Lossless properties; Cauer-I 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).

Valid LC functions

• $Z(s) = \dfrac{(s^2+1)(s^2+3)}{s(s^2+2)}$: even/odd; p 0, z 1, p 1.414, z 1.732, p ∞ — alternate. Valid.
• $Z(s) = \dfrac{s(s^2+2)}{(s^2+3)(s^2+4)}$: z 0, z 1.414, p 1.732, p 2, z ∞ — zeros adjacent and poles adjacent. Invalid.

Cauer-I realization

$Z(s)$ 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. $Z_1(s) = \frac{s^{4} + 4 s^{2} + 3}{s^{3} + 2 s}$ → quotient $s$ gives series $L_1$ = 1 H; remainder $\frac{2 s^{2} + 3}{s \left(s^{2} + 2\right)}$.
2. $Y_2(s) = \frac{s^{3} + 2 s}{2 s^{2} + 3}$ → quotient $\frac{s}{2}$ gives shunt $C_2$ = 1/2 F (≈ 0.5 F); remainder $\frac{s}{2 \left(2 s^{2} + 3\right)}$.
3. $Z_3(s) = \frac{4 s^{2} + 6}{s}$ → quotient $4 s$ gives series $L_3$ = 4 H; remainder $\frac{6}{s}$.
4. $Y_4(s) = \frac{s}{6}$ → quotient $\frac{s}{6}$ gives shunt $C_4$ = 1/6 F (≈ 0.1667 F); remainder $0$.

o--[L1]----+---[L3]----+
           |           |
         [C2]        [C4]
           |           |
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. $Z_1(s) = \frac{3 + 4 s^{2} + s^{4}}{2 s + s^{3}}$ → quotient $\frac{3}{2 s}$ gives series $C_1$ = 2/3 F (≈ 0.6667 F); remainder $\frac{s \left(2 s^{2} + 5\right)}{2 \left(s^{2} + 2\right)}$.
2. $Y_2(s) = \frac{4 + 2 s^{2}}{5 s + 2 s^{3}}$ → quotient $\frac{4}{5 s}$ gives shunt $L_2$ = 5/4 H (≈ 1.25 H); remainder $\frac{2 s}{5 \left(2 s^{2} + 5\right)}$.
3. $Z_3(s) = \frac{25 + 10 s^{2}}{2 s}$ → quotient $\frac{25}{2 s}$ gives series $C_3$ = 2/25 F (≈ 0.08 F); remainder $5 s$.
4. $Y_4(s) = \frac{1}{5 s}$ → quotient $\frac{1}{5 s}$ gives shunt $L_4$ = 5 H; remainder $0$.

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