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| − | <math>x(t)=cos(2t)</math> would yield <math>y(t)=tx(-t)=tcos(-2t)</math>. | + | <math>x(t)=cos(2t)</math> would yield <math>y(t)=tx(-t)=tcos(-2t)=tcos(2t)</math>. |
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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 | + | 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 | + | <math>\frac{1}{2}(e^{2jt}+e^{-2jt})\rightarrow system\rightarrow t(\frac{1}{2}e^{-2jt}+\frac{1}{2}e^{2jt})</math> |
| − | <math>t(\frac{1 | + | <math>=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}(2cos(2t))</math> | |
| − | + | <math>=tcos(2t)</math> | |
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Latest revision as of 10:05, 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)=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}+e^{-2jt})\rightarrow system\rightarrow 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) $
