G21 and α21 of two-port terminated in 2 Ω

For the terminated network, G21 = V2/V1 and α21 = I2/I1, where I2 is taken as the current delivered into the 2 Ω load (I2 = V2/2, opposite to the arrow drawn at port 2). With the arrow as drawn the sign of α21 is reversed.

Figure (assumption). Series arm: 1 Ω then 2 H, with a 1/3 F capacitor bridging from the port 1 side to the port 2 side across both; a 5 Ω shunt at the middle node M. Admittances: bridge capacitor Yb = s/3, 2 H: 1/(2s), shunt 1/5 S, load 1/2 S.

      +-------||-------+   1/3 F
      |                |
 o----+--/\/\--+--mm---+----o
      1Ω       |  2H       |
              5Ω          2Ω (load)
 o-------------+-----------+--o

Node equations

M: (VM − V1)/1 + (VM − V2)/(2s) + VM/5 = 0
O: (V2 − VM)/(2s) + (V2 − V1)(s/3) + V2/2 = 0
I1 = (V1 − VM)/1 + (V1 − V2)(s/3)

From M: VM(1 + 1/(2s) + 1/5) = V1 + V2/(2s). Substituting in O and simplifying (done with Python) gives:

G21 = V2/V1 = 2(12s² + 5s + 15) / (24s² + 46s + 51)

Current ratio

I2 = V2/2,   I1 from the third equation
α21 = I2/I1 = (12s² + 5s + 15) / (16s² + 13s + 21)

Check at s = 0: capacitor open, so V2/V1 = (5∥2 = 10/7)/(1 + 10/7) = 10/17 = 30/51, which matches G21(0). Also α21(0) = 15/21 = 5/7, matching the current divider 5/(5 + 2).

Answer: (i) G21(s) = 2(12s² + 5s + 15)/(24s² + 46s + 51); (ii) α21(s) = (12s² + 5s + 15)/(16s² + 13s + 21).