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LC impedance properties; Foster I of valid function

What are the properties of LC driving point impedance function? Which of the following is valid lossless impedance function and why? a) Z(s) = 2(s²+1)(s²+9)/(s(s²+4)) b) Z(s) = (s+4)(s+6)/((s+3)(s+5)) c) Z(s) = (s²+1)(s²+4)/(s(s²+9)). Also, find the Foster I form of valid lossless impedance function.

Properties of LC driving-point impedance

An LC (lossless) driving-point immittance Z(s)Z(s) or Y(s)Y(s) has these properties:

  1. It is the ratio of an even polynomial to an odd polynomial, or odd to even, with real, positive coefficients.
  2. All poles and zeros are simple and lie on the jωj\omega axis (they occur in conjugate pairs ±jωi\pm j\omega_i).
  3. Poles and zeros alternate (interlace) along the jωj\omega axis.
  4. The highest powers of numerator and denominator differ by exactly 1; so do the lowest powers.
  5. There is always either a pole or a zero at s=0s=0, and either a pole or a zero at s=∞s=\infty.
  6. The residues at all poles are real and positive.
  7. On the jωj\omega axis Z(jω)=jX(ω)Z(j\omega)=jX(\omega) is purely reactive and dX/dω>0dX/d\omega>0 (Foster's reactance theorem).

Checking the functions

FunctionCritical frequenciesResult
(a) 2(s2+1)(s2+9)s(s2+4)\dfrac{2(s^2+1)(s^2+9)}{s(s^2+4)}p 0, z 1, p 2, z 3, p ∞Even/odd, simple jωj\omega-axis roots that alternate → valid
(b) (s+4)(s+6)(s+3)(s+5)\dfrac{(s+4)(s+6)}{(s+3)(s+5)}on negative real axis: p 3, z 4, p 5, z 6Not on jωj\omega axis, not even/odd → not lossless (RC-impedance type)
(c) (s2+1)(s2+4)s(s2+9)\dfrac{(s^2+1)(s^2+4)}{s(s^2+9)}p 0, z 1, z 2, p 3Zeros 1 and 2 adjacent → invalid

Foster-I form of (a)

Expand Z(s)Z(s) in partial fractions (each term is an impedance; the elements are in series):

k∞=[Z(s)s]s→∞=2k0=[s Z(s)]s=0=922k3=[(s2+4) Z(s)s]s2=−4=152\begin{aligned} k_\infty &= \left[\frac{Z(s)}{s}\right]_{s\to\infty} = 2 \\ k_0 &= \left[s\,Z(s)\right]_{s=0} = \frac{9}{2} \\ 2k_{3} &= \left[\frac{(s^2+4)\,Z(s)}{s}\right]_{s^2=-4} = \frac{15}{2} \end{aligned} Z(s)=2s+92s+15s2(s2+4)Z(s) = 2 s + \frac{9}{2 s} + \frac{15 s}{2 \left(s^{2} + 4\right)}
TermRuleElement value
L1L_1 (series, pole at ∞\infty)k∞k_\infty2 H
C2C_2 (series, pole at 0)1/k01/k_02/9 F (≈ 0.2222 F)
L3∥C3L_3 \parallel C_3 (tank, ω2=4\omega^2=4)C=1/2kC=1/2k, L=2k/ω2L=2k/\omega^2C3C_3 = 2/15 F (≈ 0.1333 F), L3L_3 = 15/8 H (≈ 1.875 H)
o--[L1]--[C2]--[L3||C3]--o

✓ Answer checked by Claude Sonnet 5.5

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