Nyquist Stability Criterion

• 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

• 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.

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

• 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

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

• 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.

• 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 \(G\circ H(\Gamma)\).

• Consider the following transfer function:

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

• It has two open-loop poles at \(-1\pm j\).
• \[F(s) = \frac{s^2+3s+2}{s^2+2s+2}\] has two zeros (close-loop poles) at \(-1\) and \(-2\).

• 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.

• By the Cauchy’s Principle of Argument: \[Z = N+P.\]
  • \(Z\): number of unstable closed-loop poles (zeros of \(F\))
  • \(P\): number of unstable open-loop poles (poles of \(F\))
  • \(N\): 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 \(N = -P\).

• 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.

• 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).

• 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 \(N = -P\).

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 \(0\).

• 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.

• 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:

\[\lim_{s\rightarrow\infty}\frac{s}{s^2+2s+2} = \lim_{s\rightarrow\infty}\frac{s}{s^2} = 0\]

• 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.

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.

• 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.

• 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.

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.

• 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.

• 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.

Consider the two loops feedback system:

Determine the range of gain \(K\) for stability of the system using Nyquist stability criterion.

• 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

$$ G(s) = \frac{K(s+0.5)}{s^3+s^2+1}.$$

• 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.

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.

• 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.

• 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.

• 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.