2. 2
Any bounded input yields bounded output, i.e.
M
t
y
N
t
u )
(
)
(
B
A
sI
C
s
q
s
p
s
T 1
)
(
)
(
)
(
)
(
For linear systems:
BIBO Stability ⇔All the poles of the transfer function lie in the LHP.
0
)
(
s
q
Characteristic Equation
Solve for poles of the transfer function T(s)
Bounded-Input Bounded-Output (BIBO) stablility
Definition: For any constant N, M >0
3. 3
Asymptotically stable ⇔ All the eigenvalues of the A matrix
have negative real parts
(i.e. in the LHP)
t
t
x
Ax
x
t
u
as
0
)
(
system
the
e.
i.
0,
)
(
When
Cx
y
Bu
Ax
x
A
sI
B
A
sI
adj
C
B
A
sI
C
s
q
s
p
s
T
]
[
)
(
)
(
)
(
)
( 1
0
A
sI Solve for the eigenvalues for A matrix
Asymptotic stablility
Note: Asy. Stability is indepedent of B and C Matrix
For linear systems:
4. 4
Asy. Stability from Model Decomposition
,n
,
i
v
Av i
i
i
i
1
,
satisfying
,
i.e.
Suppose that all the eigenvalues of A are distinct.
n
n
R
A
CT
C
B
T
B
AT
T
A
1
1
Du
CTz
y
Bu
T
ATz
T
z
1
1
AT
T 1
]
[ 2
1 n
,v
,
, v
v
T
Coordinate Matrix
n
n
n
2
1
2
1
2
1
0
0
0
0
0
0
0
0
0
Hence, system Asy. Stable ⇔ all the eigenvales of A at lie in the LHP
,
)
0
(
)
0
(
)
0
(
)
(
)
( 2
2
1
1
2
1
n
t
n
t
t
ξ
e
v
ξ
e
v
ξ
e
v
t
T
t
x n
)
0
(
)
0
( 1
x
T
ξ
i
v
Let the eigenvector of matrix A with respect to eigenvalue i
6. 6
In the absence of pole-zero cancellations, transfer function poles are
identical to the system eigenvalues. Hence BIBO stability is
equivalent to asymptotical stability.
Conclusion: If the system is both controllable and observable, then
BIBO Stability ⇔ Asymptotical Stability
Asymptotic Stablility versus BIBO Stability
• Asymptotically stable
• All the eigenvalues of A lie in the LHP
• BIBO stable
• Routh-Hurwitz criterion
• Root locus method
• Nyquist criterion
• ....etc.
Methods for Testing Stability
8. 8
Definition: An equilibrium state of an autonomous system is
stable in the sense of Lyapunov if for every , exist a
such that for
e
x
0
0
)
(
e
e x
x
t
x
x
x )
,
( 0
0 0
t
t
1
x
2
x
e
x
0
x
)
(t
x
9. 9
Definition: An equilibrium state of an autonomous system is
asymptotically stable if
(i) it is stable
(ii) there exist a such that
e
x
0
e
t
x
t
x
x
x e
e
e as
,
0
)
(
0
1
x
2
x
e
x
0
x
e
)
(t
x
10. 10
Lyapunov Theorem
)
(x
f
x
0
:
State
Eq.
e
x
0
)
0
(
f
0
,
0
)
(
)
1
(
x
x
V
0
for
0
)
0
(
)
2
(
x
V
0
)
(
)
(
)
(
)
3
(
x
f
dx
x
dV
dt
x
dV
0
,
0
)
(
x
dt
x
dV 0
0
)
(
x
dt
x
dV
Consider the system (1)
A function V(x) is called a Lapunov fuction V(x) if
Then eq. state of the system (1) is stable.
Then eq. state of the system (1) is asy. stable.
Moreover, if the Lyapunov function satisfies
and
11. 11
Explanation of the Lyapunov Stability Theorem
1. The derivative of the Lyapunov function along the trajectory is negative.
2. The Lyapunov function may be consider as an energy function of the system.
2
x
1
x
))
(
( t
x
V
)
(t
x
0
12. 12
Lyapunov’s method for Linear system: 0
where
A
Ax
x
Px
x
x
V T
)
(
Proof: Choose
0
for
0,
)
(
)
(
x
Qx
x
x
PA
P
A
x
PAx
x
Px
A
x
x
P
x
Px
x
x
V
T
T
T
T
T
T
T
T
The eq. state is asymptotically stable.
⇔
For any p.d. matrix Q , there exists a p.d. solution of the
Lyapunov equation
Q
PA
P
AT
0
x
Q
PA
P
AT
Hence, the eq. state x=0 is asy. stable by Lapunov theorem.
13. 13
Asymptotically stable in the large
( globally asymptotically stable)
(1) The system is asymptotically stable for all the initial states .
(2) The system has only one equilibrium state.
(3) For an LTI system, asymptotically stable and globally
asymptotically stable are equivalent.
)
( 0
t
x
)
(
, x
V
x
)
(
, x
V
x
Lyapunov Theorem (Asy. Stability in the large)
If the Lyapunov function V(x) further satisfies
(i)
(ii)
Then, the (asy.) stability is global.
14. 14
Sylvester’s criterion
A symmetric matrix Q is p.d. if and only if all its n
leading principle minors are positive.
n
n
Definition
The i-th leading principle minor of an
matrix Q is the determinant of the matrix extracted from
the upper left-hand corner of Q.
n
i
Qi ,
,
3
,
2
,
1
n
n
i
i
Q
Q
q
q
q
q
Q
q
Q
q
q
q
q
q
q
q
q
q
Q
3
22
21
21
11
2
11
1
33
32
31
23
22
21
13
12
11
15. 15
Remark:
(1) are all negative Q is n.d.
(2) All leading principle minors of –Q are positive Q is n.d.
n
Q
Q
Q
,
, 2
1
𝑉(𝑥)
= 2𝑥1
2
+ 4𝑥1𝑥3 + 3𝑥1
2
+ 6𝑥2𝑥3 + 𝑥3
2
0
24
0
6
0
2
3
2
1
Q
Q
Q
Q is not p.d.
Example:
17. Q.1 Check the stability of the equilibrium state of the system
described by
X1˙ = X2
17
1
𝑋˙2 = −𝑋1 − 𝑋2𝑋2
2
Solution:
Select Lyapunov function
V X = 𝑋2+𝑋2
1
Which is positive definite function
Its derivative is
V Ẋ =
𝜕V dX1
+
𝜕V dX2
𝜕X1 dt 𝜕X2 dt
18. 1
V X
˙ = 2X1X2 + 2X2 −X1 − X2X2
18
1 2
V X
˙ = 2X1X2 − 2X2𝑋1 − 2𝑋2𝑋2
1 2
V X
˙ = −2X2X2
V X
˙ is always negative definite.
V X → ∞ as X → ∞ Then system is asymptotically stable in-the-large.
19. Q.2 Consider the nonlinear system described by the equations
19
X1˙ = X2
˙ =
X2 1 − X1 X2 − X1
Find the region in the state plane for which the equilibrium state of
the systems asymptotically stable.
Solution:
Select Lyapunov function
1 2
V X = 𝑋2+𝑋2
Which is positive definite function
Its derivative is
V Ẋ =
𝜕V dX1
+
𝜕V dX2
𝜕X1 dt 𝜕X2 dt
20. V X
˙ = 2X1X2 + 2X2 − 1 − X1
20
2
V X
˙ = 2X1X2 − 2𝑋2 1 − X1
X2 − X1
− 2𝑋1𝑋2
2
V X
˙ = −2𝑋2 1 − X1
Case1:
V X
˙ to be negative definite if 1 − X1 > 0 , X2 ≠ 0
or 𝑋1 < 1
If these conditions are fulfil then system remain asymptotically
stable.
Case2:
V X
˙ = 0 if 𝑋1 = 1 at this point rate of change of energy will
become zero. At this point there is no trajectory. But energy is not
zero at this point.
21. Case3:
V X
˙ is not negative definite if 𝑋1 > 1
In this region system will become unstable.
21
Possibly unstable
Stable Stable
Possibly unstable
X1
X2
