Question

Fourier Transform of \(e^{-|x|} \, is \) \( \frac{2}{1+p^2} \). Then what is the fourier transform of \( e^{-2|x|} \)?

(a) \(\frac{4}{(4+p^2)} \)

(b) \(\frac{2}{(4+p^2)} \)

(c) \(\frac{2}{(2+p^2)} \)

(d) \(\frac{4}{(2+p^2)} \)

The question was asked in exam.

Query is from Fourier Transform and Convolution topic in chapter Fourier Integral, Fourier Transforms and Integral Transforms of Engineering Mathematics

Answer 1

Correct answer is (a) \(\frac{4}{(4+p^2)} \)

Explanation: \(e^{-2|x|}= e^{-|2x|}= F(2x) \)

\(F\{F(2x)\} = \frac{1}{2} f(\frac{p}{2}) \)

\( = \frac{1}{2} \frac{2}{(1+\frac{p^2}{4})} \)

\( = \frac{4}{(4+p^2 )}.\)