Transfer Function of Notch Filter
To find the transfer function of a circuit you need to be able to convert the circuit to a frequency model with the Laplace transform.

Capacitors become \$ Z = \frac{1}{Cs} \$

Inductors become \$ Z =Ls \$

Resistors become \$ Z = R \$

Then you can apply circuit theory and reduce the circuit just like they were resistors.

The high and low pass section treated like a resistor divider  \$ Z = \frac{Z_{bot}}{Z_{top} + Z_{bot}} \$
(Zbot is the circuit element connected to ground)

With the starting values of the schematic I posted the low pass is: \$  \frac{R2}{R2 + \frac{1}{C2*s}}=\frac{s}{1 + R2*C2*s}\$ 

With the starting values of the schematic I posted the high pass is: \$  \frac{\frac{1}{C1*s}}{\frac{1}{C1*s}+R1}=\frac{s}{s+\frac{1}{R1*C1}}\$ 

If you want to further simplify this circuit the parralell resistor rule can be used:

\$ R_{total} =\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}\$

\$ Z_{total} =\frac{1}{\frac{1}{Z_1}+\frac{1}{Z_2}}\$

Z1 would be the lowpass and Z2 would be the high pass

\$ Z_{total} =\frac{1}{\frac{1}{\frac{s}{s+\frac{1}{R1*C1}}}+\frac{1}{\frac{s}{s+\frac{1}{R1*C1}}}}=\frac{C1*C2*R1*R2*s}{C1*R1 + C2*R2 + 2*C1*C2*R1*R2*s}\$

then you can substitute \$ s=j\omega = j2\pi f \$ to find the frequency parameters you want. 







simulate this circuit &ndash; Schematic created using CircuitLab