(New page: Here's the problem: <math>\,Input \rightarrow SYSTEM \rightarrow Output</math> :* <math>\,e^{2jt} \rightarrow SYSTEM \rightarrow te^{-2jt}</math> :* <math>\,e^{-2jt} \rightarrow SYSTEM \...)
 
 
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:* <math>\,e^{2jt} \rightarrow SYSTEM \rightarrow te^{-2jt}</math>
 
:* <math>\,e^{2jt} \rightarrow SYSTEM \rightarrow te^{-2jt}</math>
 
:* <math>\,e^{-2jt} \rightarrow SYSTEM \rightarrow te^{2jt}</math>
 
:* <math>\,e^{-2jt} \rightarrow SYSTEM \rightarrow te^{2jt}</math>
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From the information given above, it seems like the system takes an input <math>\,x(t)</math> and transforms in into an output <math>\,y(t) = tx(-t)</math>
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For an input <math>\,x(t) = cos(2t)</math>
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:<math>\,cos(2t) \rightarrow SYSTEM \rightarrow tcos(-2t)</math>
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Thus, <math>\,y(t) = tcos(-2t)</math>

Latest revision as of 05:17, 17 September 2008

Here's the problem:

$ \,Input \rightarrow SYSTEM \rightarrow Output $

  • $ \,e^{2jt} \rightarrow SYSTEM \rightarrow te^{-2jt} $
  • $ \,e^{-2jt} \rightarrow SYSTEM \rightarrow te^{2jt} $


From the information given above, it seems like the system takes an input $ \,x(t) $ and transforms in into an output $ \,y(t) = tx(-t) $



For an input $ \,x(t) = cos(2t) $

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

Thus, $ \,y(t) = tcos(-2t) $

Alumni Liaison

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

Buyue Zhang