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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?

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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.

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  • $\begingroup$ So, when there's an eigenvalue on the imaginary axis, the system is not BIBO when it's a minimal realization (no lower order state space form). Is this only true for minimal realizations? And I thought that the minimal state space model A matrix has the same eigenvalues as the poles of the minimal transfer function. Is this not true? Thanks very much for your answers Kwin. You've helped me a lot on these subjects, especially on BIBO stability. $\endgroup$ – user463102 Aug 6 '18 at 22:09
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    $\begingroup$ @user463102 When the system is not minimal means that it can have uncontrollable or unobservable states. If the uncontrollable states are the only states associated with not stable eigenvalues, then the system is still BIBO, since BIBO assumes zero initial state and those possible unstable states will remain at zero since they can't be actuated. If the unobservable states are the only states associated with not stable eigenvalues, then the system is also still BIBO, since the possible unstable states do not appear in the output so the output remains bounded. $\endgroup$ – Kwin van der Veen Aug 6 '18 at 22:29
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    $\begingroup$ @user463102 If I remember correctly then indeed the poles of the transfer function will be the same as the eigenvalues on the state space model (the zeros are a bit more tricky), however you can not deduce the multiplicity directly from the transfer function and thus whether a system might be Lyapunov stable or not. $\endgroup$ – Kwin van der Veen Aug 6 '18 at 22:42

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