(New page: First, rewrite <math> cos(2t) \,</math> as a complex exponentional :<br> :<math> cos(2t) =\frac{[e^{j2x}+e^{-j2x}]}{2}</math><br> Then, applying the system to the exponentials gives respon...) |
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First, rewrite <math> cos(2t) \,</math> as a complex exponentional :<br> | First, rewrite <math> cos(2t) \,</math> as a complex exponentional :<br> | ||
:<math> cos(2t) =\frac{[e^{j2x}+e^{-j2x}]}{2}</math><br> | :<math> cos(2t) =\frac{[e^{j2x}+e^{-j2x}]}{2}</math><br> | ||
| − | Then, applying the system | + | Then, applying the system gives response <math> y(t)\!</math>:<br> |
:<math> y(t) = \frac{[te^{-j2x}+te^{j2x}]}{2}</math>, (because the system is linear, the <math>\frac{1}{2}</math> factor remains)<br> | :<math> y(t) = \frac{[te^{-j2x}+te^{j2x}]}{2}</math>, (because the system is linear, the <math>\frac{1}{2}</math> factor remains)<br> | ||
Finally, factoring out the <math> t\!</math> and simplifying back into a sinusoid function yields: | Finally, factoring out the <math> t\!</math> and simplifying back into a sinusoid function yields: | ||
:<math> y(t) = tcos(2t)\!</math> | :<math> y(t) = tcos(2t)\!</math> | ||
Latest revision as of 07:41, 19 September 2008
First, rewrite $ cos(2t) \, $ as a complex exponentional :
- $ cos(2t) =\frac{[e^{j2x}+e^{-j2x}]}{2} $
Then, applying the system gives response $ y(t)\! $:
- $ y(t) = \frac{[te^{-j2x}+te^{j2x}]}{2} $, (because the system is linear, the $ \frac{1}{2} $ factor remains)
Finally, factoring out the $ t\! $ and simplifying back into a sinusoid function yields:
- $ y(t) = tcos(2t)\! $
