0
$\begingroup$

I have been trying to attempt this question for quite a while and I am not sure how to approach it. Would anyone be able to help out with the working?

Discrete Time convolution

$\endgroup$
4
$\begingroup$

Hint

The simplest way is to use the Z transform property "convolution in time domain is multiplication in z domain". See Z transform convolution

$$\mathrm{Z}(x[n]*h[n]) = \mathrm{Z}(x[n]) \times \mathrm{Z}(h[n])$$

Then you just need to do inverse Z transform.

Some typical Z transforms (including what is useful for you) can be found at Z transform pairs

Be careful with region of convergence (ROC).

It seems a homework?

$\endgroup$
1
$\begingroup$

First replace $x[k]$ and $h[n-k]$ with their respective expressions in $y[n]$, then take out of the sum anything that does not depend on $k$, then try simplifying what you have inside your sum. You should get something of the form:

$$y[n]=\alpha\sum_{k=-\infty}^{+\infty}\beta[k]u[k]u[n-k-3].$$

The limits of the sum (lets call them $l_1$ and $l_2$) are determined by the product $u[k]u[n-k-3]$ (where the product is nonzero). You should then obtain something of the form:

$$y[n]=\alpha\sum_{k=l_1}^{l_2}\beta[k].$$

Hint: $\beta[k]$ should be of the form $\gamma^k$, where $\gamma <1$ and the sum is of a geometric series.

$\endgroup$

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.