Question

Referring to the single-phase two-wire lossless line shown in Figure 13.3, the receiving end is terminated by an inductor with$\mathbf{2}{\mathit{L}}_{\mathbf{R}}$Henries. The source voltage at the sending end is a step ${\mathit{e}}_{\mathbf{G}}\left(t\right)\mathbf{=}\mathit{E}{\mathit{u}}_{\mathbf{-}\mathbf{1}}\left(t\right)$with . Both the line and inductor are initially unenergized. Determine and plot the voltage at the centre of the line $\mathit{v}\left(l/2,t\right)$versus time $\mathit{t}$.

Short answer

The voltage at the centre of the line is .

$v\left(\frac{l}{2},t\right)=\frac{E}{2}{u}_{-1}\left(t-\frac{\tau }{2}\right)+\frac{E}{2}\left[2e^{-\frac{Z_c}{2L_R}\left(t-\frac{3\tau }{2}\right)}-1\right]u_{-1}\left(t-\frac{3\tau }{2}\right)$

The voltage with respect to time .

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,$Z_G\left(s\right)=Z_c$

The receiving end impedance,$Z_R\left(s\right)=2L_R$

Determine the equation to calculate the voltage at the centre of the linev(l/2,t) versus time t . 

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 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 draw the voltage at the centre of the line vl/2,tversus time t.

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

Calculate the receiving end voltage reflection.

Substitute $2s{L}_R$ for$Z_R\left(s\right)$into equation (1).

$\begin{array}{l}{\Gamma }_R\left(s\right)=\frac{\frac{2s{L}_R}{Z_c}-1}{\frac{2s{L}_R}{Z_c}+1}\\ {\Gamma }_R\left(s\right)=\frac{s-\frac{Z_c}{2L_R}}{s+\frac{Z_c}{2L_R}}\end{array}$

Calculate the sending end voltage reflection.

Substitute$Z_c$for $Z_G\left(s\right)$into equation (2).

$\begin{array}{l}{\Gamma }_S\left(s\right)=\frac{\frac{Z_c}{Z_c}-1}{\frac{Z_c}{Z_c}+1}\\ {\Gamma }_S\left(s\right)=\frac{1-1}{1+1}\\ {\Gamma }_S\left(s\right)=0\end{array}$

Calculate the Laplace of voltage at distance $x$.

Substitute $Z_c$for$Z_G\left(s\right)$,$\frac{E}{s}$for$E_G$$\frac{s-\frac{Z_c}{2L_R}}{s+\frac{Z_c}{2L_R}}$, for ${\Gamma }_R$and 0 for ${\Gamma }_S$into equation (3).

$\begin{array}{l}V\left(x,s\right)=\frac{E}{s}\left[\frac{Z_c}{Z_c+Z_c}\right]\left[\frac{e^{-\frac{sx}{v}}+\left(\frac{s-\frac{Z_c}{2L_R}}{s+\frac{Z_c}{2L_R}}\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1-\frac{s-\frac{Z_c}{2L_R}}{s+\frac{Z_c}{2L_R}}\left(0\right)e^{-2s\tau }}\right]\\ V\left(x,s\right)=\frac{E}{s}\left[\frac{1}{2}\right]\left[\frac{e^{-\frac{sx}{v}}+\left(\frac{s-\frac{Z_c}{2L_R}}{s+\frac{Z_c}{2L_R}}\right)e^{s\left[\frac{x}{v}-2\tau \right]}}{1}\right]\\ V\left(x,s\right)=\frac{E}{2}\left(\frac{1}{s}\right)e^{-\frac{sx}{v}}+\frac{E}{2}\left(\frac{1}{s}\right)\left(\frac{s-\frac{Z_c}{2L_R}}{s+\frac{Z_c}{2L_R}}\right)e^{s\left[\frac{x}{v}-2\tau \right]}\end{array}$

Use the partial fraction to expand the term $\left(\frac{1}{s}\right)\left(\frac{s-\frac{Z_c}{2L_R}}{s+\frac{Z_c}{2L_R}}\right)$which is $\left(\frac{2}{s+\frac{Z_s}{2L_R}}-\frac{1}{s}\right)$

Substitute$\left(\frac{2}{s+\frac{Z_s}{2L_R}}-\frac{1}{s}\right)$for $\left(\frac{1}{s}\right)\left(\frac{s-\frac{Z_c}{2L_R}}{s+\frac{Z_c}{2L_R}}\right)$into voltage equation.

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

Take the Laplace inverse of the above equation as,

$v\left(x,t\right)=\frac{E}{2}{u}_{-1}\left(t-\frac{x}{v}\right)+\frac{E}{2}\left[2e^{-\frac{Z_c}{2L_R}\left(t+\frac{x}{v}-2\tau \right)}-1\right]u_{-1}\left(t+\frac{x}{v}-2\tau \right)$ …… (5)

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

$v\left(\frac{l}{2},t\right)=\frac{E}{2}{u}_{-1}\left(t-\frac{l}{2v}\right)+\frac{E}{2}\left[2e^{-\frac{Z_c}{2L_R}\left(t+\frac{l}{2v}-2\tau \right)}-1\right]u_{-1}\left(t+\frac{l}{2v}-2\tau \right)$

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

$\begin{array}{l}v\left(\frac{l}{2},t\right)=\frac{E}{2}{u}_{-1}\left(t-\frac{\tau }{2}\right)+\frac{E}{2}\left[2e^{-\frac{Z_c}{2L_R}\left(t+\frac{\tau }{2}-2\tau \right)}-1\right]u_{-1}\left(t+\frac{\tau }{2}-2\tau \right)\\ v\left(\frac{l}{2},t\right)=\frac{E}{2}{u}_{-1}\left(t-\frac{\tau }{2}\right)+\frac{E}{2}\left[2e^{-\frac{Z_c}{2L_R}\left(t-\frac{3\tau }{2}\right)}-1\right]u_{-1}\left(t-\frac{3\tau }{2}\right)\end{array}$

Hence the voltage at the centre of the line is .

$v\left(\frac{l}{2},t\right)=\frac{E}{2}{u}_{-1}\left(t-\frac{\tau }{2}\right)+\frac{E}{2}\left[2e^{-\frac{Z_c}{2L_R}\left(t-\frac{3\tau }{2}\right)}-1\right]u_{-1}\left(t-\frac{3\tau }{2}\right)$

Plot the voltage with respect to time$t$.