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Matlab Exercise

MATLAB/Simulinkによる現代制御入門



Chapter 3&4: 線形システムの時間応答&状態フィードバックによる制御



Asymptotic stability and BIBO (Bounded-Input Bounded-Output) stability
The asymptotic stability of system is equivalent to BIBO stability, if there are no common factors, zeros and poles cannot be canceled.
-> Example:

P:{˙x(t)=Ax(t)+bu(t)y(t)=cx(t), A=[-2001], b=[10], c=[10]               (1)

Eigenvalues of A are -2, 1. Therefore system equation (1) is not asymptotic stable . However, transfer function of equation (1) is:

P(s)=c(sI-A)-1b=s-1(s+2)(s-1)=1s+2                       (2)

System equation (1) is BIBO stable . No matter how state x2(t) vibrates unstably, there is no effect on y(t)=x1(t) . That is, y(t)=x1(t) will not disperse at all.
*If c is set to be [01]-1 , transfer function will turn out to be zero. That means the system is uncontrollable.


Transition matrix, Heaviside's method and inverse matrix
For state function as ˙x(t)=Ax(t) , the root of which may be obtained by Laplace transform [ref0.1] as pp.40 :

sX(s)-x(0)=AX(s)X(s)=x(0)sI-Ax(t)=eAtx(0)                                                          (2.1)

The definition of transition matrix eAt based on Taylor's series is:

eAt:=I+tA+t22!A2++tkk!Ak+                            (3)

The most common measure (some other method called diagonalization , see pp.44-45, another using solution of ODE, see [ref0.2] ) to work out transition matrix is Heaviside's method , which is

eAt=L-1[(sI-A)-1]                                                              (4)

By the way, the inverse of matrix under third order may be derived as [ref1] :

[abcd]-1=1ad-bc[d-b-ca]                                                 (5)

For matrix above second order, see pp.19-20 Theorem 1.2 of The Schur Complement and Its Applications.pdf using Schur complement . Or using adjoint matrix [ref2] .
-> Example (pp. 53):
Second order system with eigenvalue of state matrix A as conjugate complex number λ=α±jβ(α<0, β>0) can be written as

P:{˙x(t)=Ax(t)+bu(t), x(0)=x0y(t)=cx(t)                            A=[01-(α2+β2)2α], b=[01], c=[10], x0=[10]                 (6)

Via equation (5) , the transition matrix may be derived as:

(sI-A)-1=[s-1α2+β2s-2a]-1=1(s-α)2+β2[s-2α1-(α2+β2)s]=s-α(s-α)2+β2[]+β(s-α)2+β2[]=s-α(s-α)2+β2[1001]+β(s-α)2+β21β[-α1-(α2+β2)α]             (7)

Then with inverse Laplace transform via Laplace table, transition matrix may be derived as:

eAt=eαt([1001]cos(βt)+1β[-α1-(α2+β2)α]sin(βt))                       (8)