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Given a signal $r(t)$ which is a result of convolution between signal $x(t)$ and a channel $h(t)$ as below :

$r(t) = h(t)*x(t); $

what I know, the time reversal convolution can be process as follows : $y(t) = r(t)*h^*(-t) = h(t)*x(t)*h^H(-t);$ where * denote to the convolution and $h^H$ is the conjugate. I think that right and clear, but what does mean $h(-t)$. is it all simply equals to $conv(r(t), -h(t))$ for example as below : 

x = randn(1,5);
h = randn(1,3);
r = conv(h,x);
y = conv(r,-h);

Is y is correct in the above example? however I think it should be continuous where the above example is discrete signal.

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I believe that h(-t) means a "time-reversed" version of h(t). Your command: 'y = conv(r,-h);' computes the convolution of 'r' and negative 'h', and you don't want that. I think you want:

y = conv(r,conj(fliplr(h)));

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  • $\begingroup$ Thank you so much. Yes I need the time-reversed. But could you please explain what's the relationship between y and x in that case ? I was thinking it they will approximately similar!! $\endgroup$ – Gze Nov 9 '19 at 9:31
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    $\begingroup$ @Gze. Off the top of my head I don't see a clear, distinct, relationship between y(t) and x(t). What is the origin of your: y(t) = h(t)*x(t)h^H(-t) expression? From where did it come? $\endgroup$ – Richard Lyons Nov 10 '19 at 16:55
  • $\begingroup$ Lyonse, There should be a relationship because time reversal in multipath MIMO channel is used in that way, and detect x(t) based on y(t) using this method. $\endgroup$ – Gze Nov 11 '19 at 1:32
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    $\begingroup$ @Gze. I'm not familiar with " time reversal in multipath MIMO." I'm wondering if your original y(t) = h(t)*x(t)*h^H(-t) expression is correct. I'll have to study this whole concept further. I'm sorry I couldn't be of more help at this point in time. $\endgroup$ – Richard Lyons Nov 12 '19 at 0:40
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    $\begingroup$ @Gze. As of 11/12/2019 I have an answer to your original question. But my Answer requires me to post a .JPG image in my Answer. I spent 15 minutes trying to get this website's 'image drag-and-drop' feature to work, but without success. Send me a private e-mail at: <R.Lyons@ieee.org>. $\endgroup$ – Richard Lyons Nov 12 '19 at 11:04

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