--- title: "Equivalent ABCD of two networks in series" subject: "Power System Analysis I" chapter: 5 key: "semester_4/power_system_analysis_i:series-abcd-equivalent" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/series-abcd-equivalent site: BCT Engineering --- # Equivalent ABCD of two networks in series Power System Analysis I · Chapter 5: Transmission Line Modeling - Asked in: 2068 Magh, 6 marks - Video: [Lecture -- Transmission Matrices & Cascading Multiple Networks](https://www.youtube.com/watch?v=rfbvmGwN_8o) — EMPossible - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/series-abcd-equivalent - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5 ## Question What will be the equivalent ABCD parameters of two systems connected in series and represented by their respective ABCD parameters? ## Short answer For two networks in series, the overall ABCD matrix is the product of the individual matrices: A = A1A2 + B1C2, B = A1B2 + B1D2, C = C1A2 + D1C2, D = C1B2 + D1D2. The order of multiplication matters as matrix multiplication is not commutative. ## Answer 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. ```text 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 ($A = D = 1$, $B = 0$, $C = Y$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the equivalent ABCD matrix of two networks connected in series (cascade) found? A. Add the two matrices B. Multiply the individual matrices in order, network 1 times network 2 C. Multiply the matrices in any order D. Add A and D, multiply B and C Answer: B. The output of network 1 is the input of network 2, so [V_S, I_S] = M1 x M2 x [V_R, I_R]. (Key point: Series combination rule) 2. What is the equivalent A constant of the cascade? A. A1A2 + B1B2 B. A1 + A2 C. A1A2 + B1C2 D. A1B2 + B1D2 Answer: C. The first row of M1 times the first column of M2 gives A = A1A2 + B1C2. (Key point: Equivalent A) 3. What is the equivalent B constant of the cascade? A. A1B2 + B1D2 B. B1 + B2 C. A1A2 + B1C2 D. C1B2 + D1D2 Answer: A. First row of M1 times second column of M2 gives B = A1B2 + B1D2. (Key point: Equivalent B) 4. Which statement about the series combination is correct? A. Order does not matter because multiplication is commutative B. AD - BC becomes 2 C. AD - BC becomes 0 D. Order matters, and AD - BC remains 1 Answer: D. Matrix multiplication is not commutative, but det(M1M2) = det(M1)det(M2) = 1. 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