Question

For a lossless open-circuited line, express the sending-end voltage, $V_1$, in terms of the receiving-end voltage, $V_2$, for the three cases of short-line model, medium-length line model, and long-line model. Is it true that the voltage at the open receiving end of a long line is higher than that at the sending end, for small $\beta l$?

Short answer

The expression of sending end voltage in terms of receiving end voltage for the short, medium, and long transmission lines are $V_1=V_2$,$V_1=\left(1-\frac{{\left(\beta l\right)}^2}{2}\right)V_2$and$V_1=V_2\cosh \left(\gamma l\right)$

respectively. The receiving end voltage is greater than the sending end voltage for long transmission line.

Step-by-step solution

Determine the equation to prove that the voltage at the open receiving end of a long line is higher than that at the sending end, for small βl.

The general equation for relationship between the sending and receiving end parameter is given as follows.

$V_1=A{V}_2+B{I}_2$…… (1)

Write the expression for the current.

$I_1=C{V}_2+D{I}_2$….. (2)

Here, $V_1$and$I_1$are the sending end voltage and current respectively, localid="1655902174724" $V_2$and $I_2$are the receiving end voltage and current respectively.

Prove that the voltage at the open receiving end of a long line is higher than that at the sending end, for small .βl

The propagation constant for the $ABCD$lossless line,

$\gamma =j\beta$

For open circuit, the receiving end current is zero.

$I_2=0$

Substitute receiving end current value in equation (1) and (2).

$\begin{array}{l}{V}_1=A{V}_2+B\left(0\right)\\ V_1=A{V}_2\end{array}$ …… (3)

For the short transmission line, the $ABCD$parameter,

$\begin{array}{l}A=D\\ A=1\\ B=Z\\ C=0\end{array}$

Substitute $1$ for$A$into equation (3).

$\begin{array}{l}{V}_1=\left(1\right)V_2\\ V_1=V_2\end{array}$

Hence, the for the short transmission line, the sending end voltage is equal to receiving end voltage.

For medium transmission line, the parameter,

$\begin{array}{l}A=D\\ A=1+\frac{YZ}{2}\\ B=Z\\ C=Y\left(1+\frac{YZ}{4}\right)\end{array}$

Substitute $1+\frac{YZ}{2}$for$A$into equation (3).

$V_1=\left(1+\frac{YZ}{2}\right)V_2$ …… (4)

Since,

$ZY={\left(\gamma l\right)}^2$

Substitute ${\left(\gamma l\right)}^2$for$ZY$into equation (4).

$V_1=\left(1+\frac{{\left(\gamma l\right)}^2}{2}\right)V_2$

Substitute$j\beta$for $\gamma$into above equation.

$\begin{array}{l}{V}_1=\left(1+\frac{{\left(j\beta l\right)}^2}{2}\right)V_2\\ V_1=\left(1-\frac{{\left(\beta l\right)}^2}{2}\right)V_2\end{array}$

Hence the expression for sending end voltage is terms of receiving end voltage is .

$V_1=\left(1-\frac{{\left(\beta l\right)}^2}{2}\right)V_2$

For long transmission line, the $ABCD$parameter,

$\begin{array}{l}A=D\\ A=\cosh \left(\gamma l\right)\\ B=Z_c\sinh \left(\gamma l\right)\\ C=\frac{1}{Z_c}\sinh \left(\gamma l\right)\end{array}$

Substitute $\cosh \left(\gamma l\right)$for $A$into equation (3).

$V_1=V_2\cosh \left(\gamma l\right)$ …… (4)

Expand the term as,$\cosh \left(\gamma l\right)$

$V_1=V_2\left(1-\frac{{\left(\beta l\right)}^2}{2}\right)$

Hence, the receiving end voltage is greater than the sending end voltage.