Hello guys, I'm having some trouble with solving the next circuit (was simplified from a more complicated one with extra resistors). I'm a bit rusty with analog devices and would be thankful for some help how to solve that circuit. Hopefully I'd be able to solve the more complicated one by myself after solving the simplified one
My main issue is how to deal with the amplifier itself (let's assume it's ideal, if it helps solving it) and using the phasors. As I said, a bit rusty..

The circuit includes 2 SINE inputs, and 2 Vout outputs.
Known parameters: capacitors C1 to C6 and the inputs.
Need to find the relation between the total output voltage (VoutP-VoutN) and the parameters above.

 

The extra resistors were there so one can calculate time constants (R*C) and the delays needed for filters. Without resistors, the caps charge and discharge instantly.

Also, the op amps will lock up over time (seconds to minutes) because there is no path to ground for the inputs because there is no current flow through a capacitor - most likely if the op amps are jFet inputs with extraordinarily high I put impedence.

 

The extra resistors were there so one can calculate time constants (R*C) and the delays needed for filters. Without resistors, the caps charge and discharge instantly.

Also, the op amps will lock up over time (seconds to minutes) because there is no path to ground for the inputs because there is no current flow through a capacitor - most likely if the op amps are jFet inputs with extraordinarily high I put impedence.

My bad, GopherT. Here is the complete circuit. Does it make more sense now and possible to solve? All capacitors and resistors are given.

 

Superposition.

Replace one source with ground, solve for the other. Then switch, and then add the two results.

 

My bad, GopherT. Here is the complete circuit. Does it make more sense now and possible to solve? All capacitors and resistors are given.
View attachment 121386

In the schematic of post #3, there is still no DC path from the amplifier inputs to ground.

 

My bad, GopherT. Here is the complete circuit. Does it make more sense now and possible to solve? All capacitors and resistors are given.
View attachment 121386

The voltage in each sub circuit that include each input can be virtually any value. Capacitive coupling transfers a signal in response to the RATE OF CHANGE of the voltage and is independent of the actual DC offset voltage.

Think about DC bias of a transistor's base voltage and an AC coupled signal input to that biased point. Your circuit is similar. You need a resistor network to ground in order to calculate the voltage. Voltages that do not connect to ground are called "floating". That untethered (floating) circuit segment has no place in a differential amplifier. There is no answer to your circuit.

 

Guys thanks for the support.
If I do superposition, let's assume I have the next circuit.
In that case I do have a resistor network to ground (R3+R4, isn't that?).

So how can I continue solving that "half" circuit? Vu is an AC input.
My main problem/question is how do deal with the amplifier itself? Don't I miss it's amplifying factor somewhere? What are the voltages in its inputs? Can I assume 0 on both amplifier inputs since it's ideal one? If so, what is the voltage difference it's supposed to amplify?
AS you can see, I'm totally in the dark in here

 

Me too. What are you trying to do?

Are you just looking for a math solution?

 

Me too. What are you trying to do?

Are you just looking for a math solution?

Yeah, I'm looking for a math solution to describe VoutP and VoutN as functions of the resistors, the capacitors and the input voltages

 

May I ask where you got the circuit? What is it's function?

 

May I ask where you got the circuit? What is it's function?

To amplify the voltage, based on the differences between the capacitors. Basically, if all capacitors are equal and so are the resistors, the gain should be 1.
Any idea how to get the equations?

 

So......we are going to measure the gain to compare capacitors?

 

So......we are going to measure the gain to compare capacitors?

Basically yes