B (t a ) ≠ 0

That is, there exists a t a

Solution 9.11

∈ (t o , t f ) such that the output response at t f to an impulse input at t a is nonzero.

rank C = = p, the state equation is output controllable if and only i From Exercise 9.10, since rank C

some fixed t f > 0, t f

∫ − BB Master your semester with Scribd ∆

A (t f t )

W y = Ce 0

is invertible. We Times will will show this holds if and only if & The New York


rank

T T A (t f −t )

e

C T dt




CB CAB CAB . . . CA n −1 B


Useful =p

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

S o lu t io n s

A (t f −t )

y aT Ce

B = y aT C

n −1

Σ α (t −t ) A B = 0 , k

k f

∈ [0, t f ] t ∈

k =0

Therefore y aT W y ya = 0, which implies that W y is not invertible. For m = p = 1 argue as in Solution 9.10 to show that a linear state equation is output controllable if only if its impulse response (equivalently, transfer function) is not identically zero.

Solution 9.17

Beginning with y (t ) = c (t ) x x (t ) . . . y (t ) = c (t ) x x (t ) + c (t ) x (t ) . = [c (t ) + c (t ) A A (t )] x (t ) + c (t )b (t )u (t ) = L 1 (t ) x (t ) + L 0 (t )b (t )u (t )

it is easy to show by induction that y

(k )

k −1

(t ) = Lk (t ) x (t ) +

Σ

j =0

Now if

k − j −1 __d ____ __ ___ _ 2, . . . [ L j (t )b (t )u (t ) ] , k = 1, 2, k − j −1 dt

__ −1 ∆ Ln (t ) M M =

α0 (t ) α1 (t ) . . . αn −1 (t )

then n −1

Σ α (t ) L L (t ) = i

i

α0 (t ) . . . αn −1 (t )

i =0

L 0 (t ) . . .

= Ln (t )

Ln −1 (t )

Thus we can write Master your semester with Scribd − − d − − _____ __ ____ __ Free For 30 Days Sign t ) ]to vote on this title (t )u (up [ L (t )bRead y (t ) − Σ α (t ) y (t ) = L (t ) x (t ) + Σ − − dt & The New York Times  Useful  Not useful (n )

n 1

i

(i )

n 1

n

i =0


j =0

n j 1

n j 1

j

n −1

n −1

i −1

i 0

i 0

j 0

− Σ αi (t ) Li (t ) x (t ) − Σ αi (t )

Σ

Cancel anytime.

d i − j −1 ______ [ L j (t )b (t )u (t ) ] dt i − j −1

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

CHAPTER 10

Solution 10.2

We show show equivalence equivalence of full-rank full-rank failure failure in the respective respective controllability controllability and observab matrices, and thus conclude that one realization is controllable and observable (minimal) if and only if the oth controllable and observable observable (minimal). (minimal). First, rank

B AB

. . . A n −1 B

if and only if there exits a nonzero, n × 1 vector q such that q T B = q T AB = . . . = q T A n −1 B = 0

This holds if and only if q T B = q T ( A+BC ) B = . . . = q T ( A+BC )n −1 B = 0

which is equivalent to rank

B ( A+BC A+BC ) B

Similarly, semester with Scribd Master your & The New York Times


rank

. . . ( A+BC )n −1 B

C CA . . .


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Solutions

C (t ) B (σ) = H (t )F (σ)

for all t ,

σ, picking an appropriate t o and t f > t o , t f

M x (t o , t f )W x (t o , t f ) =

t f

∫ C (t ) H (t ) dt ∫ F (σ) B (σ) d σ T

T

t o

t o

where the left side is a product of invertible matrices by minimality. Therefore the two matrices on the right are invertible. Let t f

P −1 = M x−1 (t o , t f )

∫ C (t ) H (t ) dt T

t o

Then multiply both sides of (*) by C T (t ) and integrate with respect to t to obtain t f

M x (t o , t f ) B (σ) =

∫ C (t ) H (t ) dt F (σ) T

t o

for all σ. That is, B (σ) = P −1 F (σ)

for all σ. Similarly, (*) gives

You're Readingt f a Preview C (t )W x (t o , t f ) = H (t ) ∫ F (σ) B T (σ) d σ Unlock full access with t o a free trial.

that is,

Download With Free Trial t f C (t ) = H (t ) ∫ F (σ) B T (σ) d σ W x−1 (t o , t f ) t o

Master your semester with Scribd − Read Free Foron 30this Days Sign up to vote title F (σ) B (σ) d σ W − (t , t ) = ∫ C (t ) H (t ) dt M (t , t ) = P ∫ & The New York Times  Useful  Not useful But (**) then gives

t f

T

t o

Special offer for students: Only $4.99/month. so we have

1

t f

x

1

o

T

f

t o

x

o

f

Cancel anytime.

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

Solutions

d ___ X (t ) X (σ) = X (t ) dt

d ___ X (σ) d σ

which implies d ___ X (t ) X (σ) dt

d ___ X (σ) = X (−t ) d σ

Integrate both sides with respect to t from a fixed t o to a fixed t f > t o to obtain t f

∫

d ___ X (t ) dt

d ___ X (σ) = X (−t ) (t f − t o ) d σ t o

dt X (σ)

Now let t f

∫

d ___ X (t ) dt

1 _____ X (−t ) A = t f −t o t o

dt

to write d ___ X (σ) = A X (σ) , X (0) = I d σ

This implies X (σ) = e A σ . (Of course thereYou're are quicker ways. a For example note that Reading Preview

d ∂ X (t+σ) = X (t ) ___ ∂ X (t+σ) = ___ ___ X (σ) Unlock full access with a free trial. d σ ∂σ ∂t . . Evaluating at σ = 0 gives X (t ) = X (t ) X (0) , which implies . Download With Free X . (0) Trial X (0)t t

X (t ) = X (0)e

=e

Also the result holds for continuous solutions of the functional equation, though the proof is much more difficu

Master your semester with Scribd  factorization unreviewed in the text Solution 10.12 If rank G = r we can write (admittedly using a matrix Read Free Foron 30this Days Sign up to vote title  & The New York Times G = C B  Useful  Not useful i

i

i

i

i

Cancel anytime.

Special offer forwhere students: C isOnly p × $4.99/month. r , B is r × m, and both have rank r . Then it is easy to check that i

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

CHAPTER 11

Solution 11.4 Since rank b Ab = rank

1 1

−1 −1

= 1

the state equation is not minimal. It is easy to compute the impulse response:

σ) = C (t )e A (t −σ) B = (t 2 + 1) e −(t −σ) You'rerealization Reading a Preview Then a factorization is obvious, giving a minimal G (t ,

.

x (t full e t u (t with ) =access ) Unlock a free trial. y (t ) = (t 2 + 1)e −t x (t )


For the given impulse response,

(t , σ) = Master your semester withΓScribd  Sign up to vote on this title  & The New York Not useful It is easy to checkTimes that rank Γ (t , σ) = 2 for all t , σ, and a little more calculation that rank Γ (t ,  Useful  shows 22

22

1+e 2t / 2+e 2σ / 2 e 2σ 0 Free For 30 Days e 2t Read Cancel anytime.

a minimal realization is, using formulas in the proof of Theorem 11.3, Special offer forThen students: Only $4.99/month.

33

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

Solutions

Γ =

1 1 1 . . .

1 1 1 . . .

1 ... 1 ... 1 ... . . . . . .

and clearly the rank condition in Theorem 11.7 is satisfied with l = k = n = 1. Then, following the pro Theorem 11.7, F = F s = F c = F r = H 1 = H s1 = 1

and a minimal (dimension-1) realization is . x (t ) = x (t ) + u (t ) y (t ) = x (t )

For the truncated sequence,

Γ =

1 1 1 0 . . .

1 1 0 0 . . .

1 0 0 0 . . .

0 0 0 0 . . .

... ... ... ... . . .

You're Reading a Preview The rank condition in Theorem 11.7 is satisfied with l = k = n = 3. Taking Unlock full access with a free trial.

1 1 1 1 1 0 1 1 0 , F s = H s3 = 1 0 0 Download With Free Trial0 0 0 1 0 0

F = H 3 =

F c =

1 1 1 , F r =

Master your semester with Scribd gives a minimal realization specified by & The New York Times 0 1 0 Special offer for students: Only $4.99/month. A = F s F −1 =

0 0 1 0 0 0

1 1 1

, B=

Read Free Foron 30this Days Sign up to vote title 1 1 1




Useful

, C=

1 0 0

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

Solutions

Gi −1 Gi . . .

= α0

Gn −2+i . . .

G0 G1 . . .

Gi −2 Gi −1 . . .

+ . . . + αi −2

Gn −1 . . .

Gn −3+i . . .

By ignoring the top entry, this linear combination shows that column i +1 is given by the same linear combin of the i −1 columns to its left, and so on. Thus by the rank assumption on Γ there cannot exist such an i , an first n columns of Γ are linearly independent. A similar argument shows that the first n columns of Γ n linearly independent, for every j ≥ 0, and thus that Γnn is invertible. It remains only to show that the given A, B, C provides a realization for G (s ), since minimality is immediate. Premultiplication by Γnn verifies

Γ −nn1

Gk . . .

= ek +1 , k = 0, . . . , n −1

Gn +k −1

Then, since A = Γ snn

Γ−nn1 , A

Gk . . . Gn +k −1

Now, CB = G 0 , and


Gk +1 . a free trial. Unlock access . = Γ snnfull ek +1 = with , k = 0, . . . , n −1 . Gn +k


CA j B = CA j −1 A

G0 . . .

Master your semester with Scribd G − & The New York Times G n 1


j

= . . . = C

. . .

= CA j −1

G1 . . . Gn up Read Free Foron 30this Days Sign to vote title




Useful

= G j , j = 1, . . . , n

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

CHAPTER 12

Solution 12.1

If the state equation is uniformly bounded-input, bounded-output stable, then it is clear from definition that given δ we can take ε = η δ. Now suppose the ε, δ condition holds. In particular we can take δ = 1 and assume ε is such that, for any

u (t ) ≤ 1 , t ≥ t o implies

You're Preview (t ) ≤ ε , at ≥ t o  y Reading

t o with sup Now suppose u (t ) is any bounded input Unlock signal.full Given let µa = u (t ). Note µ > 0 can be assumed access free trial. t ≥ t o otherwise we have a trivial case. Then u (t ) / µ ≤ 1 for all t ≥ t o , and the zero-state response to u (t ) satisfie t Download With Free Trial  y (t ) =  ∫ G (t , σ)u (σ) d σ t o t

= µ  ∫ G (t , σ)u (σ) / µ d σ Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title & The New York Times  Not useful ≤ µε = ε sup u (t ) ≥ t , t Useful t o

Special offer for students: Only $4.99/month. Thus we have

t ≥ t o

Cancel anytime. o

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

Solutions

rank

B AB . . . A n −1 B

=n

Then given δ we can take δ

ε = λmin

∫ e

e − A δ

T T BB T e A τ d τ e − A δ

A τ

0

For a time-varying example, take scalar a (t ) = 0, b (t ) = e −t / 2 . Then W (t −δ, t ) = e −t (e δ −1)

Given any δ > 0, W (t −δ, t ) > 0 for all t , but there exists no ε > 0 such that W (t −δ, t ) ≥ ε

for all t .

Solution 12.9

Consider a scalar state equation . x (t ) = b (t )u (t ) y (t ) = x (t )

where b (t ) is a ‘smooth bump function’ described as follows. It is a continuous, nonnegative function that is You're Reading a Preview for t ∉ [0, 1], and has unit area on [0, 1]. Then for any input signal the zero-state response satisfies Unlock full 1access with a free trial. y (t ) ≤ b (σ)u (σ) d σ

∫ 0

for any t . Thus for any t o and any t ≥ t o ,

Download With Free Trial 1

∫

 y (t ) ≤ b (σ) d σ . sup u (t ) t ≥ t o

Master your semester with Scribd ≤ sup u (t ) ≥ & The New York Times 0

t t o





Useful

 

the state equation is uniformly bounded-input, bounded-output stable with η = 1. However if we consi Special offer forand students: Only $4.99/month. bounded input that is continuous and satisfies

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Solutions

u (t ) ≤

ε ___ , t ≥ T 2 2ρ

Let T = 2 max [T 1 , T 2 ]. Then for t ≥ T , t

∫

 y (t ) ≤ G (t −σ)u (σ) d σ 0 t

T

ε ___ G (t −σ) d σ = µ G (t −σ) d σ + 2ρ T 0

∫

∫

Changing the variables of integration gives t

ε ___  y (t ) ≤ µ G (τ) d τ + 2ρ t −T

∫

t −T

∫ G (τ) d τ 0

ε

ε ___ ρ =ε + ≤ µ ___ 2µ

2ρ

This shows that y (t ) → 0 as t → ∞.

Solution 12.11

The hypotheses imply that given ε > 0 there exist δ1 ,

δ2 > 0 such that if You're Reading a Preview  xo  < δ1 ; u (t ) < δ2 , t ≥ t o Unlock full access with a free trial.

where u (t ) is n × 1, then the solution of . x (t ) = A (t ) x (t ) + u (t ) , x (t o ) = xo


satisfies

 x (t ) < ε , t ≥ t o

In particular, with x = 0, this shows that if u (t ) < δ for t ≥ t , then the corresponding zero-state solutio Master your semester with Scribd  the state equation Read Free Foron 30this Days Sign up to vote title  . x (t ) = A (t ) x (t ) + u (t  ) Useful  Not useful & The New York Times o

2

Special offer for students: Only $4.99/month. satisfies  ( )

for

o

Cancel anytime.

y (t ) = x (t )

But this implies uniform bounded-input, bounded

ut stabilit

by Exe

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Solutions

∞

∞

∫

y (t )e −η t dt =

0

=

∞

t

∫ ∫

G (t −σ)e −λ σ d σ

0

0

∞

∞

∞

∫ ∫ G (t −σ)e −λ σ d σ 0

∫ ∫ G (t −σ)e −η 0

e −η t dt =

t

e −η t dt

0

dt e −λ σ d σ

0

where all integrals are well-defined because of the stability assumption, and λ , η > 0. Changing the varia integration in the inner integral from t to γ = t −σ gives ∞ ∞ ∞ η t − y (t )e dt = G (γ )e −η γ d γ e −η σ e −λ σ d σ

∫

∫ ∫

0

0

0

∞

=

G(s ) s = η

∫ e − η λ σ d σ ( + )

0

=

1 _____ G(η) η+λ

Without the stability assumption we can say that U (s ) = 1 / (s+ λ) for Re [s ] > −λ, and the integral for converges for Re [s ] > Re [ p 1 ], . . . , Re [ pn ], where p 1 , . . . , pn are the poles of G (s ). Thus ∞ G (s ) _____ = y t )e −st dt (Preview Y (s ) = Reading You're a s+ λ 0

∫

Unlock full access with a free trial. is valid for Re [s ] > −λ, Re [ p 1 ], . . . , Re [ p n ]. This implies that if

then

η > −λ, Re [ p 1 ], . . . , Re [ pn ] Download With Free Trial ∞

G (η) ∫ y (t )e −η dt = _____ η+λ t

Master your semester with Scribd even though y (t ) may be unbounded. & The New York Times 0





Useful

 

Special offer forSolution students: 12.14 Only $4.99/month. Given u (t ), t ≥ 0, and xo , suppose x (t ) is a solution of the given state equation. Then v (t ) = y (t ) = C x (t ) we have

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

Solutions

we get t

x (t ) − z (t ) =

∫ α (t −σ) Bu (σ) d σ 0

0

Then . w (t ) = −(CB )−1 CAP z (t ) − (CB )−1 CAB (CB )−1 C x (t ) + (CB )−1 C x (t ) = −(CB )−1 CAP z (t ) − (CB )−1 CAB (CB )−1 C x (t ) + (CB )−1 CA x (t ) + (CB )−1 CBu (t ) = −(CB )−1 CAP z (t ) + (CB )−1 CA [ − B (CB )−1 C + I ] x (t ) + u (t ) = (CB )−1 CAP [ x (t ) − z (t ) ] + u (t ) t

= (CB )−1 CAP

∫ α (t −σ) Bu (σ) d σ + u (t ) 0

0

Again using PB = 0 gives w (t ) = u (t ) , t ≥ 0

To address stability, since PB = 0 we see that P is not invertible. Thus AP is not invertible, which implie second state equation is never exponentially stable. The scalar case with A = −1, B = C = 1 is unifo bounded-input, bounded-output stable, but the resulting You're Reading a Preview . z (t ) = −v (t ) Unlock full access with. a free trial. w (t ) = v (t ) + v (t ) is not, as the bounded input v (t ) = cos (e t )Download shows. With Free Trial



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

CHAPTER 13

Solution 13.1 Suppose n = 2 and A has complex eigenvalues. Let A =

a 11 a 12 a 21 a 22

b=

,

b1 b2

Then A has eigenvalues


a 11 +a 22 ± √(a 11 +a 22 )2 −4(a 11 a 22 −a 12 a 21 ) ____________________________________ Unlock full access 2 with a free trial.

and since the eigenvalues are complex,


(a 11 +a 22 )2 − 4(a 11 a 22 −a 12 a 21 ) = (a 11 −a 22 )2 + 4a 12 a 21 < 0 Supposing that det [ b Ab ] = 0, we will show that if b ≠ 0 we get a contradiction. For Ab ] = a b − a 0 = det [ bScribd Master your semester with implies & The New York Times (a −a ) b b = (a 21


11

22

2

2 2 1 2

2 1

2 12 b 2

21

− (a 11 −a 22 )b 1 b 2


 Useful  Not useful b 2 −a b 2 )2 Cancel anytime. 1

12

 

2

If b 1 = 0, b 2 ≠ 0, then (**) implies a 12 = 0, which contradicts (*). If b 1 ≠ 0, b 2 = 0, then (**) implies a

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Solution 13.4



Solutions

We need to show that rank

B AB . . . A n −1 B

=n,

rank

if and only if the ( n +p )-dimensional state equation . A 0 B z (t ) = z (t ) + C 0 D

A B C D

= n+p

u (t )

is controllable. First suppose (+) holds but (++) is not controllable. Then there exists a complex so such that so I n − A 0 −C so I p

rank

B D

< n+p

Since rank [ so I − A B ] = n, this implies rank

−C

so I p D

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


rank

Cancel anytime.

o n

−C

so I p

D

< n +p

so = 0

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

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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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. 






Useful



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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

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


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˜


1

1


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

Cancel anytime.

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

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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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


A + BK =

2





Useful

A 11 +B 1 K 1 A 12 +B 1 K 2 0 A 22

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

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 )


= [ 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.

Read Free Foron 30this Days w (t ) up Sign to vote title ) z (t Useful   Not useful Cancel anytime.

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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.)


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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

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


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.

Cancel anytime.

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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

Cancel anytime.

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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

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


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

η 1

ρ

η

ρ

η 1

ρ 1

secondOnly equation gives Special offer forThe students: $4.99/month.

P

P

Q Q −1

η

Cancel anytime.

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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 ,


˜ ] ≤ . . . ≤ 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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

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,


(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





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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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),

Master your semester with Scribd A ⊂ A ( A − & The New York Times Therefore A ⊂ . V

V

W


1


∩ Im [ A ]  Useful  Not useful Cancel anytime.

W ) = W

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

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


2

o

N

N

N

o

N

Cancel anytime.

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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


Ker

z T = Ker

T

z , x T



 Not useful Cancel anytime. for all z ∈ S Useful

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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 ) ⊥ .


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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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

Master your semester with Scribd Let K = −u . . . − u & The New York Times 1


q

Special offer forThen students: Only $4.99/month.

( E + BK )

= Ew

BK

−1 w1 . . . w q Useful

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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


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


12 b


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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

FT


T u (k )T

Read Free Foron 30this Days Sign up to vote title τ τ sgn [u ( e F  G d Useful k )] Not useful

 

Cancel anytime.

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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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 )


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,

Special offer for students: Only $4.99/month. However, for j 1 ≥ k

T

T

T

Cancel anytime.

. . . A (−k+ 2) A (−k+1)

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

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 )


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


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

Cancel anytime.

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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.


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.

Read Free Foron 30this Days Sign up to vote title T

∫

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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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.


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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

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.

You're Reading a Preview Unlock full access with a free trial.




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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


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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

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 λk = k ( e lnλ ) , k ≥ 0 Let η = −lnλ, so that η > 0 since λ < 1. Then max k λk ≤ 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


________ ∆ = β , k ≥ 0 Download −e lnWith λ Free Trial

k λk  ≤

1

To get a decaying exponential bound, write

k λk = 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

Special offer for students: Only $4.99/month. For j > 1 write

k =0


2β ≤ _______ 1− √λ 


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

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


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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

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

=


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

Special offer forgives students: Only $4.99/month.

2

−1

−1

Read Free Foron 30this Days Sign0 up to vote title 1  −Useful  Not useful Cancel anytime.

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

Solutions

p H A T QAp − p H Qp = − p H Mp That is,

( λ2 −1 ) p H Qp = − p H Mp If p H Mp > 0, then λ2 −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


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

Master your semester with Scribd x (k +1) = & The New York Times

0 1 / 9 Free Foron 30this Days Sign to vote title ) up x (Read k 4 0




Useful

eigenvalues are ± 2 / 3, so the state equation is uniformly stable, but clearly (#) fails. Special offer forThe students: Only $4.99/month.

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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 ]

Master your semester with Scribd ≤ γ e  z (k ) , & The New York Times This implies uniform stability. γα

Special offer for students: Only $4.99/month. For the scalar example

o

j =k o

Read ≥ k o upFree k ≥ Foron 30this Days Sign to vote title 


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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,

Master your semester with Scribd M (k , k ) x = O (k , k ) & The New York Times o


f

o

T

o

f

y (k o ) . up Read Free Foron 30this Days Sign to vote title . .  Useful  Not useful y (k f f −1) Cancel anytime.

If M M (k k ) is invertible, then the initial state is uniquely determined by

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

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

Master your semester with Scribd Therefore & The New York Times y C (k )Φ(k , j +1) B ( j ) = 0 ,

2

j =k o


T a

f

f


 Not useful . . , . ,Cancel j = k o , .Useful k f −1anytime.

But by output reachability, with y f = ya , there exists an input u a (k ) such that

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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


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.

Cancel anytime.

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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)

=


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

Special offer for students: Only $4.99/month. has the transfer function

1 n

Cancel anytime.

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

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 )


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

z (k +1) = Fz (k ) + gu (k ) Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title y (k ) = hz (k ) & The New  Useful  Not useful and we York then have Times


Cancel anytime.

. adj ( zI − A ) . b h . adj ( zI −F ) . g ______________ _c____________ = det ( zI F ) det ( zI A )

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

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


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 )


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

ij


Cancel anytime.

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

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






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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


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


W 11 = (V 11 −V 12 V −221 V 21 )−1

1 22

21

11

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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 . . . . . . . . .


Similar to Solution 2.8. Unlock full access with a free trial.

Solution 28.8


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


ˆ

N = [C ( I − A − BK )−1 B ]−1 Λ

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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 )


full access with a free trial. ∞. Now Therefore e b (k ) → 0 exponentially as k →Unlock

∆ ˆ (k ) e (k ) = x (k ) − x


= 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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Linear System Theory 2 e Sol

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