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?

## Best 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}\$.