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I've been trying to wrap my head around the synthesis techniques for passive reactive filters. I've been able to understand the Cauer synthesis technique for one-port networks. The book I'm reading (Зевеке Г.В., Ионкин П.А., Нетушил А.В., Страхов С.В., "Основы теории цепей", 1989, chapter 19) then dives into the synthesis of two-port networks with the examples based on the Butterworth and Chebyshev filters. It goes something like this:

Let's make a current source-driven filter with a load \$Z_l\$ on the other side. If we analyze our two-port network with the help of the impedance matrix, we get $$ \frac{I_s(s)}{U_l(s)} = \frac{Z_{21}(s)Z_l}{Z_{22}(s) + Z_l(s)}. $$ Let's now normalize everything by the load impedance. We then get $$ K(s) = \frac{Z_{21}(s)}{Z_{22}(s) + 1}.\tag{1} $$ We now try to transform the rational transfer function into the same shape as (1). For this (this being a case of an odd-ordered denominator) we divide the numerator and the denominator by a polynomial which composed of the odd elements of the initial denominator: $$ K(s) = \frac{a_0}{s^m + b_{m - 1}s^{m - 1} + \dots + b_0} = \frac{{a_0}/(s^m + \dots + b_1s)}{(b_{m - 1}s^{m - 1} + \dots + b_0)/(s^m + \dots + b_1s) + 1}. $$ Due to the fact that \$Z_{22}\$ is the load point impedance for a shirt-circuited input side, we can use the Cauer approach to find out the values of the elements based on the fact that by comparison $$ Z_{22}(s) = \frac{b_{m - 1}s^{m - 1} + \dots + b_0}{s^m + \dots + b_1s}. $$

This is all nice and good, but then the book just basically pats itself on the back and goes on to the next chapter while completely disregarding the fact that we haven't verified that we've obtained the correct topology for the required \$Z_{21}\$ as well. This is where my question lies: is there a way to prove that the \$Z_{21}\$ value is satisfied automatically if we construct a network which will provide the required \$Z_{22}\$ in general case or to at least verify that it will be correct for the Cauer topology?

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For a lossless network, input power is either reflected (S11) or transmitted (S21), it has nowhere else to go. Therefore if we get the reflected power to be correct, the transmitted power will automatically be correct.

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  • \$\begingroup\$ Yeah. I've decided to pick up Pozar and read about the insertion loss method in the meantime. The derivation in that book is interesting to me because it seems to not rely on the energy approach and instead works purely with lumped-element currents and voltages which I've never seen before. I really want to find out if it can be proven in its entirety in this way. \$\endgroup\$ Commented 12 hours ago

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