What is the time reversal convolution

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.

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)));

• 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!!
– Gze
Nov 9 '19 at 9:31
• @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? Nov 10 '19 at 16:55
• 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.
– Gze
Nov 11 '19 at 1:32
• @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. Nov 12 '19 at 0:40
• @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>. Nov 12 '19 at 11:04