GATE 2025 EE Q.22: solution and answer

Official answer: BSource: Official answer key, GATE 2025 EE (IIT Roorkee)

1. A stable first-order system that starts from rest has the step response $y(t)=y_{ss}\left(1-e^{-t/T}\right)$, where $T$ is the time constant. The $t=\infty$ column gives the final value $y_{ss}=3$.

2. Rearrange to get $T$ from any single reading: $$T=\frac{-t}{\ln \left(1-\frac{y(t)}{y_{ss}}\right)}$$

3. At $t=0.6\text{s}$: $1-\frac{0.78}{3}=0.74$, so $T=\frac{0.6}{-\ln 0.74}=\frac{0.6}{0.301}\approx 1.99\text{s}$.

4. At $t=1.6\text{s}$: $1-\frac{1.65}{3}=0.45$, so $T=\frac{1.6}{-\ln 0.45}=\frac{1.6}{0.799}\approx 2.00\text{s}$.

5. At $t=2.6\text{s}$: $1-\frac{2.18}{3}=0.273$, so $T=\frac{2.6}{-\ln 0.273}=\frac{2.6}{1.297}\approx 2.00\text{s}$. At $t=10\text{s}$: $1-\frac{2.98}{3}=0.00667$, so $T=\frac{10}{-\ln 0.00667}=\frac{10}{5.01}\approx 2.00\text{s}$.

6. Quick check with the 63.2% rule. At $t=T$ the output is $1-e^{-1}\approx 63.2\%$ of its final value: $0.632\times 3=1.896$. That value falls between the readings at $1.6\text{s}$ ($1.65$) and $2.6\text{s}$ ($2.18$), so $T$ lies between 1.6 s and 2.6 s. Only option (B) is in that range.

7. All four data points give the same value, $T\approx 2\text{s}$. That agreement also confirms the model assumed in step 1 (first order, starting from zero output). Answer: option (B).