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A Multi-Input Multi-Output system with stable transfer matrix $L(s)$ is considered.

The theorem states that if $\rho(L(j\omega)<1$ then the closed-loop system is stable.

In place of the actual theorem its reverse is proved: if the closed-loop is unstable (that is the Nyquist plot of $\det(I+L(s))$ does encircle the origin) then there exists an eigenvalue $\lambda_i(L(j\omega))$ which is larger than $1$ at some $\omega$.

The textbook says: if $\det(I+L(s))$ does encircle the origin then there must exists a gain  $\epsilon \in (0, 1]$ and a frequency $\omega'$ such that: $$\det(I+\epsilon L(j\omega'))=0$$

I really cannot figure out the reason why this holds.

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  • $\begingroup$ what is $L$ in terms of the representation $\dot{x} = Ax + Bu$? I'm assuming the system is LTI because otherwise the Nyquist criterion doesn't apply. $\endgroup$ – Mortified Through Math Feb 20 '17 at 17:08
  • $\begingroup$ @ALB $L$ is the loop gain of the MIMO feedback system, that the product of all the transfer functions (Laplace transforms, so LTI system of course) in the loop (feedforward and feedback paths). For the sake of simplicity, it is assumed the $L(s)$ is the transfer function of the direct path, while the feedback is unitary: in this case the (matrix) transfer function between the reference input and the output is $\frac{Y(s)}{R(s)} = L(s)\left( I+L(s) \right)^{-1}$ $\endgroup$ – Vexx23 Feb 21 '17 at 19:01
  • $\begingroup$ I'm in the process of writing a answer, but just to be 100% clear, the system is LTI with A, B ,C ,D as usual and introduces the state feedback $u(t) = K(r-x)$, making $L = (sI-A)^{-1}BK$. This is the state realization of the diagram which closes the loop around unity inverting FB in the feedback path and preamplifies it by the LTI gain $K$ before feeding into the plant. $\endgroup$ – Mortified Through Math Feb 22 '17 at 20:15
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This is the standard homotopy argument. The closed curve $$ \gamma_\epsilon(\omega)=\det(I+\epsilon L(j\omega)) $$ depends continuously on $\epsilon$ and $\omega$ (the property of determinant). We know that for $\epsilon=1$ it encircles the origin at least once, and for $\epsilon=0$ it reduces to the point $(1,0)$ which does not encircle the origin. Therefore, due to continuity, the origin must escape from the area inside the curve to the outside at some $\epsilon'\in(0,1)$, that is for this $\epsilon'$ the origin must lie exactly on the curve $$\gamma_{\epsilon'}(\omega')=0.$$

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  • $\begingroup$ Very clever, I was starting to think to something similar. $\endgroup$ – Vexx23 May 16 '17 at 16:07

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