Transmission line equations

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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 $\Delta x$ (i.e an infinitesimal section with a very small length).

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

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

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

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

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

from EQNS (I) and(II)

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

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

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

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

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

differentiating the above equation w.r.to x

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

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

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

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

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

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

differentiating the above equation w.r.to x

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

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

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

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

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

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

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

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 $\gamma = \alpha +j\beta$ in the V and I equations

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

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

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

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

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

differentiating V w.r.to x

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

then the EQNs (3) and (4)

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

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

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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Author: Lakshmi Prasanna Ponnala

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

Transmission line equations

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