Deriviation of RMS value of full wave and half wave rectifier signal

Discussion in 'Homework Help' started by Niero, Nov 18, 2011.

  1. Niero

    Thread Starter New Member

    Nov 11, 2009
    1
    0
    "Deriviation of RMS value of full wave and half wave rectifier signal ???"

    i do a couple of google search and just can't find an answer

    waiting for response and :) any answers will be appreciated thankz in advance.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    The traditional math approach ....

    For the full-wave

    V_{rms}=\sqr{\frac{2}{T}\int_0^{\frac{T}{2}}{(V_msin(\omega t))}^2}dt=\sqr{\frac{2}{T}\int_0^{\frac{T}{2}}{V_m}^2sin^2(\omega t)}dt

    For the half-wave


    V_{rms}=\sqr{\frac{1}{T}\int_0^{\frac{T}{2}}{(V_msin(\omega  t))}^2}dt=\sqr{\frac{1}{T}\int_0^{\frac{T}{2}}{V_m}^2sin^2(\omega  t)}dt

    \omega=\frac{2\pi}{T}

    Use

    sin^2(\omega t)=\frac{1-cos(2\omega t)}{2} to do the integration.

    \int \frac{1-cos(2\omega t)}{2}dt=\frac{t}{2}-\frac{sin(2\omega t)}{4\omega}

    You should try to do some of the math work yourself to prove

    Vrms[full-wave]=Vm/√2
    Vrm[half-wave]=Vm/2
     
    inamch270 likes this.
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