Skip to main content

Nyquist plot of (s+2)/[s(s+1)(s+3)]

Short answer

No open-loop RHP poles (P=0). Nyquist plot does not cross negative real axis, so -1 is not encircled (N=0). Closed-loop stable with infinite gain margin.

Draw the Nyquist plot for the following open loop transfer function G(s)H(s) = (s+2)/[s(s+1)(s+3)].

G(s)H(s)=s+2s(s+1)(s+3)=23(1+s2)s(1+s)(1+s3)G(s)H(s) = \frac{s+2}{s(s+1)(s+3)} = \frac{\tfrac23\left(1 + \tfrac{s}{2}\right)}{s(1+s)\left(1 + \tfrac{s}{3}\right)}

Open-loop poles: 0,−1,−30, -1, -3, so P = 0. There is one pole at the origin, so the contour is indented there.

1. Magnitude and phase

∣G(jω)∣=ω2+4ωω2+1ω2+9,∠G=−90∘+tan⁡−1ω2−tan⁡−1ω−tan⁡−1ω3|G(j\omega)| = \frac{\sqrt{\omega^2 + 4}}{\omega\sqrt{\omega^2 + 1}\sqrt{\omega^2 + 9}}, \qquad \angle G = -90^\circ + \tan^{-1}\frac{\omega}{2} - \tan^{-1}\omega - \tan^{-1}\frac{\omega}{3}

2. Real and imaginary parts

G(jω)=(2+jω)[−4ω2−j(3ω−ω3)]16ω4+(3ω−ω3)2G(j\omega) = \frac{(2 + j\omega)\left[-4\omega^2 - j(3\omega - \omega^3)\right]}{16\omega^4 + (3\omega - \omega^3)^2} Re=−5ω2−ω416ω4+(3ω−ω3)2,Im=−6ω−2ω316ω4+(3ω−ω3)2\text{Re} = \frac{-5\omega^2 - \omega^4}{16\omega^4 + (3\omega - \omega^3)^2}, \qquad \text{Im} = \frac{-6\omega - 2\omega^3}{16\omega^4 + (3\omega - \omega^3)^2}
  • Im < 0 for every ω > 0, so the plot never crosses the real axis at a finite frequency. The phase approaches −180° only as ω → ∞.
  • Low-frequency asymptote: as ω → 0, Re → −5/9=−0.556-5/9 = -0.556. (Equivalently K∑T=23(1+13−12)=0.556K\sum T = \tfrac23(1 + \tfrac13 - \tfrac12) = 0.556.)

3. Table (ω > 0)

ω|G|∠GReIm
0⁺∞−90°−0.556−∞
0.16.64−94.8°−0.55−6.62
0.51.21−112.0°−0.45−1.12
10.50−126.9°−0.30−0.40
20.175−142.1°−0.14−0.11
50.036−159.5°−0.03−0.01
∞0−180°00

4. Mapping of the full contour

  • ω = 0⁺ → +∞: from ∞∠−90∘\infty\angle-90^\circ (along Re = −0.556) down to the origin, arriving at −180° (third quadrant).
  • ω = −∞ → 0⁻: mirror image in the second quadrant.
  • Infinite semicircle: maps to the origin.
  • Small indentation at s = 0: G≈23/sG \approx \tfrac23/s maps to an infinite arc, clockwise through the right half-plane (from +90° to −90°).
                       ..              |
                        .              |
                        ..             |
                         ..            |
                          ..           |
                           ...         |
                             ...       |
                               .....   |
-------@---------------------------*****----------
                               *****   |
                             ***       |
                           ***         |
                          **           |
                         **            |
                        **             |
                        *              |
                       **              |
                  -0.556               0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
asymptote Re = -0.556; no negative real crossing

5. Stability

Nyquist criterion: Z=P+NZ = P + N, where PP = number of open-loop poles in the RHP, NN = net clockwise encirclements of (−1,j0)(-1, j0) by the Nyquist plot of G(s)H(s)G(s)H(s), and ZZ = number of closed-loop poles in the RHP. The closed loop is stable only if Z=0Z = 0.

The plot never cuts the negative real axis, so (−1,j0)(-1, j0) is not encircled: N = 0, and Z=0+0=0Z = 0 + 0 = 0.

The closed-loop system is stable. Also, GM = ∞ and PM = 64.8° (at ωgc\omega_{gc} = 0.588 rad/s).

Check: s3+4s2+4s+2=0s^3 + 4s^2 + 4s + 2 = 0 gives Routh first column 1,4,3.5,21, 4, 3.5, 2, all positive. ✓

✓ Answer checked by Claude Opus 5.5 (high effort)

All Control Systems chapter 6 questions →