Which is the correct property of an RC driving-point admittance?

Nearest critical frequency to the origin is a zero (no pole at 0); Y(infinity) >= Y(0)

Nearest critical frequency to the origin is a pole; Y(0) > Y(infinity)

Why is Y(s) = (s+1)(s+4)/(s(s+2)) not an RC admittance?

It has a pole at the origin, so the lowest critical frequency is a pole

For Y(s) = 3(s+1)(s+4)/((s+3)(s+6)), the term k0 = Y(0) gives which element?

A shunt resistor of 3/2 ohm (k0 = 2/3)

Why does the Cauer-I ladder of this Y start with a series resistor?

Z(infinity) = 1/3 is a finite constant, so the first removal is a constant (series R)

2083 Baisakh · 2+3+3 marks

RC admittance check; parallel Foster and RC ladder

Short answer

Y(s)=3(s+1)(s+4)/((s+3)(s+6)) is valid RC admittance because poles/zeros alternate, lowest is zero, highest pole, Y(∞)>Y(0). Foster: shunt 1.5 Ω, series 1.5 Ω with 0.222 F, series 0.6 Ω with 0.278 F. Ladder: series 0.333 Ω, shunt 0.75 F, series 0.889 Ω, shunt 1.35 F, series 0.278 Ω.

Determine whether the following functions are RC driving point admittance functions or not. State with reason. a) Y(s) = (s+1)(s+4)/(s(s+2)) b) Y(s) = 3(s+1)(s+4)/((s+3)(s+6)). Pick one of the valid RC admittance functions and realize it in parallel Foster form as well as RC ladder form.

Checking the functions

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

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

Function	Critical frequencies (on $-\sigma$ axis)	Result
(a) $Y=\dfrac{(s+1)(s+4)}{s(s+2)}$	pole at 0, z 1, p 2, z 4	Pole at the origin (lowest critical frequency is a pole) → not RC admittance (it is an RC impedance form)
(b) $Y=\dfrac{3(s+1)(s+4)}{(s+3)(s+6)}$	z 1, p 3, z 4, p 6	Lowest is a zero, highest a pole, alternate; $Y(0) = 2/3 < Y(\infty) = 3$ → valid

Parallel Foster form of (b)

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

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

Y(s) = \frac{2}{3} + \frac{2 s}{3 \left(s + 3\right)} + \frac{5 s}{3 \left(s + 6\right)}

Term	Rule	Element value
$R_1$ (shunt)	$1/k_0$	3/2 Ω (≈ 1.5 Ω)
$R_2$ – $C_2$ in series (pole at $-3$ )	$R=1/k$ , $C=k/\sigma$	$R_2$ = 3/2 Ω (≈ 1.5 Ω), $C_2$ = 2/9 F (≈ 0.2222 F)
$R_3$ – $C_3$ in series (pole at $-6$ )	$R=1/k$ , $C=k/\sigma$	$R_3$ = 3/5 Ω (≈ 0.6 Ω), $C_3$ = 5/18 F (≈ 0.2778 F)

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

Check: $Y(\infty) = \tfrac{2}{3}+\tfrac{2}{3}+\tfrac{5}{3} = 3$ ✓.

RC ladder (Cauer-I) form of (b)

Use $Z(s) = 1/Y(s) = \dfrac{(s+3)(s+6)}{3(s+1)(s+4)}$ , which has $Z(\infty) = 1/3$ , 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 $C$ ):

$Z_1(s) = \frac{s^{2} + 9 s + 18}{3 s^{2} + 15 s + 12}$ → remove $\frac{1}{3}$ , giving series $R_1$ = 1/3 Ω (≈ 0.3333 Ω); remainder $\frac{2 \left(2 s + 7\right)}{3 \left(s + 1\right) \left(s + 4\right)}$ .
$Y_2(s) = \frac{3 s^{2} + 15 s + 12}{4 s + 14}$ → remove $\frac{3 s}{4}$ , giving shunt $C_2$ = 3/4 F (≈ 0.75 F); remainder $\frac{3 \left(3 s + 8\right)}{4 \left(2 s + 7\right)}$ .
$Z_3(s) = \frac{8 s + 28}{9 s + 24}$ → remove $\frac{8}{9}$ , giving series $R_3$ = 8/9 Ω (≈ 0.8889 Ω); remainder $\frac{20}{9 \left(3 s + 8\right)}$ .
$Y_4(s) = \frac{27 s + 72}{20}$ → remove $\frac{27 s}{20}$ , giving shunt $C_4$ = 27/20 F (≈ 1.35 F); remainder $\frac{18}{5}$ .
$Z_5(s) = \frac{5}{18}$ → remove $\frac{5}{18}$ , giving series $R_5$ = 5/18 Ω (≈ 0.2778 Ω); remainder $0$ .

Element	Position	Value
$R_1$	series	1/3 Ω (≈ 0.3333 Ω)
$C_2$	shunt	3/4 F (≈ 0.75 F)
$R_3$	series	8/9 Ω (≈ 0.8889 Ω)
$C_4$	shunt	27/20 F (≈ 1.35 F)
$R_5$	series	5/18 Ω (≈ 0.2778 Ω)

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

The final $R_5$ closes the ladder across $C_4$ .

✓ Answer checked by Claude Sonnet 5.5

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