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G21(S) and Z21(S) of two-port with 0.5 F, 1 H

Short answer

With port 2 open, series impedance Za=2(s+3)/(s+2), shunt Zb=s+1. Z21=Zb=s+1, Z11=Za+Zb=(s²+5s+8)/(s+2). Thus G21=(s+1)(s+2)/(s²+5s+8).

For the 2-port network shown in figure below, find voltage ratio transfer function, G21(S) and transfer admittance, Z21(S). [Figure: series arm at port 1: a 0.5 F capacitor in parallel with a 1 Ω resistor, followed by a 2 Ω series resistor. Shunt branch: a 1 Ω resistor in series with a 1 H inductor to the bottom wire. Ports: V1, I1 and V2, I2. The scan prints "transfer admittance, Z21(S)".]

G21 = V2/V1 and Z21 = V2/I1, both with port 2 open (I2 = 0). The scan calls Z21 a "transfer admittance", but Z21 is the transfer impedance; the transfer admittance Y21 = I2/V1 (V2 = 0) is also found below.

         0.5F
      +---||---+     2Ω
 I1   |        +---/\/\---+-------o
 o----+        |          |
      +--/\/\--+         1Ω   (1 Ω + 1 H in series)
          1Ω              |
                         1H     V2
 V1                       |
 o------------------------+-------o

Series arm

Za = (1 ∥ 0.5F) + 2 = [1·(2/s)] / [1 + 2/s] + 2 = 2/(s + 2) + 2 = 2(s + 3)/(s + 2)

Shunt arm

Zb = 1 + s

Open circuit functions

Z11 = Za + Zb = [2(s + 3) + (s + 1)(s + 2)] / (s + 2) = (s² + 5s + 8)/(s + 2)
V2  = I1·Zb   →   Z21 = Zb = s + 1
G21 = Z21/Z11 = (s + 1)(s + 2) / (s² + 5s + 8)

Transfer admittance (short-circuited port 2)

With V2 = 0 the shunt arm is shorted, so the current through the series arm is I2 = −V1/Za:

Y21 = −1/Za = −(s + 2) / [2(s + 3)]

Answer: G21(s) = (s + 1)(s + 2)/(s² + 5s + 8); Z21(s) = s + 1 (and Y21(s) = −(s + 2)/[2(s + 3)]).

✓ Answer checked by Claude Opus 5.5 (high effort)

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