I am reading a paper about the Hunt problem:

Consider the iput given by $$e(t) = \mathrm{e}^{-\left(\frac{t-400}{75}\right)^2} - \mathrm{e}^{-\left(\frac{t-600}{75}\right)^2},\quad 0\leq t \leq 1025$$ and the impuse response of the system is given by $$h(t) = \begin{cases}1,\,t\leq 250\\0,\,t > 250 \end{cases}$$ then the output response $r(t)$ is: $$r(t) = e(t) \otimes h(t)$$

and I can code the process out:

t = 0:.01:1025;
e = exp(-((t-400)/75).^2) + exp(-((t-600)/75).^2);
h = (t>0) - (t>250);
f = conv(e,h); f = f*.01;
k = 2*length(t) - 1;
k2 = linspace(2*t(1),2*t(end),k); 

subplot 221
xlim([0 1000]);
title('input e(t)')
grid on
subplot 222
ylim([0 1.2]);
grid on
xlim([0 600])
title('impulse response h(t)')
subplot (2,2,3:4)
plot(k2,f,'LineWidth',1.5); grid on
title('result r(t) = e(t) * h(t)')

The result is enter image description here

which is the same with the paper: enter image description here

And now I am facing the deconvolution problem, how could I rebuilt the INPUT from the OUTPUT discrete data $r[n]$ and the $h(t)$.

More specifically, how can I get the figure (c) from (d)? It seems that I need to find the Discrete-Time Impulse Response $h[n]$ from the Impulse response $h(t)$ and then build a Toeplitz matrix.

What is the matrix, thanks!

  • $\begingroup$ I just e-mailed a fellow who knows how to do this. He posts here. And there are lots of others here who might answer. $\endgroup$
    – Ed V
    Oct 14 at 2:27
  • $\begingroup$ Thanks for your help @Ed V ! $\endgroup$
    – ZYX
    Oct 14 at 2:40
  • $\begingroup$ If you recall the length of convolution is N1+N2-1. If you keep track the length, you can divide the FT of convolution product by the FT of the impulse function. Then take the inverse transform. You should recover the exact original function. You may need to denoise by a window. $\endgroup$
    – M. Farooq
    Oct 14 at 2:41

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