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<math> x(t+4) = \sum_{k = -\infty}^\infty \frac{1}{(t+2(\frac{1}{2}+k))^{2}+1} </math> | <math> x(t+4) = \sum_{k = -\infty}^\infty \frac{1}{(t+2(\frac{1}{2}+k))^{2}+1} </math> | ||
| − | Then we set r = \frac{1}{2}+k | + | Then we set <math> r = \frac{1}{2}+k </math> to yield, |
| − | <math> = \sum_{k = -\infty}^\infty \frac{1}{(t+ | + | <math> = \sum_{k = -\infty}^\infty \frac{1}{(t+2r)^{2}+1} </math> |
| − | Since this is equivalent to x(t) | + | Since this signal is equivalent to x(t), then x(t) is periodic. |
Latest revision as of 18:41, 15 October 2008
EXAM 1
Problem 1.
is
$ x(t) = \sum_{k = -\infty}^\infty \frac{1}{(t+2k)^{2}+1} $
periodic?
We know that for a signal to be periodic
$ x(t) = x(t + T) $
So we shift the function by a arbitrary number to try to prove the statement above
$ x(t+1) = \sum_{k = -\infty}^\infty \frac{1}{(t+1+2k)^{2}+1} $
$ x(t+4) = \sum_{k = -\infty}^\infty \frac{1}{(t+2(\frac{1}{2}+k))^{2}+1} $
Then we set $ r = \frac{1}{2}+k $ to yield,
$ = \sum_{k = -\infty}^\infty \frac{1}{(t+2r)^{2}+1} $
Since this signal is equivalent to x(t), then x(t) is periodic.
