19.5 Transmission Parameters

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19.5 Transmission Parameters

Because there are no restrictions on which terminal voltages and currents should be considered independent and which should be dependent v ariables, we expect to be able to generate many sets of parameters. Another

V¹ V²

(a)

1/s s

I1

• ‒ I I^{1 2} = 0

1/s s

1 Ω

Determining the g parameters in the s domain for the circuit in Fig. 19.28.

Practice Problem 19.7

For Practice Prob. 19.7.

‒

set of parameters relates the variables at the input port to those at the output port. Thus,

$$\mathbf{V}_1 = \mathbf{A}\mathbf{V}_2 - \mathbf{B}\mathbf{I}_2$$ $$\mathbf{I}_1 = \mathbf{C}\mathbf{V}_2 - \mathbf{D}\mathbf{I}_2$$

(19.22)

[ V1 ^I1 ] ⁼[ A B C D] [ V2 −I² ] = [T] [ V2 −I² ] (19.23)

Equations (19.22) and (19.23) relate the input variables (V1 and I1) to the output variables (V2 and −I2). Notice that in computing the transmission parameters, −I2 is used rather than I2, because the current is considered to be leaving the network, as shown in Fig. 19.31, as opposed to enter ing the network as in Fig. 19.1(b). This is done merely for conventional reasons; when you cascade two-ports (output to input), it is most logical to think of I2 as leaving the two-port. It is also customary in the power industry to consider I2 as leaving the two-port.

The two-port parameters in Eqs. (19.22) and (19.23) provide a measure of how a circuit transmits v oltage and current from a source to a load. They are useful in the analysis of transmission lines (such as cable and fiber) because they e xpress sending-end v ariables ( V1 and I1) in terms of the receiving-end variables (V2 and −I2). For this reason, the y are called transmission parameters. They are also known as ABCD parameters. They are used in the design of telephone systems, micro wave networks, and radars.

The transmission parameters are determined as

$$\mathbf{A} = \frac{\mathbf{V}_1}{\mathbf{V}_2} \Big|_{\mathbf{I}_2 = 0}, \qquad \mathbf{B} = -\frac{\mathbf{V}_1}{\mathbf{I}_2} \Big|_{\mathbf{V}_2 = 0}$$

\n

$$\mathbf{C} = \frac{\mathbf{I}_1}{\mathbf{V}_2} \Big|_{\mathbf{I}_2 = 0}, \qquad \mathbf{D} = -\frac{\mathbf{I}_1}{\mathbf{I}_2} \Big|_{\mathbf{V}_2 = 0}$$

\n(19.24)

Thus, the transmission parameters are called, specifically,

A = Open-circuit voltage ratio B = Negative short-circuit transfer impedance C = Open-circuit transfer admittance (19.25) D = Negative short-circuit current ratio

A and D are dimensionless, B is in ohms, and C is in siemens. Because the transmission parameters provide a direct relationship between input and output variables, they are very useful in cascaded networks.

Our last set of parameters may be defined by expressing the variables at the output port in terms of the variables at the input port. We obtain

$$\mathbf{V}_2 = \mathbf{a}\mathbf{V}_1 - \mathbf{b}\mathbf{I}_1$$ $$\mathbf{I}_2 = \mathbf{c}\mathbf{V}_1 - \mathbf{d}\mathbf{I}_1$$

(19.26)

Figure 19.31 Terminal variables used to define the ADCB parameters.

$$\begin{bmatrix} \mathbf{V}_2 \\ \mathbf{I}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{a} & \mathbf{b} \\ \mathbf{c} & \mathbf{d} \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \\ -\mathbf{I}_1 \end{bmatrix} = [\mathbf{t}] \begin{bmatrix} \mathbf{V}_1 \\ -\mathbf{I}_1 \end{bmatrix}$$

(19.27)

The parameters a, b, c, and d are called the inverse transmission, or t, parameters. They are determined as follows:

$$\mathbf{a} = \frac{\mathbf{V}_2}{\mathbf{V}_1} \bigg|_{\mathbf{I}_1 = 0}, \qquad \mathbf{b} = -\frac{\mathbf{V}_2}{\mathbf{I}_1} \bigg|_{\mathbf{V}_1 = 0}$$

\n

$$\mathbf{c} = \frac{\mathbf{I}_2}{\mathbf{V}_1} \bigg|_{\mathbf{I}_1 = 0}, \qquad \mathbf{d} = -\frac{\mathbf{I}_2}{\mathbf{I}_1} \bigg|_{\mathbf{V}_1 = 0}$$

\n(19.28)

From Eq. (19.28) and from our experience so far, it is evident that these parameters are known individually as

a = Open-circuit voltage gain ^b = Negative short-circuit transfer impedance (19.29) c = Open-circuit transfer admittance d = Negative short-circuit current gain

While a and d are dimensionless, b and c are in ohms and siemens, respectively.

In terms of the transmission or in verse transmission parameters, a network is reciprocal if

AD – BC = 1,

$$ad - bc = 1$$

(19.30)

These relations can be pro ved in the same w ay as the transfer imped ance relations for the z parameters. Alternatively, we will be able to use Table 19.1 a little later to deri ve Eq. (19.30) from the f act that z12 = z²¹ for reciprocal networks.

Find the transmission parameters for the two-port network in Fig. 19.32. Example 19.8

Solution:

To determine A and C, we leave the output port open as in Fig. 19.33(a) so that I2 = 0 and place a voltage source V1 at the input port. We have

$$V_1 = (10 + 20)I_1 = 30I_1$$

and $V_2 = 20I_1 - 3I_1 = 17I_1$

Thus,

$$A = \frac{V_1}{V_2} = \frac{30I_1}{17I_1} = 1.765, \qquad C = \frac{I_1}{V_2} = \frac{I_1}{17I_1} = 0.0588 \text{ S}$$

To obtain B and D, we short-circuit the output port so that V2 = 0 as shown in Fig. 19.33(b) and place a voltage source V1 at the input port. At node a in the circuit of Fig. 19.33(b), KCL gives

$$\frac{\mathbf{V}_1 - \mathbf{V}_a}{10} - \frac{\mathbf{V}_a}{20} + \mathbf{I}_2 = 0$$

(19.8.1)

Figure 19.33

For Example 19.8: (a) finding A and C, (b) finding B and D.

But Va = 3I1 and I1 = (V1 − Va)∕10. Combining these gives

$$V_a = 3I_1 \t V_1 = 13I_1 \t (19.8.2)$$

Substituting Va = 3I1 into Eq. (19.8.1) and replacing the first term with I1,

$$\mathbf{I}_1 - \frac{3\mathbf{I}_1}{20} + \mathbf{I}_2 = 0 \qquad \Rightarrow \qquad \frac{17}{20}\mathbf{I}_1 = -\mathbf{I}_2$$

Therefore,

$$\mathbf{D} = -\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{20}{17} = 1.176, \qquad \mathbf{B} = -\frac{\mathbf{V}_1}{\mathbf{I}_2} = \frac{-13\mathbf{I}_1}{(-17/20)\mathbf{I}_1} = 15.29 \text{ }\Omega$$

Practice Problem 19.8 Find the transmission parameters for the circuit in Fig. 19.16 (see Practice Prob. 19.3).

Answer: A = 1.5, B = 11 Ω, C = 250 mS, D = 2.5.

Figure 19.34 For Example 19.9.

Example 19.9 The ABCD parameters of the two-port network in Fig. 19.34 are

4	20 Ω
[	]
0.1 S	2

The output port is connected to a v ariable load for maximum po wer transfer. Find RL and the maximum power transferred.

Solution:

What we need is to find the Thevenin equivalent (ZTh and VTh) at the load or output port. We find ZTh using the circuit in Fig. 19.35(a). Our goal is to get ZTh = V2∕I2. Substituting the given ABCD parameters into Eq. (19.22), we obtain

$$V_1 = 4V_2 - 20I_2 \tag{19.9.1}$$ $$I_1 = 0.1V_2 - 2I_2 \tag{19.9.2}$$

At the input port, V1 = −10I1. Substituting this into Eq. (19.9.1) gives

$$-10\mathbf{I}_1 = 4\mathbf{V}_2 - 20\mathbf{I}_2$$ $$I_1 = -0.4V_2 + 2I_2 \tag{19.9.3}$$

Figure 19.35

Solution of Example 19.9: (a) finding ZTh, (b) finding VTh, (c) finding RL for maximum power transfer.

Setting the right-hand sides of Eqs. (19.9.2) and (19.9.3) equal,

$$0.1\mathbf{V}_2 - 2\mathbf{I}_2 = -0.4\mathbf{V}_2 + 2\mathbf{I}_2 \implies 0.5\mathbf{V}_2 = 4\mathbf{I}_2$$

Hence,

$$Z_{\text{Th}} = \frac{V_2}{I_2} = \frac{4}{0.5} = 8 \ \Omega$$

To find VTh, we use the circuit in Fig. 19.35(b). At the output port I2 = 0 and at the input port V1 = 50 − 10I1. Substituting these into Eqs. (19.9.1) and (19.9.2),

$$50 - 10I_1 = 4V_2 \tag{19.9.4}$$ $$\mathbf{I}_1 = 0.1 \mathbf{V}_2 \tag{19.9.5}$$

Substituting Eq. (19.9.5) into Eq. (19.9.4),

$$50 - V_2 = 4V_2 \qquad \Rightarrow \qquad V_2 = 10$$

Thus,

$$\mathbf{V}_{\mathrm{Th}} = \mathbf{V}_2 = 10 \mathrm{V}$$

The equivalent circuit is shown in Fig. 19.35(c). For maximum power transfer,

$$R_L = \mathbf{Z}_{\text{Th}} = 8 \ \Omega$$

From Eq. (4.24), the maximum power is

$$P = I^2 R_L = \left(\frac{\mathbf{V}_{\text{Th}}}{2R_L}\right)^2 R_L = \frac{\mathbf{V}_{\text{Th}}^2}{4R_L} = \frac{100}{4 \times 8} = 3.125 \text{ W}$$

Find I1 and I2 if the transmission parameters for the two-port in Fig. 19.36 Practice Problem 19.9 are

For Practice Prob. 19.9.

Answer: 1 A, −0.2 A.

19.6 † Relationships Between Parameters

Because the six sets of parameters relate the same input and output terminal variables of the same two-port network, they should be interrelated. If two sets of parameters exist, we can relate one set to the other set. Let us demonstrate the process with two examples.

Given the z parameters, let us obtain the y parameters. From Eq. (19.2),

$$\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{z}_{11} & \mathbf{z}_{12} \\ \mathbf{z}_{21} & \mathbf{z}_{22} \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} = [\mathbf{z}] \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix}$$

(19.31)

or

$$\begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} = [\mathbf{z}]^{-1} \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix}$$

(19.32)

Also, from Eq. (19.9),

$$\begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{y}_{11} & \mathbf{y}_{12} \\ \mathbf{y}_{21} & \mathbf{y}_{22} \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = [\mathbf{y}] \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix}$$

(19.33)

Comparing Eqs. (19.32) and (19.33), we see that

[

$$[y] = [z]^{-1}$$

(19.34)

The adjoint of the [z] matrix is

$$\begin{bmatrix} \mathbf{z}_{22} & -\mathbf{z}_{12} \\ -\mathbf{z}_{21} & \mathbf{z}_{11} \end{bmatrix}$$

and its determinant is

$$\Delta_z = \mathbf{z}_{11}\mathbf{z}_{22} - \mathbf{z}_{12}\mathbf{z}_{21}$$

Substituting these into Eq. (19.34), we get

$$\begin{bmatrix} \mathbf{y}_{11} & \mathbf{y}_{12} \\ \mathbf{y}_{21} & \mathbf{y}_{22} \end{bmatrix} = \frac{\begin{bmatrix} \mathbf{z}_{22} & -\mathbf{z}_{12} \\ -\mathbf{z}_{21} & \mathbf{z}_{11} \end{bmatrix}}{\Delta_z}$$

(19.35)

Equating terms yields

$$y_{11} = \frac{z_{22}}{\Delta_z}$$

, $y_{12} = -\frac{z_{12}}{\Delta_z}$ , $y_{21} = -\frac{z_{21}}{\Delta_z}$ , $y_{22} = \frac{z_{11}}{\Delta_z}$ (19.36)

As a second example, let us determine the h parameters from the z parameters. From Eq. (19.1),

$$\mathbf{V}_1 = \mathbf{z}_{11}\mathbf{I}_1 + \mathbf{z}_{12}\mathbf{I}_2 \tag{19.37a}$$ $$V_2 = z_{21}I_1 + z_{22}I_2 \tag{19.37b}$$