RC properties; lossless check; Foster series and Cauer II

Properties of RC driving-point impedance

An RC driving-point impedance $Z_{RC}(s)$ has these properties:

1. All poles and zeros are simple and lie on the negative real axis of the $s$-plane (including the origin).
2. Poles and zeros alternate along the negative real axis.
3. The critical frequency nearest the origin (lowest) is a pole; it may be at $s=0$.
4. The critical frequency farthest from the origin is a zero; it may be at $s=\infty$.
5. $Z_{RC}(0) > Z_{RC}(\infty)$, and $Z_{RC}(\infty)\ge 0$ is a constant (no pole at infinity).
6. Residues of $Z_{RC}(s)$ at its poles are real and positive; on the real axis $dZ(\sigma)/d\sigma<0$.
7. The same form describes an RL admittance $Y_{RL}(s)$.

Testing for lossless (LC) functions

(If the intended denominator is $s^4+4s^2+3 = (s^2+1)(s^2+3)$, then $Z_3 = \dfrac{s(s^2+4)}{(s^2+1)(s^2+3)}$ has z 0, p 1, p 1.732, z 2 — two adjacent poles, so it is still invalid.)

Only $Z_1(s) = \dfrac{2s^4+18s^2+16}{s^3+4s}$ 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-II realization

$Z(s)$ 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. $Z_1(s) = \frac{16 + 18 s^{2} + 2 s^{4}}{4 s + s^{3}}$ → quotient $\frac{4}{s}$ gives series $C_1$ = 1/4 F (≈ 0.25 F); remainder $\frac{2 s \left(s^{2} + 7\right)}{s^{2} + 4}$.
2. $Y_2(s) = \frac{4 + s^{2}}{14 s + 2 s^{3}}$ → quotient $\frac{2}{7 s}$ gives shunt $L_2$ = 7/2 H (≈ 3.5 H); remainder $\frac{3 s}{14 \left(s^{2} + 7\right)}$.
3. $Z_3(s) = \frac{98 + 14 s^{2}}{3 s}$ → quotient $\frac{98}{3 s}$ gives series $C_3$ = 3/98 F (≈ 0.03061 F); remainder $\frac{14 s}{3}$.
4. $Y_4(s) = \frac{3}{14 s}$ → quotient $\frac{3}{14 s}$ gives shunt $L_4$ = 14/3 H (≈ 4.667 H); remainder $0$.

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