Question

Find the Fourier Cosine Transform of F(x) = 2x for 0<x<4.

(a) \(fc(p) = \frac{32}{(p^2 π^2)} (cos(pπ)-1)p \) not equal to 0 and if equal to 0 \( fc(p) = 16 \)

(b) \(fc(p) = \frac{32}{(p^2 π^2)} (cos(pπ)-1)p \) not equal to 0 and if equal to 0 \( fc(p) = 32 \)

(c) \(fc(p) = \frac{64}{(pπ^2)} (cos(pπ)-1)p \) not equal to 0 and if equal to 0 \( fc(p) = 16 \)

(d) \(fc(p) = \frac{32}{(pπ^2)} (cos(pπ)-1)p \) not equal to 0 and if equal to 0 \( fc(p) = 64 \)

I have been asked this question during an online interview.

Question is from Fourier Transform and Convolution topic in section Fourier Integral, Fourier Transforms and Integral Transforms of Engineering Mathematics

Answer 1

Correct option is (a) \(fc(p) = \frac{32}{(p^2 π^2)} (cos(pπ)-1)p \) not equal to 0 and if equal to 0 \( fc(p) = 16 \)

For explanation: \(fc(p) = \int_0^4 2 xcos(\frac{pπx}{4})dx \)

\( = 2\bigg[\frac{4xsin(\frac{pπx}{4})}{pπ} + \frac{16cos(\frac{pπx}{4})}{p^2 π^2}\bigg] \) from 0 to 4

\( = \frac{32}{(p^2 π^2)} (cos(pπ)-1) \)

When \(p=0, fc(p) = \int_0^4 2 xdx = 16. \)