I'm talking about this op-amp. In the solution it finds an expression for the transfer function using a virtual short circuit without considering the open loop gain \$ A_{vol} \$ at all. I wanted to approach it using the following expression which takes in consideration the open loop gain:$$ V_{u} = A_{vol}(V^+ - V^-)$$values for \$V^+\$ and \$V^-\$ are:$$V^+ = \frac{R_1C_1s}{R_1C_1s+1}V_s$$$$V^-=\frac{R_3C_2s+R_3}{R_3C_2s+1}V_u $$then I wanted to proceed plugging them into the first expression.
The solution however completely neglects \$ A_{vol} \$ and makes use of a virtual short circuit, so \$V^+ = V^-\$ and it proceeds from there to find the transfer function. I can't understand why it's being neglected, there are many other circuits where it's neglected and I don't know if it's just negligible for these circuits in my book, or if it doesn't matter when finding the transfer function.
I also have another question, how do I calculate gain? I'm assuming that once I have the transfer function I can calculate gain at low frequencies and as for higher frequencies I'd have to plot the Bode diagram since gain would be dependent on gain?
Thanks for the help.