5 ms·

You can also think of the Fourier Transform as a projection (dot product) of a signal onto the space of all sinusoids. That's the explanation that made everythi

• flatline 14y ago
  It is both the least squares approximation using periodic functions of this sort (i.e. the projection you mentioned), and an interpolant - a very nice combination of properties.
• NickPollard 14y ago
  That's a really great way of thinking about it, actually - particularly if you're used to vector mathematics.
• noahl 14y ago
  I like this way of thinking about it, but I think it is not quite accurate for discrete Fourier Transforms. In this case, we're not projecting onto the space of all sinusoids, only the space of sinusoids whose period is a multiple of (1/N). We could probably prove (if we wanted to try) that those form a basis for the vector space of N-long complex vectors, so using any more sinusoids would be redundant. However, I believe the continuous Fourier Transform works exactly like that.
  • dropdownmenu 14y ago
    Very true, and if you want to be more specific my definition only works for finite duration continuous time signals with finite second moment. I use this definition as it also works well for understanding other transforms such as the Laplace transform.
• revelation 14y ago
  As is explained, you don't even need to project. What you do is a change of basis, which is to say no information is lost.