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Nyquist contour mapping of 200/[(s+3)(s+1)(s+2)]

Short answer

The Nyquist contour encloses the RHP and maps G(s) to a plot crossing at –3.33, giving two clockwise encirclements of –1. With P=0, N=2, so Z=2 and the closed-loop system is unstable.

What is Nyquist Contour? Map the Nyquist contour of open loop transfer function G(s) = 200/[(s+3)(s+1)(s+2)] into G(s) plane and apply Nyquist criterion to check the stability of closed loop system.

Nyquist contour

The Nyquist contour is a closed path in the s-plane that encloses the entire right half-plane. It consists of:

  1. the whole jωj\omega-axis from −j∞-j\infty to +j∞+j\infty;
  2. a semicircle of infinite radius (s=Rejθs = Re^{j\theta}, R→∞R \to \infty, θ from +90° to −90°), traversed clockwise;
  3. small semicircular indentations (radius ε → 0) around any open-loop poles that lie on the jωj\omega-axis, so the path avoids them.

Mapping this contour through G(s)H(s)G(s)H(s) gives the Nyquist plot. By the argument principle, the encirclements of (−1,j0)(-1, j0) count the RHP zeros of 1+G(s)H(s)1 + G(s)H(s), i.e. the RHP closed-loop poles.

Mapping G(s)=200(s+1)(s+2)(s+3)G(s) = \dfrac{200}{(s+1)(s+2)(s+3)}

Open-loop poles: −1,−2,−3-1, -2, -3, so P = 0. No indentation is needed.

(jω+1)(jω+2)(jω+3)=(6−6ω2)+j(11ω−ω3)(j\omega+1)(j\omega+2)(j\omega+3) = (6 - 6\omega^2) + j(11\omega - \omega^3)
SectionMapping
ω = 0G=200/6=33.3∠0∘G = 200/6 = 33.3\angle0^\circ
ω = 120∠−90∘20\angle -90^\circ (crosses the −j axis)
ω = 28.77∠−142.1∘8.77\angle -142.1^\circ
ω = √11 = 3.317Im of denominator = 0, Re = 6−66=−606 - 66 = -60, so G=−3.33G = -3.33
ω = 51.25∠−205.9∘1.25\angle -205.9^\circ (second quadrant)
ω → +∞0∠−270∘0\angle -270^\circ
ω < 0mirror image about the real axis
infinite semicirclemaps to the origin
         |........................
      .....                      ......
    ...  |                            ....
  ...    |                               ...
 ..      |                                 ...
 .       |                                   ..
 ..      |                                    ..
  ..     |                                     .
---*****@*-------------------------------------*--
  **     |                                     *
 **      |                                    **
 *       |                                   **
 **      |                                 ***
  ***    |                               ***
    ***  |                            ****
      *****                      ******
         |************************
   -3.33                                     33.3
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
w = 0 at 33.3; crosses at -3.33 (w = 3.317)

Nyquist criterion

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 crosses the negative real axis at −3.33, to the left of −1. The closed curve (ω from −∞ to +∞) goes round the point (−1,j0)(-1, j0) twice clockwise, so N = 2.

Z=P+N=0+2=2Z = P + N = 0 + 2 = 2

The closed-loop system is unstable, with two closed-loop poles in the RHP.

  • Gain margin: GM=1/3.33=0.3=−10.5GM = 1/3.33 = 0.3 = -10.5 dB (negative).
  • The gain would have to be reduced below 200/3.33=60200/3.33 = 60 for stability.
  • Check (Routh): s3+6s2+11s+206s^3 + 6s^2 + 11s + 206 gives 6×11=66<2066 \times 11 = 66 < 206, so it is unstable. ✓

✓ Answer checked by Claude Opus 5.5 (high effort)

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