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Homework Help: Root-mean-square voltage of the waveform

  1. May 3, 2010 #1
    given the below graph of the waveform, what is the rootmeansquare voltage.

    [PLAIN]http://img41.imageshack.us/img41/5296/voltage.png [Broken]

    Please help, i have no idea how to do this. Note that the waveform repeats itself every 3 seconds. thank you
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 3, 2010 #2

    diazona

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    What's the formula for RMS voltage? Or if you don't know the formula, what do you know about how, in general, one calculates an RMS voltage?
     
  4. May 3, 2010 #3
    i think its like the average voltage. so i would integrate to get the areas of those triangles and divide the period..which is 3 seconds, but i would get zero since the triangles negate themselves. but 0 is not the correct answer.

    i have no idea honestly, an answer will suffice though, thanks.
     
  5. May 3, 2010 #4

    diazona

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    No it will not, because you won't learn anything if you don't work through the problem yourself.

    The formula for root mean square voltage is
    [tex]V_\text{RMS} = \sqrt{\int (V - \bar{V})^2 \mathrm{d}t}[/tex]
    It's just like standard deviation, except for a continuous signal: you subtract the mean voltage, square it, integrate, and take the square root. Try that and see what you get.
     
  6. May 3, 2010 #5
    ok, but what exactly is the mean voltage.

    i got 2.3 volts, using mean voltage = 2..but thats not correct
     
  7. May 3, 2010 #6

    diazona

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    . . .
     
  8. May 3, 2010 #7
    with mean voltage = 0 i come up with 3.26599 volts when i integrate the first and second triangles separately, add those two values, and root them using your formula. this is also wrong - dude can you just show me please
     
  9. May 3, 2010 #8

    diazona

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    oh wait - just realized I left out a part of the formula, sorry about that. This should be correct.
    [tex]V_\text{RMS} = \sqrt{\frac{1}{\Delta t}\int_t^{t+\Delta t} (V - \bar{V})^2 \mathrm{d}t}[/tex]
     
  10. May 3, 2010 #9
    ok..i got it lol
     
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