(New page: 1. Is the signal :<math>x(t) = \sum_{k = -\infty}^\infty \frac{1}{(t+2k)^{2}+1}\,</math> periodic? The answer is yes because :<math>x(t+2) = \sum_{k = -\infty}^\infty \frac{1}{(t+2+2k...) |
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let <math>r</math> = <math>k+1</math> | let <math>r</math> = <math>k+1</math> | ||
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| + | :<math>x(t+2) = \sum_{r = -\infty}^\infty \frac{1}{(t+2r)^2+1} = x(t)\,</math> | ||
Latest revision as of 21:51, 14 October 2008
1. Is the signal
- $ x(t) = \sum_{k = -\infty}^\infty \frac{1}{(t+2k)^{2}+1}\, $
periodic? The answer is yes because
- $ x(t+2) = \sum_{k = -\infty}^\infty \frac{1}{(t+2+2k)^2+1}\, $
- $ x(t+2) = \sum_{k = -\infty}^\infty \frac{1}{(t+2(k+1))^2+1}\, $
let $ r $ = $ k+1 $
- $ x(t+2) = \sum_{r = -\infty}^\infty \frac{1}{(t+2r)^2+1} = x(t)\, $
