(New page: == Problem == Given that the response to <math>x(t)=exp(2jt)</math> is <math>y(t)=t*exp(-2jt)</math>, and its response to <math>x(t)=exp(-2jt)</math> is <math>y(t)=t*exp(2jt)</math>. Wha...) |
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== Solution == | == Solution == | ||
| + | If a system is linear we know that: | ||
| + | |||
| + | <math>ax(t)+bx(t)=ay(t)+by(t)</math> | ||
| + | |||
| + | Using Euler's Method we know that: | ||
| + | |||
| + | <math>exp(jt)=cos(t)+jsin(t)</math> | ||
| + | |||
| + | Given this: | ||
| + | |||
| + | <math>exp(2jt)+exp(-2jt)=cos(2t)+j*sin(2t)+cos(-2t)+j*sin(-2t)</math> | ||
| + | |||
| + | <math>exp(2jt)+exp(-2jt)=cos(2t)+j*sin(2t)+cos(2t)-j*sin(2t)</math> | ||
| + | |||
| + | <math>exp(2jt)+exp(-2jt)=2cos(2t)</math> | ||
| + | |||
| + | So now: | ||
| + | |||
| + | <math>.5*2cos(2t)=cos(2t)</math> | ||
| + | |||
| + | <math>cos(2t)-->.5*(t*exp(2jt)+t*exp(-2jt))</math> | ||
| + | |||
| + | <math>cos(2t)--> t*cos(2t)</math> | ||
Latest revision as of 05:59, 18 September 2008
Problem
Given that the response to $ x(t)=exp(2jt) $ is $ y(t)=t*exp(-2jt) $, and its response to $ x(t)=exp(-2jt) $ is $ y(t)=t*exp(2jt) $. What is the systems response to $ x(t)=cos(2t) $?
Solution
If a system is linear we know that:
$ ax(t)+bx(t)=ay(t)+by(t) $
Using Euler's Method we know that:
$ exp(jt)=cos(t)+jsin(t) $
Given this:
$ exp(2jt)+exp(-2jt)=cos(2t)+j*sin(2t)+cos(-2t)+j*sin(-2t) $
$ exp(2jt)+exp(-2jt)=cos(2t)+j*sin(2t)+cos(2t)-j*sin(2t) $
$ exp(2jt)+exp(-2jt)=2cos(2t) $
So now:
$ .5*2cos(2t)=cos(2t) $
$ cos(2t)-->.5*(t*exp(2jt)+t*exp(-2jt)) $
$ cos(2t)--> t*cos(2t) $
