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Driving point impedance and voltage ratio, two 0.5 F

Short answer

With port 2 open, cascade two RC sections. Driving-point impedance Z11(s)=(s²+6s+4)/[s(s+4)] and voltage ratio G21(s)=4/(s²+6s+4).

For the given two port network, determine the driving point impedance and voltage ratio transfer function. [Figure: port 1 (V1, I1) with a 1 Ω series resistor, a 0.5 F shunt capacitor, another 1 Ω series resistor and a second 0.5 F shunt capacitor at port 2 (V2, I2)]

Driving point impedance Z11 = V1/I1 and G21 = V2/V1, both with port 2 open (I2 = 0). Each 0.5 F capacitor has impedance 1/(0.5s) = 2/s.

 I1   1Ω       1Ω      I2=0
 o---/\/\--+--/\/\--+---o
           |        |
          0.5F     0.5F      V2 (open)
           |        |
 o---------+--------+---o

Driving point impedance

Looking from port 1, the second section is 1 Ω in series with 2/s, in parallel with the first capacitor:

Z_B = 2/s ∥ (1 + 2/s) = (2/s)(s + 2)/s ÷ [(s + 4)/s]
    = 2(s + 2) / [s(s + 4)]

Z11 = 1 + Z_B = [s(s + 4) + 2(s + 2)] / [s(s + 4)]

Z11(s) = (s² + 6s + 4) / [s(s + 4)]

Voltage ratio

Two voltage dividers in cascade, because no current flows into the open port 2:

V2/Vx = (2/s) / (1 + 2/s) = 2/(s + 2)          (Vx = voltage across first C)
Vx/V1 = Z_B / (1 + Z_B) = 2(s + 2) / (s² + 6s + 4)

G21 = V2/V1 = 2/(s+2) · 2(s+2)/(s²+6s+4)

G21(s) = 4 / (s² + 6s + 4)

(Check: at s = 0, G21 = 1, as expected because capacitors are open at d.c. and no current flows.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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