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Z and T parameters with CCCS 4V1

Short answer

Form node equations to obtain Y parameters, then invert to Z: z11 = 4/15 Ω, z12 = 1/15 Ω, z21 = −1/5 Ω, z22 = 7/10 Ω. Convert to T: A = z11/z21 = −4/3, B = Δz/z21 = −1 Ω, C = 1/z21 = −5 S, D = z22/z21 = −7/2.

Find Z-parameters and hence the T-parameters for the 2 port network shown in figure below. [Figure: port 1 (V1, I1); a 2 Ω shunt resistor at port 1; a 1 Ω series resistor and a 2 Ω series resistor in the top wire; a dependent current source 4V1 (arrow pointing down) as a shunt at the junction of the two series resistors; a 1 Ω shunt resistor at port 2 (V2, I2)]

Z-parameters: V1 = z11 I1 + z12 I2, V2 = z21 I1 + z22 I2. Transmission parameters: V1 = A V2 − B I2, I1 = C V2 − D I2.

Circuit. 2 Ω shunt at port 1; 1 Ω then 2 Ω in the series arm; a dependent current source 4V1 (arrow down) in shunt at the junction J of the two series resistors; 1 Ω shunt at port 2.

Because the sources and the series resistors form a ladder, it is easiest to write the node equations and solve for the currents in terms of voltages (Y form), then invert.

Node equations

Port 1 node:   I1 = V1/2 + (V1 − VJ)/1
Node J:        (V1 − VJ)/1 = (VJ − V2)/2 + 4V1
Port 2 node:   I2 + (VJ − V2)/2 = V2/1

From node J: −3V1 − 1.5 VJ + 0.5 V2 = 0, so VJ = V2/3 − 2V1.

I1 = 1.5 V1 − VJ = 3.5 V1 − V2/3
I2 = 1.5 V2 − VJ/2 = V1 + (4/3) V2

Thus y11 = 7/2, y12 = −1/3, y21 = 1, y22 = 4/3, and Δy = (7/2)(4/3) + 1/3 = 5.

Z-parameters (Z = Y⁻¹)

z11 = y22/Δy = 4/15 Ω       z12 = −y12/Δy = 1/15 Ω
z21 = −y21/Δy = −1/5 Ω      z22 = y11/Δy = 7/10 Ω

T-parameters

A = z11/z21 = −4/3          B = Δz/z21 = −1 Ω     (Δz = 1/5)
C = 1/z21   = −5 S          D = z22/z21 = −7/2
Check: AD − BC = 14/3 − 5 = −1/3 ≠ 1 → not reciprocal.

Answer: Z = [4/15 1/15; −1/5 7/10] Ω; T: A = −4/3, B = −1 Ω, C = −5 S, D = −7/2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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