Electronics > Beginners

Flat topping sine wave and RMS value?

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rjardina:
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?  :-//  |O

jlmoon:
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 you

Jon

David Hess:
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.

T3sl4co1l:
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

Benta:
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.

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