Electronic – Oscillating active filter

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 αVout, where α is some scaling factor between 0 and 1. The T-network still acts as a voltage divider, dividing between Vin and αVout. 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

The first thing to do when looking to design a filter for a signal is to obtain the spectrum of the signal. This can be done in different ways. For example, you can measure or capture the signal with a scope. Most modern scopes can calculate an FFT of captured data and give a spectrum.

For a design though, it is nice to make an idealized waveform and then write the Fourier series of it. This way you get only the signal and no unwanted stuff, and so get a clean spectrum which will just be the coefficients of the harmonics. Then you can truncate or alter harmonic content (coefficient amplitudes) according to some filter type. Reconstruct a filtered signal from those altered coefficients and the filter effects will be evident.

From the comments of having a 5MHz carrier with ~1MHz BPSK, it is likely that many harmonics will be needed if a good representation of the original signal is desired. It wouldn't be surprising if 20 to 40 harmonics were needed for good representation. That would be a filter that started to roll off at > 20MHz, or maybe 40MHz. That's only 2 or 3 octaves from the 2 meter band (~150MHz). A filter with 1st order roll off isn't going to do that. You're going to need a filter with at least 3rd order performance for something like that. It's possible to get 2nd order performance out of a single stage Sallen-Key low pass filter.

As to the design shown in the schematic, it looks like the amplifiers are OK for what's being asked of them. LC filters are sensitive to termination. Best performance is when they are terminated into their characteristic impedance. The LC shown has \$Z_o\$ of 45 Ohms, and when used alone shows a resonant lobe of about 6dB. You can see the effects of impedance matching in case 4 when the LC is terminated into U2 with 100 Ohms. For a better match, R8 and R7 values could be halved, or L1 could be doubled and C5 could be halved.

Usually when active filters are used, inductors are not. It's because inductors are not needed to get complex poles and high Q roll offs, amplifier gain (and sometimes a little positive feedback, as in Sallen-Key) give that. Passive filters are good if the filter will be someplace where there is no bias voltage, or where frequencies and or filter order is high.