0
$\begingroup$

I would really appreciate any help in understanding this.

If $$y[n] = x[n]*r[n]\tag{1}$$ then what is $y[n-n_0]$ for any system?

Is it $$y[n-n_0] = x[n-n_0]*r[n-n_0]\tag{2}$$ or $$y[n-n_0] = x[n-n_0]*r[n]\tag{3}$$ or $$y[n-n_0] = x[n]*r[n-n_0]\tag{4}$$

Will that result (answer to the above) change when we now have an LTI system with $h[n]=r[n]$ being its impulse response?

Thank you very much

$\endgroup$
3
$\begingroup$

Since $y[n-n_0] = y[n] \ast \delta [n - n_0]$ and convolution is both associative and distributive, is it not the case that

$$ y[n-n_0] = \left(x[n] \ast r[n] \right) \ast \delta[n-n_0] = x[n] \ast r[n-n_0] = x[n-n_0] \ast r[n]? $$

|improve this answer|||||
$\endgroup$
  • $\begingroup$ +1 nice answer .... $\endgroup$ – Isma007 Feb 3 at 17:47
2
$\begingroup$

Why not taking the definition of convolution and see what happens?

$$y[n-n_0] = \sum_k x[k]r[n-n_0-k]$$ which gives $$y[n-n_0] = x[n]*r[n-n_0]$$

Also, since $x[n]*r[n] = r[n]*x[n]$, $$y[n-n_0] = \sum_k r[k]x[n-n_0-k]$$ which gives $$y[n-n_0] = r[n] * x[n-n_0]$$

So both (3) and (4) are correct.

|improve this answer|||||
$\endgroup$
0
$\begingroup$

I will go with $(4)$, since it is possible to build a system with $h[n-n_0]$

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Thank you very much @jomegaA, a follow-up question will be why is (3) not true as well since the convolution operation is commutative. $\endgroup$ – site Feb 2 at 23:52
  • $\begingroup$ Practically I may have no influence on the input but I can influence the design of my system. $\endgroup$ – jomegaA Feb 3 at 4:37

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.