GATE 2020 EE Q.37: solution and answer

Official answer: CSource: Official answer key, GATE 2020 EE (IIT Delhi)

1. The question compares $y(t)$ and $z(t)$ with $\le$ and $\ge$, so the signals are real-valued. Write both outputs as convolution integrals: $$y(t)={\int }_{-\infty }^{\infty }x(\tau )h(t-\tau )d\tau ,z(t)={\int }_{-\infty }^{\infty }|x(\tau )||h(t-\tau )|d\tau$$

2. For every $\tau$, $|x(\tau )h(t-\tau )|=|x(\tau )||h(t-\tau )|$. The magnitude of an integral is at most the integral of the magnitude (the triangle inequality for integrals), so $$|y(t)|=\left|{\int }_{-\infty }^{\infty }x(\tau )h(t-\tau )d\tau \right|\le {\int }_{-\infty }^{\infty }|x(\tau )||h(t-\tau )|d\tau =z(t)$$

3. A real number is never larger than its magnitude, so $y(t)\le |y(t)|$. Combining this with step 2: $$y(t)\le |y(t)|\le z(t)\text{for every }t$$ No condition on $x$ or $h$ was used, so $z(t)\ge y(t)$ for all $t\in (-\infty ,\infty )$, for every input and every impulse response. This is option (C).

4. (D) says $z(t)\ge y(t)$ holds for some but NOT all $t$. Step 3 shows it holds at every $t$, so the 'not all' part is never true and (D) is always false.

5. (A) is not always true. Take $x(t)=u(t)$ and $h(t)=-u(t)$. For $t\ge 0$, $$y(t)={\int }_0^t(1)(-1)d\tau =-t,z(t)={\int }_0^t(1)(1)d\tau =t$$ and both are zero for $t<0$. So $y(t)=-tu(t)$ and $z(t)=tu(t)$. At $t=1$, $z=1>y=-1$, so $z(t)\le y(t)$ fails and (A) is false.

6. (B) is not always true either. Take $x(t)=h(t)=u(t)$. Both signals are non-negative, so $|x|=x$ and $|h|=h$, giving $z(t)=y(t)=tu(t)$. Then $z(t)\le y(t)$ holds at EVERY $t$ (with equality), so 'some but not all' is false. (B) does happen to hold for the pair in step 5, which is why one example cannot settle this. The question asks which statement is true for every $x$ and $h$, and only (C) is. Answer: (C).