The following question is from a System Theory test without answers or solutions.
Let a continuous-time LTI system be given by the following differential equations:
$\frac{d^2}{dt^2}y_1(t)+4\frac{d}{dt}y_1(t)-\frac{d}{dt}y_2(t)+4y_1(t)-2y_2(t)=0\quad (1)$
$\frac{d}{dt}y_2(t)+y_2(t)-\frac{d}{dt}u_1(t)+u_2(t)=0\qquad \qquad \quad \qquad (2)$
Here, $u_i (t) \in R$ are inputs and $y_i(t) \in R$ are outputs, $i = 1,2$ at time $t \in R^+$. a minimal state-space representation of this system is given by
$\frac{d}{dt}x(t)=Ax(t)+Bu(t), \quad y(t)=Cx(t)+Du(t), \quad t \in R^+$
$\qquad a) \quad$ What is the order of this system?
$\qquad b) \quad$ What can be said about the stability of this system?
$\qquad \qquad b_A)$ Only Lyapunov stable.
$\qquad \qquad b_B)$ Both BIBO stable and Lyapunov stable but not asymptotically stable.
$\qquad \qquad b_C)$ BIBO stable, Lyapunov stable and asymptotically stable.
$\qquad \qquad b_D)$ It is BIBO stable, but there is not enough information to know wether it is Lyapunov $\qquad$ $\qquad$ $\quad$ or asymptotically stable.
$\qquad \qquad b_E)$ Not Lyapunov, not BIBO and not asymptotically stable.
My approach:
First take the Laplace transforms of $(1)$ and $(2)$. This gives:
$(3)\quad s^2y_1+4sy_1-sy_2+4y_1-2y_2=0$
$(4)\quad sy_2+y_2-su_1+u_2=0$
Rewriting $(2)$ gives:
$y_2=\frac{su_1-u_2}{s+1}$
Substituting this in $(3)$ we get, after simplifying:
$y_1=\frac{u_1s-u_2}{(s+1)(s+2)}$
The order of this system is 2 and looking at the pole locations we can conclude that this system is BIBO, Lyapunov and asymptotically stable.
Is this approach correct?
Best Answer
This conclusion is correct since all poles have a negative real part. However if the transfer function would have had more then one pole on the imaginary axis it would not have been possible to conclude whether the system is Lyapunov stable or not.
A more general approach would be to calculate the minimal state-space model. If its $A$ matrix has only eigenvalues with negative real then the system is Lyapunov, BIBO and asymptotically stable. Since it is a minimal realization the system would not be BIBO or asymptotically stable whenever the $A$ matrix has an eigenvalue along the imaginary axis, but the system still could be Lyapunov stable. Namely the system is still Lyapunov stable whenever the geometric multiplicity of the eigenvalues on the imaginary axis is at most one.