GATE 2024 EE Q.15: solution and answer

Official answer: ASource: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Let $z(t)=x(t)\ast y(t)$. By the convolution integral, $$z(t)={\int }_{-\infty }^{\infty }x(\tau )y(t-\tau )d\tau ={\int }_{-\infty }^{\infty }x(\tau )x(\tau -t)d\tau$$ since $y(t-\tau )=x(-(t-\tau ))=x(\tau -t)$. For a real $x(t)$ this is the autocorrelation of $x(t)$ at lag $t$.

2. Test for even symmetry by replacing $t$ with $-t$: $$z(-t)={\int }_{-\infty }^{\infty }x(\tau )x(\tau +t)d\tau$$

3. Substitute $\sigma =\tau +t$ (so $d\sigma =d\tau$, and the limits stay $-\infty$ to $\infty$): $$z(-t)={\int }_{-\infty }^{\infty }x(\sigma -t)x(\sigma )d\sigma =z(t)$$ No property of $x(t)$ was used, so $z(t)$ is even for every $x(t)$ whose convolution exists. (A) is always true.

4. Rule out (B): an odd signal must have $z(0)=-z(0)$, i.e. $z(0)=0$. But $z(0)={\int }_{-\infty }^{\infty }{x}^2(\tau )d\tau$ is the energy of $x(t)$, which is positive for any nonzero real $x(t)$. So $z(t)$ is not odd in general.

5. Rule out (C) and (D) with one counterexample: take $x(t)=u(t)-u(t-1)$, a unit pulse on $0\le t\le 1$, so $x(-t)$ is a unit pulse on $-1\le t\le 0$. Their convolution is the triangle $z(t)=1-|t|$ for $|t|\le 1$ (zero elsewhere), which is nonzero for both $t<0$ and $t>0$. So it is neither causal nor anti-causal.

6. Frequency-domain check: $x(-t)\leftrightarrow X(-j\omega )$, so $Z(j\omega )=X(j\omega )X(-j\omega )$, which is unchanged when $\omega \to -\omega$, i.e. an even spectrum, and an even spectrum belongs to an even time signal. For real $x(t)$, $X(-j\omega )=X^{\ast }(j\omega )$, so $Z(j\omega )=|X(j\omega ){|}^2$, which is also real. Answer: (A).