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 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: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.