--- title: "Butterworth characteristics and 5th order TF" subject: "Filter Design" chapter: 2 key: "semester_5/filter_design:butterworth-fifth-order-tf" source: https://www.bctengineer.com/q/semester_5/filter_design/butterworth-fifth-order-tf site: BCT Engineering --- # Butterworth characteristics and 5th order TF Filter Design · Chapter 2: Approximation Methods - Asked in: 2073 Chaitra, 3+4 marks - Video: [Butterworth Filter : Design of Low Pass and High Pass Filters](https://www.youtube.com/watch?v=lc6QT8VjqVc) — ALL ABOUT ELECTRONICS - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/butterworth-fifth-order-tf - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-2 ## Question What are the characteristics of butterworth response? Calculate the transfer function of 5th order Butterworth filter. ## Short answer Characteristics: maximally flat, monotonic, |T(jω)|^2=1/(1+ε^2 ω^(2n)), -3 dB at ω=1 for ε=1, roll-off 20n dB/decade, all-pole, poles on a circle. Normalized 5th-order transfer function: T(s)=1/[(s+1)(s^2+0.618s+1)(s^2+1.618s+1)] = 1/(s^5+3.236s^4+5.236s^3+5.236s^2+3.236s+1). ## Answer #### Characteristics of the Butterworth response The **Butterworth (maximally flat)** low-pass response has magnitude $$ |T(j\omega)|^2 = \frac{1}{1 + \varepsilon^2\omega^{2n}} $$ (normalized so that $\omega_p$ = 1 rad/s; with $\varepsilon = 1$ the half-power frequency is 1 rad/s). Its characteristics are: - **Maximally flat at $\omega = 0$:** the first $2n-1$ derivatives of $|T(j\omega)|^2$ are zero at $\omega = 0$, so the passband is as flat as possible. - **Monotonic:** no ripple in the passband or the stopband; the gain falls continuously with frequency. - $|T(j0)| = 1$; at $\omega = 1$ (when $\varepsilon = 1$) $|T| = 1/\sqrt{2}$ (−3 dB) for every order. - **Roll-off** of $20n$ dB/decade far into the stopband; the transition is wider than Chebyshev or elliptic for the same $n$. - **All-pole** function: all transmission zeros are at $\omega = \infty$. - **Poles lie on a circle** of radius $\omega_o = \varepsilon^{-1/n}$ in the left-half s-plane, equally spaced by $\pi/n$, symmetric about the real axis. - Phase is fairly linear (better than Chebyshev), with a moderate delay peak near the cut-off. - As $n \to \infty$ the response approaches the ideal brick-wall. ```text |T| 1 |----.__ | `. n = 2 0.707 - - - - :\. n = 4 | : \ \. | : `. `-._ +---------+---------`----> w 1 ``` #### Fifth-order Butterworth transfer function For the normalized ($\varepsilon$ = 1, $\omega_c$ = 1 rad/s) case the poles lie on the unit circle at $$ p_k = -\sin\theta_k \pm j\cos\theta_k, \qquad \theta_k = \frac{(2k-1)\pi}{2n} = 18°, 54°, 90° $$ | $\theta_k$ | Pole(s) | Factor | |---|---|---| | 18° | $-0.309 \pm j0.951$ | $s^2 + 0.618s + 1$ | | 54° | $-0.809 \pm j0.588$ | $s^2 + 1.618s + 1$ | | 90° | $-1$ | $s + 1$ | (Each pair gives $s^2 + 2\sin\theta_k\, s + 1$.) $$ \begin{aligned} T(s) &= \frac{1}{(s+1)(s^2 + 0.618s + 1)(s^2 + 1.618s + 1)} \\ &= \frac{1}{s^5 + 3.236s^4 + 5.236s^3 + 5.236s^2 + 3.236s + 1} \end{aligned} $$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is true of the Butterworth magnitude response in the passband and stopband? A. Ripple in the passband only B. No ripple in either band; gain falls monotonically C. Ripple in the stopband only D. Equal ripple in both bands Answer: B. Butterworth is monotonic with no ripple anywhere; Chebyshev type I has passband ripple. (Key point: Monotonic response) 2. Which statement about Butterworth poles and zeros is correct? A. Poles on the jw axis B. Poles on a circle in the right-half plane C. Zeros on the jw axis and poles on an ellipse D. All-pole: zeros at infinity; poles on a circle in the left-half plane Answer: D. The Butterworth function is all-pole with transmission zeros at infinity; poles lie on a circle, equally spaced, in the left half-plane. (Key point: Poles and zeros) 3. Which angles theta_k = (2k-1)pi/(2n) give the 5th-order poles? A. 30, 60, 90 degrees B. 18, 54, 90 degrees C. 15, 45, 75 degrees D. 36, 72, 108 degrees Answer: B. For n = 5, theta_k = 18, 54, 90 degrees; the 90 degree pole is the real pole at s = -1. (Key point: Pole angles for n = 5) 4. Which denominator is correct for the normalized 5th-order Butterworth? A. (s+1)(s^2+1.414s+1)^2 B. (s^2+0.618s+1)(s^2+1.618s+1) C. (s+1)(s^2+0.765s+1)(s^2+1.848s+1) D. (s+1)(s^2+0.618s+1)(s^2+1.618s+1) Answer: D. Each pair gives s^2 + 2 sin(theta) s + 1; the 90 degree pole gives s+1, producing s^5 + 3.236s^4 + 5.236s^3 + 5.236s^2 + 3.236s + 1. 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