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From the first two statments we can deduce that the general behavior of the system is
 
From the first two statments we can deduce that the general behavior of the system is
  
  <math>/ x(t) \rightarrow SYSTEM \rightarrow ty(-t)</math>
+
  <math>\ x(t) \rightarrow SYSTEM \rightarrow tx(-t)</math>
 +
 
 +
So, we apply the known behavior of the system to the response for cos(2t)
 +
 
 +
<math>\ \frac{e^{2jt} + e^{-2jt}}{2} \rightarrow SYSTEM \rightarrow \frac{te^{-2jt} + te^{2jt}}{2} </math>
 +
 
 +
Which simplifies to
 +
 
 +
<math>\ t\frac{e^{-2jt} + e^{2jt}}{2} </math>
 +
 
 +
Finally, we substitute cos(2t) into the formula yielding
 +
 
 +
<math>\ tcos(2t) </math>
 +
 
 +
So the the behavior of input cos(2t) in this system will be the following
 +
 
 +
<math>\ cos(2t) \rightarrow SYSTEM \rightarrow tcos(2t)</math>

Latest revision as of 00:50, 19 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

$ \ 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

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood