GATE 2024 EE Q.21: solution and answer

Official answer: BSource: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Pole geometry. For $0<\zeta <1$ the poles are $p,p^{\ast }=-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}$. Their distance from the origin is $\sqrt{{\zeta }^2{\omega }_n^2+{\omega }_n^2(1-{\zeta }^2)}={\omega }_n$, so they lie on the circle of radius ${\omega }_n$. The figure measures $\theta$ from the negative real axis, so $\cos \theta =\frac{\zeta {\omega }_n}{{\omega }_n}=\zeta$.

2. What sets the settling time. The underdamped step response is $1-\frac{e^{-\zeta {\omega }_{n}t}}{\sqrt{1-{\zeta }^2}}\sin ({\omega }_{d}t+\theta )$. How fast it settles depends on the decay rate of the envelope, $\sigma =\zeta {\omega }_n={\omega }_n\cos \theta$, which is how far the poles sit from the $j\omega$ axis. With the 2% criterion, $t_s\approx \frac{4}{\zeta {\omega }_n}$.

3. System 1: ${\zeta }_1=\cos 60^{\circ }=0.5$ (underdamped), so ${\sigma }_1=3\times 0.5=1.5{\text{s}}^{-1}$ and $t_{s1}\approx \frac{4}{1.5}\approx 2.67\text{s}$.

4. System 2: ${\zeta }_2=\cos 70^{\circ }\approx 0.342$ (underdamped), so ${\sigma }_2=1\times 0.342=0.342{\text{s}}^{-1}$ and $t_{s2}\approx \frac{4}{0.342}\approx 11.7\text{s}$.

5. Comparison. ${\sigma }_2<{\sigma }_1$, so System 2's poles are closer to the imaginary axis. Its envelope $e^{-\sigma t}$ decays more slowly and it takes longer to settle. The ratio is $t_{s2}/t_{s1}={\sigma }_1/{\sigma }_2=1.5/0.342\approx 4.4$. Under the 5% criterion ($t_s\approx 3/\zeta {\omega }_n$) the ratio is the same, so the choice of criterion does not change the answer. Option (D) is wrong because ${\omega }_n$ and $\theta$ together fix $\zeta {\omega }_n$, and $\zeta {\omega }_n$ is all the settling time needs. So the settling time of System 2 is more than that of System 1.