(New page: Is the signal <math>\ x(t) = \sum_{k = - \infty}^{\infty} \frac{1}{(t + 2k)^{2} + 1}</math> periodic? Answer yes/no and justify your answer mathematically. Yes, because <math>x(t + 2)...) |
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Yes, because <math>x(t + 2) = \sum_{k = - \infty}^{\infty} \frac{1}{(t + 2 + 2k)^{2} + 1} = \sum_{k = - \infty}^{\infty} \frac{1}{(t + 2 (k + 1))^{2} + 1}</math> | Yes, because <math>x(t + 2) = \sum_{k = - \infty}^{\infty} \frac{1}{(t + 2 + 2k)^{2} + 1} = \sum_{k = - \infty}^{\infty} \frac{1}{(t + 2 (k + 1))^{2} + 1}</math> | ||
| − | let <math>\ r = k + 1, \sum_{ | + | let <math>\ r = k + 1, \sum_{r = - \infty}^{\infty} \frac{1}{(t + 2r)^{2} + 1} = x(t)</math> |
Latest revision as of 13:52, 15 October 2008
Is the signal
$ \ x(t) = \sum_{k = - \infty}^{\infty} \frac{1}{(t + 2k)^{2} + 1} $
periodic? Answer yes/no and justify your answer mathematically.
Yes, because $ x(t + 2) = \sum_{k = - \infty}^{\infty} \frac{1}{(t + 2 + 2k)^{2} + 1} = \sum_{k = - \infty}^{\infty} \frac{1}{(t + 2 (k + 1))^{2} + 1} $
let $ \ r = k + 1, \sum_{r = - \infty}^{\infty} \frac{1}{(t + 2r)^{2} + 1} = x(t) $
