GATE 2022 EE Q.17: solution and answer

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

1. First order and stable. A first-order system has exactly one pole, and because the system is stable that pole is in the LHP. Options (C) and (D) both have two poles, so neither is first order, and (D) is unstable as well. That leaves one LHP pole at $s=-a$ ($a>0$) and one zero.

2. Flat magnitude. Let the zero be at $s=\pm b$ with $b>0$. Then $$|G(j\omega )|=K\frac{\sqrt{{\omega }^2+b^2}}{\sqrt{{\omega }^2+a^2}}$$ At $\omega \to 0$ this is $Kb/a$, and as $\omega \to \infty$ it tends to $K$. It is constant at every frequency only when $b=a$, so the zero sits at the same corner frequency as the pole. What remains to decide is which half-plane the zero is in.

3. Case (B), LHP zero: $G(s)=K\frac{s+a}{s+a}=K$. The pole and zero cancel, so the phase is $0^{\circ }$ at every frequency. That contradicts the plot, where the phase falls to $-180^{\circ }$.

4. Case (A), RHP zero (all-pass): $G(s)=K\frac{a-s}{a+s}$ with $K>0$, so $$G(j\omega )=K\frac{a-j\omega }{a+j\omega },|G(j\omega )|=K\frac{\sqrt{a^2+{\omega }^2}}{\sqrt{a^2+{\omega }^2}}=K\text{ for all }\omega$$

5. Phase of the all-pass. The numerator $a-j\omega$ has angle $-{\tan }^{-1}(\omega /a)$. The denominator $a+j\omega$ has angle $+{\tan }^{-1}(\omega /a)$, and dividing by it subtracts that angle. So $$\angle G(j\omega )=-2{\tan }^{-1}\left(\frac{\omega }{a}\right)$$

6. Check the limits. As $\omega \to 0$ the phase is $0^{\circ }$. At $\omega =a$ it is $-90^{\circ }$. As $\omega \to \infty$ it tends to $-2\times 90^{\circ }=-180^{\circ }$. The magnitude is flat and the phase goes from $0^{\circ }$ to $-180^{\circ }$, exactly as in the plot. The system therefore has one LHP pole and one RHP zero at the same frequency, which is option (A).