GATE 2023 IN Q.50: solution and answer

Official answer: -2 to 2Source: Official answer key, GATE 2023 IN (IIT Kanpur)

1. Write the frequency response: $G(j\omega )=\frac{2(1-j\omega )}{(1+j\omega {)}^2}$. The zero is at $s=+1$, in the right half-plane, so the system is non-minimum phase.

2. Magnitude. The RHP zero has the same magnitude as an LHP zero: $|1-j\omega |=\sqrt{1+{\omega }^2}$. So $$|G(j\omega )|=\frac{2\sqrt{1+{\omega }^2}}{1+{\omega }^2}=\frac{2}{\sqrt{1+{\omega }^2}}$$

3. Gain crossover frequency, from $|G(j{\omega }_{gc})|=1$: $\sqrt{1+{\omega }_{gc}^2}=2$, so ${\omega }_{gc}^2=3$ and ${\omega }_{gc}=\sqrt{3}\text{rad/s}$.

4. Phase. The RHP zero $(1-j\omega )$ contributes $-{\tan }^{-1}\omega$, a lag like a pole's. The double pole adds $-2{\tan }^{-1}\omega$. The gain $2$ is positive, so it adds no phase. Therefore $\angle G(j\omega )=-3{\tan }^{-1}\omega$.

5. At gain crossover: $\angle G(j\sqrt{3})=-3{\tan }^{-1}\sqrt{3}=-3(60^{\circ })=-180^{\circ }$. Direct check: $1-j\sqrt{3}=2\angle -60^{\circ }$ and $(1+j\sqrt{3}{)}^2=4\angle 120^{\circ }$, so $G(j\sqrt{3})=\frac{2\cdot 2\angle -60^{\circ }}{4\angle 120^{\circ }}=1\angle -180^{\circ }=-1$.

6. Phase margin: $$\text{PM}=180^{\circ }+\angle G(j{\omega }_{gc})=180^{\circ }-180^{\circ }=0^{\circ }$$ The Nyquist plot passes through $-1$ at $\omega =\sqrt{3}\text{rad/s}$, so the gain and phase crossover frequencies coincide and the closed loop is marginally stable.