Question

In Finite Fourier Cosine Transform, if the upper limit l = π, then its inverse is given by ________

(a) \(F(x) = \frac{2}{π} ∑_{p=1}^∞ fc (p)cos(px)+ \frac{1}{π} fc(0) \)

(b) \(F(x) = \frac{2}{π} ∑_{p=1}^∞ fc (p)cos(px) \)

(c) \(F(x) = \frac{2}{π} ∑_{p=1}^∞ fc (p)cos(\frac{px}{π}) \)

(d) \(F(x) = \frac{2}{π} ∑_{p=0}^∞ fc (p)cos(px)+ \frac{1}{π} fc(0) \)

I got this question by my school teacher while I was bunking the class.

The above asked question is from Fourier Transform and Convolution topic in chapter Fourier Integral, Fourier Transforms and Integral Transforms of Engineering Mathematics

Answer 1

Correct choice is (a) \(F(x) = \frac{2}{π} ∑_{p=1}^∞ fc (p)cos(px)+ \frac{1}{π} fc(0) \)

Explanation: Now since we have fourier cosine transform, we have to use the constant \(\frac{2}{π}\). And since while writing as sum of series it also has a term if p=0. Hence, \(F(x) = \frac{2}{π} ∑_{p=1}^∞ fc (p)cos(px)+ \frac{1}{π} fc(0) \)