Lossless impedance check; Foster I and Cauer I

Testing the functions

Tests for a valid LC impedance: (1) even/odd or odd/even ratio, (2) simple poles and zeros only on the $j\omega$ axis, (3) poles and zeros alternate, (4) a pole or a zero at both $s=0$ and $s=\infty$ (degrees differ by exactly 1), (5) positive residues.

Synthesize (c): $Z(s) = \dfrac{2s^4+20s^2+18}{s^3+4s}$.

Foster-I form

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

$$\begin{aligned} k_\infty &= \left[\frac{Z(s)}{s}\right]_{s\to\infty} = 2 \\ k_0 &= \left[s\,Z(s)\right]_{s=0} = \frac{9}{2} \\ 2k_{3} &= \left[\frac{(s^2+4)\,Z(s)}{s}\right]_{s^2=-4} = \frac{15}{2} \end{aligned}$$ $$Z(s) = 2 s + \frac{9}{2 s} + \frac{15 s}{2 \left(s^{2} + 4\right)}$$

o--[L1]--[C2]--[L3||C3]--o

Cauer-I form

$Z(s)$ has a pole at $s=\infty$ (numerator degree is higher), so the first element is a series inductor. Divide repeatedly with polynomials in descending powers (continued fraction about $s=\infty$), inverting the remainder each time:

1. $Z_1(s) = \frac{2 s^{4} + 20 s^{2} + 18}{s^{3} + 4 s}$ → quotient $2 s$ gives series $L_1$ = 2 H; remainder $\frac{6 \left(2 s^{2} + 3\right)}{s \left(s^{2} + 4\right)}$.
2. $Y_2(s) = \frac{s^{3} + 4 s}{12 s^{2} + 18}$ → quotient $\frac{s}{12}$ gives shunt $C_2$ = 1/12 F (≈ 0.08333 F); remainder $\frac{5 s}{12 \left(2 s^{2} + 3\right)}$.
3. $Z_3(s) = \frac{24 s^{2} + 36}{5 s}$ → quotient $\frac{24 s}{5}$ gives series $L_3$ = 24/5 H (≈ 4.8 H); remainder $\frac{36}{5 s}$.
4. $Y_4(s) = \frac{5 s}{36}$ → quotient $\frac{5 s}{36}$ gives shunt $C_4$ = 5/36 F (≈ 0.1389 F); remainder $0$.

o--[L1]----+---[L3]----+
           |           |
         [C2]        [C4]
           |           |
o----------+-----------+