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RC impedance properties; Foster form

Short answer

RC impedance: poles and zeros on negative real axis, simple, alternating; lowest is a pole, highest a zero; Z(0)>Z(∞). Valid: Z(s)=(s+3)(s+6)/((s+1)(s+5)). Foster I: series 1 Ω, then parallel branches R=2.5 Ω, C=0.4 F and R=0.1 Ω, C=2 F.

What are the properties of RC driving point impedance function? Which of the following functions are valid RC driving point impedance function and why? Z(s) = (s+3)(s+6)/((s+1)(s+5)); Z(s) = 2(s+1)(s+3)/((s+2)(s+4)). Find Foster form of valid RC driving point impedance function.

Properties of RC driving-point impedance

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 functions

FunctionCritical frequenciesResult
(a) (s+3)(s+6)(s+1)(s+5)\dfrac{(s+3)(s+6)}{(s+1)(s+5)}p 1, z 3, p 5, z 6Lowest is a pole, highest a zero, alternate; Z(0)=3.6>Z(∞)=1Z(0)=3.6 > Z(\infty)=1 → valid RC
(b) 2(s+1)(s+3)(s+2)(s+4)\dfrac{2(s+1)(s+3)}{(s+2)(s+4)}z 1, p 2, z 3, p 4Lowest critical frequency is a zero and Z(0)=0.75<Z(∞)=2Z(0)=0.75 < Z(\infty)=2 → not RC impedance (it is an RL impedance / RC admittance)

Foster form of (a)

Foster-I (series) form: Expand Z(s)Z(s) in partial fractions; every residue must be positive:

k∞=Z(∞)=1k2=[(s+1)Z(s)]s=−1=52k3=[(s+5)Z(s)]s=−5=12\begin{aligned} k_\infty &= Z(\infty) = 1 \\ k_{2} &= [(s+1)Z(s)]_{s=-1} = \frac{5}{2} \\ k_{3} &= [(s+5)Z(s)]_{s=-5} = \frac{1}{2} \end{aligned} Z(s)=1+52(s+1)+12(s+5)Z(s) = 1 + \frac{5}{2 \left(s + 1\right)} + \frac{1}{2 \left(s + 5\right)}
TermRuleElement value
R1R_1 (series)k∞k_\infty1 Ω
R2∥C2R_2 \parallel C_2 (pole at −1-1)C=1/kC=1/k, R=k/σR=k/\sigmaR2R_2 = 5/2 Ω (≈ 2.5 Ω), C2C_2 = 2/5 F (≈ 0.4 F)
R3∥C3R_3 \parallel C_3 (pole at −5-5)C=1/kC=1/k, R=k/σR=k/\sigmaR3R_3 = 1/10 Ω (≈ 0.1 Ω), C3C_3 = 2 F
o--[R1]--[R2||C2]--[R3||C3]--o

Foster-II (parallel) form: Take Y(s)=1/Z(s)=s2+6s+5s2+9s+18Y(s)=1/Z(s) = \frac{s^{2} + 6 s + 5}{s^{2} + 9 s + 18}. Expand Y(s)/sY(s)/s in partial fractions and multiply back by ss:

k0=Y(0)=518k2=[(s+3)Y(s)s]s=−3=49k3=[(s+6)Y(s)s]s=−6=518\begin{aligned} k_0 &= Y(0) = \frac{5}{18} \\ k_{2} &= \left[\frac{(s+3)Y(s)}{s}\right]_{s=-3} = \frac{4}{9} \\ k_{3} &= \left[\frac{(s+6)Y(s)}{s}\right]_{s=-6} = \frac{5}{18} \end{aligned} Y(s)=518+4s9(s+3)+5s18(s+6)Y(s) = \frac{5}{18} + \frac{4 s}{9 \left(s + 3\right)} + \frac{5 s}{18 \left(s + 6\right)}
TermRuleElement value
R1R_1 (shunt)1/k01/k_018/5 Ω (≈ 3.6 Ω)
R2R_2–C2C_2 in series (pole at −3-3)R=1/kR=1/k, C=k/σC=k/\sigmaR2R_2 = 9/4 Ω (≈ 2.25 Ω), C2C_2 = 4/27 F (≈ 0.1481 F)
R3R_3–C3C_3 in series (pole at −6-6)R=1/kR=1/k, C=k/σC=k/\sigmaR3R_3 = 18/5 Ω (≈ 3.6 Ω), C3C_3 = 5/108 F (≈ 0.0463 F)
o--+-------+-------+
   |       |       |
  [R1]    [R2]    [R3]
   |      [C2]    [C3]
   |       |       |
o--+-------+-------+

✓ Answer checked by Claude Sonnet 5.5

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