1. Modern Control Systems (MCS)
Lecture-39-40
Design of Control Systems in Sate Space
Second Method of Liapunov
Dr. Imtiaz Hussain
URL :

2. Outline Introduction to second method of Liapunov
Liapunov Energy Function
Liapunov Stability Theorem
Examples
Liapunov Stability analysis of LTI systems

3. Introduction
A vibratory system is stable if its total energy (A positive definite function) is continuously decreasing until an equilibrium state is reached.
That is, the time derivative of total energy must be negative definite.
The second method of Liapunov is based on the generalization of this fact.
If the system has asymptotically stable equilibrium state then the stored energy of the system displaced within the domain of attraction decays with increase in time until it finally assumes its final value at equilibrium state.
For purely mathematical systems, however, there is no way of defining an “energy function”.
To circumvent this difficulty, Liapunov introduced the :Liapunov Function”, a fictitious energy function.

4. Liapunov Function
Any scalar function satisfying the hypothesis of Liapunov’s stability theorem can serve as a Liapunov function.
Liapunov function depend on 𝒙 1 , 𝒙 2 , … 𝒙 𝑛 and t.
In second method of Liapunov the sign behavior of V(𝒙) and that of its time derivative 𝑉 (𝒙) gives us information as to the stability, asymptotic stability or instability of an equilibrium state without the solution of system state equations.

5. Liapunov Stability Theorem
Suppose that a system is described by
Where
If there exists a scalar function 𝑉 𝒙 having continuous first derivative and satisfying the following conditions,
𝑉 𝒙 is positive definite
𝑉 (𝒙) is negative definite
then the equilibrium state at the origin is uniformly asymptotically stable.
If in addition 𝑉 𝒙 →∞ as 𝒙 →∞, then the equilibrium state at the origin is uniformly asymptotically stable in large.
𝒙 =𝑓 𝒙,𝑡
𝑓 𝟎,𝑡 =𝟎

6. Example-1 Consider a system described by
Using Liapunov stability theorem determine whether the equilibrium state of the system described by above state equations is asymptotically stable, asymptotically stable in large or unstable.
𝑥 1 = 𝑥 2 − 𝑥 1 ( 𝑥 𝑥 2 2 )
𝑥 2 = −𝑥 1 − 𝑥 2 ( 𝑥 𝑥 2 2 )

7. Example-1
𝑥 1 = 𝑥 2 − 𝑥 1 ( 𝑥 𝑥 2 2 )
𝑥 2 = −𝑥 1 − 𝑥 2 ( 𝑥 𝑥 2 2 )
Clearly the origin of state space is the only equilibrium state.
If we define a scalar function 𝑉 𝒙 by (satisfying Liapunov stability Theorem)
Which is positive definite. Then the time derivative of 𝑉 𝒙 is
Which is negative definite. Hence 𝑉 𝒙 is a Liapunov Function.
𝑉 𝒙 = 𝑥 𝑥 2 2
𝑉 𝒙 =2 𝑥 1 𝑥 𝑥 2 𝑥 2
𝑉 𝒙 =2 𝑥 1 [ 𝑥 2 − 𝑥 1 𝑥 𝑥 2 2 ]+2 𝑥 2 [ −𝑥 1 − 𝑥 2 ( 𝑥 𝑥 2 2 )]
𝑉 𝒙 =−2 𝑥 𝑥

8. Example-1
𝑉 𝒙 = 𝑥 𝑥 2 2
Since 𝑉 𝒙 becomes infinite with infinite deviation from the equilibrium state (i.e. 𝑉 𝒙 →∞ as 𝒙 →∞). The equilibrium state at the origin of the system is said to be asymptotically stable in large.

9. Example-2 Consider the following system
Using Liapunov stability theorem determine whether the equilibrium state of the system described by above state equations is asymptotically stable, asymptotically stable in large or unstable.
𝑥 𝑥 2 = 0 1 −1 −1 𝑥 1 𝑥 2

10. Example-2
𝑥 1 = 𝑥 2
𝑥 2 = −𝑥 1 − 𝑥 2
Clearly the origin of state space is the only equilibrium state.
If we define a scalar function 𝑉 𝒙 by
Which is positive definite. Then the time derivative of 𝑉 𝒙 is
Which is indefinite. Hence 𝑉 𝒙 is not a Liapunov Function.
𝑉 𝒙 = 2𝑥 𝑥 2 2
𝑉 𝒙 =4 𝑥 1 𝑥 𝑥 2 𝑥 2
𝑉 𝒙 =2 𝑥 1 𝑥 2 −2 𝑥 2 2

11. Liapunov Stability analysis of LTI systems
Consider a LTI system
Where 𝒙 is a state vector (n-vector) and A is an nxn constant matrix. We assume that A is nonsingular. Then the only equilibrium state is the origin 𝒙=𝟎.
The stability of the equilibrium state of the LTI system can be investigated by use of the second method of Liapunov.
For the system described above let us choose a possible Liapunov function as
Where P is positive definite real symmetric matrix.
𝒙 =𝑨𝒙
𝑉(𝒙)= 𝒙 𝑇 𝑷𝒙

12. Liapunov Stability analysis of LTI systems
The time derivative of 𝑉(𝒙) along any trajectory is given as
Substituting 𝒙 =𝑨𝒙 in above equation
Taking 𝒙 𝑇 pre-common and 𝒙 post common yields
Since 𝑉(𝒙) was chosen to be positive definite, we require, for asymptotic stability, that 𝑉 𝒙 be negative definite. Therefore we require that
Where
𝑉(𝒙)= 𝒙 𝑇 𝑷𝒙
𝑉 𝒙 = 𝒙 𝑇 𝑷𝒙+ 𝒙 𝑇 𝑷 𝒙
𝑉 𝒙 = (𝑨𝒙) 𝑇 𝑷𝒙+ 𝒙 𝑇 𝑷(𝑨𝒙)
𝑉 𝒙 = 𝒙 𝑇 𝑨 𝑇 𝑷𝒙+ 𝒙 𝑇 𝑷𝑨𝒙
𝑉 𝒙 = 𝒙 𝑇 ( 𝑨 𝑇 𝑷+𝑷𝑨)𝒙
𝑉 𝒙 = −𝒙 𝑇 𝑸𝒙
𝑸=−(𝑨 𝑇 𝑷+𝑷𝑨)

13. Liapunov Stability analysis of LTI systems
Hence for the asymptotic stability of the system it is sufficient that 𝑸 be positive definite.
So for LTI system 𝒙 =𝑨𝒙 necessary and sufficient condition that the equilibrium state 𝒙=𝟎 be asymptotically stable in the large is that, given any positive real symmetric matrix Q, there exists a positive definite real symmetric matrix P such that
In determining whether or not there exists a positive definite real symmetric matrix P, it is convenient to choose 𝐐=𝑰.
𝑸=−(𝑨 𝑇 𝑷+𝑷𝑨)
𝑨 𝑇 𝑷+𝑷𝑨=−𝑸
𝑨 𝑇 𝑷+𝑷𝑨=−𝑰

14. Liapunov Stability analysis of LTI systems
Comments:
If 𝑉 𝒙 = −𝒙 𝑇 𝑸𝒙 does not vanish identically along any trajectory, then 𝑸 may be chosen as positive semi-definite.
𝑉 𝒙 does not vanish along any trajectory if positive semi-definite matrix Q satisfy the following rank condition.
𝑟𝑎𝑛𝑘 𝑸 𝑸 𝑨 ⋮ 𝑸 𝑨 𝑛−1 =𝑛

15. Liapunov Stability analysis of LTI systems
Comments:
If we choose an arbitrary positive definite Matrix as Q or positive semi-definite matrix as Q and solve the matrix equation
to determine P, then the positive definiteness of P is a necessary and sufficient condition for the asymptotic stability of the equilibrium state 𝒙=𝟎.
The final result does not depend on a particular Q matrix chosen as long as it is positive definite (or positive semi-definite, as the case may be).
𝑨 𝑇 𝑷+𝑷𝑨=−𝑸

16. Example-3 Consider the following system
Determine whether the equilibrium state of the system described by above state equations is asymptotically stable, asymptotically stable in large or unstable.
𝑥 𝑥 2 = 0 1 −1 −1 𝑥 1 𝑥 2

17. Example-3 Consider the following system
The only equilibrium state of the system is at origin of state space.
Let us assume a tentative Liapunov function
Where P is to be determine from
𝑥 𝑥 2 = 0 1 −1 −1 𝑥 1 𝑥 2
𝑉(𝒙)= 𝒙 𝑇 𝑷𝒙
𝑨 𝑇 𝑷+𝑷𝑨=−𝑰

18. Example-3 𝑨 𝑇 𝑷+𝑷𝑨=−𝑰 Substituting the values yields
By expanding the matrix equations we obtain three simultaneous equations as follows
Solving for 𝑝 11 ,𝑝 12 and 𝑝 22 we obtain
𝑨 𝑇 𝑷+𝑷𝑨=−𝑰
0 1 −1 −1 𝑇 𝑝 11 𝑝 12 𝑝 12 𝑝 𝑝 11 𝑝 12 𝑝 12 𝑝 −1 −1 =−
−2𝑝 12 =−1
𝑝 11 −𝑝 12 − 𝑝 22 =0
2𝑝 12 −2 𝑝 22 =−1
𝑝 11 𝑝 12 𝑝 12 𝑝 22 =

19. Example-3
𝑝 11 𝑝 12 𝑝 12 𝑝 22 =
To test the positive definiteness of P we can use Sylvester's criterion as follows
Clearly P is a positive definite. Hence the equilibrium state at the origin is asymptotically stable in large.
Liapunov function is
>0
3 2 >0
𝑉 𝒙 = 𝒙 𝑇 𝑷𝒙= 𝒙 𝟏 𝒙 𝟐 𝒙 𝟏 𝒙 𝟐

20. Example-3 Which is further simplified as and
𝑉 𝒙 = 𝒙 𝑇 𝑷𝒙= 𝒙 𝟏 𝒙 𝟐 𝒙 𝟏 𝒙 𝟐
Which is further simplified as
and
𝑉 𝒙 = 1 2 (3 𝑥 𝑥 1 𝑥 2 +2 𝑥 2 2 )
𝑉 𝒙 =−( 𝑥 𝑥 2 2 )

21. Example-4 Consider the following system
Determine whether the equilibrium state of the system described by above state equations is asymptotically stable, asymptotically stable in large or unstable.
𝑥 𝑥 2 = 0 1 −1 −2 𝑥 1 𝑥 2

22. Example-4
The only equilibrium state of the system is at origin of state space.
Let us choose Q a positive semi-definite function as
Then the rank of
is 2. Hence we ay use Q matrix and solve equation
𝑥 𝑥 2 = 0 1 −1 −4 𝑥 1 𝑥 2 𝑄=
𝑸 𝑸 𝑨 =
𝑨 𝑇 𝑷+𝑷𝑨=−𝑸

23. Example-4
𝑨 𝑇 𝑷+𝑷𝑨=−𝑸
0 1 −1 −2 𝑇 𝑝 11 𝑝 12 𝑝 12 𝑝 𝑝 11 𝑝 12 𝑝 12 𝑝 −1 −2 =−
−2𝑝 𝑝 11 −2𝑝 12 − 𝑝 𝑝 11 −2𝑝 12 − 𝑝 𝑝 12 −4 𝑝 22 = −
From which we obtain
Matrix P is positive definite. Hence the equilibrium state 𝒙=𝟎 is asymptotically stable.
𝑝 11 =5
𝑝 12 =2
𝑝 22 =1 𝐏=

24. Example-5
Determine the stability range for the gain K the system given below.
𝑥 𝑥 𝑥 3 = −2 1 −𝐾 0 −1 𝑥 1 𝑥 2 𝑥 𝐾 𝑢

25. Example-5
In determining the stability range for the gain K we assume the input u to be zero.
We find that the origin of state space is equilibrium state.
Let us choose the positive semi-definite real symmetric matrix Q to be
𝑥 1 = 𝑥 2
𝑥 2 =−2 𝑥 2 + 𝑥 3
𝑥 3 =−𝐾 𝑥 1 − 𝑥 3 𝑸=

26. Example-5 Then the rank of 𝑸= 0 0 0 0 0 0 0 0 1𝑸=
Then the rank of
𝑟𝑎𝑛𝑘 𝑸 𝑸 𝑨 𝑸 𝑨 3 =3
𝑟𝑎𝑛𝑘 𝑸 𝑸 𝑨 𝑸 𝑨 3 =𝑟𝑎𝑛𝑘 −𝐾 0 − 𝐾 −𝐾 1 =3

27. Example-5 Hence we ay use Q matrix and solve equation
Solving the equation for the elements of P we obtain
For P to be positive definite or
𝑨 𝑇 𝑷+𝑷𝑨=−𝑸
−2 1 −𝐾 0 −1 𝑇 𝑝 11 𝑝 12 𝑝 13 𝑝 12 𝑝 22 𝑝 𝑝 13 𝑝 23 𝑝 𝑝 11 𝑝 12 𝑝 13 𝑝 12 𝑝 22 𝑝 𝑝 13 𝑝 23 𝑝 −2 1 −𝐾 0 −1 =−
𝑃= 𝐾 2 +12𝐾 12−2𝐾 6𝐾 12−2𝐾 0 6𝐾 12−2𝐾 3𝐾 12−2𝐾 𝐾 12−2𝐾 0 𝐾 12−2𝐾 −2𝐾
12−2𝐾>0
and
𝐾>0
0<𝐾<6

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End of Lectures-39-40