(New page: Problem 1 is the problem I received the least amount of points on, therefore I will solve it. Is the signal <math>x(t) = \sum^{- \infty}_{\infty} \frac{1}{(t+2k)^2 + 1)}</math> periodic...) |
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Is the signal | Is the signal | ||
| − | <math>x(t) = \sum^{ | + | <math>x(t) = \sum^{\infty}_{k = - \infty} \frac{1}{(t+2k)^2 + 1)}</math> |
periodic? Answer yes/no and justify your answer mathematically. | periodic? Answer yes/no and justify your answer mathematically. | ||
| Line 9: | Line 9: | ||
Yes, because: | Yes, because: | ||
| − | <math>x(t+2) = \sum^{ | + | <math>x(t+2) = \sum^{\infty}_{k = - \infty} \frac{1}{(t+2+2k)^2 + 1}</math> |
| − | <math>= \sum^{ | + | <math>= \sum^{\infty}_{k = - \infty} \frac{1}{(t+2(k+1))^2 + 1}</math> |
let r = k + 1 | let r = k + 1 | ||
| − | <math>= \sum^{ | + | <math>= \sum^{\infty}_{r = - \infty} \frac{1}{(t+2r)^2 + 1} = x(t)</math> |
Latest revision as of 14:19, 15 October 2008
Problem 1 is the problem I received the least amount of points on, therefore I will solve it.
Is the signal
$ x(t) = \sum^{\infty}_{k = - \infty} \frac{1}{(t+2k)^2 + 1)} $
periodic? Answer yes/no and justify your answer mathematically.
Yes, because:
$ x(t+2) = \sum^{\infty}_{k = - \infty} \frac{1}{(t+2+2k)^2 + 1} $
$ = \sum^{\infty}_{k = - \infty} \frac{1}{(t+2(k+1))^2 + 1} $
let r = k + 1
$ = \sum^{\infty}_{r = - \infty} \frac{1}{(t+2r)^2 + 1} = x(t) $
