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i need to evaluate the convolution sum of x[n] * h[n].
x[n] is the step function u[n]. I know how the output should look like but i don't know how i can calculate it.
I think the lower border is 0, cause the step function is 1 for n >= 0.
But the upper border is infinite, it's only approaching to 0.

\begin{aligned} g[n] & = \sum_{m=0}^{\infty} x[m] h[n-m]\\ x[n] & = u[n] \\ h[n] & = (\frac{1}{2})^n u[n] \\ \end{aligned}

Hope someone can help me...

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1 Answer 1

up vote 3 down vote accepted

\begin{equation} g[n] = \sum_{m=0}^{\infty} x[m] h[n-m]\ \end{equation}

i.e., we have \begin{equation} g[n] = \sum_{m=0}^\infty u[m] \left(\frac{1}{2}\right)^{n-m} u[n-m]\ \end{equation}

Since $u[m]>0$ when $m \geq 0$ and $u[n-m]>0$ when $n-m \geq 0$, we have

\begin{equation} g[n]= \sum_{m=0}^{n} \left(\frac{1}{2}\right)^{n-m}\ \end{equation}

This is just a simple geometric series. The answer is $2(1-0.5^{n+1})$.

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Thank you! So you can ignore u[n-m] because it is always 1 in this case? –  madmax Apr 15 '11 at 19:43
    
@madmax I'm not ignoring $u[n-m]$. The summation is reduced from a sum with infinitely many terms to one with only a finite number of terms because I look only at those terms where both $u[m]$ and $u[n-m]$ are non zero. –  svenkatr Apr 15 '11 at 19:52
    
If you found the answer useful, I'd appreciate an upvote :) –  svenkatr Apr 15 '11 at 19:53
    
I would upvote, but i need 15 points to do that ;-) But when i set n=0, i get g[0]=2, shouldn't it be 1? h[0] = 1, x[0] = 1, 1*1=1 ;) –  madmax Apr 15 '11 at 20:09
    
@madmax When $n=0$, $g[0] = 2(1-0.5) = 1$ –  svenkatr Apr 15 '11 at 20:16

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