Equivalent ABCD of two networks in series

When two two-port networks are connected in series (cascade), the output of the first is the input of the second. The overall ABCD matrix is the product of the individual matrices, taken in order.

 VS  +-------------+  V2, I2  +-------------+  VR
 o---| A1 B1 C1 D1 |----------| A2 B2 C2 D2 |---o
 IS  |  network 1  |          |  network 2  |  IR
 o---+-------------+----------+-------------+---o

Derivation

For network 2:

$$\begin{bmatrix} V_2 \\ I_2 \end{bmatrix} = \begin{bmatrix} A_2 & B_2 \\ C_2 & D_2 \end{bmatrix}\begin{bmatrix} V_R \\ I_R \end{bmatrix}$$

For network 1:

$$\begin{bmatrix} V_S \\ I_S \end{bmatrix} = \begin{bmatrix} A_1 & B_1 \\ C_1 & D_1 \end{bmatrix}\begin{bmatrix} V_2 \\ I_2 \end{bmatrix}$$

Substituting:

$$\begin{bmatrix} V_S \\ I_S \end{bmatrix} = \begin{bmatrix} A_1 & B_1 \\ C_1 & D_1 \end{bmatrix}\begin{bmatrix} A_2 & B_2 \\ C_2 & D_2 \end{bmatrix}\begin{bmatrix} V_R \\ I_R \end{bmatrix}$$

Equivalent constants

$$\begin{aligned} A &= A_1A_2 + B_1C_2 \\ B &= A_1B_2 + B_1D_2 \\ C &= C_1A_2 + D_1C_2 \\ D &= C_1B_2 + D_1D_2 \end{aligned}$$

Notes

• The order matters: matrix multiplication is not commutative, so swapping the two networks generally changes the result. (For two symmetrical networks, B and C stay the same, but A and D swap values.)
• $AD - BC = (A_1D_1 - B_1C_1)(A_2D_2 - B_2C_2) = 1$, so the combination is still reciprocal.
• Example uses: a line with a transformer at each end; a line with a series capacitor ($A = D = 1$, $B = -jX_C$, $C = 0$) or a shunt reactor ($A = D = 1$, $B = 0$, $C = Y$).