| Line 29: | Line 29: | ||
We know | We know | ||
| − | <math>a_k e^{jkw_0t}</math> | + | <math>a_k e^{jkw_0t}</math> corresponds to <math>\frac{3}{2}e^{j2t}</math> and <math>\frac{3}{2}e^{-j2t}</math> |
| − | + | ||
| − | corresponds to | + | |
| − | + | ||
| − | <math>\frac{3}{2}e^{j2t}</math> and <math>\frac{3}{2}e^{-j2t}</math> | + | |
and since <math>w_0 = 1</math>, then we can conclude | and since <math>w_0 = 1</math>, then we can conclude | ||
| Line 45: | Line 41: | ||
and | and | ||
| − | <math>a_k = 0</math> for <math>K \neq 2, -2</math> | + | <math>a_k = 0 \!</math> for <math>K \neq 2, -2</math> |
Revision as of 16:37, 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} $
and
$ a_k = 0 \! $ for $ K \neq 2, -2 $
