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Prerequisite information:

Note that $u[n]$ is the Heaviside or unit step function

$y_1[n] = x_1[n] * x_2[n[ $

$x_1[n] = (0.5)^n u[n] $

$x_2[n] = u[n+3]$

$$\sum_{k=0}^{\infty} (0.5)^k \; u (n+3-k)$$

This evaluates to

$$ y_1(n) = x_1(n)*x_2(n) = \begin{cases} 2(1-(1/2)^{n+4}), & n \ge -3 \\ 0, & \text{otherwise} \end{cases} $$

I know that a geometric summation of $(0.5)^n$ goes to $2$, but I don't understand the other term or how it is split up.

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What is that $\,u[n+3-k]\,$ thing? – DonAntonio Feb 25 '13 at 0:21
I think its the notation that it's using for a function of a value. – AJMansfield Feb 25 '13 at 0:23
I edited your question to use $\LaTeX$, just check it to make sure I interpreted it correctly. – AJMansfield Feb 25 '13 at 0:25
@DonAntonio I think I fixed it. – AJMansfield Feb 25 '13 at 0:28
But what are these $u, y_1, x_1,$ and $x_2$? – AJMansfield Feb 25 '13 at 0:34
up vote 2 down vote accepted

OK, let's compute $y_1(2)$ and see if that helps you see what's going on.

$$y_1(2)=\sum_0^{\infty}(1/2)^ku(5-k)=u(5)+(1/2)u(4)+(1/4)u(3)+(1/8)u(2)+(1/16)u(1)+(1/32)u(0)+\cdots$$ But this is $1+(1/2)+(1/4)+(1/8)+(1/16)+(1/32)$ because Heaviside is $1$ for nonnegative arguments, and zero for negative arguments. Those numbers add up to $1{31\over32}=2-{1\over32}=2(1-{1\over64})$, as required.

Now, do you see how to reproduce these calculations for other values of $n$?

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I hope the OP can do the calculations as you suggested. +1 – Babak S. Feb 27 '13 at 16:30

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