Modern Control Systems (MCS) Lecture-39-40 Design of Control Systems in Sate Space Second Method of Liapunov Dr. Imtiaz Hussain email: imtiaz.hussain@faculty.muet.edu.pk URL :http://imtiazhussainkalwar.weebly.com/

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

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.

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.

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

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 1 2 + 𝑥 2 2 ) 𝑥 2 = −𝑥 1 − 𝑥 2 ( 𝑥 1 2 + 𝑥 2 2 )

Example-1 𝑥 1 = 𝑥 2 − 𝑥 1 ( 𝑥 1 2 + 𝑥 2 2 ) 𝑥 2 = −𝑥 1 − 𝑥 2 ( 𝑥 1 2 + 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 1 2 + 𝑥 2 2 𝑉 𝒙 =2 𝑥 1 𝑥 1 +2 𝑥 2 𝑥 2 𝑉 𝒙 =2 𝑥 1 [ 𝑥 2 − 𝑥 1 𝑥 1 2 + 𝑥 2 2 ]+2 𝑥 2 [ −𝑥 1 − 𝑥 2 ( 𝑥 1 2 + 𝑥 2 2 )] 𝑉 𝒙 =−2 𝑥 1 2 + 𝑥 2 2 2

Example-1 𝑉 𝒙 = 𝑥 1 2 + 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.

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 1 𝑥 2 = 0 1 −1 −1 𝑥 1 𝑥 2

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 1 2 + 𝑥 2 2 𝑉 𝒙 =4 𝑥 1 𝑥 1 +2 𝑥 2 𝑥 2 𝑉 𝒙 =2 𝑥 1 𝑥 2 −2 𝑥 2 2

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

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)

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

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 1 2 𝑸 1 2 𝑨 ⋮ 𝑸 1 2 𝑨 𝑛−1 =𝑛

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

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 1 𝑥 2 = 0 1 −1 −1 𝑥 1 𝑥 2

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 1 𝑥 2 = 0 1 −1 −1 𝑥 1 𝑥 2 𝑉(𝒙)= 𝒙 𝑇 𝑷𝒙 𝑨 𝑇 𝑷+𝑷𝑨=−𝑰

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 22 + 𝑝 11 𝑝 12 𝑝 12 𝑝 22 0 1 −1 −1 =− 1 0 0 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 = 3 2 1 2 1 2 1

Example-3 𝑝 11 𝑝 12 𝑝 12 𝑝 22 = 3 2 1 2 1 2 1 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 3 2 1 2 1 2 1 >0 3 2 >0 𝑉 𝒙 = 𝒙 𝑇 𝑷𝒙= 𝒙 𝟏 𝒙 𝟐 3 2 1 2 1 2 1 𝒙 𝟏 𝒙 𝟐

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

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 1 𝑥 2 = 0 1 −1 −2 𝑥 1 𝑥 2

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 1 𝑥 2 = 0 1 −1 −4 𝑥 1 𝑥 2 𝑄= 4 0 0 0 𝑸 1 2 𝑸 1 2 𝑨 = 2 0 0 0 0 2 0 0 𝑨 𝑇 𝑷+𝑷𝑨=−𝑸

Example-4 𝑨 𝑇 𝑷+𝑷𝑨=−𝑸 0 1 −1 −2 𝑇 𝑝 11 𝑝 12 𝑝 12 𝑝 22 + 𝑝 11 𝑝 12 𝑝 12 𝑝 22 0 1 −1 −2 =− 4 0 0 0 −2𝑝 12 𝑝 11 −2𝑝 12 − 𝑝 22 𝑝 11 −2𝑝 12 − 𝑝 22 2𝑝 12 −4 𝑝 22 = −4 0 0 0 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= 5 2 2 1

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

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= 0 0 0 0 0 0 0 0 1

Example-5 Then the rank of 𝑸= 0 0 0 0 0 0 0 0 1 𝑸= 0 0 0 0 0 0 0 0 1 Then the rank of 𝑟𝑎𝑛𝑘 𝑸 1 2 𝑸 1 2 𝑨 𝑸 1 2 𝑨 3 =3 𝑟𝑎𝑛𝑘 𝑸 1 2 𝑸 1 2 𝑨 𝑸 1 2 𝑨 3 =𝑟𝑎𝑛𝑘 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 −𝐾 0 −1 0 0 0 0 0 0 𝐾 −𝐾 1 =3

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 0 1 0 0 −2 1 −𝐾 0 −1 𝑇 𝑝 11 𝑝 12 𝑝 13 𝑝 12 𝑝 22 𝑝 123 𝑝 13 𝑝 23 𝑝 33 + 𝑝 11 𝑝 12 𝑝 13 𝑝 12 𝑝 22 𝑝 123 𝑝 13 𝑝 23 𝑝 33 0 1 0 0 −2 1 −𝐾 0 −1 =− 0 0 0 0 0 0 0 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 6 12−2𝐾 12−2𝐾>0 and 𝐾>0 0<𝐾<6

End of Lectures-39-40 To download this lecture visit http://imtiazhussainkalwar.weebly.com/ End of Lectures-39-40