ISRO EC 2025 Q.53: solution and answer

Official answer: DSource: ISRO's final answer key of 15 May 2026

1. The forward path is $G(s)=\frac{K}{s(s+1)(s+4)}$ and the feedback block is $H(s)=s+u$, so $\frac{C(s)}{R(s)}=\frac{G}{1+GH}$.

2. Since $H(0)=u\ne 1$, the feedback is not unity at DC. The steady state error is therefore taken between input and output, $E(s)=R(s)-C(s)=R(s)\frac{1+GH-G}{1+GH}$.

3. Multiply numerator and denominator by $s(s+1)(s+4)$: $$E(s)=R(s)\frac{s(s+1)(s+4)+K(s+u)-K}{s(s+1)(s+4)+K(s+u)}$$

4. For a unit step, $R(s)=1/s$. By the final value theorem (assuming the closed loop is stable), $$e_{ss}=\underset{s\to 0}{\lim }sE(s)=\frac{K(u-1)}{Ku}=\frac{u-1}{u}=1-\frac{1}{u}.$$

5. The sensitivity of $e_{ss}$ to the parameter $u$ is $S=\frac{u}{e_{ss}}\frac{\partial e_{ss}}{\partial u}$. Here $\frac{\partial e_{ss}}{\partial u}=\frac{1}{u^2}$.

6. Therefore $$S=\frac{u}{(u-1)/u}\cdot \frac{1}{u^2}=\frac{u^2}{u-1}\cdot \frac{1}{u^2}=\frac{1}{u-1}.$$

7. Option (a) is $e_{ss}$ itself, not its sensitivity. Options (b) and (c) do not equal $\frac{u}{e_{ss}}\frac{d{e}_{ss}}{du}$ either. Only $\frac{1}{u-1}$ does.