---
title: "4th order Butterworth LPF with two MFB biquads"
subject: "Filter Design"
chapter: 8
key: "semester_5/filter_design:cascade-mfb-fourth-order"
source: https://www.bctengineer.com/q/semester_5/filter_design/cascade-mfb-fourth-order
site: BCT Engineering
---

# 4th order Butterworth LPF with two MFB biquads

Filter Design · Chapter 8: Design of High-Order Active Filters

- Asked in: 2081 Bhadra, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/filter_design/cascade-mfb-fourth-order
- Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8

## Question

Design a 4th order Butterworth LPF using cascade two MFB biquads with dc gain equal to unity and half power frequency at 1000rad/sec. Make the largest capacitance equal to 0.1 μF in your final circuit.

## Short answer

Two MFB biquads with R1=R2=R3, C1=3Q, C2=1/(3Q) scaled to ω0=1000 rad/s, C1=0.1 μF. Section 1 (Q=0.5412): R=16.24 kΩ, C1=0.1 μF, C2=37.94 nF. Section 2 (Q=1.3066): R=39.2 kΩ, C1=0.1 μF, C2=6.509 nF. DC gain +1.

## Answer

**Butterworth poles (n = 4)** lie at ±22.5° and ±67.5° from the negative real axis on the unit circle:

$$
T_N(s) = \frac{1}{(s^2 + 1.8478s + 1)(s^2 + 0.7654s + 1)}
$$

so the two biquads have $Q_1 = 1/1.8478 = 0.5412$ and $Q_2 = 1/0.7654 = 1.3066$, with $\omega_0$ = 1000 rad/s.

#### MFB lowpass biquad

```text
              +----------[R2]-----------+
              |                         |
              |       +----[C2]----+    |
              |       |            |    |
 Vi o--[R1]---+--[R3]-+----(-)     |    |
              |              A ----+----+--o Vo
             [C1]     GND---(+)
              |
             GND
```

For this multiple-feedback (MFB) lowpass biquad:

$$
T(s) = \frac{V_o}{V_i} = \frac{-\dfrac{1}{R_1R_3C_1C_2}}{s^2 + \dfrac{s}{C_1}\left(\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\right) + \dfrac{1}{R_2R_3C_1C_2}}, \qquad T(0) = -\frac{R_2}{R_1}
$$

**Normalised design** ($\omega_0 = 1$, unity DC gain): take $R_1 = R_2 = R_3 = 1\ \Omega$. Then $\omega_0^2 = 1/(C_1C_2) = 1$ and $\omega_0/Q = 3/C_1$, so

$$
C_1 = 3Q, \qquad C_2 = \frac{1}{3Q}
$$

#### Scaling

$\omega_0$ = 1000 rad/s. In each section set $C_1$ (the larger capacitor) to 0.1 μF: $k_z = C_{1n}/(\omega_0 \times 10^{-7})$; then $R_1 = R_2 = R_3 = k_z$ and $C_2 = C_{2n}/(k_z\omega_0)$.

| Section | Q | $C_{1n}$ | $C_{2n}$ | $R_1 = R_2 = R_3$ | $C_1$ | $C_2$ |
|---|---|---|---|---|---|---|
| $s^2 + 1.8478s + 1$ | 0.5412 | 1.6236 F | 0.6159 F | 16.24 kΩ | 100 nF | 37.94 nF |
| $s^2 + 0.7654s + 1$ | 1.3066 | 3.9197 F | 0.2551 F | 39.2 kΩ | 100 nF | 6.509 nF |

Check for section 2: $\omega_0 = 1/\sqrt{R^2C_1C_2} = 1/\sqrt{(39.2\text{k})^2(10^{-7})(6.509 \times 10^{-9})} \approx 1000$ rad/s.

#### Final circuit

```text
 Vi -->[MFB 1, Q=0.5412]-->[MFB 2, Q=1.3066]--> Vo
         R = 16.24k             R = 39.2k
         C1 = 0.1uF             C1 = 0.1uF
         C2 = 37.94nF           C2 = 6.509nF
```

Each section has DC gain −1, so the cascade has DC gain +1 (unity) and half-power frequency 1000 rad/s.

**Answer:** section 1: $R_1 = R_2 = R_3$ = 16.24 kΩ, $C_1$ = 0.1 μF, $C_2$ = 37.94 nF; section 2: $R_1 = R_2 = R_3$ = 39.2 kΩ, $C_1$ = 0.1 μF, $C_2$ = 6.509 nF.

## Check yourself

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

1. For a 4th order Butterworth filter, what are the Q values of the two biquads?
   A. 0.5412 and 1.3066
   B. 0.7071 and 0.7071
   C. 1.8478 and 0.7654
   D. 1.0 and 1.414
   Answer: A. The denominator factors are s^2+1.8478s+1 and s^2+0.7654s+1, so Q = 1/1.8478 = 0.5412 and Q = 1/0.7654 = 1.3066. (Key point: Q of each biquad)

2. With R1 = R2 = R3 = 1 ohm and w0 = 1 in the MFB lowpass biquad, how are the capacitors found?
   A. C1 = 2Q, C2 = 1/(2Q)
   B. C1 = Q/3, C2 = 3/Q
   C. C1 = 3Q, C2 = 1/(3Q)
   D. C1 = C2 = 1/Q
   Answer: C. w0/Q = 3/C1 gives C1 = 3Q, and C1C2 = 1 gives C2 = 1/(3Q). (Key point: Normalised MFB capacitors)

3. After normalised design, how are the components scaled so that the larger capacitor is 0.1 uF at w0 = 1000 rad/s?
   A. Multiply all resistors by w0 only
   B. Choose k_z = C1n/(w0 x 10^-7); resistors = k_z and C2 = C2n/(k_z w0)
   C. Divide all capacitors by Q
   D. Keep resistors at 1 ohm and scale only capacitors
   Answer: B. Impedance scaling factor k_z is set by the requirement C1 = C1n/(k_z w0) = 0.1 uF; then resistors become k_z and C2 = C2n/(k_z w0). (Key point: Impedance scaling)

4. Why does the cascade of two MFB sections have unity DC gain?
   A. Each section has gain +1 because of the op-amp follower
   B. The second section has gain 1/R2
   C. Each section has T(0) = -R2/R1 = -1, and two inversions multiply to +1
   D. The capacitors block DC so gain is set by Q
   Answer: C. With R1 = R2, each MFB section has DC gain -1; (-1)(-1) = +1. (Key point: DC gain of cascade)

## Related in Filter Design

- [4th order Butterworth LPF with two Sallen-Key biquads (10 kHz)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Acascade-sallen-key-fourth-order-10khz) — Filter Design, chapter 8
- [5th order Butterworth LPF with first-order and MFB sections](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Acascade-mfb-fifth-order) — Filter Design, chapter 8
- [4th order Butterworth LPF with two Sallen-Key biquads (1 kHz)](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Acascade-sallen-key-fourth-order-1khz) — Filter Design, chapter 8
- [FDNR 4th order Butterworth LPF at 20,000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Afdnr-fourth-order-20000) — Filter Design, chapter 8
- [Leapfrog 4th order Butterworth LPF at 1000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aleapfrog-fourth-order-1000) — Filter Design, chapter 8, asked 4×
- [Leapfrog 4th order Butterworth LPF at 10,000 rad/s](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aleapfrog-fourth-order-10000) — Filter Design, chapter 8, asked 6×