fourier series of full wave rectifier.

Discussion in 'Homework Help' started by sandy92, Jul 29, 2010.

  1. sandy92

    Thread Starter New Member

    Jun 1, 2010
    1
    0
    Full wave rectifier:the function f(t)=AsinΩt.[​IMG]
    0 pi 2pi 3pi
    timeperiod T=pi.
    Ω=2pi/T=2pi/pi=2
    Thereforef(t)=Asin2t. if i tried with this function a0,an,bn is zero in fourier series.But if i tried with the function f(t)=Asint i get the following series:
    f(t)=2A/pi+4A/pisummation of(cos2nt/1-4npow(2))
    i.e.a0=2A/pi an=4a/pi(1-4nsquare)
    Which one is a right procedure?
     
  2. Fraser_Integration

    Member

    Nov 28, 2009
    142
    5
    Hi.

    The method of solution is to first state that the function can be described by:

    Asin(t) for 0 < t < pi
    and
    -Asin(t) for -pi < t < 0

    Therefore the function has a period of 2pi, so omega is 1.
     
    sandy92 likes this.
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