Lossless properties; Foster series and Cauer I

Properties of lossless one-port function

An LC (lossless) driving-point immittance $Z(s)$ or $Y(s)$ 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 $j\omega$ axis (they occur in conjugate pairs $\pm j\omega_i$).
3. Poles and zeros alternate (interlace) along the $j\omega$ 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 $s=0$, and either a pole or a zero at $s=\infty$.
6. The residues at all poles are real and positive.
7. On the $j\omega$ axis $Z(j\omega)=jX(\omega)$ is purely reactive and $dX/d\omega>0$ (Foster's reactance theorem).

Testing the functions

Only $Z_1(s) = \dfrac{2(s^2+1)(s^2+8)}{s(s^2+4)}$ is lossless.

Foster series (Foster-I) realization

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

$$\begin{aligned} k_\infty &= \left[\frac{Z(s)}{s}\right]_{s\to\infty} = 2 \\ k_0 &= \left[s\,Z(s)\right]_{s=0} = 4 \\ 2k_{3} &= \left[\frac{(s^2+4)\,Z(s)}{s}\right]_{s^2=-4} = 6 \end{aligned}$$ $$Z(s) = 2 s + \frac{4}{s} + \frac{6 s}{s^{2} + 4}$$

o--[L1]--[C2]--[L3||C3]--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. $Z_1(s) = \frac{2 s^{4} + 18 s^{2} + 16}{s^{3} + 4 s}$ → quotient $2 s$ gives series $L_1$ = 2 H; remainder $\frac{2 \left(5 s^{2} + 8\right)}{s \left(s^{2} + 4\right)}$.
2. $Y_2(s) = \frac{s^{3} + 4 s}{10 s^{2} + 16}$ → quotient $\frac{s}{10}$ gives shunt $C_2$ = 1/10 F (≈ 0.1 F); remainder $\frac{6 s}{5 \left(5 s^{2} + 8\right)}$.
3. $Z_3(s) = \frac{25 s^{2} + 40}{6 s}$ → quotient $\frac{25 s}{6}$ gives series $L_3$ = 25/6 H (≈ 4.167 H); remainder $\frac{20}{3 s}$.
4. $Y_4(s) = \frac{3 s}{20}$ → quotient $\frac{3 s}{20}$ gives shunt $C_4$ = 3/20 F (≈ 0.15 F); remainder $0$.

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