---
title: "Active vs passive; non-inverting (s+8)/(s+2)"
subject: "Filter Design"
chapter: 6
key: "semester_5/filter_design:active-passive-difference-noninverting"
source: https://www.bctengineer.com/q/semester_5/filter_design/active-passive-difference-noninverting
site: BCT Engineering
---

# Active vs passive; non-inverting (s+8)/(s+2)

Filter Design · Chapter 6: Active Filters

- Asked in: 2076 Asoj, 4+3 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/active-passive-difference-noninverting
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-6

## Question

What are the differences between active and passive filter? Design an active filter using non-inverting op-amp configuration with following transfer function. T(s) = (s+8)/(s+2)

## Short answer

Active filters use op-amps, require power, can provide gain >1, and avoid inductors; passive filters use only R, L, C with gain ≤1. For T(s)=(s+8)/(s+2) using the non-inverting topology: R1=16.67 kΩ, R2=50 kΩ, C2=10 µF (DC gain 4, HF gain 1).

## Answer

#### Differences between active and passive filters

| Point | Active filter | Passive filter |
|---|---|---|
| Elements | R, C and op-amps/transistors | R, L, C only |
| Inductors | Not needed | Needed (bulky at low f) |
| Gain | Can be > 1 | Always ≤ 1 (loss) |
| Power supply | Required | Not required |
| Loading | Op-amp buffers; easy cascading | Stages load each other |
| Frequency range | Limited by op-amp bandwidth (up to ~MHz) | Works up to very high (RF) frequencies |
| Signal level | Limited by supply (saturation) | Handles large voltage/current |
| Noise | Op-amp adds noise | Low noise (only R noise) |
| Size/IC | Small, can be integrated | Large; inductors cannot be integrated |
| Sensitivity | Often higher | Low (doubly terminated ladders) |

#### Design of $T(s) = (s+8)/(s+2)$
**Non-inverting configuration:** $Z_1$ from the inverting input to ground, $Z_2$ from output to inverting input:
$$
T(s) = 1 + \frac{Z_2}{Z_1}
$$
Zero 8 > pole 2, high-frequency gain 1, DC gain 4. Use $Z_1 = R_1$, $Z_2 = R_2 \parallel C_2$:
$$
T(s) = 1 + \frac{R_2}{R_1(1 + sR_2C_2)} = \frac{s + \frac{1}{R_2C_2}\left(1+\frac{R_2}{R_1}\right)}{s + \frac{1}{R_2C_2}}
$$
```text
          +---R2---+
          |        |
          +---||---+
          |   C2   |
   +------+-|-\    |
   |        |  >---+---- Vo
  R1  Vin---|+/
   |
  GND
```
**Matching:** $\dfrac{1}{R_2C_2} = 2$ and $1 + \dfrac{R_2}{R_1} = \dfrac{8}{2} = 4$, so $R_2 = 3R_1$.

Normalized: $C_2 = 1$ F, $R_2 = 0.5\ \Omega$, $R_1 = 0.1667\ \Omega$. With $C_2 = 10\ \mu$F: $R_2 = 50\ \text{k}\Omega$, $R_1 = 16.67\ \text{k}\Omega$.

**Answer:** $R_1 = 16.67\ \text{k}\Omega$, $R_2 = 50\ \text{k}\Omega$, $C_2 = 10\ \mu$F (DC gain 4, HF gain 1).

## Check yourself

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

1. How do active and passive filter gains compare?
   A. Both can exceed 1
   B. Passive can exceed 1
   C. Active can exceed 1; passive is always <= 1
   D. Both always <= 1
   Answer: C. Passive filters only lose signal. (Key point: Gain)

2. Which statement is correct?
   A. Active works to RF; passive only audio
   B. Both limited to 1 kHz
   C. Passive limited by op-amp
   D. Active limited by op-amp bandwidth; passive works up to RF
   Answer: D. Op-amp bandwidth limits active filters. (Key point: Frequency range)

3. For (s+8)/(s+2), DC and HF gains are:
   A. DC 4, HF 1
   B. DC 1, HF 4
   C. DC 8, HF 2
   D. DC 4, HF 4
   Answer: A. T(0) = 8/2 and T(inf) = 1. (Key point: DC and HF gain)

4. With C2 = 10 uF, R2 and R1 are:
   A. R2 = 16.67 kΩ, R1 = 50 kΩ
   B. R2 = 50 kΩ, R1 = 16.67 kΩ
   C. R2 = 5 kΩ, R1 = 1.67 kΩ
   D. R2 = R1 = 50 kΩ
   Answer: B. R2 = 1/(2×10^-5) = 50 kΩ; R1 = R2/3. (Key point: Values)

## Close variants set in other exams

- [Non-inverting realization of (s+8)/(s+2)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Anoninverting-realize-s-plus-8) — Filter Design, chapter 6, asked 5×
- [Active filter; non-inverting (s+8)/(s+2)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aactive-filter-noninverting-s-plus-8) — Filter Design, chapter 6, asked 5×
- [Active vs passive; realize 4(s+2)/(s+1)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aactive-passive-noninverting-4s-plus-2) — Filter Design, chapter 6, asked 5×

## Related in Filter Design

- [Non-inverting filter: zero at 8, pole at 4, gain 2](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Anoninverting-zero-8-pole-4) — Filter Design, chapter 6, asked 2×
- [Active filter advantages; bilinear zero 800, pole 400](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aactive-advantages-bilinear-800-400) — Filter Design, chapter 6
- [Insertion gain/loss, design steps, active vs passive](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Ainsertion-loss-steps-active-passive) — Filter Design, chapter 1
- [Non-inverting realization of 4(s+200)/(s+100)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Anoninverting-realize-4-s-plus-200) — Filter Design, chapter 6, asked 4×
- [Non-inverting: zero at -4, pole at -8, HF gain 2](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Anoninverting-zero-4-pole-8) — Filter Design, chapter 6, asked 4×
- [Non-inverting: zero at -5, pole at -3, HF gain 2](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Anoninverting-zero-5-pole-3) — Filter Design, chapter 6, asked 3×

## Related in other subjects

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