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<math>x(t)=cos(3*pi*t)cos(6*pi*t)\!</math> | <math>x(t)=cos(3*pi*t)cos(6*pi*t)\!</math> | ||
| − | + | <br> | |
| − | + | <br> | |
| − | <math> | + | <math>x(t)=[1/2*e^{j*2*pi*t}+1/2*e^{-j*2*pi*t}]*[1/2*e^{j*4*pi*t}+1/2*e^{-j*4*pi*t}]</math> |
| − | + | <br> | |
| − | + | <math> =1/4*e^{j6pit}+1/4*e^{-j2pit}+1/4*e^{j2pit}+1/4*e^{-j6pit}</math> | |
| − | + | <br> | |
| − | <math> | + | <br> |
| − | + | The fundamental frequency is 2*pi | |
| − | + | <br> | |
| − | + | a3=1/4 | |
| − | < | + | a-1=1/4 |
| + | a1=1/4 | ||
| + | a3=1/4 | ||
| + | all other ak=0 | ||
Revision as of 09:56, 26 September 2008
Equations
Fourier series of x(t):
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
Signal Coefficients:
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $
From Phil Cannon
Input Signal
$ x(t)=cos(3*pi*t)cos(6*pi*t)\! $
$ x(t)=[1/2*e^{j*2*pi*t}+1/2*e^{-j*2*pi*t}]*[1/2*e^{j*4*pi*t}+1/2*e^{-j*4*pi*t}] $
$ =1/4*e^{j6pit}+1/4*e^{-j2pit}+1/4*e^{j2pit}+1/4*e^{-j6pit} $
The fundamental frequency is 2*pi
a3=1/4
a-1=1/4
a1=1/4
a3=1/4
all other ak=0
