why my deconvolution want fit my analytique form?

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Rabih Sokhen
Rabih Sokhen 2022-2-4
评论： Rabih Sokhen 2022-2-7
hy guys,
i am trying to deconvolute a signal to refind my 2 initial signals, however when a = sinc(f) my deconvolution want fit the analytic but if i replace my sinc with sin it fit very well, any idea about this or did i miss something in the code?
thank you in advance
code:
clear all
clc
dt=0.1;
t=-10:dt:10;
z=numel(t);
df = 1/dt;
f = linspace(-df/2, df/2, z);
ff = linspace(-df/2, df/2, z/2);
to=1;
a=to*sinc(to*f);
b=1./(1+1i*2*pi*f);
y_f=conv(a,b);
[bb,r]=deconv(real(y_f),a); %bb should be similar to b
[aa,r]=deconv((y_f),real(b)); %aa should be similar to a
subplot(131)
plot(real(y_f),'r')
title('y_f= \tau*sinc(\tau*f) * 1./(1+1i*2*pi*f)')
subplot(132)
plot(f,real(aa),'r')
hold on
plot(f,a,'*k')
subplot(133)
plot(f,real(bb),'r')
hold on
plot(f,real(b),'*k')
ylim([0 2])

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采纳的回答

Paul
Paul 2022-2-4
Ran in:
For a real and b complex, the theoretical equations should be (illusrated with short sequences)
a = rand(1,5);
b = rand(1,5) + 1j*rand(1,5);
y_f = conv(a,b);
aa = deconv(real(y_f),real(b)); % or aa = deconv(imag(yf),imag(b))
bbreal = deconv(real(y_f),a);
bbimag = deconv(imag(y_f),a);
% check
a - aa
ans = 1×5
1.0e+-15 * 0 -0.1110 0.2776 0 -0.3331
real(b) - bbreal
ans = 1×5
1.0e+-15 * 0 -0.0555 0.0555 -0.2220 0
imag(b) - bbimag
ans = 1×5
1.0e+-15 * 0 0 0 -0.1110 0.2220
As we can see,, even for these "simple" sequences deconv is not exact (which I gues is not surprising). Trying these equaations on the a and b in the Question resulted in a good reconstruction of a, but not of b, which blew up very quickly. Maye deconv is very numerically sensitive operation, at least for the given inputs.

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