Hi,
This is from an old thread:
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
This is from an old thread:
I'm having a little bit of trouble with understanding the derivation for the Vrms of a half wave sinusoidal voltage.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
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