Chapter 11, Problem 30P

To determine

Show that the matrix ABCD is $\left[\begin{array}{cc}\cosh \gamma \mathcal{l}& Z_{\text{o}}\sinh \gamma \mathcal{l}\\ \frac{1}{Z_{\text{o}}}\sinh \gamma \mathcal{l}& \cosh \gamma \mathcal{l}\end{array}\right]$.

Explanation of Solution

Calculation:

Refer to the mentioned Figure given in the textbook.

The relation between the input and output variables matrix is,

$\left[\begin{array}{c}{V}_1\\ I_1\end{array}\right]=\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]\left[\begin{array}{c}{V}_2\\ -I_2\end{array}\right]$

Consider the general expression for terminal conditions.

$V_1=V_s\left(z=0\right)=V_{\text{o}}^{+}+V_{\text{o}}^{-}$        (1)

$V_2=V_s\left(z=\mathcal{l}\right)=V_{\text{o}}^{+}{e}^{-\gamma \mathcal{l}}+V_{\text{o}}^{-}{e}^{\gamma \mathcal{l}}$        (2)

$I_1=I_s\left(z=0\right)=\frac{V_{\text{o}}^{+}}{Z_{\text{o}}}-\frac{V_{\text{o}}^{-}}{Z_{\text{o}}}$        (3)

$I_2=-I_s\left(z=\mathcal{l}\right)=-\frac{V_{\text{o}}^{+}}{Z_{\text{o}}}{e}^{-\gamma \mathcal{l}}+\frac{V_{\text{o}}^{-}}{Z_{\text{o}}}{e}^{\gamma \mathcal{l}}$        (4)

Adding Equation (1) and (3). Therefore,

$V_{\text{o}}^{+}=\frac{1}{2}\left(V_1+Z_{\text{o}}{I}_1\right)$        (5)

Subtracting Equation (1) and (3). Therefore,

$V_{\text{o}}^{-}=\frac{1}{2}\left(V_1-Z_{\text{o}}{I}_1\right)$        (6)

Substitute Equation (5) and (6) in Equation (2).

$\begin{array}{c}{V}_2=\left[\frac{1}{2}\left(V_1+Z_{\text{o}}{I}_1\right)\right]e^{-\gamma \mathcal{l}}+\left[\frac{1}{2}\left(V_1-Z_{\text{o}}{I}_1\right)\right]e^{\gamma \mathcal{l}}\\ =\frac{1}{2}\left(V_1{e}^{-\gamma \mathcal{l}}+Z_{\text{o}}{I}_1{e}^{-\gamma \mathcal{l}}\right)+\frac{1}{2}\left(V_1{e}^{\gamma \mathcal{l}}-Z_{\text{o}}{I}_1{e}^{\gamma \mathcal{l}}\right)\\ =\frac{1}{2}\left(e^{\gamma \mathcal{l}}+e^{-\gamma \mathcal{l}}\right)V_1+\frac{1}{2}{Z}_{\text{o}}\left(e^{-\gamma \mathcal{l}}-e^{\gamma \mathcal{l}}\right)I_1\end{array}$

$\begin{array}{l}{V}_2=\cosh \gamma \mathcal{l}{V}_1-Z_{\text{o}}\sinh \gamma \mathcal{l}{I}_1\\ \left\{\begin{array}{l}\because \cosh \left(ix\right)=\frac{1}{2}\left(e^{ix}+e^{-ix}\right),\\ \sinh \left(ix\right)=\frac{1}{2}\left(e^{ix}-e^{-ix}\right)\end{array}\right\}\end{array}$        (7)

Substitute Equation (5) and (6) in Equation (4).

$\begin{array}{c}{I}_2=-\frac{\left[\frac{1}{2}\left(V_1+Z_{\text{o}}{I}_1\right)\right]}{Z_{\text{o}}}{e}^{-\gamma \mathcal{l}}+\frac{\left[\frac{1}{2}\left(V_1-Z_{\text{o}}{I}_1\right)\right]}{Z_{\text{o}}}{e}^{\gamma \mathcal{l}}\\ =-\frac{1}{2Z_{\text{o}}}\left(V_1+Z_{\text{o}}{I}_1\right)e^{-\gamma \mathcal{l}}+\frac{1}{2Z_{\text{o}}}\left(V_1-Z_{\text{o}}{I}_1\right)e^{\gamma \mathcal{l}}\\ =-\frac{1}{2Z_{\text{o}}}\left(V_1{e}^{-\gamma \mathcal{l}}+Z_{\text{o}}{I}_1{e}^{-\gamma \mathcal{l}}\right)+\frac{1}{2Z_{\text{o}}}\left(V_1{e}^{\gamma \mathcal{l}}-Z_{\text{o}}{I}_1{e}^{\gamma \mathcal{l}}\right)\\ =-\frac{1}{2Z_{\text{o}}}{V}_1{e}^{-\gamma \mathcal{l}}-\frac{1}{2Z_{\text{o}}}{Z}_{\text{o}}{I}_1{e}^{-\gamma \mathcal{l}}+\frac{1}{2Z_{\text{o}}}{V}_1{e}^{\gamma \mathcal{l}}-\frac{1}{2Z_{\text{o}}}{Z}_{\text{o}}{I}_1{e}^{\gamma \mathcal{l}}\end{array}$

The above equation becomes,

$I_2=\frac{1}{2Z_{\text{o}}}\left(e^{\gamma \mathcal{l}}-e^{-\gamma \mathcal{l}}\right)V_1-\frac{1}{2}\left(e^{\gamma \mathcal{l}}+e^{-\gamma \mathcal{l}}\right)I_1$

$I_2=\frac{1}{Z_{\text{o}}}\sinh \gamma \mathcal{l}{V}_1-\cosh \gamma \mathcal{l}{I}_1$        (8)

From Equation (7) and (8),

$\left[\begin{array}{c}{V}_2\\ I_2\end{array}\right]=\left[\begin{array}{cc}\cosh \gamma \mathcal{l}& -Z_{\text{o}}\sinh \gamma \mathcal{l}\\ -\frac{1}{Z_{\text{o}}}\sinh \gamma \mathcal{l}& \cosh \gamma \mathcal{l}\end{array}\right]\left[\begin{array}{c}{V}_1\\ I_1\end{array}\right]$

But,

${\left[\begin{array}{cc}\cosh \gamma \mathcal{l}& -Z_{\text{o}}\sinh \gamma \mathcal{l}\\ -\frac{1}{Z_{\text{o}}}\sinh \gamma \mathcal{l}& \cosh \gamma \mathcal{l}\end{array}\right]}^{-1}=\left[\begin{array}{cc}\cosh \gamma \mathcal{l}& Z_{\text{o}}\sinh \gamma \mathcal{l}\\ \frac{1}{Z_{\text{o}}}\sinh \gamma \mathcal{l}& \cosh \gamma \mathcal{l}\end{array}\right]$

Therefore,

$\left[\begin{array}{c}{V}_1\\ I_1\end{array}\right]=\left[\begin{array}{cc}\cosh \gamma \mathcal{l}& Z_{\text{o}}\sinh \gamma \mathcal{l}\\ \frac{1}{Z_{\text{o}}}\sinh \gamma \mathcal{l}& \cosh \gamma \mathcal{l}\end{array}\right]\left[\begin{array}{c}{V}_2\\ -I_2\end{array}\right]$

Compare the above equation with the given input and output variables matrix. Therefore, the matrix ABCD is,

$\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]=\left[\begin{array}{cc}\cosh \gamma \mathcal{l}& Z_{\text{o}}\sinh \gamma \mathcal{l}\\ \frac{1}{Z_{\text{o}}}\sinh \gamma \mathcal{l}& \cosh \gamma \mathcal{l}\end{array}\right]$

Conclusion:

Thus, the matrix ABCD is $\left[\begin{array}{cc}\cosh \gamma \mathcal{l}& Z_{\text{o}}\sinh \gamma \mathcal{l}\\ \frac{1}{Z_{\text{o}}}\sinh \gamma \mathcal{l}& \cosh \gamma \mathcal{l}\end{array}\right]$ and it is proved.