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when I read papers on signal processing, I often see the term 'coefficient' being used to refer directly to 'samples', i.e. "We can reduce the number of coefficients, by simply subsampling the even samples." or in presentations on the lifting scheme (wavelets), it seems the output of the low pass filter is also called coefficients.

To me, the output of a low pass filter is a 'smoothed' version of the original input sequence. Further, to me, a coefficient is e.g. a, b in y[n] = ax[n] + bx[n-1]. In this sense, the coefficients a and b certainly is a representation of the signal-- but i would not call y[n] the coefficients.

I am probably very confused about the terminology. Could someone clarify this with some intuition?

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There's really not much I can clarify for you here:

A coefficient really is just a factor in series. Whether that series is e.g. a Fourier series, a sum over powers of 2, filter delay taps or just a series of sampling instants can't be said. Context!

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  • $\begingroup$ Thanks for commenting. So if you need context, I am struggling to understand wavelets, or rather, how lifting is used to 'construct' wavelets. I fear I am vastly out of my league when it comes to the maths, but I'm not the worst programmer in the world. Anyway, I am reading this: pdfs.semanticscholar.org/f56b/… and there are phrases like "subsampled coefficients" and "odd coefficients" (with respect to the 'lazy wavelet' which splits in even and odd samples). $\endgroup$ – OddPodd Feb 24 '18 at 12:44
  • $\begingroup$ then these are coefficients to the wavelets basis, probably. $\endgroup$ – Marcus Müller Feb 24 '18 at 12:45
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It turns out that at high enough resolutions, the samples of a signal in fact turns out to be very close to the coefficients in the wavelet expansion of a signal.

That is-- finding the coefficients of a wavelet expansion partly boils down to doing inner product between the scaling function and the function under analysis (same for wavelet function). The scaling function sorta-kinda goes towards a dirac delta, and as such, the coefficients are simply samples of the original signal.

This is not a very obvious result to me, but it falls out of the orthogonality requirement of the wavelet system amongst other things. I guess this is sorta obvious to DSP gurus.

This matches up well with the common 'trick' i've seen that extracts the scaling coefficients of a signal by use of a low pass filter, as doing convolution does indeed seem to be very close to calculating inner products.

I can't work the maths, but i'm fairly sure a theorem such as 'with certain rules, an inner product between a function and a set of orthogonal bases are equal to low pass filtering the signal (probably FIR)' exists somewhere in the literature. Similarily, i expect the same to be true for high pass filters and wavelet (i.e. differences) functions.

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    $\begingroup$ an FIR filter is the same as an inner product of the most current set of input samples and the set of "coefficients". not all filters are LPF. i don't think all wavelets are LPF. $\endgroup$ – robert bristow-johnson Feb 25 '18 at 11:29
  • $\begingroup$ welp, i didn't quite get you. i know i can express a FIR filter as a linear set of coefficients and some samples (an inner product, indeed!) -- but i didn't know (or rather, didn't understand how) a LPF could produce coefficients for the scaling function in a wavelet expansion. this was essentially the root of my question-- but now i have confused myself again of course, because there are now two sets of coefficients: the ones internal to the FIR filter and the ones that are being produced by the filter. I guess the coefficients of the LPF somehow determine the shape of the scaling function. $\endgroup$ – OddPodd Feb 25 '18 at 12:34
  • $\begingroup$ Found a good reference, in particular for my last "theorem". I turns out it is not completely wrong. "Multirate Digital Filters, Filter Banks, Polyphase Networks, and Applications: A Tutorial" authors.library.caltech.edu/6798/1/VAIprocieee90.pdf $\endgroup$ – OddPodd Feb 25 '18 at 12:42

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