simple signal compression

Mathematics — Alex @ 1:04 am

Question

Suppose you have a (periodic) function f : [0, 2\pi) \rightarrow \C whose support is limited to the interval [0,\frac{2\pi}{m}], where m>2. Is there a more compact representation than the entire fourier series? It seems like a good possibility, since the information in the function is actually restricted to a smaller region.

Answer

Yes. Let \tilde f : [0,2\pi] \rightarrow \C be defined as f : \theta \rightarrow f(\tilde \theta) where  \tilde \theta = \theta + \frac{2\pi k}{m}, where  k \in \Z and \tilde \theta \in [0,\frac{2\pi}{m}]. Then consider the fourier coefficients for \tilde f:


\displaystyle
\tilde c_k = \frac{1}{2\pi} \int_0^{2\pi} \tilde f(\theta) e^{-i k \theta} d\theta = \frac{1}{2\pi} \sum_{n=0}^{m-1} \int_{n\left(\frac{2\pi}{m}\right)}^{(n+1)\left(\frac{2\pi}{m}\right)} \tilde f(\theta) e^{-i k \theta } d\theta

\displaystyle
= \frac{1}{2\pi} \sum_{n=0}^{m-1} \int_0^{\frac{2\pi}{m}} \tilde f(\theta + \frac{n2\pi}{m}) e^{-i k \left(\theta + \frac{n 2 \pi}{m}\right)} d\theta
= \left(\sum_{n=0}^{m-1} e^{- \left( i k \frac{2\pi}{m} \right) n } \right) \int_0^{\frac{2\pi}{m}} \tilde f(\theta) e^{-i k \theta} d\theta
= c_k \left(\sum_{n=0}^{m-1} e^{- \left( i k \frac{2\pi}{m} \right) n } \right)

Since the sum of exponentials is m when m \mid k and 0 otherwise, we can throw out all the zero coefficients and represent \tilde f or f using just the coefficients \{m c_{mk}\}_{k \in\Z}.

Comments

This is not directly useful, but considering where I got the idea from (minus the nice proof, all thought out by me :) it probably ends up figuring into quadrature mirror filter theory somewhere. Maybe this has something to do with the ubiquitous downsampling I keep hearing about in relation to wavelet transforms.

Anyhow, it is an interesting result. Notice that as m \rightarrow \infty, the sequence \{m c_{mk}\}_{k\in \Z} approaches a delta pulse, in a certain sense. Which, in a certain sense, is as expected.

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