Vrms derivation: half wave

Thread Starter

makepeace

Joined Nov 10, 2012
1
Hi,

This is from an old thread:

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
I'm having a little bit of trouble with understanding the derivation for the Vrms of a half wave sinusoidal voltage.

As I see it, you cannot use Vp*sin(wt) as your V(t) if you are using 2pi/w as your period, because for half the period, V(t)=0.

The function V(t) is a piece function where:

V(t)=Vp*sin(wt) where 0 < t < pi/w
and
V(t)=0 where pi/w < t < 2pi/w

Could someone please explain why the V(t) = 0 part is neglected?

Thanks
 

The Electrician

Joined Oct 9, 2007
2,971
You'll notice that in the expression for full-wave, there is a factor of 2/T in front of the integral; for half-wave, there is a factor of 1/T in front of the integral. Since the value of the integral would be the same for the interval 0<t<T/2 as for T/2<t<T (by symmetry), t_n_K simply did the integration over half the interval and multiplied by 2 to take care of the full-wave case.
 
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