(New page: == EXAM 1== Problem 1. is <math> x(t) = \sum_{k = -\infty}^\infty \frac{1}{(t+2k)^{2}+1} </math> periodic?) |
(→EXAM 1) |
||
| (3 intermediate revisions by the same user not shown) | |||
| Line 3: | Line 3: | ||
Problem 1. | Problem 1. | ||
| − | is <math> x(t) = \sum_{k = -\infty}^\infty \frac{1}{(t+2k)^{2}+1} </math> periodic? | + | is |
| + | |||
| + | <math> x(t) = \sum_{k = -\infty}^\infty \frac{1}{(t+2k)^{2}+1} </math> | ||
| + | |||
| + | periodic? | ||
| + | |||
| + | We know that for a signal to be periodic | ||
| + | |||
| + | <math> x(t) = x(t + T) </math> | ||
| + | |||
| + | So we shift the function by a arbitrary number to try to prove the statement above | ||
| + | |||
| + | <math> x(t+1) = \sum_{k = -\infty}^\infty \frac{1}{(t+1+2k)^{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 <math> r = \frac{1}{2}+k </math> to yield, | ||
| + | |||
| + | <math> = \sum_{k = -\infty}^\infty \frac{1}{(t+2r)^{2}+1} </math> | ||
| + | |||
| + | 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.
