Solution:

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According to Eq. (19.3), we obtain z11 and z21 by connecting a voltage V1 (or a current source I1) to port 1 with port 2 open-circuited as in Fig. 19.3(a) and finding I1 and V2; we then get

$$z_{11} = \frac{V_1}{I_1}, \qquad z_{21} = \frac{V_2}{I_1}$$

(19.5)

Similarly, we obtain z12 and z22 by connecting a voltage V2 (or a current source I2) to port 2 with port 1 open-circuited as in Fig. 19.3(b) and finding I2 and V1; we then get

$$z_{12} = \frac{V_1}{I_2}, \qquad z_{22} = \frac{V_2}{I_2}$$

(19.6)

The above procedure pro vides us with a means of calculating or mea suring the z parameters.

Sometimes z11 and z22 are called driving-point impedances, while z²¹ and z12 are called transfer impedances. A driving-point impedance is the input impedance of a two-terminal (one-port) device. Thus, z11 is the input driving-point impedance with the output port open-circuited, while z22 is the output driving-point impedance with the input port open-circuited.

When z11 = z22, the two-port network is said to be symmetrical. This implies that the network has mirrorlike symmetry about some center line; that is, a line can be found that divides the network into two similar halves.

When the two-port network is linear and has no dependent sources, the transfer impedances are equal (z12 = z21), and the two-port is said to be reciprocal. This means that if the points of excitation and response are interchanged, the transfer impedances remain the same. As illustrated in Fig. 19.4, a tw o-port is reciprocal if interchanging an ideal v oltage source at one port with an ideal ammeter at the other port gives the same ammeter reading. The reciprocal netw ork yields V = z12I according to Eq. (19.1) when connected as in Fig. 19.4(a), but yields V = z21I when connected as in Fig. 19.4(b). This is possible only if z12 = z21. Any twoport that is made entirely of resistors, capacitors, and inductors must be reciprocal. A reciprocal network can be replaced by the T-equivalent circuit in Fig. 19.5(a). If the netw ork is not reciprocal, a more general equivalent network is shown in Fig. 19.5(b); notice that this figure follows directly from Eq. (19.1).

It should be mentioned that for some tw o-port netw orks, the z parameters do not exist because they cannot be described by Eq. (19.1). As an example, consider the ideal transformer of Fig. 19.6. The defining equations for the two-port network are:

$$V_1 = \frac{1}{n} V_2, \qquad I_1 = -nI_2 \tag{19.7}$$

Observe that it is impossible to express the voltages in terms of the currents, and vice versa, as Eq. (19.1) requires. Thus, the ideal transformer has no z parameters. Ho wever, it does ha ve hybrid parameters, as we shall see in Section 19.4.

Determine the z parameters for the circuit in Fig. 19.7. Example 19.1

Solution:

■ METHOD 1 To determine z11 and z21, we apply a v oltage source V1 to the input port and lea ve the output port open as in Fig. 19.8(a). Then,

$$\mathbf{z}_{11} = \frac{\mathbf{V}_1}{\mathbf{I}_1} = \frac{(20 + 40)\mathbf{I}_1}{\mathbf{I}_1} = 60 \ \Omega$$

that is, z11 is the input impedance at port 1.

$$z_{21} = \frac{V_2}{I_1} = \frac{40I_1}{I_1} = 40 \Omega$$

To find z12 and z22, we apply a voltage source V2 to the output port and leave the input port open as in Fig. 19.8(b). Then,

$$\mathbf{z}_{12} = \frac{\mathbf{V}_1}{\mathbf{I}_2} = \frac{40\mathbf{I}_2}{\mathbf{I}_2} = 40 \ \Omega, \qquad \mathbf{z}_{22} = \frac{\mathbf{V}_2}{\mathbf{I}_2} = \frac{(30 + 40)\mathbf{I}_2}{\mathbf{I}_2} = 70 \ \Omega$$

Thus,

$$[\mathbf{z}] = \begin{bmatrix} 60 \,\Omega & 40 \,\Omega \\ 40 \,\Omega & 70 \,\Omega \end{bmatrix}$$

■ METHOD 2 Alternatively, as there is no dependent source in the given circuit, z12 = z21 and we can use Fig. 19.5(a). Comparing Fig. 19.7 with Fig. 19.5(a), we get

$$\mathbf{z}_{12} = 40 \ \Omega = \mathbf{z}_{21}$$

\n

$$\mathbf{z}_{11} - \mathbf{z}_{12} = 20 \qquad \Rightarrow \qquad \mathbf{z}_{11} = 20 + \mathbf{z}_{12} = 60 \ \Omega$$

\n

$$\mathbf{z}_{22} - \mathbf{z}_{12} = 30 \qquad \Rightarrow \qquad \mathbf{z}_{22} = 30 + \mathbf{z}_{12} = 70 \ \Omega$$

Find the z parameters of the two-port network in Fig. 19.9. Practice Problem 19.1

Answer: z11 = 12 Ω, z12 = z21 = z22 = 4 Ω.

Figure 19.6

An ideal transformer has no z parameters.

Figure 19.7 For Example 19.1.

Figure 19.8

For Example 19.1: (a) finding z11 and z21, (b) finding z12 and z22.

Figure 19.9 For Practice Prob. 19.1.

Solution:

This is not a reciprocal netw ork. We may use the equi valent circuit in Fig. 19.5(b) b ut we can also use Eq. (19.1) directly . Substituting the given z parameters into Eq. (19.1),

$$V_1 = 40I_1 + j20I_2 \tag{19.2.1}$$ $$V_2 = j30I_1 + 50I_2 \tag{19.2.2}$$

Because we are looking for I1 and I2, we substitute

$$V_1 = 100/0^\circ$$

, $V_2 = -10I_2$

into Eqs. (19.2.1) and (19.2.2), which become

$$100 = 40I_1 + j20I_2 \tag{19.2.3}$$ $$-10I_2 = j30I_1 + 50I_2 \qquad \Rightarrow \qquad I_1 = j2I_2 \tag{19.2.4}$$

Substituting Eq. (19.2.4) into Eq. (19.2.3) gives

$$100 = j80I_2 + j20I_2 \qquad \Rightarrow \qquad I_2 = \frac{100}{j100} = -j$$

From Eq. (19.2.4), I1 = j2(−j) = 2. Thus,

$$I_1 = 2/0^\circ A$$

, $I_2 = 1/-90^\circ A$

Practice Problem 19.2 Calculate I1 and I2 in the two-port of Fig. 19.11.

Figure 19.11 For Practice Prob. 19.2.

Answer: 800⧸30° mA, 400⧸120° mA.

19.3 Admittance Parameters

In the previous section we saw that impedance parameters may not exist for a tw o-port network. So there is a need for an alternati ve means of describing such a netw ork. This need may be met by the second set of parameters, which we obtain by expressing the terminal currents in terms of the terminal voltages. In either Fig. 19.12(a) or (b), the terminal cur rents can be expressed in terms of the terminal voltages as

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

\n(19.8)

or in matrix form as

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

(19.9)

The y terms are kno wn as the admittance par ameters (or , simply , y parameters) and have units of siemens.

The values of the parameters can be determined by setting V1 = 0 (input port short-circuited) or V2 = 0 (output port short-circuited). Thus,

$$\mathbf{y}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1} \Big|_{\mathbf{V}_2=0}, \quad \mathbf{y}_{12} = \frac{\mathbf{I}_1}{\mathbf{V}_2} \Big|_{\mathbf{V}_1=0}$$

\n

$$\mathbf{y}_{21} = \frac{\mathbf{I}_2}{\mathbf{V}_1} \Big|_{\mathbf{V}_2=0}, \quad \mathbf{y}_{22} = \frac{\mathbf{I}_2}{\mathbf{V}_2} \Big|_{\mathbf{V}_1=0}$$

\n(19.10)

Because the y parameters are obtained by short-circuiting the input or output port, they are also called the short-circuit admittance parameters. Specifically,

• y11 = Short-circuit input admittance
• y12 = Short-circuit transfer admittance from port 2 to port 1
• y21 = Short-circuit transfer admittance from port 1 to port 2 (19.11)
• y22 = Short-circuit output admittance

Following Eq. (19.10), we obtain y11 and y21 by connecting a current I1 to port 1 and short-circuiting port 2 as in Fig. 19.12(a), finding V1 and I2, and then calculating

$$\mathbf{y}_{11} = \frac{\mathbf{I}_1}{\mathbf{V}_1}, \qquad \mathbf{y}_{21} = \frac{\mathbf{I}_2}{\mathbf{V}_1}$$

(19.12)

Similarly, we obtain y12 and y22 by connecting a current source I2 to port 2 and short-circuiting port 1 as in Fig. 19.12(b), finding I1 and V2, and then getting

$$y_{12} = \frac{I_1}{V_2}, \qquad y_{22} = \frac{I_2}{V_2}$$

(19.13)