GATE 2025 IN Q.48: solution and answer

Official answer: ASource: Official answer key, GATE 2025 IN (IIT Roorkee)

1. Definition of cross-correlation for real signals: $$R_{xy}(\tau )={\int }_{-\infty }^{\infty }x(t)y(t-\tau )dt={\int }_{-\infty }^{\infty }x(t+\tau )y(t)dt$$ Here $x(t)=f(t)$ and $y(t)=g(-t)$.

2. Substitute $y(t-\tau )=g(-(t-\tau ))=g(\tau -t)$: $$h(\tau )={\int }_{-\infty }^{\infty }f(t)g(\tau -t)dt=f(\tau )\ast g(\tau )$$ Correlating with a time-reversed signal is the same as convolving with the original signal.

3. Convolution property of the Fourier transform: $f\ast g⟷F(j\omega )G(j\omega )$, so $H(j\omega )=F(j\omega )G(j\omega )$.

4. Check with the correlation theorem: $R_{xy}(\tau )=x(\tau )\ast y(-\tau )⟷X(j\omega )Y(-j\omega )=X(j\omega )Y^{\ast }(j\omega )$ for real $y$. Time reversal gives $Y(j\omega )=G(-j\omega )$. Because $g$ is real, $G(-j\omega )=G^{\ast }(j\omega )$, so $Y^{\ast }(j\omega )=G(j\omega )$ and $H(j\omega )=F(j\omega )G(j\omega )$ again.

5. Option (C), $F(j\omega )G(-j\omega )$, is the transform of the cross-correlation of $f$ with $g(t)$ itself, not with $g(-t)$. Option (B), $F(-j\omega )G(j\omega )$, is the transform of the correlation taken in the other order, $g$ with $f$. Option (D) is (C) with a minus sign that has no source. The answer is (A).