1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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]
    [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
    this doesn't feel right though, am i doing something wrong?
  5. Oct 28, 2013 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted