Chapter 19, Problem 69P

The circuit in Fig. 19.116 may be regarded as two two-ports connected in parallel. Obtain the y parameters as functions of s.

Figure 19.116

To determine

Find the admittance parameters for the given two-port network in Figure 19.116 in the textbook.

Answer to Problem 69P

The admittance parameters for the given two-port network are $\underset{\_}{\left[\begin{array}{cc}\frac{s+1}{s+2}& \frac{-\left(3s+2\right)}{2\left(s+2\right)}\\ \frac{-\left(3s+2\right)}{2\left(s+2\right)}& \frac{5s^2+4s+4}{2s\left(s+2\right)}\end{array}\right]}$.

Explanation of Solution

Given Data:

Refer to Figure 19.116 in the textbook for the given two-port network.

Formula used:

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

$I_{\text{1}}=y_{\text{11}}{V}_1+y_{\text{12}}{V}_{\text{2}}$        (1)

$I_{\text{2}}=y_{\text{21}}{V}_{\text{1}}+y_{\text{22}}{V}_{\text{2}}$        (2)

Calculation:

Note that the dimensions are neglected for simple calculations.

Consider upper network as network $N_a$ and lower network as $N_b$.

As the interconnection network is a parallel combination of two networks, it is easier to obtain the required objective in terms of admittance parameters. Find the admittance parameters for upper and lower network and add them to obtain the overall admittance parameters.

The admittance parameters $y_{\text{11}}$ and $y_{\text{21}}$ 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 V for $V_{\text{2}}$ as follows:

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

$y_{\text{11}}=\frac{I_{\text{1}}}{V_1}$        (3)

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

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

Redraw the upper network by short circuiting the port-2 as shown in Figure 1.

Write the expression for equivalent impedance of secondary winding capacitance when the transformer is referred to the primary winding as follows:

${Z^{\prime }}_2=\frac{X_C}{n^2}$        (5)

Here,

$X_C$ is the secondary winding reactance, which is $\frac{1}{s}$ and

$n$ is the transformer ratio, which is $\frac{1}{2}$.

Substitute $\frac{1}{s}$ for $X_C$ and $\frac{1}{2}$ for $n$ in Equation (5) to find the value of ${Z^{\prime }}_2$.

$\begin{array}{c}{Z^{\prime }}_2=\frac{\frac{1}{s}}{{\left(\frac{1}{2}\right)}^2}\\ =\frac{4}{s}\end{array}$

From Figure 1, write the expression for $V_1$ as follows:

$V_{\text{1}}=\left(2+{Z^{\prime }}_2\right)I_1$

Substitute $\frac{4}{s}$ for ${Z^{\prime }}_2$ as follows:

$V_{\text{1}}=\left(2+\frac{4}{s}\right)I_1$

$V_{\text{1}}=\left(\frac{2s+4}{s}\right)I_1$        (6)

Rearrange the expression as follows:

$\begin{array}{c}\frac{I_{\text{1}}}{V_{\text{1}}}=\frac{s}{2s+4}\\ =\frac{s}{2\left(s+2\right)}\end{array}$

Substitute $\frac{s}{2\left(s+2\right)}$ for $\frac{I_{\text{1}}}{V_{\text{1}}}$ in Equation (3) to obtain the value of $y_{\text{11}}$ for upper network.

$y_{\text{11}}=\frac{s}{2\left(s+2\right)}$

From Figure 1, write the expression for $I_2$ as follows:

$I_2=-\frac{I_1}{n}$

From Equation (6), substitute $\left(\frac{s}{2s+4}\right)V_{\text{1}}$ for $I_1$ and $\frac{1}{2}$ for $n$ as follows:

$\begin{array}{c}{I}_2=-\frac{\left(\frac{s}{2s+4}\right)V_{\text{1}}}{\frac{1}{2}}\\ =-\left(\frac{2s}{2s+4}\right)V_{\text{1}}\\ =-\left(\frac{s}{s+2}\right)V_{\text{1}}\end{array}$

Rearrange the expression as follows:

$\frac{I_2}{V_{\text{1}}}=-\frac{s}{s+2}$

Substitute $\left(-\frac{s}{s+2}\right)$ for $\frac{I_2}{V_{\text{1}}}$ in Equation (4) to obtain the value of $y_{\text{21}}$ for upper network.

$y_{\text{21}}=-\frac{s}{s+2}$

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

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

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

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

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

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

Write the expression for equivalent impedance of primary winding resistance when the transformer is referred to the secondary winding as follows:

${Z^{\prime }}_1=n^{2}R$        (9)

Here,

$R$ is the primary winding resistance.

Substitute $2\Omega$ for $R$ and $\frac{1}{2}$ for $n$ in Equation (9) to find the value of ${Z^{\prime }}_1$.

$\begin{array}{c}{Z^{\prime }}_1={\left(\frac{1}{2}\right)}^2\left(2\Omega \right)\\ =0.5\Omega \end{array}$

From Figure2, write the expression for $V_2$ as follows:

$V_2=\left({Z^{\prime }}_1+\frac{1}{s}\right)I_2$

Substitute 0.5 for ${Z^{\prime }}_1$ as follows:

$V_2=\left(0.5+\frac{1}{s}\right)I_2$

$V_2=\left(\frac{0.5s+1}{s}\right)I_2$        (10)

Rearrange the expression as follows:

$\begin{array}{c}\frac{I_{\text{2}}}{V_{\text{2}}}=\frac{s}{0.5s+1}\\ =\frac{s}{0.5\left(s+\frac{1}{0.5}\right)}\\ =\frac{2s}{s+2}\end{array}$

Substitute $\frac{2s}{s+2}$ for $\frac{I_{\text{2}}}{V_{\text{2}}}$ in Equation (8) to obtain the value of $y_{\text{22}}$ for upper network.

$y_{\text{22}}=\frac{2s}{s+2}$

From Figure 1, write the expression for $I_1$ as follows:

$I_1=-n{I}_2$

From Equation (10), substitute $\left(\frac{s}{0.5s+1}\right)V_2$ for $I_2$ and $\frac{1}{2}$ for $n$ as follows:

$\begin{array}{c}{I}_1=-\left(\frac{1}{2}\right)\left(\frac{s}{0.5s+1}\right)V_2\\ =-\left(\frac{s}{s+2}\right)V_2\end{array}$

Rearrange the expression as follows:

$\frac{I_1}{V_2}=-\frac{s}{s+2}$

Substitute $\left(-\frac{s}{s+2}\right)$ for $\frac{I_1}{V_{\text{2}}}$ in Equation (7) to obtain the value of $y_{\text{12}}$ for upper network.

$y_{\text{12}}=-\frac{s}{s+2}$

From the calculations, write the y-parameters for upper network as follows:

$\left[y_a\right]=\left[\begin{array}{cc}\frac{s}{2\left(s+2\right)}& -\frac{s}{s+2}\\ -\frac{s}{s+2}& \frac{2s}{s+2}\end{array}\right]$

Redraw the lower network by short circuiting the port-2 as shown in Figure 3.

As the current bypasses through the short circuit path, redraw the circuit in Figure 3 as shown in Figure 4.

From Figure 4, write the expression for $I_{\text{1}}$ as follows:

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

Rearrange the expression as follows:

$\begin{array}{c}\frac{I_{\text{1}}}{V_{\text{1}}}=\frac{1}{2}\\ =0.5\end{array}$

Substitute 0.5 for $\frac{I_{\text{1}}}{V_{\text{1}}}$ in Equation (3) to obtain the value of $y_{\text{11}}$ for lower network.

$y_{\text{11}}=0.5$

From Figure 1, the expression for current $I_{\text{2}}$ is written as follows:

$I_{\text{2}}=-I_{\text{1}}$

From Equation (11), substitute $\frac{V_{\text{1}}}{2}$ for $I_{\text{1}}$ as follows:

$I_{\text{2}}=-\frac{V_{\text{1}}}{2}$

Rearrange the expression as follows:

$\begin{array}{c}\frac{I_{\text{2}}}{V_{\text{1}}}=-\frac{1}{2}\\ =-0.5\end{array}$

Substitute $\left(-0.5\right)$ for $\frac{I_{\text{2}}}{V_{\text{1}}}$ in Equation (4) to obtain the value of $y_{\text{21}}$ for lower network.

$y_{\text{21}}=-0.5$

Redraw the lower network by short circuiting the port-1 as shown in Figure 5.

From Figure 5, write the expression for $I_{\text{2}}$ as follows:

$\begin{array}{c}{I}_{\text{2}}=\frac{V_{\text{2}}}{\left(2\Vert s\right)}\\ =\frac{V_{\text{2}}}{\frac{\left(2\right)\left(s\right)}{2+s}}\end{array}$

$I_{\text{2}}=\left(\frac{s+2}{2s}\right)V_{\text{2}}$        (12)

Rearrange the expression as follows:

$\frac{I_{\text{2}}}{V_{\text{2}}}=\frac{s+2}{2s}$

Substitute $\frac{s+2}{2s}$ for $\frac{I_{\text{2}}}{V_{\text{2}}}$ in Equation (8) to obtain the value of $y_{\text{22}}$ for lower network.

$y_{\text{22}}=\frac{s+2}{2s}$

From Figure 5, obtain the current $I_{\text{1}}$ by using current division rule as follows:

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

From Equation (12), substitute $\left(\frac{s+2}{2s}\right)V_{\text{2}}$ for $I_{\text{2}}$ as follows:

$\begin{array}{c}{I}_{\text{1}}=-\left[\left(\frac{s+2}{2s}\right)V_{\text{2}}\right]\left(\frac{s}{s+2}\right)\\ =-0.5V_{\text{2}}\end{array}$

Rearrange the expression as follows:

$\frac{I_{\text{1}}}{V_{\text{2}}}=-0.5$

Substitute $-0.5$ for $\frac{I_1}{V_{\text{2}}}$ in Equation (7) to obtain the value of $y_{\text{12}}$ for lower network.

$y_{\text{12}}=-0.5$

From the calculations, write the y-parameters for lower network as follows:

$\left[y_b\right]=\left[\begin{array}{cc}0.5& -0.5\\ -0.5& \frac{s+2}{2s}\end{array}\right]$

As the upper and lower networks are connected in parallel, write the expression for overall admittance parameters as follows:

$\left[y\right]=\left[y_a\right]+\left[y_b\right]$

Substitute $\left[\begin{array}{cc}\frac{s}{2\left(s+2\right)}& -\frac{s}{s+2}\\ -\frac{s}{s+2}& \frac{2s}{s+2}\end{array}\right]$ for $\left[y_a\right]$ and $\left[\begin{array}{cc}0.5& -0.5\\ -0.5& \frac{s+2}{2s}\end{array}\right]$ for $\left[y_b\right]$ to obtain the required objective.

$\begin{array}{c}\left[y\right]=\left[\begin{array}{cc}\frac{s}{2\left(s+2\right)}& -\frac{s}{s+2}\\ -\frac{s}{s+2}& \frac{2s}{s+2}\end{array}\right]+\left[\begin{array}{cc}0.5& -0.5\\ -0.5& \frac{s+2}{2s}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{s}{2\left(s+2\right)}+0.5& -\frac{s}{s+2}+\left(-0.5\right)\\ -\frac{s}{s+2}+\left(-0.5\right)& \frac{2s}{s+2}+\frac{s+2}{2s}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{s+s+2}{2\left(s+2\right)}& \frac{-s-0.5s-1}{s+2}\\ \frac{-s-0.5s-1}{s+2}& \frac{\left(2s\right)\left(2s\right)+\left(s+2\right)\left(s+2\right)}{\left(s+2\right)\left(2s\right)}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{s+1}{s+2}& \frac{-\left(1.5s+1\right)}{s+2}\\ \frac{-\left(1.5s+1\right)}{s+2}& \frac{4s^2+s^2+4s+4}{2s\left(s+2\right)}\end{array}\right]\end{array}$

Simplify the expression as follows:

$\left[y\right]=\left[\begin{array}{cc}\frac{s+1}{s+2}& \frac{-\left(3s+2\right)}{2\left(s+2\right)}\\ \frac{-\left(3s+2\right)}{2\left(s+2\right)}& \frac{5s^2+4s+4}{2s\left(s+2\right)}\end{array}\right]$

Conclusion:

Thus, the admittance parameters for the given two-port network are $\underset{\_}{\left[\begin{array}{cc}\frac{s+1}{s+2}& \frac{-\left(3s+2\right)}{2\left(s+2\right)}\\ \frac{-\left(3s+2\right)}{2\left(s+2\right)}& \frac{5s^2+4s+4}{2s\left(s+2\right)}\end{array}\right]}$.