(New page: == Signal == <math> x(t) = 5cos(2t) + 3sin(4t)\!</math> == Fourier Series == <math> x(t) = 5(\frac{e^{j2t} + e^{-j2t}}{2}) + 3(\frac{e^{j4t} - e^{-j4t}}{2j}) \!</math> We take <math>...) |
(→Fourier Coefficients) |
||
| Line 17: | Line 17: | ||
From the Fourier Series, we determine the coefficients to be: | From the Fourier Series, we determine the coefficients to be: | ||
| − | <math> | + | <math> a_2 = a_{-2} = \frac{5}{2}\!</math> |
| − | <math> | + | <math> a_4 = a_{-4} = \frac{3}{2j}\!</math> |
== Other Coefficients == | == Other Coefficients == | ||
Revision as of 16:57, 24 September 2008
Signal
$ x(t) = 5cos(2t) + 3sin(4t)\! $
Fourier Series
$ x(t) = 5(\frac{e^{j2t} + e^{-j2t}}{2}) + 3(\frac{e^{j4t} - e^{-j4t}}{2j}) \! $
We take $ \omega_0 = 2\! $
$ x(t) = \frac{5}{2}e^{j2t} + \frac{5}{2}e^{-j2t} + \frac{3}{2j}e^{2j2t} - \frac{3}{2j}e^{-2j2t}\! $
Fourier Coefficients
From the Fourier Series, we determine the coefficients to be:
$ a_2 = a_{-2} = \frac{5}{2}\! $
$ a_4 = a_{-4} = \frac{3}{2j}\! $
Other Coefficients
$ w_0 = 2\! $
$ x(t) = \sum^{\infty}_{k = -\infty} a_ke^{jKw_0t}\! $
where $ a_k = \frac{1}{T}\int{0}^{T} x(t)e^{-jKw_0t}dt\! $
According to the Formula, $ a_k = 0\! $ whenever $ K \neq \pm2, \pm 4\! $
