Chapter 19, Problem 17P

To determine

Calculate z and y parameters for the two-port network in Figure 19.78 in the textbook.

Answer to Problem 17P

The z and y parameters for the given two-port network are $\underset{\_}{\left[\begin{array}{cc}9.6& -0.8\\ -0.8& 8.4\end{array}\right]\Omega }$ and $\underset{\_}{\left[\begin{array}{cc}0.105& 0.01\\ 0.01& 0.12\end{array}\right]\text{S}}$, respectively.

Explanation of Solution

Given Data:

Refer to Figure 19.78 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)

Refer to the TABLE 19.1 in the textbook and write the expression for admittance parameters in terms of z parameters as follows:

$\left[y\right]=\left[\begin{array}{cc}\frac{z_{22}}{{\Delta }_z}& -\frac{z_{12}}{{\Delta }_z}\\ -\frac{z_{21}}{{\Delta }_z}& \frac{z_{11}}{{\Delta }_z}\end{array}\right]$        (3)

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

${\Delta }_z=z_{11}{z}_{22}-z_{12}{z}_{21}$        (4)

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:

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

$z_{\text{11}}=\frac{V_{\text{1}}}{I_{\text{1}}}$        (5)

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

$z_{\text{21}}=\frac{V_{\text{2}}}{I_{\text{1}}}$        (6)

Draw the given circuit as shown in Figure 1 to obtain the parameters $z_{11}$ and $z_{\text{21}}$.

In Figure 1, $4\Omega$ and $12\Omega$ resistors are in series connection as the same current of $I_a$ passes through them. Similarly, $8\Omega$ and $16\Omega$ resistors are in series connection as the same current of $I_b$ passes through them.

From Equation (5), $z_{\text{11}}=\frac{V_{\text{1}}}{I_{\text{1}}}$. Therefore, the parameter $z_{\text{11}}$ is the equivalent resistance of Figure 1. Find the parameter $z_{\text{11}}$ as follows:

$\begin{array}{c}{z}_{\text{11}}=\left(8\Omega +16\Omega \right)\Vert \left(4\Omega +12\Omega \right)\\ =\left(24\Omega \right)\Vert \left(16\Omega \right)\\ =\frac{\left(24\Omega \right)\left(16\Omega \right)}{24\Omega +16\Omega }\\ =\frac{\left(24\Omega \right)\left(16\Omega \right)}{40\Omega }\end{array}$

$z_{\text{11}}=9.6\Omega$

From Figure 1, write the expression for $I_a$ using current division rule as follows:

$\begin{array}{c}{I}_a=\frac{\left(8\Omega +16\Omega \right)}{\left(8\Omega +16\Omega \right)+\left(4\Omega +12\Omega \right)}{I}_1\\ =\frac{24\Omega }{40\Omega }{I}_1\\ =\frac{3}{5}{I}_1\end{array}$

From Figure 1, write the expression for $I_b$ using current division rule as follows:

$\begin{array}{c}{I}_b=\frac{\left(4\Omega +12\Omega \right)}{\left(8\Omega +16\Omega \right)+\left(4\Omega +12\Omega \right)}{I}_1\\ =\frac{16\Omega }{40\Omega }{I}_1\\ =\frac{2}{5}{I}_1\end{array}$

From Figure 1, write the expression for $V_{\text{2}}$ as follows:

$V_{\text{2}}=-8I_b+4I_a$

Substitute $\left(\frac{3}{5}{I}_1\right)$ for $I_a$ and $\left(\frac{2}{5}{I}_1\right)$ for $I_b$ as follows:

$\begin{array}{c}{V}_{\text{2}}=\left(-8\right)\left(\frac{2}{5}{I}_1\right)+\left(4\right)\left(\frac{3}{5}{I}_1\right)\\ =-\frac{16}{5}{I}_1+\frac{12}{5}{I}_1\\ =-\frac{4}{5}{I}_1\end{array}$

Rearrange the expression as follows:

$\begin{array}{c}\frac{V_{\text{2}}}{I_1}=-\frac{4}{5}\Omega \\ =-0.8\Omega \end{array}$

Substitute $-0.8\Omega$ for $\frac{V_{\text{2}}}{I_1}$ in Equation (6) to obtain the value of $z_{\text{21}}$.

$z_{\text{21}}=-0.8\Omega$

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:

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

$z_{\text{12}}=\frac{V_{\text{1}}}{I_{\text{2}}}$        (7)

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

$z_{\text{22}}=\frac{V_{\text{2}}}{I_{\text{2}}}$        (8)

Draw the given circuit as shown in Figure 2 to obtain the parameters $z_{12}$ and $z_{\text{22}}$.

In Figure 2, $8\Omega$ and $4\Omega$ resistors are in series connection as the same current of $I_d$ passes through them. Similarly, $12\Omega$ and $16\Omega$ resistors are in series connection as the same current of $I_c$ passes through them.

From Equation (8), $z_{\text{22}}=\frac{V_{\text{2}}}{I_{\text{2}}}$. Therefore, the parameter $z_{\text{22}}$ is the equivalent resistance of Figure 2. Find the parameter $z_{\text{22}}$ as follows:

$\begin{array}{c}{z}_{\text{22}}=\left(8\Omega +4\Omega \right)\Vert \left(16\Omega +12\Omega \right)\\ =\left(12\Omega \right)\Vert \left(28\Omega \right)\\ =\frac{\left(12\Omega \right)\left(28\Omega \right)}{12\Omega +28\Omega }\\ =\frac{\left(12\Omega \right)\left(28\Omega \right)}{40\Omega }\end{array}$

$z_{\text{22}}=8.4\Omega$

From Figure 2, write the expression for $I_c$ using current division rule as follows:

$\begin{array}{c}{I}_c=\frac{\left(8\Omega +4\Omega \right)}{\left(8\Omega +4\Omega \right)+\left(16\Omega +12\Omega \right)}{I}_2\\ =\frac{12\Omega }{40\Omega }{I}_2\\ =\frac{3}{10}{I}_2\end{array}$

From Figure 2, write the expression for $I_d$ using current division rule as follows:

$\begin{array}{c}{I}_d=\frac{\left(16\Omega +12\Omega \right)}{\left(16\Omega +12\Omega \right)+\left(4\Omega +8\Omega \right)}{I}_2\\ =\frac{28\Omega }{40\Omega }{I}_2\\ =\frac{7}{10}{I}_2\end{array}$

From Figure 2, write the expression for $V_1$ as follows:

$V_1=-8I_d+16I_c$

Substitute $\left(\frac{3}{10}{I}_2\right)$ for $I_c$ and $\left(\frac{7}{10}{I}_2\right)$ for $I_d$ as follows:

$\begin{array}{c}{V}_1=\left(-8\right)\left(\frac{7}{10}{I}_2\right)+\left(16\right)\left(\frac{3}{10}{I}_2\right)\\ =-\frac{56}{10}{I}_2+\frac{48}{10}{I}_2\\ =-\frac{8}{10}{I}_2\\ =-0.8I_2\end{array}$

Rearrange the expression as follows:

$\frac{V_1}{I_2}=-0.8\Omega$

Substitute $-0.8\Omega$ for $\frac{V_1}{I_2}$ in Equation (7) to obtain the value of $z_{\text{12}}$.

$z_{\text{12}}=-0.8\Omega$

From the analysis, the impedance parameters for the given two-port network are written as follows:

$\left[z\right]=\left[\begin{array}{cc}9.6& -0.8\\ -0.8& 8.4\end{array}\right]\Omega$

Convert the obtained z parameters into admittance parameters to obtain the admittance parameters for the given two-port network.

Substitute $9.6\Omega$ for $z_{11}$, $-0.8\Omega$ for $z_{12}$, $-0.8\Omega$ for $z_{21}$, and $8.4\Omega$ for $z_{22}$ in Equation (4) to obtain the value of ${\Delta }_z$.

$\begin{array}{c}{\Delta }_z=\left(9.6\Omega \right)\left(8.4\Omega \right)-\left(-0.8\Omega \right)\left(-0.8\Omega \right)\\ =80.64{\Omega }^2-0.64{\Omega }^2\\ =80{\Omega }^2\end{array}$

Substitute $9.6\Omega$ for $z_{11}$, $-0.8\Omega$ for $z_{12}$, $-0.8\Omega$ for $z_{21}$, $8.4\Omega$ for $z_{22}$, and $80{\Omega }^2$ for ${\Delta }_z$ in Equation (3) to obtain the admittance parameters.

$\begin{array}{c}\left[y\right]=\left[\begin{array}{cc}\frac{8.4\Omega }{80{\Omega }^2}& -\frac{-0.8\Omega }{80{\Omega }^2}\\ -\frac{-0.8\Omega }{80{\Omega }^2}& \frac{9.6\Omega }{80{\Omega }^2}\end{array}\right]\\ =\left[\begin{array}{cc}0.105\frac{1}{\Omega }& 0.01\frac{1}{\Omega }\\ 0.01\frac{1}{\Omega }& 0.12\frac{1}{\Omega }\end{array}\right]\\ =\left[\begin{array}{cc}0.105\text{S}& 0.01\text{S}\\ 0.01\text{S}& 0.12\text{S}\end{array}\right]\left\{\because \frac{1}{\Omega }=1\text{S}\right\}\\ =\left[\begin{array}{cc}0.105& 0.01\\ 0.01& 0.12\end{array}\right]\text{S}\end{array}$

Conclusion:

Thus, the z and y parameters for the given two-port network are $\underset{\_}{\left[\begin{array}{cc}9.6& -0.8\\ -0.8& 8.4\end{array}\right]\Omega }$ and $\underset{\_}{\left[\begin{array}{cc}0.105& 0.01\\ 0.01& 0.12\end{array}\right]\text{S}}$, respectively.