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We know that:
 
We know that:
  
<math>\, x(t)= e^{2jt} --> |SYSTEM| --> te^{-2jt)
+
<math>\ e^{2jt} \rightarrow SYSTEM \rightarrow te^{-2jt}</math>
 +
 
 +
<math>\ e^{-2jt} \rightarrow SYSTEM \rightarrow te^{2jt}</math>
 +
 
 +
We also know that the response for
 +
 
 +
<math>\ cos(2t) = \frac{e^{2jt} + e^{-2jt}}{2} </math>
 +
 
 +
From the first two statments we can deduce that the general behavior of the system is
 +
 
 +
<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 23:50, 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

$ \ 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. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang