# RMS bias calculation?

Discussion in 'General Electronics Chat' started by tom123, Oct 30, 2010.

1. ### tom123 Thread Starter New Member

Sep 30, 2006
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Vrms=Vpeak x 0.707.
Note Vpeak=Vpp / 2.
Is this only correct for a full sinusoidal about 0V (max. voltage = +Vpeak min. and min. voltage = -Vpeak)?

How is the is the Vrms of a full sinusoidal with a +10V dc offset (max. voltage = +20V and min. voltage = 0V)?

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Vrms_total=√(7.071^2+10^2)=√150=12.247V

The process would be

1. Remove the DC offset from the total waveform and calculate the RMS of the remaining AC sinusoidal component.
2. Total RMS=√(RMS AC component)^2+(DC offset)^2)

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3. ### tom123 Thread Starter New Member

Sep 30, 2006
6
0
Hi t_n_k

With the same wave form (AC sinusoidal 20Vpp, with 10v DC offset) but this time the wave form has been Half rectified.

So, applying the above (if I am correct in doing so):
rms of the AC sinusoidal component => Vpk / 2 (10/2 =5vrms)

Total RMS=√(RMS AC component)^2+(DC offset)^2) =>
√(5)^2+(10)^2) =√(25 +100) = 11.18Vrms (but the answer is 13.735Vrms).

Were is the mistake?

4. ### The Electrician AAC Fanatic!

Oct 9, 2007
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"Vrms=Vpeak x 0.707.
Note Vpeak=Vpp / 2."

Your forgot something; what has happened to the 0.707 factor in this new calculation?

You also asked "Is this only correct for a full sinusoidal about 0V (max. voltage = +Vpeak min. and min. voltage = -Vpeak)?"

Yes, the Vrms=Vpeak x 0.707 only applies to a sinusoidal waveshape with no DC component. Once a sine wave (without a DC component) wave is rectified, the wave shape is no longer "sinusoidal with no DC component".

However, your given waveform seems to me to be unchanged after being "half rectified" (possibly less a diode drop), depending on which way the diode is facing.

But, perhaps you mean that the original sine wave was "half rectified" before the offset was added. Can you post a plot of the waveform for the second question?

Have you learned to calculate the RMS value of any waveform?

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I've only ever applied the relationship to pure DC and sinusoids.

It would work if you derived the harmonic spectrum for the half-wave rectified part and used all the harmonic rms values.

In this case you'll need to do it the tried & trusted way.

$V_{rms}=\sqrt{\frac{1}{2 \pi}$\int^{\pi}_{0}(Vd+V_msin(\theta))^2 d\theta+\int^{2\pi}_{\pi}V_d^2 d\theta$}$

$V_{rms}=\sqrt{\frac{1}{2\pi}$\int^{\pi}_{0}(V_d^2+V_m^2 sin^2(\theta)+2V_dV_m sin(\theta)) d\theta+\int^{2 \pi}_{\pi}V_d^2 d\theta$$

$V_{rms}=\sqrt{\frac{1}{2 \pi}\{$V_d^2 \theta +\frac{V_m^2} {2 }\theta +4V_dV_m$^{\pi}_{0} +$V_d^2 \theta$^{2\pi}_{\pi}\}}$

$V_{rms}=\sqrt{(V_d^2+\frac{V_m^2}{4}+\frac{2V_dV_m}{\pi})}$

$V_{rms}=\sqrt{(100+25+\frac{200}{\pi})}$

$V_{rms}=\sqrt{188.662}=13.735$

Sorry Electrician - I didn't realise you had picked this up. I was in the middle of all the Latex stuff

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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My assumption was a waveform like this

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7. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I see that the waveform image you posted is the second option I mentioned, but when he said "With the same wave form (AC sinusoidal 20Vpp, with 10v DC offset) but this time the wave form has been Half rectified." I was puzzled.

If the words "wave form" are used to denote the same thing in that one sentence, then one would take the wave form he described for the first question and "half rectify" it. Since the first waveform was always positive, the rectification wouldn't change a thing. That didn't make sense to do that, so I guessed he must have meant the wave you showed. :-(

One could also just apply the fundamental RMS definition to the "rectified" waveform. I'll give it a go.

Edit:

Here's the calculation from first principles. Perform the integration over 2 cycles of the waveform. I allowed the sine waves to be 60Hz power line waveforms to show how to handle the issue of a different frequency.

• ###### RMScalc.png
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Last edited: Oct 31, 2010