(→CT Signal & its Fourier coefficients) |
(→CT Signal & its Fourier coefficients) |
||
| Line 8: | Line 8: | ||
<math>\ x(t) = (1+2j)(\frac{e^{jt} + e^{-jt}}{2}) + 5(\frac{e^{j4t} - e^{-j4t}}{2j}) </math> | <math>\ x(t) = (1+2j)(\frac{e^{jt} + e^{-jt}}{2}) + 5(\frac{e^{j4t} - e^{-j4t}}{2j}) </math> | ||
| + | |||
| + | We simplify | ||
| + | |||
| + | <math>\ x(t) = \frac{e^{jt} + e^{-jt}}{2} | ||
Revision as of 20:35, 23 September 2008
CT Signal & its Fourier coefficients
Lets define the signal
$ \ x(t) = (1+2j)cos(t)+5sin(4t) $
Knowing that its Fourier series is
$ \ x(t) = (1+2j)(\frac{e^{jt} + e^{-jt}}{2}) + 5(\frac{e^{j4t} - e^{-j4t}}{2j}) $
We simplify
$ \ x(t) = \frac{e^{jt} + e^{-jt}}{2} $
