Equivalent T from Y parameters of series two-ports

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.

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:

−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:

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

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].