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The Fourier Series

  1. Nov 26, 2003 #1
    The Fourier Series!!

    Hi guys, i'm having a bit a trouble helping my daughter with this question on the fourier series approximation:

    The fourier series for a real, odd function, f(t) can be written as:
    f(t) = [SUM to infinity, n=1, of]: b[subscipt n] sin(nwt)
    where f(t=T)=f(t) and w=(2[pie])/T

    Prove that b[subscipt n] = 2/T [integral between T/2 & -T/2 of]: f(t)sin(nwt)dt

    Sorry about not knowing how to do all the symbols etc, but if you write it out i'm pretty sure it makes sence.

    Anyway, if anyone could give me a hand with how to go about proving this, it would be greatly appreciated.

    Many thanks, Robert.
     
  2. jcsd
  3. Nov 26, 2003 #2

    chroot

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    Staff Emeritus
    Science Advisor
    Gold Member

    Hey Robert,

    You have a function of t represented by the series

    [tex]f(t) = \sum_{n=1}^\infty b_n \sin{n \omega t}[/tex]

    And you'd like to know what the [tex]\inline{b_n}[/tex] are. You're also given the condition [tex]\inline{f(t + T) = f(t)}[/tex], which just indicates the function's period is [tex]\inline{T}[/tex].

    You're not even required to figure out what the [tex]\inline{b_n}[/tex] are; you're just asked to demonstrate that they are, in fact

    [tex]b_n = \int_{-T/2}^{T/2} f(t) \sin{n \omega t}\ dt[/tex].

    To do this, first note the following fact:

    [tex]\begin{equation*}
    \begin{split}
    \int_{-T/2}^{T/2} \sin^2{\left( \frac{2 \pi n t}{T} \right) }\ dt &= \frac{T}{2}
    \end{split}
    \end{equation*}
    [/tex]

    for all n, and use it to perform the following steps:

    [tex]\begin{equation*}
    \begin{split}
    \int_{-T/2}^{T/2} f(t) \sin{n \omega t}\ dt &= \int_{-T/2}^{T/2} \left[ \sum_{n=1}^\infty b_n \sin{n \omega t} \right] \sin{n \omega t}\ dt\\
    &= \sum_{n=1}^\infty \int_{-T/2}^{T/2} b_n \sin^2{n \omega t}\ dt\\
    &= \frac{T}{2} b_n
    \end{split}
    \end{equation*}
    [/tex]

    Does this make sense?

    - Warren
     
  4. Nov 27, 2003 #3
    Thanks mate, yep i essentially understand, just 1 question though:

    How did you get from the second line to the third line in the following:

    [tex]\begin{equation*}
    \begin{split}
    \int_{-T/2}^{T/2} f(t) \sin{n \omega t}\ dt &= \int_{-T/2}^{T/2} \left[ \sum_{n=1}^\infty b_n \sin{n \omega t} \right] \sin{n \omega t}\ dt\\
    &= \sum_{n=1}^\infty \int_{-T/2}^{T/2} b_n \sin^2{n \omega t}\ dt\\
    &= \frac{T}{2} b_n
    \end{split}
    \end{equation*}
    [/tex]

    Could you work me though that bit in a little more detail mate?

    Thanks a lot,
    Robert.
     
  5. Nov 27, 2003 #4
    i think chroot pulled a fast one on you. there is a bit more to the trick that you have to realize to make the infinite summation turn into a single term, with one integration.

    to see this, first let me remind you of some trigonometric identities:

    [tex]
    \cos(a\pm b)=\cos a\cos b \mp\sin a\sin b
    [/tex]
    so
    [tex]
    \begin{align}
    &\frac{1}{2}(\cos(a-b)-\cos(a+b))\\
    &=-\frac{1}{2}\cos a\cos b+\frac{1}{2}\sin a\sin b\\
    & -\frac{1}{2}\cos a\cos b+\frac{1}{2}\sin a\sin b\\
    &= \sin a\sin b
    \end{align}
    [/tex]

    we need to use this identity to evaluate some integrals:

    [tex]
    \begin{align}
    &\int_{-T/2}^{T/2}\sin(\frac{2\pi mt}{T})\sin(\frac{2\pi nt}{T})\ dt=\\
    &\frac{1}{2}\int_{-T/2}^{T/2}\cos\frac{2\pi(n-m)t}{T}\ dt-\frac{1}{2}\int_{-T/2}^{T/2}\cos\frac{2\pi(n+m)t}{T}\ dt=\\
    &\left\frac{T}{4\pi(n-m)}\sin\frac{2\pi(n-m)t}{T}\right]^{T/2}_{-T/2}\\
    &-\left\frac{T}{4\pi(n+m)}\sin\frac{2\pi(n+m)t}{T}\right]^{T/2}_{-T/2}\\
    &=0
    \end{align}
    [/tex]
    note that this integration is not valid when n=m, since it involves a division by zero. in the n=m case, i have:
    [tex]
    \begin{align}
    \int_{-T/2}^{T/2}\sin^2\frac{2\pi nt}{T}\! dt\\
    =\frac{1}{2}\int_{-T/2}^{T/2}dt-\frac{1}{2}\int_{-T/2}^{T/2}\cos\frac{4\pi nt}{T}dt\\
    =T/2
    \end{align}
    [/tex]

    so, now, armed with these integrals, we can do the problem.

    i am going to rename the dummy index n, in the infinite summation to m. you will see why.
    [tex]
    f(t) = \sum_m^\infty b_m \sin\frac{2\pi mt}{T}
    [/tex]

    then


    [tex]\begin{equation*}
    \begin{split}
    \int_{-T/2}^{T/2} f(t) \sin{n \omega t}\ dt &= \int_{-T/2}^{T/2} \left[ \sum_m^\infty b_m \sin\frac{2\pi mt}{T}\right] \sin\frac{2\pi nt}{T}dt\\
    &= \sum_m^\infty b_m\int_{-T/2}^{T/2} \sin\frac{2\pi nt}{T}\sin\frac{2\pi mt}{T}\ dt
    \end{split}
    \end{equation*}
    [/tex]

    this summation contains an infinite number of terms, one for every value of m, but only when m=n is the term nonzero, so i will drop the summation, and change m to n, so i am only considering one integral, and the value of that integral is T/2. this is the only nonzero term of the infinite summation.

    [tex]
    \int_{-T/2}^{T/2} f(t) \sin{n \omega t}\ dt =\frac{T}{2}b_n
    [/tex]
    or
    [tex]
    b_n=\frac{2}{T}\int_{-T/2}^{T/2} f(t) \sin{n \omega t}\ dt
    [/tex]

    and that s it. any questions?
     
    Last edited: Nov 29, 2003
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