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i would like some help with this:

given: $$ h[n]=2^{-n} u[n] $$ calculate the series: $$ \sum_{r=- \infty }^ \infty h[n+4r] $$

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    $\begingroup$ Welcome to DSP.SE! You need to put more information in: what have you tried? Why didn't that work for you? What don't you understand? $\endgroup$ – Peter K. Jun 30 '16 at 11:17
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let me start from the general form $$ \sum_{ k = M}^{ N } r^{k} = \frac{r^M - r^{N+1}}{ 1 - r}$$

then if you multiply $r^k$ by a unit step $u[n]$ the above series become

$$ \sum_{ k = M}^{ N } r^{k} \cdot u[k] = \sum_{ k = 0}^{ N } r^{k} $$

As you can notice the lower index become zero, because the unit step is equal to one only if $k\geq0$, otherwise is equal to zero.

Therefore in this specific case we have $u[n+4r]$ that is equal to one if the index $n+4r$ is greater or equal to zero, then: $$ r \geq - \frac{n}{4}$$

At this point I think that you have all the necessary information to resolve the series

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  • $\begingroup$ Where do you know from that $u[\cdot]$ is the unit-step function? Is this only an assumption based on some nomenclature? $\endgroup$ – M529 Jun 30 '16 at 11:51
  • $\begingroup$ @M529 actually it is the unit-step function. $\endgroup$ – Codevan Jun 30 '16 at 12:09
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    $\begingroup$ @M529 Because in every book I read the unit step is written as $u[n]$. In this link are list all the symbols of the step function and in this other link at page 1020 seems to be the first time that $u[n]$ was used. Anyway is a standard in digital signal processing community using $u[n]$ as unit step function $\endgroup$ – user66350 Jun 30 '16 at 15:14
  • $\begingroup$ @user66350 Thanks for the clarification. $\endgroup$ – M529 Jun 30 '16 at 17:21

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