Nominal-π model ABCD derivation

In the nominal-Π model of a medium line (80–250 km), the total series impedance $Z = R + jX$ is lumped in the middle, and half of the total shunt admittance, $Y/2 = j\omega C/2$, is placed at each end.

 IS       IL    Z = R + jX        IR
 o-->--+-----[ R ]--[ jX ]-----+-->--o
       |                       |
 VS  [Y/2] Ic1           Ic2 [Y/2]  VR
       |                       |
 o-----+-----------------------+-----o

Derivation (per phase)

Step 1: Current in the receiving-end capacitor:

$$I_{C2} = \frac{Y}{2}V_R$$

Step 2: Current in the series impedance:

$$I_L = I_R + I_{C2} = I_R + \frac{Y}{2}V_R$$

Step 3: Sending-end voltage:

$$\begin{aligned} V_S &= V_R + ZI_L = V_R + Z\left(I_R + \frac{Y}{2}V_R\right) \\ V_S &= \left(1 + \frac{ZY}{2}\right)V_R + ZI_R \end{aligned}$$

Step 4: Sending-end current:

$$\begin{aligned} I_S &= I_L + \frac{Y}{2}V_S \\ &= I_R + \frac{Y}{2}V_R + \frac{Y}{2}\left[\left(1 + \frac{ZY}{2}\right)V_R + ZI_R\right] \\ &= \left(Y + \frac{ZY^2}{4}\right)V_R + \left(1 + \frac{ZY}{2}\right)I_R \\ I_S &= Y\left(1 + \frac{ZY}{4}\right)V_R + \left(1 + \frac{ZY}{2}\right)I_R \end{aligned}$$

Step 5: Compare with $V_S = AV_R + BI_R$, $I_S = CV_R + DI_R$:

$$A = D = 1 + \frac{ZY}{2}, \quad B = Z\ \Omega, \quad C = Y\left(1 + \frac{ZY}{4}\right)\ \text{S}$$

Check

$$AD - BC = \left(1 + \frac{ZY}{2}\right)^2 - ZY\left(1 + \frac{ZY}{4}\right) = 1 + ZY + \frac{Z^2Y^2}{4} - ZY - \frac{Z^2Y^2}{4} = 1$$

$A = D$ shows the network is symmetrical, and $AD - BC = 1$ shows it is reciprocal (passive).