(→Define a periodic CT signal and compute its Fourier series coefficients.) |
(→Define a periodic CT signal and compute its Fourier series coefficients.) |
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<math>x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}</math> | <math>x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}</math> | ||
| − | + | where | |
<math>a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt</math>. | <math>a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt</math>. | ||
| Line 17: | Line 17: | ||
<math> a_2 = a_-2 = 1 </math> | <math> a_2 = a_-2 = 1 </math> | ||
| + | |||
| + | a_k = 0 elsewhere | ||
Revision as of 09:03, 25 September 2008
Define a periodic CT signal and compute its Fourier series coefficients.
For CT,
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
where
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
Let the signal be
y(t) = 2*sin(2t)+2*cos(4t)
$ y(t) = 2(\frac{e^{j2t} - e^{-j2t}}{2j}) + 2(\frac{e^{2j2t} + e^{-2j2t}}{2}) \! $
$ a_1 = a_-1 = (\frac{1}{j}) $
$ a_2 = a_-2 = 1 $
a_k = 0 elsewhere
