Long line voltage and current at intermediate point

In a long line the parameters are distributed, so voltage and current vary from point to point. Let $x$ be the distance of the intermediate point from the receiving end.

 IS      x <------ measured from receiving end
 o-->--z.dx--+--z.dx--+-- ... --z.dx--+--> IR
             |        |               |
 VS        y.dx     y.dx            y.dx    VR
             |        |               |
 o-----------+--------+-- ... --------+----o
 sending                          receiving

Derivation

For an element $dx$ at distance $x$ from the receiving end, with $z = r + j\omega L$ and $y = g + j\omega C$ per km:

$$\frac{dV}{dx} = zI, \qquad \frac{dI}{dx} = yV$$

Differentiating the first and substituting the second:

$$\frac{d^2V}{dx^2} = zy\,V = \gamma^2V, \qquad \gamma = \sqrt{zy} = \alpha + j\beta$$

The general solution and the current (from $I = \frac{1}{z}\frac{dV}{dx}$) are:

$$\begin{aligned} V &= K_1e^{\gamma x} + K_2e^{-\gamma x} \\ I &= \frac{1}{Z_c}\left(K_1e^{\gamma x} - K_2e^{-\gamma x}\right), \quad Z_c = \sqrt{z/y} \end{aligned}$$

At $x = 0$: $V = V_R$ and $I = I_R$, so $K_1 + K_2 = V_R$ and $K_1 - K_2 = Z_cI_R$:

$$K_1 = \frac{V_R + Z_cI_R}{2}, \qquad K_2 = \frac{V_R - Z_cI_R}{2}$$

Substituting:

$$\begin{aligned} V_x &= \frac{V_R + Z_cI_R}{2}e^{\gamma x} + \frac{V_R - Z_cI_R}{2}e^{-\gamma x} \\ I_x &= \frac{V_R/Z_c + I_R}{2}e^{\gamma x} - \frac{V_R/Z_c - I_R}{2}e^{-\gamma x} \end{aligned}$$

Using $\cosh\gamma x = \frac{e^{\gamma x} + e^{-\gamma x}}{2}$ and $\sinh\gamma x = \frac{e^{\gamma x} - e^{-\gamma x}}{2}$:

$$\begin{aligned} V_x &= V_R\cosh\gamma x + Z_cI_R\sinh\gamma x \\ I_x &= \frac{V_R}{Z_c}\sinh\gamma x + I_R\cosh\gamma x \end{aligned}$$

In matrix form:

$$\begin{bmatrix} V_x \\ I_x \end{bmatrix} = \begin{bmatrix} \cosh\gamma x & Z_c\sinh\gamma x \\ \frac{1}{Z_c}\sinh\gamma x & \cosh\gamma x \end{bmatrix}\begin{bmatrix} V_R \\ I_R \end{bmatrix}$$

Notes

• If the point is given as distance $d$ from the sending end, use $x = l - d$.
• Putting $x = l$ gives the sending-end relations $V_S = V_R\cosh\gamma l + Z_cI_R\sinh\gamma l$ and $I_S = \frac{V_R}{Z_c}\sinh\gamma l + I_R\cosh\gamma l$.
• The terms with $e^{\gamma x}$ form the incident wave and those with $e^{-\gamma x}$ the reflected wave. Their sum changes with $x$, so $V_x$ and $I_x$ are different at every point.