# Solving Entries in a Four Port Transmission Matrix

The well-known transmission (ABCD) matrix for a two-port network has entries which can be solved by setting open or short boundary conditions on the opposite terminals.

Mathematically, the two-port network is expressed as

$$\left| \begin{array}{c} V_1 \\ I_1 \\ \end{array} \right| = \left| \begin{array}{cc} A & B \\ C & D \\ \end{array} \right| \left| \begin{array}{c} V_2 \\ I_2 \\ \end{array} \right|$$

And the entries are solved as $$A = \left. \frac{V_1}{V_2}\right|_{I_2 = 0} \ \ \ \ B = \left. \frac{V_1}{I_2}\right|_{V_2 = 0} \ \ \ \ , etc.$$

However, when we extend this concept to a four-port network, in which there is one input port and three output ports, such that

$$\left| \begin{array}{c} V_1 \\ I_1 \\ \end{array} \right| = \left| \begin{array}{ccc} T_{11} & T_{12} & T_{13} & T_{14} & T_{14} & T_{15} \\ T_{21} & T_{22} & T_{23} & T_{24} & T_{24} & T_{25} \\ \end{array} \right| \left| \begin{array}{c} V_2 \\ I_2 \\ V_3 \\ I_3 \\ V_4 \\ I_4 \\ \end{array} \right|$$

It no longer becomes straight forward to solve for the entries, as we will require that pairs of voltages and currents are equal to zero simultaneously.

How can the transmission matrix of a four-port network be solved?

What you can do for your four-port network is modeling it with a scattering matrix (https://en.wikipedia.org/wiki/Scattering_parameters) such as: $$\begin{pmatrix} V_1 \\ V_2 \\ V_3 \\ V_4 \end{pmatrix} = \begin{pmatrix} S_{11} & S_{12} & S_{13} & S_{14} \\ S_{21} & S_{22} & S_{23} & S_{24} \\ S_{31} & S_{32} & S_{33} & S_{34} \\ S_{41} & S_{42} & S_{43} & S_{44} \end{pmatrix} \begin{pmatrix} V_1 \\ V_2 \\ V_3 \\ V_4 \end{pmatrix}.$$ With this approach, you can simply superpose each input (considering all the others as 0) to determine its contribution to any output (including itself): $S_{ij}=\frac{V_i}{V_j}\Big|_{V_k=0,\ k\neq j}$.