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The problem that I found to be the most confusing was problem 2: | The problem that I found to be the most confusing was problem 2: | ||
| − | Is the signal | + | Is the following signal periodic? |
<math> x(t) = \sum_{k = -\infty}^\infty \frac{1}{(t+2k)^{2}+1}\,</math> | <math> x(t) = \sum_{k = -\infty}^\infty \frac{1}{(t+2k)^{2}+1}\,</math> | ||
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| + | <math> x(t+4) = \sum_{k = -\infty}^\infty \frac{1}{(t+4+2k)^{2}+1}\,</math> | ||
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| + | <math> x(t+4) = \sum_{k = -\infty}^\infty \frac{1}{(t+2(2+k))^{2}+1}\,</math> | ||
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| + | Simply set w = 2 + k to obtain: | ||
| + | <math> = \sum_{k = -\infty}^\infty \frac{1}{(t+2w)^{2}+1}\,</math> | ||
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| + | Since this is equivalent to x(t) the signal is periodic. | ||
Latest revision as of 07:50, 15 October 2008
Exam 1
The problem that I found to be the most confusing was problem 2:
Is the following signal periodic?
$ x(t) = \sum_{k = -\infty}^\infty \frac{1}{(t+2k)^{2}+1}\, $
$ x(t+4) = \sum_{k = -\infty}^\infty \frac{1}{(t+4+2k)^{2}+1}\, $
$ x(t+4) = \sum_{k = -\infty}^\infty \frac{1}{(t+2(2+k))^{2}+1}\, $
Simply set w = 2 + k to obtain: $ = \sum_{k = -\infty}^\infty \frac{1}{(t+2w)^{2}+1}\, $
Since this is equivalent to x(t) the signal is periodic.
