Lossless one-port properties; Foster II and Cauer 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).

Identifying the LC function

• $Z(s) = \dfrac{(s^2+1)(s^2+9)}{s(s^2+4)}$: even/odd; p 0, z 1, p 2, z 3, p ∞ — alternate. Valid LC.
• $Z(s) = \dfrac{s(s^2+4)(s^2+5)}{(s^2+3)(s^2+6)}$: z 0, p 1.732, z 2, z 2.236, p 2.449, p ∞ — zeros at 2 and 2.236 are adjacent (and poles at 2.449 and ∞). Not LC.

Realize $Z(s) = \dfrac{(s^2+1)(s^2+9)}{s(s^2+4)} = \dfrac{s^4+10s^2+9}{s^3+4s}$.

Foster-II form

Take $Y(s) = 1/Z(s) = \frac{s^{3} + 4 s}{s^{4} + 10 s^{2} + 9}$ and expand $Y(s)$ in partial fractions (each term is an admittance; the branches are in parallel):

$$\begin{aligned} 2k_{1} &= \left[\frac{(s^2+1)\,Y(s)}{s}\right]_{s^2=-1} = \frac{3}{8} \\ 2k_{2} &= \left[\frac{(s^2+9)\,Y(s)}{s}\right]_{s^2=-9} = \frac{5}{8} \end{aligned}$$ $$Y(s) = \frac{3 s}{8 \left(s^{2} + 1\right)} + \frac{5 s}{8 \left(s^{2} + 9\right)}$$

o--+-------+
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  [L1]    [L2]
  [C1]    [C2]
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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. $Z_1(s) = \frac{9 + 10 s^{2} + s^{4}}{4 s + s^{3}}$ → quotient $\frac{9}{4 s}$ gives series $C_1$ = 4/9 F (≈ 0.4444 F); remainder $\frac{s \left(4 s^{2} + 31\right)}{4 \left(s^{2} + 4\right)}$.
2. $Y_2(s) = \frac{16 + 4 s^{2}}{31 s + 4 s^{3}}$ → quotient $\frac{16}{31 s}$ gives shunt $L_2$ = 31/16 H (≈ 1.938 H); remainder $\frac{60 s}{31 \left(4 s^{2} + 31\right)}$.
3. $Z_3(s) = \frac{961 + 124 s^{2}}{60 s}$ → quotient $\frac{961}{60 s}$ gives series $C_3$ = 60/961 F (≈ 0.06243 F); remainder $\frac{31 s}{15}$.
4. $Y_4(s) = \frac{15}{31 s}$ → quotient $\frac{15}{31 s}$ gives shunt $L_4$ = 31/15 H (≈ 2.067 H); remainder $0$.

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