Lossless check; Cauer-I and Foster-I expansions

Which function is lossless

• $Z(s) = \dfrac{s^2+10s+24}{s^2+8s+15} = \dfrac{(s+4)(s+6)}{(s+3)(s+5)}$: numerator and denominator contain both odd and even powers (not an even/odd ratio); poles and zeros lie on the negative real axis, not the $j\omega$ axis; there is no pole or zero at $0$ or $\infty$. Not lossless (it has the form of an RC impedance: p 3, z 4, p 5, z 6).
• $Z(s) = \dfrac{s^5+10s^3+24s}{s^4+6s^2+5} = \dfrac{s(s^2+4)(s^2+6)}{(s^2+1)(s^2+5)}$: odd/even; zeros at $\omega = 0, 2, 2.449$, poles at $\omega = 1, 2.236, \infty$; order z 0, p 1, z 2, p 2.236, z 2.449, p ∞ — they alternate. Valid lossless (LC) function.

Cauer-I expansion

$Z(s)$ 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. $Z_1(s) = \frac{s^{5} + 10 s^{3} + 24 s}{s^{4} + 6 s^{2} + 5}$ → quotient $s$ gives series $L_1$ = 1 H; remainder $\frac{s \left(4 s^{2} + 19\right)}{\left(s^{2} + 1\right) \left(s^{2} + 5\right)}$.
2. $Y_2(s) = \frac{s^{4} + 6 s^{2} + 5}{4 s^{3} + 19 s}$ → quotient $\frac{s}{4}$ gives shunt $C_2$ = 1/4 F (≈ 0.25 F); remainder $\frac{5 \left(s^{2} + 4\right)}{4 s \left(4 s^{2} + 19\right)}$.
3. $Z_3(s) = \frac{16 s^{3} + 76 s}{5 s^{2} + 20}$ → quotient $\frac{16 s}{5}$ gives series $L_3$ = 16/5 H (≈ 3.2 H); remainder $\frac{12 s}{5 \left(s^{2} + 4\right)}$.
4. $Y_4(s) = \frac{5 s^{2} + 20}{12 s}$ → quotient $\frac{5 s}{12}$ gives shunt $C_4$ = 5/12 F (≈ 0.4167 F); remainder $\frac{5}{3 s}$.
5. $Z_5(s) = \frac{3 s}{5}$ → quotient $\frac{3 s}{5}$ gives series $L_5$ = 3/5 H (≈ 0.6 H); remainder $0$.

o--[L1]----+---[L3]----+---[L5]-+
           |           |        |
         [C2]        [C4]       |
           |           |        |
o----------+-----------+--------+

Foster-I expansion

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

$$\begin{aligned} k_\infty &= \left[\frac{Z(s)}{s}\right]_{s\to\infty} = 1 \\ 2k_{2} &= \left[\frac{(s^2+1)\,Z(s)}{s}\right]_{s^2=-1} = \frac{15}{4} \\ 2k_{3} &= \left[\frac{(s^2+5)\,Z(s)}{s}\right]_{s^2=-5} = \frac{1}{4} \end{aligned}$$ $$Z(s) = s + \frac{15 s}{4 \left(s^{2} + 1\right)} + \frac{s}{4 \left(s^{2} + 5\right)}$$

o--[L1]--[L2||C2]--[L3||C3]--o