GATE 2020 IN Q.55: solution and answer

Official answer: 65.4 to 65.6Source: Official answer key, GATE 2020 IN (IIT Delhi)

1. Put $s=j\omega$: $G(j\omega )H(j\omega )=\frac{2(1+j\omega )}{(j\omega {)}^2}=-\frac{2(1+j\omega )}{{\omega }^2}$.

2. Magnitude: $|GH|=\frac{2\sqrt{1+{\omega }^2}}{{\omega }^2}$. Phase: the double pole at the origin contributes $-180^{\circ }$ and the zero at $s=-1$ adds ${\tan }^{-1}\omega$, so $\angle GH=-180^{\circ }+{\tan }^{-1}\omega$.

3. Gain crossover, where $|GH|=1$: $2\sqrt{1+{\omega }^2}={\omega }^2\Rightarrow 4(1+{\omega }^2)={\omega }^4\Rightarrow {\omega }^4-4{\omega }^2-4=0$.

4. Solve the quadratic in ${\omega }^2$ and keep the positive root: $${\omega }_{gc}^2=\frac{4+\sqrt{16+16}}{2}=2+2\sqrt{2}=4.828\Rightarrow {\omega }_{gc}=2.197\text{rad/s}$$ Check: $\frac{2\sqrt{1+4.828}}{4.828}=\frac{2(2.414)}{4.828}=1$.

5. Phase margin definition: $\text{PM}=180^{\circ }+\angle GH(j{\omega }_{gc})=180^{\circ }-180^{\circ }+{\tan }^{-1}{\omega }_{gc}={\tan }^{-1}(2.197)$.

6. $\text{PM}={\tan }^{-1}(2.197)=65.53^{\circ }\approx 65.5^{\circ }$. (The phase stays above $-180^{\circ }$ for every $\omega >0$, so there is no phase crossover; consistently, the closed-loop characteristic equation $s^2+2s+2=0$ has both roots in the left half-plane.)