Generate two signals of different lengths. Compare their circular convolution and their linear convolution. Use the default value for n.

a = [1 2 -1 1];
b = [1 1 2 1 2 2 1 1];

c = cconv(a,b);            % Circular convolution
cref = conv(a,b);          % Linear convolution

dif = norm(c-cref)

dif = 9.7422e-16

The resulting norm is virtually zero, which shows that the two convolutions produce the same result to machine precision.

Generate two vectors and compute their modulo-4 circular convolution.

a = [2 1 2 1];
b = [1 2 3 4];
c = cconv(a,b,4)

c = 1×4

    14    16    14    16

Generate two complex sequences. Use cconv to compute their circular cross-correlation. Flip and conjugate the second operand to comply with the definition of cross-correlation. Specify an output vector length of 7.

a = [1 2 2 1]+1i;
b = [1 3 4 1]-2*1i;
c = cconv(a,conj(fliplr(b)),7);

Compare the result to the cross-correlation computed using xcorr.

cref = xcorr(a,b);
dif = norm(c-cref)

dif = 3.3565e-15

Generate two signals: a five-sample triangular waveform and a first-order FIR filter with response $$H(z)=1-z^{-1}$$.

x1 = conv([1 1 1],[1 1 1])

x1 = 1×5

     1     2     3     2     1

x2 = [-1 1]

x2 = 1×2

    -1     1

Compute their circular convolution with the default output length. The result is equivalent to the linear convolution of the two signals.

ccnv = cconv(x1,x2)

ccnv = 1×6

   -1.0000   -1.0000   -1.0000    1.0000    1.0000    1.0000

lcnv = conv(x1,x2)

lcnv = 1×6

    -1    -1    -1     1     1     1

The modulo-2 circular convolution is equivalent to splitting the linear convolution into two-element arrays and summing the arrays.

ccn2 = cconv(x1,x2,2)

ccn2 = 1×2

    -1     1

nl = numel(lcnv);
mod2 = sum(reshape(lcnv,2,nl/2)')

mod2 = 1×2

    -1     1

Compute the modulo-3 circular convolution and compare it to the aliased linear convolution.

ccn3 = cconv(x1,x2,3)

ccn3 = 1×3

     0     0     0

mod3 = sum(reshape(lcnv,3,nl/3)')

mod3 = 1×3

     0     0     0

If the output length is smaller than the convolution length and does not divide it exactly, pad the convolution with zeros before adding.

c = 5;
z = zeros(c*ceil(nl/c),1);
z(1:nl) = lcnv;

ccnc = cconv(x1,x2,c)

ccnc = 1×5

    0.0000   -1.0000   -1.0000    1.0000    1.0000

modc = sum(reshape(z,c,numel(z)/c)')

modc = 1×5

     0    -1    -1     1     1

If the output length is equal to or larger than the convolution length, pad the convolution and do not add.

d = 13;
z = zeros(d*ceil(nl/d),1);
z(1:nl) = lcnv;

ccnd = cconv(x1,x2,d)

ccnd = 1×13

   -1.0000   -1.0000   -1.0000    1.0000    1.0000    1.0000    0.0000   -0.0000    0.0000    0.0000    0.0000   -0.0000   -0.0000

modd = z'

modd = 1×13

    -1    -1    -1     1     1     1     0     0     0     0     0     0     0

The following example requires Parallel Computing Toolbox™ software. Refer to GPU Support by Release (Parallel Computing Toolbox) to see what GPUs are supported.

Create two signals consisting of a 1 kHz sine wave in additive white Gaussian noise. The sample rate is 10 kHz

Fs = 1e4;
t = 0:1/Fs:10-(1/Fs);
x = cos(2*pi*1e3*t)+randn(size(t));
y = sin(2*pi*1e3*t)+randn(size(t));

Put x and y on the GPU using gpuArray. Obtain the circular convolution using the GPU.

x = gpuArray(x);
y = gpuArray(y);
cirC = cconv(x,y,length(x)+length(y)-1);

Compare the result to the linear convolution of x and y.

linC = conv(x,y);
norm(linC-cirC,2)

ans =

   1.4047e-08

Return the circular convolution, cirC, to the MATLAB® workspace using gather.

cirC = gather(cirC);