Infinite Sum

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  • #1
cepheid
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I'm trying to figure out how to do this sum and not sure where to start:

[tex] \sum_{n = -\infty}^{-1} \left(\frac{1}{2}e^{-j \omega} \right)^n [/tex]

I tried rewriting it like this:

[tex] \sum_{m = 1}^{\infty} (2e^{j \omega})^m [/tex]

m = -n

but I'm not sure if that helps.
 

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  • #2
0rthodontist
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What are j and omega? Unless I'm missing something you have an infinite power series with ratio r = 2e^(j * omega) whose sum can be found by the formula r/(1-r).
 
  • #3
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0rthodontist said:
What are j and omega? Unless I'm missing something you have an infinite power series with ratio r = 2e^(j * omega) whose sum can be found by the formula r/(1-r).

That will work provided that (j*omega) < ln(1/2) because in order for an infinite geometric series to converge the common ratio must be less than 1.
 
  • #4
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d_leet said:
That will work provided that (j*omega) < ln(1/2) because in order for an infinite geometric series to converge the common ratio must be less than 1.
True.
whitespace
 
  • #5
cepheid
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0rthodontist said:
What are j and omega?

Tsk tsk. You guys don't instantly recognize the combination jω?

[tex] j = \sqrt{-1} [/tex]

I know I know. Those crazy electrical engineers...and omega is of course the angular frequency. What you are looking at is the calculation of the Fourier transform of a discrete time signal x[n], n in the set of integers.

Anyway, you must be wondering, if that's a complex exponential, then how could the series possibly converge? Well it's a moot point, because I made a mistake. Originally I had:

[tex] X(e^{j \omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j \omega n} = \sum_{n=-\infty}^{-1}\left(\frac{1}{2}\right)^{-n}e^{-j \omega n} [/tex]


Which I should have expressed like so:

[tex] \sum_{n=-\infty}^{-1}\left[\left(\frac{1}{2}\right)^{-1}e^{-j \omega }\right]^n [/tex]

which is decidedly different from what I started out with in my first post. It becomes (setting m = -n):

[tex] \sum_{m=1}^{\infty}\left(\frac{1}{2}e^{j \omega }\right)^m = \frac{\frac{1}{2}e^{j \omega }}{1 - \frac{1}{2}e^{j \omega }}[/tex]

Correct?
 

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