# Wavelets and bank filters

The wavelet transform has a problem as it gives poor time resolution for low frequencies and poor frequency resolution for high frequencies according to uncertainty conditions.

This appears well while using the window notation. But when using bank filters, I can't imagine this problem. So, do bank filters solve this problem?

In addition, if there is a signal with maximum frequency equal to 1000 hz, how are low and high pass filters designed to decompose the signal according to a certain mother wavelet?

• All transforms have their problems. Filter banks can implement either time-scale or time-frequency transforms. I do not understand the "window notation" when talking about wavelets. Could you please clarify as well if you are talking about continuous/discrete/orthogonal/redundant wavelets? The last question does not seem related to a specific frequency. High pass/low pass 2-band wavelets are kind of agnostic to the actual sampling Feb 16 '18 at 20:10
• By window notation I meant STFT. Small window gives poor frequency resolution. As in STFT we calculate the fourier for the product of the signal and the window function which becomes the convolution between the fourier transform of the signal and the window( which becomes sinc function). The width of the sinc function is inverse to the width of the window.
– EWF
Feb 22 '18 at 19:00
• But in continous wavelet transform, we calculate the correlation between the signal and a mother wavelet with a certain scale. It's easy to detect the time resolution problem with high scales, but I can't imagin the reason of the frequency resolution problem. The same when using DWT.
– EWF
Feb 22 '18 at 19:01

## 1 Answer

Filter banks are one way implement specific discretized versions of a given continuous wavelet transform (or a given SFTF). Their performance depends a lot on how one designs wavelet voices and their subsampling.

if you to the dyadic version (or DWT), then the highest subband spans a $[1/2,1]$ (in reduced frequencies), the next one $[1/4,1/2]$ and so on. So the frequency resolution is usually poor on the highest subband. Alternatives consist in further dividing each subband (wavelet packets), or in already splitting the data in $M$ subbands instead of $2$ ($M$-band wavelets), or a combination of the two.

Last, the filter banks are quite generic, so nothing prevents your to use an augmented one, with both wavelet-filters and STFT-filers.