Chapter 16, Problem 59E

(a) Obtain y, z, h, and t parameters for the network shown in Fig. 16.67 using either the defining equations or mesh/nodal equations. (b) Verify your answers using the relationships in Table 16.1.

To determine

The $y$, $z$, $h$ and $t$ parameters for the given network.

Answer to Problem 59E

The $y$ parameter is $\left[\begin{array}{cc}0.145& -0.120\\ -0.120& 0.183\end{array}\right]\text{S}$, $z$ parameter is $\left[\begin{array}{cc}15& 10\\ 10& 12\end{array}\right]\Omega$, $h$ parameter is $\left[\begin{array}{cc}6.666\Omega & 0.833\\ -0.833& 0.083\text{S}\end{array}\right]$ and $t$ parameter is $\left[\begin{array}{cc}1.5& 8\Omega \\ 0.1\text{S}& 1.2\end{array}\right]$.

Explanation of Solution

Given data:

The given diagram is shown in Figure 1.

Calculation:

Mark the branch currents and open circuit voltages.

The required diagram is shown in Figure 2.

Apply KVL at the input side.

$\begin{array}{c}{V}_1=5I_1+10\left(I_1+I_2\right)\\ V_1=15I_1+10I_2\end{array}$        (1)

Apply KVL at the output side.

$\begin{array}{c}{V}_2=2I_2+10\left(I_2+I_1\right)\\ V_2=10I_1+12I_2\end{array}$        (2)

The standard equations for $z$ parameter is written as,

$V_1=z_{11}{I}_1+z_{12}{I}_2$        (3)

$V_2=z_{21}{I}_1+z_{22}{I}_2$        (4)

Compare equation (1) with equation (3).

$z_{11}=15$

$z_{12}=10$

Compare equation (2) with equation (4).

$z_{21}=10$

$z_{22}=12$

The $z$ parameter for a circuit is written as,

$\left[z\right]=\left[\begin{array}{cc}{z}_{11}& z_{12}\\ z_{21}& z_{22}\end{array}\right]\Omega$

Substitute $15$ for $z_{11}$, $10$ for $z_{12}$, $10$ for $z_{21}$ and $12$ for $z_{22}$ in the above matrix.

$\left[z\right]=\left[\begin{array}{cc}15& 10\\ 10& 12\end{array}\right]\Omega$

Rearrange equation (1) as,

$I_1=\frac{V_1-10I_2}{15}$        (5)

Rearrange equation (2) as,

$I_2=\frac{V_2-10I_1}{12}$        (6)

Substitute $\frac{V_2-10I_1}{12}$ for $I_2$ in equation (5).

$\begin{array}{l}{I}_1=\frac{V_1-10\left(\frac{V_2-10I_1}{12}\right)}{15}\\ I_1=\frac{12V_1-10V_2+100I_1}{12\times 15}\\ 0.455I_1=0.066V_1-0.055V_2\\ I_1=0.145V_1-0.120V_2\end{array}$        (7)

Substitute $0.145V_1-0.120V_2$ for $I_1$ in equation (6).

$\begin{array}{c}{I}_2=\frac{V_2-10\left(0.145V_1-0.120V_2\right)}{12}\\ I_2=\frac{-1.45V_1+2.20V_2}{12}\\ I_2=-0.120V_1+0.183V_2\end{array}$        (8)

The standard equations for $y$ parameter is written as,

$I_1=y_{11}{V}_1+y_{12}{V}_2$        (9)

$I_2=y_{21}{I}_1+y_{22}{I}_2$        (10)

Compare equation (7) with equation (9).

$y_{11}=0.145$

$y_{12}=-0.120$

Compare equation (8) with equation (10).

$y_{21}=-0.120$

$y_{22}=0.183$

The $y$ parameter for a circuit is written as,

$\left[y\right]=\left[\begin{array}{cc}{y}_{11}& y_{12}\\ y_{21}& y_{22}\end{array}\right]\text{S}$

Substitute $0.145$ for $y_{11}$, $-0.120$ for $y_{12}$, $-0.120$ for $y_{21}$ and $0.183$ for $y_{22}$ in the above matrix.

$\left[y\right]=\left[\begin{array}{cc}0.145& -0.120\\ -0.120& 0.183\end{array}\right]\text{S}$

Substitute $\frac{V_2-10I_1}{12}$ for $I_2$ in equation (1).

$\begin{array}{c}{V}_1=15I_1+10\left(\frac{V_2-10I_1}{12}\right)\\ V_1=15I_1+\frac{10}{12}{V}_2-\frac{100}{12}{I}_1\\ V_1=6.666I_1+0.833V_2\end{array}$        (11)

The standard equations for $h$ parameter is written as,

$V_1=h_{11}{I}_1+h_{12}{V}_2$        (12)

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

Compare equation (11) with equation (12).

$h_{11}=6.666$

$h_{12}=0.833$

Compare equation (6) with equation (13).

$\begin{array}{c}{h}_{21}=-\frac{10}{12}\\ =-0.833\end{array}$

$\begin{array}{c}{h}_{22}=\frac{1}{12}\\ =0.083\end{array}$

The $h$ parameter for a circuit is written as,

$\left[h\right]=\left[\begin{array}{cc}{h}_{11}\Omega & h_{12}\\ h_{21}& h_{22}\text{S}\end{array}\right]$

Substitute $6.666$ for $h_{11}$, $0.833$ for $h_{12}$, $-0.833$ for $h_{21}$ and $0.083$ for $h_{22}$ in the above matrix.

$\left[h\right]=\left[\begin{array}{cc}6.666\Omega & 0.833\\ -0.833& 0.083\text{S}\end{array}\right]$

Rearrange equation (6) as,

$\begin{array}{l}{I}_2=\frac{1}{12}{V}_2-\frac{10}{12}{I}_1\\ I_2=0.083V_2-0.833I_1\\ 0.833I_1=0.083V_2-I_2\\ I_1=0.1V_2-1.2I_2\end{array}$        (14)

Substitute $0.1V_2-1.2I_2$ for $I_1$ in equation (1).

$\begin{array}{c}{V}_1=15\left(0.1V_2-1.2I_2\right)+10I_2\\ =1.5V_2-8I_2\end{array}$        (15)

The standard equations for $t$ parameter is written as,

$V_1=t_{11}{V}_2-t_{12}{I}_2$        (16)

$I_1=t_{21}{V}_2-t_{22}{I}_2$        (17)

Compare equation (15) with equation (16).

$t_{11}=1.5$

$t_{12}=8$

Compare equation (14) with equation (17).

$t_{21}=0.1$

$t_{22}=1.2$

The $t$ parameter for the first part of the circuit is written as,

$\left[t\right]=\left[\begin{array}{cc}{t}_{11}& t_{12}\Omega \\ t_{21}\text{S}& t_{22}\end{array}\right]$

Substitute $1.5$ for $t_{11}$, $8$ for $t_{12}$, $0.1$ for $t_{21}$ and $1.2$ for $t_{22}$ in the above matrix.

$\left[t\right]=\left[\begin{array}{cc}1.5& 8\Omega \\ 0.1\text{S}& 1.2\end{array}\right]$

Conclusion:

Therefore, the $y$ parameter is $\left[\begin{array}{cc}0.145& -0.120\\ -0.120& 0.183\end{array}\right]\text{S}$, $z$ parameter is $\left[\begin{array}{cc}15& 10\\ 10& 12\end{array}\right]\Omega$, $h$ parameter is $\left[\begin{array}{cc}6.666\Omega & 0.833\\ -0.833& 0.083\text{S}\end{array}\right]$ and $t$ parameter is $\left[\begin{array}{cc}1.5& 8\Omega \\ 0.1\text{S}& 1.2\end{array}\right]$.

To determine

To verify: The value of $y$, $z$, $h$ and $t$ parameters for the given network using conversions.

Explanation of Solution

Calculation:

The relation between $z$ and $y$ parameters is given by,

$y=\left[\begin{array}{cc}\frac{z_{22}}{{\Delta }_z}& \frac{-z_{12}}{{\Delta }_z}\\ \frac{-z_{21}}{{\Delta }_z}& \frac{z_{11}}{{\Delta }_z}\end{array}\right]$        (18)

The determinant ${\Delta }_z$ is calculated as,

$\begin{array}{c}\left|z\right|=\left|\begin{array}{cc}15& 10\\ 10& 12\end{array}\right|\\ =180-100\\ =80\end{array}$

Substitute $15$ for $z_{11}$, $10$ for $z_{12}$, $10$ for $z_{21}$, $12$ for $z_{22}$ and $80$ for ${\Delta }_z$ in equation (18).

$\begin{array}{c}y=\left[\begin{array}{cc}\frac{12}{80}& \frac{-10}{80}\\ \frac{-10}{80}& \frac{15}{80}\end{array}\right]\\ =\left[\begin{array}{cc}0.15& -0.125\\ -0.125& 0.1875\end{array}\right]\end{array}$

The relation between $z$ and $h$ parameters is given by,

$h=\left[\begin{array}{cc}\frac{{\Delta }_z}{z_{22}}& \frac{z_{12}}{z_{22}}\\ \frac{-z_{21}}{z_{22}}& \frac{1}{z_{22}}\end{array}\right]$        (19)

Substitute $15$ for $z_{11}$, $10$ for $z_{12}$, $10$ for $z_{21}$, $12$ for $z_{22}$ and $80$ for ${\Delta }_z$ in equation (18).

$\begin{array}{c}h=\left[\begin{array}{cc}\frac{80}{12}& \frac{10}{12}\\ \frac{-10}{12}& \frac{1}{12}\end{array}\right]\\ =\left[\begin{array}{cc}6.666\Omega & 0.833\\ -0.833& 0.083\text{S}\end{array}\right]\end{array}$

The relation between $z$ and $t$ parameters is given by,

$t=\left[\begin{array}{cc}\frac{z_{11}}{z_{21}}& \frac{\Delta z}{z_{21}}\\ \frac{1}{z_{21}}& \frac{z_{22}}{z_{21}}\end{array}\right]$        (20)

Substitute $15$ for $z_{11}$, $10$ for $z_{12}$, $10$ for $z_{21}$, $12$ for $z_{22}$ and $80$ for ${\Delta }_z$ in equation (18).

$\begin{array}{c}t=\left[\begin{array}{cc}\frac{15}{10}& \frac{80}{10}\\ \frac{1}{10}& \frac{12}{10}\end{array}\right]\\ =\left[\begin{array}{cc}1.5& 8\Omega \\ 0.1\text{S}& 1.2\end{array}\right]\end{array}$

Conclusion:

Thus, within the limits of error the value of $y$, $z$, $h$ and $t$ parameters for the given network using conversions is verified.