GATE 2023 IN Q.44: solution and answer

Official answer: B, CSource: Official answer key, GATE 2023 IN (IIT Kanpur)

1. Let $x(t)\leftrightarrow X(j\omega )$. The transform exists because $x(t)$ has finite energy. Time reversal gives $x(-t)\leftrightarrow X(-j\omega )$. Because $x(t)$ is real, conjugate symmetry gives $X(-j\omega )=X^{\ast }(j\omega )$, so $x(-t)\leftrightarrow X^{\ast }(j\omega )$.

2. (A) $x(t)+x(-t)\leftrightarrow X(j\omega )+X^{\ast }(j\omega )=2\text{Re}\{X(j\omega )\}$. This is real, not imaginary. (A real, even signal has a real, even transform.)

3. (B) $x(t)-x(-t)\leftrightarrow X(j\omega )-X^{\ast }(j\omega )=j2\text{Im}\{X(j\omega )\}$. This is purely imaginary. (A real, odd signal has an imaginary, odd transform.)

4. (C) By linearity, $j\left(x(t)+x(-t)\right)\leftrightarrow j2\text{Re}\{X(j\omega )\}$. This is $j$ times a real function, so it is purely imaginary.

5. (D) By linearity, $j\left(x(t)-x(-t)\right)\leftrightarrow j\cdot j2\text{Im}\{X(j\omega )\}=-2\text{Im}\{X(j\omega )\}$. This is real.

6. Why the causality and positive-energy conditions are given: since $x(t)=0$ for $t<0$, for every $t>0$ we have $x(-t)=0$, so the even part $x_e(t)=\frac{1}{2}\left(x(t)+x(-t)\right)$ and the odd part $x_o(t)=\frac{1}{2}\left(x(t)-x(-t)\right)$ both equal $x(t)/2$ there. All the energy of $x(t)$ sits in $t\ge 0$ and is positive, so neither part is the zero signal, and by uniqueness neither $2\text{Re}\{X(j\omega )\}$ nor $2\text{Im}\{X(j\omega )\}$ is identically zero. So the transforms in (A) and (D) are genuinely nonzero real functions, not trivially zero (a zero function would count as both real and imaginary). Without causality an even $x(t)$ could make (B) and (D) vanish.

7. Hence the transforms of (B) and (C) are purely imaginary. Answer: B, C.