# How to solve this question by using convolution?

I had previously posted this question but my method was wrong in it. Two signals are given:

$$x[n]=2^nu[n]$$ $$h[n]=u[n]$$

I have to find $$y[n]$$. I have spent hours on this question but I am unable to solve it. Kindly please tell me the way to solve it.

• I cannot see any $y[n]$ in your equations. Looks like a homework, maybe this tag is needed here Commented Sep 29, 2018 at 13:30
• I assume from your title you need to solve for the convolution of x[n] and h[n]. It would help direct the best answer if you please write out the base equation for convolution in terms of x[n] and h[n] and then show the steps up to the point where you are confused or stuck. Commented Sep 29, 2018 at 13:52

Assuming you mean that $$y[n] = h[n] * x[n]$$, then $$$$\begin{split} y[n] &\stackrel{(a)}{=} h[n] * x[n]\\ &\stackrel{(b)}{=} \sum_{m=-\infty}^{\infty} h[m]x[n-m]\\ &\stackrel{(c)}{=} \sum_{m=-\infty}^{\infty} u[m]2^{n-m}u[n-m]\\ &\stackrel{(d)}{=}2^n \sum_{m=-\infty}^{\infty} u[m]2^{-m}u[n-m]\\ &\stackrel{(e)}{=}2^n \sum_{m=0}^{\infty} 2^{-m}u[n-m]\\ &\stackrel{(f)}{=}2^n \sum_{m=0}^{n} 2^{-m}\\ &\stackrel{(g)}{=}2^n \frac{1 - (\frac{1}{2})^{n+1}}{1 - \frac{1}{2}}\\ &\stackrel{(h)}{=}2^{n+1} - 1 \end{split}$$$$

(a) is the convolution symbol.

(b) is using the definition of discrete convolution.

(c) replacing the quantities.

(d) $$2^n$$ is independent of the summation index.

(e) $$u[m]=1$$ if $$m\geq 0$$ and $$0$$ otherwise.

(f) if $$n-m<0$$, or if $$n then the summation is zero. So we have (for non-zero quantities), $$m = 0 \ldots n$$.

(g) Using geometric series

(h) rearrangements.

• You're videos are REALLY NICE. I mostly use R and Rcpp but was tinking about delving into python and C++. Do you happen to have pdfs of your videos ? Commented Sep 30, 2018 at 3:29
• Hello @markleeds .. thank you for your interest. I'm glad you found them useful. Unfortunately, I do not have any specific "pdfs" other than the official C++ and Python-related modules websites, that I mention in the description of the videos. Commented Sep 30, 2018 at 11:37