When you perform the Haar wavelet transform, you take the sums and differences, then at each stage, you multiply the entire signal by $\small\sqrt2$.

When taking the inverse transform, you multiply the signal by $\frac{1}{\sqrt2}$ for each iteration.

What does this "normalization" really represent?


As I understand it, the normalization is because the Haar wavelet conserves energy of the signal. In that, when you take signal from one domain to another, you aren't supposed to add energy to it, (although conceivably you might lose energy).

The normalization is just a way to ensure that the energy of your Haar-transformed signal in the Haar-domain has the exact same energy as your signal in the original domain.

Intuitively speaking, Haar, Fourier, etc, are all just basis transformations, which intuitively just mean that you are looking at the signal in a different way, (technically, through a different set of bases). Therefore, if all you are doing is looking at a signal differently, its energy cannot/should not change.

  • $\begingroup$ Ok, this makes sense. If you try it with an array of numbers, eg [ 2 1 3 4 9 7 0 4 ] -> 1 step sum/diff -> [ 1.5 3.5 8 2 | .5 -.5 1 -2 ]. The squared norm of the first signal is 176, the second one is 88. Multiplying the second signal by √2 makes it's squared norm 176 as well. $\endgroup$
    – bobobobo
    Mar 17 '12 at 20:57
  • $\begingroup$ @bobobobo Yup! You got it. Now I seem to remember that a loss of energy can in fact be possible with some transforms, (and this would also be conceivable), but I cannot recall any such cases at the moment. $\endgroup$
    – Spacey
    Mar 17 '12 at 21:01
  • 1
    $\begingroup$ Projection on an incomplete basis would lose energy, trivially: the projection is no longer identical to the original, but has lost all information(energy) orthogonal to the incomplete basis. $\endgroup$
    – MSalters
    Jan 17 '14 at 16:06

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