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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 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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theory. 2 edition.
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Khalil - Nonlinear
Systems Slides
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Linear System
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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 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
(a) For k = 2, (A + B )2 = A 2 + AB + BA + B 2 . If AB = BA, the
then (A + B )2 = A 2 + 2AB + B 2 . In ge
k
AB = BA, then the k-fold product ( A + B ) can be written as a sum of terms of the form A j B k − j , j = 0, . . . ,
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
A j B k− j
(b) Write
det
[λ I − A (t )] = λn + an −1 (t )λn −1 + . . . + a 1 (t )λ + a 0 (t )
where invertibility of A (t ) implies a 0 (t ) ≠ 0. The Cayley-Hamilton theorem implies
A n (t ) + an −1 (t )A n −1 (t ) + . . . + a 0 (t )I = 0
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a (t )I − . . . − a − (t )A −Read
A
(t ) −up
(tFor
) on
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A − (t ) =
a (t )  Useful  Not useful
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for all t . Multiplying through by A −1 (t ) yields
1
1
n 1
0
n 2
n 1
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all t . Since
0 (t ) = det [ −A (t )], a 0 (t ) = det A (t ). Assume ε > 0 is such that  det A (t ) ≥ ε for all t
A (t ) ≤ α we have aij (t ) ≤ α, and thus there exists a γ such that a j (t ) ≤ γ for all t. Then, for all t
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Linear System Theory, 2/E

Solutions
__________
_
det (λ I − A ) = det (λ I − A ) = det (λ I − A )
_
H
H
(e) α A has eigenvalues α λ1 , . . . , αλn since Ap = λp 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 TA =
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 k1 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 )
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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)
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1 0
0 0
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det (λI − A T A ) = (λ − 16)(λ − 4)
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Linear System Theory, 2/E
Solution 1.6

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
Since this holds for any
x  = 1
A x 
_______
x 
αA x 
A α x 
__________
________
= max
x  = 1/α αx 
α x 
α ≠ 0,
A x 
A x 
_______
_______
= max
x≠0
x 
x  ≠ 0 x 
A  = max
Therefore
A  ≥
for any x
A x 
_______
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
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A 
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Khalil - Nonlinear
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Linear System Theory, 2/E

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 
You're Reading a Preview
Since Q = Q T , Q T Q = Q 2 , and the eigenvalues of Q 2 are λ 21 , . . . ,
Solution 1.10
λ 2n . Therefore
access2 with a free trial.
Q  = full
Unlock
(Q ) = max λi 
√λmax
1≤i≤n
For the other equality Cauchy-Schwarz gives
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x T Qx |
≤ x T Q x  = Qx x 
≤ Q x 2 = [ max λi  ] x T x
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x . Choosing x
Thereforesemester
| x Qx | ≤ Q  forwith
all unity-norm
the eigenvalue that yields max λ  gives
≤ ≤
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T
1
i
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Thus

T
Qx 
Q 
n
i
1≤i≤n
a
as a unity-norm eigenvector of Q correspondi
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x Ta Qxa  = x Ta [ max λi  ] xa = max λi 
1≤i≤n

1≤i≤n
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Linear System Theory, 2/E
Solutions
max x T A T A x =
x  = 1
so we have A  =
Solution 1.12

λmax (A T A )
√λmax (A T A ) .
Since A T A > 0 we have λ i (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 λ i (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,
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y T A x  ≤ y T A x 
≤ y 
A Free
x  =Trial
A 
Download
With
Therefore
max
y T A x  ≤ A 
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x be suchwith
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that A xScribd
 = A , and let
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x , y  = 1
a
a
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a
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Linear System Theory, 2/E

Solutions
example
A (t ) =
t 0
0 t2
,
A (t ) =
t, 0≤t
≤1
t2 , 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
You're Reading a Preview ≠ 0,
0 < λmin (Q ) x T x ≤ x T Q x ≤ λ max (Q ) x T x
Clearly Q is positive definite, and by Rayleigh-Ritz if x
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λ min (Q ) (respectively, λ max (Q )) shows that these inequalitie
Choosing x as an eigenvector corresponding to
tight. Thus
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ε1 ≤ λmin (Q ) , λmax (Q ) ≤ ε2
Therefore
1
1
λmin (Q −1 ) = _______ ≥ ___
λmax (Q ) ε2
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Thus Rayleigh-Ritz for the positive definite matrix Q −1 gives
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Linear System Theory, 2/E
Solution 1.17

Solutions
Using the product rule to differentiate A (t ) A −1 (t ) = I yields
.
d
___
A −1 (t ) = 0
A (t ) A −1 (t ) + A (t )
dt
which gives
.
d
___
A −1 (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 T
d
___
___
x (t ) x (t ) =  x (t ) x (t ) + x T (t ) x (t ) = 2x T (t ) x (t )
x (t )2  = 
dt
dt
.
a Preview
≤ 2x (tYou're
)x (t )Reading

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For t such that x (t ) ≠ 0, comparing these expressions
gives with a free trial.
d
___
.
x (t ) ≤ x (t )

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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
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Solution 1.19 To prove the contrapositive claim, suppose for each i, j there is a constant βij such that
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Linear System Theory, 2/E
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, tb
tb
tb
ta
ta
∫ Q (σ) d σ = ∫ λ
max [Q (
≥ ta ,
tb
tb
ta
ta
σ)] d σ ≤ ∫ tr [Q (σ)] d σ = tr ∫ Q (σ) d σ
Note that
tb
You're Reading
≥ 0Preview
Q (σ ) d σ a
∫
ta
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since for every x
tb
tb
ta
ta
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x ∫ Q (σ ) d σ x = ∫ x T Q (σ ) x d σ ≥ 0
T
Thus, using a property of the trace on page 8 of Chapter 1, we have
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d σvote
) d σ ≤ tr ∫ Q (σ) d σ ≤ n Read
) to
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Khalil - Nonlinear
Systems Slides
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
CHAPTER 2
Solution 2.3
.
The nominal solution for ũ (t ) = sin (3t ) is ỹ (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
You're Reading a Preview
0
0
1
x δ (t ) +
u δ (t )
2
−4 sin
−1trial.
/3
0
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x δ (t ) =
y δ (t ) =
where
x δ (t ) = x (t ) −
sin t
cos t
,
1
0 x δ (t )
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u δ (t ) = u (t ) − sin (3t ) ,
y δ (t ) = y (t ) − sin t , x δ (0) = x (0) −
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Solution
2.5 For
ũ = 0 constant nominal solutions are solutions 
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0 = x̃ 2 − 2x̃ 1 x̃ 2 = x̃ 2 (1−2x̃ 1 )
0
1
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
rank A = rank [ A b ].
Also, x̃ is a constant nominal with c x̃ = 0 if and only if
0 = A x̃ + bũ
0 = c x̃
that is, if and only if
−bũ
A
x̃ =
c
0
As above, this holds if and only if
A
c
rank
= rank
A b
c 0
Finally, x̃ is a constant nominal with c x̃ = ũ if and only if
0 = A x̃ + bũ = ( A + bc ) x̃
and this holds if and only if
x̃
∈ Ker [ A + bc ]
(If A is invertible, we can be more explicit. For any ũ the unique constant nominal is x̃ = −A −1 bũ. Then ỹ
Reading
−1.)
ũ ≠ 0 if and only if c A −1 b = 0, and ỹ = ũ ifYou're
and only
if c A −1 ba=Preview
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Solution 2.8
(a) Since
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A B
C 0
is invertible, for any K
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C
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A + BK B
A B
C 0
R1 R2
I 0
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Linear System Theory, 2/E
Solution 2.10

Solutions
For u (t ) = ũ, x̃ is a constant nominal if and only if
0 = (A + Dũ ) x̃ + bũ
This holds if and only if bũ
∈ Im [ A + Dũ ], that is, if and only if
rank ( A + Dũ ) = rank
A+Dũ
bũ
If A +Dũ is invertible, then
x̃ = −(A + Dũ )−1 bũ
If A is invertible, then by continuity of the determinant det (A + Dũ ) ≠ 0 for all ũ such that ũ  is suffici
small, and (+) defines a corresponding constant nominal. The corresponding linearized state equation is
.
x δ (t ) = (A + Dũ ) x δ (t ) + [ b − D (A + Dũ )−1 bũ ] 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
You're
Reading
x̃ (t ) =
x̃ 1 (t ) − x̃ 3 (t ) a, Preview
x̃ (0) = −3
−2
x̃ (t ) − 2 x̃ (t )
2
3
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1 = x̃ 2 (t ) − 2 x̃ 3 (t )
Integrating for x̃ 1 (t ) and then x̃ 3 (t ) easilyDownload
gives the nominal
solution
With Free
Trialx̃ 1 (t ) = t, x̃ 2 (t ) = 2 t − 3, and x̃ 3 (t )
The corresponding linearized state equation is specified by
A=
0 0 0
1 0 −1
0 1 −2
, B (t )=
0
t
0
, C=
0 1
−2
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Khalil - Nonlinear
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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
σk
∫τ A (σ ) ∫τ A (σ ) ∫τ
1
...
2
σ2
t
∫τ A (σ
k +1 )
d σk +1 . . . d σ1
σk
τ
τ
d
_
. . . dσ
a__
Preview
t − A (τ) A (σ2 ) . . . d σk +1 . . . d σ2
A (t ) A (σ2 ) . . . A (σk +1You're
) d σk +1 Reading
2
dτ
τ
τ
τ
τ
τ
∫
=
∫
∫
∫
∫
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t
∫τ
+ A (σ 1 )
t
∂
___
∫τ A (σ ) ∫τ
∂τ
∂
___
σ2
σ1
2
d σk +1 . . . d σ1
∫τ A (σk +1 ) With
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...
σ2
σ1
∂
τ ∫τ
semester
∫τ
= A (σ1 )
σk
A (σ 2 )
∫τ
σk
...
∫τ A (σ
k +1 )
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∂
___
∂τ
1
1
σk
2
∫ A (σ ) ∫ A (σ ) ∫
2
d σk +1 . . . d σ1
...
∫ A (σ
k +1 )
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Khalil - Nonlinear
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Linear System Theory, 2/E
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 + t2
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 + t2
The complete solution formula gives, with some help from Mathematica,
.
1
________
x 2o +
x 2 (t ) =
(1 + t 2 )2
t
∫
0
(1 + σ2 )3/ 2
_________
d σ x 1o
(1 + t 2 )2
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.
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Solution 3.7
From the hint, letting
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t
r (t ) = v (σ)φ(σ) d σ
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.
we have r (t ) = v (t )φ(t ), and
φ(t ) ≤ ψ (t ) + r (t )
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v (t ) gives
Multiplying
(*) through by thewith
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Khalil - Nonlinear
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Linear System Theory, 2/E

Solutions
t
∫ v (τ) d τ
t
eo
gives
t
∫ v (τ) d τ
t
r (t ) ≤
∫
v (σ)ψ(σ)e σ
dσ
to
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
≥ to
i =1 j =1
At each t
≥ to let
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a (t ) = 2n 2 max aij (t )
1 ≤ i, j ≤ n
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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 ≤ to
z (t )2 ≤ aWith
(t )z (Free
dt
Multiplying through by the positive quantity
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− ∫ a (σ) d σ
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t
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d
___
dt
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e
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z (t )2
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Khalil - Nonlinear
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Linear System
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Linear System Theory, 2/E

Solutions
z (t ) =
t
t t
to
to
∫ A (σ) z (σ) d σ + ∫ σ∫ E (τ, σ) d τ z(σ) d σ
t
=
t
∫
∫σ
A (σ) + E (τ, σ ) d τ
to
∆
=
z (σ) d σ
t
∫ Â(t, σ) z (σ) d σ
to
Thus
t
t
∫
∫
z (t ) =  Â (t, σ) z (σ) d σ ≤ Â (t, σ)z (σ) d σ
to
to
By continuity, given T > 0 there exists a finite constant
α such that Â (t, σ) ≤ α for to ≤ σ ≤ t ≤ to + T. Thus
t
∫
z (t ) ≤ α z (t ) d σ , t ∈ [to , t o +T ]
to
and the Gronwall-Bellman inequality gives z (t ) = 0 for t
than one solution.
∈ [to , to +T ], implying that there can be no m
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Solution 3.13
From the Peano-Baker series,
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σ1
σk −1
t
t
.
.
.
.
.
.
. . . dσ
Φ(t, τ) − I + A (σ1 ) dDownload
+ With
A (σ1Free
)
σ1 +
1
Trial A (σk ) d σk
∫τ
σ1
t
∞
=
∫τ
Σ ∫τ A (σ ) ∫τ
∫τ
∫τ
σ j −1
...
∫τ
A (σ j ) d σ j . . . d σ1
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Σ ∫τ
j =k +1
1
∫τ
σj−1
∫τ
A (σ j ) d σ j . . . d σ1  ≤
∞
t
1
Σ ∫τ A (σ ) ∫τ
1
j =k +1
Cancelσanytime.
j−1
...
∫τ
A (σ j ) d σ j . . . d σ1 
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Khalil - Nonlinear
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Linear System
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
 Search document
Linear System Theory, 2/E
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)!
Because of the factorial in the denominator, given
Solution 3.15
(α2T )k +1
1
__________________
_________
=
(k −1)!(k +1)(k −α2T )
1 − α2T/k
ε > 0 there exists a K > α2T such that (*) holds.
Writing the complete solution of the state equation at t f , we need to satisfy
tf
Ho x o + H f
Φ(t f , t o ) xo + ∫ Φ(t f , σ ) f ( σ) d σ
=h
to
Thus there exists a solution that satisfies the boundary conditions if and only if
tf
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to
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h − H f ∫ Φ(t f , σ ) f ( σ) d σ ∈ Im[ Ho + H f Φ (t f , t o ) ]
There exists a unique solution that satisfies the boundary conditions if Ho + H f
a solution x (t ) satisfying the boundary conditions:
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(1) Compute Φ(t, to ) for t
(2) Compute Ho + H f
Φ(t f , to ) is invertible.
To com
∈ [to , t f ]
Φ(t f , to )
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(3) Compute ∫ Φ(t , σ) f ( σ) d σ
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Khalil - Nonlinear
Systems Slides
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1
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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 ) =
This A (t ) commutes with its integral, so we can write
− 2t − 1
1 − 2t
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
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Solution 4.4
A linear state equation corresponding to the n th -order differential equation is
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.
x (t ) =
...
...
.
.
.
...
0
0
−a 0 (t ) −a 1 (t ) . . .
0
0
.
.
.
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The corresponding
adjoint state equation is
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0
1
0
.
.
.
...
0
0
0
.
.
.
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x (t )
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Khalil - Nonlinear
Systems Slides
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Linear System
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Linear System Theory, 2/E

Solutions
.
zn −2 (t ) = −zn −3 (t ) + an −3 (t ) zn (t )
gives
.
d2
d
d3
____
___
____
[ 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 ) −
d2
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
dn
_____
_____
____
[ 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 (to ) = XTo
This has the unique solution X T (t ) = ΦA T (t )You're
(t, to )X ToReading
, so that a Preview
T
X (full
t ) =access
Xo Φ TAwith
t o ) trial.
(t ) (ta, free
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In the second matrix differential equation, let
to verify (Leibniz rule) that a solution is
Φk (t, τ) be the transition matrix for
Ak (t ), k = 1, 2. Then it is
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t
X (t ) = Φ1 (t, t o )Xo Φ
T
2 (t,
to) +
∫ Φ (t, σ)F (σ)Φ (t, σ) d σ
1
T
2
to
Or, one can generate this expression by using the obvious integrating factors on the left and right sides of
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Zon
t )this
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Khalil - Nonlinear
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Linear System
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Linear System Theory, 2/E

Solutions
Replacing τ as noted above gives
Φ (t,
=
cos τ 0
0 cos τ
=
−sin τ
−sin τ
sin τ
0
cos τ sin τ
cos τ
0).
Solution 4.10
For sufficiency, suppose
differentiable. Let z (t ) = T −1 (t ) x (t ) so that
Φz (t,
0
+
Φ x (t,
0) = T (t )e Rt .
Then T (0) = I and T (t ) is continu
0) = T −1 (t )Φx (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,
0) = e
Ra t
= P −1 (t )Φx (t, 0)P (0)
that is,
Φx (t,
0) = P (t )e
Ra t
P −1 (0)
You're Reading a Preview
Let T (t ) = P (t )P −1 (0) and R = P (0)Ra P −1 (0). Then
Φ x (t,
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(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)
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= T (t )e Rt
Solution semester
4.11 Suppose
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A2t
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___
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Khalil - Nonlinear
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Linear System
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Linear System Theory, 2/E
Solutions
d
___
dt
we have that
Φ (t,
Solution 4.13
0) = e

A1t A2t
e
e
A1t
e
A2t
= A (t )e
A1t A2t
e
, e
A10
e
A20
=I
.
Writing
∂
___
∂t
ΦA (t, τ) = A (t )ΦA (t, τ ) , Φ(τ, τ) = I
in partitioned form shows that
∂
___
∂t
Φ21 (t, τ) = A 22 (t )Φ21 (t, τ ) , Φ21 (τ, τ) = 0
Thus Φ21 (t, τ) is identically zero. But then
∂
___
∂t
Φii (t, τ) = Aii (t )Φii (t, τ ) , Φii (τ, τ) = I
for i = 1, 2, and
∂
___
∂t
Φ12 (t, τ) = A 11 (t )Φ12 (t, τ ) + A 12 (t )Φ22 (t, τ) , Φ12 (τ, τ) = 0
Using Exercise 4.6 with F (t ) = A 12 (t ) Φ22 You're
(t, τ) gives
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t
∫τ
with(σ
a)free
(t, σ ) A
(σ, τ) d σ
τ) = full
Φ12 (t,Unlock
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Φ22trial.
12
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Solution 4.17
We need to compute a continuously-differentiable, invertible P (t ) such that
t 1
1 t
.
0
1
P (t ) − P −1 (t )P (t )
2
2− t 2 t
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Multiplying on the left by P (t ), the result can be written as a dimension-4 linear state equation. Choos
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initial condition corresponding to P (0) = I, some clever guessing gives
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P (t ) =
1 0
t 1
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Linear System Theory, 2/E
Solutions
∂
___
∂t
Since Φ(−τ,

 Search document
Φ TA (−τ, −t ) = A T (−t )Φ TA (−τ, −t )
−τ) = I, we have F (t ) = A T (−t ).
Or we can use the result of Exercise 3.2 to compute:
∂
___
∂t
∂ Φ (−τ, −t ) = Φ (−τ, −t )A (−t )
ΦA (−τ, −t ) =− _____
A
∂ (−t ) A
This implies
∂
___
∂t
Since Φ(−τ,
Φ TA (−τ, −t ) = A T (−t )ΦA (−τ, −t )
−τ) = I, we have F (t ) = A T (−t ).
Solution 4.25
We can write
t+ σ
Φ(t + σ, σ ) = I +
∫σ
A (τ) d τ +
∞
τ1
t +σ
Σ ∫σ
τk −1
∫σ
∫σ
A (τ1 ) A (τ2 ) . . .
k =2
A (τk ) d τk . . . d τ1
and
_k
_
∞a Preview
You're Reading
1
___
At (σ)t k
e
= I + At (σ)t +
_
At (σ)t
Σ
!
k =2 akfree
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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
∫
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k
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Systems Slides
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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
e
At
=I+
=
Using the Laplace transform method,
0 t3
t2 0
0 t
1
1
___
___
+
+
t3 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
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full access with=a free
trial.
t
t
t
t
−
−
(e −e )/ 2 (e +e )/ 2
sinh t cosh t
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(sI − A )−1 =
s −1 −1
=
−1 s
1
_____
2
s −1
s
_____
s 2 −1
s
_____
2
s −1
1
_____
2
s −1
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cosh t sinh t
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e At =
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sinh t cosh t
Using the diagonalization method, computing eigenvectors for A and letting
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Khalil - Nonlinear
Systems Slides
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Linear System Theory, 2/E

Solutions
t
Φ (t,
∫ A (σ) d σ
0) = e 0
t2/2 t
t t2/2
= exp
And since
t2 / 2 0
0 t2/2
,
0 t
t 0
t2/2
. exp
commute,
Φ(t,
1 0
0 1
0) = exp
0 1
1 0
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
A Reading
e A σ d σ = eaAtPreview
−I
∫
0
Unlock full access with a free trial.
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 Aσ d σ = − I
∫
0
that is,
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−
A =−∫e
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Linear System
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Linear System Theory, 2/E
Solutions
t
.
x (t ) = bu (t ) + A
t
∫e
A (t −σ)
D
e
∫σ u (τ) d τ
t
t
∫
bu (σ) d σ + Du (t ) e A (t −σ) e
0
D
∫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
βk (t )’s:
Σ
k =0
You're Reading a Preview
.
λ1 β0 (with
P 0Unlock
t ) =access
t ) a free trial.
: β0 (full
.
P 1 : β1 (t ) = β0 (t ) + λ2 β1 (t )
.
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With Free Trial
.
. .
Pn −1 : βn −1 (t ) = βn −2 (t ) + λn βn −1 (t )
that is,
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λ 0
β. (t )
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1
0
2
1
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.
.
.
.
.
0
0
...
...
.
.
.
...
0
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.
.
.
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Linear System Theory, 2/E
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 (to ) = P −1 (t )P (to ) e
S (t −to )
P −1 (to )P (to )
S (t −to )
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
You're Reading a Preview
+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
= Φ(t +T, TDownload
)e −St = Φ(t, With
0)e −StFree 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.)
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Solution 5.22 The solution will be T -periodic for initial state x if and only if x satisfies (see text equ
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Linear System Theory, 2/E

Solutions
to +T
0=
∫
to +T
z To
Φ(to , σ ) f (σ) d σ =
to
∫
z T (σ) f ( σ) d σ
to
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
T
(σ) f ( σ) d σ = 0
0
holds vacuously. Thus there will exist corresponding periodic solutions.
Case 2: If ω = 1, then
T
You're
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T Aσ
z
f
d
=
z
(
)
(
)
σ
σ
σ
∫
∫ o e f ( σ) d σ
T
T
0 full access with a free trial.
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0
T
T
= −zo 1 ∫ sin (σ) d σ + zo 2 ∫ cos σ sin σ d σ
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0
0
2
≠0
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so there is no periodic solution.
Case 3: If ω = 1/k, k = 2, 3, . . . , then since
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T
T
0
0
the condition (+) will hold, and there exist periodic solutions.
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
CHAPTER 6
Solution 6.1
If the state equation is uniformly stable, then there exists a positive
the corresponding solution satisfies
γ such that for any t o
x (t ) ≤ γxo  , t ≥ to
Given a positive ε, take δ = ε / γ. Then, regardless of to , xo  ≤ δ implies
x (t ) ≤ γ δ = ε , t ≥ to
You're Reading a Preview
Conversely, given a positive ε suppose positive δ is such that, regardless of to , xo  ≤ δ implies x (
t
≥ to . For any ta ≥ to let xa be such that
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xa  = 1 , Φ(ta , to )xa  = Φ(ta , to )
With
FreeatTrial
Then xo = δ xa satisfies xo  = δ, and the Download
corresponding
solution
t = ta satisfies
x (ta ) = Φ(ta , to )xo  = δΦ(ta , to ) ≤ ε
Therefore
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Φ(t , t ) ≤ ε / δ
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a
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for all t and t with t
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, and we can take
a
o
o
Φ(t, t o ) ≤ ε / δ
/ δ to obtain
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Linear System Theory, 2/E

Solutions
t
σ1
t
∫τ
∫τ
Φ(t, τ) = I + A (σ) d σ + A (σ1 )
t
∫τ A (σ ) d σ d σ
2
∫τ
2
2
1
σ1
t
∫τ
+ . . . 
∫τ A (σ ) d σ d σ  + . . .
= I  +  A (σ) d σ +  A (σ1 )
t
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
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Solution 6.8
See the proof of Theorem 15.2.
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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
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Linear System Theory, 2/E

Solutions
∞
∫
0
∞
j +( j −1)+ . . . +1
j
_2___________
e −(η / 2 )t dt
t e  dt ≤
j
(η e )
0
.
.
.
j +( j −1)+
+1
2j
. ___
≤ _2___________
η
(η e ) j
...
22 j +( j −1)+ +1
_____________
= j
e Re [λ] j +1
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
Wkj
k =1 j =1
where λ 1 , . . . ,
t j −1
λt
______
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
λt
λt
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
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bounded for t ≥ 0. Thus (*) is a sufficient condition
for uniform
stability.
A necessary condition for uniform stability is
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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
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between this necessary condition and the sufficient condition is illustrated by the two cases
A=
0 0
0 0
,
A=
→ ∞. The
0 1
0 0
Both satisfy
the necessary condition,
neither
satisfy the sufficient condition, and the first case is uniformly st
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while the second case is not (unbounded solutions exist, as shown by
easy
computation
oftitle
the transition
mat
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Linear System Theory, 2/E

Solutions
x (ta ) = Φ(ta , to )xa  = Φ(ta , t o ) ≤ γ e
−λ(ta −to )
Since such an xa can be selected for any to and ta > to , 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
__________
ΦA (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 ΦA (t, τ) ≤ γ for all t and τ. Let P (t ) = ΦA (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 )
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1
for all t . Using P (t ) ≥ 1/P − (t ) gives
1
__________
det P (t ) With
≥ Free
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n
P −1 (t )Trial
and since P −1 (t ) = ΦA (0, t ) ≤ γ,
det P (t ) ≥
1
___
γ
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.
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Khalil - Nonlinear
Systems Slides
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
CHAPTER 7
Solution 7.3
Let  = FA, and take Q = F −1 , which is positive definite since F is positive definite. Then sin
is symmetric,
T
 Q +Q = A T FF −1 + F −1 FA = A T + A < 0
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
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−a (t ) −a (t )/ 2
A (t )Q +QA (t ) =
−a (t ) / 2 −1
T
≤0
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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
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1
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(t ) 0
, A T (t )Q (t ) + Q (t )A (t ) + Q (t ) =
0
−4
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Linear System Theory, 2/E

Solutions
η ≤ a (t ) ≤ 1/ (2η)
for all t . Then
2a ( t ) + 1 − η ≥ η + 1 > 1
(t )+1
_a
______
a (t )
and Q (t )−ηI
1
= 1+η > 1
− η ≥ 1 + ______
≥ 0, for all t, follows easily.
1/ (2η)
Similarly, with ρ = (2η+1)/ η we can show ρI −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 a 2 (t ) − 2a 3 (t ) ≤ a (t ) ≤ a (t )−ν/2
ν such th
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for all t . For example, a (t ) = 1 satisfies these conditions.
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Solution 7.11
Suppose that for every symmetric, positive-definite M there exits a unique, sym
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positive-definite Q such that
A T Q + QA + 2µQ = −M
that is,
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(A + µ I ) Q + Q (A + µ I ) = −M
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Then byYork
the argument
above Theorem 7.11 we conclude that all eigenvalues
have negative real p
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0 = det [ λI − (A +µ I ) ] = det [ ( λ − µ)I − A ]
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Linear System Theory, 2/E
Solutions
∞
∫
∞
T
x Ta e A σ Me A σ xa
dσ ≤
t
∫x e
T A Tσ
Me A σ xa
a
dσ
0
= x Ta Qxa
≤ λmax (Q ) = Q 
Also, using a change of integration variable from σ to τ = σ − t,
∞
∞
T
T ATσ
Aσ
xa e
Me xa d σ = x Ta e A (t + τ) Me A(t + τ) xa d τ
∫
t
∫
0
T
= x Ta e A t Qe At xa
At
e 
≥ λmin (Q )e At xa 2 = _______
−1
2
Q 
Therefore
e At 2
_______
Q −1 
≤ Q 
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
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e Ft = e At e (µ−ε)t
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Thus
e At  = e −(µ − ε)t e Ft 
By Theorem 7.11 the unique solution of F T Q +QF = −I is
∞
T
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d
___
T
T
x T e F σ e Fσ x = x T e F σ [ F T + F ] e F σ x
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Linear System Theory, 2/E

Solutions
T
x T e F t e Ft x
≤ 2 (A +µ−ε) x T Qx ,
t
≥0
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. ) ≤ −νI
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
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.
−2q (t )+q (t )+1
a (t )q (t )
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−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
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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
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Thus we have proven uniform exponential stability.
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To show A (t ) is not uniformly exponentially stable, write the state equation as two scalar equation
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T
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Φ
T
(t 0) =
e −t
0
t
≥0
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Systems Slides
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
CHAPTER 8
Solution 8.3
No. The matrix
A=
−2
0
√8
−1
has negative eigenvalues, but
8
−4a √Preview
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A + Reading
AT =
√8
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has an eigenvalue at zero.
Solution 8.6
−2
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Viewing F (t )x (t ) as a forcing term, for any t o , xo , and t
≥ to we can write
t
x (t ) = ΦA +F (t, t o ) xo =
ΦA (t, t o ) xo + ∫ ΦA (t, σ)F (σ) x (σ) d σ
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which gives, for suitable constants γ, λ > 0,
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to )
xo  + γ e −λ(t −σ) F (σ)x (σ) d σ
t
∫
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Linear System Theory, 2/E

Solutions
t
x (t ) ≤ γ e
−λ(t −to )
∫ γF (σ) d σ
t
xo 
eo
∞
≤ γ e −λ(t−t ) e
o
∫ γF (σ) d σ
to
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
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. trial.
F (t ) = A (t ) − 1⁄2Q −1 (t )Q (t )
an easy calculation shows
Download With Free Trial
.
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 )
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Solution 8.9 As in Exercise 8.8 we have, for all t,
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Linear System Theory, 2/E

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
to
and the result of Exercise 8.8 implies that there exists positive constants γ,
x (t ) ≤ γ e
−λ(t −to )
t
∫
xo  + γ e −λ(t −σ)
to
βρ2
____
η
λ such that, for any to and t ≥ to
x (σ) d σ
Therefore
t
e x (t ) ≤ γ e
λt
λto
xo  +
____ λσ
e x (σ) d σ
∫ _γβρ
η
2
to
and the Gronwall-Bellman inequality (Lemma 3.2) implies
t
∫ γβρ /η d σ
2
e x (t ) ≤ γ e
λt
λto
xo e
to
Thus
x (t ) ≤ γ e
−(λ−γβρ 2 /η)(t −to )
xo 
You're Reading a Preview
Now, writing the left side as ΦA (t, to )xo  and for any to and t ≥ to choosing the appropriate unity-norm x
−(λ−γβρ 2 /η)(t −to )
Unlock
t o )access
ΦA (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
.
used to conclude that uniform exponential Download
stability of x With
t ) implies uniform exponential stability o
(t ) = FFree
(t ) x (Trial
.
.
x (t ) = [ F (t ) + 1⁄2Q −1 (t )Q (t ) ] x (t ) = A (t ) x (t )
for β sufficiently small.)
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t ) =Not
A (tuseful
With F (t ) = A (t ) + (µ / 2)I we have that F (t ) ≤ α Useful
), and the eigenvalu
Solution
8.10 Times
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 + µ / 2, F (
F (t ) satisfy Re [λ (t )] ≤ −µ / 2. The unique solution of
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F T (t )Q (t ) + Q (t )F (t ) = −I
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Linear System Theory, 2/E
Solutions
∞
−z
T F T (t )τ
e
e
F (t )τ
z=
d
∫τ ___
dσ
zTe F
T
(t )σ
e F (t )σ z
dσ
∞
≥ −(2α + µ) ∫ z T e F
T
(t )σ
e F (t )σ z d σ
T
(t )σ
e F (t )σ z d σ
τ
∞
≥ −(2α + µ) ∫ z T e F
0
≥ −(2α + µ) z T Q (t ) z
Thus
eF
T
(t )τ
e F (t )τ
≤ (2α + µ) Q (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
You're Reading a Preview
Solution 8.11
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u (t ) iswith
Write (the chain rule is valid
since
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 ))
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Free
Trial
du
du
.
∆
= −B̂ (t )u (t )
Then
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x (t ) = A (u (t )) x (t ) + b (u (t ))
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t )) Not useful
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Khalil - Nonlinear
Systems Slides
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CHAPTER 9
Solution 9.7
Write
B (A −βI )B (A −βI )2 B . . .
=
B
AB −βB
A 2 B −2βAB+ β2 B . . .
2
You're=Reading
B AB aAPreview
B ...
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Im
0
0
0
.
.
.
−βIm β2 Im
I m − 2β I m
0
0
.
.
.
Im
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
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Q = ∫ e BB e
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is well defined, symmetric, and
0
T A Tt
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Khalil - Nonlinear
Systems Slides
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Linear System Theory, 2/E
Solutions
0=
dj
___
dt j
x T e At B
= x TA j B
t =0
for j = 0, 1, 2, . . . . But this implies
B AB . . . A n −1 B
xT
=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
p TA =
This implies both
≠ 0, and
λ pT
_
p HA =
λ p H , A T p = λp
Now suppose Q is as claimed. Then
_
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
2Re [λ] p H Q p = −p H BB T p
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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
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which contradicts the controllability assumption. Therefore Re [λ] < 0.
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Solution 9.10 Let
n 1
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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 [ to , t f ], but that W y (to , t f ) is not invert
invertible.
ible.
there exists a p × 1 vector ya ≠ 0 such that y aT Wy (to , t f )ya = 0. Using by now familiar
familiar arguments, this gives
y aT C (t f )Φ(t f , t )B (t ) = 0 , t
∈ [t o , t f ]
Consider the initial state
xo =
Φ(to , t f )C T (t f )[ C (t f )C T (t f ) ]−1 ya
which is well defined and nonzero since rank C (t f ) = p. There exists an input ua (t ) such that
tf
0 = C (t f )Φ(t f , t o ) xo +
∫ C (t )Φ(t , σ)B (σ)u (σ) d σ
f
f
a
to
tf
= ya +
∫ C (t )Φ(t , σ)B (σ)u (σ) d σ
f
f
a
to
Premultiplying by y aT gives
0= y aT ya
This contradicts y a ≠ 0, and thus Wy (to , 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 Wy (to , t f ) is equivalent to existence of a ta ∈ (to , t f ) such that
C (t f )Φ(t f , t a )B (ta ) ≠ 0
That is, there exists a ta
Solution 9.11
∈ (to , t f ) such that the output response at t f to an impulse input at ta is nonzero.
From Exercise 9.10, since rank C = p, the state equation is output controllable if and only i
some fixed t f > 0,
tf
−
∫
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∆
Wy = Ce
0
A (t f t )
is invertible.
We Times
will
will show this holds if and only if
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rank
BB T e
A T (t f −t )
C T dt
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CB CAB
CAB . . . CA n −1 B
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Khalil - Nonlinear
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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
y aT Ce
A (t f −t )
B = y aT C
n −1
Σ α (t − t ) A B = 0 ,
k
k
f
t
∈ [0, t f ]
k =0
Therefore y aT Wy ya = 0, which implies that Wy 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 (t )
.
.
.
y (t ) = c (t )x (t ) + c (t )x (t )
.
= [c (t ) + c (t )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 =
α0 (t ) α1 (t ) . . . αn −1 (t )
then
n −1
Σ α (t )L (t ) =
i
i
α0 (t ) . . . αn −1 (t )
i =0
L 0 (t )
.
.
.
= Ln (t )
Ln −1 (t )
Thus we can write
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− d −−
_____
__ ____
__
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(t )u (up
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y (t ) − Σ α (t )y (t ) = L (t )x (t ) + Σ
−
−
dt
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n 1
i
(i )
n 1
n
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j =0
n j 1
n j 1
j
n −1
n −1
i −1
i 0
i 0
j 0
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Σ
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d i − j −1
______
[ L j (t )b (t )u (t ) ]
dt i − j −1
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Khalil - Nonlinear
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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
if and only if there exits a nonzero, n
B AB
. . . A n −1 B
<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 )B
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rank
. . . (A+BC )n −1 B
C
CA
.
.
.
<n
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Linear System Theory, 2/E

Solutions
C (t )B (σ) = H (t )F (σ)
for all t ,
σ, picking an appropriate t o and t f > to ,
tf
Mx ( t o , t f ) W x ( t o , t f ) =
∫C
tf
T
(t )H (t ) dt
to
∫ F (σ ) B (σ ) d σ
T
to
where the left side is a product of invertible matrices by minimality. Therefore the two matrices on the right
are invertible. Let
tf
P −1 = M −x 1 (to , t f )
∫C
T
(t )H (t ) dt
to
Then multiply both sides of (*) by C T (t ) and integrate with respect to t to obtain
tf
Mx (to , t f )B (σ) =
∫C
T
(t )H (t ) dt F (σ)
to
for all σ. That is,
B (σ) = P −1 F (σ)
for all σ. Similarly, (*) gives
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C (t )Wx (to , t f ) = H (t ) ∫ F (σ ) B T (σ ) d σ
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C (t ) = H (t ) ∫ F (σ)B T (σ) d σ W −x 1 (to , t f )
to
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Khalil - Nonlinear
Systems Slides
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Linear System Theory, 2/E

Solutions
d
___
X (t ) X (σ ) = X (t )
dt
d
___
X (σ)
dσ
which implies
d
___
X (t ) X (σ)
dt
d
___
X (σ) = X (−t )
dσ
Integrate both sides with respect to t from a fixed to to a fixed t f > to to obtain
tf
∫
d
___
X (t )
dt
d
___
X (σ) = X (−t )
(t f − to )
dσ
to
dt X (σ)
Now let
1
_____
A=
t f −to
tf
∫ X (−t )
d
___
X (t )
dt
to
dt
to write
d
___
X (σ) = A X (σ) , X (0) = I
dσ
This implies X (σ) = e A σ . (Of course thereYou're
are quicker
ways. a
For
example note that
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d
∂ X (t+σ) = X (t ) ___
∂ X (t+σ) = ___
___
X (σ)
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dσ
∂σ
∂t
.
.
Evaluating at σ = 0 gives X (t ) = X (t )X (0), which implies
.
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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
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factorization
unreviewed
in the text
Solution 10.12 If rank G = r we can write (admittedly using a matrix
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r , B is r × m, and both have rank r . Then it is easy to check that
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This is the solution manual of Rugh W.J Linear system
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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
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Then a factorization is obvious, giving a minimal
G (t,
.
x (tfull
e t u (t with
) =access
)
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y (t ) = (t 2 + 1)e −t x (t )
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Solution 11.7
For the given impulse response,
(t, σ ) =
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It is easy
to checkTimes
that rank Γ (t, σ) = 2 for all t , σ, and a little more
calculation
that rank Γ (t,
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22
22
1+e 2t / 2+e 2σ / 2 e 2σ
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e 2t
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a minimal
realization is, using formulas in the proof of Theorem 11.3,
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Khalil - Nonlinear
Systems Slides
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Table of Contents
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Linear System
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Linear System Theory, 2/E

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 = Fc = Fr = 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
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1 1 1
1 1 0
1 1 0 , Fs = H s3 = 1 0 0
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1 0 0
F = H3 =
Fc =
1 1 1 , Fr =
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0 1 0
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0 0 1
0 0 0
1
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1
, B=
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Linear System Theory, 2/E

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
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= with
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.
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CA j B = CA j −1 A
G0
.
.
.
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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 ≥ to
implies
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(t ) ≤ ε , at ≥
to
y Reading
to with
sup
Now suppose u (t ) is any bounded input Unlock
signal.full
Given
let µa =
u (t ). Note µ > 0 can be assumed
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t ≥ to
otherwise we have a trivial case. Then u (t ) / µ ≤ 1 for all t ≥ to , and the zero-state response to u (t ) satisfie
t
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y (t ) =  ∫ G (t, σ)u (σ) d σ
to
t
= µ  ∫ G (t, σ)u (σ)/µ d σ
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Linear System Theory, 2/E

Solutions
rank
B AB . . . A n −1 B
=n
Then given δ we can take
δ
ε = λmin
e −A δ
∫e
Aτ
T
T
BB T e A τ d τ e −A δ
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
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for t ∉ [0, 1], and has unit area on [0, 1]. Then for any input signal the zero-state response satisfies
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y (t ) ≤ b (σ)u (σ) d σ
∫
0
for any t. Thus for any to and any t
≥ to ,
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1
∫
y (t ) ≤ b (σ) d σ . sup u (t )
t ≥ to
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the state
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is uniformly bounded-input, bounded-output stable with η = 1. However if we consi
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bounded input that is continuous and satisfies
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Khalil - Nonlinear
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Linear System Theory, 2/E
Solutions
u (t ) ≤
Let T = 2 max [T 1 , T 2 ]. Then for t
ε
___
, t ≥ T2
2ρ
≥ 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
This shows that y (t ) → 0 as t
Solution 12.11
where u (t ) is n
2ρ
→ ∞.
The hypotheses imply that given ε > 0 there exist
δ1 , δ2 > 0 such that if
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xo  < δ1 ; u (t ) < δ2 , t ≥ to
× 1, then the solution of
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.
x (t ) = A (t ) x (t ) + u (t ) , x (to ) = xo
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satisfies
x (t ) < ε , t ≥ to
In particular, with x = 0, this shows that if u (t ) < δ for t ≥ t , then the corresponding zero-state solutio
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y (t ) = x (t )
But this implies uniform bounded-input, bounded
ut stabilit
by Exe
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Linear System Theory, 2/E
Solutions
∞
∫
∞
0
∞
t
∫ ∫
y (t )e −η t dt =
0
0
∞
∞
G (t −σ)e −λ σ d σ
0
∞
∫ ∫ G (t −σ)e −λ σ d σ
0
∫ ∫ G (t −σ)e −η
=
e −η t dt =
t
e −η t dt
0
e −λ σ d σ
dt
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
∞
∞ ∞
ηt
−
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 dσ
( + )
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 )
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Linear System Theory, 2/E

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
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w (t ) = v (t ) + v (t )
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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
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a 11 +a 22 ± √(a 11 +a 22 )2 −4(a 11 a 22 −a 12 a 21 )
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and since the eigenvalues are complex,
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(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
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If b 1 = 0, b 2
≠ 0, then (**) implies
11
22
2
2 2
1 2
2
1
2
12 b 2
21
− (a 11 −a 22 )b 1 b 2
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1
12
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2
a 12 = 0, which contradicts (*). If b 1
≠ 0, b 2 = 0, then (**) implies a
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Linear System Theory, 2/E
Solution 13.4

Solutions
We need to show that
rank
B AB . . . A n −1 B
if and only if the ( n +p )-dimensional state equation
.
A 0
z (t ) =
C 0
=n,
A B
C D
rank
B
D
z (t ) +
= n+p
u (t )
is controllable. First suppose (+) holds but (++) is not controllable. Then there exists a complex so such that
so In −A 0
−C so Ip
rank
Since rank [ so I −A
B
D
< n+p
B ] = n, this implies
rank
−C
so Ip D
−A
−C
0 B
0 D
<p
In turn, this implies so = 0, so that (*) becomes
rank
< n+p
and this contradicts the second rank condition in (+).
Conversely, supposing (++) is controllable, then
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2
...
B AB A B
. . . = n+p
D full
CBaccess
CAB
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rank
This implies
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B AB
rank
A
B
=n
in other words, the first rank condition in (+) holds. Now suppose
< n+p
rank
C D
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o n
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so Ip
D
< n +p
so = 0
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Linear System Theory, 2/E
Solution 13.10

Solutions
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 rk 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 rk 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, rk −rk −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 rk ’s are generated by left-to-right column se
in [ BG ABG . . . A n −1 BG ].
Solution 13.11
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p T A = p Tλ , p T B = 0
implies p = 0, then
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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 [ to , t
f ]. Thus g
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z (t ) = [ A (t ) + B (t )K (t ) ] z (t ) + B (t )v (t )
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Linear System Theory, 2/E
Solutions
B̂b̂ =
0
0
.
.
.
0
1
Using × to denote various unimportant entries, set
B̂b̂ = Bo Rb̂ = block diagonal
0
0
.
.
.
ρi × 1 , i = 1, . . . , m
1
0
.
.
.
.
0
1
×
1
.
.
.
0 0
0 0
...
...
.
.
.
...
...
×
×
.
.
.
b̂ =
×
1
0
0
.
.
.
0
1
This gives a set of equations of the form
m
0 = b̂ 1 +
Σ × b̂
i
i
i =2
m
0 = b̂ 2 +
Σ × b̂
i
i
i =3
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0 = b̂m −1 + × b̂m
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Clearly there is a solution for the entries of b̂, 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 = b̂. For an example to show that b alone does not suffice, take Exercise 13.11 with all ×’s zero.
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Linear System Theory, 2/E

Solutions
P −1 (P−a 1 APa )P =
à 11
à 13
à 21 à 22 à 23
0
CPa P =
0
0
, P −1 (P−a 1 B ) =
à 33
B̃ 1
B̃ 2
0
˜ 0 C˜
C
1
2
˜ = Cˆ . It is easy to see that the state equation formed
where à 11 is l × l, and in fact à 33 =  22 , C
2
2
˜
C 1 , Ã 11 , B̃ 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
˜ e
C
1
A˜ 11 t
B̃ 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
=
˜ Ã
C
1 11
.
.
.
˜ Ã n −1
C
1 11
n −1
B̃ 1 Ã 11 B̃ 1 . . . Ã 11 B̃ 1
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˜
C
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1
˜
C 1 Ã 11
˜ . . . A˜ n −1 B̃
.
B˜ 1 A˜ 11 B
1
11
1
.
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.
˜ Ãn −1
C
1 11
rank
=s
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.
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
CHAPTER 14
Solution 14.2
For any t f > 0,
tf
W=
∫ e−
At
BB T e −A
T
t
dt
0
is symmetric and positive definite by controllability, and
tf
You're
Reading
a−AtPreview
d
___
T −A T t
T
∫
e BB e
AW + WA = −
dt
0
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=−e
−Atf
BB T e
−A T tf
dt
+ BB T
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Letting K = −B T W −1 , we have
(A + BK )W + W (A + BK )T = − ( e
−Atf
−A T tf
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Suppose λ is an eigenvalue of A +BK. Then λ is an eigenvalue of ( A+BK
)Free
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weon
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
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Linear System Theory, 2/E

Solutions
Solution 14.5
(a) For any n
× 1 vector x,
x H (A + A T ) x = x H A x + x H A T x
≥ − 2αm x H x
If λ is an eigenvalue of A, and x is a unity-norm eigenvector corresponding to λ, then
_
H T
A x = λ x , x A = λ xH
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−
(A +α I )t
T
BB T e −(A +α I )t dt
0
Reading a Preview
Clearly Q is positive semidefinite. If x QxYou're
= 0, then
T
(A +α I )t
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x T e −full
B =with
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positive definite.
(c) Now consider the linear state equation
.
z (t ) = ( A+ α I −BB T Q −1 )z (t )
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(d) Invoking Lemma 14.6 gives that
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Linear System Theory, 2/E
Solutions
A+BK +Bbk = A +B (K+bk )
has the specified characteristic polynomial. Thus for the original state equation, the feedback law
u (t ) = (K+bk ) x (t )
yields a closed-loop state equation with specified characteristic polynomial.
Solution 14.8
Without loss of generality we can assume the change of variables in Theorem 13.1 has
performed so that
A=
where A 11 is q
A 11 A 12
0 A 22
, B=
B1
0
× q, and
λ I −A 11
rank
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
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K = [ K 1 0 ] we have that
A 11 +B 1 K 1 A 12
0
A 22
A + BK =
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A + BK =
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A 11 +B 1 K 1 A 12 +B 1 K 2
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A 22
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Linear System Theory, 2/E

Solutions
tr [A+BLC ] = tr [A ] + tr [BLC ]
= tr [A ] + tr [CBL ]
= tr [A ]
>0
Thus at least one eigenvalue of A+BLC has positive real part, regardless of L.
Solution 14.12
Write the k th -row of G(s ) in terms of the k th -row Ck of C as
Ck (sI − A )−1 B =
∞
Σ C A Bs −
k
j
( j+1)
j =0
The k th -relative degree
κk is such that, since L Aj [Ck ](t )B (t ) = Ck A j B,
κ −2
Ck B = . . . = Ck A k B = 0
Ck A
κk −1
B≠0
Thus in the k th -row of G(s ), the minimum You're
difference
betweenathe
numerator and denominator polynomial deg
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Preview
among the entries Gk1 (s ), . . . , Gkm (s ) is κk .
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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
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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 )
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= [A −HC ]w (t )
.
z (t ) = Fz(t ) +GCw (t )
Thus the closed-loop state equation in matrix form is
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A −HC
w (t )
=
.
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z (t )
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Linear System Theory, 2/E
Solutions
τ+δ
∫τ
τ+δ
B (σ
)2
dσ ≤ γ
2
∫τ
Φ(τ, σ )B (σ)B T (σ)ΦT (τ, σ ) d σ
≤ 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
dσ ≤
B (σ)2 d σ
k
τ+( j +1)δ
j =0
τ+j δ
≤Σ
∫
B (σ)2 d σ
≤ (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.)
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Write the given state equation
in the partitioned form
.
za (t )
A 11 A 12
za (t )
B1
=
+
.
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TrialB
A
A
t
(
)
21
22
b
2
zb (t )
Solution 15.6
y (t ) =
Ip
c
2
u (t )
za (t )
zb (t )
0
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A −H A 
A ] z (t )
) = [ A − H A ] z (t ) + [ B − HB ]u (t ) + [ A 
)HNot
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22
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Linear System Theory, 2/E

Solutions
Y ( s ) = Ip
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
Solution 15.10
Similar in style to Solution 14.8.
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
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 =
A+BL JC 2 Unlock full
BLH
BLJD 21
access with a free trial.
, B̂ =
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
= DFree
L JDTrial
D LH , D̂
1
1
21
These expressions can be rewritten using
L = (I − JD 22 )−1 = I + J (I −D 22 J )−1 D 22
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which follows from Exercise 28.2 or is easily verified using the identity
Exercise
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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 ) = Ip
Na (s ) A (s ) + Da (s )B (s ) = Ip
Since D −1 (s )N (s ) = D −a1 (s )Na (s ), Na (s ) = Da (s )D −1 (s )N (s ). Substituting this into (**) gives
(s ) A (s ) +aDPreview
Da (s )You're
D −1 (s )NReading
a (s )B (s ) = Ip
that is,
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N (s ) A (s ) + D (s )B (s ) = D (s )D −a1 (s )
Similarly, N (s ) = D (s )D −a 1 (s )Na (s ), and substituting
into
(*) gives
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Na (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.
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Solution 16.5 From the given equality,
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and since N (s ) and D (s ) are right coprime there exist polynomial matrices X (s ) and Y (s ) such that
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Khalil - Nonlinear
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Linear System Theory, 2/E

Solutions
I
0
−[X (s )B (s )+Y (s )A (s )]
I
gives
X (s ) Y (s )
NL (s ) DL (s )
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 )
=I
That is
X (s ) Y (s )
NL (s ) DL (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 )
NL (s ) DL (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
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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
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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 s ρ + . . . + P ) (Q s η + . . . + Q )
ρ
η
0
0
= P ρ Q η s η+ρ + (P ρ Q η−1 +P ρ−1 Q η )s η+ρ−1 + . . .
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Linear System Theory, 2/E
Solutions
Dk (s ) = D̃ 1 (s )u 1,k (s ) + . . . + D̃J (s )uJ,k (s ) + . . . + D̃m (s )um,k (s ) , k = 1, . . . , m
Using a similar column notation for D hc and D l (s ) gives
D hc
k s
ck [D ]
˜]
hc c 1 [D
+ D lk (s ) = [D̃ 1 s
˜
l
hc c [D ]
l
+D̃ 1 (s )] u 1,k (s ) + . . . + [D̃ J s J +D̃ J (s )] uJ,k (s )
˜
hc c [D ]
l
+ . . . + [D̃ m s m +D̃ m (s )] um,k (s ) , k = 1, . . . , m
We claim that
c k [D ] =
max
j = 1, . . . , m
{ c j [D̃ ]+degree u j,k (s ) }
hc
hc
This is shown by a an argument using linear independence of D̃ 1 , . . . , D̃ m as follows. Let
c̃ =
and let µ j,k be the coefficient of s
term on the right side is
˜]
c̃ −c j [D
max
j = 1, . . . , m
{ c j [D̃ ]+degree u j,k (s ) }
in u j,k (s ). Then not all the
m
Σµ
µ j,k are zero, and the vector coefficient of th
hc
j,k
D̃ j
j =1
By linear independence this sum is nonzero, which implies ck [D ] = c̃.
Now, using the definition of J,
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ck [D ] < cJ [D̃ ] ≤ . . . ≤ cm [D̃ ] , k = 1, . . . , J −1
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and this implies uJ,k (s ) = . . . = um,k (s ) = 0. Thus U (s ) has the form
Ua (s )
U b (s )
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U (s ) =
0(m−J+1) × J Uc (s )
where U a (s ) is (J −1) × J, from which rank U (s ) ≤ m −1 for all values of s . This contradicts unimodularity,
cJ [D ] = cJ [D̃ ]. The proof is complete since the roles of D (s ) and D̃ (s ) can be reversed.
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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 ) +aCPreview
v (t )
You're
w (t ) = B T z (t )
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is a realization for G (s ) since
T
T
) ] = [ C (sIWith
− 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
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X T (s ) N T (s ) + Y T (s ) D T (s ) = I
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which implies that N T ( ) and D T ( ) are right coprime. Also, since D ( ) is row reduced, D T ( ) is column redu
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Linear System Theory, 2/E

Solutions
Therefore
SB To [ sI − A To − Q −1 VBTo ]
−1
= D −1 (s )ΨT (s )
Using the definition of N (s ),
−1
D −1 (s )N (s ) = SBTo [ sI − (A To + Q −1 VBTo ) ] Q −1 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
dσ = 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
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(s ) D
−so )−1 = 0
hC (sI −Unlock
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+ hN
uo (strial.
that is,
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(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
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hN (sOnly
o ) = 0, that is rank N (so ) < p < m, which implies that so is a transmission zero.
Conversely, suppose s
is a transmission zero that is not a pole of G (s ). Then for a right-
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Linear System Theory, 2/E
Solution 17.9

Solutions
Using a coprime right polynomial fraction description
G (s ) = N (s )D −1 (s ) =
N (s ) adj D (s )
____________
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 ) = Im
for all s. Therefore
[ X (s )G (s ) + Y (s ) ] D (s ) = Im
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
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det [ X (so )G (so ) + Y (so ) ] = ∞
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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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Systems Slides
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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
+KerTrial
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
Conversely, V ⊂ A −1 W implies, using (a),
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Khalil - Nonlinear
Systems Slides
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Linear System Theory, 2/E
Solution 18.9

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Solutions
Clearly C <A | B> = Y if and only if
B AB . . . A n −1 B
C
rank
=p
and thus the proof involves showing that the rank condition is equivalent to positive definiteness of
tf
∫ Ce
A (t f −t )
BB T e
A T (t f −t )
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
(A + BF )V ⊂ V
Selecting 0 ≠ xo
≠ V ⊂ Ker [C ] is a controlled inva
⊂ Ker [C ]
∈ V, this gives
e (A +BF )t xo
∈V ,
t
≥0
and thus
)t
You're
Reading
Ce (A +BF
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
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With Free Trial
n −1
k
N = ∩ Ker [C (A + BF ) ] ≠ 0
k =0
Thus 0≠xo
∈ N implies, using the Cayley-Hamilton theorem,
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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⊥ .
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( 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
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all z ∈ S, then x T y = 0. Equivalently, if z T y = 0 for all z ∈ S, then xRead
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Ker
zT
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Linear System Theory, 2/E
Solutions
0
=K
k +1
=K
V
V
∩ 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 ⊥ ]
A T (W K
∩ B ⊥)
⊥
∩
A T [ (V K ) ⊥
∩ B⊥ ]
⊥
= A −1
(V K ) ⊥
= (W K ) ⊥
=VK
∩
But further use of Exercise 19. 1 gives
(V K ) ⊥
AT
⊥
∩ B⊥
∩ B⊥
⊥
= A −1 (V K + B)
Thus
(W K +1 ) ⊥ = V K
∩ A −1 (V K + B) = V K +1
This completes the induction proof, and gives V * = V n = (W n ) ⊥ .
You're Reading a Preview
Solution 19.4
full access
a freeoftrial.
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
= R1
Assume now that for some positive integer K we have
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B
V
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R
K
K −1
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(A + BF ) j −1 (B
K
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Σ (A + BF) − (
j 1
B
V
*)
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Linear System Theory, 2/E
Solution 19.7

Solutions
The closed-loop state equation
.
x (t ) = (A + BF )x (t ) + (E + BK )w (t ) +BGv (t )
y (t ) = Cx (t )
is disturbance decoupled if and only if
C (sI − A − BF )−1 (E + BK ) = 0
That is, if and only if
<A +BFIm [E +BK ]> ⊂ Ker [C ]
Thus we want to show that there exist F and K such that (*) holds if and only if Im [E ] ⊂ V * + B, where V
maximal controlled invariant subspace contained in Ker [C ] for the plant.
First suppose F and K are such that (*) holds. Since <A +BF Im [E +BK ]> is invariant under ( A +
is a controlled invariant subspace contained in Ker [C ] for the plant. Then
Im [E +BK ] ⊂ <A +BF Im [E +BK ]> ⊂ V *
∈ X there is a v ∈ V * such that (E + BK )x = v. Therefore
Ex = v + B (−K x )
You're Reading a Preview
which implies Im [E ] ⊂ V * + B.
Conversely, suppose Im [E ] ⊂ V * + B, where V * is the maximal controlled invariant subspace conta
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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
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]> ⊂
<A +BF
V * ⊂ Ker [C ]
Im [E +BK
If w 1 , . . . , wq is a basis for W, then there exist v 1 , . . . , vq
∈ V * and u 1 , . . . , uq ∈ U such that
Ew j = v j + Bu j , j = 1, . . . , q
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(E + BK )
= Ew
BK
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Linear System Theory, 2/E
Since Im [BG 1 ] ⊂ B ∩ R 1 *
Solutions
ˆ =C P=
C
1
1
ˆ
C
11 0 0
ˆ =C P=
C
2
2
ˆ
0 C
11 0
⊂ R 1 * and BG 1 = PB̂ 1 we have
B̂ 1 =
Similarly, Im [BG 2 ] ⊂ B ∩ R 2 *
B̂ 11
0
B̂ 13
⊂ R 2 * gives
B̂ 2 =
Finally, (A + BF)R i *

0
B̂ 22
ˆ
B
23
⊂ R i *, i = 1, 2, and (A + BF)P = PÂ give
 11
0
0
0
 22
0
 =
 31  32  33
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That is, with z (t ) = P −1 x (t ), the closed-loop state equation takes the partitioned form
.
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za (t ) = Â 11 za (t ) + B̂ 11 r 1 (t )
.
zb (t ) = Â 22 zb (t ) + B̂ 22 r 2 (t )
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.
zc (t ) = Â 31 za (t ) + Â 32 zb (t ) + Â 33 zc (t ) + B̂ 13 r 1 (t ) + B̂ 23 r 2 (t )
ˆ z (t )
y 1 (t ) = C
11 a
ˆ z (t )
t) = C
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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
F (t −to )
t
∫
z (to ) + e F (t −σ) Gv (σ) d σ
to
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Evaluate this at t = (k +1)T and t o = kT to You're
get
(k +1)T
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(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
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T
z [(k +1)T ] = e
FT
∫
z (kT) + e F τ Gv (kT +T−τ) d τ
0
Then the special
form of v (t ) gives
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The integral term is not linear in the input sequence u (k ), so we approximate the integral when  u (k ) is s
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Linear System Theory, 2/E
Solutions
ũ
x̃ =
ũ
2
,
ỹ = ũ
2
Easy calculation gives the linearized state equation
x δ (k +1) =
y δ (k ) =
−1
0
0
−1
2ũ
x δ (k ) +
−1
2
u δ (k )
4ũ
x δ (k ) + 2ũ u δ (k )
Since A k = (−1)k I and CB = 0, the zero-state solution formula easily gives
y δ (k ) = 2ũ u δ (k )
Thus the zero-state behavior of the linearized state equation is that of a pure gain.
Solution 20.10
Φ(k,
Computing
Φ ( j +q,
j ) for the first few values of q
≥ 0 easily leads to the general formul
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
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a 1 (k −1)a 2 (k −2)a 1 (k −3)a 2 (k −4) . . . a 2 ( j )
0
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0
a 2 (k −1)a 1 (k −2)a 2 (k −3)a 1 (k −4) . . . a 1 ( j )
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Solution 20.11
≥ j +1,
j ) = F (k −1)F (k −2)
By definition, for k
ΦF (k,
. . . F ( j +1)F ( j )
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. . . A (−k+ 2)A (−k+1)
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Linear System Theory, 2/E
Solutions
Φ(k, ko ) ≤
Solution 20.16

k −1
1
_____
Φ(k, j ) Φ( j, k ) , k
k −k 1 j =k 1
Σ
≥ k 1 +1 ≥ k o +1
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 ) = ΦA (k, 0)F −k
is the correct choice. Obviously if F = I, then the variable change is P (k ) = ΦA (k, 0). Using this in Exa
20.19, where
1 a (k )
0 1
A (k ) =
gives
k −1
P (k ) = ΦA (k, 0) =
1
Σ a (i )
i =0
0
, k
≥1
1
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and
k −1 trial.
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P −1 (k +1) = ΦA (0, k +1) =
1
− Σ a (i )
i =0
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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 +7z+ 12
z +7 1
−12 z
and
Y (z ) = zc (zI −A )−1 xo + c (zI −A )−1 b U (z )
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z
z −1
____
z
_________
1/ 20
_________
−
z −19 z −1 1/ 20 + 2
= 2
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z +7z+ 12 z −1
z +7z+ 12
=0
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Therefore the complete solution is y (k ) = 0, k
≥ 0.
Solution 21.4 First compute the corresponding discrete-time state equation
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A 2 =Only
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Linear System Theory, 2/E
Solutions
x (k ) = (1+r/l )k xo +
k −1
Σ (1+r/l ) − − b
k j 1
j =0
= (1+r/l )k xo + b (1+r/l )k −1
1−1/(1+r/l )k
____________
1−1/(1+r/l )
= (1+r/l )k (xo +bl/r ) − bl/r
(b) In one year a deposit x o yields
x (l ) = (1+r/l )l xo
so
(1+r/l )l xo − xo
_____________
effective interest rate =
xo
× 100% = [(1+r/l )l − 1] × 100%
For r = 0.05, l = 2, the effective interest rate is 5.06%. For r = 0.05, l = 12, the effective interest rate is 5.12%
(c) Set
50,000
(−50,000)
_______
_________
+
You're Reading a0.05
Preview 0.05
0 = x (19) = (1.05)19
xo +
and solve to obtain xo = $604,266. Of course this means you have actually won only $654,26
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congratulations remain appropriate.
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Solution 21.9
With T = Td / 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−Td ) d σ , t ≥ T
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T
∫
z [(k +1)T ] =e FT z (kT) + e F τ
0
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d τ G v [(k −Cancel
l )T ] anytime.
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Linear System Theory, 2/E
x (k +1) =
yˆ (k ) =
Solutions
0 .. .
1 .. .
. .
. .
. .
0 0 0 .. .
0 0 0 .. .
A
0
.
.
.
B
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
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M
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Ma has a square root, with
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√Ma = Ma , but M b does not.
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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
ko +K −1
Φ(ko +K, j +1) f ( j )
[I − Φ(ko +K, ko )]xo =
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o
+K k )] is invertible. This implies that for
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Linear System Theory, 2/E

Solutions
we have by linear algebra that there exits a nonzero, n
× 1 vector p such that
[I − Φ(ko +K, ko )]T p = 0
and
pT
ko +K −1
Σ
Φ(ko +K,
∆
j +1) f ( j ) = q ≠ 0
j =ko
Now pick any xo . Then it is easy to show that the corresponding solution satisfies p T x (ko +jK ) = p
j = 1, 2, . . . . This shows that the solution is unbounded.
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CHAPTER 22
Solution 22.1
Similar to Solution 6.1.
Solution 22.4
If the state equation is uniformly exponentially stable, then there exist
γ ≥ 1 and 0 ≤ λ <
that
Φ(k, j ) ≤ γ λk − j , k ≥ j
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Φ(k +j , k ) ≤ γ λ , j ≥ 0
which implies
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φ j = sup Φ(k +j , k ) ≤ γ λ j
k
Then
φ = lim (γ
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1/j
j
1/j
j
<1
λ) = λ lim γ1/j
j→∞
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Linear System Theory, 2/E

Solutions
Similarly, for j > J,
Φ(k +j , k ) ≤ sup Φ(k +j , j )
k
= φj
< (1−ε) j = λ j
≤ γ λj
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
k≥0
≤ max t e −η t
t≥0
and a simple maximization argument (as in Exercise 6.10) gives
1
___
You'remax
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te −η t ≤ a Preview
ηe
t≥0
Therefore
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________ ∆
=β , k ≥ 0
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−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
Then
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Linear System Theory, 2/E

Solutions
which is equivalent to
ΦA 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
ΦA (k ) (3, 0) =
1/ 2 0
0 1/ 2
and it is easy to conclude uniform exponential stability. However
ΦA
T
(k ) (3,
0) =
2 0
0 1/ 8
and it is easy to see that there will be unbounded solutions.
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
CHAPTER 23
With Q = q I, where q > 0 we compute A T (k )QA (k )−Q to get the sufficient condition
uniform exponential stability:
Solution 23.1
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 (ka)Preview
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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 ) ≤
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These conclusions
show uniform exponential stability under weaker
one bounded coeffic
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sup | a 2 (k ) = α < ∞. Then we can take q 1 = α2 +0.01, q 2 = 1, and ν = 0.01 to conclude uniform expone
k
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Linear System Theory, 2/E

Solutions
holds if a 21 (k ), a 22 (k ) ≤ α < 1 for all k, but it also holds under weaker conditions. For example suppose th
bound is violated only for k = 0, and
a 21 (0) > 1 , a 21 (0)a 22 (1) < α
Then we can conclude uniform exponential stability. (More sophisticated analyses should be possible . . .
Solution 23.6
If the state equation is exponentially stable, then by Theorem 23.7 there is for any symmetr
a unique symmetric Q such that
A T QA − Q = −M
Write
M=
m1 m2
m2 m3
Q=
,
q1 q2
q2 q3
and write the discrete-time Lyapunov equation as the vector equation
−1
0
1
The condition
0
− 1− a 0
−2
a 20
a0
0
q1
q2
q3
−m 1
−m 2
−m 3
=
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0 − 1− a 0 a 0 ≠ 0
−2 Free
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0
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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
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Linear System Theory, 2/E
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
_
k
H
2
k
0 = λ p Mp = λ p H Mp λk = p H (A T )k MA k p
≥ 0,
= (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
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Thus Corollary 24.3 states that the state equation is uniformly stable if there exists a constant
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γ such that
Π max [ a 1 (i ), a 2 (i ) ] ≤ γ
i =j
for all k , j with k
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≥ j. (Note that this condition
holds if
max [ a 1 (k ), a 2 (k ) ] ≤ 1
for all but a finite number of values of k .) Of course the condition (#) is not necessary. Consider
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eigenvalues
are ± 2 / 3, so the state equation is uniformly stable, but clearly (#) fails.
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Khalil - Nonlinear
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L in e a r S y s t e m T h e o r y , 2 / E
k
r (k +1)
Π
j =ko

S o lu t io n s
1
__________
1+η( j )ν( j )
k −1
≤ r (k ) Π
j =ko
1
__________
1+η( j )ν( j )
k
Π
+ ν (k )ψ(k )
j =ko
1
__________
1+η( j )ν( j )
, k
≥ ko +
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 =ko
and substituting this into (*) yields the result.
Solution 24.7
By assumption ΦA (k, j ) ≤ γ for k
≥ j. Treating
f ( k, z (k )) as an input, the complete solu
formula is
z (k ) = ΦA (k, ko )z (ko ) +
k −1
Σ Φ (k, j +1) f ( j, z ( j )) ,
A
k ≥ k o +1
j =ko
This gives
z (k ) ≤ γ z (ko ) +
k −1
Σ γ f ( j, z ( j ))
j =ko
≤ γ z (ko ) +
k −1
Σ γ α z ( j ) ,
k ≥ k o +1
j
j =ko
Applying Lemma 24.5,
z (k ) ≤ γ z (ko ) exp [ γ
k −1
Σα]
j
j =ko
≤ γ z (ko ) exp [ γ
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
CHAPTER 25
Solution 25.1
If M (ko , k f ) is not invertible, then there exists a nonzero, n
× 1 vector x a such that
0 = x Ta M (ko , k f )xa
k f −1
=
x Ta ΦT ( j, ko )C T ( j )C ( j )Φ( j, ko )xa
Σ
j =ko
k f −1
=
Σ C ( j )Φ( j, k )x 
o
a
2
j =ko
This implies
C ( j )Φ( j, ko )xa = 0 , j = k o , . . . , k f −1
which shows that the nonzero initial state x a yields the same output on the interval as does the zero initial s
Therefore the state equation is not observable.
On the other hand, for any initial state x o we can write, just as in the proof of Theorem 25.9,
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k ) is invertible, then the initial state is uniquely determined by
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L in e a r S y s t e m T h e o r y , 2 / E

S o lu t io n s
The claim is true if A (k ) is invertible at each k . Let k f = n so that
n −1
Φ(n, j +1)b ( j )b T ( j )ΦT (n, j +1)
W (0, n ) =
Σ
j =0
Since Φ(n, j +1) is invertible for j = 0, . . . , n −1, let
b (k ) = Φ−1 (n, k +1)ek +1 , k = 0, . . . , n −1
where e k is the k th -column of I n . Then
n −1
W (0, n ) =
Σe
T
j +1 e j +1
= In
j =0
and the state equation is reachable on [0, n ].
Solution 25.7
Suppose W O (ko , k f ) is invertible. Given a p
× 1 vector y f , let
− 1 (k , k )y , k = k , . . . , k − 1
u (k ) = B T (k )ΦT (k f , k +1)C T (k f )W O
o
f f
o
f
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 [ ko , k f ].
Conversely, suppose the state equation is output reachable on [ ko , k f ]. If W O (ko , k f ) is not invertible,
there exists a nonzero p × 1 vector y a such that
0 = y Ta WO (ko , 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
T
f
T
f
T
f
a
j =ko
k f −1
=
Σ y C (k )Φ(k , j +1)B ( j )
T
a
f
f
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Linear System Theory, 2/E
Solution 25.13

Solutions
We will prove that the state equation is reconstructible if and only if
C
CA
.
.
.
z = 0 implies A n z = 0
CA n −1
That is, if and only if the null space of the observability matrix is contained in the null space of A n .
First, suppose the state equation is not reconstructible. Then there exist n × 1 vectors xa and xb such
xa ≠ xb and
C
.
.
.
xa =
CA n −1
C
.
.
.
xb ,
A n xa
≠ A n xb
CA n −1
That is
C
.
.
.
(x a −x b ) = 0 ,
A n (xa −xb ) ≠ 0
CA n −1
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Thus the condition (*) fails.
Now suppose the condition (*) fails Unlock
and z isfull
such
thatwith a free trial.
access
C
.
.
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0 and
z =With
A nTrial
z≠0
.
CA n −1
Obviously z
≠ 0. Then for x (0) = z the zero-input response is
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y (k ) = 0 , k = 0, . . . , n −1
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
CHAPTER 26
Solution 26.2
For the linear state equation
x (k +1) =
1 k
1 1
x (k ) +
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)
=
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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 2 k −1
1 1 k
From the respective ranks the state equation
is 3-step reachable,
but Trial
not 2-step reachable.
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Solution 26.4
The (n +1)-dimensional state equation
A 0
b
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1 n
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Linear System Theory, 2/E

Solutions
m
G (z ) =
σl
ΣΣG
lr
l =1 r =1
z
______
(z −λl )r
_
Here λ1 , . . . , λm are distinct complex numbers
__ such that if λL is complex, then λM = λL for some M . Furtherm
the p × m complex matrices satisfy GMr = GLr for r = 1, . . . , σL . 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 = λL for some M . Furthermore the p × m complex matrices satisfy GMr =
r = 1, . . . , σL .
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 )
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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
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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 )
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. adj (zI −A ) . b
h . adj (zI −F ) . g
______________
_c____________
=
det (zI F )
det (zI A )
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Linear System Theory, 2/E

Solutions
From the latter approach, setting
cb = 0, cAb = 1, cA 2 b = 1/ 2, cA 3 b = 1/ 2
easily yields c 1 = 0, c 0 = 1, a 0 = 1/ 4, a 1 = 1/ 2.
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
CHAPTER 27
Solution 27.1
Similar to Solution 12.1.
Solution 27.4
Suppose the entry G ij (z ) has one pole at z = 1, that is
Gij (z ) =
Nij (z )
__________
(z −1)Dij (z )
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where all roots of the polynomial D ij (z ) have magnitude less than unity (so D ij (1) ≠ 0), and the polynomial
satisfies Nij (1) ≠ 0. Suppose that the m ×Unlock
1 U (zfull
) has
all components
zero except for U j (z ) = z /(z −1). The
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with a free trial.
i th -component of the output is given by
(z ) Trial
z N ijFree
___________
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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
_____
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Since this
term is Times
unbounded, every realization of G (z ) fails to be uniform
bounded-input,
bounded-output sta
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Linear System Theory, 2/E

Solutions
µ = sup u (k ) ,
k≥0
η=
∞
Σ G (k )
k =0
The first constant is finite for a well-defined sequence that goes to zero, and the second is finite by unif
bounded-input, bounded-output stability. Then there is a positive integer K 1 such that
∞
ε
ε
___
___
, k ≥ K1
,
G (k ) ≤
u (k ) ≤
2µ
2η
k =K 1
Σ
Let K = 2K 1 . Then for k
≥ K we have
y (k ) ≤ µ
K 1 −1
Σ
j =0
≤µ
k
Σ
k
ε
___
G (k − j )
G (k − j ) +
2 η k =K 1
q =k −K 1
Σ
ε k −K 1
___
G (q )
G (q ) +
2 η q =0
Σ
ε
2η
ε
___
η=ε
+
≤ µ ___
2µ
Solution 27.8
Similar to Solution 12.12.
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CHAPTER 28
Solution 28.2
Lemma 16.18 gives that if V 11 and V are invertible, then
V −1 =
V 11 V 12
V 21 V 22
−1
V −111 +V−111 V 12 V −a1 V 21 V−111
−V−a1 V 21 V−111
=
−V−111 V 12 V−a1
−1
V −a1
where Va = V 22 −V 21 V −111 V 12 . From the expression V V −1 = I, written as
You're Reading a Preview
V 11 V 12
W 11 W 12
=I
V 21 full
V 22access
Wwith
21 W
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a 22
free trial.
we obtain
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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
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22
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1
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1
11
12
21
1
22
21
11
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V −1 V
22
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21 )
21
11
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Linear System Theory, 2/E

Solutions
T
T
K = −Bˆ (Aˆ )n
n
Σ
k
T
T
Aˆ Bˆ Bˆ (Aˆ )k
−1
Aˆ
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
(α A )n +1
T k
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 KB KAB+(KB )2 . . .
.. .
0 I
KB
.. .
0 0
I
. .
.
.
. .
.
.
. .
.
.
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Solution 28.6
Similar to Solution 2.8. Unlock full access with a free trial.
Solution 28.8
Download With Free Trial
Supposing that the linear state equation is reachable, there exists K such that all eigenvalu
A+BK have magnitude less than unity. Therefore ( I −A −BK ) is invertible, and if we suppose
A −I B
C 0
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titlem × m matrix
is invertible, then C (I −A −BK ) B is invertible from Exercise 28.6. Read
Then
given
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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 )
You're Reading a Preview
= F˜ (k )eb (k )
Therefore e b (k ) → 0 exponentially as k
full access with a free trial.
→Unlock
∞. Now
∆
e (k ) = x (k ) − xˆ (k )
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= x (k )−H (k )C (k )x (k )−J (k )z (k )
= −J (k )eb (k ) + [I −H (k )C (k )−J (k )Pb (k )]x (k )
(k )e (k )
= −JScribd
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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
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