LC impedance check; Foster series and Cauer II

Testing the functions

Tests for a valid LC impedance: (1) even/odd or odd/even ratio, (2) simple poles and zeros only on the $j\omega$ axis, (3) poles and zeros alternate, (4) a pole or a zero at both $s=0$ and $s=\infty$ (degrees differ by exactly 1), (5) positive residues.

• $Z(s) = \dfrac{2s(s^2+4)(s^2+16)}{(s^2+1)(s^2+9)}$: odd/even; z 0, p 1, z 2, p 3, z 4, p ∞ — alternate. Valid LC.
• $Z(s) = \dfrac{4(s+2)(s+5)}{(s+1)(s+4)}$: poles and zeros on the negative real axis (p 1, z 2, p 4, z 5), not on the $j\omega$ axis, and not an even/odd ratio. Not LC (it is an RC impedance).

Realize $Z(s) = \dfrac{2s^5+40s^3+128s}{s^4+10s^2+9}$.

Foster series (Foster-I) form

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

$$\begin{aligned} k_\infty &= \left[\frac{Z(s)}{s}\right]_{s\to\infty} = 2 \\ 2k_{2} &= \left[\frac{(s^2+1)\,Z(s)}{s}\right]_{s^2=-1} = \frac{45}{4} \\ 2k_{3} &= \left[\frac{(s^2+9)\,Z(s)}{s}\right]_{s^2=-9} = \frac{35}{4} \end{aligned}$$ $$Z(s) = 2 s + \frac{45 s}{4 \left(s^{2} + 1\right)} + \frac{35 s}{4 \left(s^{2} + 9\right)}$$

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

Cauer-II form

$Z(s)$ 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. $Y_1(s) = \frac{9 + 10 s^{2} + s^{4}}{128 s + 40 s^{3} + 2 s^{5}}$ → quotient $\frac{9}{128 s}$ gives shunt $L_1$ = 128/9 H (≈ 14.22 H); remainder $\frac{5 s \left(11 s^{2} + 92\right)}{128 \left(s^{2} + 4\right) \left(s^{2} + 16\right)}$.
2. $Z_2(s) = \frac{8192 + 2560 s^{2} + 128 s^{4}}{460 s + 55 s^{3}}$ → quotient $\frac{2048}{115 s}$ gives series $C_2$ = 115/2048 F (≈ 0.05615 F); remainder $\frac{128 s \left(23 s^{2} + 284\right)}{115 \left(11 s^{2} + 92\right)}$.
3. $Y_3(s) = \frac{10580 + 1265 s^{2}}{36352 s + 2944 s^{3}}$ → quotient $\frac{2645}{9088 s}$ gives shunt $L_3$ = 9088/2645 H (≈ 3.436 H); remainder $\frac{7245 s}{2272 \left(23 s^{2} + 284\right)}$.
4. $Z_4(s) = \frac{645248 + 52256 s^{2}}{7245 s}$ → quotient $\frac{645248}{7245 s}$ gives series $C_4$ = 7245/645248 F (≈ 0.01123 F); remainder $\frac{2272 s}{315}$.
5. $Y_5(s) = \frac{315}{2272 s}$ → quotient $\frac{315}{2272 s}$ gives shunt $L_5$ = 2272/315 H (≈ 7.213 H); remainder $0$.

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