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The Fourier Series can be easily found by treating | The Fourier Series can be easily found by treating | ||
| − | <math> Asin(\omega_0t) = | + | <math> Asin(\omega_0t) = \frac{A*(e^{j\omega_0t} - e^{-j\omega_0t})}{2j} </math> |
and | and | ||
| − | <math> Acos(\omega_0t) = | + | <math> Acos(\omega_0t) = \frac{A*(e^{j\omega_0t} + e^{-j\omega_0t})}{2} </math> |
This alows us to to put x(t) in the form of | This alows us to to put x(t) in the form of | ||
| Line 18: | Line 18: | ||
which gives us | which gives us | ||
| − | <math> x(t) = | + | <math> x(t) = \frac{4*(e^{j3t} - e^{-j3t})}{2j} + \frac{8*(e^{j7t} + e^{-j7t})}{2} </math> |
Simplifying and distributing | Simplifying and distributing | ||
| − | <math> x(t) = | + | <math> x(t) = \frac{2*e^{j3t} }{j}- \frac{2*e^{j3t} }{j} + 4e^{j7t} + 4e^{-j7t} </math> |
| − | <math>\ a_{-3} = 2 | + | <math>\ a_{-3} = \frac{2}{j} </math> |
| − | <math>\ a_{3}= -2 | + | <math>\ a_{3}= \frac{-2}{j} </math> |
<math>\ a_{-7} = 4 </math> | <math>\ a_{-7} = 4 </math> | ||
Revision as of 14:12, 26 September 2008
Define a Periodic CT Signal and Compute its Fourier Series Coefficients
Let's start this process by defining our signal. For simplicities sake lets use the the signal
$ x(t) = 4sin(3t) + 8cos(7t) $
The Fourier Series can be easily found by treating
$ Asin(\omega_0t) = \frac{A*(e^{j\omega_0t} - e^{-j\omega_0t})}{2j} $
and
$ Acos(\omega_0t) = \frac{A*(e^{j\omega_0t} + e^{-j\omega_0t})}{2} $
This alows us to to put x(t) in the form of
$ x(t) = \sum_{k=- \infty }^ \infty a_ke^{jk\omega_0t} $
which gives us
$ x(t) = \frac{4*(e^{j3t} - e^{-j3t})}{2j} + \frac{8*(e^{j7t} + e^{-j7t})}{2} $
Simplifying and distributing
$ x(t) = \frac{2*e^{j3t} }{j}- \frac{2*e^{j3t} }{j} + 4e^{j7t} + 4e^{-j7t} $
$ \ a_{-3} = \frac{2}{j} $
$ \ a_{3}= \frac{-2}{j} $
$ \ a_{-7} = 4 $
$ \ a{_7} = 4 $
all other $ \ a_k = 0 $
