GATE 2021 EE – Question 42
Let $f(t)$ be an even function, i.e. $f(-t)=f(t)$ for all $t$. Let the Fourier transform of $f(t)$ be defined as $F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt$. Suppose $\dfrac{dF(\omega)}{d\omega}=-\omega F(\omega)$ for all $\omega$, and $F(0)=1$. Then
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Correct answer: (A) $f(0)<1$
Explanation
The equation gives $F(\omega)=e^{-\omega^2/2}$. Then $f(0)=\frac{1}{2\pi}\int F(\omega)d\omega=\frac{\sqrt{2\pi}}{2\pi}=0.399<1$.