# Question on sketching nyquist plot of transfer function with poles on imaginary axis

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?

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 $\theta \in[-\frac{\pi }{2},\frac{\pi }{2}]$.

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:

• 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. Nov 20 '16 at 11:45