Admittance matrix for Fliege Notch Filter

Thread Starter

Skar

Joined Jun 11, 2015
5
Hello,

I search for the admittance Matrix for a Fliege notch filter. Actually I am interested in the transfer function of the filter but did some research using google and ended up in this forum http://forum.allaboutcircuits.com/t...s-filter-transfer-functions-derivation.22962/
I duplicated what they did there. But I cannot set up the admittance matrix for a Fliege notch filter like this one: http://earmark.net/gesr/opamp/notch.htm

Actually I am a physics student and all I know about admittance is that it is the inverse of the impedance.

Any help will be much appreciated!
 

The Electrician

Joined Oct 9, 2007
2,970
What I also do not understand is how the 4th row in the matrix in http://forum.allaboutcircuits.com/attachments/fliege-gif.9171/ is calculated. Why do i take into account the output of the top opamp but not anything tht comes over R3?

And why is it -A/R4 for the second element and A/R4 for the third and not vice versa?
The method shown by Jony130 is perfectly valid; I've used it before. Here's a variation that may be easier to use even though it does increase the size of the admittance matrix in the end. Since the solution is obtained using modern mathematical software, the increase in the size of the system of equations doesn't matter. Solving this system by hand would be very tedious and very prone to mistakes in the algebra. Hooray for mathematical software.

First set up an admittance matrix for the circuit without the opamps--passive components only. Using the node numbering scheme Jony130 used, but without numbering the nodes where the opamp outputs will be connected



Now, to deal with the opamps we need a constraint equation for each opamp, relating the voltages at the inputs (Vp and Vn) to the voltage at the output (Vo). Each constraint equation looks like this:

A*Vp - A*Vn = Vo, or A*Vp - A*Vn - Vo = 0

We add two more nodes to the circuit, one for each opamp output; let the output of U2 be node 5 and the output of U1 be node 6. This adds two more rows and columns (shown in red; rows 5 & 6 are the constraint equations) to the admittance matrix:



This formulation allows for the opamp gain A to be anything. For many problems, A is taken to be infinite. We could solve the system and then let A become infinite, or we can let A→∞ before we solve the system. So let's divide rows 5 and 6 by A; we get:



Then let A→∞; we get:



Solving this system, we can obtain the transfer function (voltage transfer ratio) and we see it's the same result Jony130 got:

 

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Thread Starter

Skar

Joined Jun 11, 2015
5
Wow! That is great! Thank you both a lot! I actually came up with something similar but made some mistakes. But I saw where I made my mistakes. So later I came up with the same result as Jony130. But I actually do not know what I am doing. I just followed the form and my intuition. From where do I know how to choose my nodes? And how do I treat the opamps in the KCL. What is the physics behind it?

And yes: A hooray for mathematical software :)

I will go to bed now. It is late. See you tomorrow!
 
Last edited:

The Electrician

Joined Oct 9, 2007
2,970
Wow! That is great! Thank you both a lot! I actually came up with something similar but made some mistakes. But I saw where I made my mistakes. So later I came up with the same result as Jony130. But I actually do not know what I am doing. I just followed the form and my intuition. From where do I know how to choose my nodes? And how do I treat the opamps in the KCL. What is the physics behind it?

And yes: A hooray for mathematical software :)

I will go to bed now. It is late. See you tomorrow!
Read this: http://web.ecs.baylor.edu/faculty/grady/EE411_Fall2011_Week_02.pdf

Then if you have any more questions, ask away.

You can find a lot of information on the web about opamps, including the eBook associated with this forum:

http://www.allaboutcircuits.com/textbook/semiconductors/
 
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