Root-mean-square voltage of the waveform

In summary, the rootmeansquare voltage is the average voltage over the time it takes to complete one cycle.
  • #1
Mugged
104
0
given the below graph of the waveform, what is the rootmeansquare voltage.

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

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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  • #2
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
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
Mugged said:
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
ok, but what exactly is the mean voltage.

i got 2.3 volts, using mean voltage = 2..but that's not correct
 
  • #6
mugged said:
ok, but what exactly is the mean voltage.
mugged said:
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
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
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
ok..i got it lol
 

1. What is the root-mean-square voltage of a waveform?

The root-mean-square (rms) voltage of a waveform is a measure of the effective voltage of an alternating current (AC) or voltage signal. It is a mathematical calculation that takes into account both the magnitude and duration of the voltage peaks in a waveform.

2. How is the root-mean-square voltage calculated?

The root-mean-square voltage is calculated by taking the square root of the mean (average) of the squares of the voltage values in a given waveform. This calculation is done over a full cycle of the waveform and is expressed in units of volts (V).

3. What is the significance of the root-mean-square voltage in electrical engineering?

The root-mean-square voltage is an important measurement in electrical engineering because it represents the equivalent heating effect of an AC voltage compared to a DC voltage of the same magnitude. This is important for understanding the potential risks and safety considerations of different voltage levels.

4. How does the root-mean-square voltage relate to the peak voltage of a waveform?

The root-mean-square voltage is always lower than the peak voltage of a waveform. The ratio between the rms voltage and the peak voltage depends on the shape of the waveform. For a sinusoidal waveform, the rms voltage is equal to the peak voltage divided by the square root of 2.

5. Can the root-mean-square voltage be applied to non-periodic waveforms?

No, the concept of root-mean-square voltage only applies to periodic waveforms, meaning they repeat in a regular pattern. For non-periodic waveforms, other measures such as peak voltage or average voltage may be used instead.

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