Chapter 16, Problem 35E

The two-port networks of Fig. 16.50 are connected in series. (a) Determine the impedance parameters for the series connection by first finding the z parameters of the individual networks. (b) If the two networks are instead connected in parallel, determine the admittance parameters of the combination by first finding the y parameters of the individual networks. (c) Verify your answer to part (b) by using Table 16.1 in conjunction with your answer to part (a).

To determine

The value of impedance parameters for the given condition.

Answer to Problem 35E

The value of total impedance when the two port networks are connected in series is $\left[z\right]=\left[\begin{array}{cc}167.73& 33.85\\ 33.85& 95.34\end{array}\right]\text{Ω}$.

Explanation of Solution

Given data:

The given diagram is shown in Figure 1.

The given diagram is shown in Figure 2.

 

Calculation:

Determine the impedance parameter $z_1$.

The required diagram is shown in Figure 3.

 

The impedance parameters can be expressed as,

$V_1=z_{11}{I}_1+z_{12}{I}_2$        (1)

$V_2=z_{21}{I}_1+z_{22}{I}_2$        (2)

Substitute $0$ for $I_2$ in equation (1).

$\begin{array}{l}{V}_{\text{1}}=z_{11}{I}_{\text{1}}+z_{12}\left(0\right)\\ V_{\text{1}}=z_{11}{I}_{\text{1}}\\ z_{11}=\frac{V_{\text{1}}}{I_{\text{1}}}\end{array}$        (3)

Substitute $0$ for $I_1$ in equation (1).

$\begin{array}{l}{V}_{\text{1}}=z_{11}\left(0\right)+z_{12}{I}_2\\ V_{\text{1}}=z_{12}{I}_2\\ z_{12}=\frac{V_{\text{1}}}{I_2}\end{array}$        (4)

Substitute $0$ for $I_2$  in equation (2).

$\begin{array}{l}{V}_2=z_{21}{I}_{\text{1}}+z_{22}\left(0\right)\\ V_2=z_{21}{I}_{\text{1}}\\ z_{21}=\frac{V_2}{I_{\text{1}}}\end{array}$        (5)

Substitute $0$ for $I_1$ in equation (2).

$\begin{array}{l}{V}_2=z_{21}\left(0\right)+z_{22}{I}_2\\ V_2=z_{22}{I}_2\\ z_{22}=\frac{V_2}{I_2}\end{array}$        (6)

Apply KVL left loop of Figure 3.

$-V_1+3I_1+6\left(I_1+I_2\right)=0$        (7)

Substitute $0$ for $I_2$ in equation (7).

$\begin{array}{l}-V_1+3I_1+6\left(I_1+0\right)=0\\ -V_1+3I_1+6I_1=0\\ V_1=9I_1\\ \frac{V_1}{I_1}=9\end{array}$

Substitute $9$  for $\frac{V_1}{I_1}$ in equation (3).

$z_{11}=9\text{Ω}$

Substitute $0$ for $I_1$ in equation (7).

$\begin{array}{l}-V_1+3\left(0\right)+6\left(0+I_2\right)=0\\ -V_1+6I_2=0\\ V_1=6I_2\\ \frac{V_1}{I_2}=6\end{array}$

Substitute $6$ for $\frac{V_1}{I_2}$ in equation (4).

$z_{12}=6\text{Ω}$

Apply KVL at the right loop.

$-V_2+2I_2+6\left(I_2+I_1\right)=0$        (8)

Substitute $0$ for $I_2$ in equation (8).

$\begin{array}{l}-V_2+2\left(0\right)+6\left(I_1+0\right)=0\\ -V_2+6I_1=0\\ V_2=6I_1\\ \frac{V_2}{I_1}=6\end{array}$

Substitute $6$ for $\frac{V_2}{I_1}$ in equation (5).

$z_{21}=6\text{Ω}$

Substitute $0$ for $I_1$ in equation (8).

$\begin{array}{l}-V_2+2I_2+6\left(0+I_2\right)=0\\ -V_2+8I_2=0\\ V_2=8I_2\\ \frac{V_2}{I_2}=8\end{array}$

Substitute $8$ for $\frac{V_2}{I_2}$ in equation (6).

$z_{22}=8\text{Ω}$

Hence the $z$ parameters are written as,

$z_1=\left[\begin{array}{cc}9& 6\\ 6& 8\end{array}\right]\text{Ω}$

Convert the delta connected network to star connected network.

The required diagram is shown in Figure 4.

 

The impedance $Z_1$ calculated from the figure is given by,

$\begin{array}{c}{Z}_1=\frac{\left(220\right)\left(470\right)}{220+470+100}\\ =\frac{103400}{790}\\ =130.886\text{Ω}\end{array}$

The impedance $Z_2$ calculated from the figure is given by,

$\begin{array}{c}{Z}_2=\frac{\left(100\right)\left(470\right)}{220+470+100}\\ =\frac{47000}{790}\\ =59.4936\text{Ω}\end{array}$

The impedance $Z_3$ calculated from the figure is given by,

$\begin{array}{c}{Z}_3=\frac{\left(100\right)\left(220\right)}{220+470+100}\\ =\frac{22000}{790}\\ =27.848\text{Ω}\end{array}$

Redraw the star connected circuit.

The required diagram is shown in Figure 5.

 

Apply KVL in the left loop.

$-V_1+130.886I_1+27.848\left(I_1+I_2\right)=0$        (9)

Substitute $0$ for $I_2$ in equation (9).

$\begin{array}{l}-V_1+130.886I_1+27.848\left(I_1+0\right)=0\\ -V_1+158.734I_1=0\\ V_1=158.734I_1\\ \frac{V_1}{I_1}=158.734\end{array}$

Substitute $158.734$  for $\frac{V_1}{I_1}$ in equation (3).

$z_{11}=158.734\text{Ω}$

Substitute $0$ for $I_1$ in equation (9).

$\begin{array}{l}-V_1+130.886\left(0\right)+27.848\left(0+I_2\right)=0\\ -V_1+27.848I_2=0\\ V_1=27.848I_2\\ \frac{V_1}{I_2}=27.848\end{array}$

Substitute $27.848$ for $\frac{V_1}{I_2}$ in equation (4).

$z_{12}=27.848\text{Ω}$

Apply KVL at right loop of Figure 5.

$-V_2+59.4936I_2+27.848\left(I_2+I_1\right)=0$        (10)

Substitute $0$ for $I_2$ in equation (10).

$\begin{array}{l}-V_2+59.4936\left(0\right)+27.848\left(I_1+0\right)=0\\ -V_2+27.848I_1=0\\ V_2=27.848I_1\\ \frac{V_2}{I_1}=27.848\end{array}$

Substitute $27.484$  for $\frac{V_2}{I_1}$ in equation (5).

$z_{21}=27.848\text{Ω}$

Substitute $0$ for $I_1$ in equation (10).

$\begin{array}{l}-V_2+59.4936I_2+27.848\left(0+I_2\right)=0\\ -V_2+87.34I_2=0\\ V_2=87.34I_2\\ \frac{V_2}{I_2}=87.34\end{array}$

Substitute $87.34$  for $\frac{V_2}{I_2}$ in equation (6).

$z_{22}=87.34\text{Ω}$

The $z$ parameters for Figure 2 is,

$z_2=\left[\begin{array}{cc}158.734& 27.848\\ 27.848& 87.34\end{array}\right]\text{Ω}$

The overall impedance matrix when the two port networks are connected in series is,

$\left[z\right]=\left[z_1\right]+\left[z_2\right]$        (11)

Substitute $\left[\begin{array}{cc}9& 6\\ 6& 8\end{array}\right]\text{Ω}$ for $\left[z_1\right]$ and $\left[\begin{array}{cc}158.734& 27.848\\ 27.848& 87.34\end{array}\right]\text{Ω}$ for $\left[z_2\right]$ in equation (11).

$\begin{array}{c}\left[z\right]=\left[\begin{array}{cc}9& 6\\ 6& 8\end{array}\right]\text{Ω}+\left[\begin{array}{cc}158.734& 27.848\\ 27.848& 87.34\end{array}\right]\text{Ω}\\ =\left[\begin{array}{cc}167.73& 33.85\\ 33.85& 95.34\end{array}\right]\text{Ω}\end{array}$

Conclusion:

Therefore, the value of total impedance when the two port networks are connected in series is $\left[z\right]=\left[\begin{array}{cc}167.73& 33.85\\ 33.85& 95.34\end{array}\right]\text{Ω}$.

To determine

The value of admittance parameters for the given condition.

Answer to Problem 35E

The value of total admittance when the two port networks are connected in parallel is $\left[y\right]=\left[\begin{array}{cc}0.2289& -0.169\\ -0.169& 0.2621\end{array}\right]\text{S}$.

Explanation of Solution

Calculation:

The standard equation for admittance parameters is given by,

$I_1=y_{11}{V}_1+y_{12}{V}_2$        (12)

$I_2=y_{21}{V}_1+y_{22}{V}_2$        (13)

Substitute $0$ for $V_2$ in equation (12).

$\begin{array}{l}{I}_1=y_{11}{V}_1+y_{12}\left(0\right)\\ I_1=y_{11}{V}_1\\ y_{11}=\frac{I_1}{V_1}\end{array}$        (14)

Substitute $0$ for $V_1$ in equation (12).

$\begin{array}{l}{I}_1=y_{11}\left(0\right)+y_{12}{V}_2\\ I_1=y_{12}{V}_2\\ y_{12}=\frac{I_1}{V_2}\end{array}$        (15)

Substitute $0$ for $V_2$ in equation (13).

$\begin{array}{l}{I}_2=y_{21}{V}_1+y_{22}\left(0\right)\\ I_2=y_{21}{V}_1\\ y_{21}=\frac{I_2}{V_1}\end{array}$        (16)

Substitute $0$ for $V_1$ in equation (13).

$\begin{array}{l}{I}_2=y_{21}\left(0\right)+y_{22}{V}_2\\ I_2=y_{22}{V}_2\\ y_{22}=\frac{I_2}{V_2}\end{array}$        (17)

Modify the given diagram for $V_2=0\text{V}$.

The required diagram is shown in Figure 6.

The resistances $2\text{Ω}$ and $6\text{Ω}$ are in parallel.

The equivalent resistance is,

$\begin{array}{c}{R}_{\text{eq}}=\frac{\left(2\text{Ω}\right)\left(6\text{Ω}\right)}{\left(2+6\right)\text{Ω}}\\ =\frac{3}{2}\text{Ω}\end{array}$

Further, the resistances $3\text{Ω}$ and $R_{\text{eq}}$ are in series.

The total equivalent resistance of the circuit is given by,

$R_T=3\text{Ω}+R_{\text{eq}}$

Substitute $\frac{3}{2}\text{Ω}$ for $R_{\text{eq}}$ in the above equation.

$\begin{array}{c}{R}_T=3\text{Ω}+\frac{3}{2}\text{Ω}\\ =\frac{9}{2}\text{Ω}\end{array}$

The input voltage calculated from the circuit is written as,

$V_1=I_1{R}_T$

Rearrange the above equation as,

$\frac{I_1}{V_1}=\frac{1}{R_T}$

Substitute $\frac{9}{2}$ for $R_T$ in the above equation.

$\begin{array}{c}\frac{I_1}{V_1}=\frac{1}{\left(\frac{9}{2}\right)}\\ =\frac{2}{9}\end{array}$

Substitute $\frac{2}{9}$ for $\frac{I_1}{V_1}$ in the above equation (14).

$y_{11}=\frac{2}{9}\text{S}$

Rearrange equation (7) as,

$\begin{array}{c}{V}_1=3I_1+6I_1+6I_2\\ =9I_1+6I_2\end{array}$        (18)

Apply current division rule in Figure 6.

$\begin{array}{l}{I}_2=-I_1\left(\frac{6}{6+2}\right)\\ I_2=-\left(\frac{3}{4}\right)I_1\\ I_1=-\left(\frac{4}{3}\right)I_2\end{array}$

Substitute $-\left(\frac{4}{3}\right)I_2$ for $I_1$ in equation (18).

$\begin{array}{l}{V}_1=9\left(-\left(\frac{4}{3}\right)I_2\right)+6I_2\\ V_1=-12I_2+6I_2\\ V_1=-6I_2\\ \frac{I_2}{V_1}=-\frac{1}{6}\end{array}$

Substitute $-\frac{1}{6}$ for $\frac{I_2}{V_1}$ in the above equation (16).

$y_{21}=-\frac{1}{6}\text{S}$

Modify the given diagram for $V_1=0\text{V}$.

The required diagram is shown in Figure 7.

Rearrange equation (8).

$\begin{array}{c}{V}_2=2I_2+6I_2+6I_1\\ =6I_1+8I_2\end{array}$        (19)

Apply current division rule in the circuit of Figure 7.

$\begin{array}{l}{I}_1=-I_2\left(\frac{6}{6+3}\right)\\ I_1=-\left(\frac{6}{9}\right)I_2\\ I_2=-\left(\frac{9}{6}\right)I_1\\ I_2=-\frac{3}{2}{I}_1\end{array}$

Substitute $-\left(\frac{3}{2}\right)I_1$ for $I_2$ in equation (19).

$\begin{array}{l}{V}_2=6I_1+8\left(-\left(\frac{3}{2}\right)I_1\right)\\ V_2=6I_1-12I_1\\ V_2=-6I_1\\ \frac{I_1}{V_2}=-\frac{1}{6}\end{array}$

Substitute $-\frac{1}{6}$ for $\frac{I_1}{V_2}$ in the above equation (15).

$y_{12}=-\frac{1}{6}\text{S}$

The resistances $3\text{Ω}$ and $6\text{Ω}$ are in parallel in Figure 7.

The equivalent resistance is,

$\begin{array}{c}{R^{\prime }}_{\text{eq}}=\frac{\left(3\text{Ω}\right)\left(6\text{Ω}\right)}{\left(3+6\right)\text{Ω}}\\ =\frac{18}{9}\text{Ω}\\ \text{=2}\text{Ω}\end{array}$

Further, the resistances $2\text{Ω}$ and ${R^{\prime }}_{\text{eq}}$ are in series.

The total equivalent resistance of the circuit is given by,

${R^{\prime }}_T=2\text{Ω}+{R^{\prime }}_{\text{eq}}$

Substitute $2\text{Ω}$ for ${R^{\prime }}_{\text{eq}}$ in the above equation.

$\begin{array}{c}{R^{\prime }}_T=2\text{Ω}+2\text{Ω}\\ =4\text{Ω}\end{array}$

The input voltage of Figure 7 can be expressed as,

$V_2=I_2{R^{\prime }}_T$

Rearrange the above equation as,

$\frac{I_2}{V_2}=\frac{1}{{R^{\prime }}_T}$

Substitute $4$ for ${R^{\prime }}_T$ in the above equation.

$\frac{I_2}{V_2}=\frac{1}{4}$

Substitute $\frac{1}{4}$ for $\frac{I_2}{V_2}$ in the above equation (17).

$y_{22}=\frac{1}{4}\text{S}$

The $y$ parameters for Figure 1 is,

$y_1=\left[\begin{array}{cc}\frac{2}{9}& -\frac{1}{6}\\ -\frac{1}{6}& \frac{1}{4}\end{array}\right]\text{S}$

Redraw the Figure 2 and show the input voltage $V_1$ and the output voltage $V_2$.

The required diagram is shown in Figure 8.

Redraw the above figure for $V_2=0\text{V}$.

The required diagram is shown in Figure 9.

 

The $100\text{Ω}$ resistor is short circuited.

Therefore, $220\text{Ω}$ resistor and $470\text{Ω}$ resistor are in parallel connection.

The equivalent resistance of the above circuit is,

$\begin{array}{c}{R^{″}}_{eq}=\frac{\left(220\right)\left(470\right)}{220+470}\\ =\frac{103400}{690}\\ =149.855\text{Ω}\end{array}$

The input voltage of Figure 9 can be expressed as,

$V_1=I_1{R^{″}}_{eq}$

Rearrange the above equation as,

$\frac{I_1}{V_1}=\frac{1}{{R^{″}}_{eq}}$

Substitute $149.855$ for ${R^{″}}_{eq}$ in the above equation.

$\begin{array}{c}\frac{I_1}{V_1}=\frac{1}{149.855}\\ =6.67\times {10}^{-3}\text{S}\end{array}$

Substitute $6.67\times {10}^{-3}\text{S}$ for $\frac{I_1}{V_1}$ in the above equation (14).

$y_{11}=6.67\times {10}^{-3}\text{S}$

The voltage across $470\text{Ω}$ resistor is $V_1$.

Therefore, the current through $470\text{Ω}$ resistor is,

$I_2=-\frac{ V_1}{470}$

Rearrange the above equation as,

$\begin{array}{c}\frac{I_2}{V_1}=-\frac{1}{470}\\ =-2.127\times {10}^{-3}\text{S}\end{array}$

Substitute $2.127\times {10}^{-3}\text{S}$ for $\frac{I_2}{V_1}$ in the above equation (1).

$y_{21}=-2.127\times {10}^{-3}\text{S}$

Redraw the above figure for $V_1=0\text{V}$.

The required diagram is shown in Figure 10.

 

The $220\text{Ω}$ resistor of Figure 10 is short circuited.

The voltage across $470\text{Ω}$ resistor is $V_2$.

Therefore, the current through $470\text{Ω}$ resistor is,

$I_1=-\frac{ V_2}{470}$

Rearrange the above equation as,

$\begin{array}{c}\frac{I_1}{V_2}=-\frac{1}{470}\\ =-2.127\times {10}^{-3}\text{S}\end{array}$

Substitute $-2.127\times {10}^{-3}\text{S}$ for $\frac{I_1}{V_2}$ in the above equation (15).

$y_{12}=-2.127\times {10}^{-3}\text{S}$

The $100\text{Ω}$ resistor and $470\text{Ω}$ resistor are in parallel connection.

The equivalent resistance of the above circuit is,

$\begin{array}{c}{R^{‴}}_{eq}=\frac{\left(100\right)\left(470\right)}{100+470}\\ =\frac{47000}{570}\\ =82.456\text{Ω}\end{array}$

The input voltage of Figure 10 can be expressed as,

$V_2=I_2{R^{‴}}_{eq}$

Rearrange the above equation as,

$\frac{I_2}{V_2}=\frac{1}{{R^{‴}}_{eq}}$

Substitute $82.456$ for ${R^{‴}}_{eq}$ in the above equation.

$\begin{array}{c}\frac{I_2}{V_2}=\frac{1}{82.456}\\ =12.13\times {10}^{-3}\end{array}$

Substitute $12.13\times {10}^{-3}\text{S}$ for $\frac{I_2}{V_2}$ in the equation (17).

$y_{22}=12.13\times {10}^{-3}\text{S}$

The $y$ parameters for Figure 2 is,

$y_2=\left[\begin{array}{cc}6.67\times {10}^{-3}& -2.13\times {10}^{-3}\\ -2.13\times {10}^{-3}& 12.13\times {10}^{-3}\end{array}\right]\text{S}$

The overall admittance matrix when the two port networks are connected in parallel is,

$\left[y\right]=\left[y_1\right]+\left[y_2\right]$        (20)

Substitute $\left[\begin{array}{cc}\frac{2}{9}& -\frac{1}{6}\\ -\frac{1}{6}& \frac{1}{4}\end{array}\right]\text{S}$ for $\left[y_1\right]$ and $\left[\begin{array}{cc}6.67\times {10}^{-3}& -2.13\times {10}^{-3}\\ -2.13\times {10}^{-3}& 12.13\times {10}^{-3}\end{array}\right]\text{S}$ for $\left[y_2\right]$ in equation (20).

$\begin{array}{c}\left[y\right]=\left[\begin{array}{cc}\frac{2}{9}& -\frac{1}{6}\\ -\frac{1}{6}& \frac{1}{4}\end{array}\right]\text{S}+\left[\begin{array}{cc}6.67\times {10}^{-3}& -2.13\times {10}^{-3}\\ -2.13\times {10}^{-3}& 12.13\times {10}^{-3}\end{array}\right]\text{S}\\ =\left[\begin{array}{cc}0.2289& -0.169\\ -0.169& 0.2621\end{array}\right]\text{S}\end{array}$

Conclusion:

Therefore, the value of total admittance when the two port networks are connected in parallel is $\left[y\right]=\left[\begin{array}{cc}0.2289& -0.169\\ -0.169& 0.2621\end{array}\right]\text{S}$.

To determine

The value of $y_1$ matrix and $y_2$ matrix after conversion.

Explanation of Solution

Calculation:

The determinant of $z_1$ matrix is,

$\begin{array}{c}{\Delta }_{z_1}=\left|\begin{array}{cc}9& 6\\ 6& 8\end{array}\right|\text{Ω}\\ \text{=}\left(\left(9\right)\left(8\right)\text{Ω}-\left(6\right)\left(6\right)\text{Ω}\right)\\ =\left(72\text{Ω}-36\text{Ω}\right)\\ \text{=36}{\text{Ω}}^2\end{array}$

The matrix for $y$ parameters in terms of $z$ parameters is,

$y=\left[\begin{array}{cc}\frac{z_{22}}{{\Delta }_{z_1}}& \frac{-z_{12}}{{\Delta }_{z_1}}\\ \frac{-z_{21}}{{\Delta }_{z_1}}& \frac{z_{11}}{{\Delta }_{z_1}}\end{array}\right]$        (21)

Substitute $9\text{Ω}$ for $z_{11}$, $6\text{Ω}$ for $z_{12}$, $6\text{Ω}$ for $z_{21}$, $8\text{Ω}$ for $z_{22}$ and $36{\text{Ω}}^{\text{2}}$ for ${\Delta }_{z_1}$ in equation (21).

$\begin{array}{c}y=\left[\begin{array}{cc}\frac{8\text{Ω}}{36{\Omega }^2}& \frac{-6\text{Ω}}{36{\Omega }^2}\\ \frac{-\text{6}\text{Ω}}{36{\Omega }^2}& \frac{9\text{Ω}}{36{\Omega }^2}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{2}{9}& \frac{-1}{6}\\ \frac{-1}{6}& \frac{1}{4}\end{array}\right]\text{S}\end{array}$

The determinant of $z_2$ matrix is,

$\begin{array}{c}{\Delta }_{z_2}=\left|\begin{array}{cc}158.734& 27.848\\ 27.848& 87.34\end{array}\right|\text{Ω}\\ \text{=}\left(\left(158.734\right)\left(87.34\right)\text{Ω}-\left(27.848\right)\left(27.848\right)\text{Ω}\right)\\ \text{=13088}\text{.316}{\text{Ω}}^2\end{array}$

Substitute $158.734\text{Ω}$ for $z_{11}$, $27.848\text{Ω}$ for $z_{12}$, $27.848\text{Ω}$ for $z_{21}$, $87.34\text{Ω}$ for $z_{22}$ and $13088.316{\text{Ω}}^{\text{2}}$ for ${\Delta }_{z_2}$ in equation (21).

$\begin{array}{c}y=\left[\begin{array}{cc}\frac{87.34\text{Ω}}{13088.316{\Omega }^2}& \frac{-27.848\text{Ω}}{13088.316{\Omega }^2}\\ \frac{-27.848\text{Ω}}{13088.316{\Omega }^2}& \frac{158.734\text{Ω}}{13088.316{\Omega }^2}\end{array}\right]\\ =\left[\begin{array}{cc}6.67\times {10}^{-3}& -2.13\times {10}^{-3}\\ -2.13\times {10}^{-3}& 12.13\times {10}^{-3}\end{array}\right]\text{S}\end{array}$

Conclusion:

Therefore, the value of $y_1$ matrix and $y_2$ matrix after conversion is same.