# RMS Derivation of Piecewise Sinusiodal Function

1. Apr 19, 2005

### j0shha1nes

Calculate the root, mean, square of v(t) where
v(t) = 0 0 < wt <= theta
v(t) = Vm*sin(wt) theta < wt <= pi()
v(t) = 0 pi() < wt <= pi()+theta
v(t) = Vm*sin(wt) pi()+theta < wt <= 2*pi()

In integral form, I am confident that it looks like this.

Vrms = SQRT(1/T*INTEGRAL(Vm^2*sin^2(wt)dt)) from theta/w to pi/w AND pi+theta/w to 2*pi/w

I believe it integrates like this. My thought was that I could integrate only one of the ranges and then double my answer due to symmetry in the sinusoidal function.

Vrms = Vm*SQRT(w/2*pi*(t/2-sin(2wt)/4w)) evaluated from theta/w to pi/w

After substitions and simplification (which I have been over about 10-15 times) I get this.

Vrms = Vm - Vm*SQRT(1/pi*(theta-sin(theta)*cos(theta)))

I know this is wrong because when I evaluate this equation at theta = pi, the value is zero. The correct value is Vrms = Vm/SQRT(2) when theta = pi. I have either defined the problem wrong somehow or I have performed the integral wrong.

Many thanks for any help.

2. Apr 20, 2005

### OlderDan

It appears to me that when theta = pi you should get zero. You have extended the domain of the V(t) = 0 function to the entire interval. Vrms = Vm/SQRT(2) should be the result for theta = 0. Your result does not appear to be giving that, but I'll let you check it. Check also at theta = pi/2. At that point your integral should be half the result for theta = 0 (assuming I'm right above), so Vrms = Vm/2.

3. Apr 21, 2005