Question

Rework Example 13.4 with $Z_R=5Z_c$and $Z_G=\frac{Z_c}{3}$.

Short answer

The voltage at the centre of the line is

$v\left(x,t\right)=\frac{3E}{4}\left[\begin{array}{l}{u}_{-1}\left(t-\frac{\tau }{2}\right)+\frac{2}{3}{u}_{-1}\left(t-\frac{3\tau }{2}\right)-\frac{1}{3}{u}_{-1}\left(t-\frac{5\tau }{2}\right)\\ -\frac{2}{9}{u}_{-1}\left(t-\frac{7\tau }{2}\right)+\frac{1}{9}{u}_{-1}\left(t-\frac{9\tau }{2}\right)+\frac{2}{27}{u}_{-1}\left(t-\frac{11\tau }{2}\right)\\ -\frac{1}{27}{u}_{-1}\left(t-\frac{13\tau }{2}\right)-\frac{2}{81}{u}_{-1}\left(t-\frac{15\tau }{2}\right)+\mathrm{...........}\end{array}\right]$

Plot voltage verses time curve.

The steady state value of the voltage is.$0.9375\text{E\hspace{0.33em}V}$

The steady state value of the voltage is.$0.9375\text{E\hspace{0.33em}V}$

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 calculate the voltage at the centre of the line.

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, $Z_R\left(s\right)$is 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 voltage at distance$x$ is given as follows.

$V\left(x,s\right)=E_G\left(s\right)\left[\frac{Z_c}{Z_c+Z_G\left(s\right)}\right]\left[\frac{e^{-\frac{sx}{v}}+{\Gamma }_R\left(s\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1-{\Gamma }_R\left(s\right){\Gamma }_S\left(s\right)e^{-2s\tau }}\right]$ …… (3)

Here, $v$is the velocity of the wave and$\tau$ is the transit time of the wave.

Calculate and plot the voltage at the centre of the line.

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 Laplace of voltage at distance .$x$

Substitute $5Z_c$for $Z_R$, $\frac{E}{s}$for $E_G$,$\frac{2}{3}$for ${\Gamma }_R$and $\frac{-1}{2}$for ${\Gamma }_S$into equation (3).

$\begin{array}{l}V\left(x,s\right)=\frac{E}{s}\left[\frac{Z_c}{Z_c+\frac{Z_c}{3}}\right]\left[\frac{e^{-\frac{sx}{v}}+\left(\frac{2}{3}\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1-\left(\frac{2}{3}\right)\left(\frac{-1}{2}\right)e^{-2s\tau }}\right]\\ V\left(x,s\right)=\frac{E}{s}\left[\frac{1}{1+\frac{1}{3}}\right]\left[\frac{e^{-\frac{sx}{v}}+\left(\frac{2}{3}\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1+\frac{1}{3}{e}^{-2s\tau }}\right]\end{array}$

$V\left(x,s\right)=\frac{3E}{4s}\left[\frac{e^{-\frac{sx}{v}}+\left(\frac{2}{3}\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1+\frac{1}{3}{e}^{-2s\tau }}\right]$ …… (5)

Expand the term ,$\frac{1}{1+\frac{1}{3}{e}^{-2s\tau }}$

$\begin{array}{l}1+\frac{1}{3}{e}^{-2s\tau }=1-\frac{1}{3}{e}^{-2s\tau }+{\left(\frac{1}{3}{e}^{-2s\tau }\right)}^2-{\left(\frac{1}{3}{e}^{-2s\tau }\right)}^3+{\left(\frac{1}{3}{e}^{-2s\tau }\right)}^4-\mathrm{.......}\\ 1+\frac{1}{3}{e}^{-2s\tau }=1-\frac{1}{3}{e}^{-2s\tau }+\frac{1}{9}{e}^{-4s\tau }-\frac{1}{27}{e}^{-6s\tau }+\frac{1}{81}{e}^{-8s\tau }-\mathrm{.....}\end{array}$

Substitute $1-\frac{1}{3}{e}^{-2s\tau }+\frac{1}{9}{e}^{-4s\tau }-\frac{1}{27}{e}^{-6s\tau }+\frac{1}{81}{e}^{-8s\tau }-\mathrm{.....}$for $\frac{1}{1+\frac{1}{3}{e}^{-2s\tau }}$ into equation (1).

$\begin{array}{l}V\left(x,s\right)=\frac{3E}{4s}\left[e^{-\frac{sx}{v}}+\left(\frac{2}{3}\right)e^{s\left[\frac{x}{v}-2\tau \right]}\right]\left[1-\frac{1}{3}{e}^{-2s\tau }+\frac{1}{9}{e}^{-4s\tau }-\frac{1}{27}{e}^{-6s\tau }+\frac{1}{81}{e}^{-8s\tau }-\mathrm{.....}\right]\\ V\left(x,s\right)=\frac{3E}{4s}\left[\begin{array}{l}{e}^{-\frac{sx}{v}}+\left(\frac{2}{3}\right)e^{s\left[\frac{x}{v}-2\tau \right]}-\frac{1}{3}\left(e^{-\frac{sx}{v}}+\left(\frac{2}{3}\right)e^{s\left[\frac{x}{v}-2\tau \right]}\right)e^{-2s\tau }+\frac{1}{9}\left(e^{-\frac{sx}{v}}+\left(\frac{2}{3}\right)e^{s\left[\frac{x}{v}-2\tau \right]}\right)e^{-4s\tau }\\ -\frac{1}{27}\left(e^{-\frac{sx}{v}}+\left(\frac{2}{3}\right)e^{s\left[\frac{x}{v}-2\tau \right]}\right)e^{-6s\tau }+\mathrm{........}\end{array}\right]\\ V\left(x,s\right)=\frac{3E}{4s}\left[\begin{array}{l}{e}^{-\frac{sx}{v}}+\left(\frac{2}{3}\right)e^{s\left[\frac{x}{v}-2\tau \right]}-\frac{1}{3}{e}^{-s\left[\frac{x}{v}+2\tau \right]}-\frac{2}{9}{e}^{s\left[\frac{x}{v}-4\tau \right]}+\frac{1}{9}{e}^{-s\left[\frac{x}{v}+4\tau \right]}+\frac{2}{27}{e}^{s\left[\frac{x}{v}-6\tau \right]}\\ -\frac{1}{27}{e}^{-s\left[\frac{x}{v}+6\tau \right]}-\frac{2}{81}{e}^{s\left[\frac{x}{v}-8\tau \right]}\end{array}\right]\end{array}$

Take the Laplace inverse of the above equation as,/

$v\left(x,t\right)=\frac{3E}{4}\left[\begin{array}{l}{u}_{-1}\left(t-\frac{x}{v}\right)+\frac{2}{3}{u}_{-1}\left(t+\frac{x}{v}-2\tau \right)\_\frac{1}{3}{u}_{-1}\left(t-\frac{x}{v}-2\tau \right)\\ -\frac{2}{9}{u}_{-1}\left(t-\frac{x}{v}-4\tau \right)+\frac{1}{9}{u}_{-1}\left(t-\frac{x}{v}-4\tau \right)+\frac{2}{27}{u}_{-1}\left(t-\frac{x}{v}-6\tau \right)\\ -\frac{1}{27}{u}_{-1}\left(t-\frac{x}{v}-6\tau \right)-\frac{2}{81}{u}_{-1}\left(t-\frac{x}{v}-8\tau \right)+\mathrm{...........}\end{array}\right]$

Substitute $\frac{l}{2}$for $x$into above equation.

$v\left(x,t\right)=\frac{3E}{4}\left[\begin{array}{l}{u}_{-1}\left(t-\frac{l}{2v}\right)+\frac{2}{3}{u}_{-1}\left(t+\frac{l}{2v}-2\tau \right)\_\frac{1}{3}{u}_{-1}\left(t-\frac{l}{2v}-2\tau \right)\\ -\frac{2}{9}{u}_{-1}\left(t-\frac{l}{2v}-4\tau \right)+\frac{1}{9}{u}_{-1}\left(t-\frac{l}{2v}-4\tau \right)+\frac{2}{27}{u}_{-1}\left(t-\frac{l}{2v}-6\tau \right)\\ -\frac{1}{27}{u}_{-1}\left(t-\frac{l}{2v}-6\tau \right)-\frac{2}{81}{u}_{-1}\left(t-\frac{l}{2v}-8\tau \right)+\mathrm{...........}\end{array}\right]$

Substitute $\frac{l}{v}$for $\tau$into above equation.

$\begin{array}{l}v\left(x,t\right)=\frac{3E}{4}\left[\begin{array}{l}{u}_{-1}\left(t-\frac{\tau }{2}\right)+\frac{2}{3}{u}_{-1}\left(t+\frac{\tau }{2}-2\tau \right)\_\frac{1}{3}{u}_{-1}\left(t-\frac{\tau }{2}-2\tau \right)\\ -\frac{2}{9}{u}_{-1}\left(t+\frac{\tau }{2}-4\tau \right)+\frac{1}{9}{u}_{-1}\left(t-\frac{\tau }{2}-4\tau \right)+\frac{2}{27}{u}_{-1}\left(t+\frac{\tau }{2}-6\tau \right)\\ -\frac{1}{27}{u}_{-1}\left(t-\frac{\tau }{2}-6\tau \right)-\frac{2}{81}{u}_{-1}\left(t+\frac{\tau }{2}-8\tau \right)+\mathrm{...........}\end{array}\right]\\ v\left(x,t\right)=\frac{3E}{4}\left[\begin{array}{l}{u}_{-1}\left(t-\frac{\tau }{2}\right)+\frac{2}{3}{u}_{-1}\left(t-\frac{3\tau }{2}\right)-\frac{1}{3}{u}_{-1}\left(t-\frac{5\tau }{2}\right)\\ -\frac{2}{9}{u}_{-1}\left(t-\frac{7\tau }{2}\right)+\frac{1}{9}{u}_{-1}\left(t-\frac{9\tau }{2}\right)+\frac{2}{27}{u}_{-1}\left(t-\frac{11\tau }{2}\right)\\ -\frac{1}{27}{u}_{-1}\left(t-\frac{13\tau }{2}\right)-\frac{2}{81}{u}_{-1}\left(t-\frac{15\tau }{2}\right)+\mathrm{...........}\end{array}\right]\end{array}$

Hence the voltage at the centre of the line is.

$v\left(x,t\right)=\frac{3E}{4}\left[\begin{array}{l}{u}_{-1}\left(t-\frac{\tau }{2}\right)+\frac{2}{3}{u}_{-1}\left(t-\frac{3\tau }{2}\right)-\frac{1}{3}{u}_{-1}\left(t-\frac{5\tau }{2}\right)\\ -\frac{2}{9}{u}_{-1}\left(t-\frac{7\tau }{2}\right)+\frac{1}{9}{u}_{-1}\left(t-\frac{9\tau }{2}\right)+\frac{2}{27}{u}_{-1}\left(t-\frac{11\tau }{2}\right)\\ -\frac{1}{27}{u}_{-1}\left(t-\frac{13\tau }{2}\right)-\frac{2}{81}{u}_{-1}\left(t-\frac{15\tau }{2}\right)+\mathrm{...........}\end{array}\right]$

By substituting the value of localid="1656148716852" $t$, calculate the respective value for .localid="1656148707657" $v\left(\frac{l}{2},t\right)$

Plot voltage verses time curve.

Calculate the steady state value of the voltage,

$v_{ss}\left(t\right)=\underset{s\to 0}{\lim }sV\left(x,s\right)$

Substitute $\frac{3E}{4s}\left[\frac{e^{-\frac{sx}{v}}+\left(\frac{2}{3}\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1+\frac{1}{3}{e}^{-2s\tau }}\right]$for $v\left(x,s\right)$into above equation.

$\begin{array}{l}{v}_{ss}\left(t\right)=\underset{s\to 0}{\lim }s\times \frac{3E}{4s}\left[\frac{e^{-\frac{sx}{v}}+\left(\frac{2}{3}\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1+\frac{1}{3}{e}^{-2s\tau }}\right]\\ v_{ss}\left(t\right)=\frac{3E}{4}\underset{s\to 0}{\lim }\left[\frac{e^{-\frac{sx}{v}}+\left(\frac{2}{3}\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1+\frac{1}{3}{e}^{-2s\tau }}\right]\\ v_{ss}\left(t\right)=\frac{3E}{4}\left[\frac{e^{-\frac{\left(0\right)x}{v}}+\left(\frac{2}{3}\right)e^{\left(0\right)\left[\frac{x}{v}-2\tau \right]}}{1+\frac{1}{3}{e}^{-2\left(0\right)\tau }}\right]\\ v_{ss}\left(t\right)=\frac{3E}{4}\left[\frac{1+\left(\frac{2}{3}\right)}{1+\frac{1}{3}}\right]\end{array}$

Solve further as,

$\begin{array}{l}{v}_{ss}\left(t\right)=\frac{3E}{4}\times \frac{5}{4}\\ v_{ss}\left(t\right)=\frac{15E}{16}\\ V_{ss}\left(t\right)=0.9375\text{E\hspace{0.33em}V}\end{array}$

Hence the steady state value of the voltage is .$0.9375\text{E\hspace{0.33em}V}$