The Nyquist criterion is one of the most powerful tools in classical Control Theory, but if not properly understood it might be “hard” to remember, at least, it was for me.

Let’s look at the formula:

$$\boxed{N=Z-P}$$

Where:

• $N$ is the number encirclement of the critical point $-1$. We count positively the clockwise encirclement and negatively the counterclockwise encirclements.
• $P$ is the number of poles in the right half plane of the open loop transfer function.
• $Z$ is the number of zeros in the right half plane of the open loop transfer function +1. Basically the denominator of the closed loop transfer function.

The formula is useful because the zeros of the open loop transfer function are directly mapped to the poles of the closed loop transfer function via a simple shift of the origin of the axes (more on this a few paragraphs below).

We use the formula to understand if the feedback system has poles in the right half plane (poles leading to divergence via exponential growth in time domain), thus, is not stable.

In short, for asymptotic stability, we want $Z=0$ wich implies $N=-P$, a negative number which means $P$ counterclockwise encirclements of the critical point.

• Why are we adding 1 to the open loop transfer function?

Given $G(s)$ the open loop transfer function, it’s closed loop transfer function will be $W(s)=\frac{Gs)}{1+G(s)}$ so we must sum 1 to $G(s)$ since whe are interested in the denominator of $W(s)$.

• Why we add 1 only when calculating $Z$ and not when calculating $P$?

Because the poles of $G(s)$ and $1+G(s)$ are the same while the zeros change: $$G(s)=\frac{N_{G}(s)}{D_{G}(s)} \Rightarrow 1+G(s)=\frac{D_{G}(s)+N_{G}(s)}{D_{G}(s)}$$

• My book calls $Z$ the number of poles of the closed loop system.

That is correct, but i prefer to refer only to the open loop system to avoid confusion in the formula, also Z stands for zeros and P for poles of course. That seems easier for me.

Where are those encirclements coming from?

Given a transfer function $L(s)$ if we choose a generic complex input $\bar{s}$.
$\angle{L(\bar{s})}$ will be the sum of all the angles of the phasors pointing from the zeros of $L(s)$ to the point $\bar{s}$ minus the sum of all the angles of the phasors pointing from the poles of $L(s)$ to the point $\bar{s}$.

You can see in these two pictures how $\angle{L(s)}$ is affected by the position of $s$ relative to the pole (on the left) or relative to the zero (on the right).

As you can see, encircling the zero in a clockwise direction leads to a clockwise revolution of the phase, while encircling the pole in a clockwise direction leads to a counterclockwise revolution of the phase. The effects on the phase of multiple poles and/or multiple zeros are summed up because of superposition principle.

• If $\angle{L(s)}$ does $X$ full clockwise revolutions there must be $X$ more zeros than poles.
• If $\angle{L(s)}$ does $X$ full counterclockwise revolutions there must be $X$ more poles than zeros.
• Else (if $L(s)$ does not encircle the origin), in the encircled domain there are an equal number of poles and zeros.

The domain of the Nyquist plot is an encirclement of all the right half plane, running through the imaginary axis and closing the encirclement at infinity.

Counting how many revolutions around $0$ the resulting Nyquist Plot does we find out $Z-P$.

So why are we counting revolutions around -1 and not around 0?

To study the stability of the closed-loop system we are interested to the poles of this transfer function:

$$W(s)=\frac{C(s)P(s)}{1+C(s)P(s)}$$

which are the zeros of this tranfer function:

$$1+C(s)P(s)$$

with:

• $P(s)$ the plant transfer function.
• $C(s)$ the controller transfer function.

Basically the open loop tranfer function plus 1.

So we should draw the Nyquist Plot of $1+C(s)P(s)$, but it’s easier to draw the Nyquist Plot for $C(s)P(s)$ and perform an unitary shift on the Real axis, which is the same as shifting the origin of the axis from $(0,0)$ to $(-1,0)$.

• In picture: Nyquist plot for $\frac{3+s^2}{(1+s)^2}$ in green and $1+\frac{3+s^2}{(1+s)^2}$ in yellow:

Counting the encirclements around the new shifted origin $(-1,0)$ gives use the information we needed.

How many poles/how many zeros?

Usually we know the open loop transfer function $G(s)=C(s)P(s)$ so we know how many instable poles it has.
Poles of $G(s)$ are the same as poles of $1+G(s)=1+C(s)P(s)$.
Having this information gives us the ability to know exactly how many instable poles (or zeros of $1+G(s)$) $W(s)$ has , buy plugging in $Z=N+P$ the known walues of $N$ and $P$.

example:

A plant is modelled by the transfer function $P(s)=\frac{1}{s-2}$ with a pole in the right half plane. By drawing its Nyquist plot: We find that $N=0$ while $P=1$ because of the unstable pole in the plant. The objective is asymptotic stability and from the plot above we see that with more gain we can achieve that, since we only need to encircle the critical point once.
A fine controller could be $C(s)=2.5$. Now the open loop transfer function is $G(s)=C(s)P(s)=\frac{2.5}{s-2}$.
By drawing its Nyquist plot: We see that now $N=-1=-P$ so the condition for asymptotic stability is satisfied.
Now it’s safe to perform a step response on our feedback system since we know it won’t diverge. The simple proportional controller won’t allow the system to track the step signal perfectly, so we replicate on the output a magnified version of it, to have zero error we would need an integrator in the controller. And that’s a task for another day :).

Future updates:

There’s more to say about this criterion and probably a lot of labor limae to do on this post, i’ll try to keep working on it in the future. Yet i am pretty content on how these notes came out.

~FL