Electronics > Beginners

Flat topping sine wave and RMS value?

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Benta:

--- Quote from: macboy on April 01, 2019, 02:52:27 pm ---
--- Quote from: Benta on March 31, 2019, 06:37:54 pm ---The RMS value is a mathematical calculation to determine the power of a voltage/current dissipated in a resistive load, regardless of the wave form.
The basis of the root-mean-square is the area under your wave form.

To obtain the RMS value, you need to integrate the wave form over a half cycle to calculate the area. This is what your Fluke will do for you.

For instance, the RMS value of a 50% duty cycle square wave will be equal to the peak voltage. You're somewhere in between a sine and a square wave.

--- End quote ---

This is very wrong. RMS is not related to area (integral). Integrating then dividing by time can give average, but not RMS.

RMS is literally the square-Root of the Mean(average) of the Square of the input.


--- End quote ---

You missed the point about power into a resistive load. RMS is equal to the DC voltage/current that will give the same dissipation in the load.
The integral is the mean, which you'll need to compute first for an arbitrary wave form, the root/square part is obvious.

macboy:

--- Quote from: Benta on April 01, 2019, 04:10:20 pm ---
--- Quote from: macboy on April 01, 2019, 02:52:27 pm ---
--- Quote from: Benta on March 31, 2019, 06:37:54 pm ---The RMS value is a mathematical calculation to determine the power of a voltage/current dissipated in a resistive load, regardless of the wave form.
The basis of the root-mean-square is the area under your wave form.

To obtain the RMS value, you need to integrate the wave form over a half cycle to calculate the area. This is what your Fluke will do for you.

For instance, the RMS value of a 50% duty cycle square wave will be equal to the peak voltage. You're somewhere in between a sine and a square wave.

--- End quote ---

This is very wrong. RMS is not related to area (integral). Integrating then dividing by time can give average, but not RMS.

RMS is literally the square-Root of the Mean(average) of the Square of the input.


--- End quote ---

You missed the point about power into a resistive load. RMS is equal to the DC voltage/current that will give the same dissipation in the load.

--- End quote ---
You were absolutely correct on that point. I should have noted that, instead of implying that your entire statement was inaccurate.


--- Quote from: Benta on April 01, 2019, 04:10:20 pm ---
The integral is the mean, which you'll need to compute first for an arbitrary wave form, the root/square part is obvious.

--- End quote ---
The key is that you do not compute the mean of the voltage waveform (nor integrate voltage waveform). Instead, you compute the mean of the Voltage squared. The 'squared' part is critical since power is not proportional to voltage, but to voltage squared. So, first square the voltage, then compute mean/average, then take square root.

Consider two signals: one is a DC voltage of 1 VDC. The other is a signal that toggles on/off at some rate, with 50 % duty cycle, and a peak voltage of 2 V (off voltage 0 V). If you integrate both for 1 second, you get 1 V*second. The area under the voltage curve is the same for either, which also implies that the average voltage for either is 1 V. This in intuitive.
The RMS voltage of the 1 VDC signal is 1 Vrms.
The RMS voltage of the 50% positive pulse is 1.4142 Vrms. It will drive twice the power into a resistive load (P=V2/R). This is not intuitive, but consider that a 2 VDC signal will drive 4x as much power than 1 VDC, by P=V2/R. Now consider that at 50% duty cycle, you halve the power to only 2x as much as 1 VDC. Manipulating P=V2/R for a constant R and two voltages V1 and V2, we can get P2/P1=V22/V12. With P2/P1 = 2, then V2 = V1*1.4142.

The "area under the voltage curve" idea gives average voltage but not RMS.

For more technical information on the non-trivial nature of measuring and calculating RMS voltage, I'd encourage you to read:
https://www.analog.com/media/en/technical-documentation/application-notes/AN-268.pdf
and
https://www.analog.com/en/education/education-library/rms-to-dc-application-guide.html

A side note:
All digital voltmeters (or ADCs) which use a dual slope or multi-slope technique do in fact actually integrate the input voltage over some period of time (usually called the "integration time" for obvious reasons) then de-integrate with a known fixed reference, back to zero, in order to measure the integrated charge. The de-integration time is proportional to the input voltage. This is results in an average voltage over the integration period, and provides effective noise reduction on a DC input, or gives the average level of a (rectified) AC input voltage.

ArthurDent:
Forgetting all the math and theory, the RMS value of an A.C. waveform is equal to a D.C. value when the heating effects of the two are equal. Fluke, Holt, HP, NIST, and others made thermal transfer standards that consisted of a low mass thermocouple on a heater wire in a vacuum. Here is a description and a diagram below.

"NIST uses thermal converters as “transfer” standards to calibrate AC voltage and current. In simplified terms, a known DC voltage is applied to a thermal converter’s heater and the thermocouple voltage is read. Then, an AC voltage is connected to the heater and adjusted for the same thermocouple voltage. Thus, the RMS value of the AC voltage is equal to the original (calibrating) DC voltage. In a sense, the DC voltage is transferred to the AC voltage."

That quote is from this link that gives a pretty complete explanation of the transfer process. I still have a Holt 6A thermal transfer standard and a set of different rated heater/thermocouples that would work as well today as they did almost 50 years ago.

https://www.nutsvolts.com/magazine/article/the-ac-volt

Nominal Animal:
Diverging to pure math:

If you have a sine wave that would range from \$-V_{p}\$ to \$V_{p}\$ but is clipped to \$\pm x V_{p}\$, it has a RMS voltage of approximately
$$ V_{RMS}(x) = \left ( x - 0.179 x^2 - 0.218 x^3 + 0.341 x^4 - 0.235 x^5 \right ) V_{p} $$
As you'd expect, \$V_{RMS}(0) = 0\$ with slope 1, \$V_{RMS}(1) \approx 0.7071\$ with slope 0; it resembles the first quarter of a sine wave scaled by 0.7071.

For example, if you have a sine wave with \$V_{p} = 5\$ (would range from -5 to +5), but it is clipped at ±4 (so \$x = 0.8\$), \$ V_{RMS} \approx 3.18\$.

In general, mathematically
$$ V_{RMS} = \sqrt{ \frac{1}{T} \int_0^T \left( v(t) \right)^2 dt } $$
by definition, where the integral is over one full period (from 0 to \$T\$).  Of course, in electronics this only applies if the current has the exact same waveform and phase as the voltage, for the reasons ArthurDent described.

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