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Fourier Series of Full Wave Rectifier

  1. Oct 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine the fourier series for the full-wave rectifier defined as
    f(t) = sinωt for 0 < ωt < pi
    -sinωt for -pi < ωt < 0


    2. Relevant equations



    3. The attempt at a solution
    This looks like an even function, so bm = 0
    Ao = 1/pi∫sinωt from 0 to pi
    = 1/pi(-cos(ωt))/ω) from 0 to pi
    = 2/piω

    An = 2/pi∫sin(ωt)cos(nt) from 0 to pi (because the function is even)
    =2/pi∫(0.5(sin(ωt-nt)+0.5(sin(ωt+nt)) from 0 to pi
    =-1/pi(cos(ωt-nt)/(ω-n) + cos(nt+ωt)/(n+ω)) from 0 to pi
    = -(cos(pi(ω-n))-1)/(n+ω) -(cos(pi(ω+n))-1)/(pi(n+ω))

    I'm stuck at this part. I don't know how to simplify those or what that equals to and I've been looking around for a very long time trying to figure it out...any help?
     
    Last edited: Oct 28, 2013
  2. jcsd
  3. Oct 28, 2013 #2

    vanhees71

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    First of all you should set [itex]\tau=\omega t[/itex]. Then your calculations become right. Further you must specify with respect to which integral you integrate. Of course you are right, that it's an even function and thus it's a pure cosine series.

    Then you have
    [tex]a_n=\frac{1}{\pi} \int_{-\pi}^{\pi} \mathrm{d} \tau f(\tau) \cos(n \tau)=\frac{2}{\pi} \int_0^{\pi} \mathrm{d} \tau \sin \tau \cos(n \tau)[/tex]
    and
    [tex]f(\tau)=\frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(n \tau).[/tex]
     
  4. Oct 28, 2013 #3
    ok, using that way I got

    Ao = 2/pi

    and An = -(cos(pi(1-n))-1)/pi(1-n) -(cos(pi(1+n))-1)/(pi(1+n))
    my problem with this part is knowing what the cosine part equals out to. it seems to oscillate back and fourth from -1 to +1 as n increases
    so it becomes
    -(((-1)^n+1)-1)/pi(1-n)-(((-1)^n+1)-1)/pi(1+n)
    this doesn't feel right though, am i doing something wrong?
     
  5. Oct 28, 2013 #4

    vanhees71

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    Your coefficients are correct. Simplified and written in LaTeX makes it a bit better readable:
    [tex]a_n=\frac{2}{\pi(1-n^2)}[1+\cos(n \pi)].[/tex]
    There's only an apparent problem at [itex]n=1[/itex], but a direct calculation of the integral gives [itex]a_1=0[/itex]. Thus you have for [itex]k \in \mathbb{N}_0[/itex]
    [tex]a_{2k}=\frac{4}{\pi(1-4 k^2)}, \quad a_{2k+1}=0.[/tex]
    The Fourier series thus is
    [tex]f(\tau)=\frac{2}{\pi} + \sum_{k=1}^{\infty} \frac{4}{1-4k^2} \cos(2 k \tau)[/tex]
    or substituting back [itex]\tau=\omega t[/itex]:
    [tex]f(t)=\frac{2}{\pi} + \sum_{k=1}^{\infty} \frac{4}{1-4k^2} \cos(2 k \omega t).[/tex]
     
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