Cauer-I and Cauer-II realization

$$Z(s) = \frac{4s^4+40s^2+36}{s^3+4s} = \frac{4(s^2+1)(s^2+9)}{s(s^2+4)}$$

This is a valid LC function (even/odd; p 0, z 1, p 2, z 3, p ∞ alternate), so it has a pole at $s=\infty$ (Cauer-I starts with series $L$) and a pole at $s=0$ (Cauer-II starts with series $C$).

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. $Z_1(s) = \frac{4 s^{4} + 40 s^{2} + 36}{s^{3} + 4 s}$ → quotient $4 s$ gives series $L_1$ = 4 H; remainder $\frac{12 \left(2 s^{2} + 3\right)}{s \left(s^{2} + 4\right)}$.
2. $Y_2(s) = \frac{s^{3} + 4 s}{24 s^{2} + 36}$ → quotient $\frac{s}{24}$ gives shunt $C_2$ = 1/24 F (≈ 0.04167 F); remainder $\frac{5 s}{24 \left(2 s^{2} + 3\right)}$.
3. $Z_3(s) = \frac{48 s^{2} + 72}{5 s}$ → quotient $\frac{48 s}{5}$ gives series $L_3$ = 48/5 H (≈ 9.6 H); remainder $\frac{72}{5 s}$.
4. $Y_4(s) = \frac{5 s}{72}$ → quotient $\frac{5 s}{72}$ gives shunt $C_4$ = 5/72 F (≈ 0.06944 F); remainder $0$.

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

Cauer-II realization

$Z(s)$ has a pole at $s=0$, so the first element is a series capacitor. Arrange numerator and denominator in ascending powers of $s$ and divide repeatedly (continued fraction about $s=0$); each quotient is $k/s$:

1. $Z_1(s) = \frac{36 + 40 s^{2} + 4 s^{4}}{4 s + s^{3}}$ → quotient $\frac{9}{s}$ gives series $C_1$ = 1/9 F (≈ 0.1111 F); remainder $\frac{s \left(4 s^{2} + 31\right)}{s^{2} + 4}$.
2. $Y_2(s) = \frac{4 + s^{2}}{31 s + 4 s^{3}}$ → quotient $\frac{4}{31 s}$ gives shunt $L_2$ = 31/4 H (≈ 7.75 H); remainder $\frac{15 s}{31 \left(4 s^{2} + 31\right)}$.
3. $Z_3(s) = \frac{961 + 124 s^{2}}{15 s}$ → quotient $\frac{961}{15 s}$ gives series $C_3$ = 15/961 F (≈ 0.01561 F); remainder $\frac{124 s}{15}$.
4. $Y_4(s) = \frac{15}{124 s}$ → quotient $\frac{15}{124 s}$ gives shunt $L_4$ = 124/15 H (≈ 8.267 H); remainder $0$.

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