(New page: ==Question== Is the signal <math>x(t) = \sum_{k = -\infty}^\infty \frac{1}{(t+2k)^{2}+1}</math> periodic? Prove your answer mathematically. ==Solution==) |
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| + | First, we time shift by 2. x(t+2) | ||
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| + | <math>x(t+2) = \sum_{k = -\infty}^\infty \frac{1}{(t+2+2k)^2+1}\,</math> | ||
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| + | Now, we can see that (t + 2 + 2k) can be factored into (t + 2(k+1)) | ||
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| + | <math>\sum_{k = -\infty}^\infty \frac{1}{(t+2(k+1))^2+1}\,</math> | ||
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| + | Finally, if we create a dummy variable r = k+1, we can see the signal is periodic. | ||
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| + | <math>x(t) = x(t+2) = \sum_{r = -\infty}^\infty \frac{1}{(t+2r)^2+1}\,</math> | ||
Latest revision as of 12:58, 15 October 2008
Question
Is the signal $ x(t) = \sum_{k = -\infty}^\infty \frac{1}{(t+2k)^{2}+1} $ periodic? Prove your answer mathematically.
Solution
First, we time shift by 2. x(t+2)
$ x(t+2) = \sum_{k = -\infty}^\infty \frac{1}{(t+2+2k)^2+1}\, $
Now, we can see that (t + 2 + 2k) can be factored into (t + 2(k+1))
$ \sum_{k = -\infty}^\infty \frac{1}{(t+2(k+1))^2+1}\, $
Finally, if we create a dummy variable r = k+1, we can see the signal is periodic.
$ x(t) = x(t+2) = \sum_{r = -\infty}^\infty \frac{1}{(t+2r)^2+1}\, $
