I have encountered some issues with sketching the nyquist plot for the following loop transfer function:

$L(s) = \frac{25K}{S^2+25}$

I know that you first take:

$L(\omega=0) = K$

$L(\omega = infinity) = |0| e^{-180}$

For the larger part of the D contour, as for the smaller section of the D counter, I sub $\omega = \epsilon e^{\gamma}$ to represent a number with small magnitude and phase, if I substitute that into the transfer function, that is $L(\epsilon e^{\gamma}) = \frac{25K}{25} = K$ as $\epsilon e^{\gamma} + 25 \approx 25 $

From this information I can do a rough sketch of the nyquist plot, but when I plot it out on matlab I get something very different. Is my thought process on this correct?


1 Answer 1


For the pole at $5 i$, the contour that has to be considered is $5 i+\epsilon e^{i \theta }$. Here $\epsilon \to 0$ and $\theta \in[-\frac{\pi }{2},\frac{\pi }{2}]$.

The denominator of the transfer function becomes $\left(5 i+\epsilon e^{i \theta } \right)^2+25$. Expanding and neglecting higher order terms, we get $-10 \epsilon \sin (\theta )+10 i \epsilon \cos (\theta )$.

The transfer function becomes $$\frac{25}{-10 \epsilon \sin (\theta )+10 i \epsilon \cos (\theta )}$$ which after simplification is $$-\frac{5 (\sin (\theta )+i \cos (\theta ))}{2 \epsilon }$$

This is basically a semicircle of infinite radius because $\epsilon \to 0$ and it lies in the lower half plane because $\theta \in[-\frac{\pi }{2},\frac{\pi }{2}]$.

enter image description here

A similar analysis can be done for the pole at $-5 i$, and the semicircle will lie in the upper half plane.

Finally, the complete Nyquist plot will loook like the following:

enter image description here

  • $\begingroup$ I would like to note that taking $\theta\in[-\frac{\pi}{2},-\frac{3\pi}{2}]$ is just as valid as taking $\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$. But when considering the closed loop stability you either have to include or exclude the open loop poles on the imaginary axis as unstable in the Nyquist criterion. $\endgroup$
    – fibonatic
    Nov 20, 2016 at 11:45

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