---
title: "RC admittance check; parallel Foster and RC ladder"
subject: "Filter Design"
chapter: 4
key: "semester_5/filter_design:rc-admittance-foster-ladder"
source: https://www.bctengineer.com/q/semester_5/filter_design/rc-admittance-foster-ladder
site: BCT Engineering
---

# RC admittance check; parallel Foster and RC ladder

Filter Design · Chapter 4: Properties and Synthesis of Passive Networks

- Asked in: 2083 Baisakh, 2+3+3 marks
- Video: [Foster 1 & Foster 2 Forms- LC,RC,LR- KTU Qn #EE201 #CIRCUITS](https://www.youtube.com/watch?v=1tnqYLnfpMw) — Base Your Btech
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/rc-admittance-foster-ladder
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-4

## Question

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.

## 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.

## Answer

#### Checking 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.

| 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) |

```text
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 $C$):

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

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

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which is the correct property of an RC driving-point admittance?
   A. Nearest critical frequency to the origin is a pole; Y(0) > Y(infinity)
   B. Nearest critical frequency to the origin is a zero (no pole at 0); Y(infinity) >= Y(0)
   C. Poles and zeros on the jw axis
   D. Equal degrees are impossible
   Answer: B. Y_RC has the same form as Z_RL: lowest critical frequency is a zero, highest a pole, and Y(infinity) >= Y(0). (Key point: RC admittance criteria)

2. Why is Y(s) = (s+1)(s+4)/(s(s+2)) not an RC admittance?
   A. Its zeros are complex
   B. It has a pole at the origin, so the lowest critical frequency is a pole
   C. Its degree is too high
   D. It has no zeros
   Answer: B. A pole at s = 0 means lowest critical frequency is a pole, which is the RC impedance form, not admittance. (Key point: Why (a) is not RC admittance)

3. For Y(s) = 3(s+1)(s+4)/((s+3)(s+6)), the term k0 = Y(0) gives which element?
   A. A shunt resistor of 3/2 ohm (k0 = 2/3)
   B. A shunt resistor of 2/3 ohm
   C. A shunt capacitor of 2/3 F
   D. A series resistor of 3 ohm
   Answer: A. Y(0) = 3*4/18 = 2/3 S, so R1 = 1/k0 = 3/2 ohm in the parallel Foster form. (Key point: Foster parallel form)

4. Why does the Cauer-I ladder of this Y start with a series resistor?
   A. Z(infinity) = 1/3 is a finite constant, so the first removal is a constant (series R)
   B. Z has a pole at infinity
   C. Y has a pole at 0
   D. Z(0) = 0
   Answer: A. Z = 1/Y has Z(infinity) = 1/3, so R1 = 1/3 ohm is removed first, then alternating shunt C and series R. (Key point: Ladder start)

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