GATE 2018 IN Q.39: solution and answer

Official answer: DSource: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Laplace transform of the step response (zero initial conditions): $$Y(s)=\frac{1}{s}-\frac{1}{s+2}=\frac{(s+2)-s}{s(s+2)}=\frac{2}{s(s+2)}$$

2. Definition of the transfer function: $G(s)=\frac{Y(s)}{R(s)}$ with zero initial conditions. The input is a unit step, so $R(s)=\frac{1}{s}$.

3. Divide by the input: $$G(s)=\frac{2}{s(s+2)}\cdot s=\frac{2}{s+2}$$ This is option D.

4. Check 1, final value theorem: $y(\infty )={\lim }_{s\to 0}s\cdot \frac{1}{s}\cdot G(s)=G(0)$. For D, $G(0)=\frac{2}{2}=1$, which matches $1-e^{-2t}\to 1$. Option C gives $G(0)=0.5$ and fails. Options A and B also give $G(0)=1$, so this check alone does not settle the answer.

5. Check 2, initial slope: the given response has $\dot{y}(t)=2e^{-2t}$, so $\dot{y}(0^{+})=2$. By the initial value theorem, $\dot{y}(0^{+})={\lim }_{s\to \infty }s\cdot sY(s)={\lim }_{s\to \infty }sG(s)$. This limit is $1$ for A, $0$ for B (a second-order system starts with zero slope), $1$ for C and $2$ for D. Only option D passes both checks.