I have two questions:
1-Why in one T the resistance is half the other resistances in the second T.
2- How would adding a voltage follower increases Q?
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The double-T RC circuit has a third-order transfer function. However, if the grounded C (lets call it C3) is identical to (C1+C2) and the grounded R (lets call it R3) is R3=(R1||R2) we have a so-called pole-zero cancellation which results in second-order filter with the desired bandstop characteristic:
wp=wz=1/RC and H(w=0)=H(w infinite).
Note that the mentioned conditions can be fulfilled for a variety of R and C combinations.
C1=C and C2=C/kc and C3=C(1+1/kc)
R1=R and R2=kr and R3=R * kr/(1+kr).
For a "tuned" double-T network we require kr=kc=k. For this case, the corresponding transfer function reveals that Qp=k/[2(1+k)].
We can show that Qp,max=0.5 for very large k values; for the simple case (as in the given circuit) with k=1 we have Qp=0.25.
This a very bad value (bad selectivity). For a much larger Q we can modify the circuit and connect R3 not to ground but to the output of a fixed gain positive amplifier having a gain A.
That means: We use positive feedback for increasing the selectivity (larger Q). Just adding a voltage follower does not have any positive influence on Q. It only provides decoupling between the filter and the load.
EDIT: The most simple solution for increasing Qp is to select A=1 (voltage follower) and to use different k factors: kc=1 and kr>1.
In this case: wp=wz=1/[RC*SQRT(kr)] and Qp=0.5*SQRT (kr)
1) because if you would R instead of R/2 in the top branch, the RC timeconstant would be different from the bottom branch. Same as there is 2 x C in the top branch but 2 C in the bottom branch.
3) Because without the follower the impedance Rl would load the filter and change its behaviour. Example: what would happen without the voltage follower so Rl is directly at the output and Rl has a value of R/10 or even R/10000. What would happen ?
1) The voltage at positive and negative terminals are almost same. In this case it is vout. Assume the input source vin=0. Now analyse the upper arm T circuit. The net capacitance seen by the resistance R/2 is , 2C (both capacitors are parallel). Time constant if this upper T part is (R/2*2C) = RC.
Consider the lower T. Net resistance seen by the capacitor with capacitance 2C is R/2 ( Same reason as said above - the resistors are in parallel). Again time constant is (2C*R/2)=RC.
Both upper and lower T has same time constant. Symmetrical contribution in charging and discharging rate.
2) voltage follower is not just where vout = vin. Remember that voltage follower has large input impedance. If you directly connect Rl(load resistance), you will have lot changes. You may try that by yourself
Other folk are explaining the 2C thing so I'll concentrate on: -
How would adding a voltage follower increases Q
I believe you may be thinking of the bootstrapped twin T notch filter. This can improve Q dramatically: -
Concentrate on where the output of U1b connects - it now connects to the junction of the "2 C" capacitor and the "0.5 R" resistor (formerly connected to ground). Increasing Q has this effect on the frequency response: -
This type of filter is commonly used to notch out 50 or 60 Hz AC pick-up on audio signals. It has minimal effects on the audio outside of a few hertz either side of the notch frequency. Here's a 60 Hz design taken from an old National semiconductor data sheet by the looks of it: -