Why use the convolution matrix in Matlab as opposed to the conv() function?

Question

I understand that if we have two vectors -say X and Y, we can calculate their convolution in Matlab using Z = conv(X, Y). There's another way to do this which is, as described on Mathworks.com, through the convolution matrix:

n = length(Y);
Z = convmtx(X,n)*Y;

I have two questions:

Why use the convolution matrix if we can rely on conv(X, Y)?
The cited docs say that the former method is more efficient but unfortunately they don't explain why. Is it mainly due to the fact that the second method requires the computation and allocation of length(Y) or is there more to it (is the matrix multiplication more efficient than a convolution operation, etc.)?

Thank you!

Edit: I contacted Mathworks.com as well and I thought I'd share their answer:

You can see for yourself that conv is indeed more efficient if you run this code.

tfs = 0; tsl = 0;
Nt = 20; sh = 500;
for kj = 1:Nt
    q = randn(10000,1); h = randn(sh,1);
    tic
    fst = conv(q,h);
    tfs = tfs+toc;
    tic
    slo = convmtx(q,sh); slo = slo*h;
    tsl = tsl+toc;
end
tfs = tfs/Nt; tsl = tsl/Nt; tsl/tfs

You can see that conv is faster by a factor of about 120. There are a few reasons why that is the case, but the most important may be that conv is (almost) a built-in compiled function, whereas convmtx is coded in the MATLAB language and is interpreted when called (i.e., it is parsed, subjected to a layer of type checking, etc.). Moreover, when you call convmtx you create many copies of the input array, even though you really need only one.

Luis Mendo Luis Mendo · Accepted Answer · 2014-09-14T23:10:52

The convolution matrix is just the matrix that, upon multiplication, would give the same result as convolution. It's always a Toeplitz matrix.

The convolution matrix is not intended for actual computation of the convolution. Its application is in cases when you need to represent the convolution as a matrix multiplication.

As an analogy, consider the discrete Fourier transform (DFT). As is well known, the DFT is nearly always computed with the very efficient FFT algorithm. But you can represent the DFT as a matrix if you want. You would never use that for the actual computation of the DFT; but conceptually it sometimes helps to express the DFT as matrix multiplication. For example, that allows one to view the DFT as a change of basis in a vector space.

Why use the convolution matrix in Matlab as opposed to the conv() function?

1 Answers