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RMS Derivation of Piecewise Sinusiodal Function

  1. Apr 19, 2005 #1
    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. jcsd
  3. Apr 20, 2005 #2

    OlderDan

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    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.
     
  4. Apr 21, 2005 #3
    Thanks for the reply.

    Not sure what you mean when you say that the domain of v(t) = 0 has been extended ?to the entire interval? (confusing part between the question marks). v(t) equal zero only for a fraction of the first half cycle (and second half cycle) of the T period (0 to 2pi radians).

    I will check some of the other quadrants that you mentioned. I have not done that yet. Here they are.

    Vrms(theta=pi/2) = Vm*(1-1/sqrt(2))
    Vrms(theta=3pi/2) = Vm*(1/sqrt(2)-1)

    One thing that I keep running into that I do not understand is that the RMS value over the entire period of an alternating signal should not be zero. I expect it to be an increasing function, not linearly, but always increasing as theta varied from 0 to 2pi radians. I would expect this even when there are equal positive and negative area(s) under the curve. Do you agree?
     
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