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I am working on a discrete convolution problem. I am comparing the solution to what the solution manual has and it just doesn't make sense to me.

what solution manual has: https://i.stack.imgur.com/ywkpg.jpg

The parts in red box doesn't make sense to me. No matter what I do I can't get the same answer either forward or backward.

I get bringing $alpha^{n}$ out of the summation. But the $alpha^{-k}$ has me stumped. because if I apply the equation given in the book http://imgur.com/Whn1E0L I don't get the same solution.

edit: I can't seem to get the more complicated equation to display, nor does embedded images show up properly for some reason on my browser. So I have just included the html links. Sorry. I didn't have this problem last time on this computer.

D.Zou
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1 Answers1

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Here are the calculations somewhat more detailed:

Note: We can use the formula of the finite geometric series starting with $N_1=0$, which is according to your link $$\sum_{k=0}^{N_2}\alpha^k=\frac{1-\alpha^{N_2+1}}{1-\alpha}$$

Let's start

\begin{align*} \sum_{k=0}^na^{n-k}&=a^n\sum_{k=0}^na^{-k}=a^n\sum_{k=0}^n\left(a^{-1}\right)^k\\ &=a^n\frac{1-\left(a^{-1}\right)^{n+1}}{1-a^{-1}}\\ &=a^n\frac{1-a^{-(n+1)}}{1-a^{-1}}\\ &=a^n\frac{1-\frac{1}{a^{n+1}}}{1-\frac{1}{a}}\\ &=a^n\frac{\frac{a^{n+1}-1}{a^{n+1}}}{\frac{a-1}{a}}\\ &=a^n\frac{a^{n+1}-1}{a-1}\frac{a}{a^{n+1}}\qquad\qquad\text{since}\quad\frac{\frac{p}{q}}{\frac{r}{s}}=\frac{p}{r}\frac{s}{q}\\ &=\frac{a^{n+1}-1}{a-1}\\ &=\frac{1-a^{n+1}}{1-a} \end{align*}

And the second expression:

\begin{align*} \sum_{k=0}^{N-1}a^{n-k}&=a^n\sum_{k=0}^{N-1}a^{-k}=a^n\sum_{k=0}^{N-1}\left(a^{-1}\right)^k\\ &=a^n\frac{1-a^{-N}}{1-\frac{1}{a}}\\ &=a^n\frac{1-a^{-N}}{1-\frac{1}{a}}\cdot\frac{a}{a}\\ &=a^{n+1}\frac{1-a^{-N}}{a-1}\\ \end{align*}

Hint: You could go twice through this answer. The first time consider each of the steps carefully and you may observe that some of them are quite simple. The second time consider which steps are completely evident to you, so that they could be skipped the next time in order to reduce the lengthy calculation.

Markus Scheuer
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  • thanks a lot. now it looks so damn obvious... :) – D.Zou Oct 05 '14 at 20:12
  • Great! You're Welcome! :-) – Markus Scheuer Oct 05 '14 at 20:14
  • hi, I am just looking over it again and I noticed something weird: why is the summation of the 2nd expression from 0 to N-1? if the 2nd expersion is for n> N why is the summation not from N to $inf$? – D.Zou Oct 06 '14 at 01:39
  • @D.Zou: Hi! Your question seems plausible. But it depends on the precise wording of this exercise. Your snapshot doesn't include it. Maybe there are some symmetry conditions which make the Ansatz $\sum_{k=0}^{N-1}$ reasonable. – Markus Scheuer Oct 06 '14 at 07:00