Electronic – How to find the transfer function of this op amp circuit

I'll use the opamp's pin numbers to name voltages.

\$ \begin{cases} V_6 = \dfrac{R3}{R3+R4} (V_O - V_{IN}) + V_{IN} \\ \\ \\ V_O - V_2 = I_{R2} R2 \\ \\ \\ I_{R2} = j \omega \text{ } C1 (V_2 - V_5) \end{cases}\$

The opamps will regulate their output so that \$ V_6 = V_5\$ and \$ V_2 = V_{REF}\$. Then, solving for \$V_O\$:

\$ V_O = \dfrac{(R3 + R4)(1 + j \omega \text{ } C1 R2) \text{ } V_{REF} + j \omega \text{ } C1 R2 (2 R3 + R4) \text{ } V_{IN}}{(R3 + R4) + j \omega \text{ } C1 R2 R3} \$

or

\$ V_O = \dfrac{j \omega \text{ } C1 R2 (2 R3 + R4)}{(R3 + R4) + j \omega \text{ } C1 R2 R3} V_{IN} +\dfrac{(R3 + R4)(1 + j \omega \text{ } C1 R2)}{(R3 + R4) + j \omega \text{ } C1 R2 R3} V_{REF} \$

Filling in the component values:

\$ V_O = \dfrac{j \omega \text{ } 4600 \text{ } \Omega s}{44500 \text{ } \Omega + j \omega \text{ } 150 \text{ } \Omega s} V_{IN} +\dfrac{44500 \text{ } \Omega + j \omega \text{ } 4450 \text{ } \Omega s}{44500 \text{ } \Omega + j \omega \text{ } 150 \text{ } \Omega s} V_{REF} \$

Some oscillators, particularly crystal VCOs, produce a sine wave or clipped sine wave output. For example, these.

You would not likely get good results using a digital phase detector on an analog signal, so you would likely want to use a comparator to turn it into a digital signal.