GATE 2021 CH – Question 63
A process has a transfer function $G(s)=\frac{Y(s)}{X(s)}=\frac{20}{90000s^2+240s+1}$. Initially the process is at steady state with $x(t=0)=0.4$ and $y(t=0)=100$. If a step change in x is given from 0.4 to 0.5, the maximum value of y that will be observed before it reaches the new steady state is _____ (round off to 1 decimal place).
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Correct answer: 101.99 to 103.01
Explanation
Interpret the transfer function in **deviation variables** about the stated operating point. The input change is $\Delta x=0.1$, so the final output increment is $K\Delta x=20(0.1)=2$ and the new steady output is 102.
Compare the denominator with $\tau^2s^2+2\zeta\tau s+1$: $\tau=300$ and $\zeta=240/(2\times300)=0.4$. The fractional overshoot is
$$M_p=e^{-\pi\zeta/\sqrt{1-\zeta^2}}=0.2538266.$$
The peak is therefore $100+2(1+M_p)=102.50765$, rounded to **102.5**. Overshoot applies to the **2-unit change**, not to the entire baseline value 100.