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<math> x(t) = (2 / j)*e^{j3t} - (2 / j)*e^{-j3t} + 4e^{j7t} + 4e^{-j7t} </math> | <math> x(t) = (2 / j)*e^{j3t} - (2 / j)*e^{-j3t} + 4e^{j7t} + 4e^{-j7t} </math> | ||
| + | |||
| + | <math>\ a_{-3} = 2/j </math> | ||
| + | |||
| + | <math>\ a_{3}= -2/j </math> | ||
| + | |||
| + | <math>\ a_{-7} = 4 </math> | ||
| + | |||
| + | <math>\ a{_7} = 4 </math> | ||
| + | |||
| + | all other <math> \ a_k = 0 </math> | ||
Revision as of 13:53, 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) = (A/ j2)*(e^{j\omega_0t} - e^{-j\omega_0t}) $
and
$ Acos(\omega_0t) = (A/2)*(e^{j\omega_0t} + e^{-j\omega_0t}) $
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) = (4/ j2)*(e^{j3t} - e^{-j3t}) + (8/2)*(e^{j7t} + e^{-j7t}) $
Simplifying and distributing
$ x(t) = (2 / j)*e^{j3t} - (2 / j)*e^{-j3t} + 4e^{j7t} + 4e^{-j7t} $
$ \ a_{-3} = 2/j $
$ \ a_{3}= -2/j $
$ \ a_{-7} = 4 $
$ \ a{_7} = 4 $
all other $ \ a_k = 0 $
