Synthesize LC impedance in Foster II and Cauer I

$Z(s) = \dfrac{(s^2+1)(s^2+3)}{s(s^2+2)} = \dfrac{s^4+4s^2+3}{s^3+2s}$ is a valid LC function: p 0, z 1, p 1.414, z 1.732, p ∞ alternate.

Foster-II realization

Take $Y(s) = 1/Z(s) = \frac{s^{3} + 2 s}{s^{4} + 4 s^{2} + 3}$ and expand $Y(s)$ in partial fractions (each term is an admittance; the branches are in parallel):

$$\begin{aligned} 2k_{1} &= \left[\frac{(s^2+1)\,Y(s)}{s}\right]_{s^2=-1} = \frac{1}{2} \\ 2k_{2} &= \left[\frac{(s^2+3)\,Y(s)}{s}\right]_{s^2=-3} = \frac{1}{2} \end{aligned}$$ $$Y(s) = \frac{s}{2 \left(s^{2} + 1\right)} + \frac{s}{2 \left(s^{2} + 3\right)}$$

o--+-------+
   |       |
  [L1]    [L2]
  [C1]    [C2]
   |       |
o--+-------+

Cauer-I realization

$Z(s)$ 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. $Z_1(s) = \frac{s^{4} + 4 s^{2} + 3}{s^{3} + 2 s}$ → quotient $s$ gives series $L_1$ = 1 H; remainder $\frac{2 s^{2} + 3}{s \left(s^{2} + 2\right)}$.
2. $Y_2(s) = \frac{s^{3} + 2 s}{2 s^{2} + 3}$ → quotient $\frac{s}{2}$ gives shunt $C_2$ = 1/2 F (≈ 0.5 F); remainder $\frac{s}{2 \left(2 s^{2} + 3\right)}$.
3. $Z_3(s) = \frac{4 s^{2} + 6}{s}$ → quotient $4 s$ gives series $L_3$ = 4 H; remainder $\frac{6}{s}$.
4. $Y_4(s) = \frac{s}{6}$ → quotient $\frac{s}{6}$ gives shunt $C_4$ = 1/6 F (≈ 0.1667 F); remainder $0$.

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