ABCD constants by nominal-T model

In the nominal-T model of a medium line, the whole shunt admittance $Y = j\omega C$ is placed at the mid-point of the line, and the series impedance $Z = R + jX$ is split into two halves, $Z/2$ on each side (per-phase values).

 IS     Z/2            Z/2     IR
 o-->--[R/2 jX/2]--+--[R/2 jX/2]-->--o
                   |
 VS              [ Y ] Ic   (VC)     VR
                   |
 o-----------------+-----------------o

Derivation

Step 1: Voltage at the mid-point (across the capacitor):

$$V_C = V_R + \frac{Z}{2}I_R$$

Step 2: Current through the shunt branch:

$$I_C = YV_C = YV_R + \frac{ZY}{2}I_R$$

Step 3: Sending-end current:

$$I_S = I_R + I_C = YV_R + \left(1 + \frac{ZY}{2}\right)I_R$$

Step 4: Sending-end voltage:

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

Step 5: Comparing with $V_S = AV_R + BI_R$ and $I_S = CV_R + DI_R$:

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

Check

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

So the network is reciprocal, and since $A = D$ it is symmetrical. Compared with the nominal-Π model, $A$ and $D$ are the same, but the factor $(1 + ZY/4)$ moves from $C$ to $B$.