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Long line sending end voltage derivation

Short answer

Derivation starts with dV/dx = zI, dI/dx = yV, leading to V = K1 e^{gamma x} + K2 e^{-gamma x}. Using receiver-end conditions, V_S = V_R cosh gamma l + Z_c I_R sinh gamma l, obtained by setting x=l. This expression is the exact distributed-parameter sending-end voltage.

Starting from a suitable point, derive the expression of sending end voltage in terms of the receiving end voltage and current for a long length transmission line.

For a long line (above about 250 km) the resistance, inductance and capacitance are distributed uniformly along the length, so the line is analysed with differential equations rather than lumped circuits. Measure xx 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

Starting point and solution

For an element dxdx at distance xx from the receiving end, with z=r+jωLz = r + j\omega L and y=g+jωCy = g + j\omega C per km:

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

Differentiating the first and substituting the second:

d2Vdx2=zy V=γ2V,γ=zy=α+jβ\frac{d^2V}{dx^2} = zy\,V = \gamma^2V, \qquad \gamma = \sqrt{zy} = \alpha + j\beta

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

V=K1eγx+K2e−γxI=1Zc(K1eγx−K2e−γx),Zc=z/y\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=0x = 0: V=VRV = V_R and I=IRI = I_R, so K1+K2=VRK_1 + K_2 = V_R and K1−K2=ZcIRK_1 - K_2 = Z_cI_R:

K1=VR+ZcIR2,K2=VR−ZcIR2K_1 = \frac{V_R + Z_cI_R}{2}, \qquad K_2 = \frac{V_R - Z_cI_R}{2}

So, at any point xx:

Vx=VR+ZcIR2eγx+VR−ZcIR2e−γxV_x = \frac{V_R + Z_cI_R}{2}e^{\gamma x} + \frac{V_R - Z_cI_R}{2}e^{-\gamma x}

Sending-end voltage

At the sending end, x=lx = l and Vx=VSV_x = V_S:

VS=VR+ZcIR2eγl+VR−ZcIR2e−γl=VR eγl+e−γl2+ZcIR eγl−e−γl2VS=VRcosh⁡γl+ZcIRsinh⁡γl\begin{aligned} V_S &= \frac{V_R + Z_cI_R}{2}e^{\gamma l} + \frac{V_R - Z_cI_R}{2}e^{-\gamma l} \\ &= V_R\,\frac{e^{\gamma l} + e^{-\gamma l}}{2} + Z_cI_R\,\frac{e^{\gamma l} - e^{-\gamma l}}{2} \\ V_S &= V_R\cosh\gamma l + Z_cI_R\sinh\gamma l \end{aligned}

Similarly, IS=VRZcsinh⁡γl+IRcosh⁡γlI_S = \frac{V_R}{Z_c}\sinh\gamma l + I_R\cosh\gamma l.

Hence A=D=cosh⁡γlA = D = \cosh\gamma l, B=Zcsinh⁡γlB = Z_c\sinh\gamma l, C=sinh⁡γlZcC = \frac{\sinh\gamma l}{Z_c}, where:

  • γ=zy=α+jβ\gamma = \sqrt{zy} = \alpha + j\beta is the propagation constant (per km),
  • Zc=z/yZ_c = \sqrt{z/y} is the characteristic impedance (Ω),
  • ll is the line length.

In the exponential form, the first term is the incident wave (travelling from sending to receiving end) and the second the reflected wave. When the line is terminated in ZcZ_c, VR=ZcIRV_R = Z_cI_R, the reflected wave is zero and VS=VReγlV_S = V_Re^{\gamma l}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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