# Fliege Notch Filter Transfer Function Help

#### jac9762

Joined Apr 23, 2022
5
Hello! I am a student who would greatly appreciate some help determining the transfer function T(s) of a Fliege Notch Filter. I have attached a general schematic below, and have listed the values I will be using.

To clarify, Voltage-in is the left side, and Voltage-out is the right side

R1 = R2 = 1M Ohms
R3 = R4 = 5k Ohms
R5 = R6 = 100k Ohms
C1 = C2 = 0.53 uC

This is one of a few steps in my circuit, and the last Transfer function T(s) I need to calculate to get the complete function and create my last Bode Plots on MATLAB. If anyone is willing to help either a generalized T(s) or an exact one with my values would be awesome, thank you in advance!

#### The Electrician

Joined Oct 9, 2007
2,887
Does your problem assume that the opamps are ideal?

You must show your own effort to determine the transfer function. We won't just do the work for you.

#### jac9762

Joined Apr 23, 2022
5
Yes I am assuming the op amps are ideal, I am sorry I forgot to mention that.

My current attempt to derive the Transfer function was to divide it into two parts T(s)1 = Vout1/Vin and T(s)2 = Vout2/Vout1. The first part, which I think I have correct, would be separating C1, R1, and R2 into their own T(s)1. Vout1 being the potential difference between the pre-defined ground pole, and the + input of the U2 opamp, and Vin being the original analog input on the left. Vout2 would be the original output on the right.

Defining C1 and R1 as Z1, I determined the impedance to be Z1 = (R1/[s*C1]) / (R1 + (1/[s*C1]))

Next I define R2 as Z2 and R2 = R1; therefore, Z2 = R1.

Vout1 equals the voltage across Z2. Therefore, Vout1 = I(s)*Z2.

Vin = (Z1+Z2)*I(s). Therefore, I(s) = Vin/(Z1+Z2)

Substitute backwards, and we get Vout1 = (Vin*Z2)/(Z2+Z1). Therefore, Vout1/Vin = T(s)1 = Z2/(Z2+Z1)

Substitute everything back in for Z1 and Z2 and you get: T(s)1 = R1/(R1 + [(R1/[s*C1]) / (R1 + (1/[s*C1]))]).

If I am not mistaken, the total T(s) of this circuit would be the product of T(s)1 and T(s)2. I unfortunately am unsure how to determine T(s)2.

I hope this helps, and once again thank you for your time!

#### The Electrician

Joined Oct 9, 2007
2,887
It would be easier to help you if you would annotate the schematic with things like Vin, Vout1, Vout2, I(s), rather than using text descriptions of where on the schematic these things exist.

Also, your derivation so far for Vout1 fails to take into account R3.

#### jac9762

Joined Apr 23, 2022
5
Vout1 is just my own personal attempt to derive the answer, it could be completely wrong.

I(s) is a general term for current in the frequency phase, I derived it from a V(s)= I(s)*Z - I apologize if I was unclear.

Also, R3 runs from the U2 + Input to the U3 positive output, so I do not think be included in what I defined as Vout1, but for all I know I am completely wrong.

My work could completely wrong, which is why I am seeking assistance - I just thought I would show the work I currently have as requested.

Unfortunately I am a student who does not specialize in EE or signal processing. I am sorry I could not be more helpful. I am attempting to derive the transfer function T(s) and am stumped.

My handwriting is really bad, and I do not have any sort of touch screen to draw on. But I can try to make a follow-up virtual image if that is helpful, I just cannot promise my hand drawn schematic will be much better. As for right now, the screenshot I have is the best representation of the circuit in question with the Analog Input being the left most node labeled, and the Analog Output being the right most node labeled, both sharing the same ground with the common ground signal.

#### The Electrician

Joined Oct 9, 2007
2,887
Vout1 is just my own personal attempt to derive the answer, it could be completely wrong.

I(s) is a general term for current in the frequency phase, I derived it from a V(s)= I(s)*Z - I apologize if I was unclear.

Also, R3 runs from the U2 + Input to the U3 positive output, so I do not think be included in what I defined as Vout1, but for all I know I am completely wrong.

My work could completely wrong, which is why I am seeking assistance - I just thought I would show the work I currently have as requested.

Unfortunately I am a student who does not specialize in EE or signal processing. I am sorry I could not be more helpful. I am attempting to derive the transfer function T(s) and am stumped.

My handwriting is really bad, and I do not have any sort of touch screen to draw on. But I can try to make a follow-up virtual image if that is helpful, I just cannot promise my hand drawn schematic will be much better. As for right now, the screenshot I have is the best representation of the circuit in question with the Analog Input being the left most node labeled, and the Analog Output being the right most node labeled, both sharing the same ground with the common ground signal.
Because R3 goes from the junction of Z1 and Z2 to the output of U3, it is a load on that junction and the voltage there is not just due to the voltage divider composed of Z1 and Z2.

Are you attending a class in a University, or is this online learning? Haven't you been taught how to derive transfer functions?

You could derive the transfer function by just solving the network using nodal analysis; do you know how to do that?

#### jac9762

Joined Apr 23, 2022
5
I am sorry, I came here to get assistance. If I knew how to easily solve the network using nodal analysis, I would not have asked for help. My project involves medicine, and so does my degree. Immunology is my speciality, not electrical engineering, which is why I came here for assistance.

If it is very simple, would you please mind assisting me? I tried my best and showed my work.

#### jac9762

Joined Apr 23, 2022
5
I greatly appreciate your time, but would also appreciate a little bit more respect. If I knew how to easily perform this doing nodal analysis, I would not have wanted to waste anyones time. I attempted to do this to the best of my ability teaching myself online and was unsuccessful. As I said, I am not an electrical engineer, so I cannot solve this network using nodal analysis.

If it's very simple to solve the network using nodal analysis for an EE, the solution or any sort of explanation would be appreciated rather than unconstructive criticism. Thank you again for your time.

#### RBR1317

Joined Nov 13, 2010
691
There is no easy way to derive the transfer function of a Fliege filter (even using nodal analysis). However, a TI application note on designing notch filters (attached) may be all you need....

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#### The Electrician

Joined Oct 9, 2007
2,887
I greatly appreciate your time, but would also appreciate a little bit more respect. If I knew how to easily perform this doing nodal analysis, I would not have wanted to waste anyones time. I attempted to do this to the best of my ability teaching myself online and was unsuccessful. As I said, I am not an electrical engineer, so I cannot solve this network using nodal analysis.

If it's very simple to solve the network using nodal analysis for an EE, the solution or any sort of explanation would be appreciated rather than unconstructive criticism. Thank you again for your time.
You misconstrue my questions to be disrespectful and critical. I am trying to gauge your skill and knowledge level with respect to the finding of transfer functions. I don't want to waste my time and your time explaining things you already know, nor do I want to fail to explain something you don't know.

You say "teaching myself online". If this is not actual homework that you will be graded on, then the help the forum rules allow me to give you is different. Is the finding of the transfer function a task that will be graded on as part of a formal class, or are you self teaching?

Sometimes students come here with a problem that hasn't been covered in class or in the text.

You said " If I knew how to easily perform this doing nodal analysis, I would not have wanted to waste anyones time". That doesn't necessarily follow; you might have just learned nodal analysis without having learned how it can be used to find transfer functions. Do you know how to do a nodal analysis of a circuit like you have here? I ask not to offend you, but to determine what I would need to do to help you.

#### LvW

Joined Jun 13, 2013
1,438
jac9762 - just two questions from my side:
1) How did you arrive at the parts values without having the transfer function?
2.) Are you interested to see the transfer function only or do you need some help to derive the function step by step?
_________________________
Here are some hints:
* The Fliege-notch is based on a symmetrical operated GIC-block (Generalized Impedance Converter). The transfer function of the whole circuit can be found by using superposition of the two output signals caused by the signal voltages at both non.inv. opamp input terminals.
* The input impedances Zin1 and Zin2 at both non-inverting opamp input terminals are known (as a function) and can be found in many books and papers.
* Therefore, the signal voltages at the two non-inv. input nodes can be found by simply applying the rules of voltage division between (a) R1||C1 and R2||Zin1 and (b) R6 and Zin2.
* Finally, there is a relatively simple relation (inverse voltage division) between the opamp output voltage and the corresponding signal voltage at the non-inv. opamp input terminal. As a background, we must remember that (for ideal opamps) the signal voltages between both opamp input terminals can be set to zero. Hence, all 4 opamp input nodes have the same voltage with respect to ground.
* For convenience, it is recommended to use transconductances Y=sC (resp. 1/R) instead of impedances Z=1/sC (resp. R).

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#### The Electrician

Joined Oct 9, 2007
2,887
There is no easy way to derive the transfer function of a Fliege filter (even using nodal analysis). However, a TI application note on designing notch filters (attached) may be all you need....
The homework help rules allow us to show the details of finding the transfer function of a similar filter which the RS can then apply to the filter of post #1. A suitable filter for this purpose would be the low pass version of the Fliege: I don't think the TS expects it to be easy, so @RBR1317 , you could show him the technique you used here: https://forum.allaboutcircuits.com/threads/transfer-function-derivation-1.161258/post-1410758

Maybe the technique described by @LvW is easier, and he could demonstrate it as an alternative.

If we post again to the thread, the TS will get a notification and maybe he will return to find what he seeks. #### RBR1317

Joined Nov 13, 2010
691
I don't think the TS expects it to be easy, so @RBR1317 , you could show him the technique you used here:
The same procedure works:
1. Write the Node Equations
2. Solve for V(-) & V(+) at Op Amp input
3. Set V(-)=V(+) & solve for Vo
4. Divide Vo by Vi to solve the for transfer function

Why would you think that a procedure meant for analyzing a single op-amp circuit would work for a multiple op-amp Fliege configuration? Of course, there are similarities: op-amp inputs are all at the same voltage (for ideal op-amps, Vb1 = Vb2 = Vb3; otherwise the output voltages would be infinite). It would seem one needs to find Vred in terms Vin & Vout. That could be difficult, but not impossible. #### The Electrician

Joined Oct 9, 2007
2,887
jac9762 - just two questions from my side:
1) How did you arrive at the parts values without having the transfer function?
2.) Are you interested to see the transfer function only or do you need some help to derive the function step by step?
_________________________
Here are some hints:
* The Fliege-notch is based on a symmetrical operated GIC-block (Generalized Impedance Converter). The transfer function of the whole circuit can be found by using superposition of the two output signals caused by the signal voltages at both non.inv. opamp input terminals.
* The input impedances Zin1 and Zin2 at both non-inverting opamp input terminals are known (as a function) and can be found in many books and papers.
* Therefore, the signal voltages at the two non-inv. input nodes can be found by simply applying the rules of voltage division between (a) R1||C1 and R2||Zin1 and (b) R6 and Zin2.
* Finally, there is a relatively simple relation (inverse voltage division) between the opamp output voltage and the corresponding signal voltage at the non-inv. opamp input terminal. As a background, we must remember that (for ideal opamps) the signal voltages between both opamp input terminals can be set to zero. Hence, all 4 opamp input nodes have the same voltage with respect to ground.
* For convenience, it is recommended to use transconductances Y=sC (resp. 1/R) instead of impedances Z=1/sC (resp. R).
@LvW, with respect to the low pass version in post #12, since the opamps are ideal, wouldn't the input impedances to the non-inverting terminals be ∞?

Perhaps you could show step by step how you would derive the transfer function?

I would especially like to understand what you call "inverse voltage division".

#### LvW

Joined Jun 13, 2013
1,438
@LvW, with respect to the low pass version in post #12, since the opamps are ideal, wouldn't the input impedances to the non-inverting terminals be ∞?

Perhaps you could show step by step how you would derive the transfer function?

I would especially like to understand what you call "inverse voltage division".
1.) No - the input impedances at the opamp terminals are not infinite because there are som other components connected. As a result - the GIC has an input impedance which can realize different functions (inductance, FDNR, ...)
2.) I think I have roughly described the different steps - nevertheless, I can try to be more detailed (within the next two days).
3.) OK - i agree that I should be more clear. Simple voltage division (nomenclature as in post'13):
Vb1=Vout*R2/(R1+R2) can be (inversely) solved for Vout=Vb1(1+R1/R2) with Vb1=Vb2=Vb3.

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#### LvW

Joined Jun 13, 2013
1,438 Upon request (The Electrician), here comes a more detailed description of the calculation:

* For convenience, only conductances are used (simpler insertion of Y=sC).

* From known (basic) GIC chatacteristics, the input conductances at both non-inv. input nodes for OP1 and OP2 are :
Yin1=Y2Y4Y6/Y3Y5 and Yin2=Y3Y5(Y0+Y1)/Y2Y4.

* Using the voltage divider rules for conductances, the voltages at the (non-inv.) opamp input nodes (all are equal) can be found by superposition:
Vp(1)=Vin*Y1/(Y1+Y0+Yin1) and Vp(2)=Vin*Y6/(Y6+Yin2).

* Using both portions of the ouput voltage, we can also write (simple voltage division):
Vp(1)=Vout(1)*Y5(Y5+Y6) and Vp(2)=Vout(2)*Y5/(Y5+Y6)+Vin*Y6/(Y5+Y6)

* Using these equations, we can eliminate Vp(1) and Vp(2) - and after some manipulations we find for Vout(1)+Vout(2) the final result:

Vout/Vin=N/D with N=Y1Y3Y5+Y2Y4Y6-Y0Y3Y6 and D=Y1Y3Y5+Y2Y4Y6+YoY3Y5.

* Parts selection for a notch filter: Two equal C and five equal R:
Y1=sC+1/R; Y3=sC; Y0=Y4=Y5=Y6=1/R; Y2=1/R2 results in

H(s)=(1+s²R*R2*C²)/(1+2sR2*C+s²R*R2*C²) with

wp=wz=1/[C*SQRT(R*R2)] and Qp=0.5*SQRT(R/R2).

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#### The Electrician

Joined Oct 9, 2007
2,887
In post A#11, you referred to " The input impedances Zin1 and Zin2 at both non-inverting opamp input terminals are known (as a function) and can be found in many books and papers. " You did not refer to the "transresistances Zin1 and Zin2".

Yet in post # 16 you refer to " the input transconductances at both non-inv. input nodes for OP1 and OP2 "; you do not refer to the "the input conductances at both non-inv. input nodes for OP1 and OP2" Do you really mean to use the "transconductances" rather than "conductances"?

The transconductance at a node is the ratio of a current injected at that node divided by the voltage appearing at another node. If you really want to use "transconductance", what is the other node?

#### LvW

Joined Jun 13, 2013
1,438
In post A#11, you referred to " The input impedances Zin1 and Zin2 at both non-inverting opamp input terminals are known (as a function) and can be found in many books and papers. " You did not refer to the "transresistances Zin1 and Zin2".
Yes - I really mean "input impedance". It is well known that the input impedance can be realized to behave as an inductor or as an FDNR.

Yet in post # 16 you refer to " the input transconductances at both non-inv. input nodes for OP1 and OP2 "; you do not refer to the "the input conductances at both non-inv. input nodes for OP1 and OP2" Do you really mean to use the "transconductances" rather than "conductances"?
You are right. Sorry for this error. Of course I mean "conductance" (rather than "transconductance"). I have corrected this false expression. Thank you for pointing out this error.

The transconductance at a node is the ratio of a current injected at that node divided by the voltage appearing at another node. If you really want to use "transconductance", what is the other node?
See my correction above.

#### The Electrician

Joined Oct 9, 2007
2,887
View attachment 266287

Upon request (The Electrician), here comes a more detailed description of the calculation:

* For convenience, only conductances are used (simpler insertion of Y=sC).

* From known (basic) GIC chatacteristics, the input transconductances at both non-inv. input nodes for OP1 and OP2 are :
Yin1=Y2Y4Y6/Y3Y5 and Yin2=Y3Y5(Y0+Y1)/Y2Y4.

* Using the voltage divider rules for transconductances, the voltages at the (non-inv.) opamp input nodes (all are equal) can be found by superposition:
Vp(1)=Vin*Y1/(Y1+Y0+Yin1) and Vp(2)=Vin*Y6/(Y6+Yin2).

* Using both portions of the ouput voltage, we can also write (simple voltage division):
Vp(1)=Vout(1)*Y5(Y5+Y6) and Vp(2)=Vout(2)*Y5/(Y5+Y6)+Vin*Y6/(Y5+Y6)

* Using these equations, we can eliminate Vp(1) and Vp(2) - and after some manipulations we find for Vout(1)+Vout(2) the final result:

Vout/Vin=N/D with N=Y1Y3Y5+Y2Y4Y6-Y0Y3Y6 and D=Y1Y3Y5+Y2Y4Y6+YoY3Y5.

* Parts selection for a notch filter: Two equal C and five equal R:
Y1=sC+1/R; Y3=sC; Y0=Y4=Y5=Y6=1/R; Y2=1/R2 results in

H(s)=(1+s²R*R2*C²)/(1+2sR2*C+s²R*R2*C²) with

wp=wz=1/[C*SQRT(R*R2)] and Qp=0.5*SQRT(R/R2).

I see that you have edited this post, but I still see "transconductances", which I have highlighted in red.

#### The Electrician

Joined Oct 9, 2007
2,887
View attachment 266287

Upon request (The Electrician), here comes a more detailed description of the calculation:

* For convenience, only conductances are used (simpler insertion of Y=sC).

* From known (basic) GIC chatacteristics, the input conductances at both non-inv. input nodes for OP1 and OP2 are :
Yin1=Y2Y4Y6/Y3Y5 and Yin2=Y3Y5(Y0+Y1)/Y2Y4.

Using the schematic shown, when I apply a 1 amp test current to the non-inverting input of OP1 and solve for the voltage at that terminal, I don't get a conductance of (Y2 Y4 Y6)/Y3 Y5; my result involves Y0 and Y1, which surely must be true. Is the expression you give for Yin1 for a simpler circuit than shown in this post? I also get a different result for Yin2 for the circuit shown here.