Deconvolution using FFT - a classical problem

Question

Vijayananda 2026-2-19，17:06

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2182722-deconvolution-using-fft-a-classical-problem

编辑： Matt J 34 minutes 前

Hello friends, I am new to signal processing and I am trying to achive deconvolution using FFT. I have an input step function u(t) applied to an impulse response given by

. The output function is

. I am trying to convolve g and u to get y as well as deconvolve y and g to get u. However, I quite cannot get the right answers. I understand that the deconvolution process is ill-posed and I have to use some kind of normalization process but I am lost. I also apply zero padding to twice the length of the input signals. Any sort of guidance will be appreciated.

After using deconvolution in the fourier domain:

Y = fft(y)

G = fft(g)

X = Y./G

x = ifft(X)

I am getting an output shown below:

Which is not the expected outcome. Can someone shead light on what is happening here? Thank you.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Matt J 2026-2-19，20:20

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2182722-deconvolution-using-fft-a-classical-problem#answer_1573874

编辑：Matt J 2026-2-19，20:49

在 MATLAB Online 中打开

dt=0.001;

N=20/dt;

t= ( (0:N-1)-ceil((N-1)/2) )*dt; %t-axis

u=(t>=0);

g=3*exp(-t).*u;

y=conv(g,u,'same')*dt;

Y = fft(y);

G = fft(g);

X = Y./G;

x = fftshift(ifft(X,'symmetric')/dt);

figure;

sub=1:0.3/dt:N;

plot(t,3*(1-exp(-t)).*u,'r.' , t(sub), y(sub),'-bo');

xlabel t

legend Theoretical Numeric Location northwest

title 'Output y(t)'

figure;

plot(t, u,'r.' , t(sub), x(sub),'-bo'); ylim([-1,4])

title Deconvolution

xlabel t

legend Theoretical Numeric Location northwest

8 个评论
显示 6更早的评论隐藏 6更早的评论

Paul about 8 hours 前

@Vijayananda

If you show you're code, it will be easier for people to help with whatever specific problem you're having.

@Matt J

"FFTs and IFFTs inherently integrate over both positive and negative times and frequencies."

I'm curious about this statement.

Did you mean "integrate" in the context of summation?

It seems to me that the DFT and IDFT, which are implemented in Matlab with fft and ifft respectively, are inherently defined only for positive times and positive frequencies. The DFT is (typically? usually? certainly in fft) defined as the summation over x[n], n = 0:N-1, where n is the discrete-time independent variable. The IDFT is defined as the summation over X[k], k = 0:N-1, where 2*pi*k/N are the discrete "frequencies" (which are actually angles). Postive times and positive "frequencies" respectively. Of course, if a finite-duration discrete-time signal is non-zero over an interval that includes n < 0, we can map it to another signal defined over n = 0:N-1, take the DFT of the mapped signal, and then manipulate that DFT (if necessary, depending on how we do the manipulation) to get a frequency domain representation of the original signal. But that manipulation is needed precisely because the DFT is only defined/implemented for signals that are non-zero for positive time. An analgous statement could be made regarding the IDFT.

"Usually FFTs are for non-causal, finite duration signal processing, whereas infinite, causal signals are usually processed with z-transforms."

See above regarding "FFTs are for non-causal" (agree on the finite-duration part)

Infinite-duration* signals can be processed (analyzed?) with the Discrete Time Fourier Transform or the z-transform, both causal and non-causal.

* More precisely, certain subsets of infinite-duration signals.

Vijayananda about 5 hours 前

编辑：Matt J about 3 hours 前

在 MATLAB Online 中打开

Spectral_deconvolution_Semi_infinite_medium.txt

Hello friends @Paul, I am trying to solve an inverse heat transfer problem. In a semi-infinite medium which is supplied with a constant heat flux

, the temperature distribution is given by:

This is an LTI system. The impulse response of the system is given by:

and the constant heat flux in the step form is q" given the signal starts from t = 0.

So we can write:

In fourier domain:

Z = G*Q

I have analytical forms for all three functions as shown below.

and

I am getting temperature by convolving g and q. No issues. However, when I try to get either g, or q back from the other signals, I am running into error. Division in the fourier domain is ill posed and there are zeros in the denominator. I am not sure how to rectify these. If you add noise to the signals, solving becomes more difficult. The code is attached to this comment.

clc

clearvars

dt = 1e-6;%sampling period

df = 1/dt;%sampling frequency

t = 0:dt:1-dt;%Time vector

qmax = 1e5;%W/m2 maximum step heat flux

alpha = 3.56E-6;%m2/s diffusivity

k = 15; %W/m-k thermal conductivity

x = 1e-5;%m junction depth

DeltaT = (((2*qmax)/k).*sqrt((alpha*t)/pi).*exp((-x*x)./(4*alpha*t)))-(((qmax*x)/k).*(1-erf(x./(2*sqrt(alpha*t)))));

g = sqrt(alpha./(pi*k*k.*t)).*exp((-x*x)./(4*alpha.*t));

g(1) = 1e-9;%to replace NaN at t = 0

q = qmax*ones(length(t),1)';

n = length(DeltaT);

m = length(g);

%Analytical function plots

% figure

% subplot(3,1,1)

% plot(t,DeltaT,LineWidth=2)

% title("Temperature")

% subplot(3,1,2)

% plot(t,g,LineWidth=2)

% title("Impulse response")

% subplot(3,1,3)

% plot(t,q,LineWidth=2)

% title("Heat flux")

%Zero padding to get linear convolution from circular convolution

qpad = paddata(q,n+m-1);%padded heat flux

gpad = paddata(g,n+m-1);%padded impulse response

temppad = paddata(DeltaT,n+m-1);%padded temperature

% Perform fourier transform

G = fft(gpad);

T = fft(temppad);

Q = fft(qpad);

% Convolve heat flux and impulse response

Z1 = dt.* G.*Q;

% Convolve temperature and impulse response

Z2 = df.* T./G;

% Convolve temperature and heat flux

Z3 = df.* T./Q;

% Back to time domain

z = (ifft(Z3)); %change Z3 to Z2 and Z1 for other outputs

znew = z(1:length(t));

plot(t,znew)

Matt J about 1 hour 前

编辑：Matt J 34 minutes 前

在 MATLAB Online 中打开

You seem to have assumed that T and Z1 are the same and can therefore be used interchangeably in,

Z2 = df.* T./G;
Z3 = df.* T./Q;

You can easily check, though, that they are not the same. It should really be,

Z2 = df.* Z1./G;
Z3 = df.* Z1./Q;

The reason T and Z1 are not the same is that in the actual continuous-time convolution of g(t) with a step, you get a temperature profile that increases on the interval

, but then gradually decays to zero on

. Conversely, in your construction of temppad, and hence T, there is no gradual decay. You simply truncate abruptly to zero once

is reached.

请先登录，再进行评论。

Answer 2

Matt J about 6 hours 前

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2182722-deconvolution-using-fft-a-classical-problem#answer_1573903

编辑：Matt J about 6 hours 前

Since you are trying to deconvolve in the presence of noise, it would make sense to use a regularized deconvolver like deconvreg or deconvwnr. These are from the Image Processing Toolbox, but there is no reason they wouldn't apply to one-dimensional signals as well.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Deconvolution using FFT - a classical problem

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

回答（2 个）

8 个评论
显示 6更早的评论隐藏 6更早的评论

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

另请参阅

标签

Community Treasure Hunt

Deconvolution using FFT - a classical problem

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

回答（2 个）

8 个评论 显示 6更早的评论隐藏 6更早的评论

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

另请参阅

标签

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

8 个评论
显示 6更早的评论隐藏 6更早的评论

0 个评论
显示 -2更早的评论隐藏 -2更早的评论