# Interconnection of Networks

A large, complex network may be divided into subnetworks for the purposes of analysis and design. The subnetworks are modeled as two-port networks, interconnected to form the original network. The two-port networks may therefore be regarded as building blocks that can be interconnected to form a complex network. The interconnection can be in series, in parallel, or in cascade. Although the interconnected network can be described by any of the six parameter sets, a certain set of parameters may have a definite advantage.
For example, when the networks are in series, their individual z parameters add up to give the z parameters of the larger network. When they are in parallel, their individual y parameters add up to give the y parameters the larger network. When they are cascaded, their individual transmission parameters can be multiplied together to get the transmission parameters of the larger network.
Fig. 1: Series connection of two two-port networks.
Consider the series connection of two two-port networks shown in Fig. 1. The networks are regarded as being in series because their input currents are the same and their voltages add. In addition, each network has a common reference, and when the circuits are placed in series, the common reference points of each circuit are connected together. For network $N_a$,
\begin{aligned} & V_{1 a}=z_{11 a} \mathbf{I}_{1 a}+z_{12 a} \mathbf{I}_{2 a} \\ & V_{2 a}=\mathbf{z}_{2 a a} \mathbf{I}_{\mathrm{la}}+\mathbf{z}_{22 a} \mathbf{I}_{2 a} \end{aligned} \tag{1}
and for network $N_b$,
\begin{aligned} & \mathbf{V}_{1 b}=\mathbf{z}_{11 b} \mathbf{I}_{1 b}+\mathbf{z}_{12 b} \mathbf{I}_{2 b} \\ & \mathbf{V}_{2 b}=\mathbf{z}_{21 b} \mathbf{I}_{1 b}+\mathbf{z}_{22 b} \mathbf{I}_{2 b} \end{aligned} \tag{2}
We notice from Fig. $1$ that
$$\mathbf{I}_1=\mathbf{I}_{1 a}=\mathbf{I}_{1 b}, \quad \mathbf{I}_2=\mathbf{I}_{2 a}=\mathbf{I}_{2 b} \tag{3}$$
and that
\begin{aligned} & \mathbf{V}_1=\mathbf{V}_{1 a}+\mathbf{V}_{1 b}=\left(\mathbf{z}_{11 a}+\mathbf{z}_{11 b}\right) \mathbf{I}_1+\left(\mathbf{z}_{12 a}+\mathbf{z}_{12 b}\right) \mathbf{I}_2 \\ & \mathbf{V}_2=\mathbf{V}_{2 a}+\mathbf{V}_{2 b}=\left(\mathbf{z}_{21 a}+\mathbf{z}_{21 b}\right) \mathbf{I}_1+\left(\mathbf{z}_{22 a}+\mathbf{z}_{22 b}\right) \mathbf{I}_2 \end{aligned} \tag{4}
Thus, the z parameters for the overall network are
$$\left[\begin{array}{ll} \mathbf{z}_{11} & \mathbf{z}_{12} \\ \mathbf{z}_{21} & \mathbf{z}_{22} \end{array}\right]=\left[\begin{array}{ll} \mathbf{z}_{11 a}+\mathbf{z}_{11 b} & \mathbf{z}_{12 a}+\mathbf{z}_{12 b} \\ \mathbf{z}_{21 a}+\mathbf{z}_{21 b} & \mathbf{z}_{22} a+\mathbf{z}_{22 b} \end{array}\right] \tag{5}$$
or
$$\bbox[10px,border:1px solid grey]{[\mathbf{z}]=\left[\mathbf{z}_i\right]+\left[\mathbf{z}_b\right]} \tag{6}$$
showing that the z parameters for the overall network are the sum of the $z$ parameters for the individual networks. This can be extended to $n$ networks in series. If two two-port networks in the [h] model, for example, are connected in series, then apply Eq. (6).
Two two-port networks are in parallel when their port voltages are equal and the port currents of the larger network are the sums of the individual port currents. In addition, each circuit must have a common reference and when the networks are connected together, they must all have their common references tied together. The parallel connection of two two-port networks is shown in Fig. 2.
Fig. 2: Parallel connection of two two-port networks.
For the two networks,
\begin{aligned} & \mathbf{I}_{1 a}=\mathbf{y}_{11 a} \mathbf{V}_{\mathrm{la}}+\mathbf{y}_{12 a} \mathbf{V}_{2 a} \\ & \mathbf{I}_{2 a}=\mathbf{y}_{21 a} \mathbf{V}_{\mathrm{la}}+\mathbf{y}_{22 a} \mathbf{V}_{2 a} \end{aligned} \tag{7}
and
\begin{aligned} & \mathbf{I}_{1 b}=\mathbf{y}_{11 b} \mathbf{V}_{1 b}+\mathbf{y}_{12 b} \mathbf{V}_{2 b} \\ & \mathbf{I}_{2 a}=\mathbf{y}_{21 b} \mathbf{V}_{1 b}+\mathbf{y}_{2 b} \mathbf{V}_{2 b} \end{aligned} \tag{8}
But from Fig. 2,
\begin{aligned} & \mathbf{V}_1=\mathbf{V}_{1 a}=\mathbf{V}_{1 b}, \quad \mathbf{V}_2=\mathbf{V}_{2 a}=\mathbf{V}_{2 b} \quad (9.a)\\ & \mathbf{I}_1=\mathbf{I}_{1 a}+\mathbf{I}_{1 b,}, \quad \mathbf{I}_2=\mathbf{I}_{2 a}+\mathbf{I}_{2 b} \quad (9.b)\\ & \end{aligned}
Substituting Eqs. (7) and (8) into Eq. (9.b) yields
$$\begin{array}{l}\mathbf{I}_{1}=\left(\mathbf{y}_{11 a}+\mathbf{y}_{11 b}\right) \mathbf{V}_{1}+\left(\mathbf{y}_{12 a}+\mathbf{y}_{12 b}\right) \mathbf{V}_{2} \\\mathbf{I}_{2}=\left(\mathbf{y}_{21 a}+\mathbf{y}_{21 b}\right) \mathbf{V}_{1}+\left(\mathbf{y}_{22 a}+\mathbf{y}_{22 b}\right) \mathbf{V}_{2}\end{array} \tag{10}$$
Thus, the $y$ parameters for the overall network are
$$\left[\begin{array}{ll}\mathbf{y}_{11} & \mathbf{y}_{12} \\\mathbf{y}_{21} & \mathbf{y}_{22}\end{array}\right]=\left[\begin{array}{ll}\mathbf{y}_{11 a}+\mathbf{y}_{11 b} & \mathbf{y}_{12 a}+\mathbf{y}_{12 b} \\\mathbf{y}_{21 a}+\mathbf{y}_{21 b} & \mathbf{y}_{22 a}+\mathbf{y}_{22 b}\end{array}\right] \tag{11}$$
or
$$\bbox[10px,border:1px solid grey]{[\mathbf{y}]=\left[\mathbf{y}_{a}\right]+\left[\mathbf{y}_{b}\right]} \tag{12}$$
showing that the $y$ parameters of the overall network are the sum of the $y$ parameters of the individual networks. The result can be extended to $n$ two-port networks in parallel.
Two networks are said to be cascaded when the output of one is the input of the other. The connection of two two-port networks in cascade is shown in Fig. 2. For the two networks,
$$\begin{array}{l}{\left[\begin{array}{l}\mathbf{V}_{1 a} \\\mathbf{I}_{1 a}\end{array}\right]=\left[\begin{array}{ll}\mathbf{A}_{a} & \mathbf{B}_{a} \\\mathbf{C}_{a} & \mathbf{D}_{a}\end{array}\right]\left[\begin{array}{c}\mathbf{V}_{2 a} \\-\mathbf{I}_{2 a}\end{array}\right]} \quad (13) \\{\left[\begin{array}{l}\mathbf{V}_{1 b} \\\mathbf{I}_{1 b}\end{array}\right]=\left[\begin{array}{ll}\mathbf{A}_{b} & \mathbf{B}_{b} \\\mathbf{C}_{b} & \mathbf{D}_{b}\end{array}\right]\left[\begin{array}{c}\mathbf{V}_{2 b} \\-\mathbf{I}_{2 b}\end{array}\right]} \quad(14) \end{array}$$
Fig. 3: Cascade connection of two two-port networks.
From Fig. 3,
$$\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{I}_{1}\end{array}\right]=\left[\begin{array}{l}\mathbf{V}_{1 a} \\\mathbf{I}_{1 a}\end{array}\right], \quad\left[\begin{array}{c}\mathbf{V}_{2 a} \\-\mathbf{I}_{2 a}\end{array}\right]=\left[\begin{array}{l}\mathbf{V}_{1 b} \\\mathbf{I}_{1 b}\end{array}\right], \quad\left[\begin{array}{c}\mathbf{V}_{2 b} \\-\mathbf{I}_{2 b}\end{array}\right]=\left[\begin{array}{c}\mathbf{V}_{2} \\-\mathbf{I}_{2}\end{array}\right]$$
Substituting these into Eqs. (13) and (14),
$$\left[\begin{array}{l}\mathbf{V}_{1} \\\mathbf{I}_{1}\end{array}\right]=\left[\begin{array}{ll}\mathbf{A}_{a} & \mathbf{B}_{a} \\\mathbf{C}_{a} & \mathbf{D}_{a}\end{array}\right]\left[\begin{array}{ll}\mathbf{A}_{b} & \mathbf{B}_{b} \\\mathbf{C}_{b} & \mathbf{D}_{b}\end{array}\right]\left[\begin{array}{c}\mathbf{V}_{2} \\-\mathbf{I}_{2}\end{array}\right]$$
Thus, the transmission parameters for the overall network are the product of the transmission parameters for the individual transmission parameters:
$$\left[\begin{array}{ll}\mathbf{A} & \mathbf{B} \\\mathbf{C} & \mathbf{D}\end{array}\right]=\left[\begin{array}{ll}\mathbf{A}_{a} & \mathbf{B}_{a} \\\mathbf{C}_{a} & \mathbf{D}_{a}\end{array}\right]\left[\begin{array}{ll}\mathbf{A}_{b} & \mathbf{B}_{b} \\\mathbf{C}_{b} & \mathbf{D}_{b}\end{array}\right]$$
or
$$[\mathbf{T}]=\left[\mathbf{T}_{a}\right]\left[\mathbf{T}_{b}\right]$$
It is this property that makes the transmission parameters so useful. Keep in mind that the multiplication of the matrices must be in the order in which the networks $N_{a}$ and $N_{b}$ are cascaded. Figure 3 Cascade connection of two two-port networks.
Example 1: Evaluate $\mathbf{V}_{2} / \mathbf{V}_{s}$ in the circuit in Fig. $4$.
Fig. 4: For Example 1.
Solution:
This may be regarded as two two-ports in series. For $N_{b}$,
$$\mathbf{z}_{12 b}=\mathbf{z}_{21 b}=10=\mathbf{z}_{11}=\mathbf{z}_{22}$$
Thus,
$$[\mathbf{z}]=\left[\mathbf{z}_{a}\right]+\left[\mathbf{z}_{b}\right]=\left[\begin{array}{cc}12 & 8 \\8 & 20\end{array}\right]+\left[\begin{array}{cc}10 & 10 \\10 & 10\end{array}\right]=\left[\begin{array}{cc}22 & 18 \\18 & 30\end{array}\right]$$
But
$$\begin{array}{l}\mathbf{V}_{1}=\mathbf{z}_{11} \mathbf{I}_{1}+\mathbf{z}_{12} \mathbf{I}_{2}=22 \mathbf{I}_{1}+18 \mathbf{I}_{2} \quad (1.1) \\\mathbf{V}_{2}=\mathbf{z}_{21} \mathbf{I}_{1}+\mathbf{z}_{22} \mathbf{I}_{2}=18 \mathbf{I}_{1}+30 \mathbf{I}_{2} \quad (1.2) \end{array}$$
Also, at the input port
$$\mathbf{V}_{1}=\mathbf{V}_{s}-5 \mathbf{I}_{1} \tag{1.3}$$
and at the output port
$$\mathbf{V}_{2}=-20 \mathbf{I}_{2} \quad \Longrightarrow \quad \mathbf{I}_{2}=-\frac{\mathbf{V}_{2}}{20} \tag{1.4}$$
Substituting Eqs. (1.3) and (1.4) into Eq. (1.1) gives
$$\mathbf{V}_{s}-5 \mathbf{I}_{1}=22 \mathbf{I}_{1}-\frac{18}{20} \mathbf{V}_{2} \quad \Longrightarrow \quad \mathbf{V}_{s}=27 \mathbf{I}_{1}-0.9 \mathbf{V}_{2} \tag{1.5}$$
while substituting Eq. (1.4) into Eq. (1.2) yields
$$\mathbf{V}_{2}=18 \mathbf{I}_{1}-\frac{30}{20} \mathbf{V}_{2} \quad \Longrightarrow \quad \mathbf{I}_{1}=\frac{2.5}{18} \mathbf{V}_{2} \tag{1.6}$$
Substituting Eq. (1.6) into Eq. (1.5), we get
$$\mathbf{V}_{s}=27 \times \frac{2.5}{18} \mathbf{V}_{2}-0.9 \mathbf{V}_{2}=2.85 \mathbf{V}_{2}$$
And so,
$$\frac{\mathbf{V}_{2}}{\mathbf{V}_{s}}=\frac{1}{2.85}=0.3509$$