I'm currently studying the Fast Wavelet Transform. As I currently understand, the Fast Wavelet Transform is implemented as a QMF filter bank where the frequency resolution decreases as the signal is low-pass filtered and sub-sampled. Pictured below:

enter image description here

However, I'm curious as to why this "works".

Let's say I have a true 750Hz signal that is sampled at 2kHz. The amplitude of this 750Hz signal is 2Vpp.

The Fast Wavelet Transform says that I can high-pass filter with a pass region of 500Hz - 1000Hz, decimate by a factor of 2, and it will output coefficients that correspond to the signal frequency. Because I know the signal is constant frequency 750Hz, I would hope for constant coefficients of magnitude 1 at Level 1.

But I'm confused. Because we're subsampling at 1KHz (half of the original sample rate), we're essentially aliasing the 750Hz signal into the 0Hz - 500Hz range. It seems like the coefficient value will vary depending on where we're sampling. However, because we know the true signal is a constant 750Hz signal, it would be desired that the wavelet coefficient is also constant.

How exactly do the coefficients of the Fast Wavelet Transform correspond to the true frequency content of a signal?

Edit: Are there any good resources for understanding the general properties of Wavelet coefficients? e.g. should a wavelet coefficient remain constant if the signal has constant frequency content?

  • $\begingroup$ Careful with "true frequency content"; I'm unfamiliar with the FWT, but DFT for one will rarely yield such information. Assuming FWT implements DWT, this might help. $\endgroup$ Commented Oct 31, 2020 at 21:51

1 Answer 1


[Specific answer] If you have a real sine or cosine, the discretization and finite support may entail issues in the possibility of having a constant: edge samples or periods that cannot be divided by powers-of-two can affect the wavelet coefficients.

But more important: let us suppose the discrete sine period is well chosen. Then a low-pass or high pass filter on a sine will yield something close to a sine (because complex exponentials are invariant under linear systems). So even after downsamling, the coefficients in a wavelet subband are likely to behave as sinusoidal values.

If you expect "constant values", you are likely looking at a complexified and absolute-valued representation of the coefficients.

With a real signal and discrete wavelet, one will likely see ripples on several subbbands, instead of a constant.

[Generic answer] The purpose of most transforms (whether fast or not) is to help in handling or interpreting the content of some data, or classes of data. Invertible transformations keep all information, sometimes with redundancy.

Discrete wavelet transforms (DWT) are meant to capture information in a non-redundant way from not-so-stationary signals. They often are really bad as capturing information from a purely periodic sine signal.

As most wavelet filter bank filters are not perfect, and the signal is of finite length, filtering and representing coefficients will suffer artifacts. So in a DWT, a sine would spread on many subbands, and suffer from aliasing, which is well known. However, despite the above caveat, the magic of wavelets can however recover the original sine, from the spread and aliased coefficients.

So :

  • don't expect DWT coefficients of a sine to be constant
  • don't expect them correspond to the true frequency content of a signal

They are not meant for that. However, if a signal has constant frequency content, but phase shifts (thus non-stationary), then wavelets (maybe continuous once) can be great at detecting their location or onsets.

  • $\begingroup$ This is good information. Do you have a source for better interpreting wavelet coefficients? $\endgroup$
    – Izzo
    Commented Nov 3, 2020 at 22:35
  • $\begingroup$ Interprétation for DWT only? $\endgroup$ Commented Nov 3, 2020 at 22:41
  • $\begingroup$ DWT would be preferred. $\endgroup$
    – Izzo
    Commented Nov 3, 2020 at 22:47
  • $\begingroup$ I have strongly edited the content, to better answer your question $\endgroup$ Commented Nov 5, 2020 at 22:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.