Testing lossless functions; Foster I and Cauer II

Testing a lossless function

A given $Z(s)$ is checked as follows:

1. Factor numerator and denominator. Each must be a product of factors $s$ and $(s^2+\omega_i^2)$ only, so that one is even and the other odd.
2. Write the critical frequencies in increasing order of $\omega$ and check that poles and zeros alternate.
3. Check that there is a pole or zero at $s=0$ and at $s=\infty$ (degrees differ by exactly 1).
4. Check that all residues of the partial-fraction expansion are positive (this follows when steps 1–3 hold).

Which functions are valid

Realize $Z(s) = \dfrac{s^4+4s^2+3}{s^5+6s^3+8s}$.

Foster-I form

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

$$\begin{aligned} k_0 &= \left[s\,Z(s)\right]_{s=0} = \frac{3}{8} \\ 2k_{2} &= \left[\frac{(s^2+2)\,Z(s)}{s}\right]_{s^2=-2} = \frac{1}{4} \\ 2k_{3} &= \left[\frac{(s^2+4)\,Z(s)}{s}\right]_{s^2=-4} = \frac{3}{8} \end{aligned}$$ $$Z(s) = \frac{3}{8 s} + \frac{s}{4 \left(s^{2} + 2\right)} + \frac{3 s}{8 \left(s^{2} + 4\right)}$$

o--[C1]--[L2||C2]--[L3||C3]--o

Cauer-II form

$Z(s)$ 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. $Z_1(s) = \frac{3 + 4 s^{2} + s^{4}}{8 s + 6 s^{3} + s^{5}}$ → quotient $\frac{3}{8 s}$ gives series $C_1$ = 8/3 F (≈ 2.667 F); remainder $\frac{s \left(5 s^{2} + 14\right)}{8 \left(s^{2} + 2\right) \left(s^{2} + 4\right)}$.
2. $Y_2(s) = \frac{64 + 48 s^{2} + 8 s^{4}}{14 s + 5 s^{3}}$ → quotient $\frac{32}{7 s}$ gives shunt $L_2$ = 7/32 H (≈ 0.2188 H); remainder $\frac{8 s \left(7 s^{2} + 22\right)}{7 \left(5 s^{2} + 14\right)}$.
3. $Z_3(s) = \frac{98 + 35 s^{2}}{176 s + 56 s^{3}}$ → quotient $\frac{49}{88 s}$ gives series $C_3$ = 88/49 F (≈ 1.796 F); remainder $\frac{21 s}{44 \left(7 s^{2} + 22\right)}$.
4. $Y_4(s) = \frac{968 + 308 s^{2}}{21 s}$ → quotient $\frac{968}{21 s}$ gives shunt $L_4$ = 21/968 H (≈ 0.02169 H); remainder $\frac{44 s}{3}$.
5. $Z_5(s) = \frac{3}{44 s}$ → quotient $\frac{3}{44 s}$ gives series $C_5$ = 44/3 F (≈ 14.67 F); remainder $0$.

o--[C1]----+---[C3]----+---[C5]-+
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         [L2]        [L4]       |
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o----------+-----------+--------+