GATE 2018 IN Q.14: solution and answer

Official answer: 180, or -180Source: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Frequency response: $G(j\omega )=\frac{2}{(1+j\omega )(2+j\omega )}$, so $$|G(j\omega )|=\frac{2}{\sqrt{{\omega }^2+1}\sqrt{{\omega }^2+4}},\angle G(j\omega )=-{\tan }^{-1}\omega -{\tan }^{-1}\frac{\omega }{2}$$

2. Gain crossover condition $|G(j{\omega }_{gc})|=1$: $$({\omega }^2+1)({\omega }^2+4)=4\Rightarrow {\omega }^4+5{\omega }^2+4=4\Rightarrow {\omega }^2({\omega }^2+5)=0$$

3. The only real, non-negative root is ${\omega }_{gc}=0\text{rad/s}$ (the other factor gives ${\omega }^2=-5$, which has no real solution). Check: the DC gain is $G(0)=\frac{2}{1\cdot 2}=1$, i.e. $0\text{dB}$, and since $({\omega }^2+1)({\omega }^2+4)$ increases for every $\omega >0$, $|G(j\omega )|$ falls strictly below $1$ after that. The magnitude curve therefore touches $0\text{dB}$ only at $\omega =0$.

4. Phase at gain crossover: $\angle G(j0)=-{\tan }^{-1}0-{\tan }^{-1}0=0^{\circ }$. (On the Nyquist plot, the curve starts at the point $1+j0$, which is exactly $180^{\circ }$ away from the critical point $-1+j0$ on the unit circle.)

5. Phase margin definition: $$\text{PM}=180^{\circ }+\angle G(j{\omega }_{gc})=180^{\circ }+0^{\circ }=180^{\circ }$$