(New page: ==Question== Is the signal ::<math> x(t) = \sum_{-\infty}^{\infty} \frac {1}{(t +2k)^2 +1} </math> periodic? Answer yes/no and justify your answer mathematically.) |
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Is the signal | Is the signal | ||
| − | ::<math> x(t) = \sum_{-\infty}^{\infty} \frac {1}{(t +2k)^2 +1} </math> | + | ::<math> x(t) = \sum_{k = -\infty}^{\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. | ||
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| + | ==Answer== | ||
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| + | Yes because: | ||
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| + | ::<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> | ||
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| + | Change of variable, let r = k+1 | ||
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| + | ::<math> => \sum_{k = -\infty}^{\infty}\frac {1}{(t+2r)^2+1} </math> | ||
| + | |||
| + | This equation is of the original form, therefore it is periodic. | ||
Latest revision as of 15:59, 15 October 2008
Question
Is the signal
- $ x(t) = \sum_{k = -\infty}^{\infty} \frac {1}{(t +2k)^2 +1} $
periodic? Answer yes/no and justify your answer mathematically.
Answer
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} $
Change of variable, let r = k+1
- $ => \sum_{k = -\infty}^{\infty}\frac {1}{(t+2r)^2+1} $
This equation is of the original form, therefore it is periodic.
