MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.0 1 1 2 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.161 Signal Processing – Continuous and Discrete Introduction to Continuous Time Filter Design 1 1 Introduction The design techniques described here are based on the creation of a prototype low-pass filter, with subsequent transformation to other filter forms (high-pass, band-pass, band-stop) if necessary. 2 Low-Pass Filter Design The prototype low-pass filter is based upon the magnitude-squared of the frequency response function |H(jΩ)|2, or the frequency response power function. The phase response of the filter is not considered. We begin by defining tolerance regions on the power frequency response plot, as shown in Fig. 1. | H ( j W ) |2 1 + e 1 + l 2 0 W c W r W ( r a d / s e c ) p a s s b a n d s t o p b a n dt r a n s i t i o n b a n d Figure 1: Tolerance regions in the frequency response plot. The filter specifications are that 1 1 ≥|H(jΩ)|2 > for |Ω| < Ωc (1)1+ 2 1D. Rowell, October 2, 2008 11 and |H(jΩ)|2 < for |Ω| > Ωr, (2)1+ λ2 where Ωc is the cut-off frequency, Ωr is the rejection frequency, and  and λ are design parameters that select the filter attenuation at the two critical frequencies. For example, if  =1, atΩc the power response response |H(jΩc)|2 =0.5, the -3 dB response frequency. In general we expect the response function to be monotonically decreasing in the transition band. The filter functions examined in this document will be of the form 1 |H(jΩ)|2 = 1+ f2(Ω) . (3) where f(Ω) → 0asΩ → 0, and f (Ω) →∞as Ω →∞to generate a low-pass filter action. 2.1 The Butterworth Filter The Butterworth filter is defined by the power gain 1 |H(jΩ)|2 = (4)1+ 2 (Ω/Ωc)2N where N is a positive integer defining the filter order. Note that λ does not appear in this formulation, but clearly N and λ are interrelated, since at Ω = Ωr 1 1 1+ λ2 ≥ 1+ 2 (Ωr/Ωc)2N (5) which may be solved to show log(λ/)N ≥ (6)log(Ωr/Ωc) The power response function of Butterworth filters for N =1 ...5 is shown in Fig. 2. Butterworth filters are also known as “maximally flat” filters because the response has the maximum number of vanishing derivatives at Ω = 0 and Ω = ∞ for filters of the form of Eq. 3. 2.1.1 The Transfer Function of the Butterworth Filter The poles of the power gain transfer function may be found from the characteristic equation � �2N 1+ 2 s = 0 (7)jΩc which yields 2N roots (poles) that lie on a circle: pn =Ωc−1/N ejπ(2n+N −1)/2N n =1 ...2N (8) with radius r =Ωc−1/N , and angular separation of π/N rad. Notice that if N is odd a pair of poles will lie on the real axis at s = ±Ωc−1/N , while if N is even the roots will form complex conjugate pairs. 20 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Power response N=1 N=5 N increasing |H(j )| 2ω 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Normalized frequency ω/ω c Figure 2: Power response of Butterworth filters for N =1 ...5( = 1). Figure 3a shows the six poles of |H(s)|2 for a third-order (N = 3) Butterworth filter. For a causal system we must have |H(jΩ)|2 = H(jΩ)H(jΩ) = H(s)H(−s)|s=jΩ which allows us to take take the N poles of |H(jΩ)|2 in the left half-plane as the poles of the filter H(s), that is the poles specified by (n =1 ...N) in Eq. (8) above, pn =Ωc−1/N ejπ(2n+N −1)/2N n =1 ...N (9) as is shown in Fig. 3b. If the filter is to have unity gain at low frequencies lim |H(jΩ)| =1 Ω→0 we require the complete Butterworth transfer function to be (−p1)(−p2) ...(−pN )H(s)= (s − p1)(s − p2) ...(s − pN ) (−1)N p1p2 ...pN = ,(s − p1)(s − p2) ...(s − pN ) where only the N stable left-half-plane poles are included. 3s x j W ) n = 1 n = 2 n = 3 x x x x x x j W 2 p / 6 r = W e - 1 / Nc s - p l a n en = 1 n = 2 n = 3 n = 4 n = 5 n = 6 s - p l a n e r = W c e - 1 / N ( a ) ( b x x s p / 3 Figure 3: The poles of (a) |H(s)|2, and (b) H(s) for a third-order (N = 3) Butterworth filter. x x j W s p / 4 r = W c s - p l a n e Figure 4: Poles of a second-order Butterworth filter (N =2,= 1). 4Example Consider a second-order (N =2,  = 1) Butterworth filter with cut-off frequency Ωc. Equation (9) generates a pole pair as shown shown in the pole-zero of Fig. 4. The transfer function is that of a second-order system with damping ratio ζ =0.707 and undamped natural frequency Ωn =Ωc. Show that the power frequency response satisfies the Butterworth specification of Eq. (4). The transfer function for the second-order filter is Ω2 H(s)= √ c (10) s2 + 2Ωcs +Ω2 c and the frequency response is Ω2 H(jΩ) = H(s)| = (Ω2 c − Ω2)+ c j √ 2ΩcΩ (11)s=jΩ Ω4 1c| H(jΩ)| 2 = H(jΩ)H(jΩ) = (Ω4 c +Ω4) = 1+(Ω/Ωc)4 (12) which is of the Butterworth form of Eq. (4). Example Design a Butterworth low-pass filter to meet the power gain specifications shown in Fig. 5. Comparing Figs. 1 and 5 1 =0.9 −→  =0.3333 1+ 2 1 =0.05 −→ λ =4.358 1+ λ2 Then log(λ/)N ≥ =3.70 log(Ωr/Ωc) we therefore select N=4. The 4 poles (p1 ...p4) lie on a circle of radius r = Ωc−1/N =13.16 and are given by | pn| =13.16 pn = π(2n +3)/8 for n =1 ...4, giving a pair of complex conjugate pole pairs p1,4 = − 5.04 ± j12.16 p2,3 = − 12.16 ± j5.04 50 1 0 . 0 5 | H ( j W ) |2 0 . 9 0 1 0 2 0 W p a s s b a n d s t o p b a n dt r a n s i t i o n b a n d Figure 5: Filter specification for Butterworth design. Bode Diagram 50 0 −50 −100 −1500 −90 −180 −270 −360 10−1 100 101 102 103 Frequency (rad/sec) Figure 6: Bode plots for fourth-order Butterworth design example. Phase (deg) Magnitude (dB) 6The transfer function, normalized to unity gain, is 29993 H(s) = (s2 + 10.07s + 173.2)(s2 + 24.32s + 173.2) and the filter Bode plots are shown in Fig. 6. 2.2 Chebyshev Filters The order of a filter required to met a low-pass specification …