(New page: ==CT Signal & its Fourier coefficients== Lets define the signal <math>\ x(t) = (1+2j)cos(t)+5sin(4t) </math> Knowing that its Fourier series is <math>\ x(t) = (1+2j)(\frac{e^{jt} + e^{...) |
(→CT Signal & its Fourier coefficients) |
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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> | <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> | ||
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| + | So, we get the coefficients: | ||
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| + | <math>\ a_{1} = \frac{1+2j}{2} </math> | ||
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| + | <math>\ a_{-1} = \frac{1+2j}{2} </math> | ||
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| + | <math>\ a_{2} = \frac{5}{2j} </math> | ||
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| + | <math>\ a_{-2} = - \frac{5}{2j}</math> | ||
Revision as of 17:09, 26 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} $
So, we get the coefficients:
$ \ a_{1} = \frac{1+2j}{2} $
$ \ a_{-1} = \frac{1+2j}{2} $
$ \ a_{2} = \frac{5}{2j} $
$ \ a_{-2} = - \frac{5}{2j} $
