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I run in to a real problem which must be a classic, only I cannot find the answer. I know Fourier transform shifts a signal from the time spectrum to the frequency spectrum. Is there a similar transform to shift a function from the time spectrum to the phase spectrum?

I.e., where Fourier translates $g(t) \rightarrow a_0 / 2 \sum_1^\infty a_n \cos(nt) +b_n \sin(nt)$ I am wondering if there is a likewise transform that translates $g(t) \rightarrow a_0 + \sum_1^\infty a_n \cos(t+2\pi/n) +b_n \sin(t+2\pi/n)$? Or anything of the likes of that?

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As $n \to \infty$, $t + 2\pi/n \to t$ so that the higher "phase" modes are very close to each other in any reasonable metric on the space of the functions. Therefore it's unlikely that such a system would form a base of all functions with useful properties. Compare this with Fourier transform where all the modes are orthogonal to each other (in the standard inner product). –  Marek Dec 6 '12 at 11:27
    
Thanks Mark. I know the function I gave is not a basis, because the functions are not orthogonal. However, is there a series of functions which can be interpreted as a transformation into the phase spectrum? –  Ran Dec 6 '12 at 12:06
    
The Fourier series may be decomposed into the product of its amplitude spectrum and its phase spectrum. Then $g(t)=\sum_n a_n \cos(nt + \phi_n)$ where $\phi_n$ is your phase spectrum. Of course it differs from your suggested transform, but it is the usual meaning of 'phase spectrum'. –  Oberdada May 13 '13 at 23:10

1 Answer 1

This would be a very bad transformation.

Imagine, you would have $$g(t) = a_0 / 2 +\sum_1^\infty a_n \cos(t+2\pi/n) +b_n \sin(t+2\pi/n)$$ Which would yield $$g''(t) =-( \sum_1^\infty a_n \cos(t+2\pi/n) +b_n \sin(t+2\pi/n) ) = -g(t) -a_0/2$$ Which isn't true for arbitary function. Also $g^{[n]}(t)$ is never the zero function for any n. So you couldn't represent any function with a vanashing derivative, e.g. polynomials.

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