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The above solution works for this case, but it does not show the properties of linearity. | The above solution works for this case, but it does not show the properties of linearity. | ||
| − | If <math>x_1(t)=e^{2jt | + | 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 | + | 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 | + | <math>\frac{1,2}e^{2jt}+\frac{1,2}e^{-2jt}\rightarrowsystem\rightarrow\t(frac{1,2}e^{-2jt}+\frac{1,2}e^{2jt})</math> |
| − | <math>t(frac{1,2}e^{-2jt | + | <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> | ||
Revision as of 10:50, 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}\rightarrowsystem\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) $
