The last two stages of operational amplifier circuits are the same, both of which are used as "followers" and are only used for signal shaping. The final output is the same as the original input, that is, Udc=Uo. UO- output of the first stage operational amplifier.

Let the potential of GA be UA and that of GN be UB.

For the first stage operational amplifier, first look at the noninverting input:

According to the "virtual break", the current at the "3" terminal is "0", so the potential at this time is: U3 = UA×10/(8× 510+10) = UA/409.

According to the concept of "virtual short", U2=U3=UA/409.

In this operational amplifier circuit, C 1 and C2 play a filtering role, and D 1 and D2 are bidirectional amplitude limiting, which plays a protective role and does not affect the operation result.

For the inverting input terminal, the current at the "2" terminal is "0", and according to KCL:

(u b-U2)/(8×5 10)=(U2-Uo)/ 10 .

Simplification: UB-U2=408U2-408Uo, 408Uo=409U2-UB.

And: 409U2=409×UA/409=UA.

So: 408Uo=UA-UB, Uo=(UA-UB)/408=(GA-GN)/408.

So: Udc=(GA-GN)/408.