EXAMPLES:
Example 1: Consider the system
x 1  x 2
1 5
x 2   x1  x1  x 2
16
Calculate the equilibrium points for the system.
Plot the phase portrait of the system.
Solution:
The equilibrium points must be stationary. Therefore for the first system we
have
0  x2
1 5
0   x1  x1  x 2
16
0  x2
1 5
0   x1  x1   x1 1  0.0625x14 
16
0  x2
1 5
0   x1  x1   x1 1  0.0625x14 
16
x1=0
roots([-1/16 0 0 0 1])
ans =
-2.0000
-0.0000 + 2.0000i
-0.0000 - 2.0000i
2.0000
The equilibrium points are
xe=[(0,0),(2,0),(-2,0)]
The jacobian matrix is defined as
 f1
 x
J 1
 f 2
 x1
f1 
0
x 2  
5 4


1

x1
f 2  
16

x 2 
1

 1

0 1
J xe1( 0, 0 )  


1

1


eig ( J ) 
0

5

4

1

(

2
)

16
J xe 2( 2, 0 )
1  0 1 

 1 4  1

eig (J ) 
The same result is
obtained for xe3 (2,0)
- 0.5000 + 0.8660i
1.5616
- 0.5000 - 0.8660i
- 2.5616
Saddle points
2
Stable node
1.5
1
2
0.5
x
[x1, x2] = meshgrid(-4:0.2:4, -2:0.2:2);
x1dot = x2;
x2dot = -x1+(1/16)*x1.^5-x2;
quiver(x1,x2,x1dot,x2dot)
xlabel('x_1')
ylabel('x_2')
0
-0.5
-1
-1.5
-2
-3
-2
-1
0
x1
1
2
3
Example 2. Show that the origin of the system is stable, using a suitable
Lyapunov function.
x 1  x 2
x 2   x13  x 32
Solution: Let us use the following Lyapunov function
1 4 1 2
V( x )  x1  x 2
4
2
 ( x )  dV  V dx  V  f ( x )
V
dt
x dt
 V V   f1 ( x ) 

,
 f ( x ) 

x

x

2 2
 1


 x2 
3

V ( x )  x1 , x 2  3
3

x

x
2
 1
 (x)  x 3x  x  x 3  x 3
V
1 2
2
1
2
 (x)  x 4  0
V


2
The system is stable in the sense of Lyapunov.
Example 3:
R(s) +
y
1
s2  1
s
y3
C(s)
3
s 1
N
Find the describing function of the nonlinear element N of the control system.
1
3 sin t   sin 3t 
sin t  
4
For a sinusoidal input
w  y3
3
0.8
ODD FUNCT ION
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-2
-1.5
-1
3
A
3 sin t   sin 3t 
w ( t )  y3 ( t )  A 3 sin 3 t  
4
w(t )  a1 cost   b1 sin t 
-0.5
0
0.5
1
a1=0
1.5
2

1   3A3
A3
b1   
sin t  
sin 3t  sin t  dt
   4
4

>>syms tet;syms A;
>>b1=‘((3*A^3/4)*sin(tet)-A^3/4*sin(3*tet))*sin(tet)’;
>>int(b1,-pi,pi)
1  3  A3  3A3
 
b1  
 4 
4
 3A3 
 sin t 
w1  
 4 
w1  NA,  y(t)  N(A, ) A sin t
 3A 2 
 A sin t 
w1  
 4 
N(A)
Example 4:
Determine whether the system in the Figure exhibits a self-sustained oscillation
(a limit cycle).
R(s) +
K
s2  3s  2
1
-1
C(s)
-
N(A,ω)
4M
4
N(A, )  NA  

A A
1  NA  G (s)  0
4
K
1
0
2
A s  3 s  2
As 2  3As  2A  4K  0
s1, 2
9 2 A 2  4A * 2A  4K 
3A


2A
2A
s1, 2
9 2 A 2  8 2 A 2  16KA
 1.5 
2A
s1, 2
 2 A 2  16KA
 1.5 
2A
Since there is always a negative real part, the system doesn’t exhibit a limit
cycle.
LYAPUNOV STABILITY FOR LINEAR TIME-INVARIANT SYSTEMS:
Given a linear system of the form
x  A x
Let us consider a quadratic Lyapunov function candidate
V  x T Px
where P is a given symmetric positive definite matrix.
Differentiating the positive definite function V along the system trajectory yields
another quadratic form
  x T Px  x T Px
V
where
T

V  Ax  Px  x T PAx 
 x T A T Px  x T P A x
 x T A T P  P A  x
A T P  P A  Q
 x TQ x
If there exists a positive definite matrix Q satisfying the equation (Lyapunov
equation), the system is said to be stable in the sense of Lyapunov (ISL).
AT P  P A  Q  0
Lyapunov equation.
A useful way of studying a given linear system using scalar quadratic functions
is to derive a positive definite matrix P from a given positive definite matrix Q,
i.e.,
•choose a positive definite matrix Q
•solve for P from the Lyapunov equation
•check whether P is positive definite
If P is positive definite, then xTPx is a Lyapunov function for the linear system
and global asymptotical stability is guaranteed.
Example:
Consider two matrices,
1
 0
1 0
A
,Q



12

8
0
1




The linear system is stable (Real parts of all eigenvalues of the system matrix A
are negative) if there is a positive definite matrix P.
Using Matlab, we can find the matrix P as
clc;clear;
A=[0 1;-12 -8];
Q=[1 0;0 1];
P=lyap(A,Q)
eig(P)
P=
0.4010 -0.5000
-0.5000 0.8125
ans =
0.0661
1.1474
The matrix P is positive definite, since the eigenvalues are
real, and the system is stable ISL.
LYAPUNOV FUNCTION FOR NONLINEAR SYSTEM:
Krasovskii’s method suggests a simple form of Lyapunov function candidate
(LFC) for autonomous nonlinear systems, namely, V=fTf. The basic idea of the
method is simply to check whether this particular choice indeed leads to a
Lyapunov function.
Theorem (Krasovskii): Consider the autonomous system defined by dx/dt=f(x),
with the equilibrium point of interest being the origin. Let J(x) denote the
Jacobian matrix of the system, i.e.,
f
J(x ) 
x
If the matrix F=J+JT is negative definite, the equilibrium point at the origin is
asymptotically stable. A Lyapunov function for this system is
V(x)  f T (x)f (x)
If V(x)  ∞ as ǁxǁ ∞, then the equilibrium point is globally asymptotically
stable.
Example:
Consider a nonlinear system
x 1  6 x1  2 x 2
x 2  2x1  6x 2  2x 32
We have
 f1
f  x1
J

x  f 2
 x1
f1 
2

x 2   6

f 2   2  6  6 x 22 
x 2 
4
 12

FJJ  
2
4

12

12
x
2

T
The matrix F is negative definite over the whole state space. Therefore, the
origin is asymptotically stable, and a Lyapunov function candidate is
-8
-8.5
-9
-9.5
2
clc;clear;
x2=-10:0.1:10;
for i=1:length(x2)
F=[-12 4;4 -12-12*x2(i)^2];
eg=eig(F)
plot(eg(1),eg(2))
hold on
end
-10
-10.5
-11
-11.5
-12
-1400
-1200
-1000
-800
-600
-400
-200
0
1


V( x )  f ( x )f ( x )   6x1  2x 2 , 2x1  6x 2  2x
T

3
2
  6x1  2x 2  
 2x  6x  2x 3 
2
2 
 1
 
V( x)  f T (x)f (x)   6x1  2x 2   2x1  6x 2  2x 32
2


2
Since V(x)  ∞ as ǁxǁ ∞, then the equilibrium point is globally asymptotically
stable.
Example (Variable Gradient Method):
Consider a nonlinear system
x 1  2x1
x 2  2x 2  2x1x 22
We assume that the gradient of the undetermined Lyapunov function has the
following form
V1  a11x1  a12 x 2
V2  a 21x1  a 22 x 2
The curl equation is
V1 V2

x 2
x1
 a12
 a 21
a12  x 2
 a 21  x1
 x2
 x1
Slotine and Li, Applied Nonlinear Control
If the coefficients are choosen to be
a11=a22=1, a12=a21=0
which leads to
V1  x1
V2  x 2
Then
  V x  2x 2  2x 2 1  x x 
V
1
2
1 2
Thus, dV/dt is locally negative definite in the region (1-x1x2)>0. the function V
can be computed as
x12  x 22
V( x )   x1dx1   x 2dx 2 
2
0
0
x1
x2
This is indeed positive definite, and therefore the asymptotic stability is
guaranteed.
Slotine and Li, Applied Nonlinear Control