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

$ \ tcos(2t)  $

So the the behavior of input cos(2t) in this system will be the following

$ \ cos(2t) \rightarrow SYSTEM \rightarrow tcos(2t) $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva