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Title: Infinite Sum (Z-transforms) Post by william wu on Dec 11th, 2002, 1:41am Studying for a signal processing exam and I'm stumped on a problem, maybe you guys would be interested. Evaluate the following (your expression will be in terms of n) sum ( (-2)n-m 4m ) from m = -inf to 3 Z-transforms could be helpful. Also notice that the summation looks very similar to a convolution. I considered interpreting the summation as going from m = -inf to +inf, and introducing a unit step into the argument of th e sigma. But I got lost shortly. |
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Title: Re: Infinite Sum (Z-transforms) Post by towr on Dec 11th, 2002, 2:20am on 12/11/02 at 01:41:44, william wu wrote:
4 = (-2)2 so 4m = (-2)2m and so (-2)n-m 4m = (-2)n+m the question is, is that usefull in anyway.. sum ( (-2)n+m) from m = -inf to 3 = sum ( (-2)n+3+m) from m = -inf to 0 = sum ( (-2)n+3-m) from m = 0 to inf = sum ( (-2)n+3 * (-2)-m) from m = 0 to inf sum(x[k]) from k =0 to n =z=> X(z)*z/(z-1) and lim n ->inf x[n] =z=> lim z->1 (z-1)/z * X(z) => lim n ->inf (sum(x[k]) from k =0 to n) =z=> lim z->1 (z-1)/z * X(z)*z/(z-1) = X(1) (?) and anu[n] =z=> z/(z-a) (here a = -1/2) => (-2)n+3 * 1/(1- - 1/2) = (-2)n+3 *1/(3/2) = (-2)n+3 * 2/3 It's been a few years, and I don't think we even had to ever do exactly this kind of thing.. So it could well be totally off, or maybe just a little.. |
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Title: Re: Infinite Sum (Z-transforms) Post by william wu on Dec 11th, 2002, 3:40am Your answer is correct; many thanks for being so prompt! It sure is cool to have fellow electrical engineers on board here. I didn't see the following step: sum ( (-2)n+m) from m = -inf to 3 = sum ( (-2)n+3+m) from m = -inf to 0 After that it was easy coasting. However, I don't understand what you were doing with Z-transforms and the final value theorem. Did you define to be x[k]? and did you change m to k? Basically I don't know where the following line came from: Anyways, I was able to solve it without using Z-transforms as follows: sum ( (-2)n+3 * (-2)-m) = (-2)n+3 * sum ( (-2)-m) = (-2)n+3 * (1 / (1 - (-1/2))) // using infinite geometric series sum formula = (-16/3) (-2)n |
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Title: Re: Infinite Sum (Z-transforms) Post by towr on Dec 11th, 2002, 9:00am on 12/11/02 at 03:40:49, william wu wrote:
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As I said, it's been a while.. I'm not an electrical engineer though. digital signal processing is just one of the courses we have to follow for cognitive science. I don't even really know how the two relate, perhaps via cognitive ergonomics (which might mean you may have to monitor heartrate, and thus analyze the heartrate signal, in which case some signal processing comes in handy) In any case, glad I could be of some help =) |
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