EEE 586
HW # 1 SOLUTIONS
Problem 1.6
Pick as states x = [q1 ; q1 ; q2 ; q2 ]. Then
_
_
0
x2
dx B M 1 (x1 )[h(x1 ; x2 )x2 + K (x1 x3 )]
=B
@ x4
dt
J 1 K (x1 x3 ) + J 1 u
1
C
C
A
Problem 1.15
Dierentiating the L cos and L sin terms we get
_
my + mL( cos 2 s
EEE 586
HW # 1 SOLUTIONS
Problem 1.6
Pick as states x = [q1 ; q1 ; q2 ; q2 ]. Then
_
_
0
x2
dx B M 1 (x1 )[h(x1 ; x2 )x2 + K (x1 x3 )]
=B
@ x4
dt
J 1 K (x1 x3 ) + J 1 u
1
C
C
A
Problem 1.15
Dierentiating the L cos and L sin terms we get
_
my + mL( cos 2 s
EEE 586
HW # 2 SOLUTIONS
Problem 4.2
_
Let V (x) = 1 x2 . Then V = axp+1 + xg (x).
2
_
1. When p is odd and a < 0, V a xp+1 + a xp+1 1
2
2
_
2. When p is odd and a > 0, V a xp+1 + a xp+1 1
2
2
2kjxj
jaj
2kjxj
a
< 0 in fjxj < jaj=2kg. Hence, 0 is AS.
> 0
EEE 586
HW # 3 SOLUTIONS
Problem 4.43
With the given control, the closed loop system is x = f (x) G(x)G> (x)P x. Using V = x> P x as a
_
Lyapunov function candidate,
_
V = 2x> P x = 2x> P f (x) 2x> P G(x)G> (x)P x x> P x W (x) V (x)
_
Hence, the origin is
EEE 586
HW # 3b SOLUTIONS
Problem 5.10
p
_
2. Using V = 1 x2 , we have V = x4 + ux4 x6 x4 ; 8jxj u. Then, by Theorem 4.19, the system
2
is I2S stable. Then, by Theorem 5.3, we conclude that the system is L1 stable . The origin of the unforced
system is no
EEE 586
HW # 4 SOLUTIONS
Problem 6.1
Dene yo = y + K1 u as the output of the transformations, and uo as the input. Then u, the input to the
nonlinearity, is u = K 1 (uo + yo ). From the sector condition < h(t; u) K1 u ; h(t; u) K2 u > 0, we have
< yo ; yo
EEE 586
HW # 5 SOLUTIONS
Problem 7.1
4. The transfer function has poles on the jw-axis. The correct semicircles at innity should be drawn
corresponding to the indentations of the Nyquist path, from wihch the allowed choices for a disk is to be
contained i
EEE 586
HW # 6 SOLUTIONS
Problem 8.16
Consider the Lyapunov function candidate
Z x1
1
1
1
1
V (x) = x2 +
(y y 3 ) dy = x2 + x2 x4
22
22 21 41
0
p
_
_
which is positive denite in the region jx1 j < 2. Then, V = x2 which is negative semidenite and V 0 )
2
3
EEE 586, TEST 1
NAME: SOLUTIONS
Closed-Book, 1 sheet of formulae allowed
Problem 1
1. Determine whether the function f : R 7 R; f (x) = jxj sin jxj is:
!
A. dierentiable; B. locally Lipschitz continuous; C. globally Lipschitz continuous.
The only potentia
EEE 586, TEST 2
NAME: SOLUTIONS
Closed-Book, 1 sheet of formulae allowed
Problem 1
Consider the system x = x + G(x)u.
_
1. Show that the system is I2S stable if G(x) is bounded.
2. Give a counterexample for I2S stability if G(x) = x3 .
p
_
1. Let V = 1 x2
EEE 586
HW # 5 SOLUTIONS
Problem 4.6 (1)
The system is:
x1
_
x2
_
= x2
2
= x2 + x2 + x1 x2
1
The linearization around the origin produces
@f
0 0
=
A=
0 1
@x x=0
which has an eigenvalue equal to zero and another equal to 1.
Using the center manifold equat
EEE 586
HW # 4 SOLUTIONS
Problem 5.10.1 For the system
x = (1 + u)x3 ;
_
1 2
2x ,
y=x
Thm.5.2 is applicable: V (x) =
is pdf and decrescent, Vx f (x; 0) = x4 < 0, jVx j jxj. Furthermore, for
x 2 B(0; r) (a ball of radius r, centered at 0) jf (x; u) f (x; 0
EEE 586
HW # 2 SOLUTIONS
Problem 3.1 1. f (x) = x2 + jxj is not continuously dierentiable at the origin (jxj has a discontinuous
derivative). It is locally Lipschitz since its derivative (left and right limits) is bounded on any compact set. It
is continu
EEE 586
HW # 2 SOLUTIONS
Problem 4.2
_
Let V (x) = 1 x2 . Then V = axp+1 + xg (x).
2
_
1. When p is odd and a < 0, V a xp+1 + a xp+1 1
2
2
_
2. When p is odd and a > 0, V a xp+1 + a xp+1 1
2
2
2kjxj
jaj
2kjxj
a
< 0 in fjxj < jaj=2kg. Hence, 0 is AS.
> 0
EEE 586
HW # 3 SOLUTIONS
Problem 4.43
With the given control, the closed loop system is x = f (x) G(x)G> (x)P x. Using V = x> P x as a
_
Lyapunov function candidate,
_
V = 2x> P x = 2x> P f (x) 2x> P G(x)G> (x)P x x> P x W (x) V (x)
_
Hence, the origin is
EEE 586
HW # 3b SOLUTIONS
Problem 5.10
p
_
2. Using V = 1 x2 , we have V = x4 + ux4 x6 x4 ; 8jxj u. Then, by Theorem 4.19, the system
2
is I2S stable. Then, by Theorem 5.3, we conclude that the system is L1 stable . The origin of the unforced
system is no
EEE 582
HW # SVD
% SVD application in the modeling of a noisy oscillatory signal
%
as the output of an autoregressive model:
%
y(n)=[y(n-1),y(n-2),.]*q
% Define the signals
t=(0:.01:10)';
% time
n=rand(size(t)-.5;
% noise
y=sin(2*t+1)+sin(10*t+1)+.02*n/2;
EEE 582, TEST 2
NAME:_SOLUTIONS_
Nov. 16, 2011, 3:30-4:45, 4 Problems, Equal Credit, Closed-book, Closed-notes, calculator and 1 sheet of formulae
allowed
1 1 / 4
Problem 1. Use the function V = x T Px , P =
, to find conditions on a such that
1 / 4 1
EEE 582, Test 1
NAME: _SOLUTIONS_
Oct. 12, 2011, 4 Problems, Equal Credit, Closed-book, Closed-notes, 1 sheet of formulae allowed
Problem 1. (Short questions)
1. Let |.| be a matrix norm; is it always true that |AB| < |A| |B|?
Not unless the norm is consi
EEE 586, TEST 2
NAME: SOLUTIONS
Closed-Book, 1 sheet of formulae allowed
Problem 1
Consider the system x = x + G(x)u.
_
1. Show that the system is I2S stable if G(x) is bounded.
2. Give a counterexample for I2S stability if G(x) = x3 .
p
_
1. Let V = 1 x2
EEE 586, TEST 1
NAME: SOLUTIONS
Closed-Book, 1 sheet of formulae allowed
Problem 1
1. Determine whether the function f : R 7 R; f (x) = jxj sin jxj is:
!
A. dierentiable; B. locally Lipschitz continuous; C. globally Lipschitz continuous.
The only potentia
EEE 586
HW # 1 SOLUTIONS
Problem 1.5 Pick as states x = [q1 ; q1 ; q2 ; q2 ]. Then
_
_
0
x2
dx B MgL sin(x1 ) k (x1 x3 )
I
I
=B
@ x4
dt
1
k
J (x1 x3 ) J u
1
C
C
A
Problem 1.7 A state space realizaton for G is
x = Ax + Bu ;
_
y = Cx
Substituting u = r z =
EEE 586
HW # 6 SOLUTIONS
Problem 8.16
Consider the Lyapunov function candidate
Z x1
1
1
1
1
V (x) = x2 +
(y y 3 ) dy = x2 + x2 x4
22
22 21 41
0
p
_
_
which is positive denite in the region jx1 j < 2. Then, V = x2 which is negative semidenite and V 0 )
2
3