# Transmission line equations

The transmission line will be analyzed in terms of voltage and current and also the relation between V and I of a line can be obtained by the simplest type of Transmission line.

i.e, a pair of parallel wires of uniform size which are spaced a small distance (S) apart in air.

Consider a short section of Transmission line $AA^{'}BB^{'}$ of length $\Delta&space;x$ (i.e an infinitesimal section with a very small length).

at $AA^{'}$ – voltage is V and current is I and at $BB^{'}$ voltage is $V+\Delta&space;V$ and current is $I+\Delta&space;I$ .

the voltage drop between $AA^{'}$ and $BB^{'}$ is

$V-(V+\Delta&space;V)&space;=&space;I(R+j\omega&space;L)&space;\Delta&space;x---------EQN(I)$

similarly, the current difference between $AA^{'}$ and $BB^{'}$ is

$I-(I+\Delta&space;I)&space;=&space;V(G+j\omega&space;C)&space;\Delta&space;X-------------EQN(II)$

from EQNS (I) and(II)

$\therefore&space;-\frac{\Delta&space;V}{\Delta&space;x}=I&space;(R+j\omega&space;L)$     and         $-\frac{\Delta&space;I}{\Delta&space;x}=V&space;(G+j\omega&space;C)$.

to neglect the transit time effect the condition to be applied is  $\Delta&space;x\rightarrow&space;0$, then the two equations will become

$\lim_{\Delta&space;x\rightarrow&space;0}&space;-\frac{\Delta&space;V}{\Delta&space;x}=\lim_{\Delta&space;x\rightarrow&space;0}I&space;(R+j\omega&space;L)$    $\Rightarrow&space;-\frac{d&space;V}{d&space;x}=I&space;(R+j\omega&space;L)$.

$\lim_{\Delta&space;x\rightarrow&space;0}&space;-\frac{\Delta&space;I}{\Delta&space;x}=\lim_{\Delta&space;x\rightarrow&space;0}V&space;(G+j\omega&space;C)$  $\Rightarrow&space;-\frac{d&space;I}{d&space;x}=V&space;(G+j\omega&space;C)$.

choose $-\frac{d&space;V}{d&space;x}=I&space;(R+j\omega&space;L)$

differentiating the above equation w.r.to x

$\Rightarrow&space;-\frac{d^{2}&space;V}{d&space;x^{2}}=\frac{dI}{dx}&space;(R+j\omega&space;L)$

replacing $\frac{dI}{dx}$ in the above equation will result

$\frac{d^{2}&space;V}{d&space;x^{2}}=&space;(R+j\omega&space;L)(G+j\omega&space;C)V$.

$\frac{d^{2}&space;V}{d&space;x^{2}}=\gamma&space;^{2}&space;V----------EQN(1)$ ,   Let $\gamma&space;^{2}&space;=(R+j\omega&space;L)(G+j\omega&space;C)$.

where $\gamma$ is the propagation constant which is a complex number.

Similarly  from $-\frac{d&space;I}{d&space;x}=V&space;(G+j\omega&space;C)$

differentiating the above equation w.r.to x

$\Rightarrow&space;-\frac{d^{2}&space;I}{d&space;x^{2}}=\frac{dV}{dx}&space;(G+j\omega&space;C)$

replacing $\frac{dV}{dx}$ in the above equation will result

$\frac{d^{2}&space;I}{d&space;x^{2}}=&space;(R+j\omega&space;L)(G+j\omega&space;C)I$.

$\frac{d^{2}&space;I}{d&space;x^{2}}=\gamma&space;^{2}&space;I------EQN(2)$

the solutions of  Equations(1) and (2) are

$V=A&space;e^{-\gamma&space;x}+Be^{\gamma&space;x}$.

$I=Ce^{-\gamma&space;X}+D&space;e^{\gamma&space;X}$.

where A, B, C and D are arbitrary constants in which A and B have dimensions of voltage and C and D have current dimensions.

since $\gamma$ is complex that is by replacing $\gamma&space;=&space;\alpha&space;+j\beta$ in the V and I equations

$V=A&space;e^{-\alpha&space;x}&space;e^{-j\beta&space;x}+Be^{\alpha&space;x}&space;e^{j\beta&space;x}$ .

$I=C&space;e^{-\alpha&space;x}&space;e^{-j\beta&space;x}+D&space;e^{\alpha&space;x}&space;e^{j\beta&space;x}$.

the  term $e^{-\alpha&space;x}&space;e^{-j\beta&space;x}$ represents waves travelling from source end  to load end and are called as incident waves .

Similarly the term $e^{\alpha&space;x}&space;e^{j\beta&space;x}$ represents reflected waves when a transmission line is terminated with any load impedance $Z_{R}$ at the output end.

$V=A&space;e^{-\gamma&space;x}+Be^{\gamma&space;x}$.

differentiating V w.r.to x

$\frac{dV}{dx}=-A\gamma&space;e^{-\gamma&space;x}+B\gamma&space;e^{\gamma&space;x}$.

but $\frac{dV}{dx}=-I(R+j\omega&space;L)$

$\therefore&space;-I(R+j\omega&space;L)=-A\gamma&space;e^{-\gamma&space;x}+B\gamma&space;e^{\gamma&space;x}$.

$I=\frac{\gamma&space;}{(R+j\omega&space;L)}(Ae^{-\gamma&space;x}-B&space;e^{\gamma&space;x})$.

$I=\frac{\sqrt{(R+j\omega&space;L)(G+j\omega&space;C)}}{(R+j\omega&space;L)}(Ae^{-\gamma&space;x}-B&space;e^{\gamma&space;x})$

since $\gamma&space;={\sqrt{(R+j\omega&space;L)(G+j\omega&space;C)}}$.

$I=\sqrt{\frac{(G+j\omega&space;C)}{(R+j\omega&space;L)}}(Ae^{-\gamma&space;x}-B&space;e^{\gamma&space;x})$.

$\therefore&space;I=\frac{1}{Z_{o}}(Ae^{-\gamma&space;x}-B&space;e^{\gamma&space;x})$,   where $Z_{o}&space;=\sqrt{\frac{(R+j\omega&space;L)}{(G+j\omega&space;C)}}$ .

AS  $\cos&space;h\gamma&space;x&space;=&space;\frac{e^{\gamma&space;x}+e^{-\gamma&space;x}}{2}$    and $\sin&space;h\gamma&space;x&space;=&space;\frac{e^{\gamma&space;x}-e^{-\gamma&space;x}}{2}$ .

this implies $\cos&space;h\gamma&space;x-\sin&space;h\gamma&space;x&space;=&space;e^{-\gamma&space;x}$      ,    $\cos&space;h\gamma&space;x+\sin&space;h\gamma&space;x&space;=&space;e^{\gamma&space;x}$.

by substituting $e^{-\gamma&space;x}$  and $e^{\gamma&space;x}$ in the Voltage and current equations , V and I results to be

$V=&space;(A+B)\cos&space;h\gamma&space;x-(A-B)\sin&space;h\gamma&space;x---EQN(3)$.

$I=&space;\frac{1}{Z_{o}}((A-B)\cos&space;h\gamma&space;x-(A+B)\sin&space;h\gamma&space;x)---EQN(4)$

at the input terminals $x=0$ , $V&space;=&space;V_{s}$  and $I&space;=&space;I_{s}$ , after substituting this condition in EQNs (1) and (2)

$V_{s}&space;=(A+B)$   and  $I_{s}&space;=\frac{1}{Z_{o}}(A-B)$.

then the EQNs (3) and (4)

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

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

The above equations are known as general equations of a transmission line for voltage and current at any point  which is located at x from the sending end.

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