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<math>x(t)=\frac{1}{2}e^{j4\pi t}+\frac{1}{2}e^{-j4\pi t}+\frac{1}{2j}e^{j2\pi t}+\frac{-1}{2j}e^{-j2\pi t}</math> | <math>x(t)=\frac{1}{2}e^{j4\pi t}+\frac{1}{2}e^{-j4\pi t}+\frac{1}{2j}e^{j2\pi t}+\frac{-1}{2j}e^{-j2\pi t}</math> | ||
| − | <math> | + | <math>a_4=\frac{1}{2}</math> |
| − | <math>a_{- | + | <math>a_{-4}=\frac{1}{2}</math> |
<math>a_2=\frac{1+j}{2j}</math> | <math>a_2=\frac{1+j}{2j}</math> | ||
Revision as of 15:22, 26 September 2008
A periodic CT signal
Fourier series of x(t):
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
, where $ a_k $ is
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
Input CT signal: $ x(t) = cos2t+sin2t $
$ \,x(t)=\frac {e^{j4\pi t}+e^{-j4 \pi t}}{2} + \frac {e^{j2 \pi t}-e^{-j2 \pi t}}{2j} $
$ x(t)=\frac{1}{2}e^{j4\pi t}+\frac{1}{2}e^{-j4\pi t}+\frac{1}{2j}e^{j2\pi t}+\frac{-1}{2j}e^{-j2\pi t} $
$ a_4=\frac{1}{2} $
$ a_{-4}=\frac{1}{2} $
$ a_2=\frac{1+j}{2j} $
$ a_{-2}=\frac{-1-j}{2j} $
All other $ \,a_k $ values are zero.
