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

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

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  1. Nov 18, 2011 #1
    Niero

    Thread Starter New Member

    Nov 11, 2009
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    "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.
     
    #1 Reply
  2. Nov 18, 2011 #2
    t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    789
    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
     
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    inamch270 likes this.
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