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| − | The Fourier series of | + | The Fourier series of a CT signal can be found by: |
<math> x(t) = \sum_{n=-\infty}^\infty a_k e^{jkw_0t}</math> | <math> x(t) = \sum_{n=-\infty}^\infty a_k e^{jkw_0t}</math> | ||
Revision as of 16:33, 26 September 2008
Problem
Find the Fourier series coefficients of a CT signal.
I chose the signal $ x(t) = 3cos(2t) $.
Fourier Series Coefficients
The Fourier series of a CT signal can be found by:
$ x(t) = \sum_{n=-\infty}^\infty a_k e^{jkw_0t} $
Fourier series coefficients can be found with the equation:
$ a_k = \frac{1}{T} \int_{0}^{T} x(t)e^{-jkw_0t} \,\ dt $
However, in the case of sinusoidal waves, the coefficients can be found more simply by applying Euler's formula to solve for the Fourier series and take the coefficients from that.
$ x(t) = 3cos(2t) = 3(\frac{e^{j2t}}{2}+\frac{e^{-j2t}}{2}) $
$ = \frac{3}{2}e^{j2t}+\frac{3}{2}e^{-j2t} $
And so, we know that our coefficients are both $ \frac{3}{2} $. Now, we need to find which $ a_k $s these belong to.
Knowing $ T $ is $ 2\pi $, we can find $ w_0 $.
$ w_0 = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1 $
Finding the corresponding $ k $s should be easy.
We know
$ a_k e^{jkw_0t} $
corresponds to
$ \frac{3}{2}e^{j2t} $ and $ \frac{3}{2}e^{-j2t} $
and since $ w_0 = 1 $, then we can conclude
$ a_2 = \frac{3}{2} $
and
$ a_{-2} = \frac{3}{2} $
