GATE 2020 IN Q.20: solution and answer

Official answer: 0.78 to 0.79Source: Official answer key, GATE 2020 IN (IIT Delhi)

1. Match the poles to the standard form $s=-\zeta {\omega }_n\pm j{\omega }_d$. This gives $\zeta {\omega }_n=3$ and ${\omega }_d=4\text{rad/s}$.

2. Check that the system is underdamped: ${\omega }_n=\sqrt{3^2+4^2}=5\text{rad/s}$ and $\zeta =3/5=0.6<1$. The step response therefore overshoots, so it has a true maximum.

3. For a standard second-order system with no zeros, the unit step response is $c(t)=1-\frac{e^{-\zeta {\omega }_{n}t}}{\sqrt{1-{\zeta }^2}}\sin ({\omega }_{d}t+\theta )$. Its derivative is $\frac{dc}{dt}=\frac{{\omega }_n}{\sqrt{1-{\zeta }^2}}{e}^{-\zeta {\omega }_{n}t}\sin ({\omega }_{d}t)$. This is zero when ${\omega }_{d}t=n\pi$, and the first maximum is at $n=1$. So $t_p=\frac{\pi }{{\omega }_d}$.

4. Substitute: $$t_p=\frac{\pi }{{\omega }_d}=\frac{\pi }{4}=0.7854\text{s}$$

5. Round to two decimal places: $t_p\approx 0.79\text{s}$.