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The formula for voltage RMS is usually given as the peak value divided by the square root of 2. I feel like maybe I am missing something: how can two AC signals with same amplitude but different frequencies have the same RMS value?

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    \$\begingroup\$ Does the peak value change with frequency? Does the value of root 2 change with frequency? But I understood RMS meant square, then mean then root... \$\endgroup\$
    – Solar Mike
    Feb 20 at 14:42
  • \$\begingroup\$ @SolarMike assuming the peak values remain static for both signals but frequency is different. Do they have the same RMS value? \$\endgroup\$ Feb 20 at 14:52
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    \$\begingroup\$ how can two AC signals with same amplitude but different frequencies have the same RMS value? Yes, you're missing something. With questions like this, it's hard to figure out what you're missing because you don't say why you expect there to be a difference. For two sinusoidal signals of the same peak amplitude, how could they have different RMS values? (They can't, assuming you average over a whole number of periods for both of them, i.e. the long-term average. Are you thinking about time-scales shorter than the period? Or something else?) \$\endgroup\$ Feb 22 at 5:42
  • \$\begingroup\$ @PeterCordes if you sample the mean squares over 1 period for example (for each AC signal respectively). I was thinking mostly about the mathematical formula for RMS which includes frequency/period ... \$\endgroup\$ Feb 22 at 12:28
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    \$\begingroup\$ @AbdelAleem When you mention "the mathematical formula for RMS" what exactly do you mean? \$\endgroup\$
    – MikeB
    Feb 22 at 15:11

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You can think of the RMS voltage as related to the power that the voltage can deliver into a resistive load.

A 240VAC 50Hz mains can supply just as much power as a 240VAC 60Hz (or 400Hz) mains to a resistive load.

The \$\sqrt{2}\$ factor is specifically for a sinusoidal waveform. A waveform that is not sinusoidal will, in general, have a different factor.

But even if the waveform is different, the RMS value of a periodic signal is independent of the frequency when it is calculated over an integral number of cycles. The power supplied to a resistive load varies throughout the cycle, of course, and the energy delivered to the load per cycle is less at higher frequencies, but when you average the energy per unit time (which is power) you get the same number regardless of frequency, all other things being equal.

Here is a simulation of sine waves feeding 1Ω resistors. One is 60Hz and one is 120Hz. Both voltages are 10V peak sine waves (~7.071V RMS). The top pane shows the voltages plotted over 100ms. The bottom pane shows the power delivered to the resistors over time.

enter image description here

In both cases the power averaged over an integral number of cycles (in this case, I used 100ms) is exactly 50W, which is \$V^2/R\$ if you use the RMS voltage for V.

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For any periodic signal, the RMS voltage is independent of frequency. The formula that you are referring to is a particular case when the signal in question is sinusoidal.

To understand the reason two AC signals with different frequencies can have the same RMS value, we have to look at the definition of "RMS" or Root-Mean-Square. This definition literally means to take the squareROOT of the MEAN of the instantaneous voltage values SQUARED. There is no frequency terms involved in this definition; only the square of the instantaneous voltage values integrated across one period. An identical signal only differing by frequency would have a one-to-one correspondence of instantaneous squared voltage values. The only difference being the period is shorter for the faster frequency signal. The formula looks like this: SQRT[(MEAN(integral(V^2(t)))] integrated over one period, or t=0 to t=2*pi

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  • \$\begingroup\$ It doesn't have to be periodic. RMS is a fine measure of random noise. \$\endgroup\$
    – John Doty
    Feb 20 at 15:43
  • \$\begingroup\$ @JohnDoty Agreed. But based on the context of the question, I was trying to keep the answer simple. Comparing noise sources is much more esoteric than comparing sinusoidals. \$\endgroup\$
    – MOSFET
    Feb 20 at 15:50
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Does RMS value of AC signal depend on frequency?

No, but the time taken to calculate RMS has to be longer for a lower frequency for it to be an effective comparison.

how can two AC signals with same amplitude but different frequencies have the same RMS value?

A 1 volt p-p sinewave of 1 kHz has exactly the same RMS value as it does when 2 kHz.

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In simplistic terms, the RMS AC voltage is equivalent to the DC voltage value that produces the same amount of power dissipation in a resistor.

In mathematical terms:

$$ V_{RMS} = \sqrt{{1 \over {\theta_2-\theta_1}} \int_{\theta_1}^{\theta_2} \left(f(x) \right)^2\;dx} $$

For \$f(x) = sin(x)\$, if \$ \theta_2 - \theta_1\$ is a integer multiple of \$\pi\$ (half cycle) the RMS voltage will be \$1 \over \sqrt{2}\$.
Notice that frequency isn't part of the calculation, but angle is.

If the data is sampled:

$$ V_{RMS} = \sqrt{{1 \over n} \sum_{i=1}^{n}x_i^2} $$

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There are some rigorous mathematically-based answers here; I'll offer my intuitive answer.

Periodic signals spend more time at every voltage as they get slower; the extension of dwell time cancels out the frequency change when calculating a mean. For RMS, which is a special kind of mean, frequency is irrelevant, as long as you perform the measurement for a long enough time to gather data (either exactly one half of a cycle or so many cycles that also including a fractional part of an extra cycle isn't significant).

This is a theoretical consideration, however; don't confuse this with real-world quantities such as reactive or apparent power, which are very much frequency-dependent. While frequency "doesn't matter" to RMS, it does matter in AC circuits for other measurements or calculations. The fact that RMS remains the same regardless of frequency does not mean that frequency doesn't change power delivery or dissipation.

Another "gotcha" is that depending upon the implementation of a meter, it may be affected by waveform shape and frequency, so you've got to understand how your equipment works and if it's applicable to the signal you're measuring.

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