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Can anyone help me in understanding of deconvolution based on toeplitz matrix?

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Junjie Zhao
Junjie Zhao 2024년 9월 25일 11:17
댓글: William Rose 2024년 9월 26일 3:41
First of all, d is a trace whose size is (1,2500), p is a trace whose size is (1,2500).
For d*w=p, which is Dw=p, where D is Toeplitz matrix made by d. Take d and p as an example [1 2 3 4 5].
I construct D that the first column is [1 2 3 4 5 0 0 0 0], the first row is [1 0 0 0 0], which means d lags 2 time step up and 2 time step down, the 0 time lag is place 3 when the first place is 1.
To find w, I use deconvolution, therefore, w=((D^T)D)^(-1)(D^T)p
Confusion comes in here. the number of row of D is 9, however, the p is (1,5) size, how should I concatenate p using 0? Just under the last element of p or two 0 before and two 0 behind?
What I really want to know is the exact form of deconvoution in programs. Really Really appraciate it if anyone could help. Please!

답변 (1개)

William Rose
William Rose 2024년 9월 25일 16:01
Here is an examle, using d=[1 2 3 4 5], which you suggested. Since d has length 5, then the convolution d*w = p (where * indicates convolution) will be 4 elements longer than w. So if w has length 5, p will have length 9, etc. For this example, I will assume w has length 6, in order to demonstrate that the length of w does not have to equal the length of d. I will do the forward convoltion (compute p=d*w) first. Then I will do the inverse convolution.
Compute Toeplitz matrix:
w=[-1 0 1 2 1 3]';
d=[1 2 3 4 5]';
%nr=length(d)+length(w)-1; % rows in Toeplitz matrix
%nc=length(w); % columns in Toeplitz matrix
c=[d;zeros(length(w)-1,1)]; % column 1 of Toeplitz matrix
r=[1,zeros(1,length(w)-1)]; % row 1 of Toeplitz matrix
D=toeplitz(c,r);
disp(D)
1 0 0 0 0 0 2 1 0 0 0 0 3 2 1 0 0 0 4 3 2 1 0 0 5 4 3 2 1 0 0 5 4 3 2 1 0 0 5 4 3 2 0 0 0 5 4 3 0 0 0 0 5 4 0 0 0 0 0 5
Use D to compute p=convolution of d with w:
p=D*w;
disp(p')
-1 -2 -2 0 3 15 22 23 17 15
Compute estimate of w, from p, using the Toeplitz matrix:
wEst=inv(D'*D)*D'*p;
disp(wEst')
-1.0000 -0.0000 1.0000 2.0000 1.0000 3.0000
The result shows that the estimate of w equals the original w.
  댓글 수: 2
Paul
Paul 2024년 9월 25일 18:17
Hi Wiliam,
Can also use convmtx for D and save a few lines of code.
w=[-1 0 1 2 1 3]';
d=[1 2 3 4 5]';
D = convmtx(d,numel(w))
D = 10×6
1 0 0 0 0 0 2 1 0 0 0 0 3 2 1 0 0 0 4 3 2 1 0 0 5 4 3 2 1 0 0 5 4 3 2 1 0 0 5 4 3 2 0 0 0 5 4 3 0 0 0 0 5 4 0 0 0 0 0 5
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William Rose
William Rose 2024년 9월 26일 3:41
@Paul, thank you. I didn't know about convmtx().

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