Flat topping sine wave and RMS value? - Page 1
Forum section: electronics / beginners

author: rjardina
author_rank: Contributor
timestamp: March 31, 2019, 12:55:46 pm
content: I tried googling and found a few things that explains this but can't seem to rap my head around it.(T)RMS = Root Mean Square, A mathematical equation.I thought a multimeter was reading the voltage and taking random-ish samples, and then throwing those samples in the RMS equation and spitting it out on the display? (in layman terms)Been reading about Crest Factor and a pure sine wave would be 1.414 (square root of 2). Once you flat top the sine wave it effects the Crest Factor.But once the sine wave has flat topped, why doesn't it effect the RMS value? How much flat topping would it take to effect the RMS value?The attached picture is of 482 RMS? and the scope shows about 611 peak. Under a prefect Crest Factor 480RMS is 678 peak.That missing / flat topping of 67-ish volts doesn't effect the RMS?Why am I not seeing at least some voltage drop in the RMS value?

author: jlmoon
author_rank: Unknown
timestamp: March 31, 2019, 03:08:16 pm
content: Morning,Hope I can shed some light on the subject as I perceive it. When the P-P signal is flat topping your not experiencing any voltage change relative to time. So if you try to include that part of the waveform into the RMS equation you get no change. The root mean square is based on the square root of a rotating value ie: 2pi is a complete circle right?, so imagine the part of a full cycle where the amplitude is not changing, what do you get?Plot it out on a X-Y chart. The RMS requires a constant change in voltage with respect to time, in other words a pure sine-wave. One other tip, square waves are the mathematical equivalent to the sum of a sine wave at that same frequency, plus an infinite series of odd-multiple frequency sine waves at diminishing amplitudes. Hope this helps youJon

author: David Hess
author_rank: Super Contributor
timestamp: March 31, 2019, 03:23:53 pm
content: For a constant peak voltage, as a sine wave is clipped the RMS value increases and the crest factor decreases up to the point where it becomes a square wave where the peak and RMS values are the same and the crest factor is 1.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 31, 2019, 03:50:29 pm
content: What are you measuring against? Is the wave becoming flattened as a result of series impedance and rectification? Are you comparing to the initial voltage? Or just the peak to RMS at a given condition?Tim

author: Benta
author_rank: Super Contributor
timestamp: March 31, 2019, 06:37:54 pm
content: 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.

author: rjardina
author_rank: Contributor
timestamp: March 31, 2019, 09:30:10 pm
content: Quote from: T3sl4co1l on March 31, 2019, 03:50:29 pmIt's a test bench for testing VFD modules. It's feeding the DC bus, the VFD is running with no load. This is the output of the transformer before rectificationQuote from: Benta on March 31, 2019, 06:37:54 pmQuote from: David Hess on March 31, 2019, 03:23:53 pmI will play with the scope and DMM again later this week. Can't right now. My mind is more mechanical in nature. But how I'm understanding it right now.Flat topping doesn't just effect the top of the waveform. but the rest of the sine wave too. The nice smooth part that connects the flat topping are more straight, and more vertical. More square-ishThe raise and the fall voltage have a quicker time base and the AREA is still the same as a pure sine wave just oddly shaped now?The peak voltage as dropped but the wave has gotten fatter making up for the lost area from flat topping?Is this correct or an okay way to view this?

author: rjardina
author_rank: Contributor
timestamp: March 31, 2019, 09:38:45 pm
content: Quote from: Benta on March 31, 2019, 06:37:54 pmHave no idea why I over read this the first time.thanks

author: vk6zgo
author_rank: Super Contributor
timestamp: April 01, 2019, 12:06:01 am
content: Quote from: Benta on March 31, 2019, 06:37:54 pmA true 50% duty cycle square wave returns to zero volts for 50% of the time.It you take samples of the voltage at regular intervals across a cycle, square the values, take the "mean" of all those samples, then find the square root of that, it willnotbe equal to the peak voltage, because half of the samples will be at zero volts.If, however, you have a "sort of" square wave which iscentredon zero volts, with each half cycle different polarities (in a similar manner to a sine wave),--- yes, it will be equal to the peak value.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 01, 2019, 07:27:07 am
content: Quote from: vk6zgo on April 01, 2019, 12:06:01 amNo one defines such an unqualified signal as having a nonzero DC component.QuoteIf we wish to calculate the RMS of such a wave, however, we can simply use Parseval's theorem. Namely, for a biased square wave ("pulsed DC") of maximum V and minimum 0, the DC component is V/2 and the RMS of the AC component is V/2. The RMS of these is sqrt((V/2)^2 + (V/2)^2) = V * sqrt(2)/2.Tim

author: macboy
author_rank: Super Contributor
timestamp: April 01, 2019, 02:52:27 pm
content: Quote from: Benta on March 31, 2019, 06:37:54 pmThis 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 theMean(average) of theSquare of the input.In some multimeters, the input is sampled at "high" speed (kHz to MHz range) and the RMS is calculated. In many multimeters, an analog computation is performed to 'square' the input, average the result, and take the square root of that (or an equivalent using anti-log and log). The DC output of this analog computation is then measured to give the RMS voltage. Analog Devices AD636/637 chips are common for this, and their documentation and app notes describe the process nicely.rjardina,Note that while your waveform might be flattened, it also seems to hold at that peak value for some time, when it should be dropping. This will increase the RMS value.

author: Benta
author_rank: Super Contributor
timestamp: April 01, 2019, 04:10:20 pm
content: Quote from: macboy on April 01, 2019, 02:52:27 pmYou missed the point about power into aresistiveload. 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.

author: macboy
author_rank: Super Contributor
timestamp: April 01, 2019, 05:04:12 pm
content: Quote from: Benta on April 01, 2019, 04:10:20 pmYou 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 pmThe key is that you do not compute themean of the voltagewaveform (nor integrate voltage waveform). Instead, you compute themean of the Voltage squared. The 'squared' part is critical since power is not proportional to voltage, but to voltage squared. So,firstsquare the voltage,thencompute 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 V1and 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.pdfandhttps://www.analog.com/en/education/education-library/rms-to-dc-application-guide.htmlA 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.

author: ArthurDent
author_rank: Super Contributor
timestamp: April 01, 2019, 06:08:06 pm
content: 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

author: Nominal Animal
author_rank: Super Contributor
timestamp: April 02, 2019, 08:28:15 am
content: 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.