Complex Fourier series of full wave rectifier

Discussion in 'Homework Help' started by guitarguy12387, Oct 4, 2008.

  1. guitarguy12387

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    I have a full wave rectified sine wave. Basically, the graph looks like abs(sinx).

    When calculating the coefficients, i have tried evaluating:

    ∫[sin(t)exp(-jnt)]dt from 0 to pi.

    and am not getting the correct answer.

    I'm wondering if i need to just look for a computation error... or if I am setting the problem up wrong.

    I'm having doubts about if i'm allowed to setup the integral as such.

    Thanks.
     
  2. Mark44

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    The function you give, f(x) = |sin x|, is an even function, so its Fourier series consists only of cosine terms. An even function is one for which f(-x) = f(x), for all x in the domain of f.

    You can also find the Fourier coefficients using the approach you showed, and the integral is c_n = 1/(2pi)∫f(t) exp(-jnt) dt, integrated from -pi to pi.

    You need to take into account the fact that your function is equal to -sin(t) on the interval -pi to 0, so the integral above can be rewritten as
    c_n = 1/(2pi) ∫ -sin(t) exp(-jnt) dt + 1/(2pi) ∫ -sin(t) exp(-jnt) dt,
    with the limits of integration for first integral are -pi to 0, and for the second, 0 to pi.

    Mark
     
    Last edited: Oct 6, 2008
  3. guitarguy12387

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    ahhh ok. Thanks Mark.

    Won't that result in integrating over TWO periods of the function though? Is that allowed?
     
  4. guitarguy12387

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    Figured it out...

    Apparently i had the right approach, but i had the wrong argument in the exponential. Oops heh. Thanks again, Mark.
     
  5. Mark44

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    Good point.
     
  6. Mark44

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    Was the work that you actually did different from what you posted? I didn't notice anything obviously wrong in what you posted.
     
  7. guitarguy12387

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    Just the exponential... since ω = 2pi/T = 2, it should have been:

    ∫[sin(t)exp(-j2nt)]dt from 0 to pi
     
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