GATE 2020 IN Q.6: solution and answer

Official answer: ASource: Official answer key, GATE 2020 IN (IIT Delhi)

1. Read off the corner frequencies from the time constants. The zero corner is at ${\omega }_z=\frac{1}{0.2}=5\text{rad/s}$ and the pole corner is at ${\omega }_p=\frac{1}{0.05}=20\text{rad/s}$. Because ${\omega }_z<{\omega }_p$, the network is a lead network.

2. The phase of the network is $\varphi (\omega )={\tan }^{-1}(0.2\omega )-{\tan }^{-1}(0.05\omega )$. Set $\frac{d\varphi }{d\omega }=\frac{0.2}{1+0.04{\omega }^2}-\frac{0.05}{1+0.0025{\omega }^2}=0$. Cross-multiplying gives $0.2+0.0005{\omega }^2=0.05+0.002{\omega }^2$, so $0.0015{\omega }^2=0.15$ and ${\omega }^2=100=\frac{1}{0.2\times 0.05}$.

3. So the maximum phase occurs at the geometric mean of the two corner frequencies: $${\omega }_m=\sqrt{{\omega }_z{\omega }_p}=\sqrt{5\times 20}=\sqrt{100}=10\text{rad/s}$$

4. Cross-check with the standard form $\frac{1+Ts}{1+\alpha Ts}$. Here $T=0.2$ and $\alpha T=0.05$, so $\alpha =0.25$. Then ${\omega }_m=\frac{1}{T\sqrt{\alpha }}=\frac{1}{0.2\times 0.5}=10\text{rad/s}$, which is option (A).

5. This was not asked, but it is a useful consistency check: ${\varphi }_m={\tan }^{-1}(0.2\times 10)-{\tan }^{-1}(0.05\times 10)={\tan }^{-1}(2)-{\tan }^{-1}(0.5)\approx 63.4^{\circ }-26.6^{\circ }=36.9^{\circ }$. This matches ${\sin }^{-1}\frac{1-\alpha }{1+\alpha }={\sin }^{-1}\frac{0.75}{1.25}={\sin }^{-1}(0.6)\approx 36.9^{\circ }$.