# Transfer Function of Notch Filter

I'm looking at designs for notch filters for an audio system. I want to filter out 10kHz and attenuate by 40dB. I found this topology online: I'm unfamiliar with transfer function calculations for something this complicated. Could somebody walk me through this please?

I believe that once I have the transfer function I can figure out what the values are.

• Here's a link with the transfer function & a pair of calculators: sim.okawa-denshi.jp/en/TwinTCRkeisan.htm Mar 8 '16 at 4:31
• Note your schematic does not show what we usually call a twin-T filter. Mar 8 '16 at 5:14

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 – Schematic created using CircuitLab