Transmission and Y parameters, reciprocity and symmetry

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

Reading the figure (assumption). At port 1 a 1 Ω resistor and a current source 2V2 (arrow up, i.e. into the top node) are in shunt. A 1 Ω resistor then a voltage source 2V1 (− on the left, + on the right) are in the series arm. A 2 Ω shunt resistor is across port 2.

 I1   +--------+------/\/\------(-+ 2V1)----+------+---- I2
 o----|        |        1Ω                   |      |
      1Ω     (2V2 ↑)                         |     2Ω
 o----+--------+-----------------------------+------+----

Y-parameters

Let Is be the current through the series 1 Ω resistor from the port 1 node towards port 2.

KCL at port 1 node:    I1 + 2V2 = V1/1 + Is
Series arm:            V2 = (V1 − 1·Is) + 2V1 = 3V1 − Is   →  Is = 3V1 − V2
KCL at port 2 node:    Is + I2 = V2/2

Therefore:

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

y11 = 4 S, y12 = −3 S, y21 = −3 S, y22 = 1.5 S

T-parameters

From I2 = y21V1 + y22V2: V1 = (1.5 V2 − I2)/3 = 0.5 V2 − (1/3) I2. Then I1 = 4V1 − 3V2 = −V2 − (4/3) I2.

A = 0.5     B = 1/3 Ω
C = −1 S    D = 4/3
(also A = −y22/y21, B = −1/y21, C = −Δy/y21, D = −y11/y21, with Δy = 6 − 9 = −3)

Reciprocity and symmetry

• Reciprocal: y12 = y21 = −3 S, and AD − BC = (0.5)(4/3) − (1/3)(−1) = 2/3 + 1/3 = 1. So the network is reciprocal (the two dependent sources happen to cancel in their effect on y12 and y21).
• Symmetrical: needs A = D (or y11 = y22). Here A = 0.5 ≠ D = 4/3 and y11 = 4 ≠ y22 = 1.5, so it is not symmetrical.

Answer: Y = [4 −3; −3 1.5] S; T = [0.5, 1/3; −1, 4/3]; reciprocal but not symmetrical.