GATE 2024 PH – Question 56
The transforms are $\tilde f(\omega)=(2\pi)^{-1/2}\int f(x)e^{i\omega x}dx$ and $f(x)=(2\pi)^{-1/2}\int\tilde f(\omega)e^{-i\omega x}d\omega$. Define $(f*g)(x)=(2\pi)^{-1/2}\int f(y)g(x-y)dy$, all integrals over the real line. Which relations hold?
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Correct answer: (A) $f*g=g*f$; (B) $\widetilde{f*g}=\widetilde{g*f}$; (D) $\widetilde{f*g}=\tilde f\tilde g$
Explanation
Definitions: $\tilde f(\omega)=\dfrac{1}{\sqrt{2\pi}}\int f(x)e^{i\omega x}dx$ and the convolution $(f*g)(x)=\dfrac{1}{\sqrt{2\pi}}\int f(y)g(x-y)\,dy$.
- **A. $f*g=g*f$.** True. Substitute $y^\prime=x-y$ in the integral. ✓
- **B. $\widetilde{f*g}=\widetilde{g*f}$.** True, because $f*g=g*f$. ✓
- **C. $\widetilde{f*g}=\widetilde{fg}$.** False. The transform of a convolution is not the transform of the pointwise product.
- **D. $\widetilde{f*g}=\tilde f\,\tilde g$.** True with this symmetric normalisation. Writing out the double integral,
$$\widetilde{f*g}(\omega)=\frac{1}{2\pi}\iint f(y)g(x-y)e^{i\omega x}\,dy\,dx=\left[\frac1{\sqrt{2\pi}}\int f(y)e^{i\omega y}dy\right]\left[\frac1{\sqrt{2\pi}}\int g(u)e^{i\omega u}du\right]=\tilde f\tilde g .$$ ✓
Answer **A, B and D**.