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I'm currently studying microwave engineering as a subject.

Everywhere on the internet and the books, they've only shown the derivation of S11 and S22 from ABCD parameters of a network. Yes, that's intuitive, since S11 is tau(in) and S22 is tau(out), and I have impedance formulas for both, making their calculation easy.

However, how do I derive S12 and S21 from ABCD parameters of a network? I've tried searching a lot, but I'm not getting the derivation, only the result.

I tried solving this on my own, but I'm not getting a hang of it, primarily because scattering parameters need transmitted and reflected wave while other parameters work on voltages/currents.

Any resources you can point me to? Thanks!

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Hewlet Packard's appnote A-95 is a classic text on this subject. Since the related divisions of HP were spun off to form Agilent in the mid 90's, this text is now hosted at http://cp.literature.agilent.com/litweb/pdf/5952-1130.pdf.

Check pages 2 and 3 for the derivations of the S parameters. To summarize, given a two-port network as shown:

enter image description here

with impedance \$Z_0\$, the normalized incident voltages \$a_1\$ and \$a_2\$ are calculated as follows:

\$a_1=\frac{V_1+I_1*Z_0}{2\sqrt{Z_0}}\$

\$a_2=\frac{V_2+I_2*Z_0}{2\sqrt{Z_0}}\$

The reflected voltages \$b_1\$ and \$b_2\$ are calculated as follows. Note the similar form:

\$b_1=\frac{V_1-I_1*Z_0}{2\sqrt{Z_0}}\$

\$b_2=\frac{V_2-I_2*Z_0}{2\sqrt{Z_0}}\$

These values are derived from the voltages and currents. They're simply normalized so that the following equations make more sense and are more consistent and compact. The S parameters can be calculated as

\$s_{11} = \left.{\frac{b_1}{a_1}} \right|_{a_2=0}\$

\$s_{22} = \left.{\frac{b_2}{a_2}} \right|_{a_1=0}\$

\$s_{21} = \left.{\frac{b_2}{a_1}} \right|_{a_2=0}\$

\$s_{12} = \left.{\frac{b_1}{a_2}} \right|_{a_1=0}\$

The text goes into a good deal more detail on the meanings of each of these values and their derivations, but those equations should allow you to make progress on your problem.

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  • \$\begingroup\$ Just what I wanted. I could derive the equations successfully now! :-) \$\endgroup\$ – AgilE May 7 '12 at 16:29

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