GATE 2026 BT Q.36: solution and answer

Official answer: BSource: Official answer key, GATE 2026 BT (IIT Guwahati)

1. Symbols: $K_p$ is the process gain (the stem prints it as $K_P$, the figure as $K_p$; they are the same quantity), ${\tau }_p$ is the characteristic time (time constant) and $M$ is the size of the input. The input $U(s)=M/s$ is a step of magnitude $M$ applied at $t=0$.

2. The process is open loop, so the output is the transfer function times the input: $$Y(s)=G_p(s)U(s)=\frac{K_p}{{\tau }_{p}s+1}\cdot \frac{M}{s}=\frac{M{K}_p}{s({\tau }_{p}s+1)}$$

3. Partial fractions: $$\frac{1}{s({\tau }_{p}s+1)}=\frac{1}{s}-\frac{{\tau }_p}{{\tau }_{p}s+1}=\frac{1}{s}-\frac{1}{s+1/{\tau }_p}$$ (check: $\frac{({\tau }_{p}s+1)-{\tau }_{p}s}{s({\tau }_{p}s+1)}=\frac{1}{s({\tau }_{p}s+1)}$).

4. Inverse Laplace, using ${\mathcal{L}}^{-1}\{1/s\}=1$ and ${\mathcal{L}}^{-1}\{1/(s+a)\}=e^{-at}$ with $a=1/{\tau }_p$: $$y(t)=M{K}_p\left(1-e^{-t/{\tau }_p}\right)$$

5. Check the limits. At $t=0$ this gives $y=0$, as it should for a first-order system starting from rest. As $t\to \infty$ it gives $y\to M{K}_p$, which matches the final value theorem (valid because the only pole of $sY(s)$ is at $s=-1/{\tau }_p<0$): ${\lim }_{s\to 0}sY(s)={\lim }_{s\to 0}\frac{M{K}_p}{{\tau }_{p}s+1}=M{K}_p$. The steady output is the gain times the step size.

6. Option (A) leaves out $M$, so it is the response to a unit step, not to a step of size $M$. Option (C) gives $y(0)=K_p(M-1)\ne 0$. Option (D) gives $y(0)=K_p(1-M)\ne 0$ and the wrong final value, $K_p$. Only (B) fits, so the answer is (B).