(New page: <font size="3">Let <math>x(t)=cos(t)</math>, then the steps to calculate its Fourier series coefficients are as follows:</font>) |
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| − | <font size="3">Let <math>x(t)=cos(t)</math>, then | + | <font size="3">Let <math>x(t)=cos(4 \pi t) + sin(6 \pi t)</math>, then its Fourier series coefficients are as follows: |
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| + | <math>x(t)=\frac{e^{4 \pi jt} + e^{-4 \pi jt}}{2} + \frac{e^{6 \pi jt} + e^{-6 \pi jt}}{2j}</math> and <math>\omega_{0} = \pi</math> | ||
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| + | <math>a_{4} = a_{-4} = \frac{1}{2}</math> | ||
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| + | <math>a_{6} = -a_{-6} = \frac{1}{2j}</math> | ||
| + | |||
| + | All other <math>a_{k}</math> values are 0 | ||
| + | </font> | ||
Revision as of 12:36, 23 September 2008
Let $ x(t)=cos(4 \pi t) + sin(6 \pi t) $, then its Fourier series coefficients are as follows:
$ x(t)=\frac{e^{4 \pi jt} + e^{-4 \pi jt}}{2} + \frac{e^{6 \pi jt} + e^{-6 \pi jt}}{2j} $ and $ \omega_{0} = \pi $
$ a_{4} = a_{-4} = \frac{1}{2} $
$ a_{6} = -a_{-6} = \frac{1}{2j} $
All other $ a_{k} $ values are 0
