Fourier Series DC Component

Discussion in 'Homework Help' started by jp1390, Mar 5, 2014.

  1. jp1390

    Thread Starter Member

    Aug 22, 2011
    45
    2
    Hello,

    ***SOLUTION FOUND***

    Good resource: MIT Lecture @ around 40:40.

    Basically all the coefficients need to be normalized and the 0th term is the only exception where it needs to be normalized by 2.



    I am reviewing Fourier Series material for a course, particularly the trigonometric expansion:

    <br />
f(t) = \sum\limits_{n=0}^\infty a_{n}cos(n\omega t) + \sum\limits_{n=0}^\infty b_{n}sin(n\omega t)<br />

    Where:

    <br />
\omega = \frac{2\pi}{T}<br />
<br />
a_{n} = \int_{\frac{-T}{2}}^{\frac{T}{2}} f(t)cos(n\omega t) dt<br />
<br />
b_{n} = \int_{\frac{-T}{2}}^{\frac{T}{2}} f(t)sin(n\omega t) dt<br />

    When solving for the a0 (DC term) it always results in being double the actual DC value based on the above formula and needs to be halved for the correct function value.

    Given a function f(t) = cos(wt) + 3 and expanding this all out using the above formula:

    a0 = 6
    a1 = 1
    bn = 0, for all n

    cos(wt) + 6 != f(t)

    Wondering why the a0 term does not fit the formula (i.e. needing to divide by 2 to equal the identical function)?

    Thank you
     
    Last edited: Mar 5, 2014
  2. jp1390

    Thread Starter Member

    Aug 22, 2011
    45
    2
    Does it have to do with the fact that in the trigonometric form the An and Bn coefficients are taking the weight of both positive and negative frequency components and since there is no such thing as negative DC, you need to account for that after the fact?
     
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