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== B == | == B == | ||
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| + | <font size="3">Let <math>x(t)=cos(4 \pi t) + sin(6 \pi t)</math> with Fourier series coefficients are as follows: | ||
| + | |||
| + | <math>a_{4} = a_{-4} = \frac{1}{2}</math> | ||
| + | |||
| + | <math>a_{6} = -a_{-6} = \frac{1}{2j}</math> | ||
| + | |||
| + | All other <math>a_{k}</math> values are 0 | ||
| + | |||
| + | Then the response of <math>x(t)</math> to the system <math>y(t)</math> based on <math>H(s)</math> and the Fouries series coefficients is: | ||
| + | |||
| + | <math>y(t)=\sum_{k=-\infty}^{\infty}a_{k}H(s)</math> | ||
| + | |||
| + | <math>=\frac{1}{s} + \frac{1}{s} + \frac{1}{sj} - \frac{1}{sj}</math> | ||
| + | |||
| + | <math>=\frac{2}{s}</math> | ||
| + | </font> | ||
Latest revision as of 11:58, 24 September 2008
A
Let $ y(t)=\int_{-\infty}^{\infty}2x(t)dt $
Then $ h(t) =2u(t) $
And $ H(s) = \int_{-\infty}^{\infty}h(t)e^{-st}dt $
$ =\int_{-\infty}^{\infty}2u(t)e^{-st}dt $
$ =\int_{0}^{\infty}2e^{-st}dt $
$ =(\frac{-2}{s}e^{-st})|_{0}^{\infty} $
$ =\frac{2}{s} $
B
Let $ x(t)=cos(4 \pi t) + sin(6 \pi t) $ with Fourier series coefficients are as follows:
$ a_{4} = a_{-4} = \frac{1}{2} $
$ a_{6} = -a_{-6} = \frac{1}{2j} $
All other $ a_{k} $ values are 0
Then the response of $ x(t) $ to the system $ y(t) $ based on $ H(s) $ and the Fouries series coefficients is:
$ y(t)=\sum_{k=-\infty}^{\infty}a_{k}H(s) $
$ =\frac{1}{s} + \frac{1}{s} + \frac{1}{sj} - \frac{1}{sj} $
$ =\frac{2}{s} $
