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Topic: Infinite Sum (Z-transforms) (Read 5622 times) |
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william wu
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Infinite Sum (Z-transforms)
« on: Dec 11th, 2002, 1:41am » |
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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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| « Last Edit: Dec 11th, 2002, 2:06am by william wu » |
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towr
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Re: Infinite Sum (Z-transforms)
« Reply #1 on: Dec 11th, 2002, 2:20am » |
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on Dec 11th, 2002, 1:41am, william wu wrote:
Evaluate the following:
sum ( (-2)n-m 4m ) from m = -inf to 3
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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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william wu
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Re: Infinite Sum (Z-transforms)
« Reply #2 on: Dec 11th, 2002, 3:40am » |
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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
(-2)n+3 * (-2)-m
to be x[k]? and did you change m to k? Basically I don't know where the following line came from:
sum(x[k]) from k =0 to n =z=> X(z)*z/(z-1)
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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| « Last Edit: Dec 11th, 2002, 3:41am by william wu » |
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towr
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Re: Infinite Sum (Z-transforms)
« Reply #3 on: Dec 11th, 2002, 9:00am » |
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on Dec 11th, 2002, 3:40am, william wu wrote:
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
(-2)n+3 * (-2)-m
to be x[k]? and did you change m to k?
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yes and yes... My textbook uses x[k], so I copied the formulas and transformations, then inserted the above formula..
Quote: Basically I don't know where the following line came from:
sum(x[k]) from k =0 to n =z=> X(z)*z/(z-1)
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From the textbook. I couldn't find anything to transform the sum from 0 to inf, but I could find it for 0 to n, and one for a limit n to inf, so I combined them.
Quote: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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Probably a much better method, since I'm sure I should have tranformed back from the z-domain, which I didn't.. (which in this case might have worked out since it was a constant, which is an impuls in time, which has the same value)
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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