Question

From the results of Example 13.2, plot the voltage and current profiles along the line at times $\mathbf{\tau }\mathbf{/}\mathbf{2}\mathbf{,}\mathbf{\tau }$, and $\mathbf{2}\mathbf{\tau }$. That is, plot $\mathbf{v}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{\tau }\mathbf{/}\mathbf{2}\mathbf{)}$and $\mathbf{i}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{\tau }\mathbf{/}\mathbf{2}\mathbf{)}$versus$\mathit{x}$ for $\mathbf{0}\mathbf{\le }\mathbf{x}\mathbf{\le }\mathbf{1}$; then plot$\mathbf{v}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{\tau }\mathbf{)}\mathbf{,}\mathbf{i}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{x}\mathbf{)}\mathbf{,}\mathbf{v}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{2}\mathbf{\tau }\mathbf{)}$ and $\mathbf{i}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{2}\mathbf{x}\mathbf{)}$ versus$\mathit{x}$ .

Short answer

The plot $v\left(x,\frac{\tau }{2}\right)$ and$i\left(x,\frac{\tau }{2}\right)$ verses$x$ .

The plot $v(x,\tau )$ and$i(x,\tau )$ verses $x$.

The plot $v(x,2\tau )$ and$i(x,2\tau )$ verses$x$ .

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 impedance,$Z_G\left(t\right)=Z_c$

The result of the example 13.2.

The Laplace of voltage at distance $x$ is $v(x,t)=\frac{E}{2}{u}_{-1}\left(t-\frac{x}{v}\right)+\frac{E}{2}{u}_{-1}\left(t+\frac{x}{v}-2\tau \right)$.

The Laplace of current at distance $x$is $i(x,t)=\frac{E}{2Z_c}{u}_{-1}\left(t-\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-1}\left(t+\frac{x}{v}-2\tau \right)$.

Plot the voltage and current profiles along the line at times τ/2,τ , and 2τ .

Let $l$is the length of the line $v$ is the velocity of the wave and $\tau$ is the transit time of the wave.

The Laplace of voltage at distance $x$,

$v(x,t)=\frac{E}{2}{u}_{-1}\left(t-\frac{x}{v}\right)+\frac{E}{2}{u}_{-1}\left(t+\frac{x}{v}-2\tau \right)$

Substitute $\frac{\tau }{2}$for $t$ into above equation.

$v\left(x,\frac{\tau }{2}\right)=\frac{E}{2}{u}_{-1}\left(\frac{\tau }{2}-\frac{x}{v}\right)+\frac{E}{2}{u}_{-1}\left(\frac{\tau }{2}+\frac{x}{v}-2\tau \right)$

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

$v\left(x,\frac{\tau }{2}\right)=\frac{E}{2}{u}_{-1}\left(\frac{l}{2v}-\frac{x}{v}\right)+\frac{E}{2}{u}_{-1}\left(\frac{l}{2v}+\frac{x}{v}-2\tau \right)$

The Laplace of current at distance$x$

$i(x,t)=\frac{E}{2Z_c}{u}_{-1}\left(t-\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-1}\left(t+\frac{x}{v}-2\tau \right)$

Substitute$\frac{\tau }{2}$ for$t$ into above equation.

$i\left(x,\frac{\tau }{2}\right)=\frac{E}{2Z_c}{u}_{-1}\left(\frac{\tau }{2}-\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-1}\left(\frac{\tau }{2}+\frac{x}{v}-2\tau \right)$

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

$i\left(x,\frac{\tau }{2}\right)=\frac{E}{2Z_c}{u}_{-1}\left(\frac{l}{2v}-\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-1}\left(\frac{l}{2v}+\frac{x}{v}-2\tau \right)$

Draw the plot $v\left(x,\frac{\tau }{2}\right)$ and$i\left(x,\frac{\tau }{2}\right)$ verses $x$.

The Laplace of voltage at distance$x$ ,

$v(x,t)=\frac{E}{2}{u}_{-1}\left(t-\frac{x}{v}\right)+\frac{E}{2}{u}_{-1}\left(t+\frac{x}{v}-2\tau \right)$

Substitute $\tau$for $t$into above equation.

$v(x,\tau )=\frac{E}{2}{u}_{-1}\left(\tau -\frac{x}{v}\right)+\frac{E}{2}{u}_{-1}\left(\tau +\frac{x}{v}-2\tau \right)$

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

$v\left(x,\frac{\tau }{2}\right)=\frac{E}{2}{u}_{-1}\left(\frac{l}{v}-\frac{x}{v}\right)+\frac{E}{2}{u}_{-1}\left(\frac{l}{v}+\frac{x}{v}-2\tau \right)$

The Laplace of current at distance$x$

$i(x,t)=\frac{E}{2Z_c}{u}_{-1}\left(t-\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-1}\left(t+\frac{x}{v}-2\tau \right)$

Substitute $\tau$for $t$into above equation.

$i(x,\tau )=\frac{E}{2Z_c}{u}_{-1}\left(\tau -\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-1}\left(\tau +\frac{x}{v}-2\tau \right)$

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

$i(x,\tau )=\frac{E}{2Z_c}{u}_{-1}\left(\frac{l}{v}-\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-1}\left(\frac{l}{v}+\frac{x}{v}-2\tau \right)$

Draw the plot$v(x,\tau )$ and$i(x,\tau )$ verses x .

The Laplace of voltage at distance$x$ ,

$v(x,t)=\frac{E}{2}{u}_{-1}\left(t-\frac{x}{v}\right)+\frac{E}{2}{u}_{-1}\left(t+\frac{x}{v}-2\tau \right)$

Substitute$2\tau$ for$t$ into above equation.

$\begin{array}{l}v(x,2\tau )=\frac{E}{2}{u}_{-1}\left(2\tau -\frac{x}{v}\right)+\frac{E}{2}{u}_{-1}\left(2\tau +\frac{x}{v}-2\tau \right)\\ v(x,2\tau )=\frac{E}{2}{u}_{-1}\left(2\tau -\frac{x}{v}\right)+\frac{E}{2}{u}_{-1}\left(\frac{x}{v}\right)\end{array}$

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

$v(x,2\tau )=\frac{E}{2}{u}_{-1}\left(2\frac{l}{v}-\frac{x}{v}\right)+\frac{E}{2}{u}_{-1}\left(\frac{x}{v}\right)$

The Laplace of current at distance$x$

$i(x,t)=\frac{E}{2Z_c}{u}_{-1}\left(t-\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-1}\left(t+\frac{x}{v}-2\tau \right)$

Substitute $2\tau$for $t$into above equation.

$\begin{array}{l}i(x,2\tau )=\frac{E}{2Z_c}{u}_{-1}\left(2\tau -\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-1}\left(2\tau +\frac{x}{v}-2\tau \right)\\ i(x,2\tau )=\frac{E}{2Z_c}{u}_{-1}\left(2\tau -\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-1}\left(\frac{x}{v}\right)\end{array}$

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

$i(x,2\tau )=\frac{E}{2Z_c}{u}_{-1}\left(2\frac{l}{v}-\frac{x}{v}\right)-\frac{E}{2Z_c}{u}_{-1}\left(\frac{x}{v}\right)$

Draw the plot $v(x,2\tau )$ and $i(x,2\tau )$verses$x$ .