Chapter 19, Problem 99CP

Assume that the two circuits in Fig. 19.135 are equivalent. The parameters of the two circuits must be equal. Using this factor and the z parameters, derive Eqs. (9.67) and (9.68).

Figure 19.135

To determine

Derive the expressions in Equations (9.67) and (9.68) in the textbook.

Explanation of Solution

Given Data:

Refer to Figure 19.135 in the textbook given circuits.

Consider the parameters of two circuits are equal.

Calculation:

Refer to Figure 19.135 (a) in the textbook and write the expression for $Z_{ab}$ as follows:

$Z_{ab}=Z_1+Z_3$        (1)

Refer to Figure 19.135 (b) in the textbook and write the expression for $Z_{ab}$ as follows:

$\begin{array}{c}{Z}_{ab}=Z_c\Vert \left(Z_a+Z_b\right)\\ =\frac{Z_c\left(Z_a+Z_b\right)}{Z_c+\left(Z_a+Z_b\right)}\\ =\frac{Z_c\left(Z_a+Z_b\right)}{Z_a+Z_b+Z_c}\end{array}$

From Equation (1), substitute $\left(Z_1+Z_3\right)$ for $Z_{ab}$ as follows:

$Z_1+Z_3=\frac{Z_c\left(Z_a+Z_b\right)}{Z_a+Z_b+Z_c}$        (2)

Refer to Figure 19.135 (a) in the textbook and write the expression for $Z_{cd}$ as follows:

$Z_{cd}=Z_2+Z_3$        (3)

Refer to Figure 19.135 (b) in the textbook and write the expression for $Z_{cd}$ as follows:

$\begin{array}{c}{Z}_{cd}=Z_a\Vert \left(Z_b+Z_c\right)\\ =\frac{Z_a\left(Z_b+Z_c\right)}{Z_a+\left(Z_b+Z_c\right)}\\ =\frac{Z_a\left(Z_b+Z_c\right)}{Z_a+Z_b+Z_c}\end{array}$

From Equation (3), substitute $\left(Z_2+Z_3\right)$ for $Z_{cd}$ as follows:

$Z_2+Z_3=\frac{Z_a\left(Z_b+Z_c\right)}{Z_a+Z_b+Z_c}$        (4)

Refer to Figure 19.135 (a) in the textbook and write the expression for $Z_{ac}$ as follows:

$Z_{ac}=Z_1+Z_2$        (5)

Refer to Figure 19.135 (b) in the textbook and write the expression for $Z_{ac}$ as follows:

$\begin{array}{c}{Z}_{ac}=Z_b\Vert \left(Z_a+Z_c\right)\\ =\frac{Z_b\left(Z_a+Z_c\right)}{Z_b+\left(Z_a+Z_c\right)}\\ =\frac{Z_b\left(Z_a+Z_c\right)}{Z_a+Z_b+Z_c}\end{array}$

From Equation (5), substitute $\left(Z_1+Z_2\right)$ for $Z_{ac}$ as follows:

$Z_1+Z_2=\frac{Z_b\left(Z_a+Z_c\right)}{Z_a+Z_b+Z_c}$        (6)

Subtract Equation (4) from Equation (2) and obtain the expression as follows:

$\begin{array}{c}{Z}_1-Z_2=\frac{Z_c\left(Z_a+Z_b\right)}{Z_a+Z_b+Z_c}-\frac{Z_a\left(Z_b+Z_c\right)}{Z_a+Z_b+Z_c}\\ =\frac{Z_a{Z}_c+Z_b{Z}_c-Z_a{Z}_b-Z_a{Z}_c}{Z_a+Z_b+Z_c}\end{array}$

$Z_1-Z_2=\frac{Z_b\left(Z_c-Z_a\right)}{Z_a+Z_b+Z_c}$        (7)

Add Equations (6) and (7) and obtain the expression as follows:

$\begin{array}{c}2Z_1=\frac{Z_b\left(Z_a+Z_c\right)}{Z_a+Z_b+Z_c}+\frac{Z_b\left(Z_c-Z_a\right)}{Z_a+Z_b+Z_c}\\ =\frac{Z_a{Z}_b+Z_b{Z}_c+Z_b{Z}_c-Z_a{Z}_b}{Z_a+Z_b+Z_c}\\ =\frac{2Z_b{Z}_c}{Z_a+Z_b+Z_c}\end{array}$

Simplify the expression as follows:

$Z_1=\frac{Z_b{Z}_c}{Z_a+Z_b+Z_c}$        (8)

Subtract Equation (8) from Equation (6) and obtain the expression as follows:

$\begin{array}{l}{Z}_1+Z_2-Z_1=\frac{Z_b\left(Z_a+Z_c\right)}{Z_a+Z_b+Z_c}-\frac{Z_b{Z}_c}{Z_a+Z_b+Z_c}\\ Z_2=\frac{Z_a{Z}_b+Z_b{Z}_c-Z_b{Z}_c}{Z_a+Z_b+Z_c}\end{array}$

$Z_2=\frac{Z_a{Z}_b}{Z_a+Z_b+Z_c}$        (9)

Subtract Equation (8) from Equation (2) and obtain the expression as follows:

$\begin{array}{l}{Z}_1+Z_3-Z_1=\frac{Z_c\left(Z_a+Z_b\right)}{Z_a+Z_b+Z_c}-\frac{Z_b{Z}_c}{Z_a+Z_b+Z_c}\\ Z_3=\frac{Z_a{Z}_c+Z_b{Z}_c-Z_b{Z}_c}{Z_a+Z_b+Z_c}\end{array}$

$Z_3=\frac{Z_a{Z}_c}{Z_a+Z_b+Z_c}$        (10)

From Equations (8), (9), and (10), the expressions in Equation (9.68) are derived.

Note that, the obtained expressions are not same as the expressions in the textbook, since the position of the impedances are changed in the given circuits.

Use expressions in Equations (8), (9), and (10) and obtain the expression as follows:

$\begin{array}{c}{Z}_1{Z}_2+Z_2{Z}_3+Z_3{Z}_1=\left[\begin{array}{l}\left(\frac{Z_b{Z}_c}{Z_a+Z_b+Z_c}\right)\left(\frac{Z_a{Z}_b}{Z_a+Z_b+Z_c}\right)+\left(\frac{Z_a{Z}_b}{Z_a+Z_b+Z_c}\right)\left(\frac{Z_a{Z}_c}{Z_a+Z_b+Z_c}\right)\\ +\left(\frac{Z_a{Z}_c}{Z_a+Z_b+Z_c}\right)\left(\frac{Z_b{Z}_c}{Z_a+Z_b+Z_c}\right)\end{array}\right]\\ =\frac{Z_a{\left(Z_b\right)}^2{Z}_c+{\left(Z_a\right)}^2{Z}_b{Z}_c+Z_a{Z}_b{\left(Z_c\right)}^2}{\left(Z_a+Z_b+Z_c\right)\left(Z_a+Z_b+Z_c\right)}\\ =\frac{Z_a{Z}_b{Z}_c\left(Z_b+Z_a+Z_c\right)}{\left(Z_a+Z_b+Z_c\right)\left(Z_a+Z_b+Z_c\right)}\end{array}$        (11)

Divide Equation (11) by Equation (8) and obtain the expression as follows:

$\begin{array}{c}\frac{Z_1{Z}_2+Z_2{Z}_3+Z_3{Z}_1}{Z_1}=\frac{\left(\frac{Z_a{Z}_b{Z}_c}{Z_a+Z_b+Z_c}\right)}{\left(\frac{Z_b{Z}_c}{Z_a+Z_b+Z_c}\right)}\\ =Z_a\end{array}$

$Z_a=\frac{Z_1{Z}_2+Z_2{Z}_3+Z_3{Z}_1}{Z_1}$        (12)

Divide Equation (11) by Equation (9) and obtain the expression as follows:

$\begin{array}{c}\frac{Z_1{Z}_2+Z_2{Z}_3+Z_3{Z}_1}{Z_2}=\frac{\left(\frac{Z_a{Z}_b{Z}_c}{Z_a+Z_b+Z_c}\right)}{\left(\frac{Z_a{Z}_b}{Z_a+Z_b+Z_c}\right)}\\ =Z_c\end{array}$

$Z_c=\frac{Z_1{Z}_2+Z_2{Z}_3+Z_3{Z}_1}{Z_2}$        (13)

Divide Equation (11) by Equation (10) and obtain the expression as follows:

$\begin{array}{c}\frac{Z_1{Z}_2+Z_2{Z}_3+Z_3{Z}_1}{Z_3}=\frac{\left(\frac{Z_a{Z}_b{Z}_c}{Z_a+Z_b+Z_c}\right)}{\left(\frac{Z_a{Z}_c}{Z_a+Z_b+Z_c}\right)}\\ =Z_b\end{array}$

$Z_b=\frac{Z_1{Z}_2+Z_2{Z}_3+Z_3{Z}_1}{Z_3}$        (14)

From Equations (12), (13), and (14), the expressions in Equation (9.67) are derived.

Note that, the obtained expressions are not same as the expressions in the textbook, since the position of the impedances are changed in the given circuits.

Conclusion:

Thus, the expressions in Equations (9.67) and (9.68) in the textbook are derived.