Nyquist stability of (s+5)/[(s−2)(s+2)] and GM

$$G(s)H(s) = \frac{s+5}{(s-2)(s+2)} = \frac{s+5}{s^2 - 4}$$

Open-loop poles: $+2, -2$, so P = 1 (one RHP pole). No poles lie on the $j\omega$-axis.

Frequency response

$$G(j\omega) = \frac{5 + j\omega}{-(\omega^2 + 4)} = \frac{-5}{\omega^2 + 4} - j\frac{\omega}{\omega^2 + 4}$$

• For ω > 0 the plot is in the third quadrant, from −1.25 to the origin (arriving along −90°).
• For ω < 0 it is the mirror image in the second quadrant.
• The infinite semicircle maps to the origin.

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              .......................    |
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         ...                            ..
---------*-----@-------------------------*--------
         ***                            **
           ****                     *****|
              ***********************    |
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       -1.25                             0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
w = 0 at -1.25; w -> +-inf at origin

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 $Z = 0$.

As ω goes from −∞ to +∞, the curve runs from the origin through the upper half to −1.25 and back through the lower half. This is a counter-clockwise loop, and it contains −1 because −1.25 < −1. So N = −1.

$$Z = P + N = 1 - 1 = 0$$

The closed-loop system is stable. Check: $s^2 - 4 + s + 5 = s^2 + s + 1 = 0$ has roots $-0.5 \pm j0.866$. ✓

Gain margin

Phase crossover is at ω = 0, where $|G| = 5/4 = 1.25$.

$$GM = \frac{1}{1.25} = 0.8 = 20\log0.8 = -1.94\ \text{dB}$$

For this open-loop unstable system, the negative GM means the gain can be reduced by at most a factor of 0.8 (gain $K > 0.8$ is needed) before the system becomes unstable.

Answer: P = 1, N = −1, Z = 0, so the system is stable. GM = 0.8 (−1.94 dB).