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I am working on a lab for school and did forgot to measure min and max of waveform to obtain the offset, but I do have the RMS and peak to peak values.

I tried using the following to calculate DC offset, but this does not agree at all with simulated results or waveform displayed in the screenshot using the divisions.

DC offset=Vpp/(2*sqrt(2))-Vrms

Is my logic off? Can this not be computed properly from a scope measured Vrms value?

enter image description here

Props to anyone that can deduce the circuit haha!

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  • \$\begingroup\$ Are you sure that the scope didn't actually filter the DC offset? Can you post a screenshot of the waveform? \$\endgroup\$ – AndrejaKo Oct 7 '11 at 8:07
  • \$\begingroup\$ Gnd should be lined up on the y axis...RMS of ch2 should be same as is the ch1 rms if DC offset was filtered? \$\endgroup\$ – user623879 Oct 7 '11 at 8:15
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Summary:

  • RMS powers can be added.
    So - RMS Volts^2 can be added.

  • Offset = Vdc = sqrt( Vrms_total_signal^2 - Vrms_ac_only^2)

    = sqrt (5.452^2 - 5.175^2) = 1.716 volt

(Voltages above are mean RMS voltages from supplied table. )

enter image description here


Calculating AC only RMS from Vpk-pk would give

  • Vrms_AC = (Vpk-pk)/2 x 1/sqrt(2) = Vpk-pk/2.sqrt(2)

    = 14.68/2/1.4142 = 5.190 V RMS.

The RMS voltage displayed is 5.175V mean suggesting that
the signal is not quite a pure sinewave
and/or that the measurement system is not 100.000% accurate (of course).

5.19 / 5.175 = 1.0029 or about 0.3% higher.


"RMS powers" add directly.
ie Power total = Power_ACV + Power_DC
Power is proportional to V^2 so
(Vrms_AC+DC)^2 = (Vrms_AC)^2 + (Vrms_DC)^2

as Vrms_DC = Vdc

Vrms_combined^2 = Vrms_AC^2 + Vdc^2

or Vdc^2 = Vrms_combined^2 - Vrnms_ac^2

So Voffset = Vdc = (Vrms_combined^2 - Vrma_ac^2)^ 0.5

Here
Vrms_ac = 5.175V average (from table)
Vrms_combined = 5.452 mean
So
Voffset = sqrt( 5.453^2 - 5.175^2) = 1.716 Volt.

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  • \$\begingroup\$ awesome, thx... \$\endgroup\$ – user623879 Oct 7 '11 at 16:48

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