butterworth filter transfer function decompostion question

Thread Starter

yef smith

Joined Aug 2, 2020
693
Hello, If i understand correctly we are approximating a general response with multisection transfer function shown bellow.
In the pozar book they give me a complete result.
but i wat to develop such table by myself.
1.How do i convert a single line of the table below into a mathematical transfer function?
2.Is there some example where they take a transfer function and aproximate it and creating such line.
I know its being stuied more in digital filters.if you could please help me with such rf example guidance.

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LvW

Joined Jun 13, 2013
1,749
(1) "Maximally flat time delay" filters are Thomson-Bessel lowpass filters (and not Butterworth).
(2) What is the meaning of g1....g11 ? I suppose these figures are normalized conductances for a typical ladder topology. right? In this case, go would be the source conductance .
 

Papabravo

Joined Feb 24, 2006
20,994
(1) "Maximally flat time delay" filters are Thomson-Bessel lowpass filters (and not Butterworth).
(2) What is the meaning of g1....g11 ? I suppose these figures are normalized conductances for a typical ladder topology. right? In this case, go would be the source conductance .
The derivation for the Butterworth transfer function is given in Van Valkenburg, M.E., Analog Filter Design, Chapter 6. It also has the derivation for the Bessel-Thomson filters in Chapter 10. The striking feature of these transfer functions is that the numerator, and the constant and linear terms in the denominator all have the same value.

For example, the 3rd order transfer function looks like

\( T(s)\;=\;\cfrac{15}{s^3+6s^2+15s+15} \)

1702129281659.png
 

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Thread Starter

yef smith

Joined Aug 2, 2020
693
Hello Papabravo,i have this table from chapter six .
suppuse i have n=3.
B(s)=S^3+1*S+2S^2+2*S^3
this polynomial is always rising.
you showed me the final result,i want to know how you built this transfer function
could you please show me how given a coefficient table i could get a transfer function like you did?
Thanks.

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Papabravo

Joined Feb 24, 2006
20,994
The one I did was for a Bessel filter. The derivation criteria is different than the one for the Butterworth polynomial.

There are a couple of ways to determine the transfer function. Probably the easiest is to know the pole locations, for the order you are interested in, which gives you the factors, and you multiply them out. The other way is to start with the maximally flat criteria and derive the transfer function from that condition. If you do that for the maximally flat in the passband low pass filter, you are led to the form of the transfer function:

\( |T_n(j\omega)|^2\;=\;\cfrac{A_0}{B_0+B_2\omega^2+B_4\omega^4+\cdot\cdot\cdot + B_{2n}\omega^{2n}} \)

We make this choice to maximize the difference in degree between the numerator and the denominator which leads to the steep rolloff that we desire. So the Butterworth response ends up having the transfer function:

\( |T_n(j\omega)|\;=\;\cfrac{1}{\sqrt{1+\omega^{2n}}} \)

It is there in Chapter Six. For a Butterworth filter, all of the poles for the normalized (ω=1) transfer function are located on the unit circle. The third order Butterworth polynomial in the left half plane is:

\( (s+1)(s^2+s+1)\;=\;(s+1)\left(s+\cfrac{1}{2}+j\cfrac{\sqrt{3}}{2}\right)\left(s+\cfrac{1}{2}-j\cfrac{\sqrt{3}}{2}\right) \)

Clearly when you multiply this out you get

\( s^3+2s^2+2s+1 \)

The coefficient of the highest order term is always 1, and the constant term is always 1 and equal to the magnitude of the numerator, so at j0 the magnitude will be 1/1 = 0dB

The coefficients of the linear term and the squared term are both 2 as table 6.2 suggests.

So, the product of a Butterworth polynomial of order n and its conjugate which gives the magnitude squared will have 2n poles spaced evenly around the unit circle.
 
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