1. Nyquist Stability Criterion

1 Nyquist Stability Criterion

1.1 Review

Frequency response
Bode plot
Frequency domain modeling

One Fundamental Question: Given transfer functions $G(s),\,H(s)$, how to determine if the system is closed-loop stable?
- Time Domain: Routh-Hurwitz stability criterion, Root Locus
- Frequency Domain: Nyquist Plot + Nyquist Stability Criterion

1.2 Contour and Plot

$G(s)H(s)$ is the open-loop transfer function
$G(s)H(s)$ is a mapping from complex $s$-plane to complex $GH$-plane.
A closed-path (contour) on $s$-plane will be mapped to $w = G(s)H(s)$ to a closed path (plot) on $GH$-plane.

1.3 Zeros and Poles

Consider the following transfer function \[G(s) = \frac{s}{s^2+2s+2} = \frac{s}{(s+1+j)(s+1-j)}.\]

1.4 Observation

If a clockwise contour does not encircle zeros nor poles, then the plot will not encircle the origin.
If a clockwise contour encircles a zero, then the plot will encircle the origin clockwise once.
If a clockwise contour encircles a pole, then the plot will encircle the origin counterclockwise once.
Cauchy's Argument Principle: If a clockwise contour encircles $Z$ zeros and $P$ poles, then the number of clockwise encirclements of the origin $N$, is given by

\[N = Z - P\Rightarrow Z = N+P.\]

1.5 Counting Closed-Loop Poles using Argument Principle

The closed-loop transfer function is given by

\[\frac{G(s)}{1+G(s)H(s)}\]

Define $F(s) = 1+G(s)H(s)$, we notice that
- The poles of $F(s)$ is the poles of $G(s)H(s)$
- The zeros of $F(s)$ will be poles for the close loop transfer function.

1.6 Counting Closed-Loop Poles using Argument Principle

By Cauchy's Argument Principle, for a clockwise contour $\Gamma$ on the $s$-plane:

\[Z = N+P\]

$Z$ is the number of zeros of $F$, i.e., number of closed-loop poles in the contour
$P$ is the number of poles of $F$, i.e., number of open-loop poles in the contour
$N$ is the number of clockwise encirclements of the origin for the plot $F(\Gamma)$
$N$ is also the number of clockwise encirclements of $-1$ for the plot $G\circ H(\Gamma)$.

1.7 Example

Consider the following transfer function:

\[ G(s)H(s) = \frac{s}{s^2+2s+2}.\]

It has two open-loop poles at $-1\pm j$.
\[F(s) = (s^2+3s+2)/(s^2+2s+2)\] has two zeros (close-loop poles) at $-1$ and $-2$.
The contour encircles 2 poles and 1 zeros. $Z = 1, P = 2$.
The $F$-plot encircles origin counterclockwise once. $N = -1$.

1.8 Nyquist Contour $\Gamma_n$

The stability of system is related to whether there exists any closed-loop poles (or zeros of $F(s)$) on the Right Half Plane (RHP).
We select a contour (Nyquist contour) consisting of
- Segment 1: The imaginary axis from 0 to $+j\infty$.
- Segment 2: A semicircle of infinite radius that encloses the entire right half $s$-plane.
- Segment 3: The imaginary axis from $-j\infty$ to $0$.
The Nyquist Contour is a ‘big’ semicircle that encloses the RHP. The direction of the encirclement is clockwise.

1.9 Nyquist Stability Criterion

By the Cauchy’s Principle of Argument: \[Z = N+P.\]
- $Z$ is the number of unstable closed-loop poles (zeros of $F$)
- $P$ is the number of unstable open-loop poles (poles of $F$)
- $N$ is the number of clockwise encirclements of $-1$ on the $GH$-plane for the plot $G\circ H(\Gamma)$.
The closed-loop system is stable, i.e. $Z = 0$, when $N = -P$.

1.10 Nyquist Stability Criterion

A feedback system is stable if and only if $N=-P$, i.e. the number of the counterclockwise encirclements of $–1$ point by the Nyquist plot in the $GH$-plane is equal to the number of the unstable poles of the open-loop transfer function.

1.11 Implication of Nyquist Stability Criterion

If the open-loop system is stable($P=0$), the closed-loop system is stable if and only if the Nyquist plot does not encircle $–1$ point
If the open-loop system has $P$ unstable poles, the closed-loop system is stable if and only if the Nyquist plot encircles $–1$ point $P$ times counterclockwise.
If the Nyquist plot passes through $-1$, then the system has a closed-loop pole on the imaginary axis (critically stable).

1.12 Procedure for Determining Stability using Nyquist Stability Criterion

Draw the Nyquist Plot
Determine the clockwise encirclement $N$.
From the open-loop transfer function, find the number of unstable open-loop poles ($P$)
Stable if $N = -P$.

1.12.1 How to Sketch the Nyquist Plot

Nyquist Contour consists of 3 segments:

Segment 1: The imaginary axis from 0 to $+j\infty$.
Segment 2: A semicircle of infinite radius that encloses the entire right half $s$-plane.
Segment 3: The imaginary axis from $-j\infty$ to $0$.

Segment 1:
- We use Bode plot to help us sketch the first segment.
- We need to find 4 types of point:
  - $\omega = 0$;
  - Real intersection: Phase = $180N^\circ$;
  - Imaginary intersection: Phase = $180N+90^\circ$;
  - $\omega = +\infty$;
- We can also deduce the trend of the plot around those points:
  - If the phase is decreasing, the plot goes clockwise
  - If the phase is increasing, the plot goes counterclockwise
- Plot those points on the $GH$-plane and draw a smooth line to connect them.
Segment 2:
- Consider the following open-loop transfer function:
\[G(s)H(s) = \begin{cases}\frac{s}{s^2+2s+2}&\text{strictly proper}\\\frac{s-1}{s+1}&\text{proper}\\s&\text{non-proper}\end{cases}\]
- For strictly proper function, the order of the denominator is greater than the order of the numerator:
\[\lim_{s\rightarrow\infty}\frac{s}{s^2+2s+2} = \lim_{s\rightarrow\infty}\frac{s}{s^2} = 0\]
- For proper function, the order of the denominator is no less than the order of the numerator:
\[\lim_{s\rightarrow\infty}\frac{s-1}{s+1} = \lim_{s\rightarrow\infty}\frac{s}{s} = 1\]
- Segment 2 is the origin point for strictly proper function. It is a constant for proper function.
- Non-proper transfer functions are not physically realizable.
Segment 3:
- From the property of Laplace transform:
\[G(s) = \overline{G(-s)}.\]
- Therefore, segment 3 is the mirror reflection of segment 1 around the real axis.

1.13 Example 1

Consider the system with open-loop transfer function: \[G(s)H(s) = \frac{1}{(s+1)(0.1s+1)}.\] Determine the stability of the closed-loop system using the Nyquist stability criterion.

1.13.1 Sketch Nyquist Plot

Segment 1:
- When $\omega \rightarrow 0$, $G(j\omega)H(j\omega)\rightarrow 1$.
- There is no real intersection for $0 < \omega < \infty$.
- There is a imaginary intersection when $\omega \approx 3$. The intersection is around $0.3\angle -90^\circ$. ($0.3\approx -10dB$)
- More precisely, the intersection is at $0.287\angle -90^\circ$ and the corresponding frequency is $\omega = \sqrt{10}$.
- When $\omega \rightarrow \infty$, $G(j\omega)H(j\omega)\rightarrow 0$.
- The phase is always decreasing, therefore the plot goes clockwise.
Segment 2: Since the system is strictly proper, Segment 2 is the origin.
Segment 3: Mirror reflection of segment 1.

1.13.2 Finding $N$ and $P$

The Nyquist plot does not encircle $-1$. Therefore $N = 0$.
The open-loop poles are $-1$, $-10$. Therefore $P = 0$.
$Z = N + P = 0$. The closed-loop system is stable.

1.14 Example 2

Consider a feedback system with open-loop transfer function \[G(s)H(s) = \frac{K(s-1)}{s^2+s+4},\,K>0\] Determine the range of $K$ such that the feedback system is stable.

1.14.1 Nyquist Plot

Assume $K=1$ first.

Segment 1:
- When $\omega \rightarrow 0$, $G(j\omega)H(j\omega)\rightarrow -0.25$.
- There is a imaginary intersection when $1<\omega <2$. The intersection is between $0.1j$ and $j$.
- There is a real intersection when $\omega \approx 2$. The intersection is around $1$.
- When $\omega \rightarrow \infty$, $G(j\omega)H(j\omega)\rightarrow 0$.
- The phase is always decreasing. Therefore the plot goes clockwise.
Segment 2: Since the system is strictly proper, Segment 2 is the origin.
Segment 3: Mirror reflection of segment 1.

1.14.2 Determine $N$ and $P$

For the open-loop system, the poles are at $-0.5\pm 1.94j$. Therefore, $P = 0$
If $K < 4$, then $-0.25K > -1$, the Nyquist plot does not encircle $-1$. Therefore $N = 0$ and the system is closed-loop stable.
If $K > 4$, then $-0.25K < -1$, the Nyquist plot encircle $-1$ clockwise once. Therefore $N = 1$ and $Z = 1$. There is one unstable pole for the closed-loop system.

1.15 Example 3

Consider the two loops feedback system:

Determine the range of gain $K$ for stability of the system using Nyquist stability criterion.

1.15.1 Sketch Nyquist Plot

We first compute the closed-loop transfer function of the inner loop.

\[G_2(s) = \frac{1/(s^3+s^2)}{1+1/(s^3+s^2)} = \frac{1}{s^3+s^2+1}.\]

The open-loop transfer function is

$$ G(s) = \frac{K(s+0.5)

Segment 1:
- When $\omega \rightarrow 0$, $G(j\omega)H(j\omega)\rightarrow 0.5K$.
- There is a imaginary intersection when $ω ≈ 0.7 $. The intersection is between $0.1j$ and $j$.
- There is a real intersection when $1 <\omega < 2$. Calculation shows that the exact frequency is $\omega = \sqrt{2}$ and the crossing is at $0.5K\angle -180^\circ$.
- When $\omega \rightarrow \infty$, $G(j\omega)H(j\omega)\rightarrow 0$. Calculation shows that the exact frequency is $\omega = 1/\sqrt{2}$ and the crossing is at $\sqrt{2}K\angle -270^\circ$.
- The phase is increasing around above points. Therefore the plot goes counterclockwise around the above points.
Segment 2: Since the system is strictly proper, Segment 2 is the origin.
Segment 3: Mirror reflection of segment 1.

1.15.2 Determine $N$ and $P$

For the open-loop system, the poles are at $-1.47,\,0.23\pm0.79j$. Therefore, $P = 2$
If $K < 2$, then $-0.5K > -1$, the Nyquist plot does not encircle $-1$. Therefore $N = 0,\,Z = 2$. The system has 2 unstable poles.
If $K > 2$, then $-0.5K < -1$, the Nyquist plot encircle $-1$ counterclockwise twice. Therefore $N = -2$ and $Z = 0$. The system is stable.

1.16 Example 4

Consider a feedback system with open-loop transfer function \[G(s)H(s) = \frac{K}{(s-1)(s+2)},\,K>0\] Determine the range of $K$ such that the feedback system is stable.

1.16.1 Sketch Nyquist Plot

Segment 1:
- When $\omega \rightarrow 0$, $G(j\omega)H(j\omega)\rightarrow -K/2$.
- There is no real intersection for $0 < \omega < \infty$.
- There is no imaginary intersection for $0 < \omega < \infty$.
- When $\omega \rightarrow \infty$, $G(j\omega)H(j\omega)\rightarrow 0$.
- The phase is always increasing at $\omega = 0$, therefore the plot goes counterclockwise at $-K/2$.
Segment 2: Since the system is strictly proper, Segment 2 is the origin.
Segment 3: Mirror reflection of segment 1.

1.16.2 Determine $N$ and $P$

For the open-loop system, the poles are at $1,\,-2$. Therefore, $P = 1$
If $K < 2$, then $-0.5K > -1$, the Nyquist plot does not encircle $-1$. Therefore $N = 0,\,Z = 1$. The system has 1 unstable poles.
If $K > 2$, then $-0.5K < -1$, the Nyquist plot encircle $-1$ counterclockwise once. Therefore $N = -1$ and $Z = 0$. The system is stable.

1.17 Summary

Cauchy's Argument Principle: $Z = N+P$.
Nyquist Stability Criterion: Closed-loop system is stable if and only if \[N+P = 0.\]
Sketch Nyquist plot from Bode plot
Determine stability of the closed-loop system using Nyquist stability criterion.

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