Fliege Bandpass Filter transfer function's derivation

The Electrician

Joined Oct 9, 2007
2,970
Lieta, TI has an app note with a lot of opamp filters described:
focus.ti.com/lit/an/sloa058/sloa058.pdf

Also, have a look at this one:
http://www.ti.com/litv/pdf/slod006b

Are these problems you're working on actual class assignments? The reason I ask is because the twin T one was first posted back in January, and you didn't ask about it again until just a few days ago.

Are you going to be doing many of these problems where you must derive transfer functions? If so, it might be more efficient for you if you were to learn systematic ways to do it.

Do you know how to solve simultaneous linear equations? Have you studied any linear algebra (matrix algebra)?
 

Thread Starter

Lieta

Joined Jan 18, 2009
12
This is for my bachelor's work. The theme is "Voltage controlled bandpass filters". I'll have to analyse a few more, like Akerberg-Moosberg and BiQuad. It's not just deriving transfer function, but also considering how it can be controlled, etc., any information about the circuit.
I know how to solve simultaneous linear equations and have studied matrix algebra as well, but it takes a lot of time that I dont't have.
 

The Electrician

Joined Oct 9, 2007
2,970
Do you have access to mathematical software that can do symbolic linear algebra, like Maple, Mathematica or Mathcad?

It really doesn't take that long. Tomorrow I'll derive the transfer function for your Fliege filter and time myself.
 

The Electrician

Joined Oct 9, 2007
2,970
Ok. It took less than 5 minutes to set up the admittance matrix for the Fliege filter, but about 1/2 hour to do the write-up.

This .pdf explains quite a bit about how and why to set up an admittance matrix:

http://web.ecs.baylor.edu/faculty/grady/EE411_Fall2011_Week_02.pdf

See the attachment for the worked out transfer function. If you're going to be doing a lot of them, you ought to learn how to do this.
 

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The Electrician

Joined Oct 9, 2007
2,970
For example, the row 2 elements are found like this:

Let Vbo be the voltage at the output of the bottom opamp. Then Vbo = A*(V4 - V3). Summing currents at node 2:

(V2 - V1)/R1 + V2 s C1 + (V2 - Vbo)/R2 = 0

substituting Vbo = A*(V4 - V3) we get:

(V2 - V1)/R1 + V2 s C1 + (V2 - A*(V4 - V3))/R2 = 0

collecting terms we have:

V1(-1/R1) + V2(1/R1 + 1/R2 + s C1) + V3(A/R2) + V4(-A/R2) = 0

The coefficients of V1, V2, V3 and V4 are the elements of row 2.

Similarly for the other rows.

This method is advantageous because it is systematic, and the individual elements of the matrix aren't very complicated. The various expressions become complicated when we solve the system (invert the matrix), but the computer does the algebra, and doesn't make any mistakes!
 
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