# 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?

## 1 Answer

Transmission matrices can be used only for two-port networks, since they relate the input on one side to the output on the other and cannot account for more than that. They were conceived to help in the modeling of cascaded two-port networks, since multiplication of transmission matrices results in the transmission matrix of an equivalent two-port network.

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{equation} \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}. \end{equation} With this approach, you can simply superpose each input (considering all the others as 0) to determine its contribution to any output (including itself): $S_{ij}=\frac{V_i}{V_j}\Big|_{V_k=0,\ k\neq j}$.

• Thanks for the response. Is your scattering matrix definition missing incident and reflected wave notation? In any case, I feel like this is avoiding the issue -- while the ABCD matrix may be specifically for two port networks, I feel there must be some method to relate the voltages and currents at each port without using other types of network characterizations. – Stuart Barth Aug 28 '16 at 19:43
• "Is your scattering matrix definition missing incident and reflected wave notation?" If I understood your question, in the scattering matrix, elements Sjj are the reflected waves at port j; the other Sij are the fraction of incident wave transmitted to other ports. – DavideM Aug 31 '16 at 14:16
• "I feel there must be some method to relate the voltages and currents at each port without using other types of network characterizations." Well, since transmission matrix works only for two-port networks, you just can't use it for a four-port one. If you want to compute the ports currents, you may use an impedance matrix (en.wikipedia.org/wiki/Impedance_parameters) to relate voltages to currents, but you do need another type of network characterization. – DavideM Aug 31 '16 at 14:19