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Finding Voltage As A Function of Peak Voltage

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data

    triangle_wave_rms_avg.gif

    http://www.rfcafe.com/references/electrical/triangle-wave-voltage-conversion.htm

    I'm trying to understand how this website figured out that V = Vpk*(2/pi)*theta

    2. Relevant equations

    Vpk*(2/pi)*theta

    3. The attempt at a solution


    The only thing that comes to mind is that they're using the formula for the area of triangle. This doesn't seem to be the case though because:

    b = pi/2 h = Vpk

    A = (Vpk*pi)/4
     
  2. jcsd
  3. Feb 22, 2012 #2

    cepheid

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    The red part is a straight line. In other words, V varies linearly with θ. That means it can be expressed it the form V = mθ + b where m is the slope of the line, and b is its y-intercept. We find b by finding the value of V when θ = 0. Since the plot clearly passes through the origin (0,0), it follows that 0 = m*0 + b. Or, 0 = b.

    That leaves the equation in the form V = mθ, where m is the slope. So all we need to do is find the slope of the red line. We can do that just by looking at it. Slope = rise/run. In this case, the rise is Vpk, because the line rises from V = 0 to V = Vpk. It does so over a horizontal distance of π/2, (since the line goes from θ = 0 to θ = π/2). So the run is π/2, and hence m = rise/run = Vpk/(π/2) = (2Vpk)/π. Substituting in this value for m, the equation of the line becomes:

    V = mθ = [(2Vpk)/π

    That's all there is to it. Really simple.
     
  4. Feb 28, 2012 #3
    Late response, but thanks tons for the response. I had spent HOURS trying to make sense of this; I think this is my lesson to reach out for help a little more often....
     
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