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RMS of Fullwave rectified sine wave.

  1. Feb 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine the RMS value of fullwave rectified sine wave.

    2. Relevant equations
    RMS = [tex]\sqrt{({1}/{b-a})\int^{b}_{a}[(fx)]^{2}dx}[/tex]

    3. The attempt at a solution

    Notes: The Period of a full wave rectified sine wave is pi.


    Let's do square root at the end.

    =[tex]1/pi \int^{pi}_{0}sin^{2}xdx[/tex]

    =1/pi [pi/2 - [sin(2pi)]/4] - 1/pi [pi/2 - [sin(2pi)]/4]

    =1/2 - 1/2 ? ? ? ? ? ?

    Where am i going wrong? Sorry I am not good with latex code, even with the reference.
    Last edited: Feb 16, 2008
  2. jcsd
  3. Feb 16, 2008 #2
    So you used the double angle identity.

    [tex]sin^2(x) =\frac{1}{2}(1-cos(2x))[/tex]

    [tex]\int_0^{\pi} \frac{1}{2}(1-cos(2x)) = \frac{\pi}{2} - \left[sin(2x)/4 \right]^{x=\pi}_{x=0}[/tex]

    [tex]\int_0^{\pi} \frac{1}{2}(1-cos(2x)) = \frac{\pi}{2}-(0)[/tex]

    I'm not sure what you were doing.
  4. Feb 16, 2008 #3
    I used integral of sin^2(u) du is = u/2 - [sin(2u)]/4 + C

    When i sub pi in the term sin2u , then this becomes sin2pi which is 0.
  5. Feb 16, 2008 #4
    Right so you have


    RMS = 1/√2
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