1
$\begingroup$

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.

Improve this question
$\endgroup$
3
$\begingroup$

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)));

Improve this answer
$\endgroup$
6
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.