how to deconvolute a array ?

hy guys

i would like to deconvolute a matrix

code:

clear all

clc

a=rand(10,3);

b=rand10,3); %b=conv2(a,c)

%suppose that b is already the convolution of the array "a" with an array "c"

% I would like to deconvulte " b " to re-obtain "a" and "c".

% any idea how to do so?

% thanks you in advance

Answer 1

[Chris Turnes]

Deconvolution is equivalent to polynomial division. You can get the polynomial division and its remainder with the deconv function.

rng('default');
% Take two vectors.
a = randn(7,1);
c = randn(10,1);
% Compute their convolution.
b = conv(a, c);
% "Recover" the first by deconvolving c from b:
[ahat,r] = deconv(b, c);
% Check the residual and the remainder polynomial
norm(a-ahat)

ans = 8.1907e-11

r'

ans = 1×16

1.0e+-9 * 0 0 0 0 0 0 0 0.2664 0.2397 -0.1352 0.2409 0.0603 -0.0114 0.0599 -0.0156 -0.0101

However, it's important to note that this is not a least-squares solution to the deconvolution, and if b isn't really the result convolving something with c, you may not get an answer that's particularly close to the least squares result. To get the least squares result, you would construct the Toeplitz system corresponding to the convolution and solve it:

% Add some "noise" to c:
bhat = conv(a, c + 1e-3*randn(size(c)));
% Solve with deconv:
ahat_deconv = deconv(bhat, c);
% Compare convolving the result with c against the vector we started with:
norm(bhat - conv(ahat_deconv, c))

ans = 6.0224e+03

% Solve with least-squares:
T = convmtx(c, length(a));
ahat_ls = T \ bhat;
% Compare convolving the least-squares result with c against the vector we
% started with:
norm(bhat - conv(ahat_ls, c))

ans = 0.0044

There are efficient algorithms to solve the Toeplitz system, though there are not any functions directly in MATLAB to do so.

[Rabih Sokhen]

thank you

[Rabih Sokhen]

is it possible to do the same in 2D , with marix "a","b" and "c"?

[Chris Turnes]

Yes:

% Generate source data.
a = randn(5, 7);
c = randn(9, 11);
% Convolve.
b = conv2(a,c);
% De-convolve via least squares.
% convmtx2 returns a sparse matrix, so cast
% to full.
ahat = full(convmtx2(c, size(a)) \ b(:));
% Reshape to a 2-D array.
ahat = reshape(ahat, size(a));
% Check the result.
norm(a-ahat, 'fro')

ans = 2.0547e-15

[Rabih Sokhen]

Wow, that's great.

Is it possible to do the same, but this time we consider that we already have the matrix "c" and "a" and c=conv2 (a, b) and we want to find " b "?Is there a way to find "b" with the minimum error possible?

thank you in advance

[Chris Turnes]

Yes, I think this is just a labeling question. If you take my example and replace "b" with "c", "a" with "b", and "c" with "a" then I think it gives exactly what you're asking.

For what it's worth, I am just building the matrix and solving the system with \ because it's convenient for illustration; in practice you may want to consider an iterative method instead that doesn't require you to explicitly build the matrix, as the convolution matrix can become quite large if the inputs are large.

There are also fast methods for solving these 2-D systems, but again, there are no functions directly in MATLAB to my knowledge that do this.

[Rabih Sokhen]

okay deal

thank you Chris

[Rabih Sokhen]

clear all

clc

a = randn(5, 7);

c = randn(13,17);

b= full(convmtx2(a, size(a)) \ c(:));

b= reshape(b, size(c)-size(a)+1);

%i tried the following code but i got a error, can you plz help me fix it?

[Chris Turnes]

Your error here is that you're not specifying the right size for the convolution matrix. The size argument is the size of the thing you are convolving with a -- so in this case, the size of b. You've alredy determined this later, so you just need to pass it into the function:

a = randn(5, 7);
c = randn(13,17);
szB = size(c) - size(a) + 1;
b= full(convmtx2(a, szB) \ c(:));
b= reshape(b, szB);