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<math>\ t\frac{e^{-2jt} + e^{2jt}}{2} | <math>\ t\frac{e^{-2jt} + e^{2jt}}{2} | ||
| + | |||
| + | Finally, we substitute cos(2t) into the formula yielding | ||
| + | |||
| + | <math>\ tcos(2t) | ||
Revision as of 23:48, 18 September 2008
We know that:
$ \ e^{2jt} \rightarrow SYSTEM \rightarrow te^{-2jt} $
$ \ e^{-2jt} \rightarrow SYSTEM \rightarrow te^{2jt} $
We also know that the response for
$ \ cos(2t) = \frac{e^{2jt} + e^{-2jt}}{2} $
From the first two statments we can deduce that the general behavior of the system is
$ \ x(t) \rightarrow SYSTEM \rightarrow tx(-t) $
So, we apply the known behavior of the system to the response for cos(2t)
$ \ \frac{e^{2jt} + e^{-2jt}}{2} \rightarrow SYSTEM \rightarrow \frac{te^{-2jt} + te^{2jt}}{2} $
Which simplifies to
$ \ t\frac{e^{-2jt} + e^{2jt}}{2} Finally, we substitute cos(2t) into the formula yielding <math>\ tcos(2t) $
