--- title: "5th order Butterworth LPF with first-order and MFB sections" subject: "Filter Design" chapter: 8 key: "semester_5/filter_design:cascade-mfb-fifth-order" source: https://www.bctengineer.com/q/semester_5/filter_design/cascade-mfb-fifth-order site: BCT Engineering --- # 5th order Butterworth LPF with first-order and MFB sections Filter Design · Chapter 8: Design of High-Order Active Filters - Asked in: 2082 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-fifth-order - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-8 ## Question Design a 5th order Butterworth LPF using cascade of 1st order active section and MFB biquads with DC gain equal to unity and half power frequency at 1 kHz. Make the largest capacitance 0.1 μF. ## Short answer First-order section: R=1.592 kΩ, C=0.1 μF. Two MFB biquads with R1=R2=R3, C1=3Q, C2=1/(3Q) scaled to ω0=6283.2 rad/s, largest C1=0.1 μF. Section 1 (Q=1.618): R=7.726 kΩ, C1=0.1 μF, C2=4.244 nF. Section 2 (Q=0.618): R=2.951 kΩ, C1=0.1 μF, C2=29.09 nF. Overall DC gain magnitude 1. ## Answer **Butterworth poles (n = 5)** lie on the unit circle at 0°, ±36° and ±72° from the negative real axis, so $$ T_N(s) = \frac{1}{(s+1)(s^2 + 0.618s + 1)(s^2 + 1.618s + 1)} $$ Each quadratic $s^2 + (1/Q)s + 1$ gives $Q = 1/0.618 = 1.618$ and $Q = 1/1.618 = 0.618$. The filter is one first-order section followed by two MFB biquads; $\omega_0 = 2\pi(1000) = 6283.2$ rad/s. #### First-order section An inverting lossy integrator: input $R_1$, feedback $R_f \parallel C$: $$ T_1(s) = -\frac{1/(R_1C)}{s + 1/(R_fC)} $$ Normalised: $R_1 = R_f = 1\ \Omega$, $C = 1$ F (pole at 1, gain 1). Scaling with $C$ = 0.1 μF: $R = 1/(\omega_0 C) = 1/(6283.2 \times 10^{-7})$ = **1.592 kΩ** for both $R_1$ and $R_f$. ```text +---[Rf]---+ +---| C |--+ Vi-[R1]--+--(-) | A -----+--o GND --(+) ``` #### MFB biquads ```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:** $\omega_0$ = 6283.2 rad/s. For each section choose $k_z$ so that its largest capacitor $C_1$ becomes 0.1 μF: $k_z = C_{1n}/(\omega_0 \times 10^{-7})$. Then all three resistors are $k_z$ and $C_2 = C_{2n}/(k_z\omega_0)$. (Each section is driven by an op-amp output, so sections may be scaled separately.) | Section | Q | $C_{1n}$ | $C_{2n}$ | $R_1 = R_2 = R_3$ | $C_1$ | $C_2$ | |---|---|---|---|---|---|---| | $s^2 + 0.618s + 1$ | 1.618 | 4.8541 F | 0.2060 F | 7.726 kΩ | 100 nF | 4.244 nF | | $s^2 + 1.618s + 1$ | 0.618 | 1.8541 F | 0.5393 F | 2.951 kΩ | 100 nF | 29.09 nF | #### Final circuit ```text Vi -->[1st order]-->[MFB, Q=1.618]-->[MFB, Q=0.618]--> Vo R=1.592k R=7.726k R=2.951k C=0.1uF C1=0.1uF C1=0.1uF C2=4.244nF C2=29.09nF ``` Each stage has DC gain −1, so the overall DC gain magnitude is 1 (sign −1; add a unity inverter if a positive output is needed). Half-power frequency 1 kHz. **Answer:** first-order: R = 1.592 kΩ, C = 0.1 μF; MFB 1 (Q = 1.618): R = 7.726 kΩ, C1 = 0.1 μF, C2 = 4.244 nF; MFB 2 (Q = 0.618): R = 2.951 kΩ, C1 = 0.1 μF, C2 = 29.09 nF. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is the correct normalised 5th-order Butterworth transfer function denominator? A. (s+1)(s^2 + 0.618s + 1)(s^2 + 1.618s + 1) B. (s^2 + s + 1)^2 (s+1) C. (s+1)^5 D. (s^2 + 1.414s + 1)^2 (s+1) Answer: A. Poles lie on the unit circle at 0, +-36 and +-72 degrees from the negative real axis, giving those factors. (Key point: Butterworth factorisation) 2. What Q does each quadratic have? A. Q = 0.618 for the first factor and 1.618 for the second, and Q = 1 for the first-order B. Q = 1/0.618 = 1.618 for the first and 1/1.618 = 0.618 for the second C. Both Q = 0.707 D. Q = 0.618 for the first and 1.618 for the second Answer: B. s^2 + (1/Q)s + 1: 1/Q = 0.618 gives Q = 1.618; 1/Q = 1.618 gives Q = 0.618. (Key point: Q of sections) 3. With R1 = R2 = R3 = 1 ohm for unity DC gain, what are C1 and C2? A. C1 = Q/3, C2 = 3/Q B. C1 = 3Q, C2 = 1/(3Q) C. C1 = C2 = Q D. C1 = 1/(3Q), C2 = 3Q Answer: B. w0^2 = 1/(C1 C2) = 1 and w0/Q = 3/C1 give C1 = 3Q and C2 = 1/(3Q). (Key point: MFB capacitor design) 4. What is the scaling for the first-order section and the overall DC gain magnitude? A. R = 1.592 kohm with C = 0.1 uF; each stage has DC gain -1 so overall magnitude is 1 B. R = 15.9 kohm; overall gain 3 C. R = 1.592 kohm; overall gain 0.5 D. R = 159 ohm; overall gain 1/3 Answer: A. R = 1/(w0 C) = 1/(6283.2 x 10^-7) = 1.592 kohm; T(0) = -R2/R1 = -1 per stage, so the product has magnitude 1. (Key point: Scaling and gain) ## Related in Filter Design - [4th order Butterworth LPF with two MFB biquads](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Acascade-mfb-fourth-order) — Filter Design, chapter 8 - [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 - [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 - [Sallen-Key 2nd order Butterworth LPF at 10 kHz](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asallen-key-butterworth-10khz) — Filter Design, chapter 6 - [Sallen-Key LPF at 10 kHz, largest C 0.01 μF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asallen-key-10khz-largest-capacitance) — Filter Design, chapter 6 - [Tow-Thomas 2nd order Butterworth LPF at 5 kHz](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-butterworth-5khz) — Filter Design, chapter 6