--- title: "Equivalent T from Y parameters of series two-ports" subject: "Electrical Circuit Theory" chapter: 7 key: "semester_3/electrical_circuit_theory:equivalent-t-parameter-two-y-networks" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/equivalent-t-parameter-two-y-networks site: BCT Engineering --- # Equivalent T from Y parameters of series two-ports Electrical Circuit Theory · Chapter 7: Two-port Parameter of Networks - Asked in: 2071 Chaitra, 8 marks - Video: [Equivalent circuit of Y-parameters and solving problems for finding Y-parameters of two port network](https://www.youtube.com/watch?v=fITFU1gi8Ck) — RF Design Basics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/equivalent-t-parameter-two-y-networks - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-7 ## Question The Y-parameters of two TPNS are given as: [1/4 −5/4; −1/4 −3/4] and [1/3 −1/3; −1/3 1/3]. If these two TPNS are connected in series. What will be the equivalent Transmission parameter of the combination? ## Short answer Given Ya and Yb, Yb is singular (Δyb=0) so direct Z addition fails. Series constraints force I2 = −I1. With network a this gives V1a = 4I1, V2a=0. Then V1 = V2 + 7I1, I1 = −I2, so equivalent T: A = 1, B = 7 Ω, C = 0 S, D = 1. ## Answer **Given.** Ya = [1/4 −5/4; −1/4 −3/4] S and Yb = [1/3 −1/3; −1/3 1/3] S, connected in series. #### Step 1: Z-parameters for the series connection For a series connection, the Z-matrices add: Z = Za + Zb, where Z = Y⁻¹. ```text Δya = (1/4)(−3/4) − (−5/4)(−1/4) = −3/16 − 5/16 = −1/2 Za = (1/Δya) [y22 −y12; −y21 y11] = −2 [−3/4 5/4; 1/4 1/4] = [3/2 −5/2; −1/2 −1/2] Ω ``` But for network b: Δyb = (1/3)(1/3) − (−1/3)(−1/3) = 0. So Yb is singular and **Zb does not exist**. (Yb is simply a 3 Ω resistor in the series arm, a network that has no Z-matrix.) The formula Z = Za + Zb therefore cannot be applied directly. #### Step 2: use the connection conditions directly In a series connection I1a = I1b = I1, I2a = I2b = I2, V1 = V1a + V1b, V2 = V2a + V2b. Network b is a series 3 Ω resistor: I1b = (V1b − V2b)/3 and I2b = −I1b. So the series connection forces I2 = −I1. For network a with I2 = −I1: ```text −V1a/4 − 3V2a/4 = −(V1a/4 − 5V2a/4) → V2a = 0 ``` so network a behaves as a 4 Ω resistor at its input: I1 = V1a/4 → V1a = 4 I1. Then: ```text V1b = V2b + 3 I1 V1 = V1a + V1b = 4 I1 + V2b + 3 I1 = V2 + 7 I1 (V2 = V2a + V2b = V2b) ``` With I1 = −I2: V1 = V2 − 7 I2 and I1 = 0·V2 − 1·I2. #### Step 3: transmission parameters ```text V1 = A V2 − B I2 → A = 1, B = 7 Ω I1 = C V2 − D I2 → C = 0, D = 1 ``` **Answer: T_eq = [A B; C D] = [1 7 Ω; 0 1]**, the same as a single 7 Ω series resistor (4 Ω + 3 Ω). Assumption note: the data make the Z-addition method impossible (Yb singular), so the result above comes from the series-connection constraints. For comparison, the individual T-matrices are Ta = [−3 4; −2 1] (using A = −y22/y21, B = −1/y21, C = −Δy/y21, D = −y11/y21) and Tb = [1 3; 0 1]. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a series connection, which parameters add? A. Y matrices B. Z matrices C. T matrices D. Y matrices inverted Answer: B. Currents are common and voltages add, so Z = Za + Zb. (Key point: Series rule) 2. Why can Z = Za + Zb not be applied directly? A. Delta yb = 0, so Yb is singular and Zb does not exist B. Ya is singular C. The networks are not reciprocal D. The ports are mismatched Answer: A. Delta yb = (1/3)(1/3) - (1/3)(1/3) = 0, so Yb cannot be inverted. (Key point: Singularity of Yb) 3. What is network b physically? A. A 3 ohm shunt resistor B. A 3 ohm series resistor C. A 1/3 ohm series resistor D. An ideal transformer Answer: B. Yb = [1/3 -1/3; -1/3 1/3] is a 3 ohm resistor in the series arm. (Key point: Network b equivalent) 4. What is the equivalent T matrix of the combination? A. [1 7; 0 1] B. [1 3; 0 1] C. [-3 4; -2 1] D. [1 0; 7 1] Answer: A. The connection forces I2 = -I1 and V1 = V2 + 7 I1, equal to a 7 ohm series resistor. 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