Nyquist plot of (s+2)/[(s+1)(s−1)]

$$G(s) = \frac{s+2}{s^2-1} = \frac{s+2}{(s-1)(s+1)}$$

The open-loop poles are $+1$ and $-1$, so P = 1 (one RHP pole). No poles lie on the $j\omega$-axis, so the Nyquist contour needs no indentation.

Frequency response

$$G(j\omega) = \frac{2 + j\omega}{-(1+\omega^2)} = \frac{-2}{1+\omega^2} - j\frac{\omega}{1+\omega^2}$$

• For $\omega > 0$ the imaginary part is always negative, so the plot lies in the third quadrant. It runs from $-2$ (at ω = 0) to the origin (at ω = ∞), arriving along $-90^\circ$.
• The part for $\omega < 0$ is the mirror image about the real axis.
• The large semicircle ($s = Re^{j\theta}$, $R \to \infty$) maps to the origin, because $G \to 1/s \to 0$.

Complete Nyquist plot

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            ......              ......   |
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        ..                              ..
--------*---------------@----------------*--------
        **                              **
         ****                        ****|
            ******              ******   |
                 ****************        |
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       -2              -1                0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
w = 0 at -2; w -> +-inf at origin

Stability by Nyquist criterion

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 $-\infty$ to $+\infty$, the plot runs from the origin through the upper half to $-2$, then through the lower half back to the origin. This is a closed curve traced counter-clockwise.
• The point $-1$ lies inside it, since the plot crosses the real axis at $-2$, to the left of $-1$.
• So $N = -1$ (one counter-clockwise encirclement).

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

The closed-loop system is stable. Check: $1 + G = 0 \Rightarrow s^2 + s + 1 = 0$, whose roots $-0.5 \pm j0.866$ are in the LHP. ✓

Gain margin

The plot cuts the negative real axis at ω = 0 (phase crossover frequency $\omega_{pc} = 0$), where $|G| = 2$.

$$GM = \frac{1}{|G(j\omega_{pc})|} = \frac{1}{2} = 0.5 \;\Rightarrow\; GM = 20\log_{10}0.5 = -6.02\ \text{dB}$$

Because the open loop is unstable, a negative GM does not mean instability here. It means the loop gain can be reduced by a factor of 2 before the system becomes unstable. (With gain $K$: $s^2 + Ks + 2K - 1 = 0$ is stable only for $K > 0.5$.)

Answer: P = 1, N = −1 (one CCW encirclement), Z = 0, so the closed loop is stable. GM = 0.5 (−6.02 dB).