Chapter 19, Problem 61P

For the bridge circuit in Fig. 19.108, obtain:

1. (a) the z parameters
2. (b) the h parameters
3. (c) the transmission parameters

Figure 19.108

To determine

Calculate z parameters for the network in Figure 19.108 in the textbook.

Answer to Problem 61P

The z parameters for the given two-port network is $\underset{\_}{\left[\begin{array}{cc}\frac{5}{3}& \frac{4}{3}\\ \frac{4}{3}& \frac{5}{3}\end{array}\right]\Omega }$.

Explanation of Solution

Given Data:

Refer to Figure 19.108 in the textbook for the given two-port network.

Formula used:

Write the expressions for impedance parameters of a two-port network as follows:

$V_1=z_{11}{I}_1+z_{12}{I}_2$ (1)

$V_{\text{2}}=z_{\text{21}}{I}_{\text{1}}+{\text{z}}_{\text{22}}{I}_{\text{2}}$ (2)

Calculation:

The impedance parameters $z_{\text{11}}$ and $z_{\text{21}}$ are obtained when the port-2 of the network is open circuited. The current $I_{\text{2}}$ becomes zero when port-2 is open-circuited. Therefore, rewrite the expressions in Equation (1) and (2) by substituting 0 A for $I_{\text{2}}$ as follows:

$\begin{array}{c}{V}_{\text{1}}=z_{\text{11}}{I}_1+z_{\text{12}}\left(0\right)\\ =z_{\text{11}}{I}_1\end{array}$

$z_{\text{11}}=\frac{V_{\text{1}}}{I_{\text{1}}}$ (3)

$\begin{array}{c}{V}_{\text{2}}=z_{\text{21}}{I}_{\text{1}}+z_{\text{22}}\left(0\right)\\ =z_{\text{21}}{I}_{\text{1}}\end{array}$

$z_{\text{21}}=\frac{V_{\text{2}}}{I_{\text{1}}}$ (4)

Redraw the given two-port network by open circuiting the port-2 as shown in Figure 1.

Apply KCL at node $V$ in Figure 1.

$\begin{array}{l}\frac{V-V_{\text{1}}}{1}+\frac{V}{1}+\frac{V-V_2}{1}=0\\ 3V-V_{\text{1}}-V_2=0\\ V=\frac{V_{\text{1}}+V_2}{3}\end{array}$

Apply KCL at node $V_2$ in Figure 1.

$\begin{array}{l}\frac{V_2-V}{1}+\frac{V_2-V_1}{1}=0\\ 2V_2-V-V_{\text{1}}=0\\ V_2=\frac{V+V_{\text{1}}}{2}\end{array}$

Substitute $\left(\frac{V_{\text{1}}+V_2}{3}\right)$ for $V$ as follows:

$\begin{array}{c}{V}_2=\frac{\left(\frac{V_{\text{1}}+V_2}{3}\right)+V_{\text{1}}}{2}\\ =\frac{V_{\text{1}}+V_2+3V_{\text{1}}}{6}\end{array}$

$5V_2=4V_{\text{1}}$

$V_2=\frac{4}{5}{V}_{\text{1}}$ (5)

Apply KCL at node $V_1$ in Figure 1.

$\begin{array}{l}{I}_{\text{1}}=\frac{V_{\text{1}}-V}{1}+\frac{V_1-V_2}{1}\\ 2V_{\text{1}}-V-V_2=I_{\text{1}}\end{array}$

Substitute $\left(\frac{V_{\text{1}}+V_2}{3}\right)$ for $V$ and $\left(\frac{4}{5}{V}_{\text{1}}\right)$ for $V_2$ as follows:

$\begin{array}{l}2V_{\text{1}}-\left(\frac{V_{\text{1}}+V_2}{3}\right)-\left(\frac{4}{5}{V}_{\text{1}}\right)=I_{\text{1}}\\ 30V_{\text{1}}-5V_{\text{1}}-5V_2-12V_{\text{1}}=15I_{\text{1}}\\ 13V_{\text{1}}-5V_2=15I_{\text{1}}\end{array}$

Substitute $\left(\frac{4}{5}{V}_{\text{1}}\right)$ for $V_2$ as follows:

$\begin{array}{l}13V_{\text{1}}-5\left(\frac{4}{5}{V}_{\text{1}}\right)=15I_{\text{1}}\\ 13V_{\text{1}}-4V_{\text{1}}=15I_{\text{1}}\\ 9V_{\text{1}}=15I_{\text{1}}\end{array}$

Simplify the expression as follows:

$\begin{array}{c}\frac{V_{\text{1}}}{I_{\text{1}}}=\frac{15}{9}\Omega \\ =\frac{5}{3}\Omega \end{array}$

Substitute $\frac{5}{3}\Omega$ for $\frac{V_{\text{1}}}{I_{\text{1}}}$ in Equation (3) to obtain the value of $z_{\text{11}}$.

$z_{\text{11}}=\frac{5}{3}\Omega$

From Equation (5), substitute $\frac{4}{5}{V}_{\text{1}}$ for $V_2$ in Equation (4) as follows:

$\begin{array}{c}{z}_{\text{21}}=\frac{\frac{4}{5}{V}_{\text{1}}}{I_{\text{1}}}\\ =\left(\frac{4}{5}\right)\left(\frac{V_{\text{1}}}{I_{\text{1}}}\right)\end{array}$

Substitute $\frac{5}{3}\Omega$ for $\frac{V_{\text{1}}}{I_{\text{1}}}$ to obtain the value of $z_{\text{21}}$.

$\begin{array}{c}{z}_{\text{21}}=\left(\frac{4}{5}\right)\left(\frac{5}{3}\Omega \right)\\ =\frac{4}{3}\Omega \end{array}$

The impedance parameters $z_{12}$ and $z_{\text{22}}$ are obtained when the port-1 of the network is open circuited. The current $I_{\text{1}}$ becomes zero when port-1 is open-circuited. Therefore, rewrite the expressions in Equation (1) and (2) by substituting 0 A for $I_{\text{1}}$ as follows:

$\begin{array}{c}{V}_{\text{1}}=z_{\text{11}}\left(0\right)+z_{\text{12}}{I}_{\text{2}}\\ =z_{\text{12}}{\text{I}}_{\text{2}}\end{array}$

$z_{\text{12}}=\frac{V_{\text{1}}}{I_{\text{2}}}$ (6)

$\begin{array}{c}{V}_{\text{2}}=z_{\text{21}}\left(0\right)+z_{\text{22}}{\text{I}}_{\text{2}}\\ =z_{\text{22}}{\text{I}}_{\text{2}}\end{array}$

$z_{\text{22}}=\frac{V_{\text{2}}}{I_{\text{2}}}$ (7)

Redraw the given two-port network by open circuiting the port-1 as shown in Figure 2.

Apply KCL at node $V$ in Figure 2.

$\begin{array}{l}\frac{V-V_{\text{1}}}{1}+\frac{V}{1}+\frac{V-V_2}{1}=0\\ 3V-V_{\text{1}}-V_2=0\end{array}$

$V=\frac{V_{\text{1}}+V_2}{3}$ (8)

Apply KCL at node $V_1$ in Figure 2.

$\begin{array}{l}\frac{V_{\text{1}}-V}{1}+\frac{V_1-V_2}{1}=0\\ 2V_{\text{1}}-V-V_2=0\end{array}$

Substitute $\left(\frac{V_{\text{1}}+V_2}{3}\right)$ for $V$ as follows:

$\begin{array}{l}2V_{\text{1}}-\frac{V_{\text{1}}+V_2}{3}-V_2=0\\ 6V_{\text{1}}-V_{\text{1}}-V_2-3V_2=0\\ 5V_{\text{1}}=4V_2\\ V_{\text{1}}=\frac{4}{5}{V}_2\end{array}$

Substitute $\frac{4}{5}{V}_2$ for $V_{\text{1}}$ in Equation (8) as follows:

$\begin{array}{c}V=\frac{\left(\frac{4}{5}{V}_2\right)+V_2}{3}\\ =\frac{4V_2+5V_2}{15}\end{array}$

$\begin{array}{l}15V=9V_2\\ V=\frac{3}{5}{V}_2\end{array}$

Apply KCL at node $V_2$ in Figure 2.

$\begin{array}{l}{I}_2=\frac{V_2-V}{1}+\frac{V_2-V_1}{1}\\ 2V_2-V-V_{\text{1}}=I_2\end{array}$

Substitute $\left(\frac{3}{5}{V}_2\right)$ for $V$ and $\left(\frac{4}{5}{V}_2\right)$ for $V_1$ as follows:

$\begin{array}{l}2V_2-\left(\frac{3}{5}{V}_2\right)-\left(\frac{4}{5}{V}_2\right)=I_2\\ 10V_2-3V_2-4V_2=5I_2\\ 3V_2=5I_2\\ \frac{V_2}{I_2}=\frac{5}{3}\Omega \end{array}$

Substitute $\frac{5}{3}\Omega$ for $\frac{V_{\text{2}}}{I_{\text{2}}}$ in Equation (7) to obtain the value of $z_{\text{22}}$.

$z_{\text{22}}=\frac{5}{3}\Omega$

Substitute $\left(\frac{4}{5}{V}_2\right)$ for $V_1$ in Equation (6) as follows:

$\begin{array}{c}{z}_{\text{12}}=\frac{\left(\frac{4}{5}{V}_2\right)}{I_{\text{2}}}\\ =\left(\frac{4}{5}\right)\left(\frac{V_2}{I_{\text{2}}}\right)\end{array}$

Substitute $\frac{5}{3}\Omega$ for $\frac{V_{\text{2}}}{I_{\text{2}}}$ to obtain the value of $z_{\text{12}}$.

$\begin{array}{c}{z}_{\text{12}}=\left(\frac{4}{5}\right)\left(\frac{5}{3}\Omega \right)\\ =\frac{4}{3}\Omega \end{array}$

From the calculations, write the z-parameters as follows:

$z=\left[\begin{array}{cc}\frac{5}{3}& \frac{4}{3}\\ \frac{4}{3}& \frac{5}{3}\end{array}\right]\Omega$

Conclusion:

Thus, the z parameters for the given two-port network is $\underset{\_}{\left[\begin{array}{cc}\frac{5}{3}& \frac{4}{3}\\ \frac{4}{3}& \frac{5}{3}\end{array}\right]\Omega }$.

To determine

Calculate hybrid [h] parameters from the obtained impedance parameters.

Answer to Problem 61P

The hybrid parameters for the given circuit are $\underset{\_}{\left[\begin{array}{cc}\frac{3}{5}\Omega & \frac{4}{5}\\ -\frac{4}{5}& \frac{3}{5}\text{S}\end{array}\right]}$.

Explanation of Solution

Formula used:

From the TABLE 19.1 in the textbook for Conversion of two-port parameters and write the expression for hybrid parameters in terms of z parameters as follows:

$\left[h\right]=\left[\begin{array}{cc}\frac{{\Delta }_z}{z_{22}}& \frac{z_{12}}{z_{22}}\\ -\frac{z_{21}}{z_{22}}& \frac{1}{z_{22}}\end{array}\right]$ (9)

Write the expression for ${\Delta }_z$ as follows:

${\Delta }_z=z_{11}{z}_{22}-z_{12}{z}_{21}$ (10)

Calculation:

From Part (a), the z parameters are written as follows:

$\begin{array}{l}{z}_{11}=\frac{5}{3}\Omega \\ z_{12}=\frac{4}{3}\Omega \\ z_{21}=\frac{4}{3}\Omega \\ z_{22}=\frac{5}{3}\Omega \end{array}$

Substitute $\frac{5}{3}\Omega$ for $z_{11}$, $\frac{4}{3}\Omega$ for $z_{12}$, $\frac{4}{3}\Omega$ for $z_{21}$, and $\frac{5}{3}\Omega$ for $z_{22}$ in Equation (10) to obtain the value of ${\Delta }_z$.

$\begin{array}{c}{\Delta }_z=\left(\frac{5}{3}\Omega \right)\left(\frac{5}{3}\Omega \right)-\left(\frac{4}{3}\Omega \right)\left(\frac{4}{3}\Omega \right)\\ =\frac{25}{9}{\Omega }^2-\frac{16}{9}{\Omega }^2\\ =1{\Omega }^2\end{array}$

Substitute $\frac{5}{3}\Omega$ for $z_{11}$, $\frac{4}{3}\Omega$ for $z_{12}$, $\frac{4}{3}\Omega$ for $z_{21}$, $\frac{5}{3}\Omega$ for $z_{22}$, and $1{\Omega }^2$ for ${\Delta }_z$ in Equation (9) to obtain the hybrid parameters.

$\begin{array}{c}\left[h\right]=\left[\begin{array}{cc}\frac{1{\Omega }^2}{\frac{5}{3}\Omega }& \frac{\frac{4}{3}\Omega }{\frac{5}{3}\Omega }\\ -\frac{\frac{4}{3}\Omega }{\frac{5}{3}\Omega }& \frac{1}{\frac{5}{3}\Omega }\end{array}\right]\\ =\left[\begin{array}{cc}\frac{3}{5}\Omega & \frac{4}{5}\\ -\frac{4}{5}& \frac{3}{5}\text{S}\end{array}\right]\end{array}$

Conclusion:

Thus, the hybrid parameters for the given circuit are $\underset{\_}{\left[\begin{array}{cc}\frac{3}{5}\Omega & \frac{4}{5}\\ -\frac{4}{5}& \frac{3}{5}\text{S}\end{array}\right]}$.

To determine

Calculate transmission [T] parameters from the obtained impedance parameters.

Answer to Problem 61P

The transmission parameters for the given circuit are $\underset{\_}{\left[\begin{array}{cc}\frac{5}{4}& \frac{3}{4}\Omega \\ \frac{3}{4}\text{S}& \frac{5}{4}\end{array}\right]}$.

Explanation of Solution

Formula used:

From the TABLE 19.1, write the expression for transmission parameters in terms of z parameters as follows:

$\left[T\right]=\left[\begin{array}{cc}\frac{z_{11}}{z_{21}}& \frac{{\Delta }_z}{z_{21}}\\ \frac{1}{z_{21}}& \frac{z_{22}}{z_{21}}\end{array}\right]$ (11)

Calculation:

Substitute $\frac{5}{3}\Omega$ for $z_{11}$, $\frac{4}{3}\Omega$ for $z_{12}$, $\frac{4}{3}\Omega$ for $z_{21}$, $\frac{5}{3}\Omega$ for $z_{22}$, and $1{\Omega }^2$ for ${\Delta }_z$ in Equation (11) to obtain the transmission parameters.

$\begin{array}{c}\left[T\right]=\left[\begin{array}{cc}\frac{\frac{5}{3}\Omega }{\frac{4}{3}\Omega }& \frac{1{\Omega }^2}{\frac{4}{3}\Omega }\\ \frac{1}{\frac{4}{3}\Omega }& \frac{\frac{5}{3}\Omega }{\frac{4}{3}\Omega }\end{array}\right]\\ =\left[\begin{array}{cc}\frac{5}{4}& \frac{3}{4}\Omega \\ \frac{3}{4}\text{S}& \frac{5}{4}\end{array}\right]\end{array}$

Conclusion:

Thus, the transmission parameters for the given circuit are $\underset{\_}{\left[\begin{array}{cc}\frac{5}{4}& \frac{3}{4}\Omega \\ \frac{3}{4}\text{S}& \frac{5}{4}\end{array}\right]}$.