GATE 2022 EE Q.46: solution and answer

Official answer: DSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

1. Take the Laplace transform with zero initial conditions. The impulse response of a causal LTI system described by a differential equation is found at initial rest, so every initial condition is zero: $$Y(s)+\frac{1}{4}sY(s)=2X(s)$$

2. Transfer function: $$H(s)=\frac{Y(s)}{X(s)}=\frac{2}{1+\frac{s}{4}}=\frac{8}{s+4}$$

3. The system is causal, so take the right-sided inverse using $e^{-at}u(t)\leftrightarrow \frac{1}{s+a}$ with $a=4$: $$h(t)=8e^{-4t}u(t)$$

4. Check with the initial-value theorem, which applies because $H(s)$ is strictly proper: $h(0^{+})={\lim }_{s\to \infty }sH(s)={\lim }_{s\to \infty }\frac{8s}{s+4}=8$. This rules out (A) and (B), which both start at 2.

5. Check the time constant. The pole at $s=-4$ gives $\tau =\frac{1}{4}\text{s}$, so the exponent is $-4t$, not $-t/4$. This rules out (C).

6. Direct check in the time domain: $\frac{dh}{dt}=8\delta (t)-32e^{-4t}u(t)$, so $h(t)+\frac{1}{4}\frac{dh}{dt}=8e^{-4t}u(t)+2\delta (t)-8e^{-4t}u(t)=2\delta (t)$. That is the right-hand side $2x(t)$ with $x(t)=\delta (t)$. The answer is (D).