Nyquist Stability

Nyquist Plots

Contour

$\Gamma$ is a contour if it is a simple, closed curve with a direction.

• Simple: No self intersections
• Closed: Starts and ends at the same point

Example 11.1

Lemma

Let $p\in \mathbb{C}$. Let $\Gamma$ be a contour. Then

$$\frac{1}{2\pi j}{\oint }_{\Gamma }\frac{1}{z-p}dz=\{\begin{array}{ll}1& \text{ if }\Gamma \text{ encloses }P\\ 0& \text{ otherwise}\end{array}$$

Lemma: Argument Principle

Let $\Gamma$ be a contour, and $G[z]$ be real, rational, and proper. then $N=Z-P$

• $N$ is the number of times $G(\Gamma )$ encloses the origin
• $Z$ is the number of zero enclosed by $\Gamma$ (counting multiplicities)
• $P$ is the number of poles enclosed by $\Gamma$ (counting multiplicities)

If we choose $\Gamma$ to be the unit circle, $Z$/$P$ are the number of stable zeros and poles.

Choose $V[z]=1+K[z]=1+L[z]$ . Then,

$$N=\text{\# of encirclements of }-1\text{ by }L[\Gamma ]$$

• Use $-\frac{1}{K}$ is varying $K$

Let $L[z]=\frac{\overline{N}[z]}{\overline{D}[z]}$, where $\overline{N}$ and $\overline{D}$ are coprime. Then, the characteristic polynomial is:

$$\Delta [z]=\overline{N}[z]+\overline{D}[z]$$

Thus,

$$V[z]=1+L[z]=\frac{\overline{N}[z]+\overline{D}[z]}{\overline{D}[z]}=\frac{\Delta [z]}{\overline{D}[z]}$$

Then,

$$\begin{array}{r}\text{zeros}(V[z])=\text{roots}(\Delta [z])=\text{closed loop poles}⟹Z=\text{\# of stable CL poles}\\ \text{poles}(V[z])=\text{roots}(\overline{D}[z])=\text{open loop poles}⟹P=\text{\# of stable OL poles}\end{array}$$

by the argument principle.

Nyquist Stability Theorem

Nyquist Stability Theorem

Given

$$N=Z-P$$

• $Z$ is the number of stable CL poles
• $P$ is the number of stable OL poles $N$ is the number of encirclements of $-1$ by $L[\Gamma ]$, where $\Gamma$ is the unit circle (traversed in the positive direction) and

$$L[z]=G[z]D[z]$$

The plot of $L[\Gamma ]$ is called the Nyquist plot.

Corollary

The feedback system is closed-loop stable if and only if $N$ = number of unstable open-loop poles.

Proof sketch – We desire CL stability, therefore:

$$\begin{array}{rl}\text{\# of stable CL poles}& =\text{\# of CL poles}\\ & =\text{\# of OL poles}\\ & =\text{\# of stable OL poles}+\text{\# of unstable OL poles}\end{array}$$

Then:

$$\begin{array}{rl}& \underset{Z}{\underbrace{\text{\# of stable CL poles}}}-\underset{P}{\underbrace{\text{\# of stable OL poles}}}\\ & =\text{\# of unstable OL poles}\end{array}$$

Therefore $N$ is the number of unstable OL poles.

Ex 11.2