Lossless one-port properties; Foster I and Cauer I

Properties of lossless (LC) one-port

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 given function

$$Z(s) = \frac{s^4+4s^2+3}{2s^3+3s} = \frac{(s^2+1)(s^2+3)}{2s(s^2+1.5)}$$

Even/odd; poles at $\omega = 0,\ 1.225,\ \infty$; zeros at $\omega = 1,\ 1.732$ — they alternate (p 0, z 1, p 1.225, z 1.732, p ∞), so it is a valid LC impedance.

Foster-I form

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

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

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

Cauer-I form

$Z(s)$ has a pole at $s=\infty$ (numerator degree is higher), so the first element is a series inductor. Divide repeatedly with polynomials in descending powers (continued fraction about $s=\infty$), inverting the remainder each time:

1. $Z_1(s) = \frac{s^{4} + 4 s^{2} + 3}{2 s^{3} + 3 s}$ → quotient $\frac{s}{2}$ gives series $L_1$ = 1/2 H (≈ 0.5 H); remainder $\frac{5 s^{2} + 6}{2 s \left(2 s^{2} + 3\right)}$.
2. $Y_2(s) = \frac{4 s^{3} + 6 s}{5 s^{2} + 6}$ → quotient $\frac{4 s}{5}$ gives shunt $C_2$ = 4/5 F (≈ 0.8 F); remainder $\frac{6 s}{5 \left(5 s^{2} + 6\right)}$.
3. $Z_3(s) = \frac{25 s^{2} + 30}{6 s}$ → quotient $\frac{25 s}{6}$ gives series $L_3$ = 25/6 H (≈ 4.167 H); remainder $\frac{5}{s}$.
4. $Y_4(s) = \frac{s}{5}$ → quotient $\frac{s}{5}$ gives shunt $C_4$ = 1/5 F (≈ 0.2 F); remainder $0$.

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