Polar and Nyquist plots, Nyquist stability criterion

Why it matters

The Nyquist criterion is the most general frequency-domain stability test: it works for loops with dead time, for open-loop unstable plants and for conditionally stable systems where Bode margins mislead. Because it uses the open-loop frequency response — something you can measure on a running plant — it tells you whether closing the loop is safe before you close it.

Key ideas

Polar plot: the locus of the complex number G(jω)H(jω) in the complex plane as ω goes from 0 to ∞. Each point has length |GH| and angle ∠GH; frequency is a parameter along the curve, not an axis. Rules of thumb for sketching:

• Start (ω → 0): a type-0 plot starts on the positive real axis at Kp; type 1 starts at infinity along −90°; type 2 at infinity along −180°.
• End (ω → ∞): for a strictly proper function the plot ends at the origin, arriving at angle −90°·(n − m) (n poles, m zeros).
• Crossings: set Im(GH) = 0 to find where the plot cuts the real axis (the phase crossover for the negative real axis), and Re(GH) = 0 for the imaginary axis.
• A type-1 plot approaches its start along a vertical asymptote whose real part is lim(ω→0) Re[GH(jω)].

Nyquist plot: the image under GH of the Nyquist contour — the whole jω-axis plus a semicircle of infinite radius enclosing the right half-plane (RHP). The part for negative ω is the mirror image (about the real axis) of the polar plot. If GH has poles on the jω-axis (e.g. at the origin), the contour detours around them by small semicircles into the RHP, and each such detour maps into a large arc in the GH-plane: for a pole of order N at the origin, the arc has infinite radius and sweeps N × 180° clockwise.

Principle of the argument: as s travels clockwise once round a closed contour, the curve F(s) encircles the origin clockwise Z − P times, where Z and P are the zeros and poles of F inside the contour. With F(s) = 1 + G(s)H(s):

• zeros of F are the closed-loop poles;
• poles of F are the open-loop poles;
• encircling the origin by 1 + GH is the same as encircling −1 + j0 by GH.

Nyquist stability criterion: Z = N + P, where N = number of clockwise encirclements of −1 by the Nyquist plot of GH, P = number of open-loop poles in the RHP, Z = number of closed-loop poles in the RHP. The closed loop is stable only if Z = 0, i.e. N = −P: the plot must encircle −1 counter-clockwise P times. For the common case P = 0 this reduces to "no encirclement of −1".

Relative stability follows directly: the gain margin is how far the negative-real-axis crossing is from −1 (as a ratio), and the phase margin is the angle between the unit-circle crossing and the negative real axis (see the next topic).

Dead time e^(−sT) simply rotates every point of the plot clockwise by ωT rad without changing its length, producing a spiral into the origin — Nyquist handles it exactly, while Routh cannot.

Formulas

• Z = N + P — Nyquist criterion (N counted clockwise; counter-clockwise encirclements count negative).
• Closed-loop stable ⇔ Z = 0 ⇔ N = −P.
• Im[G(jω)H(jω)] = 0 → phase-crossover frequency ωpc; real-axis crossing point x = Re[GH(jωpc)].
• GM = 1/|GH(jωpc)|.
• G(jω) = K/[jω(1 + jωT)] has Re → −K·T as ω → 0 (low-frequency asymptote).
• GH(jω)·e^(−jωT) — dead time adds phase −ωT rad with no change of magnitude.

Symbols: ω rad/s; N, P, Z integers; K loop gain; T time constant or dead time (s). The criterion assumes an LTI loop and counts poles and zeros with multiplicity.

Worked examples

Example 1 — polar plot of a type-1 loop Given: G(s)H(s) = 10/[s(1 + 0.5s)].

1. G(jω) = 10/[jω(1 + j0.5ω)] = −5/(1 + 0.25ω²) − j·10/[ω(1 + 0.25ω²)].
2. As ω → 0: magnitude → ∞, phase → −90°, and the real part → −5. The plot comes in from infinity along the vertical asymptote Re = −5.
3. At ω = 1 rad/s: G = −4 − j8.
4. As ω → ∞: magnitude → 0, phase → −180°; the plot reaches the origin tangent to the negative real axis.
5. Im(G) = 0 only as ω → ∞, so the plot never crosses the negative real axis at a finite point.

Result: the polar plot lies in the third quadrant, from infinity along Re = −5 to the origin at −180°. With P = 0 and no encirclement of −1, the closed loop is stable for every K > 0 (infinite gain margin).

Example 2 — Nyquist test for a third-order type-1 loop (GATE level) Given: G(s)H(s) = K/[s(s + 1)(s + 2)]. Find where the plot crosses the negative real axis, the critical K, and the stability for K = 10.

1. Phase crossover: −90° − tan⁻¹ω − tan⁻¹(ω/2) = −180° → tan⁻¹ω + tan⁻¹(ω/2) = 90° → ω·(ω/2) = 1 → ωpc = √2 = 1.414 rad/s.
2. Magnitude there: |G| = K/[√2 × √3 × √6] = K/6. The crossing point is −K/6.
3. Critical gain: crossing at −1 → K = 6.
4. For K = 10: crossing at −10/6 = −1.67, to the left of −1. P = 0 (open-loop poles at 0, −1, −2; the origin pole is bypassed by the contour). Closing the plot with the infinite clockwise arc from the pole at the origin, −1 is encircled twice clockwise: N = 2.
5. Z = N + P = 2 + 0 = 2.
6. Check with Routh: s³ + 3s² + 2s + 10: first column 1, 3, −4/3, 10 → two sign changes.

Result: ωpc = 1.414 rad/s, crossing at −K/6, critical K = 6; for K = 10 the closed loop has 2 RHP poles (unstable).

Common mistakes

• Using "no encirclement means stable" when the open loop has RHP poles (P ≠ 0).
• Mixing sign conventions: decide whether N counts clockwise or counter-clockwise and keep Z = N + P consistent with it.
• Forgetting the infinite arc produced by open-loop poles at the origin, which closes the plot and decides the encirclement count.
• Counting encirclements of the origin instead of −1 + j0.
• Plotting G alone when the loop has a non-unity H; the test is on G·H.
• Thinking dead time changes the magnitude: it only rotates the plot.

For GATE IN

Expect questions that give G(s)H(s) and ask for the negative-real-axis intercept, the phase-crossover frequency, the critical gain, or the number of closed-loop RHP poles from a described Nyquist plot with given P. Also common: identifying the start and end behaviour of polar plots by system type and order, and the effect of dead time. Practise the Im = 0 method and the Z = N + P bookkeeping.

Quick check

1. Where does the polar plot of 1/(1 + jωT) lie?
2. An open loop has P = 1. How many counter-clockwise encirclements of −1 are needed for closed-loop stability?
3. For K/[s(s + 1)(s + 2)], at what point does the plot cut the negative real axis when K = 3?
4. Does adding dead time change the magnitude of G(jω)?

Answers: 1. A semicircle of diameter 1 in the fourth quadrant, from 1 to 0; 2. One; 3. At −3/6 = −0.5; 4. No — only the phase.