Nyquist stability of 50/[(s+1)(s+2)] and phase margin

$$G(s)H(s) = \frac{50}{(s+1)(s+2)}, \qquad G(j\omega) = \frac{50}{\sqrt{(1+\omega^2)(4+\omega^2)}}\angle\left(-\tan^{-1}\omega - \tan^{-1}\tfrac{\omega}{2}\right)$$

Open-loop poles

$-1$ and $-2$, so P = 0. There are no poles on the $j\omega$-axis.

Polar (ω: 0 → ∞) data

• The plot starts at 25 on the positive real axis, goes through the 4th and 3rd quadrants, and reaches the origin along $-180^\circ$.
• The phase reaches $-180^\circ$ only at ω = ∞, so the plot never cuts the negative real axis at a finite point.
• The $\omega < 0$ part is its mirror image. The infinite semicircle of the s-plane maps to the origin.

        |      ................
        | ......              .......
       ....                         ....
     ...|                              ....
    ..  |                                 ..
   ..   |                                  ...
   ..   |                                    ..
    ..  |                                     .
-----*@**-------------------------------------*---
    **  |                                     *
   **   |                                    **
   **   |                                  ***
    **  |                                 **
     ***|                              ****
       ****                         ****
        | ******              *******
        |      ****************
        0                                    25
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
w = 0 at 25; ends at origin along -180 deg

Nyquist stability

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

The closed curve meets the real axis only at the origin and at $+25$, so the point $-1$ lies outside it and $N = 0$.

$$Z = P + N = 0 + 0 = 0$$

The closed-loop system is stable for this gain (and in fact for any $K > 0$, because the plot never crosses the negative real axis, so GM = ∞).

Phase margin

Gain crossover: $|G| = 1$:

$$\begin{aligned} (1+\omega^2)(4+\omega^2) &= 2500 \\ \omega^4 + 5\omega^2 - 2496 &= 0 \\ \omega^2 &= \frac{-5 + \sqrt{25 + 9984}}{2} = 47.52 \\ \omega_{gc} &= 6.894\ \text{rad/s} \end{aligned}$$ $$\begin{aligned} \angle G(j\omega_{gc}) &= -\tan^{-1}6.894 - \tan^{-1}3.447 = -81.75^\circ - 73.82^\circ = -155.57^\circ \\ PM &= 180^\circ - 155.57^\circ = 24.43^\circ \end{aligned}$$

Answer: N = 0, P = 0, so Z = 0 and the closed loop is stable. PM ≈ 24.4° at $\omega_{gc}$ = 6.89 rad/s (GM = ∞).