# Twin-T Active Notch Filter Analysis

Could anyone give me a hint in analyzing the Twin-T Active Notch Filter? I tried a delta-star transform, followed by nodal analysis, but ended up with conflicting equations. For an example, look at Figure 1 from the Texas Instruments application note "An audio circuit collection, part 2":

In the more general example I am studying, I remove C4/C5 and R6/R7 (and that Vcc) and treat the T passive components as matched conductances as follows:

R1 and R2 become Y1, R3 becomes 2Y1, C1 and C2 become Y2, C3 becomes 2Y2, R4 and R5 generic voltage divider with resistances R1 and R2

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This sounds like a question that dsp.stackexchange.com thinks should be on-topic there. What do others think? –  Kellenjb Sep 9 '11 at 17:02
@Kellenjb - It is on-topic here as well, but might get a better response there. If the OP or the DSP guys want it migrated, we can do that - it certainly could deal with a bit more attention. Alternatively, draw up a schematic and upload the image to bump this to the front page where it should get more attention....not sure how it got missed the first time around. –  Kevin Vermeer Oct 8 '11 at 21:09

The Delta-Star transform can be used to analyze the Twin-T network using the following procedure:

1. The two T networks can be converted into twin Delta networks in parallel:
2. Condense these two Delta networks into a single Delta network
3. Convert the resulting Delta network back into a T network.

4. To see the notch behavior of the passive twin T, assume node 2 is tied to ground, and treat the Delta network you got in step 3 as a voltage divider.

You'll find a transfer function of $$H(s) =\frac{s^2 + {\omega_0}^2}{s^2 + 4s\omega_0 + {\omega_0}^2}$$.

5. To see the effect of bootstrapping, assume that node 2 is held at a voltage α*Vout, where α is some scaling factor between 0 and 1. The T-network still acts as a voltage divider, dividing between Vin and α*Vout. To find the behavior of the system, we need to solve the equation $$v_\textrm{out} = \alpha \cdot v_\textrm{out} + H(s) ( v_\textrm{in} - \alpha\cdot v_\textrm{out} )$$, where $H(s)=Z_2/(Z_1 + Z_2)$ is the transfer function without feedback. Doing this, we find a new transfer function: $$G(s) = \frac{1}{(1-\alpha)\frac{1}{H(s)} + \alpha}$$. Note that for $\alpha=0$ (no feedback), we have $G(s)=H(s)$, as expected. For $\alpha=1$, the system becomes unstable. Plotting this function for values of alpha between 0 and 1, we find a huge increase in the Q of the notch.

The resulting transfer function is: $$G(s) =\frac{s^2 + {\omega_0}^2}{s^2 + 4s\omega_0(\alpha - 1) + {\omega_0}^2}$$.

Here's what the frequency response looks like, as the feedback gain $\alpha$ is changed:

The algebra of the various transforms is a bit tedious. I used Mathematica to do it:

(* Define the delta-star and star-delta transforms *)

deltaToStar[{z1_,z2_,z3_}]:={z2 z3, z1 z3, z1 z2}/(z1+z2+z3)
starToDelta[z_]:=1/deltaToStar[1/z]

(* Check the definition *)
deltaToStar[{Ra,Rb,Rc}]

(* Make sure these transforms are inverses of each other *)
starToDelta[deltaToStar[{z1,z2,z3}]]=={z1,z2,z3}//FullSimplify
deltaToStar[starToDelta[{z1,z2,z3}]]=={z1,z2,z3}//FullSimplify

(* Define impedance of a resistor and a capacitor *)
res[R_]:=R
cap[C_]:=1/(s C)

(* Convert the twin T's to twin Delta's *)
starToDelta[{res[R], cap[2C], res[R]}]//FullSimplify
starToDelta[{cap[C], res[R/2], cap[C]}]//FullSimplify

(* Combine in parallel *)
1/(1/% + 1/%%)//FullSimplify

(* Convert back to a T network *)
deltaToStar[%]//FullSimplify

starToVoltageDivider[z_]:=z[[2]]/(z[[1]]+z[[2]])
starToVoltageDivider[%%]//FullSimplify

% /. {s-> I ω, R ->  1/(ω0 C)} // FullSimplify

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The feedback circuit as shown in the TI datasheet shouldn't be that much more complicated, since the output is buffered by U1A and U1B then you could create a similar current source equivalent circuit; instead of R2 and C2 in my first diagram going to ground they would be connected to a voltage source with a value of $Vo*\alpha$, where alpha is the voltage division ratio.