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I'm confused about the Vrms value of a clipped rectified sine:

enter image description here

The reason it's clipped and distorted, is because it's the output from a tiny transformer (0.7W), and the filter bank charges at the peaks of the waveform, pulling down the voltage.

Anyway, it measures as 3.88V peak, yet only has 1.39V RMS. I don't understand that. If it were a sine with the same peak of 3.88V, Vrms would be 2.74V. If anything, I expect Vrms to be higher than that (like driving an amplifier into clipping increases power to the load).

Can this really be the correct Vrms value?

The scope itself seems to behave OK; sines, quares, triangles with and without DC offset measure correctly. Edit: actually, it doesn't... It doesn't include DC offset in the Vrms value.

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    \$\begingroup\$ You're swapping peak (amplitude) and peak-to-peak values. The cursors measure peak-to-peak, which is double of the sine amplitude. RMS values are often calculated from amplitude. \$\endgroup\$ – Hans Oct 10 '15 at 14:16
  • \$\begingroup\$ @Hans The middle of this wave is not 0V. But, I see now that the RMS calculation of the scope actually doesn't consider the DC offset. If I take a function generator and change DC offset, Vrms stays the same. Seems a bug in the scope to me. \$\endgroup\$ – Halfgaar Oct 10 '15 at 14:25
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As Hans mentioned in the comments, your scope might be ignoring the DC offset of your signal.

You described that your signal has a peak-to-peak voltage of \$3.88~V\$. This is an amplitude (from the center) of $$ A = \frac{3.88~V}{2} = 1.94~V $$ so if your signal was a perfect sine wave, it would have an RMS amplitude of $$ A_{RMS} = \frac{A}{\sqrt{2}} = \frac{1.94~V}{\sqrt{2}} = 1.38~V $$ This is almost exactly what your scope has measured, so I think it's right.

Note: to add a DC bias to an RMS level, you can use the relationship $$ A_{shifted} = \sqrt{A_{RMS}^2 + A_{DC}^2} $$ In your case, this would be $$ A_{shifted} = \sqrt{(1.39~V)^2 + (1.94~V)^2} = 2.38~V $$

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