On Sat, 12 Nov 2005, Charles Henry wrote:
The Fourier transform is at the heart of DSP for a specific reason. Any signal can be *exactly* represented through a change of basis (change of basis here means a transformation from one set of numbers to another through a one-to-one function that preserves energy (the metric))
What you're describing is not just a change of basis, it's an isometric one. An ordinary change of basis could change the definition of energy in any possible way, as long as it has an inverse change of basis that can go back exactly to the original basis.
Okay, so suppose we have sqrt(x) where x is R. (I'll use "is" to denote "belongs to")
That would be confusing. It's much better to use "is in" or even "is a".
To show why a signal can be exactly represented by its transform is even more tricky, and requires the definition of Hilbert spaces, the set of continuous functions C0, the polynomials, trigonometric identities on cos(n*x), and the Lebesgue integral.....I don't want to go through all the details, right now. It really is a *long* explanation.
It's only so if you really want to posit the existence of infinitely detailed signals. In practice you have to do away with all kinds of infinity when dealing with actual samples and so you can safely dismiss most of the artifacts of so-called "Real" numbers.
Using [fft~] in Pd is little more than converting from a 128-dimensional space into another 128-dimensional space, which may sound scary and alien, but not as much as any flavour of infinite-dimensional spaces that we may have to introduce in the case of pretending to have infinite precision.
They are only indicators of phases of sinusoids in the frequency domain.
Right. The real vs imaginary parts correspond to cosine amplitude vs sine amplitude. A polar transform (or a complex-log transform) can separate the phase from the total amplitude.
Mathieu Bouchard - tél:+1.514.383.3801 - http://artengine.ca/matju Freelance Digital Arts Engineer, Montréal QC Canada