Infinite line: surge impedance and forward/backward waves

A lossless line has series inductance $L$ and shunt capacitance $C$ per unit length. When a source $V$ is switched on, a voltage wave and a current wave travel along the line together.

Impedance offered to the travelling wave

Let the wavefront travel with velocity $v$. In a time $dt$ it covers a length $dx = v\,dt$, and this new length is charged to voltage $V$:

• Charge added: $dq = C\,dx\,V$, so the current is $I = \dfrac{dq}{dt} = CVv$ ... (1)
• Flux set up in the new length: $d\psi = L\,dx\,I$, so the voltage is $V = \dfrac{d\psi}{dt} = LIv$ ... (2)

Dividing (2) by (1):

$$\frac{V}{I} = \frac{LIv}{CVv} \;\Rightarrow\; \left(\frac{V}{I}\right)^2 = \frac{L}{C} \;\Rightarrow\; \frac{V}{I} = \sqrt{\frac{L}{C}} = Z_0$$

Multiplying (1) and (2): $VI = LC\,VI\,v^2$, so $v = 1/\sqrt{LC}$.

For an infinite line (or before the wave reaches the open end) there is no reflected wave. The source therefore sees a pure resistance $Z_0 = \sqrt{L/C}$, the surge impedance, even though the line contains only $L$ and $C$. Typical values are 300–500 Ω for overhead lines and 30–60 Ω for cables.

The waves consist of forward and backward components

For a lossless line the telegraph equations are

$$\frac{\partial v}{\partial x} = -L\frac{\partial i}{\partial t}, \qquad \frac{\partial i}{\partial x} = -C\frac{\partial v}{\partial t}$$

Eliminating $i$ gives the wave equation:

$$\frac{\partial^2 v}{\partial x^2} = LC\frac{\partial^2 v}{\partial t^2} = \frac{1}{v_p^2}\frac{\partial^2 v}{\partial t^2}$$

Its general (d'Alembert) solution is

$$v(x,t) = f_1(x - v_p t) + f_2(x + v_p t)$$

• $f_1(x - v_p t)$ keeps its shape while $x$ increases with $t$. It is the forward wave.
• $f_2(x + v_p t)$ moves towards decreasing $x$. It is the backward wave.

Substituting into the first equation gives the current:

$$i(x,t) = \frac{1}{Z_0}\left[f_1(x - v_p t) - f_2(x + v_p t)\right]$$

So $v_f/i_f = +Z_0$ for the forward wave and $v_b/i_b = -Z_0$ for the backward wave.

With the receiving end open, the current there must be zero. When the forward wave arrives it is reflected with $v_b = +v_f$ and $i_b = -i_f$. The voltage doubles to $2V$ and the current becomes zero. The total voltage and current on the line are then the sums of the forward and backward waves. Only on an infinite line, where no reflection exists, is the ratio $v/i$ always $Z_0$.