(→Ao) |
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<math>Ao =\int_0^{2\pi}[(1+j)cos(4t) + 14sin(6t)]e^{0}dt</math> | <math>Ao =\int_0^{2\pi}[(1+j)cos(4t) + 14sin(6t)]e^{0}dt</math> | ||
==A1== | ==A1== | ||
| + | <math>Ao =\int_0^{2\pi}[(1+j)cos(4t) + 14sin(6t)]e^{1}dt</math> | ||
==A2== | ==A2== | ||
| + | <math>Ao =\int_0^{2\pi}[(1+j)cos(4t) + 14sin(6t)]e^{2}dt</math> | ||
==A-1== | ==A-1== | ||
| + | <math>Ao =\int_0^{2\pi}[(1+j)cos(4t) + 14sin(6t)]e^{-1}dt</math> | ||
==A-2== | ==A-2== | ||
| + | <math>Ao =\int_0^{2\pi}[(1+j)cos(4t) + 14sin(6t)]e^{-2}dt</math> | ||
Revision as of 07:55, 26 September 2008
Contents
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)=(1+j)cos(3t)+14sin(6t)\! $
Ao
$ Ao =\int_0^{2\pi}[(1+j)cos(4t) + 14sin(6t)]e^{0}dt $
A1
$ Ao =\int_0^{2\pi}[(1+j)cos(4t) + 14sin(6t)]e^{1}dt $
A2
$ Ao =\int_0^{2\pi}[(1+j)cos(4t) + 14sin(6t)]e^{2}dt $
A-1
$ Ao =\int_0^{2\pi}[(1+j)cos(4t) + 14sin(6t)]e^{-1}dt $
A-2
$ Ao =\int_0^{2\pi}[(1+j)cos(4t) + 14sin(6t)]e^{-2}dt $
