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Fourier series: relation of coefficients

  1. Jan 12, 2013 #1

    The fourier series can (among others) expressed in terms of sines and cosines with coefficients [itex]a_n[/itex] and [itex]b_n[/itex] and solely by sines using amplitudes [itex]A_n[/itex] and phase [itex]\phi_n[/itex].

    I want to express the latter using [itex]a_n[/itex] and [itex]b_n[/itex]. Using

    a_n = A_n \sin(\phi_n) \\
    b_n = A_n \cos(\phi_n)

    I quickly found [itex]A_n[/itex] by expressing the arccos and arcsin. For [itex]\phi_n[/itex] I would get

    [tex]\phi_n = \arccos \frac{a_n}{\sqrt{a_n^2 + b_n^2}}[/tex]

    However, according to the German Wikipedia (http://de.wikipedia.org/wiki/Fourier-Reihe) this seems not so trivial. One option there is (for n \neq 0):

    [tex]\phi_n = 2 \arctan \frac{b_n}{A_n + a_n}[/tex]

    or using arctan and signum function. What I am missing or is my approach also correct?

  2. jcsd
  3. Jan 12, 2013 #2

    Simon Bridge

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    You are starting with the two styles of the fourier transforms:
    $$\sum_n a_n\cos nx + b_n\sin nx = \sum_m A_m\sin(mx+\phi_m)$$
    ... and you want to relate ##\{a_n,b_n\}## to ##\{A_m,\phi_m\}## ... is that correct?
  4. Jan 12, 2013 #3
    yes, exactly. And I relate them with [itex]\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)[/itex] to get:

    A_0 = a_0/2\\
    a_0 = 2A_0

    a_n = A_n \sin\phi_n \\
    b_n = A_n \cos\phi_n

    For represeting [itex]A_n[/itex] I get using both equations above:

    A_n = a_n/\sin\phi_n \\
    cos\phi_n = b_n/A_n \\
    \phi_n = \arccos b_n/A_n = \arcsin \sqrt{1-b_n^2/A_n^2} \\
    A_n = \frac{a_n}{\sqrt{1-\frac{b_n^2}{A_n^2}}} \\
    \dots \\
    A_n = \sqrt{a_n^2 + b_n^2}

    I think this is correct, right?
    In the same spirit, I can derive

    \phi_n = \arccos\frac{b_n}{A_n} = \arccos\frac{b_n}{\sqrt{a_n^2 + b_n^2}}

    ... but according to the German Wikipedia this is wrong. Why?
  5. Jan 13, 2013 #4

    Simon Bridge

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    That works ...

    To put it another way - you can expand each term in the sin-only expansion in terms of the sin-cos expansion ... i.e.

    $$A_n\sin(nx+\phi_n)=\sum_m a_m\cos(mx)+b_m\sin(mx)$$

    Then you apply the identity.
    Since you are expanding an arbitrary sine in terms of sines and cosines, the sum should only have one term in it... (for each value of n on the RHS.)

    Leads you to:



    ... sure. Nicely done.
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