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There is a famous paper from Milton Dishal called 'Two New Equations on the Design of Filters'. The paper was published in the journal 'electrical communication' and can be found here

http://www.americanradiohistory.com/Archive-ITT/50s/ITT-Vol-30-1953-04.pdf

on page 65. The equations 10A and 10B as well as Figure 7 allow to calculate the normalised coupling coefficients of Butterworth and Chebyshev filters. What I don't understand is the following: using the K and Q values, I would like to create the coupling matrix as it is shown in many papers on filter theory. I found out (by try and error), that the coupling matrix must be (say, for a filter having 3 poles) as follows:

$$ M = \begin{pmatrix} 0 & \sqrt{1/q} & 0 & 0 & 0 \\ \sqrt{1/q} & 0 & k_{12} & 0 & 0 \\ 0 & k_{12} & 0 & k_{23} & 0 \\ 0 & 0 & k_{23} & 0 & \sqrt{1/q} \\ 0 & 0 & 0 & \sqrt{1/q} & 0 \end{pmatrix} $$

Then, from this, I calculate the frequency response of the filter as follows:

$$ Y = G + s\,L + j\,M $$

with the abbreviations:

$$ G = \begin{pmatrix} 1 & 0 & 0 & \ldots \\ 0 & 0 & 0 & \ldots \\ \vdots \\ 0 & 0 & \ldots & 1 \end{pmatrix} \text{,}\quad L = \begin{pmatrix} 0 & 0 & 0 & \ldots \\ 0 & 1 & 0 & \ldots \\ 0 & 0 & 1 & \ldots \\ \vdots \\ 0 & 0 & \ldots & 0 \\ \end{pmatrix}$$

For a bandpass filter, the frequency transformation

$$ s = \frac{1}{FBW} \cdot \left( \frac{j\omega}{\omega_0} + \frac{\omega_0}{j\omega} \right) $$

is used. The scattering parameters are then found with $$ S_{11} = \left(Y^{-1}\right)_{1, 1} $$ and $$ S_{21} = \left( Y^{-1} \right)_{N+2,1} $$.

However, in many books and papers, the coupling matrix M has completely different values in it; it is kind of denormalized with the coupling coefficients being multiplied with 1/FBW. Why is this, and what is the difference to the coupling coefficients found with dishals equations? And why is S11 and S21 found with these equations, how can this be derived?

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    \$\begingroup\$ I don't know. Could it be that the state-of-the-art has advanced since 1953? \$\endgroup\$ – Enric Blanco May 8 '17 at 20:56
  • \$\begingroup\$ Of course, but the theory is probably still the same as I think phsics has not changed that much since then :-) \$\endgroup\$ – T. Pluess May 9 '17 at 15:09

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