# open circuit line and Short Circuit line

open circuit line:-

In order to observe the properties of an open circuited Transmission Line, chose a Transmission line of length l and is open circuited at the load end.

from the figure

$V_{S}$  –  Source voltage (at sending end (or) source end).

$I_{S}$ –  Source current (at sending end (or) source end).

$V_{R}$ –  Load voltage (at receiving end (or) Load end).

$I_{R}$ –  Load current (at receiving end (or) Load end).

at Load end $I_{R}&space;=0$  ,  $V_{R}&space;-$  maximum voltage  and $Z_{R}=\infty$ .

We knew  that  the input impedance of a Transmission line is given by  $Z_{S}&space;=&space;Z_{O}&space;\frac{(Z_{R}+Z_{O}&space;\tan&space;h\gamma&space;l)}{(Z_{O}+Z_{R}&space;\tan&space;h\gamma&space;l)}$.

after substituting the conditions of a open circuit conditions  that is $Z_{R}=\infty$

$Z_{S}&space;=\frac{Z_{O}}{\tan&space;h\gamma&space;l}$.

$Z_{S}&space;=Z_{O}\cot&space;h\gamma&space;l$.

if the Transmission line is a loss-less line then $\alpha&space;=0$   and  $\gamma&space;=&space;j\beta$ .

$Z_{S}&space;=Z_{O}\cot&space;(h&space;j\beta&space;l)$ .

$Z_{S}&space;=-jZ_{O}\cot&space;\beta&space;l$.     since $\cot&space;(hj\beta&space;l)=-j\cot&space;\beta&space;l$ .

alternative method:-

The alternative way to derive the input impedance is by using voltage and current equations of a basic Transmission line

$V=&space;V_{s}\cos&space;h\gamma&space;x-I_{s}Z_{o}\sin&space;h\gamma&space;x$.

$I=&space;I_{s}\cos&space;h\gamma&space;x-\frac{V_{s}}{Z_{o}}\sin&space;h\gamma&space;x$.

at $x=l$   ,  $V=V_{R}$    and  $I=I_{R}$ then the equations changes to

$V_{R}=&space;V_{s}\cos&space;h\gamma&space;l-I_{s}Z_{o}\sin&space;h\gamma&space;l-----EQN(1)$.

$I_{R}=&space;I_{s}\cos&space;h\gamma&space;l-\frac{V_{s}}{Z_{o}}\sin&space;h\gamma&space;l-----EQN(2)$.

by substituting $I_{R}&space;=0$  in equation(2)  and also choosing the line as loss-less line

$0=&space;I_{s}\cos&space;(j\beta&space;l)-\frac{V_{s}}{Z_{o}}\sin&space;(j\beta&space;l)$ .

$I_{s}\cos&space;\beta&space;l=\frac{V_{s}}{Z_{o}}&space;j\sin&space;\beta&space;l$ .     since  $\cos&space;(hj\beta&space;l)&space;=&space;\cos&space;\beta&space;l$   and  $\sin&space;(hj\beta&space;l)&space;=&space;j\sin&space;\beta&space;l$ .

$Z_{S}&space;=\frac{V_{s}}{I_{s}}=Z_{o}&space;\frac{\cos&space;\beta&space;l}{j\sin&space;\beta&space;l}$.

$Z_{S}=Z_{OC}&space;=-jZ_{o}&space;\cot&space;\beta&space;l$.

short circuit line:-

In order to observe the properties of an short circuited Transmission Line, chose a Transmission line of length l and is short circuited at the load end.

from the figure

$V_{S}$  –  Source voltage (at sending end (or) source end).

$I_{S}$ –  Source current (at sending end (or) source end).

$V_{R}$ –  Load voltage (at receiving end (or) Load end).

$I_{R}$ –  Load current (at receiving end (or) Load end).

at Load end $V_{R}&space;=0$  ,  $I_{R}&space;-$  maximum current and $Z_{R}=0$ .

We knew  that  the input impedance of a Transmission line is given by  $Z_{S}&space;=&space;Z_{O}&space;\frac{(Z_{R}+Z_{O}&space;\tan&space;h\gamma&space;l)}{(Z_{O}+Z_{R}&space;\tan&space;h\gamma&space;l)}$.

after substituting the conditions of a short circuit conditions  that is $Z_{R}=0$

$Z_{S}&space;=Z_{O}\tan&space;h\gamma&space;l$.

if the Transmission line is a loss-less line then $\alpha&space;=0$   and  $\gamma&space;=&space;j\beta$ .

$Z_{S}&space;=Z_{O}\tan&space;(h&space;j\beta&space;l)$ .

$Z_{S}&space;=jZ_{O}\tan&space;\beta&space;l$.     since $\tan&space;(hj\beta&space;l)=j\tan&space;\beta&space;l$ .

alternative method:-

The alternative way to derive the input impedance is by using voltage and current equations of a basic Transmission line

$V=&space;V_{s}\cos&space;h\gamma&space;x-I_{s}Z_{o}\sin&space;h\gamma&space;x$.

$I=&space;I_{s}\cos&space;h\gamma&space;x-\frac{V_{s}}{Z_{o}}\sin&space;h\gamma&space;x$.

at $x=l$   ,  $V=V_{R}$    and  $I=I_{R}$ then the equations changes to

$V_{R}=&space;V_{s}\cos&space;h\gamma&space;l-I_{s}Z_{o}\sin&space;h\gamma&space;l-----EQN(1)$.

$I_{R}=&space;I_{s}\cos&space;h\gamma&space;l-\frac{V_{s}}{Z_{o}}\sin&space;h\gamma&space;l-----EQN(2)$.

by substituting $V_{R}&space;=0$  in equation(1)  and also choosing the line as loss-less line

$0=&space;V_{s}\cos&space;(j\beta&space;l)-I_{s}Z_{o}\sin&space;(j\beta&space;l)$ .

$V_{s}\cos&space;\beta&space;l=I_{s}Z_{o}&space;j\sin&space;\beta&space;l$ .     since  $\cos&space;(hj\beta&space;l)&space;=&space;\cos&space;\beta&space;l$   and  $\sin&space;(hj\beta&space;l)&space;=&space;j\sin&space;\beta&space;l$ .

$Z_{S}&space;=\frac{V_{s}}{I_{s}}=Z_{o}&space;\frac{j\sin&space;\beta&space;l}{\cos&space;\beta&space;l}$.

$Z_{S}=Z_{SC}&space;=jZ_{o}&space;\tan&space;\beta&space;l$.

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