19.7 Interconnection of Networks

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19.7 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 netw ork can be described by an y of the six parameter sets, a certain set of parameters may have a definite advantage. For example, when the netw orks are in series, their indi vidual z parameters add up to gi ve the z parameters of the larger network. When they are in parallel, their individual y parameters add up to gi ve the y parameters of the lar ger network. When they are cascaded, their individual transmission parameters can be multiplied together to get the transmission parameters of the larger network.

Figure 19.37 For Example 19.11.

Figure 19.39 Series connection of two two-port networks.

Consider the series connection of tw o two-port networks shown in Fig. 19.39. The networks are re garded 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 Na,

$$\mathbf{V}_{1a} = \mathbf{z}_{11a}\mathbf{I}_{1a} + \mathbf{z}_{12a}\mathbf{I}_{2a}$$

\n

$$\mathbf{V}_{2a} = \mathbf{z}_{21a}\mathbf{I}_{1a} + \mathbf{z}_{22a}\mathbf{I}_{2a}$$

(19.44)

and for network Nb,

$$V_{1b} = z_{11b}I_{1b} + z_{12b}I_{2b}$$

\n

$$V_{2b} = z_{21b}I_{1b} + z_{22b}I_{2b}$$

(19.45)

We notice from Fig. 19.39 that

I1 = I¹a = I¹b, I2 = I²a = I^2b (19.46)

and that

$$V_1 = V_{1a} + V_{1b} = (z_{11a} + z_{11b})I_1 + (z_{12a} + z_{12b})I_2$$

\n

$$V_2 = V_{2a} + V_{2b} = (z_{21a} + z_{21b})I_1 + (z_{22a} + z_{22b})I_2$$

(19.47)

Thus, the z parameters for the overall network are

$$\begin{bmatrix} \mathbf{z}_{11} & \mathbf{z}_{12} \\ \mathbf{z}_{21} & \mathbf{z}_{22} \end{bmatrix} = \begin{bmatrix} \mathbf{z}_{11a} + \mathbf{z}_{11b} & \mathbf{z}_{12a} + \mathbf{z}_{12b} \\ \mathbf{z}_{21a} + \mathbf{z}_{21b} & \mathbf{z}_{22a} + \mathbf{z}_{22b} \end{bmatrix}$$

(19.48)

or

$$[\mathbf{z}] = [\mathbf{z}_a] + [\mathbf{z}_b] \tag{19.49}$$

showing that the z parameters for the o verall 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, we use Table 19.1 to convert the h to z and then apply Eq. (19.49). We finally convert the result back to h using Table 19.1.

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 ha ve a common reference and when the netw orks are connected together , they must all have their common references tied together. The parallel connection of two two-port networks is shown in Fig. 19.40. For the two networks,

$$\mathbf{I}_{1a} = \mathbf{y}_{11a}\mathbf{V}_{1a} + \mathbf{y}_{12a}\mathbf{V}_{2a} \n\mathbf{I}_{2a} = \mathbf{y}_{21a}\mathbf{V}_{1a} + \mathbf{y}_{22a}\mathbf{V}_{2a}$$

\n(19.50)

and

$$\mathbf{I}_{1b} = \mathbf{y}_{11b} \mathbf{V}_{1b} + \mathbf{y}_{12b} \mathbf{V}_{2b}$$

\n

$$\mathbf{I}_{2a} = \mathbf{y}_{21b} \mathbf{V}_{1b} + \mathbf{y}_{22b} \mathbf{V}_{2b}$$

(19.51)

But from Fig. 19.40,

$$V_1 = V_{1a} = V_{1b}, \qquad V_2 = V_{2a} = V_{2b} \tag{19.52a}$$

I1 = I¹a + I¹b, I2 = I²a + I^2b (19.52b)

Figure 19.40 Parallel connection of two two-port networks.

Substituting Eqs. (19.50) and (19.51) into Eq. (19.52b) yields

$$\mathbf{I}_1 = (\mathbf{y}_{11a} + \mathbf{y}_{11b})\mathbf{V}_1 + (\mathbf{y}_{12a} + \mathbf{y}_{12b})\mathbf{V}_2 \n\mathbf{I}_2 = (\mathbf{y}_{21a} + \mathbf{y}_{21b})\mathbf{V}_1 + (\mathbf{y}_{22a} + \mathbf{y}_{22b})\mathbf{V}_2$$

\n(19.53)

Thus, the y parameters for the overall network are

[

$$\begin{bmatrix}\n\mathbf{y}_{11} & \mathbf{y}_{12} \\ \mathbf{y}_{21} & \mathbf{y}_{22}\n\end{bmatrix} = \begin{bmatrix}\n\mathbf{y}_{11a} + \mathbf{y}_{11b} & \mathbf{y}_{12a} + \mathbf{y}_{12b} \\ \mathbf{y}_{21a} + \mathbf{y}_{21b} & \mathbf{y}_{22a} + \mathbf{y}_{22b}\n\end{bmatrix}$$

\n(19.54)

or

$$[y] = [y_a] + [y_b]$$

(19.55)

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. 19.41. For the two networks,

$$\begin{bmatrix} \mathbf{V}_{1a} \\ \mathbf{I}_{1a} \end{bmatrix} = \begin{bmatrix} \mathbf{A}_a & \mathbf{B}_a \\ \mathbf{C}_a & \mathbf{D}_a \end{bmatrix} \begin{bmatrix} \mathbf{V}_{2a} \\ -\mathbf{I}_{2a} \end{bmatrix}$$

(19.56)

$$\begin{bmatrix} \mathbf{V}_{1b} \\ \mathbf{I}_{1b} \end{bmatrix} = \begin{bmatrix} \mathbf{A}_b & \mathbf{B}_b \\ \mathbf{C}_b & \mathbf{D}_b \end{bmatrix} \begin{bmatrix} \mathbf{V}_{2b} \\ -\mathbf{I}_{2b} \end{bmatrix}$$

(19.57)

From Fig. 19.41,

$$\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_1 \end{bmatrix} = \begin{bmatrix} \mathbf{V}_{1a} \\ \mathbf{I}_{1a} \end{bmatrix}, \quad \begin{bmatrix} \mathbf{V}_{2a} \\ -\mathbf{I}_{2a} \end{bmatrix} = \begin{bmatrix} \mathbf{V}_{1b} \\ \mathbf{I}_{1b} \end{bmatrix}, \quad \begin{bmatrix} \mathbf{V}_{2b} \\ -\mathbf{I}_{2b} \end{bmatrix} = \begin{bmatrix} \mathbf{V}_2 \\ -\mathbf{I}_2 \end{bmatrix}, (19.58)$$

Substituting these into Eqs. (19.56) and (19.57),

$$\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_1 \end{bmatrix} = \begin{bmatrix} \mathbf{A}_a & \mathbf{B}_a \\ \mathbf{C}_a & \mathbf{D}_a \end{bmatrix} \begin{bmatrix} \mathbf{A}_b & \mathbf{B}_b \\ \mathbf{C}_b & \mathbf{D}_b \end{bmatrix} \begin{bmatrix} \mathbf{V}_2 \\ -\mathbf{I}_2 \end{bmatrix}$$

(19.59)

Thus, the transmission parameters for the overall network are the product of the transmission parameters for the individual transmission parameters:

$$\begin{bmatrix} A & B \\ C & D \end{bmatrix} = \begin{bmatrix} A_a & B_a \\ C_a & D_a \end{bmatrix} \begin{bmatrix} A_b & B_b \\ C_b & D_b \end{bmatrix}$$

(19.60)

or

$$[\mathbf{T}] = [\mathbf{T}_a][\mathbf{T}_b]$$

(19.61)

I1	I1a		I2a	I1b		I2b	I2
+	+		+	+		+	+
V1	V1a	Na	V2a	V1b	Nb	V2b	V2
‒	‒		‒	‒		‒	‒

Figure 19.41

Cascade connection of two two-port networks.

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 Na and Nb are cascaded.

Example 19.12 Evaluate V2∕Vs in the circuit in Fig. 19.42.

Figure 19.42 For Example 19.12.

Solution:

This may be regarded as two two-ports in series. For Nb,

$$\mathbf{z}_{12b} = \mathbf{z}_{21b} = 10 = \mathbf{z}_{11b} = \mathbf{z}_{22b}$$

Thus,

$$[\mathbf{z}] = [\mathbf{z}_a] + [\mathbf{z}_b] = \begin{bmatrix} 12 & 8 \\ 8 & 20 \end{bmatrix} + \begin{bmatrix} 10 & 10 \\ 10 & 10 \end{bmatrix} = \begin{bmatrix} 22 & 18 \\ 18 & 30 \end{bmatrix}$$

But

$$V_1 = z_{11}I_1 + z_{12}I_2 = 22I_1 + 18I_2 \qquad (19.12.1)$$ $$\mathbf{V}_2 = \mathbf{z}_{21}\mathbf{I}_1 + \mathbf{z}_{22}\mathbf{I}_2 = 18\mathbf{I}_1 + 30\mathbf{I}_2 \tag{19.12.2}$$

Also, at the input port

$$\mathbf{V}_1 = \mathbf{V}_s - 5\mathbf{I}_1 \tag{19.12.3}$$

and at the output port

$$V_2 = -20I_2$$

$\Rightarrow$ $I_2 = -\frac{V_2}{20}$ (19.12.4)

Substituting Eqs. (19.12.3) and (19.12.4) into Eq. (19.12.1) gives

$$\mathbf{V}_s - 5\mathbf{I}_1 = 22\mathbf{I}_1 - \frac{18}{20}\mathbf{V}_2 \qquad \Rightarrow \qquad \mathbf{V}_s = 27\mathbf{I}_1 - 0.9\mathbf{V}_2 \tag{19.12.5}$$

while substituting Eq. (19.12.4) into Eq. (19.12.2) yields

$$\mathbf{V}_2 = 18\mathbf{I}_1 - \frac{30}{20}\mathbf{V}_2 \qquad \Rightarrow \qquad \mathbf{I}_1 = \frac{2.5}{18}\mathbf{V}_2 \tag{19.12.6}$$

Substituting Eq. (19.12.6) into Eq. (19.12.5), 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,

$$\frac{\mathbf{V}_2}{\mathbf{V}_s} = \frac{1}{2.85} = 0.3509$$

Find V2∕Vs in the circuit in Fig. 19.43. Practice Problem 19.12

Answer: 0.6799⧸−29.05°.

Find the y parameters of the two-port in Fig. 19.44. Example 19.13

Solution:

Let us refer to the upper network as Na and the lower one as Nb. The two networks are connected in parallel. Comparing Na and Nb with the circuit in Fig. 19.13(a), we obtain

$$y_{12a} = -j4 = y_{21a}$$

, $y_{11a} = 2 + j4$ , $y_{22a} = 3 + j4$

or

$$[\mathbf{y}_a] = \begin{bmatrix} 2+j4 & -j4 \\ -j4 & 3+j4 \end{bmatrix} \text{S}$$

and

$$y_{12b} = -4 = y_{21b}$$

, $y_{11b} = 4 - j2$ , $y_{22b} = 4 - j6$

or

$$[\mathbf{y}_b] = \begin{bmatrix} 4 - j2 & -4 \\ -4 & 4 - j6 \end{bmatrix} \text{S}$$

The overall y parameters are

$$[\mathbf{y}] = [\mathbf{y}_a] + [\mathbf{y}_b] = \begin{bmatrix} 6+j2 & -4-j4 \\ -4-j4 & 7-j2 \end{bmatrix} \text{S}$$

Figure 19.44 For Example 19.13.

Practice Problem 19.13 Obtain the y parameters for the network in Fig. 19.45.

Answer:

$$\begin{bmatrix} 27 - j15 & -25 + j10 \ -25 + j10 & 27 - j5 \end{bmatrix}$$

S.

Figure 19.45 For Practice Prob. 19.13.

Figure 19.46 For Example 19.14.

Example 19.14 Find the transmission parameters for the circuit in Fig. 19.46.

Solution:

We can regard the given circuit in Fig. 19.46 as a cascade connection of two T networks as shown in Fig. 19.47(a). We can show that a T network, shown in Fig. 19.47(b), has the following transmission parameters [see Prob. 19.52(b)]:

$$\mathbf{A} = 1 + \frac{R_1}{R_2}, \qquad \mathbf{B} = R_3 + \frac{R_1(R_2 + R_3)}{R_2}$$ $$\mathbf{C} = \frac{1}{R_2}, \qquad \mathbf{D} = 1 + \frac{R_3}{R_2}$$

Applying this to the cascaded networks Na and Nb in Fig. 19.47(a), we get

$$\mathbf{A}_a = 1 + 4 = 5, \qquad \mathbf{B}_a = 8 + 4 \times 9 = 44 \text{ }\Omega$$ $$\mathbf{C}_a = 1 \text{ S}, \qquad \mathbf{D}_a = 1 + 8 = 9$$

or in matrix form,

$$[\mathbf{T}_a] = \begin{bmatrix} 5 & 44 \ \Omega \\ 1 \ \mathrm{S} & 9 \end{bmatrix}$$ $$[1_a] = \begin{bmatrix} 1 & 0 & 9 \end{bmatrix}$$ $$A_b = 1$$

, $B_b = 6 \Omega$ , $C_b = 0.5 S$ , $D_b = 1 + \frac{6}{2} = 4$

i.e.,

and

Thus, for the total network in Fig. 19.46,

$$\begin{aligned} [\mathbf{T}] &= [\mathbf{T}_a][\mathbf{T}_b] = \begin{bmatrix} 5 & 44 \\ 1 & 9 \end{bmatrix} \begin{bmatrix} 1 & 6 \\ 0.5 & 4 \end{bmatrix} \\ &= \begin{bmatrix} 5 \times 1 + 44 \times 0.5 & 5 \times 6 + 44 \times 4 \\ 1 \times 1 + 9 \times 0.5 & 1 \times 6 + 9 \times 4 \end{bmatrix} \\ &= \begin{bmatrix} 27 & 206 \ \Omega \\ 5.5 \ \text{S} & 42 \end{bmatrix} \end{aligned}$$

Figure 19.47

For Example 19.14: (a) Breaking the circuit in Fig. 19.46 into two two-ports, (b) a general T two-port.

Notice that

$$\Delta_{T_a}=\Delta_{T_b}=\Delta_T=1$$

showing that the network is reciprocal.