GATE 2023 IN Q.43: solution and answer

Official answer: BSource: Official answer key, GATE 2023 IN (IIT Kanpur)

1. Read the magnitude plot. The gain is $0\text{dB}$ at low frequency and there is one sharp resonant peak at $\omega \approx 100\text{rad/s}$. Every option has $G(0)=1$, which is $0\text{dB}$. For $\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$ with small $\zeta$, the peak is at ${\omega }_r={\omega }_n\sqrt{1-2{\zeta }^2}\approx {\omega }_n$. So ${\omega }_n=100\text{rad/s}$ and ${\omega }_n^2=10000$. Option (D) has ${\omega }_n^2=100$, so ${\omega }_n=10\text{rad/s}$ and its peak would be at $10\text{rad/s}$. (D) is out.

2. Check the peak height and the high-frequency end. For $s^2+2s+10000$: $2\zeta {\omega }_n=2$, so $\zeta =0.01$. $$M_r=\frac{1}{2\zeta \sqrt{1-{\zeta }^2}}\approx 50\Rightarrow 20{\log }_{10}50\approx 34\text{dB},$$ which matches the plotted peak of about 33 to 34 dB. At $\omega =200$: $$|G(j200)|=\frac{10000}{\left|10000-40000+j400\right|}\approx \frac{10000}{30003}\approx 0.333\Rightarrow \approx -9.5\text{dB},$$ which matches the curve ending just above $-10\text{dB}$. Because $|e^{-j\omega T}|=1$ for every real $\omega$, options (A), (B) and (C) have exactly the same magnitude plot. The magnitude cannot tell them apart, so the phase has to.

3. Phase of the quadratic alone. $$\angle \frac{10000}{10000-{\omega }^2+j2\omega }=-{\tan }^{-1}\frac{2\omega }{10000-{\omega }^2},$$ taken between $0^{\circ }$ and $-180^{\circ }$. It starts at $0^{\circ }$, equals $-90^{\circ }$ at $\omega =100\text{rad/s}$ and approaches $-180^{\circ }$, but it never goes past $-180^{\circ }$. With $\zeta =0.01$, almost all of that change happens close to $100\text{rad/s}$: about $-11^{\circ }$ at $95\text{rad/s}$ and about $-168^{\circ }$ at $105\text{rad/s}$. This accounts for the steep drop at $100\text{rad/s}$. The plotted phase reaches $-720^{\circ }$, though, so option (A) cannot produce it.

4. The rest of the phase is a transport delay. Away from $100\text{rad/s}$ the phase falls along a straight line on this linear frequency axis. A pure delay does exactly that: $\angle e^{-j\omega T}=-\omega T\text{rad}$ is linear in $\omega$, has no lower limit and leaves the magnitude unchanged. So $G(s)$ is the quadratic multiplied by $e^{-sT}$, and its total phase is $$\varphi (\omega )=-{\tan }^{-1}\frac{2\omega }{10000-{\omega }^2}-\frac{180}{\pi }\omega T\text{degrees}.$$

5. Estimate $T$ at a point far from resonance. The curve leaves the bottom of the axes, at $-720^{\circ }$, at about $\omega \approx 189\text{rad/s}$. At that frequency the quadratic has almost finished its swing: $$-\left(180^{\circ }-{\tan }^{-1}\frac{378}{35721-10000}\right)\approx -\left(180^{\circ }-0.8^{\circ }\right)\approx -179^{\circ }.$$ That leaves about $-720^{\circ }+179^{\circ }\approx -541^{\circ }\approx -9.44\text{rad}$ for the delay, so $$T\approx \frac{9.44}{189}\approx 0.05\text{s}.$$

6. Check $T=0.05\text{s}$ at the low-frequency end, and rule out (C). At the left edge of the plot, $\omega =10\text{rad/s}$, the delay gives $-0.5\text{rad}\approx -28.6^{\circ }$ and the quadratic about $-0.1^{\circ }$. The total is about $-29^{\circ }$, which matches the curve starting near $-30^{\circ }$. Near $\omega =62\text{rad/s}$ the delay gives $-3.1\text{rad}\approx -178^{\circ }$ and the quadratic about $-1^{\circ }$. That puts the $-180^{\circ }$ crossing a little past $60\text{rad/s}$, which is also where the plot crosses it. Option (C) has $T=0.5\times 10^{-12}\text{s}$. Even at $200\text{rad/s}$ its delay adds only $200\times 0.5\times 10^{-12}=10^{-10}\text{rad}\approx 6\times 10^{-9}$ degrees. Its phase would look the same as (A)'s and stay above $-180^{\circ }$, so (C) is out.

7. Conclusion. $$G(s)=\frac{10000}{s^2+2s+10000}{e}^{-0.05s},$$ which is option (B).