How to solve the Fast Fourier transform?

Does Fourier Transform imply that delta function value divided by infinity is 1?

• In some explanations of Fourier Transform, you can tell that if you use Fourier Transform on a cyclic functions, the value of the function is 2Ï€*(delta-function)*the value at the point. This explanation relates to how you prove that fourier transform is similar to taking a Fourier series and expand the area from [-Ï€,Ï€], to infinity, (as you can see on page 6, here:http://www.slideshare.net/darvind/signal-16181198, thanks ) Then, the Fourier series become Fourier Transform, but that implies that the normalization by infinity would turn all the values of the frequencies to zero, except maybe the delta function, so does this mean that delta function value divided by infinity is one?

• Answer:

  The Fourier transform is like the limit of the Fourier series with the interval made very long, except WITHOUT dividing by the length of the interval, so that the frequencies are no longer normalized. This is required so that you get a sensible limit out, as you noticed. The result is that you need to integrate the modes, not sum, and the difference between an integral and a sum is a differential factor that effectively normalizes each mode back to the tiny value it should have to reproduce the function you are transforming. The value of a delta function at the origin is infinite, but it is not well defined as a value--- if you make a lattice approximation to a delta-function on a lattice of size epsilon, the value at the origin is 1/epsilon and everywhere else it's zero.