Electronic circuit determination output voltage

Good evening I have an exercise to achieve, but I'm struggling to terminate the output voltage! I tried to isolate on one side v1 'of the first equation and v2' on the second equation.
Thank you in advance for your assistance

Equation 1: ((v1-v1 ') / R3) + (v2'-v1') / zc1 + (v0-v1') / (2R4) = 0
Equation 2: ((v2-v1 ') / R3) + ((v1'-v2') / zc1) + (v2 '/ (2R4)) = 0
V + = V- that is ((v2') / 2) = (v0 + v1') / (2)
Vo = v2 '- v1'
Vo being the output voltage

We must obtain at the end:
v0 = (- R4) / (R3 + R4) = (v1-v2) / (1 + f / fc) with fc = (R3 + R4) / (4.pi * R3 * R4 * C2)

Hi,

There are some different approaches.

The simplest method (for me) is.
* it is a difference amplifier (therefore the difference input voltage is: V2 - V1
* it's output is GND referenced (no need to add a constant voltage to the output)
* gain is : R4 / (R4 + R3)

Therefore Vo = ( V2 - V1) × ( R4 / (R3 + R4))

For random Opamp circuit calculations:
* first calculate the voltage at V+ input
* then decide that V- input has the same voltage, as V+, no Opamp input current
* with this information you can calculate the output voltage

Klaus

Hi,
You can arrive at the correct answer by applying superposition and Kirchoff's first law. That is: you calculate Vo(let's call this Vo1) by first considering V1 only (V2 grounded).

Than you calculate Vo(let's call this Vo2) considering V2 only (V1 grounded).

At the end to obtain the overall Vo, you just sum the two partial answers (Vo = Vo1 + Vo2).

Be careful, the superposition principle works only on linear circuits.

The set of ecuations considering V2 grounded is:

(V1-V1')/R3=(V1'-V2')/zC1+(V1'-V-)/R4
(V1'-V-)/R4+(Vo-V-)/R4=0
-V2'/R3+(V1'-V2')/zC1=(V2'-V+)/R4
(V2'-V+)/R4=V+/R4
V+=V- -> Operational amplifier in negative feedback configuration

by solving this ecuations you get the output voltage as a function of V1 (Vo1).

Next by grounding V1 and solving a similar set of ecuations you get Vo2.

Summing Vo1 and Vo2 you get the wanted answer.