# Finding Q for wien bridge notch filter

Discussion in 'Homework Help' started by onebird, Nov 1, 2010.

1. ### onebird Thread Starter New Member

Mar 27, 2010
16
1
I am currently working on the design of a 2nd order wien bridge notch filter whose general circuit is shown in the attached file. I derived the transfer function to be:
H(s)=-[(R/(3R-3))(s^2+(1/RC)^2)]/[s^2+(3/RC)s+(1/RC)^2}

from this I could deduce that my passband gain is -R/(3R-3)

and also that 2∏f=(1/RC)

However, I am tempted to think that my rejection quality,Q, is = (RC/3) from the term in s in the denominator but I doubt if my assumptions are correct.
If anyone can confirm this, I would be very grateful. I was could not find anything on the internet which talked about a formula for rejection quality,Q, for a wien bridge notch filter though a lot of websites promised that it was simple to make

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I think your H(s) is almost correct except for the R/(3R-3) part.

One obvious check is ask what is the DC Gain of the circuit - when 's'=0 that is.

It should have (minus) unity gain i.e Av(DC)=-1.

I think it should just be

$H(s)=-\frac{s^2+\frac{1}{R^2C^2}}{s^2+\frac{3}{RC}s + \frac{1}{R^2C^2}}$

I'll have to give the Q factor some further thought. Maybe another member will have a direct answer.

An initial observation for the denominator is that it corresponds to an over-damped case.

If the second order form is s^2+2as+ω^2

then a=3/(2RC) and ω^2=1/(R^2C^2)

a^2=9/(4R^2C^2)

So a^2>ω^2 - making it an over-damped case.

Choosing different (rather than equal) C values will presumably have an effect on damping.

3. ### onebird Thread Starter New Member

Mar 27, 2010
16
1
Very very late reply, I must apologise.
When I first wrote this thread I was not completely sure about what the design was going to involve. So for the sake of anyone who might be doing something similar. The tranfer function given by tnk is correct assuming that the potential divider at the non-inverting input from the input and from the output has resistors in the ratio 1:2. that is, only a third of the voltage from the input and the output is fed to the non-inverting input.

However this circuit would have a shallow notch at the desired frequency. This is because the transfer function has conjugate ZEROs which lie on the imaginary axis(critical damping).

By making all the resistors equal, the transfer function will have a numerator in the form of
s^2+(1/RC)s+(1/RC)^2
simple mathematics would show that this is an overdamped case of the general equation.
This way the notch is sharper and deeper. But Q is generally around 0.3(from practical experiments)

Ordinarily text books would reserve such analyses for the poles only. But it is through working with this particular filter that I have obseved this. My post is open to correction