a Transmission Line is any thing used to transmit a electric signal.

in a AC circuit, the voltage does not only variate on time but also on space. For low frequency, the wavelength is long and this can be neglected. But at hight frequency, the variation is significant.

since the voltage is changing from different location. we cut the transmission line in a small sector, and each sector is analog to some circuit elements, no matter the shape of the line. this gives us a easy understanding of what is going on for the voltage and current. But this analogy neglected the effect of temperature, material non-linearity and magnetic hysteresis effect.

the equation of voltage across a section is

$V(x) = ( R + i \omega L )\Delta x I(x)+ V(x+\Delta x )$

since the resistance and inductance have unit per length. rearrange and take limit of x.

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

the equation of current is

$I(x) = ( Q + i \omega C )\Delta x V(x+ \Delta x) + I(x+ \Delta x )$

The Q is conductance, which is NOT an invert of resistance in case of AC. both Q and C will draw some current away in AC circuit. take limit gives

$- \frac { d I(x) }{dx} = (Q + i \omega C ) V(x)$

now we have 2 coupled equations. If we de-couple them, we will have

$\frac {d^2 V(x)}{dx^2} = k^2 V(x)$

$k = \sqrt{ (R+i \omega L ) ( Q + i \omega C ) }$

and the current share the same equation. notice that k is a complex number.

the solution are:

$V(x) = V_f Exp( - k x ) + V_b Exp( k x)$

$I(x) = I_f Exp( - k x) + I_b Exp( k x )$

where the subscript means forward and backward. from the coupled equation of Current, we can related the voltage and current and find out the impedance.

$V(x) = \sqrt{ \frac { R+ i \omega L } { Q + i \omega C } } (I_f Exp(-kx) - I_b Exp(k x)$

Thus, we define the Characteristic Impedance for forward wave.

$Z_0 = \sqrt{ \frac { R+ i \omega L } { Q + i \omega C }}$

the Characteristic Impedance for backward wave is a minus sign. there fore, we can rewrite the current in term of voltage.

$I(x) = \frac {1}{Z_0} ( V_f Exp(-kx) - V_b Exp(kx)$

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impedance matching

for a load at the end of the transmission line, the wave will get reflected. to see this, we have to consider the load, which imposed another equation. the voltage across the load is:

$V(L) = V_f Exp( - k L ) + V_b Exp( k L)$

and the current input to the load is:

$I(L) = \frac {1}{Z_0} ( V_f Exp(-kL) - V_b Exp(kL)$

the current and the voltage is related by:

$V(L) = Z_L I(L)$

solve it, and find the ratio of :

$\frac { V_b }{V_f} = Exp( - 2 k L ) \frac { Z_L - Z_0 }{ Z_L + Z_0 }$

since we can do nothing on the exponential, thus, we define a reflection coefficient:

$\Gamma_0 = \frac { Z_L - Z_0} {Z_L - Z_0}$

for no reflected wave, the impedance of the load and the transmission line should be equal and it is called impedance matching.

the power of the load is:

$P = V(L)I*(L) = \frac { 1}{Z_0} ( V_f^2 Exp( - 2 k L ) - V_b^2 Exp( 2k L) + V_f V_b* -V_f* V_b )$

when the impedance is matched, the power is maximum.