$R(s) = \frac{1}{s^2+2s+5}$ and $R(i\omega) = \frac{5-\omega^2-2i\omega}{(5-\omega^2)^2 + 4\omega^2}$.

The Nyquist criterion states:

Let $R$ be strictly proper and stable. Suppose that the contour $R(i\omega)$ does not pass through or wind around -1. Then $T$ has all its poles in the open left half plane, so is also strictly proper and stable.

How can I show that $R(i\omega)$ does not cross or wind around -1?

Hint: We were meant to find the real part of $R(i\omega)$ which is: $$\frac{5-\omega^2}{(5-\omega^2)^2 + 4\omega^2}$$

I am not sure if this helps or not.

Thank you!

  • 1
    $\begingroup$ Niquist criterion is often shown by plotting. $\endgroup$ – Arash May 16 '18 at 1:56

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