Chapter 19, Problem 75P

For the individual two-ports shown in Fig. 19.121 where,

1. (a) Determine the y parameters of the overall two-port.
2. (b) Find the voltage ratio V_o∕V_i when Z_L = 2 Ω.

Figure 19.121

To determine

Calculate y parameters for the overall two-port network in Figure 19.121 in the textbook.

Answer to Problem 75P

The y parameters for the overall two-port network are $\underset{\_}{\left[\begin{array}{cc}0.3015& -0.1765\\ 0.0588& 10.9412\end{array}\right]\text{S}}$.

Explanation of Solution

Given Data:

Refer to Figure 19.121 in the textbook for the given two-port network.

The impedance parameters of network $N_a$ and admittance parameters of network $N_b$ are given as follows:

$\begin{array}{l}\left[z_a\right]=\left[\begin{array}{cc}8& 6\\ 4& 5\end{array}\right]\Omega \\ \left[y_b\right]=\left[\begin{array}{cc}8& -4\\ 2& 10\end{array}\right]\text{S}\end{array}$

Formula used:

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

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

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

${\Delta }_z=z_{11}{z}_{22}-z_{12}{z}_{21}$        (2)

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

$\left[T\right]=\left[\begin{array}{cc}-\frac{y_{22}}{y_{21}}& -\frac{1}{y_{21}}\\ -\frac{{\Delta }_y}{y_{21}}& -\frac{y_{11}}{y_{21}}\end{array}\right]$        (3)

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

${\Delta }_y=y_{11}{y}_{22}-y_{12}{y}_{21}$        (4)

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

$\left[y\right]=\left[\begin{array}{cc}\frac{D}{B}& -\frac{{\Delta }_T}{B}\\ -\frac{1}{B}& \frac{A}{B}\end{array}\right]$        (5)

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

${\Delta }_T=AD-BC$        (6)

Calculation:

As the two networks $N_a$ and $N_b$ are in cascade connection, convert the impedance parameters of network $N_a$ and admittance parameters of network $N_b$ into transmission parameters and then perform the multiplications to obtain the overall transmission parameters. Then, convert the overall transmission parameters into admittance parameters to attain the required objective.

From impedance parameters of network $N_a$, the z parameters are written as follows:

$\begin{array}{l}{z}_{11}=8\Omega \\ z_{12}=6\Omega \\ z_{21}=4\Omega \\ z_{22}=5\Omega \end{array}$

Substitute $8\Omega$ for $z_{11}$, $6\Omega$ for $z_{12}$, $4\Omega$ for $z_{21}$, and $5\Omega$ for $z_{22}$ in Equation (2) to obtain the value of ${\Delta }_z$.

$\begin{array}{c}{\Delta }_z=\left(8\Omega \right)\left(5\Omega \right)-\left(6\Omega \right)\left(4\Omega \right)\\ =40{\Omega }^2-24{\Omega }^2\\ =16{\Omega }^2\end{array}$

Substitute $8\Omega$ for $z_{11}$, $6\Omega$ for $z_{12}$, $4\Omega$ for $z_{21}$, $5\Omega$ for $z_{22}$, and $16{\Omega }^2$ for ${\Delta }_z$ in Equation (1) to obtain the transmission parameters for network $N_a$.

$\begin{array}{c}\left[T_a\right]=\left[\begin{array}{cc}\frac{8\Omega }{4\Omega }& \frac{16{\Omega }^2}{4\Omega }\\ \frac{1}{4\Omega }& \frac{5\Omega }{4\Omega }\end{array}\right]\\ =\left[\begin{array}{cc}2& 4\Omega \\ 0.25\text{S}& 1.25\end{array}\right]\end{array}$

From admittance parameters of network $N_b$, the y parameters are written as follows:

$\begin{array}{l}{y}_{11}=8\text{S}\\ y_{12}=-4\text{S}\\ y_{21}=2\text{S}\\ y_{22}=10\text{S}\end{array}$

Substitute $8\text{S}$ for $y_{11}$, $-4\text{S}$ for $y_{12}$, $2\text{S}$ for $y_{21}$, and $10\text{S}$ for $y_{22}$ in Equation (4) to obtain the value of ${\Delta }_y$.

$\begin{array}{c}{\Delta }_y=\left(8\text{S}\right)\left(10\text{S}\right)-\left(-4\text{S}\right)\left(2\text{S}\right)\\ =80{\text{S}}^2+8{\text{S}}^2\\ =88{\text{S}}^2\end{array}$

Substitute $8\text{S}$ for $y_{11}$, $-4\text{S}$ for $y_{12}$, $2\text{S}$ for $y_{21}$, $10\text{S}$ for $y_{22}$, and $88{\text{S}}^2$ for ${\Delta }_y$ in Equation (3) to obtain the transmission parameters for network $N_b$.

$\begin{array}{c}\left[T_b\right]=\left[\begin{array}{cc}-\frac{10\text{S}}{2\text{S}}& -\frac{1}{2\text{S}}\\ -\frac{88{\text{S}}^2}{2\text{S}}& -\frac{8\text{S}}{2\text{S}}\end{array}\right]\\ =\left[\begin{array}{cc}-5& -0.5\Omega \\ -44\text{S}& -4\end{array}\right]\end{array}$

Write the expression to find the overall transmission parameters for the cascaded network as follows:

$\left[T\right]=\left[T_a\right]\left[T_b\right]$

Use the expression and write the MATLAB code to obtain the overall transmission parameters for the cascaded network as follows:

Ta=[2 4;0.25 1.25];

>> Tb=[-5 -0.5;-44 -4];

>> T=Ta*Tb

The MATLAB output is obtained as follows:

T =

 -186.0000  -17.0000

  -56.2500   -5.1250

From the MATLAB output, the overall transmission parameters for the cascaded network are written as follows:

$\left[T\right]=\left[\begin{array}{cc}-186& -17\Omega \\ -56.25\text{S}& -5.125\end{array}\right]$

Convert the obtained transmission parameters into admittance parameters to obtain the required objective.

From the obtained transmission parameters the values of $A$, $B$, $C$, and $D$ are written as follows:

$\begin{array}{l}A=-186\\ B=-17\Omega \\ C=-56.25\text{S}\\ D=-5.125\end{array}$

Substitute $-186$ for $A$, $-5.125$ for $D$, $-17\Omega$ for $B$, and $-56.25\text{S}$ for $C$ in Equation (6) to obtain the value of ${\Delta }_T$.

$\begin{array}{c}{\Delta }_T=\left(-186\right)\left(-5.125\right)-\left(-17\Omega \right)\left(-56.25\text{S}\right)\\ =953.25-956.25\\ =-3\end{array}$

Substitute $-186$ for $A$, $-5.125$ for $D$, $-17\Omega$ for $B$, $-56.25\text{S}$ for $C$, and $-3$ for ${\Delta }_T$ in Equation (5) to obtain the overall admittance parameters for the cascaded network.

$\begin{array}{c}\left[y\right]=\left[\begin{array}{cc}\frac{-5.125}{-17\Omega }& -\frac{-3}{-17\Omega }\\ -\frac{1}{-17\Omega }& \frac{-186}{-17\Omega }\end{array}\right]\\ =\left[\begin{array}{cc}0.3015\frac{1}{\Omega }& -0.1765\frac{1}{\Omega }\\ 0.0588\frac{1}{\Omega }& 10.9412\frac{1}{\Omega }\end{array}\right]\\ \cong \left[\begin{array}{cc}0.3015& -0.1765\\ 0.0588& 10.9412\end{array}\right]\text{S}\end{array}$

Conclusion:

Thus, the y parameters for the overall two-port network are $\underset{\_}{\left[\begin{array}{cc}0.3015& -0.1765\\ 0.0588& 10.9412\end{array}\right]\text{S}}$.

To determine

Calculate the voltage ratio $V_o/V_i$.

Answer to Problem 75P

The value of voltage ratio $V_o/V_i$ is $\underset{\_}{-0.0051}$.

Explanation of Solution

Given Data:

$Z_L=2\Omega$

Formula used:

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

$\left[z\right]=\left[\begin{array}{cc}\frac{A}{C}& \frac{{\Delta }_T}{C}\\ \frac{1}{C}& \frac{D}{C}\end{array}\right]$        (7)

Calculation:

Substitute $-186$ for $A$, $-5.125$ for $D$, $-17\Omega$ for $B$, $-56.25\text{S}$ for $C$, and $-3$ for ${\Delta }_T$ in Equation (7) to obtain the overall impedance parameters for the cascaded network.

$\begin{array}{c}\left[z\right]=\left[\begin{array}{cc}\frac{-186}{-56.25\text{S}}& \frac{-3}{-56.25\text{S}}\\ \frac{1}{-56.25\text{S}}& \frac{-5.125}{-56.25\text{S}}\end{array}\right]\\ =\left[\begin{array}{cc}3.3067\frac{1}{\text{S}}& 0.0533\frac{1}{\text{S}}\\ -0.0178\frac{1}{\text{S}}& 0.0911\frac{1}{\text{S}}\end{array}\right]\\ =\left[\begin{array}{cc}3.3067& 0.0533\\ -0.0178& 0.0911\end{array}\right]\Omega \end{array}$

Refer to Figure 19.5 (b) in the textbook for general equivalent circuit for z parameters and draw the equivalent circuit for the given network as shown in Figure 1.

From Figure 1, write the expression for $V_i$ as follows:

$V_i=z_{11}{I}_1+z_{12}{I}_2$        (8)

From Figure 1, write the expression for $V_o$ as follows:

$V_o=z_{21}{I}_1+z_{22}{I}_2$        (9)

Also write the expression for $V_o$ in terms of load impedance as follows:

$V_o=-I_2{Z}_L$

Rearrange the expression as follows:

$I_2=-\frac{V_o}{Z_L}$

Substitute $\left(-\frac{V_o}{Z_L}\right)$ for $I_2$ in Equation (9) as follows:

$V_o=z_{21}{I}_1+z_{22}\left(-\frac{V_o}{Z_L}\right)$

Simplify the expression as follows:

$\begin{array}{l}{z}_{21}{I}_1=\left(1+\frac{z_{22}}{Z_L}\right)V_o\\ I_1=\left(\frac{1}{z_{21}}+\frac{z_{22}}{z_{21}{Z}_L}\right)V_o\end{array}$

Substitute $\left(-\frac{V_o}{Z_L}\right)$ for $I_2$ and $\left(\frac{1}{z_{21}}+\frac{z_{22}}{z_{21}{Z}_L}\right)V_o$ for $I_1$ in Equation (8) as follows:

$\begin{array}{c}{V}_i=z_{11}\left(\frac{1}{z_{21}}+\frac{z_{22}}{z_{21}{Z}_L}\right)V_o+z_{12}\left(-\frac{V_o}{Z_L}\right)\\ =\left(\frac{z_{11}}{z_{21}}+\frac{z_{11}{z}_{22}}{z_{21}{Z}_L}-\frac{z_{12}}{Z_L}\right)V_o\\ =\left(\frac{z_{11}{Z}_L+z_{11}{z}_{22}-z_{12}{z}_{21}}{z_{21}{Z}_L}\right)V_o\end{array}$

Rearrange the expression as follows:

$\frac{V_o}{V_i}=\frac{z_{21}{Z}_L}{z_{11}{Z}_L+z_{11}{z}_{22}-z_{12}{z}_{21}}$

From the obtained overall z parameters, substitute $3.3067\Omega$ for $z_{11}$, $0.0533\Omega$ for $z_{12}$, $-0.0178\Omega$ for $z_{21}$, $0.0911\Omega$ for $z_{22}$, and $2\Omega$ for $Z_L$ to obtain the required voltage ratio.

$\begin{array}{c}\frac{V_o}{V_i}=\frac{\left(-0.0178\Omega \right)\left(2\Omega \right)}{\left(3.3067\Omega \right)\left(2\Omega \right)+\left(3.3067\Omega \right)\left(0.0911\Omega \right)-\left(0.0533\Omega \right)\left(-0.0178\Omega \right)}\\ =\frac{-0.0356{\Omega }^2}{6.6134{\Omega }^2+0.3012{\Omega }^2+\left(9.4874\times {10}^{-4}{\Omega }^2\right)}\\ =\frac{-0.0356{\Omega }^2}{6.9155{\Omega }^2}\\ =-0.0051\end{array}$

Conclusion:

Thus, the value of voltage ratio $V_o/V_i$ is $\underset{\_}{-0.0051}$.