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If <math>x_1(t)=e^{2jt}</math> and <math>x_2(t)=e^{-2jt}</math>,
 
If <math>x_1(t)=e^{2jt}</math> and <math>x_2(t)=e^{-2jt}</math>,
Take the input signal to be <math>\frac{1 2}(x_1(t)+x_2(t))</math>.  
+
Take the input signal to be <math>\frac{1}{2}(x_1(t)+x_2(t))</math>.  
The response to the input is <math>y(t)=\frac{1 2}y_1(t)+\frac{1 2}y_2(t)</math> because the system is linear.
+
The response to the input is <math>y(t)=\frac{1}{2}y_1(t)+\frac{1}{2}y_2(t)</math> because the system is linear.
  
<math>\frac{1 2}(e^{2jt}+\frac{1 2}e^{-2jt})\rightarrow system\rightarrow t(\frac{1 2}e^{-2jt}+\frac{1 2}e^{2jt})</math>
+
<math>\frac{1}{2}(e^{2jt}+\frac{1 2}e^{-2jt})\rightarrow system\rightarrow t(\frac{1}{2}e^{-2jt}+\frac{1}{2}e^{2jt})</math>
  
<math>t(\frac{1 2}e^{-2jt}+\frac{1 2}e^{2jt})=t(\frac{1 2}(e^{-2jt}+e^{2jt})</</math>
+
<math>t(\frac{1}{2}e^{-2jt}+\frac{1}{2}e^{2jt})=t(\frac{1}{2}(e^{-2jt}+e^{2jt})</math>
  
                             <math>=t(\frac{1 2}(cos(-2t)+jsin(-2t)+cos(2t)+jsin2(t))</math>
+
                             <math>=t(\frac{1}{2}(cos(-2t)+jsin(-2t)+cos(2t)+jsin2(t))</math>
  
                             <math>=t(\frac{1 2}(cos(2t)-jsin(2t)+cos(2t)+jsin2(t))</math>
+
                             <math>=t(\frac{1}{2}(cos(2t)-jsin(2t)+cos(2t)+jsin2(t))</math>
 
        
 
        
                             <math>=t(\frac{1 2}(2cos(2t))</math>
+
                             <math>=t(\frac{1}{2}(2cos(2t))</math>
  
 
                             <math>=tcos(2t)</math>
 
                             <math>=tcos(2t)</math>

Revision as of 09:58, 18 September 2008

since $ x(t)=e^{2jt} $ yields $ y(t)=te^{-2jt} $ and $ x(t)=e^{-2jt} $ yields $ x(t)=te^{2jt} $, it is easy to see that x(t) yields y(t)=t*x(-t).

Based on the above, $ x(t)=cos(2t) $ would yield $ y(t)=tx(-t)=tcos(-2t) $.


The above solution works for this case, but it does not show the properties of linearity.

If $ x_1(t)=e^{2jt} $ and $ x_2(t)=e^{-2jt} $, Take the input signal to be $ \frac{1}{2}(x_1(t)+x_2(t)) $. The response to the input is $ y(t)=\frac{1}{2}y_1(t)+\frac{1}{2}y_2(t) $ because the system is linear.

$ \frac{1}{2}(e^{2jt}+\frac{1 2}e^{-2jt})\rightarrow system\rightarrow t(\frac{1}{2}e^{-2jt}+\frac{1}{2}e^{2jt}) $

$ t(\frac{1}{2}e^{-2jt}+\frac{1}{2}e^{2jt})=t(\frac{1}{2}(e^{-2jt}+e^{2jt}) $

                            $ =t(\frac{1}{2}(cos(-2t)+jsin(-2t)+cos(2t)+jsin2(t)) $
                            $ =t(\frac{1}{2}(cos(2t)-jsin(2t)+cos(2t)+jsin2(t)) $
      
                            $ =t(\frac{1}{2}(2cos(2t)) $
                            $ =tcos(2t) $

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