Question

Let $f(t)$ be an even function. The Fourier transform is $F(\omega)=\int_{-\infty}^\infty f(t)e^{-j\omega t}dt$. Suppose $\frac{dF(\omega)}{d\omega} = -\omega F(\omega)$ for all $\omega$, and $F(0)=1$. Then
 

• A. $f(0) < 1$
• B. $f(0) > 1$
• C. $f(0)=1$
• D. $f(0)=0$

Hint:

If $F(\omega)$ is Gaussian, then $f(t)$ is also Gaussian, and $f(0)$ is always less than 1.

Answer 1

The Correct Option is A

Given the differential equation \[ \frac{dF}{d\omega} = -\omega F(\omega), \] we separate variables: \[ \frac{dF}{F} = -\omega\, d\omega. \]
Integrating, \[ \ln F(\omega) = -\frac{\omega^{2}}{2} + C. \]
Using $F(0)=1$: \[ C = 0. \] Thus, \[ F(\omega) = e^{-\omega^{2}/2}. \]
For even $f(t)$, \[ f(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty} F(\omega)\, d\omega. \] Since \[ \int_{-\infty}^{\infty} e^{-\omega^{2}/2} d\omega = \sqrt{2\pi}, \] we get \[ f(0)=\frac{\sqrt{2\pi}}{2\pi}=\frac{1}{\sqrt{\pi}} < 1. \]
Final Answer: $f(0) < 1$