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Provided that: <br>
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Provided that:<br>
 
(1) <math> e^{j2t}\ </math> ----------> System ----------> <math> te^{-2jt}\ </math><br>
 
(1) <math> e^{j2t}\ </math> ----------> System ----------> <math> te^{-2jt}\ </math><br>
 
(2) <math> e^{-j2t}\ </math>----------> System ----------> <math> te^{2jt}\ </math><br>
 
(2) <math> e^{-j2t}\ </math>----------> System ----------> <math> te^{2jt}\ </math><br>
(3) The System is Linear. <br>
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(3) The System is Linear. <br><br>
  
The following should hold true:
+
The following should hold true:<br><br>
<math> e^{j2t} + e^{-j2t}\ </math> ----------> System -----------> <math> te^{-2jt} + te^{2jt}\ </math><br><br>
+
(1)<math> e^{j2t} + e^{-j2t}\ </math> ----------> System -----------> <math> te^{-2jt} + te^{2jt}\ </math><br>
 +
(2)<math> {e^{j2t} + e^{-j2t}\over 2} </math> ----------> System -----------> <math> {te^{-2jt} + te^{2jt}\over 2} </math><br><br>
  
The Key to approach this problem is: What is <math> {e^{j2t} + e^{-j2t} \over 2} </math>?
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The Key to approach this problem is: What is <math> \cos 2t\ </math>?
 
<br>
 
<br>
 
(1) <math> \cos 2t\ = {e^{j2t} + e^{-j2t} \over 2} </math> by Euler's Formalas.<br>
 
(1) <math> \cos 2t\ = {e^{j2t} + e^{-j2t} \over 2} </math> by Euler's Formalas.<br>
 +
(2) The response to (1) is <math>{te^{-2jt} + te^{2jt}\over 2} </math><br><br>
 +
It is unnecessary to say this but it is <math> t\ \cos 2t\ </math>

Latest revision as of 07:10, 19 September 2008

Provided that:
(1) $ e^{j2t}\ $ ----------> System ----------> $ te^{-2jt}\ $
(2) $ e^{-j2t}\ $----------> System ----------> $ te^{2jt}\ $
(3) The System is Linear.

The following should hold true:

(1)$ e^{j2t} + e^{-j2t}\ $ ----------> System -----------> $ te^{-2jt} + te^{2jt}\ $
(2)$ {e^{j2t} + e^{-j2t}\over 2} $ ----------> System -----------> $ {te^{-2jt} + te^{2jt}\over 2} $

The Key to approach this problem is: What is $ \cos 2t\ $?
(1) $ \cos 2t\ = {e^{j2t} + e^{-j2t} \over 2} $ by Euler's Formalas.
(2) The response to (1) is $ {te^{-2jt} + te^{2jt}\over 2} $

It is unnecessary to say this but it is $ t\ \cos 2t\ $

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