Deconvolution of a polynomial and exponential function

Question

larry liu 2022-3-11

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1669074-deconvolution-of-a-polynomial-and-exponential-function

评论： yicong 2024-9-6

I got the same problem with following link:

https://www.mathworks.com/matlabcentral/answers/602500-deconvolution-of-a-polynomial-and-exponential-function

I got trouble using Bayesian deconvolution

here I already got wt(z) and a'(z) with same one dimension size array

I have no idea how to got iterative.

If need my data, I will give my data. thanks a lot ><

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高飞支 2023-12-17

移动：John D'Errico 2023-12-17

I have the same trouble have you solved it?can you give me some advices?

Rui 2024-8-6

hello, larry liu

i am doing the same work as you did, did u use the Bayes iteration that you showed in the picture? cuz i try to deal with time constant in this way. It doesn't work..

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Answer 1

Yash 2023-11-24

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1669074-deconvolution-of-a-polynomial-and-exponential-function#answer_1359222

编辑：Yash 2023-11-24

在 MATLAB Online 中打开

Hi Larry,

I understand that you are facing issues while using the "deconv" function in MATLAB. "deconv(y,h)" deconvolves a vector "h" out of a vector "y" using polynomial long division, and returns the quotient "x" and remainder "r" such that "y = conv(x,h) + r".

Given that you already have a deconvolution function in your case, you may not need to use the "deconv" function. Instead, you can make an initial guess for "R(z)" and then obtain the approximation iteratively using a loop as follows:

n = length(df2);
w = Wzinstead';
a = df2';
R = randi(1000,n,1)/1000;     % Initial guess, random values
nIter = 10;
for i=1:1:nIter
    R1=R*(corr(w, a./(conv(w,R,'same'))));
    R=R1;
end

Hope this helps you address the issue.

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yicong 2024-9-6

% Define initial parameters

R1 = 6.3;

R2 = 6.3;

R3 = 6.3;

tau1 = 1e-5;

tau2 = 1e-2;

tau3 = 1e1;

% Define the time range

t_min = 1e-6;

t_max = 1e2;

% Calculate the number of interpolation points

num_points_per_decade = 20;

num_decades = log10(t_max) - log10(t_min);

num_points = round(num_points_per_decade * num_decades);

% Generate logarithmically spaced time points

t_interp = logspace(log10(t_min), log10(t_max), num_points);

z_interp = log(t_interp);

% Calculate the analytical model a(t)

a_t = R1 * (1 - exp(-exp(log(t_interp))/tau1)) + ...

R2 * (1 - exp(-exp(log(t_interp))/tau2)) + ...

R3 * (1 - exp(-exp(log(t_interp))/tau3));

% Use gradient to compute da/dz

da_dz = gradient(a_t, z_interp);

% Define W(z) = exp(z - exp(z))

W_z = exp(z_interp - exp(z_interp));

% Uncomment to normalize each column of W_z

% for l = 1:num_points

% W_z(:, l) = W_z(:, l) / sum(W_z(:, l)); % Normalize each column

% end

max_iterations = 100; % Maximum number of iterations

tolerance = 1e-6; % Convergence threshold

% Initialize Psi(z) with da/dz

Psi_z = da_dz';

w=W_z';

a=da_dz';

% Iterative correction

for iter = 1:1:max_iterations

R1=Psi_z*(corr(w, a./(conv(w,Psi_z,'same'))));

Psi_z=R1;

end

% Plot the results

figure;

subplot(3,1,1);

plot(z_interp, da_dz);

title('da/dz vs z');

xlabel('z');

ylabel('da/dz');

subplot(3,1,2);

plot(z_interp, W_z);

title('W(z)');

xlabel('z');

ylabel('W(z)');

subplot(3,1,3);

plot(z_interp, Psi_z);

title('\Psi(z) after Van Cittert Deconvolution');

xlabel('z');

ylabel('\Psi(z)');

-----------

I apply your code here while get wrong time constant result.

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Deconvolution of a polynomial and exponential function

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