Current and voltage ratio transfer functions, 2 H and 3 H

The current ratio transfer function is α21 = I2/I1 and the voltage ratio transfer function is G21 = V2/V1, with the 1 Ω load connected across port 2. Take I2 as the current into the load (I2 = V2/1 = V2, opposite to the arrow drawn); with the arrow as drawn, α21 changes sign.

Figure (assumption). Series arm: 2 H then 3 H, with a 2 Ω resistor bridging across both (port 1 side to port 2 side); a ½ F capacitor shunt at the junction M of the two inductors (the scan reads ½ F); 1 Ω load across port 2.

      +---------/\/\---------+   2Ω
      |                      |
 o----+----mmm---+----mmm----+----o
           2H    |     3H        |
                ½F              1Ω load
 o---------------+---------------+--o

Node equations

Admittances: 2H → 1/(2s), 3H → 1/(3s), bridge 2 Ω → 1/2, capacitor → s/2, load → 1.

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

Eliminating VM (solved in Python):

G21 = V2/V1 = (3s³ + 5s + 2) / (9s³ + 2s² + 15s + 2)

Current ratio

α21 = I2/I1 = V2/I1 = 2(3s³ + 5s + 2) / [(2s + 1)(3s² + 4s + 4)]
            = (6s³ + 10s + 4) / (6s³ + 11s² + 12s + 4)

Answer: G21(s) = (3s³ + 5s + 2)/(9s³ + 2s² + 15s + 2); α21(s) = (6s³ + 10s + 4)/(6s³ + 11s² + 12s + 4).

Check: at s = 0, G21 = 2/2 = 1 and α21 = 4/4 = 1, correct because at d.c. the inductors are short-circuits.