I am trying to do some (de)convolution with audio samples. I have one sample s and the same sample with some filters added on top of it s_f
The rank(x)
returns the rank of matrix. In other words number of dimensions it contains. Please check whether ranks of s
and f
are the same before call to signal.convolve()
. Otherwise you will receive an exception you quote.
I have no idea why deconvolution may return something with more dimensions than given input. This requires deeper investigation I don't have time for.
deconvolve
returns two arrays, the quotient and the remainder. So try:
f, r = signal.deconvolve(s, s_f)
For a long time, deconvolve
has not had a proper docstring, but it has one in the master branch on github: https://github.com/scipy/scipy/blob/master/scipy/signal/signaltools.py#L731
The docstring shows an example of the use of deconvolve
. Here's another (sig
is scipy.signal
and np
is numpy
):
The signal to be deconvolved is z
, and the filter coefficients are in filter
:
In [9]: z
Out[9]:
array([ 0.5, 2.5, 6. , 9.5, 11. , 10. , 9.5, 11.5, 10.5,
5.5, 2.5, 1. ])
In [10]: filter = np.array([0.5, 1.0, 0.5])
Apply deconvolve
:
In [11]: q, r = sig.deconvolve(z, filter)
In [12]: q
Out[12]: array([ 1., 3., 5., 6., 5., 4., 6., 7., 1., 2.])
Apply the filter to q
to verify that we get back z
:
In [13]: sig.convolve(q, filter)
Out[13]:
array([ 0.5, 2.5, 6. , 9.5, 11. , 10. , 9.5, 11.5, 10.5,
5.5, 2.5, 1. ])
By construction, this is a very clean example. The remainder is zero:
In [14]: r
Out[14]: array([ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.])
Of course, you won't always get such nice results.