Lossless one-port properties; Cauer I and Foster II

Properties of lossless one-port function

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).

Given $Z(s) = \dfrac{s(s^2+4)}{(s^2+2)(s^2+6)} = \dfrac{s^3+4s}{s^4+8s^2+12}$: z 0, p 1.414, z 2, p 2.449, z ∞ — alternating, odd/even, so it is a valid LC function.

Cauer-I realization

$Z(s)$ has a zero at $s=\infty$, so $Y(s)=1/Z(s)$ has a pole at infinity and the first element is a shunt capacitor. Divide repeatedly with polynomials in descending powers (continued fraction about $s=\infty$), inverting the remainder each time:

1. $Y_1(s) = \frac{s^{4} + 8 s^{2} + 12}{s^{3} + 4 s}$ → quotient $s$ gives shunt $C_1$ = 1 F; remainder $\frac{4 \left(s^{2} + 3\right)}{s \left(s^{2} + 4\right)}$.
2. $Z_2(s) = \frac{s^{3} + 4 s}{4 s^{2} + 12}$ → quotient $\frac{s}{4}$ gives series $L_2$ = 1/4 H (≈ 0.25 H); remainder $\frac{s}{4 \left(s^{2} + 3\right)}$.
3. $Y_3(s) = \frac{4 s^{2} + 12}{s}$ → quotient $4 s$ gives shunt $C_3$ = 4 F; remainder $\frac{12}{s}$.
4. $Z_4(s) = \frac{s}{12}$ → quotient $\frac{s}{12}$ gives series $L_4$ = 1/12 H (≈ 0.08333 H); remainder $0$.

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

Foster-II realization

Take $Y(s) = 1/Z(s) = \frac{s^{4} + 8 s^{2} + 12}{s^{3} + 4 s}$ and expand $Y(s)$ in partial fractions (each term is an admittance; the branches are in parallel):

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

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