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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.

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    $\begingroup$ I cannot see any $y[n]$ in your equations. Looks like a homework, maybe this tag is needed here $\endgroup$ – Laurent Duval Sep 29 '18 at 13:30
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    $\begingroup$ 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. $\endgroup$ – Dan Boschen Sep 29 '18 at 13:52
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Assuming you mean that $y[n] = h[n] * x[n]$, then \begin{equation} \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} \end{equation}

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

(g) Using geometric series

(h) rearrangements.

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  • $\begingroup$ 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 ? $\endgroup$ – mark leeds Sep 30 '18 at 3:29
  • $\begingroup$ 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. $\endgroup$ – Ahmad Bazzi Sep 30 '18 at 11:37
  • $\begingroup$ okay. no problem. thanks for great videos. $\endgroup$ – mark leeds Sep 30 '18 at 16:38

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