4 ms·
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
by 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 agoVery 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.