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G21(s) and Y21(s) of ladder with bridging capacitor

Short answer

Use nodal analysis with port 2 open for G21 and shorted for Y21. Results: G21(s)=2s(3s²+50s+102)/(3s⁴+66s³+328s²+488s+44); Y21(s)=−s(3s²+50s+102)/(15s⁴+236s³+1006s²+1384s+220).

Find the forward voltage ratio transfer function G21(s) and forward transfer admittance Y21(s) in the following circuit. [Figure: port 1 (V1, I1) followed by a 1 Ω series resistor to node A. From A to node B: a 10 Ω resistor in series with a 1 H inductor. From B to the port 2 terminal: a 10 Ω resistor in series with a 1 H inductor. A 1/2 F capacitor in series with a 10 Ω resistor runs from node A to the port 2 terminal (as drawn, passing across node B). Shunt elements: a 1/2 F capacitor at node A and a 1 H inductor at node B to the bottom wire. Port 2 has V2, I2.]

G21 = V2/V1 with port 2 open (I2 = 0). Y21 = I2/V1 with port 2 short-circuited (V2 = 0), I2 being the current entering port 2.

Reading of the figure (assumption). Name the node after the 1 Ω port resistor P, the node at the 1 H shunt N and the port 2 top terminal O. The elements are: P–N: 10 Ω + 1 H in series; N–O: 10 Ω + 1 H in series; shunt at P: ½ F; shunt at N: 1 H; and the ½ F capacitor in series with the 10 Ω resistor runs from P to O. This lower branch only crosses the vertical wire of N (no dot at the crossing), so it is not connected to N.

Branch admittances

yPN = yNO = 1/(s + 10)           (10 Ω + 1 H)
yPO = 1/(10 + 2/s) = s/(10s + 2) (10 Ω + ½ F)
shunt at P: s/2                  shunt at N: 1/s

Node equations (V1 applied through the 1 Ω resistor)

P: (VP − V1)·1 + (s/2)VP + (VP − VN)yPN + (VP − VO)yPO = 0
N: (VN − VP)yPN + VN/s + (VN − VO)yNO = 0
O: (VO − VN)yNO + (VO − VP)yPO = I2

G21 (I2 = 0)

Solving the three equations (Cramer's rule, checked in Python):

G21 = 2s(3s² + 50s + 102) / (3s⁴ + 66s³ + 328s² + 488s + 44)

Y21 (VO = 0)

Put VO = 0, solve P and N for VP and VN, then I2 = −[VN·yNO + VP·yPO] (current flowing into the shorted port, so it is negative of the current leaving it):

Y21 = I2/V1 = − s(3s² + 50s + 102) / (15s⁴ + 236s³ + 1006s² + 1384s + 220)

Answer: G21(s) = 2s(3s² + 50s + 102)/(3s⁴ + 66s³ + 328s² + 488s + 44); Y21(s) = −s(3s² + 50s + 102)/(15s⁴ + 236s³ + 1006s² + 1384s + 220).

Both have a zero at s = 0 (the inductors short d.c. at the output side) and share the same numerator polynomial.

✓ Answer checked by Claude Opus 5.5 (high effort)

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