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| − | Provided that: <br> | + | 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> | + | (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> | + | The Key to approach this problem is: What is <math> \cos 2t\ </math>? |
| − | + | <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 06:10, 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\ $
