ABCD with shunt capacitor at line end

Given: $A = D = 0.9955\angle 0.086^\circ$, $B = 31.552\angle 71.608^\circ\ \Omega$, $C = 3\times10^{-4}\angle 90^\circ$ S; shunt capacitor $Y_{sh} = 2.5\times10^{-4}\angle 90^\circ$ S at the receiving end of the line.

Method

The shunt capacitor is a two-port with constants $\begin{bmatrix} 1 & 0 \\ Y_{sh} & 1 \end{bmatrix}$. It is in cascade (series) after the line, so the overall matrix is the product:

 VS    +------------+         VR
 o-----|  Line      |----+-----o
       |  A B C D   |    |
       +------------+  [Ysh]
 o-----------------------+-----o

$$\begin{bmatrix} A' & B' \\ C' & D' \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix}\begin{bmatrix} 1 & 0 \\ Y_{sh} & 1 \end{bmatrix} = \begin{bmatrix} A + BY_{sh} & B \\ C + DY_{sh} & D \end{bmatrix}$$

Calculation

$$\begin{aligned} BY_{sh} &= 31.552\angle 71.608^\circ\times2.5\times10^{-4}\angle 90^\circ = 7.888\times10^{-3}\angle 161.608^\circ \\ &= -0.007485 + j0.002489 \\ A' &= (0.995499 + j0.001494) + (-0.007485 + j0.002489) \\ &= 0.98801 + j0.00398 = 0.9880\angle 0.231^\circ \\ B' &= B = 31.552\angle 71.608^\circ\ \Omega \\ DY_{sh} &= 0.9955\angle 0.086^\circ\times2.5\times10^{-4}\angle 90^\circ = 2.4888\times10^{-4}\angle 90.086^\circ \\ C' &= j3\times10^{-4} + (-3.7\times10^{-7} + j2.4887\times10^{-4}) \\ &= 5.489\times10^{-4}\angle 90.04^\circ\ \text{S} \\ D' &= D = 0.9955\angle 0.086^\circ \end{aligned}$$

Check: $A'D' - B'C' = 1.000$ (to 5 decimal places).

Answer: $A' = 0.9880\angle 0.231^\circ$, $B' = 31.552\angle 71.608^\circ\ \Omega$, $C' = 5.489\times10^{-4}\angle 90.04^\circ$ S, $D' = 0.9955\angle 0.086^\circ$.

Note: the combined network is no longer symmetrical ($A' \ne D'$). If the capacitor were at the sending end, $A$ and $D$ would swap roles: $A' = 0.9955\angle 0.086^\circ$, $D' = 0.9880\angle 0.231^\circ$.