# 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
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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.