fourier transform

Hello snipped-for-privacy@gmail.com

Thats nice, hope the answer finds you :o)

Cheers IBM

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

Maybe this will help in your journey.

F[f^n] =3D 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] =3D int(f^n*e^(iwx))

can be reduced by parts

u =3D f^n du =3D n*f^(n-1)*f'dx

dv =3D e^(iwx)dx v =3D 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 =3D f' du =3D f''dx

dv =3D f^(n-1)*e^(iwx)dx v =3D I[n-1] so

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

or

I[n] =3D f^n*e^(iwx)/iw - n/iw*int(f^(n-1)*f'*e^(iwx)) =3D 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.