Identify LC lossless function; Foster-I and Cauer-I

Checking the functions

An LC function must be an even/odd (or odd/even) ratio with simple poles and zeros on the $j\omega$ axis that alternate, and it must have a pole or zero at both $s=0$ and $s=\infty$ (degrees differ by exactly 1).

So $Z_2(s) = \dfrac{(s^2+3)(s^2+6)}{s(s^2+4)(s^2+9)} = \dfrac{s^4+9s^2+18}{s^5+13s^3+36s}$ is realized.

Foster-I realization

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

$$\begin{aligned} k_0 &= \left[s\,Z(s)\right]_{s=0} = \frac{1}{2} \\ 2k_{2} &= \left[\frac{(s^2+4)\,Z(s)}{s}\right]_{s^2=-4} = \frac{1}{10} \\ 2k_{3} &= \left[\frac{(s^2+9)\,Z(s)}{s}\right]_{s^2=-9} = \frac{2}{5} \end{aligned}$$ $$Z(s) = \frac{1}{2 s} + \frac{s}{10 \left(s^{2} + 4\right)} + \frac{2 s}{5 \left(s^{2} + 9\right)}$$

o--[C1]--[L2||C2]--[L3||C3]--o

Cauer-I realization

$Z(s)$ has a zero at $s=\infty$, so $Y(s)=1/Z(s)$ has a pole at infinity and the first element is a shunt capacitor. Divide repeatedly with polynomials in descending powers (continued fraction about $s=\infty$), inverting the remainder each time:

1. $Y_1(s) = \frac{s^{5} + 13 s^{3} + 36 s}{s^{4} + 9 s^{2} + 18}$ → quotient $s$ gives shunt $C_1$ = 1 F; remainder $\frac{2 s \left(2 s^{2} + 9\right)}{\left(s^{2} + 3\right) \left(s^{2} + 6\right)}$.
2. $Z_2(s) = \frac{s^{4} + 9 s^{2} + 18}{4 s^{3} + 18 s}$ → quotient $\frac{s}{4}$ gives series $L_2$ = 1/4 H (≈ 0.25 H); remainder $\frac{9 \left(s^{2} + 4\right)}{4 s \left(2 s^{2} + 9\right)}$.
3. $Y_3(s) = \frac{8 s^{3} + 36 s}{9 s^{2} + 36}$ → quotient $\frac{8 s}{9}$ gives shunt $C_3$ = 8/9 F (≈ 0.8889 F); remainder $\frac{4 s}{9 \left(s^{2} + 4\right)}$.
4. $Z_4(s) = \frac{9 s^{2} + 36}{4 s}$ → quotient $\frac{9 s}{4}$ gives series $L_4$ = 9/4 H (≈ 2.25 H); remainder $\frac{9}{s}$.
5. $Y_5(s) = \frac{s}{9}$ → quotient $\frac{s}{9}$ gives shunt $C_5$ = 1/9 F (≈ 0.1111 F); remainder $0$.

o----+---[L2]----+---[L4]----+
     |           |           |
   [C1]        [C3]        [C5]
     |           |           |
o----+-----------+-----------+

Both networks have five reactive elements, the minimum (canonical) number for this fifth-order function.