how to deconvolute a array ?
显示 更早的评论
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
采纳的回答
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.
8 个评论
thank you
is it possible to do the same in 2D , with marix "a","b" and "c"?
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
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
2022-2-11
编辑:Chris Turnes
2022-2-11
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.
okay deal
thank you Chris
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?
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);
更多回答(0 个)
类别
在 帮助中心 和 File Exchange 中查找有关 Linear Algebra 的更多信息
另请参阅
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!选择网站
选择网站以获取翻译的可用内容,以及查看当地活动和优惠。根据您的位置,我们建议您选择:。
您也可以从以下列表中选择网站:
如何获得最佳网站性能
选择中国网站(中文或英文)以获得最佳网站性能。其他 MathWorks 国家/地区网站并未针对您所在位置的访问进行优化。
美洲
- América Latina (Español)
- Canada (English)
- United States (English)
欧洲
- Belgium (English)
- Denmark (English)
- Deutschland (Deutsch)
- España (Español)
- Finland (English)
- France (Français)
- Ireland (English)
- Italia (Italiano)
- Luxembourg (English)
- Netherlands (English)
- Norway (English)
- Österreich (Deutsch)
- Portugal (English)
- Sweden (English)
- Switzerland
- United Kingdom (English)
