Gain of an active filter without calculating the transfer function with two OP-AMPs
I realize that I derived the transfer function of this bandpass filter in Transfer function of a bandpass filter without documenting how I got \$H_0\$. Let's consider the circuit in which I replaced the resistive divider made of \$R_4\$ and \$R_5\$ by a ratio \$k\$ equal to \$\frac{R_5}{R_5+R_4}\$. The schematic is below:



First, considering an open-loop gain \$A_{OL}\$, I can determine the voltage at the low-side op amp output: \$V_{o2}=V_{out}(k-1)A_{OL}\$. Then, by using superposition, I can get the voltage at node (2), the junction of \$R_1\$ and \$R_2\$: \$V_{(2)}=V_{in}\frac{R_2}{R_2+R_1}+V_{o2}\frac{R_1}{R_2+R_1}\$. Finally, the output voltage is the voltage at node (2) minus \$V_{out}\$ times the open-loop gain because no current flows through \$R_3\$ considering infinite input resistances for both op amps. Substitute and rearrange to obtain the definition of \$H_0\$:

\$H_0=\frac{R_2}{\frac{R_2}{A_{OL}}+\frac{R_1}{A_{OL}}+R_1+R_2+R_1A_{OL}(1-k)}\$

With a 100-dB (100k) open-loop gain for the op amps and a 1-V bias, the output is 64 µV as confirmed by simulation and Mathcad:



The dc gain in this case, when \$s=0\$ is -83.9 dB. As \$A_{OL}\$ approaches infinity, the output voltage is 0 V. A tricky little circuit! : )