Chapter 19, Problem 71P

Determine the z parameters for the network in Fig. 19.118.

Figure 19.118

To determine

Find the impedance parameters for the given two-port network in Figure 19.118 in the textbook.

Answer to Problem 71P

The impedance parameters for the given two-port network are $\underset{\_}{\left[\begin{array}{cc}2& -3.334\\ 3.334& 20.22\end{array}\right]\Omega }$.

Explanation of Solution

Given Data:

Refer to Figure 19.118 in the textbook for the given two-port network.

Formula used:

Write the expressions for transmission (ABCD) parameters of a two-port network as follows:

$V_{\text{1}}=A{V}_2-B{I}_{\text{2}}$        (1)

$I_{\text{1}}=C{V}_2-D{I}_{\text{2}}$        (2)

Refer to Figure 19.106 in the textbook and write the expression for transmission parameters for a T-network as follows:

$\left[T\right]=\left[\begin{array}{cc}1+\frac{R_1}{R_2}& R_3+\frac{R_1}{R_2}\left(R_2+R_3\right)\\ \frac{1}{R_2}& 1+\frac{R_3}{R_2}\end{array}\right]$        (3)

Refer to TABLE 19.1 in the textbook, write the expression for g parameters in terms of transmission parameters as follows:

$\left[g\right]=\left[\begin{array}{cc}\frac{C}{A}& -\frac{{\Delta }_T}{A}\\ \frac{1}{A}& \frac{B}{A}\end{array}\right]$        (4)

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

${\Delta }_T=AD-BC$        (5)

From TABLE 19.1 in the textbook, write the expression for impedance parameters in terms of g parameters as follows:

$\left[z\right]=\left[\begin{array}{cc}\frac{1}{g_{11}}& -\frac{g_{12}}{g_{11}}\\ \frac{g_{21}}{g_{11}}& \frac{{\Delta }_g}{g_{11}}\end{array}\right]$        (6)

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

${\Delta }_g=g_{11}{g}_{22}-g_{12}{g}_{21}$        (7)

Calculation:

The given interconnected network is a series-parallel combination of two two-port networks. Find g parameters for each network and add them for overall g parameters of entire network. And then convert the overall g parameters into z parameters.

Consider upper network (T-network) as network $N_a$ and lower network (series combination of one resistive network with the transformer) as $N_b$.

Compare upper network with the Figure 19.106 in the textbook and write the resistance values as follows:

$\begin{array}{l}{R}_1=8\Omega \\ R_2=10\Omega \\ R_3=6\Omega \end{array}$

Substitute $8\Omega$ for $R_1$, $10\Omega$ for $R_2$, and $6\Omega$ for $R_3$ in Equation (3) to obtain the transmission parameters for the upper network.

$\begin{array}{c}\left[T_a\right]=\left[\begin{array}{cc}1+\frac{8\Omega }{10\Omega }& 6\Omega +\frac{8\Omega }{10\Omega }\left(10\Omega +6\Omega \right)\\ \frac{1}{10\Omega }& 1+\frac{6\Omega }{10\Omega }\end{array}\right]\\ =\left[\begin{array}{cc}1.8& 18.8\Omega \\ 0.1\text{S}& 1.6\end{array}\right]\end{array}$

Convert the obtained transmission parameters into g parameters as follows:

Substitute 1.8 for $A$, 1.6 for $D$, $18.8\Omega$ for $B$, and $0.1\text{S}$ for $C$ in Equation (5) to obtain the value of ${\Delta }_T$.

$\begin{array}{c}{\Delta }_T=\left(1.8\right)\left(1.6\right)-\left(18.8\Omega \right)\left(0.1\text{S}\right)\\ =2.88-1.88\\ =1\end{array}$

Substitute 1.8 for $A$, 1.6 for $D$, $18.8\Omega$ for $B$, $0.1\text{S}$ for $C$, and 1 for ${\Delta }_T$ in Equation (4) to obtain the g parameters for the upper network.

$\begin{array}{c}\left[g_a\right]=\left[\begin{array}{cc}\frac{0.1\text{S}}{1.8}& -\frac{1}{1.8}\\ \frac{1}{1.8}& \frac{18.8\Omega }{1.8}\end{array}\right]\\ =\left[\begin{array}{cc}0.0556\text{S}& -0.5556\\ 0.5556& 10.4444\Omega \end{array}\right]\end{array}$

As the lower network is a series combination of resistive network and transformer, find the transmission parameters for each network and product them to get the transmission parameters for lower network.

Consider resistive network as $N_{b1}$ and transformer network as $N_{b2}$.

Find the transmission parameters for network $N_{b1}$ as follows:

The transmission parameters $A$ and $C$ 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 for $I_{\text{2}}$ as follows:

$\begin{array}{c}{V}_{\text{1}}=A{V}_2-B\left(0\right)\\ =A{V}_2\end{array}$

$A=\frac{V_{\text{1}}}{V_2}$        (8)

$\begin{array}{c}{I}_{\text{1}}=C{V}_2-D\left(0\right)\\ =C{V}_2\end{array}$

$C=\frac{I_{\text{1}}}{V_2}$        (9)

Redraw the network $N_{b1}$ by open circuiting the port-2 as shown in Figure 1.

From Figure 1, write the expression for $V_2$ using voltage division rule as follows:

$V_2=\left(\frac{5\Omega }{4\Omega +5\Omega }\right)V_{\text{1}}$

$V_2=\left(\frac{5}{9}\right)V_{\text{1}}$        (10)

Rearrange the expression as follows:

$\begin{array}{c}\frac{V_{\text{1}}}{V_2}=\frac{9}{5}\\ =1.8\end{array}$

Substitute 1.8 for $\frac{V_{\text{1}}}{V_2}$ in Equation (8) to obtain the value of $A$.

$A=1.8$

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

$\begin{array}{c}{V}_{\text{1}}=\left(4\Omega +5\Omega \right)I_{\text{1}}\\ =\left(9\Omega \right)I_{\text{1}}\end{array}$

Substitute $\left(9\Omega \right)I_{\text{1}}$ for $V_{\text{1}}$ in Equation (10) as follows:

$\begin{array}{c}{V}_2=\left(\frac{5}{9}\right)\left(9\Omega \right)I_{\text{1}}\\ =\left(5\Omega \right)I_{\text{1}}\end{array}$

Rearrange the expression as follows:

$\begin{array}{c}\frac{I_{\text{1}}}{V_2}=\frac{1}{5\Omega }\\ =0.2\text{S}\end{array}$

Substitute $0.2\text{S}$ for $\frac{I_1}{V_2}$ in Equation (9) to obtain the value of $C$ as follows:

$C=0.2\text{S}$

The transmission parameters $B$ and $D$ are obtained when the port-2 of the network is short circuited. The voltage $V_{\text{2}}$ becomes zero when port-2 is short-circuited. Therefore, rewrite the expressions in Equation (1) and (2) by substituting 0 for $V_{\text{2}}$ as follows:

$\begin{array}{c}{V}_{\text{1}}=A\left(0\right)-B{I}_{\text{2}}\\ =-B{I}_{\text{2}}\end{array}$

Rearrange the expression as follows:

$B=-\frac{V_{\text{1}}}{I_{\text{2}}}$        (11)

$\begin{array}{c}{I}_{\text{1}}=C\left(0\right)-D{I}_{\text{2}}\\ =-D{I}_{\text{2}}\end{array}$

$D=-\frac{I_{\text{1}}}{I_{\text{2}}}$        (12)

Redraw the network $N_{b1}$ by short circuiting the port-2 as shown in Figure 2.

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

$I_{\text{2}}=-\left(\frac{5\Omega }{5\Omega +2\Omega }\right)I_{\text{1}}$

$I_{\text{2}}=-\left(\frac{5}{7}\right)I_{\text{1}}$        (13)

Rearrange the expression as follows:

$\begin{array}{c}\frac{I_{\text{1}}}{I_{\text{2}}}=-\frac{7}{5}\\ =-1.4\end{array}$

Substitute $\left(-1.4\right)$ for $\frac{I_{\text{1}}}{I_{\text{2}}}$ in Equation (12) to obtain the value of $D$.

$\begin{array}{c}D=-\left(-1.4\right)\\ =1.4\end{array}$

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

$\begin{array}{c}{V}_{\text{1}}=\left[\left(2\Omega \Vert 5\Omega \right)+4\Omega \right]I_{\text{1}}\\ =\left[\frac{\left(2\Omega \right)\left(5\Omega \right)}{2\Omega +5\Omega }+4\Omega \right]I_{\text{1}}\\ =\left(5.4286\Omega \right)I_{\text{1}}\end{array}$

From Equation (13), substitute $\left[-\left(\frac{5}{7}\right)I_{\text{1}}\right]$ for $I_{\text{2}}$ and $\left(5.4286\Omega \right)I_{\text{1}}$ for $V_{\text{1}}$ in Equation (11) to obtain the value of $B$.

$\begin{array}{c}B=-\frac{\left(5.4286\Omega \right)I_{\text{1}}}{\left[-\left(\frac{5}{7}\right)I_{\text{1}}\right]}\\ =7.6\Omega \end{array}$

From the calculations, write the transmission parameters for the network $N_{b1}$  as follows:

$\left[T_{b1}\right]=\left[\begin{array}{cc}1.8& 7.6\Omega \\ 0.2\text{S}& 1.4\end{array}\right]$

Find the transmission parameters for network $N_{b2}$ (transformer) as follows:

From the given transformer network $N_{b2}$, write the transformer ratio in terms of voltage ratio as follows:

$\frac{V_{\text{1}}}{V_2}=\frac{1}{2}$

Rearrange the expression as follows:

$V_{\text{1}}=\left(\frac{1}{2}\right)V_2-\left(0\right)I_2$        (14)

From the given transformer network $N_{b2}$, write the transformer ratio in terms of current ratio as follows:

$\frac{I_{\text{1}}}{\left(-I_2\right)}=2$

Rearrange the expression as follows:

$I_{\text{1}}=\left(0\right)V_2-\left(2\right)I_2$        (15)

Compare Equation (14) with Equation (1) and obtain the parameters $A$ and $B$ for network $N_{b2}$.

$\begin{array}{c}A=\frac{1}{2}\\ =0.5\\ B=0\Omega \end{array}$

Compare Equation (15) with Equation (2) and obtain the parameters $C$ and $D$ for network $N_{b2}$.

$\begin{array}{c}C=0\text{S}\\ D=2\end{array}$

From the calculations, write the transmission parameters for the network $N_{b2}$  as follows:

$\left[T_{b2}\right]=\left[\begin{array}{cc}0.5& 0\Omega \\ 0\text{S}& 2\end{array}\right]$

As the networks $N_{b1}$ and $N_{b2}$ are connected in series, write the expression for transmission parameters for lower network as follows:

$\left[T_b\right]=\left[T_{b2}\right]\left[T_{b1}\right]$

Substitute $\left[\begin{array}{cc}1.8& 7.6\Omega \\ 0.2\text{S}& 1.4\end{array}\right]$ for $\left[T_{b1}\right]$ and $\left[\begin{array}{cc}0.5& 0\Omega \\ 0\text{S}& 2\end{array}\right]$ for $\left[T_{b2}\right]$ to obtain the transmission parameters for lower network.

$\begin{array}{c}\left[T_b\right]=\left[\begin{array}{cc}0.5& 0\Omega \\ 0\text{S}& 2\end{array}\right]\left[\begin{array}{cc}1.8& 7.6\Omega \\ 0.2\text{S}& 1.4\end{array}\right]\\ =\left[\begin{array}{cc}0.9& 3.8\Omega \\ 0.4\text{S}& 2.8\end{array}\right]\end{array}$

Convert the obtained transmission parameters into g parameters for lower network as follows:

Substitute 0.9 for $A$, 2.8 for $D$, $3.8\Omega$ for $B$, and $0.4\text{S}$ for $C$ in Equation (5) to obtain the value of ${\Delta }_T$.

$\begin{array}{c}{\Delta }_T=\left(0.9\right)\left(2.8\right)-\left(3.8\Omega \right)\left(0.4\text{S}\right)\\ =1\end{array}$

Substitute 0.9 for $A$, 2.8 for $D$, $3.8\Omega$ for $B$, $0.4\text{S}$ for $C$, and 1 for ${\Delta }_T$ in Equation (4) to obtain the g parameters for the lower network.

$\begin{array}{c}\left[g_b\right]=\left[\begin{array}{cc}\frac{0.4\text{S}}{0.9}& -\frac{1}{0.9}\\ \frac{1}{0.9}& \frac{3.8\Omega }{0.9}\end{array}\right]\\ =\left[\begin{array}{cc}0.4444\text{S}& -1.1111\\ 1.1111& 4.2222\Omega \end{array}\right]\end{array}$

Write the expression for overall g parameters for the given network as follows:

$\left[g\right]=\left[g_a\right]+\left[g_b\right]$

Substitute $\left[\begin{array}{cc}0.0556\text{S}& -0.5556\\ 0.5556& 10.4444\Omega \end{array}\right]$ for $\left[g_a\right]$ and $\left[\begin{array}{cc}0.4444\text{S}& -1.1111\\ 1.1111& 4.2222\Omega \end{array}\right]$ for $\left[g_b\right]$ to obtain the overall g parameters.

$\begin{array}{c}\left[g\right]=\left[\begin{array}{cc}0.0556\text{S}& -0.5556\\ 0.5556& 10.4444\Omega \end{array}\right]+\left[\begin{array}{cc}0.4444\text{S}& -1.1111\\ 1.1111& 4.2222\Omega \end{array}\right]\\ =\left[\begin{array}{cc}0.5\text{S}& -1.6667\\ 1.6667& 14.6666\Omega \end{array}\right]\end{array}$

Convert the obtained g parameters into impedance parameters to attain the required objective.

Substitute $0.5\text{S}$ for $g_{11}$, $\left(-1.6667\right)$ for $g_{12}$, 1.6667 for $g_{21}$, and $14.6666\Omega$ for $g_{22}$ in Equation (7) to obtain the value of ${\Delta }_g$.

$\begin{array}{c}{\Delta }_g=\left(0.5\text{S}\right)\left(14.6666\Omega \right)-\left(-1.6667\right)\left(1.6667\right)\\ =10.1112\end{array}$

Substitute $0.5\text{S}$ for $g_{11}$, $\left(-1.6667\right)$ for $g_{12}$, 1.6667 for $g_{21}$, $14.6666\Omega$ for $g_{22}$, and 10.1112 for ${\Delta }_g$ in Equation (5) to obtain the impedance parameters for given network.

$\begin{array}{c}\left[z\right]=\left[\begin{array}{cc}\frac{1}{0.5\text{S}}& -\frac{\left(-1.6667\right)}{0.5\text{S}}\\ \frac{1.6667}{0.5\text{S}}& \frac{10.1112}{0.5\text{S}}\end{array}\right]\\ =\left[\begin{array}{cc}2\frac{1}{\text{S}}& 3.334\frac{1}{\text{S}}\\ 3.334\frac{1}{\text{S}}& 20.22\frac{1}{\text{S}}\end{array}\right]\\ =\left[\begin{array}{cc}2& 3.334\\ 3.334& 20.22\end{array}\right]\Omega \end{array}$

Conclusion:

Thus, the impedance parameters for the given two-port network are $\underset{\_}{\left[\begin{array}{cc}2& -3.334\\ 3.334& 20.22\end{array}\right]\Omega }$.