Nominal-T phasor diagram and ABCD parameters

In the nominal-T model of a medium transmission line, the total series impedance $Z = R + jX$ is split into two halves $Z/2$ at each end, and the total shunt admittance $Y = j\omega C$ is placed at the middle.

Circuit

 Is   Z/2        Vc        Z/2   Ir
 o---[R/2 X/2]---+---[R/2 X/2]---o
 +               |               +
 Vs            Ic| Y             Vr
 -               |               -
 o---------------+---------------o

Phasor diagram (lagging load)

Take $V_R$ as reference; $I_R$ lags it by $\phi_R$.

Currents:

        Ic (leads Vc by ~90 deg)
        ^
        |    Is = Ir + Ic
        |   /
        |  /
   O----+-+--------------> Vr (reference)
         \
          \  phi_R (lag)
           v Ir

Voltages (drops added head to tail):

                               Vs
                              /|
                             / | Is.X/2 (90 deg
                            /  |  ahead of Is)
                           /---+ Is.R/2 (along Is)
                       Vc /|
                         / | Ir.X/2 (90 deg
                        /  |  ahead of Ir)
                       /---+ Ir.R/2 (along Ir)
                      /
   O-----------------+ Vr
   Vs leads Vr by the load angle delta

Construction steps:

1. Draw $V_R$ along the reference axis; draw $I_R$ lagging by $\phi_R$.
2. Add $I_R R/2$ parallel to $I_R$ and $I_R X/2$ perpendicular (leading $I_R$ by 90°) to get $V_C$.
3. Draw $I_C$ leading $V_C$ by 90°; $I_S = I_R + I_C$.
4. Add $I_S R/2$ parallel to $I_S$ and $I_S X/2$ perpendicular to it to get $V_S$.

Derivation of ABCD constants

Voltage across the shunt branch:

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

Shunt current and sending-end current:

$$\begin{aligned} I_C &= YV_C = YV_R + \frac{YZ}{2}I_R \\ I_S &= I_R + I_C = YV_R + \left(1 + \frac{YZ}{2}\right)I_R \end{aligned}$$

Sending-end voltage:

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

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

Check: $AD - BC = \left(1 + \frac{YZ}{2}\right)^2 - YZ\left(1 + \frac{YZ}{4}\right) = 1$, and $A = D$ since the circuit is symmetrical.