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<math> ak = \frac{1}{T} \int_{0}^{T} x(t) * e^{-j*k} * \frac{2*\pi}{T} *dt </math> | <math> ak = \frac{1}{T} \int_{0}^{T} x(t) * e^{-j*k} * \frac{2*\pi}{T} *dt </math> | ||
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| + | |||
<math> x(t) = cos(2* \pi * t) * cos(4* \pi * t) </math> | <math> x(t) = cos(2* \pi * t) * cos(4* \pi * t) </math> | ||
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<math> = \frac{e^{j*2*\pi*t} + e^{-j*2*\pi*t}}{2} </math> | <math> = \frac{e^{j*2*\pi*t} + e^{-j*2*\pi*t}}{2} </math> | ||
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<math> = \frac{1*e^{j*6*\pi*t}}{4} + \frac{e^{-j*2*\pi*t}}{4} + \frac{e^{j*2*\pi*t}}{4} + \frac{e^{-j*6*\pi*t}}{4} </math> | <math> = \frac{1*e^{j*6*\pi*t}}{4} + \frac{e^{-j*2*\pi*t}}{4} + \frac{e^{j*2*\pi*t}}{4} + \frac{e^{-j*6*\pi*t}}{4} </math> | ||
| + | |||
| + | |||
| + | K = 3 K= -1 K= 1 K= -3 | ||
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| + | |||
| + | <math> a3 = \frac{1}{4} a-1 = \frac{1}{4} a1 = \frac{1}{4} a-3 = \frac{1}{4} </math> | ||
Revision as of 17:37, 23 September 2008
Define a Periodic CT signal and compute its Fourier series coefficients
Consider the following CT signal:
x(t) such that
$ ak = \frac{1}{T} \int_{0}^{T} x(t) * e^{-j*k} * \frac{2*\pi}{T} *dt $
$ x(t) = cos(2* \pi * t) * cos(4* \pi * t) $
$ = \frac{e^{j*2*\pi*t} + e^{-j*2*\pi*t}}{2} $
$ = \frac{1*e^{j*6*\pi*t}}{4} + \frac{e^{-j*2*\pi*t}}{4} + \frac{e^{j*2*\pi*t}}{4} + \frac{e^{-j*6*\pi*t}}{4} $
K = 3 K= -1 K= 1 K= -3
$ a3 = \frac{1}{4} a-1 = \frac{1}{4} a1 = \frac{1}{4} a-3 = \frac{1}{4} $
