grlee77
commented
7 years ago Yes, the edge mode is periodization for the SWT transforms. This is basically like periodic, but handles odd-sized inputs a bit differently. (IIRC, for odd input the last sample is duplicated to make an even sized signal and then the even sized signal is filtered using periodic boundary conditions).
The DWT routines have a variety of other edge modes implemented, but the SWT routines do not.
JoostJM
When performing swt using a Haar wavelet, the last index in the L-filter is calculated as 1/sqrt(2) * (index[-1] + index[0]), i.e. it seems as if the function wraps around. Is this intentional? I'm certainly no expert on wavelet transforms. Here is the output from some simple tests I did in jupyter:
Test1, length = 8 In: [5 4 3 0 0 0 0 0] Out-H: [ 6.364 4.9497 2.1213 0. 0. 0. 0. 3.5355] Out-L: [ 0.7071 0.7071 2.1213 0. 0. 0. 0. -3.5355]
Test2, length = 6 In: [5 4 3 0 0 0] Out-H: [ 6.364 4.9497 2.1213 0. 0. 3.5355] Out-L: [ 0.7071 0.7071 2.1213 0. 0. -3.5355]
Test3, reverse In: [0 0 0 0 0 3 4 5] Out-H: [ 0. 0. 0. 0. 2.1213 4.9497 6.364 3.5355] Out-L: [ 0. 0. 0. 0. -2.1213 -0.7071 -0.7071 3.5355]
Test4, reverse with extra padding In: [0 0 0 0 0 3 4 5 0 0] Out-H: [ 0. 0. 0. 0. 2.1213 4.9497 6.364 3.5355 0. 0. ] Out-L: [ 0. 0. 0. 0. -2.1213 -0.7071 -0.7071 3.5355 0. 0. ]
Test5, discreet instead of stationary wavelet transform In: [5 4 3 0 0 0 0 0] Out-H: [ 6.364 2.1213 0. 0. ] Out-L: [ 0.7071 2.1213 0. 0. ]
3.5355 is equal to 1/sqrt(2) * 5