scipy.linalg.

convolution_matrix#

scipy.linalg.convolution_matrix(a, n, mode='full')[source]#

Construct a convolution matrix.

Constructs the Toeplitz matrix representing one-dimensional convolution [1]. See the notes below for details.

Parameters:
a(…, m) array_like

The 1-D array to convolve. N-dimensional arrays are treated as a batch: each slice along the last axis is a 1-D array to convolve.

nint

The number of columns in the resulting matrix. It gives the length of the input to be convolved with a. This is analogous to the length of v in numpy.convolve(a, v).

modestr

This is analogous to mode in numpy.convolve(v, a, mode). It must be one of (‘full’, ‘valid’, ‘same’). See below for how mode determines the shape of the result.

Returns:
A(…, k, n) ndarray

The convolution matrix whose row count k depends on mode:

=======  =========================
 mode    k
=======  =========================
'full'   m + n -1
'same'   max(m, n)
'valid'  max(m, n) - min(m, n) + 1
=======  =========================

For batch input, each slice of shape (k, n) along the last two dimensions of the output corresponds with a slice of shape (m,) along the last dimension of the input.

See also

toeplitz

Toeplitz matrix

Notes

The code:

A = convolution_matrix(a, n, mode)

creates a Toeplitz matrix A such that A @ v is equivalent to using convolve(a, v, mode). The returned array always has n columns. The number of rows depends on the specified mode, as explained above.

In the default ‘full’ mode, the entries of A are given by:

A[i, j] == (a[i-j] if (0 <= (i-j) < m) else 0)

where m = len(a). Suppose, for example, the input array is [x, y, z]. The convolution matrix has the form:

[x, 0, 0, ..., 0, 0]
[y, x, 0, ..., 0, 0]
[z, y, x, ..., 0, 0]
...
[0, 0, 0, ..., x, 0]
[0, 0, 0, ..., y, x]
[0, 0, 0, ..., z, y]
[0, 0, 0, ..., 0, z]

In ‘valid’ mode, the entries of A are given by:

A[i, j] == (a[i-j+m-1] if (0 <= (i-j+m-1) < m) else 0)

This corresponds to a matrix whose rows are the subset of those from the ‘full’ case where all the coefficients in a are contained in the row. For input [x, y, z], this array looks like:

[z, y, x, 0, 0, ..., 0, 0, 0]
[0, z, y, x, 0, ..., 0, 0, 0]
[0, 0, z, y, x, ..., 0, 0, 0]
...
[0, 0, 0, 0, 0, ..., x, 0, 0]
[0, 0, 0, 0, 0, ..., y, x, 0]
[0, 0, 0, 0, 0, ..., z, y, x]

In the ‘same’ mode, the entries of A are given by:

d = (m - 1) // 2
A[i, j] == (a[i-j+d] if (0 <= (i-j+d) < m) else 0)

The typical application of the ‘same’ mode is when one has a signal of length n (with n greater than len(a)), and the desired output is a filtered signal that is still of length n.

For input [x, y, z], this array looks like:

[y, x, 0, 0, ..., 0, 0, 0]
[z, y, x, 0, ..., 0, 0, 0]
[0, z, y, x, ..., 0, 0, 0]
[0, 0, z, y, ..., 0, 0, 0]
...
[0, 0, 0, 0, ..., y, x, 0]
[0, 0, 0, 0, ..., z, y, x]
[0, 0, 0, 0, ..., 0, z, y]

Added in version 1.5.0.

References

Examples

>>> import numpy as np
>>> from scipy.linalg import convolution_matrix
>>> A = convolution_matrix([-1, 4, -2], 5, mode='same')
>>> A
array([[ 4, -1,  0,  0,  0],
       [-2,  4, -1,  0,  0],
       [ 0, -2,  4, -1,  0],
       [ 0,  0, -2,  4, -1],
       [ 0,  0,  0, -2,  4]])

Compare multiplication by A with the use of numpy.convolve.

>>> x = np.array([1, 2, 0, -3, 0.5])
>>> A @ x
array([  2. ,   6. ,  -1. , -12.5,   8. ])

Verify that A @ x produced the same result as applying the convolution function.

>>> np.convolve([-1, 4, -2], x, mode='same')
array([  2. ,   6. ,  -1. , -12.5,   8. ])

For comparison to the case mode='same' shown above, here are the matrices produced by mode='full' and mode='valid' for the same coefficients and size.

>>> convolution_matrix([-1, 4, -2], 5, mode='full')
array([[-1,  0,  0,  0,  0],
       [ 4, -1,  0,  0,  0],
       [-2,  4, -1,  0,  0],
       [ 0, -2,  4, -1,  0],
       [ 0,  0, -2,  4, -1],
       [ 0,  0,  0, -2,  4],
       [ 0,  0,  0,  0, -2]])

>>> convolution_matrix([-1, 4, -2], 5, mode='valid')
array([[-2,  4, -1,  0,  0],
       [ 0, -2,  4, -1,  0],
       [ 0,  0, -2,  4, -1]])
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