# fourier transform

Discussion in 'Electronic Design' started by [email protected], Jan 16, 2009.

1. ### Guest

I have been trying to solve fourier transform of f(t)^n...where f(t)
is any random function.I require an answer in terms of fourier
transform of f(t)... Thanks a lot.

Hello
Thats nice, hope the answer finds you )

Cheers
IBM

3. ### oopereGuest

If f(t) is a random process then you will be having trouble computing
it's Fourier Transform.

If you mean that f(t) could be any function, then you should pay more
attention in class:

The Fourier transform of f(t)·g(t) is F(f)*G(f), where * stands for
convolution. Next, work out f(t)·f(t), f(t)·f(t)·f(t) and so on...

Pere

4. ### JamieGuest

Maybe this will help in your journey.

http://groovit.disjunkt.com/analog/time-domain/fft.html

5. ### Guest

F[f^n] = F[f*f^(n-1)]

the fourier transform of a product is a convolution

F[f] * F[f^(n-1)]

Hence you end up with something like F[f] * .... * F[f]

which is in terms of f only(where f is not the original f but the
transform of f).

also

I[n] = int(f^n*e^(iwx))

can be reduced by parts

u = f^n
du = n*f^(n-1)*f'dx

dv = e^(iwx)dx
v = 1/iw*e^(iwx)

so
f^n*e^(iwx)/iw - n/iw*int(f^(n-1)*f'*e^(iwx))

the last integral, by parts again

u = f'
du = f''dx

dv = f^(n-1)*e^(iwx)dx
v = I[n-1]
so

I[n-1]*f' - int(I[n-1]*f')

or

I[n] = f^n*e^(iwx)/iw - n/iw*int(f^(n-1)*f'*e^(iwx)) = f^n*e^(iwx)/iw
- n/iw*(I[n-1]*f' - int(I[n-1]*f''))

or something similar.

The point here is that you can find a recursive formula for computing
it. You can then potentially reduce it even farther to a close
formula. At least for polynomials you can easily do this since the
formula will definitely terminate after a finite number of steps. Also
with sinusoids the formula should simplify to a closed form after a
few steps.

I'll leave you to do the rest of the work since it's difficult to show
mathematical derivations in ascii.