Root-mean-square voltage of the waveform

  • Thread starter Mugged
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  • #1
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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
 
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Answers and Replies

  • #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?
 
  • #3
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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.
 
  • #4
diazona
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i have no idea honestly, an answer will suffice though, thanks.
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.
 
  • #5
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ok, but what exactly is the mean voltage.

i got 2.3 volts, using mean voltage = 2..but thats not correct
 
  • #6
diazona
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ok, but what exactly is the mean voltage.
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.
. . .
 
  • #7
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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
 
  • #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]
 
  • #9
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ok..i got it lol
 

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