Discrete Wavelets - Question on decimation of high pass

[TheWang]

Hi, I'm trying to grasp the concept of discrete wavelets and can't seem to find an answer to my question.
In the decomposition of a signal using wavelets filter banks, the signal goes through a low pass and high pass filter. The output of the low pass and high pass is decimated by 2. I can understand the low pass output being decimated by 2 since only the lower half of the bandwidth is kept. However, how is it possible to decimate the high pass output by 2 given Nyquist theory? I'm imagining if the original signal was sampled at a rate slightly higher than twice the highest frequency in the signal, the output of the high pass filter bank can't be downsampled without losing the high frequency information. Can anyone help me see this clearly? Thanks alot

 

[Valerie Park]

! The key to understanding the decimation of the high pass output is to understand that the wavelet filters used in the decomposition of a signal are designed to be bandpass filters, meaning they have a finite bandwidth. This means that the high pass filter will only pass frequencies above a certain cutoff frequency. The cutoff frequency of the high pass filter is determined by the sampling rate of the original signal. So, when the signal is downsampled, it is still possible to discard the lower half of the bandwidth of the high pass filter without losing any of the high frequency information because the highest frequency in the signal will still be above the cutoff frequency of the high pass filter.