GATE 2022 IN Q.43: solution and answer

Official answer: ASource: Official answer key, GATE 2022 IN (IIT Kharagpur)

1. Gain crossover is at $P$. $$|P|=\sqrt{0.77^2+0.64^2}=\sqrt{0.5929+0.4096}=\sqrt{1.0025}\approx 1.00$$ The small excess over 1 comes from rounding the given coordinates to two decimals, so $P$ is the point where $|G(j{\omega }_{gc})|=1$. The curve is outside the unit circle from $Q$ ($|G|=14.40$) up to $P$ and inside it afterwards, so $P$ is the only gain-crossover point.

2. Phase at $P$: follow the curve from $\omega =0$. At $Q$ the phase is $0^{\circ }$. Moving clockwise, the curve passes through the fourth quadrant (phase $0^{\circ }$ to $-90^{\circ }$) and into the third quadrant (phase $-90^{\circ }$ to $-180^{\circ }$) before it reaches $P$. So the continuous phase at $P$ is a third-quadrant angle lying between $-90^{\circ }$ and $-180^{\circ }$: $$\angle G(j{\omega }_{gc})=-180^{\circ }+{\tan }^{-1}\left(\frac{0.64}{0.77}\right)=-180^{\circ }+{\tan }^{-1}(0.831)=-180^{\circ }+39.7^{\circ }=-140.3^{\circ }$$

3. Phase margin. $$\text{PM}=180^{\circ }+\angle G(j{\omega }_{gc})=180^{\circ }-140.3^{\circ }=39.7^{\circ }$$ Geometrically, PM is the angle between the negative real axis and the line $OP$. It is positive because the phase at gain crossover has not yet reached $-180^{\circ }$: $P$ comes before the curve's crossing of the negative real axis. If the curve had met the unit circle only after crossing that axis, the phase there would be past $-180^{\circ }$ and PM would be negative.

4. Phase crossover is at $R$. After $P$ the curve reaches the negative real axis at $R$, where $\angle G(j{\omega }_{pc})=-180^{\circ }$ and $|G(j{\omega }_{pc})|=0.21$. It is the only crossing of the negative real axis for $0<\omega <\infty$, because after $R$ the curve stays in the second quadrant and reaches the origin only as $\omega \to \infty$.

5. Gain margin. $$\text{GM}=\frac{1}{|G(j{\omega }_{pc})|}=\frac{1}{0.21}\approx 4.76$$ In decibels this is $20{\log }_{10}4.76\approx 13.6\text{dB}$. The options give GM as a ratio, so $\text{GM}=4.76$.

6. Check with the Nyquist criterion. The open loop is stable ($P_{ol}=0$). The plot meets the negative real axis only at $-0.21$, which is to the right of $-1$, so $-1$ is not encircled and the closed loop is stable. That agrees with both margins being positive ($\text{GM}>1$, $\text{PM}>0^{\circ }$).

7. Pick the option. $\text{PM}=39.7^{\circ }$ and $\text{GM}=4.76$, which is option (A). The value $0.07$ in (B) and (D) is $1/14.40$. It uses $Q$, the DC value $G(0)$, which has phase $0^{\circ }$ and plays no part in either margin. The $-39.7^{\circ }$ in (C) and (D) has the wrong sign.