Question

The phase margin of the transfer function $G(s)=\dfrac{2(1-s)}{(1+s)^2}$ is ______ degrees (rounded off to the nearest integer).

Hint:

For $G(s)=K\frac{(1-\tfrac{s}{z})}{(1+\tfrac{s}{p})^2}$, at $|G|=1$ the phase is the sum of zero/ pole phase lags. Here it becomes exactly $-180^\circ$, giving PM $=0^\circ$.

Answer 1

Correct Answer: -2

Evaluate on $s=j\omega$: \; $G(j\omega)=\dfrac{2(1-j\omega)}{(1+j\omega)^2}$.
Magnitude: \[ |G(j\omega)|=\frac{2|1-j\omega|}{|(1+j\omega)^2|}=\frac{2\sqrt{1+\omega^2}}{1+\omega^2}=\frac{2}{\sqrt{1+\omega^2}}. \] Gain crossover: $|G|=1 \Rightarrow \sqrt{1+\omega_g^2}=2 \Rightarrow \omega_g^2=3 \Rightarrow \omega_g=\sqrt{3}$.
[2mm] Phase: \[ \angle G(j\omega)=\angle(1-j\omega)-2\,\angle(1+j\omega) =-\tan^{-1}\!\omega-2\tan^{-1}\!\omega=-3\tan^{-1}\!\omega. \] At $\omega_g=\sqrt{3}$: $\tan^{-1}(\sqrt{3})=60^\circ \Rightarrow \angle G(j\omega_g)=-180^\circ$.
Phase margin: $\text{PM}=180^\circ+\angle G(j\omega_g)=180^\circ-180^\circ=0^\circ$ (nearest integer $=0$).