Infinite Line equivalent circuit

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Consider the basic form of Transmission line with some impedance $Z_{o}$ at the Load end.

an infinite line can be approximated by an equivalent finite line with load impedance $Z_{o}$ as shown in the above figure, then the input impedance can be calculated from the voltage and current equations.

now at x= l , $V=V_{R} \ and \ I=I_{R}$ .

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

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

The load voltage is given by the equation $V_{R}=I_{R}Z_{o}$ .

$V_{s}\cos h\gamma l-I_{s}Z_{o}\sin h\gamma l = Z_{o}(I_{s}\cos h\gamma l-\frac{V_{s}}{Z_{o}}\sin h\gamma l)$

$V_{s}Z_{O}\cos h\gamma l-I_{s}Z_{o}^{2}\sin h\gamma l = Z_{o}(I_{s}Z_{o}\cos h\gamma l-V_{S}\sin h\gamma l)$

$V_{s}(Z_{O}\cos h\gamma l+Z_{R}\sin h\gamma l) = I_{s}Z_{o}(Z_{o}\sin h\gamma l+Z_{R}\cos h\gamma l)$

$Z_{S}=\frac{V_{S}}{I_{S}}=Z_{o} \frac{(Z_{o}\cos h\gamma l+Z_{o}\sin h\gamma l)}{(Z_{o}\cos h\gamma l+Z_{o}\sin h\gamma l)}$

$Z_{S} \ (or) \ Z_{in}=Z_{o}$ .

represents the source (or) input impedance of an infinite Transmission Line.

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Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems. View all posts by Lakshmi Prasanna Ponnala

Infinite Line equivalent circuit

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