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Alternate expressions of Fourier series formula

  1. Oct 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that the Fourier series formula [tex]F(t)=\frac{1}{2}a_{0}+\sum^{\infty}_{n=1}(a_{n}cos(nwt)+b_{n}sin(nwt))[/tex] can be expressed as [tex]F(t)=\frac{1}{2}a_{0}+\sum^{\infty}_{n=1}c_{n}cos(nwt-\phi_{n})[/tex]. Relate the coefficients [tex]c_{n}[/tex] to [tex]a_{n}[/tex] and [tex]b_{n}[/tex].

    2. Relevant equations

    We have the usual equations for the coefficients of a Fourier series.

    3. The attempt at a solution

    I'm really just checking the integrity of my solution here. I want to be sure I did not misunderstand the nature of [tex]\phi_{n}[/tex] or anything else.

    Let [tex]n\in Z^{+}[/tex]. Pick [tex]\phi_{n}[/tex] such that [tex]\frac{a_{n}}{cos(\phi_{n})}=\frac{b_{n}}{sin(\phi_{n})}[/tex]. We know we can do this since we could just choose [tex]\phi_{n}=nw_{0}t[/tex], where [tex]w_{0}=2 \pi f[/tex], the fundamental frequency of the Fourier series. Let [tex]c_{n}=\frac{a_{n}}{cos(\phi_{n})}=\frac{b_{n}}{sin(\phi_{n})}[/tex].

    [tex]F(t)=\frac{1}{2}a_{0}+\sum^{\infty}_{n=1}(a_{n}cos(nwt)+b_{n}sin(nwt))[/tex]
    [tex]F(t)=\frac{1}{2}a_{0}+\sum^{\infty}_{n=1}(c_{n}cos(\phi_{n})cos(nwt)+c_{n}sin(\phi_{n})sin(nwt))[/tex]
    [tex]F(t)=\frac{1}{2}a_{0}+\sum^{\infty}_{n=1}(c_{n}cos(nwt-\phi_{n}))[/tex]

    Which completes the problem.
     
  2. jcsd
  3. Oct 9, 2010 #2

    diazona

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    Homework Helper

    Your proof seems fine, but the one thing I would point out is that φn is supposed to be a constant. So you shouldn't pick φn = nω0t, because that would make the phase constant a linear function of time. Instead, just solve the equation
    [tex]\frac{a_n}{\cos\phi_n} = \frac{b_n}{\sin\phi_n}[/tex]
    for φn, getting
    [tex]\phi_n = \tan^{-1}\frac{b_n}{a_n}[/tex]
    (I'll leave the investigation of the case an=0 to you :wink:) Using that value of φn, you can define cn the same way you did, and the rest of the proof should be unchanged.
     
  4. Oct 10, 2010 #3
    Thanks!
     
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