GATE 2023 EE Q.36: solution and answer

Official answer: DSource: Official answer key, GATE 2023 EE (IIT Kanpur)

1. Flat magnitude means no poles or zeros. $|G(j\omega )|$ stays at 8 dB at every frequency, so $G(s)$ is a constant gain times a factor of unit magnitude. That rules out option (B): its $\frac{1}{s+1}$ factor gives 0 dB at $\omega =0$ (not 8 dB) and rolls the magnitude off above $\omega =1\text{rad/s}$.

2. Gain from decibels: $$20{\log }_{10}K=8\Rightarrow K=10^{8/20}=10^{0.4}\approx 2.51$$ That rules out option (C), whose gain of 1.04 would plot at only about 0.34 dB.

3. A phase that falls linearly against $\omega$ itself means a pure time delay. Note the phase plot's axis is $\omega$, not ${\log }_{10}\omega$. For $G(s)=K{e}^{-sT}$, putting $s=j\omega$ gives $|G(j\omega )|=K$ and $\angle G(j\omega )=-\omega T$ rad: constant magnitude and a straight-line phase through the origin, exactly as plotted.

4. Delay from the marked point: at $\omega =1\text{rad/s}$ the phase is $-60^{\circ }=-\frac{\pi }{3}$ rad, so $$\omega T=\frac{\pi }{3}\Rightarrow T=\frac{\pi /3}{1}\approx 1.047\text{s}$$

5. Transfer function: $G(s)=2.51e^{-1.047s}$, which is option (D). As a check on the others: option (A)'s $T=0.032\text{s}$ would give only about $-1.8^{\circ }$ at $\omega =1\text{rad/s}$, and option (C)'s $T=2.514\text{s}$ would give about $-144^{\circ }$.