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Fourier Coefficients

by dimensionless
Tags: coefficients, fourier
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dimensionless
#1
Aug1-07, 09:52 AM
P: 464
A function [tex]f(t)[/tex] can be represented by the expansion

[tex]
f(t) = \frac{1}{2}A_{0} + A_{1}cos(\omega t) + A_{2}cos(2 \omega t) + A_{3}cos(3 \omega t) + ....
B_{1}sin(\omega t) + B_{2}sin(2 \omega t) + B_{3}sin(3 \omega t) + ....
[/tex]

Do the constants [tex]A_{n}[/tex] and [tex]B_{n}[/tex] the same thing as the real and imaginary components of the Fourier transform? If so, why is there no imaginary component in the zeroth term?
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mathman
#2
Aug1-07, 04:26 PM
Sci Advisor
P: 6,057
In computing the Fourier transform, the kernel is of the form einwt. For A0, the kernel is simply 1, so there is no imaginary part.


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