Transmission and y parameters, prove not reciprocal or symmetrical

Transmission parameters: V1 = A V2 − B I2, I1 = C V2 − D I2. Admittance parameters: I1 = y11V1 + y12V2, I2 = y21V1 + y22V2.

Reading the figure (assumption). 2 Ω series from port 1 to node a; 1 Ω shunt at node a; a dependent voltage source 2I1 in the series arm with + on the port 1 side (Va − Vb = 2I1), reaching node b; 3 Ω shunt at node b; 1 Ω series resistor to port 2.

Equations

Let Is be the current through the source (from a to b).

Va = V1 − 2 I1                       (2 Ω drop)
Vb = Va − 2 I1 = V1 − 4 I1           (controlled source)
Node a:  I1 = Va/1 + Is    →  Is = I1 − (V1 − 2 I1) = 3 I1 − V1
Node b:  Is = Vb/3 + (Vb − V2)/1
Port 2:  I2 = V2 − Vb                (current through the 1 Ω into port 2)

Solving (first for Z)

From port 2: V2 = Vb + I2. From node b: Is = Vb/3 − I2, so Vb = 3 Is + 3 I2 = 9 I1 − 3 V1 + 3 I2. Equate with Vb = V1 − 4 I1:

V1 − 4 I1 = 9 I1 − 3 V1 + 3 I2   →   V1 = (13/4) I1 + (3/4) I2
V2 = Vb + I2 = V1 − 4 I1 + I2     →   V2 = −(3/4) I1 + (7/4) I2

So Z = [13/4 3/4; −3/4 7/4] Ω and Δz = (13/4)(7/4) + (3/4)(3/4) = 25/4.

y-parameters (Y = Z⁻¹)

y11 = z22/Δz = 7/25 = 0.28 S       y12 = −z12/Δz = −3/25 = −0.12 S
y21 = −z21/Δz = 3/25 = 0.12 S      y22 = z11/Δz = 13/25 = 0.52 S

Transmission parameters

A = −y22/y21 = −13/3          B = −1/y21 = −25/3 Ω
C = −Δy/y21 = −4/3 S          D = −y11/y21 = −7/3          (Δy = 4/25)

Reciprocity and symmetry

• Reciprocal needs y12 = y21 (or AD − BC = 1). Here y12 = −0.12 ≠ y21 = 0.12, and AD − BC = (−13/3)(−7/3) − (−25/3)(−4/3) = 91/9 − 100/9 = −1 ≠ 1. Not reciprocal.
• Symmetrical needs A = D (or y11 = y22): −13/3 ≠ −7/3. Not symmetrical.

Answer: T = [−13/3 −25/3 Ω; −4/3 S −7/3]; Y = [0.28 −0.12; 0.12 0.52] S; the network is neither reciprocal nor symmetrical.