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Short note: Nyquist stability criterion

Short answer

Nyquist criterion maps the Nyquist contour (encircling the RHP) through G(s)H(s). Z=P+N, where P=open-loop RHP poles, N=clockwise encirclements of –1. For stability Z=0, so N=–P. It handles delays, experimental data, and open-loop unstable systems.

Write a short note on Nyquist stability criterion.

The Nyquist stability criterion is a graphical frequency-domain method that finds the number of closed-loop poles in the right half ss-plane from the plot of the open-loop transfer function G(jω)H(jω)G(j\omega)H(j\omega). It is based on Cauchy's principle of the argument.

Statement

If the Nyquist contour (the whole jωj\omega axis plus an infinite semicircle enclosing the RHP, traversed clockwise) is mapped through G(s)H(s)G(s)H(s), then

N=Z−PorZ=P+NN = Z - P \quad\text{or}\quad Z = P + N
  • PP = number of open-loop poles in the RHP
  • ZZ = number of closed-loop poles (zeros of 1+GH1+GH) in the RHP
  • NN = number of clockwise encirclements of the critical point −1+j0-1+j0

The closed-loop system is stable only if Z=0Z=0, i.e. N=−PN=-P (the plot encircles −1-1 anticlockwise PP times). For an open-loop stable system (P=0P=0): stable if the Nyquist plot does not encircle −1+j0-1+j0.

Steps

  1. Find PP from the open-loop poles.
  2. Plot G(jω)H(jω)G(j\omega)H(j\omega) for ω=0→∞\omega=0\to\infty, add its mirror image, and close it (poles at origin give infinite clockwise arcs).
  3. Count NN around −1+j0-1+j0, compute ZZ.

Merits

  • Works with experimental frequency-response data.
  • Handles time delay and open-loop unstable systems.
  • Gives relative stability (gain and phase margins) as well.

Example: GH=60(s+1)(s+2)(s+5)GH=\dfrac{60}{(s+1)(s+2)(s+5)} cuts the real axis at −0.476-0.476; P=0P=0, N=0N=0, so Z=0Z=0: stable.

✓ Answer checked by Claude Opus 5.5 (high effort)

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