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| − | <math></math> | + | <math>x(t)=\sum_{k=-\infty}^\infty k = \frac{1}{(t+2k)^2+1}</math> periodic? |
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| + | Yes it is periodic | ||
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| + | <math>x(t+2)=\sum_{k=-\infty}^\infty k \frac{1}{(t+2+2k)^2+1}</math> | ||
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| + | <math>=\sum_{k=-\infty}^\infty k \frac{1}{(t+2+2k)^2+1}</math> | ||
| + | |||
| + | <math>=\sum_{k=-\infty}^\infty k \frac{1}{(t+2(k+1))^2+1}</math> | ||
| + | (Assume R = k+1) | ||
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| + | <math>=\sum_{k=-\infty}^\infty k = \frac{1}{(t+2R)^2+1}</math> | ||
| + | |||
| + | <math>=\text{x(t+2)}</math> | ||
Latest revision as of 06:29, 17 October 2008
$ x(t)=\sum_{k=-\infty}^\infty k = \frac{1}{(t+2k)^2+1} $ periodic?
Yes it is periodic
$ x(t+2)=\sum_{k=-\infty}^\infty k \frac{1}{(t+2+2k)^2+1} $
$ =\sum_{k=-\infty}^\infty k \frac{1}{(t+2+2k)^2+1} $
$ =\sum_{k=-\infty}^\infty k \frac{1}{(t+2(k+1))^2+1} $ (Assume R = k+1)
$ =\sum_{k=-\infty}^\infty k = \frac{1}{(t+2R)^2+1} $
$ =\text{x(t+2)} $
