(→CT Signal & its Fourier coefficients) |
(→CT Signal & its Fourier coefficients) |
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<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> | ||
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| + | We simplify | ||
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| + | <math>\ x(t) = \frac{1+2j}{2} e^{jt} + \frac{1+2j}{2} e^{-jt} + \frac{5}{2j}e^{j4t} - \frac{5}{2j}e^{-j4t} </math> | ||
Latest revision as of 21:38, 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{1+2j}{2} e^{jt} + \frac{1+2j}{2} e^{-jt} + \frac{5}{2j}e^{j4t} - \frac{5}{2j}e^{-j4t} $
