LC impedance properties; Cauer second form

Properties of LC driving-point impedance

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).

Checking the functions

• $Z(s) = \dfrac{8s^3+10s}{s^4+6s^2+5} = \dfrac{8s(s^2+1.25)}{(s^2+1)(s^2+5)}$: odd/even; zeros at $\omega = 0,\ 1.118,\ \infty$ (degree 3 over 4), poles at $\omega = 1,\ 2.236$; order z 0, p 1, z 1.118, p 2.236, z ∞ — they alternate, with a zero at both $s=0$ and $s=\infty$. Valid LC impedance.
• $Z(s) = \dfrac{(s^2+4)(s^2+9)}{(s^2+16)(s^2+25)}$: even/even with equal degrees, so no pole or zero at $s=0$ or $s=\infty$; zeros at 2, 3 and poles at 4, 5 do not alternate (two zeros, then two poles). Not LC.

Cauer second form of the valid function

$Z(s)$ has a zero at $s=0$, so $Y(s)=1/Z(s)$ has a pole at the origin and the first element is a shunt inductor. Arrange numerator and denominator in ascending powers of $s$ and divide repeatedly (continued fraction about $s=0$); each quotient is $k/s$:

1. $Y_1(s) = \frac{5 + 6 s^{2} + s^{4}}{10 s + 8 s^{3}}$ → quotient $\frac{1}{2 s}$ gives shunt $L_1$ = 2 H; remainder $\frac{s \left(s^{2} + 2\right)}{2 \left(4 s^{2} + 5\right)}$.
2. $Z_2(s) = \frac{10 + 8 s^{2}}{2 s + s^{3}}$ → quotient $\frac{5}{s}$ gives series $C_2$ = 1/5 F (≈ 0.2 F); remainder $\frac{3 s}{s^{2} + 2}$.
3. $Y_3(s) = \frac{2 + s^{2}}{3 s}$ → quotient $\frac{2}{3 s}$ gives shunt $L_3$ = 3/2 H (≈ 1.5 H); remainder $\frac{s}{3}$.
4. $Z_4(s) = \frac{3}{s}$ → quotient $\frac{3}{s}$ gives series $C_4$ = 1/3 F (≈ 0.3333 F); remainder $0$.

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