T and Y parameters of bridged-T network

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

Circuit (from the figure). A 1 Ω resistor and a 2 Ω resistor in series in the top wire, a 3 Ω resistor connected across the pair (from port 1 node to port 2 node), and a 5 Ω shunt resistor at the junction M of the 1 Ω and 2 Ω resistors.

      +-------------/\/\-------------+   3Ω
      |                              |
 o----+----/\/\----+----/\/\---------+----o
           1Ω      |      2Ω
                   5Ω
 o-----------------+----------------------o

Node M

(VM − V1)/1 + (VM − V2)/2 + VM/5 = 0
VM (1 + 1/2 + 1/5) = V1 + V2/2
VM = (10 V1 + 5 V2)/17

Y-parameters

I1 = (V1 − VM)/1 + (V1 − V2)/3
   = (1 − 10/17 + 1/3) V1 + (−5/17 − 1/3) V2 = (38/51) V1 − (32/51) V2
I2 = (V2 − VM)/2 + (V2 − V1)/3
   = (−5/17 − 1/3) V1 + (1/2 − 5/34 + 1/3) V2 = −(32/51) V1 + (35/51) V2

y11 = 38/51 = 0.745 S    y12 = −32/51 = −0.627 S
y21 = −32/51 = −0.627 S  y22 = 35/51 = 0.686 S
Δy = (38·35 − 32²)/51² = 306/2601 = 2/17

T-parameters (from Y)

A = −y22/y21 = 35/32 = 1.094        B = −1/y21 = 51/32 = 1.594 Ω
C = −Δy/y21 = 3/16 = 0.1875 S       D = −y11/y21 = 19/16 = 1.1875

Reciprocity and symmetry

• y12 = y21 = −32/51 and AD − BC = (35/32)(19/16) − (51/32)(3/16) = (665 − 153)/512 = 1. The network is reciprocal (it contains only resistors).
• Symmetry needs y11 = y22 (or A = D). Here y11 = 38/51 ≠ y22 = 35/51 and A = 1.094 ≠ D = 1.1875, so it is not symmetrical (the 3 Ω bridge is symmetric, but the 1 Ω and 2 Ω arms are different).

Answer: Y = (1/51)[38 −32; −32 35] S; T: A = 35/32, B = 51/32 Ω, C = 3/16 S, D = 19/16; reciprocal, not symmetrical.