Step-by-step Nyquist plot example. Part I

We went over how to sketch straight-line approximations in Bode plots in a series of posts. In this post we continue those examples by going from the Bode plot to a Nyquist plot.

Consider for example the second-order minimum-phase transfer-function:

$$G(s) = \frac{s+2}{s^2+11 s + 10}$$

and the corresponding sketches for the magnitude and phase of its frequency response:

discussed earlier in this and that post.

The key for sketching the Nyquist plot is to focus on the phase and not on the magnitude. In particular, focus on the crossings of the real and imaginary axis, say the phases $0^\circ $, $90^\circ$, $180^\circ$, $270^\circ$, etc.

In the above example, there is only two relevant angles that we need to pay attention to: $0^\circ$ and $-90^\circ$. $0^\circ$ is associated with a magnitude of $-13$dB which corresponds to a radius of $0.2$. From there, the phase decreases, that is it becomes negative. This can be sketched as in the following figure:

The next relevant angle is $-90^\circ$, which is associated with a magnitude of $-\infty$dB, that is, a radius of $0$, as in the next figure:

This is it for this plot, so just connect the two ends to get:

Because we read this information from the Bode plot sketch, only positive frequencies have been considered. The negative frequencies correspond to the mirror image of the current plot, that is:

As for closed-loop stability for $L(s) = G(s)$, since this plot is completely on the right-hand side of the complex plane, there will be no encirclements of $-1/\alpha$, that is

$$\frac{1}{2\pi} \Delta_{\Gamma}^{-1/\alpha} L(s) = 0.$$

Because $L(s)$ is asymptotically stable, $P_\Gamma = 0$ and

$$Z_\Gamma = P_\Gamma \, – \, \frac{1}{2\pi} \Delta_{\Gamma}^{-1/\alpha} L(s) = 0,$$

the closed-loop is asymptotically stable for all values of $\alpha > 0$.

Concerning stability margins, since the Nyquist diagram never reaches $180^\circ$, the gain margin is infinite. The phase margin depends on the value of $\alpha$ selected. For $\alpha = 1$ the Nyquist diagram never crosses the unit circle and therefore it also has infinite gain margin.