Valid RC admittance; Foster II and RC ladder

Testing the functions

An RC driving-point admittance $Y_{RC}(s)$ (same form as $Z_{RL}$) must have:

1. Simple poles and zeros on the negative real axis, alternating.
2. The critical frequency nearest the origin is a zero (or $Y(0)$ is a finite constant); there is no pole at $s=0$.
3. The critical frequency farthest from the origin is a pole (it may be at $s=\infty$).
4. $Y_{RC}(\infty) \ge Y_{RC}(0)$.
5. Residues of $Y_{RC}(s)/s$ are real and positive.

Realize $Y(s) = \dfrac{(s+1)(s+3)}{(s+2)(s+4)} = \dfrac{s^2+4s+3}{s^2+6s+8}$.

Foster-II (parallel) form

Expand $Y(s)/s$ in partial fractions and multiply back by $s$:

$$\begin{aligned} k_0 &= Y(0) = \frac{3}{8} \\ k_{2} &= \left[\frac{(s+2)Y(s)}{s}\right]_{s=-2} = \frac{1}{4} \\ k_{3} &= \left[\frac{(s+4)Y(s)}{s}\right]_{s=-4} = \frac{3}{8} \end{aligned}$$ $$Y(s) = \frac{3}{8} + \frac{s}{4 \left(s + 2\right)} + \frac{3 s}{8 \left(s + 4\right)}$$

o--+-------+-------+
   |       |       |
  [R1]    [R2]    [R3]
   |      [C2]    [C3]
   |       |       |
o--+-------+-------+

Check: $Y(\infty) = \tfrac{3}{8}+\tfrac{1}{4}+\tfrac{3}{8} = 1$ ✓.

RC ladder (Cauer-I) form

Use $Z(s) = 1/Y(s) = \dfrac{(s+2)(s+4)}{(s+1)(s+3)}$; $Z(\infty) = 1$, so the ladder starts with a series resistor.

Divide with descending powers (continued fraction about $s=\infty$). For an RC impedance the quotients alternate between a constant (series $R$) and $ks$ (shunt $C$):

1. $Z_1(s) = \frac{s^{2} + 6 s + 8}{s^{2} + 4 s + 3}$ → remove $1$, giving series $R_1$ = 1 Ω; remainder $\frac{2 s + 5}{\left(s + 1\right) \left(s + 3\right)}$.
2. $Y_2(s) = \frac{s^{2} + 4 s + 3}{2 s + 5}$ → remove $\frac{s}{2}$, giving shunt $C_2$ = 1/2 F (≈ 0.5 F); remainder $\frac{3 \left(s + 2\right)}{2 \left(2 s + 5\right)}$.
3. $Z_3(s) = \frac{4 s + 10}{3 s + 6}$ → remove $\frac{4}{3}$, giving series $R_3$ = 4/3 Ω (≈ 1.333 Ω); remainder $\frac{2}{3 \left(s + 2\right)}$.
4. $Y_4(s) = \frac{3 s + 6}{2}$ → remove $\frac{3 s}{2}$, giving shunt $C_4$ = 3/2 F (≈ 1.5 F); remainder $3$.
5. $Z_5(s) = \frac{1}{3}$ → remove $\frac{1}{3}$, giving series $R_5$ = 1/3 Ω (≈ 0.3333 Ω); remainder $0$.

o--[R1]----+---[R3]----+---[R5]-+
           |           |        |
         [C2]        [C4]       |
           |           |        |
o----------+-----------+--------+