Which describes a valid RC impedance function?

Lowest critical frequency is a pole, highest is a zero, alternating, with Z(0) > Z(infinity)

Lowest critical frequency is a zero, Z(0) < Z(infinity)

Why is 2(s+1)(s+3)/((s+2)(s+4)) not an RC impedance?

The lowest critical frequency is a zero and Z(0)=0.75 < Z(infinity)=2

What are the first two elements of the Cauer-I ladder of (a)?

Series R1 = 1 ohm, then shunt C2 = 1/3 F

Shunt C 1 F, then series R 3 ohm

Series R = 3 ohm, then shunt C = 1/3 F

Asked 3/26
2082 Bhadra · 2+3+3 marks

Valid RC impedance; Foster-II and Cauer-I

Short answer

Z(s)=(s+3)(s+6)/((s+1)(s+5)) is valid RC impedance because poles/zeros alternate, lowest is pole, highest zero, Z(0)=3.6 > Z(∞)=1. Foster-II: shunt 3.6 Ω, series 2.25 Ω with 0.148 F, series 3.6 Ω with 0.0463 F. Cauer-I: series 1 Ω, shunt 0.333 F, series 1.8 Ω, shunt 0.417 F, series 0.8 Ω.

Which of the following functions are valid RC driving point impedance function and why? Pick the valid RC function and synthesize it with Foster-II and Cauer-I form. a) Z(s) = (s+3)(s+6)/((s+1)(s+5)) b) Z(s) = 2(s+1)(s+3)/((s+2)(s+4))

Checking the functions

An RC impedance has simple, alternating poles and zeros on the negative real axis, with a pole nearest the origin, a zero farthest from it, and $Z(0) > Z(\infty)$ .

Function	Critical frequencies	Result
(a) $\dfrac{(s+3)(s+6)}{(s+1)(s+5)}$	p 1, z 3, p 5, z 6	Lowest is a pole, highest a zero, alternate; $Z(0)=3.6 > Z(\infty)=1$ → valid RC
(b) $\dfrac{2(s+1)(s+3)}{(s+2)(s+4)}$	z 1, p 2, z 3, p 4	Lowest critical frequency is a zero and $Z(0)=0.75 < Z(\infty)=2$ → not RC impedance (it is an RL impedance / RC admittance)

Foster-II (parallel) form of (a)

Take $Y(s)=1/Z(s) = \frac{s^{2} + 6 s + 5}{s^{2} + 9 s + 18}$ . Expand $Y(s)/s$ in partial fractions and multiply back by $s$ :

\begin{aligned} k_0 &= Y(0) = \frac{5}{18} \\ k_{2} &= \left[\frac{(s+3)Y(s)}{s}\right]_{s=-3} = \frac{4}{9} \\ k_{3} &= \left[\frac{(s+6)Y(s)}{s}\right]_{s=-6} = \frac{5}{18} \end{aligned}

Y(s) = \frac{5}{18} + \frac{4 s}{9 \left(s + 3\right)} + \frac{5 s}{18 \left(s + 6\right)}

Term	Rule	Element value
$R_1$ (shunt)	$1/k_0$	18/5 Ω (≈ 3.6 Ω)
$R_2$ – $C_2$ in series (pole at $-3$ )	$R=1/k$ , $C=k/\sigma$	$R_2$ = 9/4 Ω (≈ 2.25 Ω), $C_2$ = 4/27 F (≈ 0.1481 F)
$R_3$ – $C_3$ in series (pole at $-6$ )	$R=1/k$ , $C=k/\sigma$	$R_3$ = 18/5 Ω (≈ 3.6 Ω), $C_3$ = 5/108 F (≈ 0.0463 F)

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

Cauer-I form of (a)

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

$Z_1(s) = \frac{s^{2} + 9 s + 18}{s^{2} + 6 s + 5}$ → remove $1$ , giving series $R_1$ = 1 Ω; remainder $\frac{3 s + 13}{\left(s + 1\right) \left(s + 5\right)}$ .
$Y_2(s) = \frac{s^{2} + 6 s + 5}{3 s + 13}$ → remove $\frac{s}{3}$ , giving shunt $C_2$ = 1/3 F (≈ 0.3333 F); remainder $\frac{5 \left(s + 3\right)}{3 \left(3 s + 13\right)}$ .
$Z_3(s) = \frac{9 s + 39}{5 s + 15}$ → remove $\frac{9}{5}$ , giving series $R_3$ = 9/5 Ω (≈ 1.8 Ω); remainder $\frac{12}{5 \left(s + 3\right)}$ .
$Y_4(s) = \frac{5 s + 15}{12}$ → remove $\frac{5 s}{12}$ , giving shunt $C_4$ = 5/12 F (≈ 0.4167 F); remainder $\frac{5}{4}$ .
$Z_5(s) = \frac{4}{5}$ → remove $\frac{4}{5}$ , giving series $R_5$ = 4/5 Ω (≈ 0.8 Ω); remainder $0$ .

Element	Position	Value
$R_1$	series	1 Ω
$C_2$	shunt	1/3 F (≈ 0.3333 F)
$R_3$	series	9/5 Ω (≈ 1.8 Ω)
$C_4$	shunt	5/12 F (≈ 0.4167 F)
$R_5$	series	4/5 Ω (≈ 0.8 Ω)

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

✓ Answer checked by Claude Sonnet 5.5

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