# Why is the high-pass filter result in a discrete wavelet transform (DWT) downsampled?

From Wikipedia's description of the Discrete Wavelet Transform, a signal yields a set of:

• approximation coefficients (low-pass: by averaging + downsampling)
• detail coefficients (high-pass: convolving with wavelet + downsampling)

This process is cascaded over approximation coefficients:

Why is the high-frequency part of a signal downsampled and how does it retain information? Can it be reconstructed? Is information lost due to the Nyquist theorem?

Sources I looked at:

## 3 Answers

Assume the original sampling rate is $$F_s$$ and that the filters are perfect brickwall Halfband filters. After the low pass filtering the frequency content is from $$0 \rightarrow \frac{F_s}{4}$$. Because of the new bandwidth, the sampling rate can be reduced to $$F_s/2$$.

For the high-pass filter, the frequency content lies in the range $$\frac{F_s}{4} \rightarrow \frac{F_s}{2}$$. There is no content in the range $$0 \rightarrow \frac{F_s}{4}$$. We can downsample by a factor of 2 in this case as well. The downsampling causes the frequency content in the $$\frac{F_s}{4} \rightarrow \frac{F_s}{2}$$ to be aliased into the range $$0 \rightarrow \frac{F_s}{4}$$, but because there is no frequency content in this range, due to the high-pass filter, we have not lost any information.

• Ah! My understanding of the Nyquist theorem was lacking: sampling rate has to be 2 x the bandwidth, and not the highest frequency. In cases where the signal is in the baseband, the two coincide. But when we have a non-baseband signal (like a high-pass), we can still get away with downsampling as long as the rate satisfies the theorem with respect to the bandwidth. – hazrmard Jun 20 at 20:26

The whole idea is that every level splits the image information in two equal halves, which can be represented by half the number of coefficient as the previous level, each.

Otherwise, you'd not be doing much of a decomposition, would you?

So, since you only need half of the coefficients (the rest is redundant), you just keep half of the coefficients. That's what the decimation stage does.

The classical discrete wavelet transform is critically decimated. In other words, it should preserve the quantity of "samples". In other words, apart from data border effects, the number of wavelet coefficients should be the same as the numer of data samples.

More generally, a critically-sampled, analysis multi-band filter bank with $$M$$ channels ($$M=2$$ for the DWT) is composed of $$M$$ filters $$H_m$$ in parallel, followed by downsampling factors $$k_m$$, such that $$\sum_{m\in\{1\ldots M\}} 1/k_m=1$$.

Invertible filter banks require the existence of a perfect synthesis filter bank. It is composed of $$M$$ filters $$G_m$$ in parallel, combined onto an output, preceded by the corresponding upsampling factors $$k_m$$. Input and output thus have the same global number of samples. But information is not lost only if certain conditions on the $$H_m$$ and $$G_m$$ are met.

Of course, with $$M=1$$, no subsampling is required (no aliasing), but if $$H_1$$ is not invertible, you will loose information. Moreover, if $$H_1$$ is FIR, is inverse is not (except for trivial cases).

The magic of wavelets is that, with $$M=2$$, there are many pairs of FIR filters such that, even if you downsample their output by $$2$$ and create aliasing, FIR synthesis filter bank exist. So 2-fold downsampling creates aliasing on both the low-pass and the high pass filter outputs (in fact, the high-pass output is shifted to the lower part of the spectrum). But the synthesis filter bank can cancel this aliasing.

Finally, for some well-chosen analysis and synthesis filters, even if aliasing occur in the middle, it is finally cancelled, Nyquist remains fulfilled.