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I have an input signal $$x(n)=\left(3,-5,4,3,-1,-2,6,8\right), n=-3,..,4$$ and impulse response $$h(n)=(1,-1,1,-1,1), n=-1,...,3.$$

The convolution between $x(n)$ and $h(n)$ is

$$x(n)*h(n)=\sum_{-\infty}^\infty x(k)h(n-k)$$ If I'm not mistaken, I can reduce this to the finite sum $$=x(-3)h(n+3)+x(-2)h(n+2)+x(-1)h(n+1)+x(0)h(n)+x(1)h(n-1)+x(2)h(n-2)+x(3)h(n-3)+x(4)h(n-4)$$

So, if I were to calculate the sum by hand, I would proceed by first evaluating

$x(-3)h(n+3)=3h(n+3) = (3,-3,3,-3,3)$ for $n=-4,...,0$ $x(-2)h(n+2)=-5h(n+3) = (-5,+5,-5,+5,-5)$ for $n=-3,...,1$
etc...

Then I would only add up the terms whose position $n$ aligns (that is, add all terms at n=-4 together, all terms at n=-3 together, etc...).

Is my approach to evaluating the convolution correct?

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I would have liked to give a longer answer, but yes, your approach is correct. I don't see any problems with what you laid out in your question.

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  • $\begingroup$ No worries... That's all I was looking for :) $\endgroup$
    – Paul
    Commented Oct 3, 2012 at 2:18
  • $\begingroup$ Is there any other approach to compute the same quantities manually? $\endgroup$
    – Paul
    Commented Oct 3, 2012 at 2:23
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    $\begingroup$ @Paul You may learn at a later date about transform methods and the like which provide a different way of computing, but there is little else to be done except what you have for a manual approach. Also, see the second half of this answer to a different question on this site, which also lays out a similar calculation using a tabular approach. $\endgroup$
    – Dilip Sarwate
    Commented Oct 3, 2012 at 2:44

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