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[[Category: 2007]]
 
[[Category: 2007]]
 
[[Category: mboutin]]
 
[[Category: mboutin]]
[[Category: Defintions]]
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[[Category:signals and systems]]
[[Category: Geometric Series]]
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The Geometric Series formulas below still hold for <math> \alpha\ </math>'s containing complex exponentials.
 
The Geometric Series formulas below still hold for <math> \alpha\ </math>'s containing complex exponentials.
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In this case, <math> \alpha=\frac{1}{2}e^{-j\omega} </math> in the above Geometric Series formula.
 
In this case, <math> \alpha=\frac{1}{2}e^{-j\omega} </math> in the above Geometric Series formula.
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[[ECE301|Back to ECE301]]
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[[ECE438|Back to ECE438]]
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[[More_on_geometric_series|More on geometric series]]
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[[Category:geometric series]]

Latest revision as of 17:40, 21 April 2013


The Geometric Series formulas below still hold for $ \alpha\ $'s containing complex exponentials.

For k from 0 to n, where $ \alpha \ne 1 $:

$ \sum_{k=0}^{n} \alpha^k = \frac{1-\alpha^{n+1}}{1-\alpha} $
(else, = n + 1)

For k from 0 to $ \infty\ $, where $ \alpha < 1\ $:

$ \sum_{k=0}^\infty \alpha^k = \frac{1}{1-\alpha} $
(else it diverges)

Example: We want to evaluate the following:

$ \sum_{k=0}^\infty (\frac{1}{2})^k e^{-j \omega k}= \sum_{k=0}^\infty (\frac{1}{2}e^{-j\omega})^k = \frac{1}{1-\frac{1}{2}e^{-j\omega}} $

In this case, $ \alpha=\frac{1}{2}e^{-j\omega} $ in the above Geometric Series formula.


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Back to ECE438

More on geometric series

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