Hi, guys.
Could somebody help me out with the subject? I've searched the net, but didn't find anything useful.
Link to circuit: http://www-k.ext.ti.com/SRVS/Data/ti/KnowledgeBases/analog/document/faqs/spttbphq.gif

 

Hello,

Take a look at the PDF in the link.
http://www.drp.fmph.uniba.sk/ESM/twin.pdf

Greetings,
Bertus

 

Assume an ideal op amp so you can say that the voltage on the non-inverting terminal (V2) equals the voltage on the inverting terminal (V1). The voltage are measured with respect to ground. First find the transfer function (relationship) between Vin and V2. Then find the transfer function between Vo and V1. Finally, because V1=V2, substitute V1 with the equation relating V2 with Vin to get the transfer function between Vo and Vin.

 

Hi, guys.
Could somebody help me out with the subject? I've searched the net, but didn't find anything useful.
Link to circuit: http://www-k.ext.ti.com/SRVS/Data/ti/KnowledgeBases/analog/document/faqs/spttbphq.gif

What do you need to know about this circuit topology?

hgmjr

 

What do you need to know about this circuit topology?

hgmjr

Judging from the subject line, I think he wants you to produce one of those lovely detailed transfer function derivations that you do so well.

But, he hasn't shown any of his own work yet.

 

Thanks, bertus, for the document. The twin-T filter without an op-amp is a band-stop filter (see fig. 4).
Denote
Vout - output of whle circuit.
Vin_pas - input of passive twin-T filter
V(-) - input of inverting input of op-amp.
V(+) - input of non-inverting input of op-amp
Vin - input of the whole circuit
Vin_pas = V(-) = V (+) = Vin*k, where k = R5/(R4+R5).
Vout/Vin_pas - band-stop filter transfer funct. and because Vin_pas = Vin*k,
Vout/Vin also band-stop filter transfer funct., but actually the circuit if band-pass filter. How to take an op-amp into account? Where am I wrong?

 

Are you still interested in deriving the overall transfer function for the opamp plus twin-T feedback network?

Are you required to use any particular method to derive it, or can you use any method you like?

This thread is several months old, so I thought I would ask if your interest level is still high.

 

I'm still very interested in deriving the overall transfer function for the opamp plus twin-T feedback network. Any method is fine.

 

You said "How to take an op-amp into account?". Does this mean that you don't want to make the usual assumption that the opamp is ideal, with infinite gain?

 

No, assume op-amp as ideal. I ment how to derive transfer function for whole circuit, including op-amp, because without op-amp it's a notch filter, but with it it's band pass filter.

 

Let V+ and V- be the voltages at the + and - inputs of the opamp.

Let f(ω) be the transfer function of the twin T network as found in the link Bertus gave you in post #2.

We'll ignore the effect of Cin.

The voltage V+ is given by V+ = Vin * R5/(R4+R5).

The voltage V- is given by V- = Vout * f(ω).

Since the opamp is ideal, V- = V+, so that Vout * f(ω) = Vin * R5/(R4+R5).

From this we have Vout/Vin = (R5/(R4+R5))/f(ω)

 

Thanks, The Electritian, now it's clear. The idea is that Vout is actally input of passive twin-T filter and V- is output.

 

Suppose the transfer function of a bandpass filter is
A*(1+Bs)/(1+Cs+Ds*s), where
s=j*ω,
A-D are constants.
How to find
f0 - mid-frequency,
K0 - module of transfer function at mid frequency
Q - filter quality?

 

The prototypical response of a standard bandpass filter is:

A*(Bs)/(1+Cs+Ds*s)

not:

A*(1+Bs)/(1+Cs+Ds*s)

Then K0 is A*B/C

I think you can find the other two items by searching on the web. For example, on this page: http://en.wikipedia.org/wiki/Q_factor

about halfway down the page, see the expression for H(s).

 

Unfortunately I have A*(1+Bs)/(1+Cs+Ds*s), and I haven't mistaken, because it matches the one in book. They have also given equations for K0, ω0 and Q, so I'm wondering how they got them. I've printed a graph |K(ω)| and it really is a bandpass filter, though it's not symmetric with respect to ω0, it has bigger values at lower frequencies than at higher frequencies.

 

Unfortunately I have A*(1+Bs)/(1+Cs+Ds*s), and I haven't mistaken, because it matches the one in book. They have also given equations for K0, ω0 and Q, so I'm wondering how they got them. I've printed a graph |K(ω)| and it really is a bandpass filter, though it's not symmetric with respect to ω0, it has bigger values at lower frequencies than at higher frequencies.

I'm not suggesting you're mistaken; I'm just saying that that response is not a standard bandpass response. You can tell that by the unsymmetrical response. You will just have to decide what frequency is the midband frequency by some method that takes into account the asymmetry.

The usual method to determine Q makes use of the 3 dB down frequencies on either side of the midband frequency.

The (1+Bs) numerator is why the twin T feedback network isn't usually used to get a bandpass response.