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Voltage ratio transfer function of two-section LC network

Short answer

Voltage ratio transfer function with port 2 open: G21(s) = (s^2+1)/(s^2+2). The first shunt inductor is directly across the source and irrelevant; the transfer is a simple voltage divider between the series parallel LC arm and the second shunt inductor.

For the given 2-port network shown in figure below, find the voltage ratio transfer function. [Figure: port 1 (V1(S), I1(S)); a 1 H inductor as a shunt branch; a series arm of 1 H in parallel with 1 F; a second 1 H inductor as a shunt branch; a second series arm of 1 H in parallel with 1 F; port 2 (V2(S), I2(S))]

G21 = V2/V1 with port 2 open (I2 = 0, assumption, since no load is given).

Notation. The 1 H and 1 F in each series arm are in parallel:

Zp = (s · 1/s) / (s + 1/s) = s / (s² + 1)

 o--+--------[ 1H ∥ 1F ]---+--------[ 1H ∥ 1F ]---o
    |                      |                       open
   1 H                    1 H                      V2
    |                      |
 o--+----------------------+------------------------o

Method

  1. The first shunt 1 H sits directly across the source V1, so it has no effect on the transfer function.
  2. With port 2 open the last series arm carries no current, so V2 equals the voltage across the second shunt 1 H.
  3. The voltage divider formed by the first series arm Zp and the second shunt inductor sL (= s) gives:
V2/V1 = s / (s + Zp) = s / [s + s/(s² + 1)]
      = (s² + 1) / (s² + 2)

Answer: G21(s) = V2/V1 = (s² + 1)/(s² + 2).

Zeros at s = ±j1: at ω = 1 rad/s the parallel 1 H ∥ 1 F arm is at resonance and has infinite impedance, so no signal reaches the output (G21 = 0). Poles at s = ±j√2 (on the jω axis because the circuit is lossless). G21(0) = 1/2 and G21 → 1 as s → ∞.

✓ Answer checked by Claude Opus 5.5 (high effort)

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