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RC impedance properties; Foster and Cauer synthesis

Short answer

RC impedances have simple negative-real-axis poles and zeros that alternate, with lowest critical a pole and highest a zero. For Z(s)=3(s+2)(s+4)/(s(s+3)), Foster-I gives series R=3Ω, C=0.125F, parallel R=0.333Ω||C=1F. Cauer-I yields series R=3Ω, shunt C=0.111F, series R=27Ω, shunt C=0.0139F.

What are the properties of RC impedance function? Synthesize the given RC impedance in Foster and Cauer form. Z(s) = 3(s+2)(s+4)/(s(s+3))

Properties of RC impedance function

An RC driving-point impedance ZRC(s)Z_{RC}(s) has these properties:

  1. All poles and zeros are simple and lie on the negative real axis of the ss-plane (including the origin).
  2. Poles and zeros alternate along the negative real axis.
  3. The critical frequency nearest the origin (lowest) is a pole; it may be at s=0s=0.
  4. The critical frequency farthest from the origin is a zero; it may be at s=∞s=\infty.
  5. ZRC(0)>ZRC(∞)Z_{RC}(0) > Z_{RC}(\infty), and ZRC(∞)≥0Z_{RC}(\infty)\ge 0 is a constant (no pole at infinity).
  6. Residues of ZRC(s)Z_{RC}(s) at its poles are real and positive; on the real axis dZ(σ)/dσ<0dZ(\sigma)/d\sigma<0.
  7. The same form describes an RL admittance YRL(s)Y_{RL}(s).

Checking the given function

Z(s)=3(s+2)(s+4)s(s+3)=3s2+18s+24s2+3sZ(s) = \dfrac{3(s+2)(s+4)}{s(s+3)} = \dfrac{3s^2+18s+24}{s^2+3s}: poles at 0,−30, -3; zeros at −2,−4-2, -4. Order p 0, z 2, p 3, z 4 — alternating, pole nearest origin, zero farthest; Z(0)=∞>Z(∞)=3Z(0)=\infty > Z(\infty)=3. Valid RC impedance.

Foster (Foster-I) form

Expand Z(s)Z(s) in partial fractions; every residue must be positive:

k∞=Z(∞)=3k0=[s Z(s)]s=0=8k3=[(s+3)Z(s)]s=−3=1\begin{aligned} k_\infty &= Z(\infty) = 3 \\ k_0 &= [s\,Z(s)]_{s=0} = 8 \\ k_{3} &= [(s+3)Z(s)]_{s=-3} = 1 \end{aligned} Z(s)=3+8s+1s+3Z(s) = 3 + \frac{8}{s} + \frac{1}{s + 3}
TermRuleElement value
R1R_1 (series)k∞k_\infty3 Ω
C2C_2 (series)1/k01/k_01/8 F (≈ 0.125 F)
R3∥C3R_3 \parallel C_3 (pole at −3-3)C=1/kC=1/k, R=k/σR=k/\sigmaR3R_3 = 1/3 Ω (≈ 0.3333 Ω), C3C_3 = 1 F
o--[R1]--[C2]--[R3||C3]--o

Cauer (Cauer-I) form

Divide with descending powers (continued fraction about s=∞s=\infty). For an RC impedance the quotients alternate between a constant (series RR) and ksks (shunt CC):

  1. Z1(s)=3s2+18s+24s2+3sZ_1(s) = \frac{3 s^{2} + 18 s + 24}{s^{2} + 3 s} → remove 33, giving series R1R_1 = 3 Ω; remainder 3(3s+8)s(s+3)\frac{3 \left(3 s + 8\right)}{s \left(s + 3\right)}.
  2. Y2(s)=s2+3s9s+24Y_2(s) = \frac{s^{2} + 3 s}{9 s + 24} → remove s9\frac{s}{9}, giving shunt C2C_2 = 1/9 F (≈ 0.1111 F); remainder s9(3s+8)\frac{s}{9 \left(3 s + 8\right)}.
  3. Z3(s)=27s+72sZ_3(s) = \frac{27 s + 72}{s} → remove 2727, giving series R3R_3 = 27 Ω; remainder 72s\frac{72}{s}.
  4. Y4(s)=s72Y_4(s) = \frac{s}{72} → remove s72\frac{s}{72}, giving shunt C4C_4 = 1/72 F (≈ 0.01389 F); remainder 00.
ElementPositionValue
R1R_1series3 Ω
C2C_2shunt1/9 F (≈ 0.1111 F)
R3R_3series27 Ω
C4C_4shunt1/72 F (≈ 0.01389 F)
o--[R1]----+---[R3]----+
           |           |
         [C2]        [C4]
           |           |
o----------+-----------+

✓ Answer checked by Claude Sonnet 5.5

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