Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier Representation of Simple Half-Wave Rectifier

  1. Nov 4, 2004 #1


    User Avatar

    I, in fact, know the correct Fourier representation
    for the following (it was given to me):

    [tex]f(t)=0 \text { if } -\pi \leq \omega t \leq 0[/tex]


    [tex]f(t)=sin(\omega t) \text { if } 0 \leq \omega t \leq \pi [/tex]

    [tex] \hrule [/tex]

    I'm curious about the derivation that led to it -- specifically how the coefficients were derived.

    I know, in general...

    [tex]A_0=\frac{1}{2\pi} \int_{-\pi}^{\pi}f(x)dx[/tex]

    [tex]A_N=\frac{1}{\pi} \int_{-\pi}^{\pi}f(x)cos(nx)dx[/tex]

    [tex]B_N=\frac{1}{\pi} \int_{-\pi}^{\pi}f(x)sin(nx)dx[/tex]

    ... but am stuck when it comes to setting-up the
    integrals (substitution rules, how integrals might be broken-up into sub-integrals, intervals, etc.)

  2. jcsd
  3. Nov 5, 2004 #2


    User Avatar
    Science Advisor

    You understand that multipication by that f(x) is equivalent to simply changing the limits of integration from [-pi, pi] to [0, pi] right.

    For the integral of sin(x) sin(nx) and sin(x) cos(nx) just use integration by parts (twice) using sin(x) = d/dx(-cos x) etc, this gets you the original integral on both LHS and RHS which makes it easy to solve.

    For example with the sine series integral this gets you something like I = n^2 I, (where I is the integral you're trying to find). This of course tells you that either n=1 or I = 0 and indeed the sine series coefficient are all zero except for the first one (n=1). The exact same procedure will work for the cosine coeficients and this time give a non trivial series.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Fourier Representation of Simple Half-Wave Rectifier