AC voltage averaging converter

The schematic diagram
shows RP1 and RP2S for adjusting the measurement accuracy (adjust before use).
Adjustment method:

1. Connect positive and negative power supplies, measure at point A with a multimeter, then adjust RP2, making the multimeter reading as close to 0 as possible; 2.
Connect positive and negative power supplies and a 1Vrms, 100Hz signal, measure at point A with a multimeter, then adjust RP1, making the multimeter reading as close to 1 as possible.

Principle analysis:
Break this down into two operational amplifier circuits, analyze them separately, and then superimpose at
point B: (Simplified)
Alternating current has positive and negative cycles
. Positive cycle: The upper diode conducts, the lower diode does not conduct. It is reverse amplified, with a magnification factor of -1, so the positive half-cycle is a waveform of equal magnitude but opposite phase to the input waveform. Negative
half-cycle: The upper diode does not conduct, the lower diode conducts. The voltage is zero.
That is: output a half-wave, and this half-wave is opposite to the input.
Point A: (simplified diagram)
Here, A and B are connected together by R4, which is not considered for now, and will be calculated when superimposed.
The blue text in the diagram is not considered here.
This is a reverse amplification, and the amplification factor is -1, so it is a waveform that is equal in size and opposite in phase to the input waveform.
The following is the waveform of point A at 1KHZ, 1V (blue line).
This waveform is the waveform with R4 and C2 disconnected. Disconnecting R4 means disconnecting the connection of the previous op-amp.
C2 is a filtering function. Disconnecting C2 is to better see the waveform
superposition analysis:
disconnect R7 and C2.
R7: AC power only enters from the first op-amp, disconnecting the connection between AC power and the second op-amp.
C2: Remove the filtering function. Disconnecting C2 is to better see the waveform.
From the previous analysis of point B, at this time, point B is a half-slope and opposite;
point B then goes to the second op-amp, forming a reverse amplification, with an amplification factor of -2;
that is to say, point B is amplified again by inversion, so after two inversions, this waveform becomes a positive half-slope and twice the size (-1*-2=2).
The waveform at point A (input blue line, output yellow line)
is shown below. Next, consider disconnecting only C2 and analyzing the waveform before filtering
: this is the superposition of the waveform shown above and the waveform analyzed at point A.
The inverted waveform superimposed with the double half-slope equals the full wave.
The waveform at point A (input blue line, output yellow line)
is shown below. This basically concludes the analysis. Finally, reconnect C2 for filtering.
This converts the 1Vrms, 100Hz AC power into 1V DC
power. (Physical demonstration)