Question

Rework Example 13.2 if the source voltage at the sending end is a ramp, ${\mathbf{e}}_{\mathbf{G}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{Eu}}_{\mathbf{-}\mathbf{2}}\mathbf{M}\mathbf{=}{\mathbf{Etu}}_{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ with ${\mathbf{Z}}_{\mathbf{G}}\mathbf{=}\mathbf{2}{\mathbf{Z}}_{\mathbf{c}}$.

Short answer

(a) The voltage$v(x,t)$ and $i(x,t)$are$\frac{E}{2}{u}_{-2}\left(t-\frac{x}{v}\right)+\frac{E}{2}{u}_{-2}\left(t+\frac{x}{v}-2\tau \right)$ and$\frac{E}{2Z_c}{u}_{-2}\left(t-\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-2}\left(t+\frac{x}{v}-2\tau \right)$ respectively.

(b) The plot between voltage and$t$ .

The plot between current and$t$ .

Step-by-step solution

Write the given data from the question.

The source voltage, $e_G\left(t\right)=E{u}_{-2}\left(t\right)=E{u}_{-1}\left(t\right)$

The Source impedance,$Z_G\left(s\right)=Z_c$

Determine the equation to calculate the v(x,t),i(x,t) and plot the voltage and current verses  t at the centre of the line.

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

${\mathbf{\Gamma }}_{\mathbf{R}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}\frac{\frac{{\mathbf{Z}}_{\mathbf{R}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}{{\mathbf{Z}}_{\mathbf{c}}}\mathbf{-}\mathbf{1}}{\frac{{\mathbf{Z}}_{\mathbf{R}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}{{\mathbf{Z}}_{\mathbf{c}}}\mathbf{+}\mathbf{1}}$ …… (1)

Here ,${\mathbf{Z}}_{\mathbf{R}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$is the receiving end impedance and ${\mathbf{z}}_{\mathbf{c}}$ is the characteristics impedance.

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

${\mathbf{\Gamma }}_{\mathbf{S}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}\frac{\frac{{\mathbf{Z}}_{\mathbf{G}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}{{\mathbf{Z}}_{\mathbf{c}}}\mathbf{-}\mathbf{1}}{\frac{{\mathbf{Z}}_{\mathbf{G}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}{{\mathbf{Z}}_{\mathbf{c}}}\mathbf{+}\mathbf{1}}$ …… (2)

The equation to calculate the voltage at distance is given as follows.

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

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

The equation to calculate the current at distance is given as follows.

$\mathbf{I}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{s}\mathbf{)}\mathbf{=}\frac{{\mathbf{E}}_{\mathbf{G}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}{{\mathbf{Z}}_{\mathbf{c}}\mathbf{+}{\mathbf{Z}}_{\mathbf{G}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}}\left[\frac{e^{-\frac{\mathrm{sx}}{v}}-{\Gamma }_R\left(s\right)e^{s[\frac{x}{v}-2\tau ]}}{1-{\Gamma }_R\left(s\right){\Gamma }_S\left(s\right)e^{-2\mathrm{s\tau }}}\right]$ …… (4)

Calculate the v(x,t),and i(x,t) .

The Laplace of the source voltage,$E_G\left(s\right)=\frac{E}{s^2}$

Receiving end is open, therefore $Z_R\left(s\right)\to \infty$

Calculate the receiving end voltage reflection.

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

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 $Z_G\left(s\right)$, $\frac{E}{s^2}$for $E_G$, $1$for ${\Gamma }_R$and $0$for ${\Gamma }_S$into equation (3).

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

Take the Laplace inverse of the above equation as,

$v(x,t)=\frac{E}{2}{u}_{-2}\left(t-\frac{x}{v}\right)+\frac{E}{2}{u}_{-2}\left(t+\frac{x}{v}-2\tau \right)$ …… (5)

Calculate the Laplace of current at distance $x$,

Substitute $Z_C$for $Z_G\left(s\right)$, $\frac{E}{s^2}$for $E_G$, 1 for ${\Gamma }_R$and $0$for ${\Gamma }_S$ into equation (4).

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

Take Laplace inverse of the above equation.

$i(x,t)=\frac{E}{2Z_c}{u}_{-2}\left(t-\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-2}\left(t+\frac{x}{v}-2\tau \right)$ …… (6)

Hence the voltage$v(x,t)$ and $i(x,t)$are$\frac{E}{2}{u}_{-2}\left(t-\frac{x}{v}\right)+\frac{E}{2}{u}_{-2}\left(t+\frac{x}{v}-2\tau \right)$ and $\frac{E}{2Z_c}{u}_{-2}\left(t-\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-2}\left(t+\frac{x}{v}-2\tau \right)$ respectively.

Plot the voltage and current verses t at the centre of the line.

At the centre of the line,$x=\frac{l}{2}$

Here,$l$ is the length of the line.

Calculate the voltage at distance of $\frac{l}{2}$.

Substitute$\frac{l}{2}$ for$x$ into equation (5).

$v(\frac{l}{2},t)=\frac{E}{2}{u}_{-2}\left(t-\frac{l}{2v}\right)+\frac{E}{2}{u}_{-2}\left(t+\frac{l}{2v}-2\tau \right)$

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

$v\left(\frac{l}{2},t\right)=\frac{E}{2}{u}_{-2}t\left(-\frac{\tau }{2}\right)+\frac{E}{2}{u}_{-2}\left(t+\frac{\tau }{2}-2\tau \right)$

Calculate the current at distance of $\frac{l}{2}$.

Substitute$\frac{l}{2}$ for$x$ into equation (6).

$i\left(\frac{l}{2},t\right)=\frac{E}{2Z_c}{u}_{-2}\left(t-\frac{l}{2v}\right)-\frac{E}{2Z_c}{u}_{-2}\left(t+\frac{l}{2v}-2\tau \right)$

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

$\begin{array}{l}i\left(\frac{l}{2},t\right)=\frac{E}{2Z_c}{u}_{-2}\left(t-\frac{\tau }{2}\right)-\frac{E}{2Z_c}{u}_{-2}\left(t+\frac{\tau }{2}-2\tau \right)\\ i\left(\frac{l}{2},t\right)=\frac{E}{2Z_c}{u}_{-2}\left(t-\frac{\tau }{2}\right)-\frac{E}{2Z_c}{u}_{-2}\left(t-\frac{3\tau }{2}\right)\end{array}$

Draw the plot between voltage and $t$.

Draw the plot between current and$t$ .