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

Thread Starter

Niero

Joined Nov 11, 2009
1
"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.
 

t_n_k

Joined Mar 6, 2009
5,447
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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