Question

Draw the Bewley lattice diagram for Problem 13.5.

Short answer

The Bewley Lattice diagram as,

Step-by-step solution

Write the given data from the question.

The source voltage,$e_G\left(t\right)=E{u}_{-1}\left(t\right)$

The Source impedance,$Z_G=\frac{Z_c}{3}$

The receiving end impedance,$Z_R=5Z_c$

Determine the equation to drawthe Bewley lattice diagram.

The equation to calculate the receiving end voltage reflection is given as follows.

${\Gamma }_R\left(s\right)=\frac{\frac{Z_R\left(s\right)}{Z_c}-1}{\frac{Z_R\left(s\right)}{Z_c}+1}$ …… (1)

Here, i$Z_R\left(s\right)$s the receiving end impedance and$Z_c$ is the characteristics impedance.

The equation to calculate the sending end voltage reflection is given as follows.

${\Gamma }_S\left(s\right)=\frac{\frac{Z_G\left(s\right)}{Z_c}-1}{\frac{Z_G\left(s\right)}{Z_c}+1}$ …… (2)

The equation to calculate the first forward travelling wave is given as follows.

$V_1\left(s\right)=E_G\left(s\right)\left[\frac{Z_c}{Z_c+Z_G}\right]$ …… (3)

Draw the Bewley lattice diagram.

The Laplace of the source voltage,$E_G\left(s\right)=\frac{E}{s}$

Calculate the receiving end voltage reflection.

Substitute $5Z_c$for $Z_R\left(s\right)$into equation (1).

$\begin{array}{l}{\Gamma }_R\left(s\right)=\frac{\frac{5Z_c}{Z_c}-1}{\frac{5Z_c}{Z_c}+1}\\ {\Gamma }_R\left(s\right)=\frac{5-1}{5+1}\\ {\Gamma }_R\left(s\right)=\frac{4}{6}\\ {\Gamma }_R\left(s\right)=\frac{2}{3}\text{\hspace{0.33em}pu}\end{array}$

Calculate the sending end voltage reflection.

Substitute $\frac{Z_c}{3}$for $Z_G\left(s\right)$into equation (2).

$\begin{array}{l}{\Gamma }_S\left(s\right)=\frac{\frac{Z_c}{3Z_c}-1}{\frac{Z_c}{3Z_c}+1}\\ {\Gamma }_S\left(s\right)=\frac{\frac{1}{3}-1}{\frac{1}{3}+1}\\ {\Gamma }_S\left(s\right)=\frac{-1}{2}\text{\hspace{0.33em}pu}\end{array}$

Calculate the first forward travelling wave.

Substitute $\frac{Z_c}{3}$for $Z_G\left(s\right)$and $\frac{E}{s}$for $E_G\left(s\right)$into equation (3).

$\begin{array}{l}{V}_1\left(s\right)=\frac{E}{s}\left[\frac{Z_c}{Z_c+\frac{Z_c}{3}}\right]\\ V_1\left(s\right)=\frac{E}{s}\times \frac{3}{4}\\ V_1\left(s\right)=\frac{0.75E}{s}\text{\hspace{0.33em}V}\end{array}$

Calculate the second backward travelling wave.

$V_2\left(s\right)={\Gamma }_R\left(s\right)V_1\left(s\right)$

Substitute $\frac{0.75E}{\text{s}}$for $V_1\left(s\right)$and $\frac{2}{3}\text{\hspace{0.33em}pu}$for ${\Gamma }_R\left(s\right)$into above equation.

$\begin{array}{l}{V}_2\left(s\right)=\frac{0.75E}{s}\times \frac{2}{3}\\ V_2\left(s\right)=\frac{0.5E}{s}\end{array}$

Calculate the third forward travelling wave.

$V_3\left(s\right)={\Gamma }_S\left(s\right)V_2\left(s\right)$

Substitute $\frac{0.5E}{\text{s}}$for $V_2\left(s\right)$and $\frac{-1}{2}\text{\hspace{0.33em}pu}$ for ${\Gamma }_S\left(s\right)$into above equation

$\begin{array}{l}{V}_3\left(s\right)=\frac{0.5E}{s}\times \frac{-1}{2}\\ V_3\left(s\right)=-\frac{0.25E}{s}\end{array}$

$\begin{array}{l}{V}_3\left(s\right)=\frac{0.5E}{s}\times \frac{-1}{2}\\ V_3\left(s\right)=-\frac{0.25E}{s}\end{array}$

Calculate the fourth backward travelling wave.

localid="1656150569335" $V_4\left(s\right)={\Gamma }_R\left(s\right)V_3\left(s\right)$

Substitute localid="1656150638820" $-\frac{0.25E}{\text{s}}$for localid="1656150642825" $V_3\left(s\right)$and localid="1656150647599" $\frac{2}{3}\text{\hspace{0.33em}pu}$for localid="1656150652317" ${\Gamma }_R\left(s\right)$into above equation.

localid="1656150656437" $\begin{array}{l}{V}_4\left(s\right)=-\frac{0.25E}{s}\times \frac{2}{3}\\ V_4\left(s\right)=-\frac{0.1667E}{s}\end{array}$

Calculate the fifth forward travelling wave.

localid="1656150573053" $V_5\left(s\right)={\Gamma }_S\left(s\right)V_4\left(s\right)$

Substitute localid="1656150661034" $\frac{0.1667E}{\text{s}}$for localid="1656150665831" $V_4\left(s\right)$and localid="1656150670182" $\frac{-1}{2}\text{\hspace{0.33em}pu}$for localid="1656150674966" ${\Gamma }_S\left(s\right)$into above equation.

localid="1656150679106" $\begin{array}{l}{V}_5\left(s\right)=-\frac{0.1667E}{s}\times \frac{-1}{2}\\ V_5\left(s\right)=\frac{0.0834E}{s}\end{array}$

Calculate the sixth backward travelling wave.

localid="1656150577658" $V_6\left(s\right)={\Gamma }_R\left(s\right)V_5\left(s\right)$

Substitutelocalid="1656150683816" $\frac{0.0834E}{\text{s}}$forlocalid="1656150687610" $V_3\left(s\right)$and localid="1656150691969" $\frac{2}{3}\text{\hspace{0.33em}pu}$for into above equation.

localid="1656150697178" $\begin{array}{l}{V}_6\left(s\right)=\frac{0.0834E}{s}\times \frac{2}{3}\\ V_6\left(s\right)=\frac{0.0556E}{s}\end{array}$

Draw the Bewley Lattice diagram as,