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MIT 2 161 - Butterworth Filter Design Example

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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 0 . 9 0 . 0 5 1 Massachusetts Institute of Technology Department of Mechanical Engineering 2.161 Signal Processing - Continuous and Discrete Fall Term 2008 Lecture 71 Reading: • Class handout: Introduction to Continuous Time Filter Design. Butterworth Filter Design Example (Same problem as in the Class Handout). Design a Butterworth low-pass filter to meet the power gain specifications shown below: | H ( j 9 ) |2 0 1 0 2 0 9 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 At the two critical frequencies 1 =0.9 −→  =0.3333 1+ 2 1 =0.05 −→ λ =4.358 1+ λ2 Then log(λ/)N ≥ =3.70 log(Ωr/Ωc) 1copyright cD.Rowell 2008 7–1we 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 The 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 below. Bode Diagram −1 0 1 2 310 10 10 10 10Frequency (rad/sec) 2 Chebyshev Filters The order of a filter required to met a low-pass specification may often be reduced by relaxing the requirement of a monotonically decreasing power gain with frequency, and allowing −150 −100 −50 0 50 Magnitude (dB) −360 −270 −180 −90 0 Phase (deg) 7–2“ripple” to occur in either the pass-band or the stop-band. The Chebyshev filters allow these conditions: Typ e 1 |H(jΩ)| 2 = 1 + �2TN 1 2 (Ω/Ωc) (1) 1 Typ e 2 |H(jΩ)| 2 = 1 + �2 (TN 2 (Ωr/Ωc)/TN 2 (Ωr/Ω)) (2) Where TN (x) is the Chebyshev polynomial of degree N. Note the similarity of the form of the Type 1 power gain (Eq. (1)) to that of the Butterworth filter, where the function TN (Ω/Ωc) has replaced (Ω/Ωc)N . The Type 1 filter produces an all-pole design with slightly different pole placement from the Butterworth filters, allowing resonant peaks in the pass-band to introduce ripple, while the Type 2 filter introduces a set of zeros on the imaginary axis above Ωr, causing a ripple in the stop-band. The Chebyshev polynomials are defined recursively as follows T0(x) = 1 T1(x) = x T2(x) = 2x 2 − 1 T3(x) = 4x 3 − 3x . . . TN (x) = 2xTN−1(x) − TN−2(x), N > 1 (3) with alternate definitions TN (x) = cos(N cos−1(x)) (4) = cosh(N cosh−1(x)) (5) The Chebyshev polynomials have the min-max property: Of all polynomials of degree N with leading coefficient equal to one, the polynomial TN (x)/2N−1 has the smallest magnitude in the interval x This “minimum maximum” amplitude is 21−N . | | ≤ 1. In low-pass filters given by Eqs. (13) and (14), this property translates to the following characteristics: Filter Pass-Band Characteristic Stop-Band Characteristic Butterworth Maximally flat Maximally flat Chebyshev Type 1 Ripple between 1 and 1/(1 + �2) Maximally flat Chebyshev Type 2 Maximally flat Ripple between 1 and 1/(1 + λ2) 7–3� � 2.1 The Chebyshev Type 1 Filter With the power response from Eq. (13) 1 |H(jΩ)| 2 = 1 + �2TN 2 (Ω/Ωc) and the filter specification from Fig. 1, the required filter order may be found as follows. At the edge of the stop-band Ω = Ωr 1 12 |H(jΩr| = 1 + �2TN 2 (Ωr/Ωc) ≤ 1 + λ2 so that λ ≤ �TN (Ωr/Ωc) = � cosh �N cosh−1 (Ωr/Ωc)� and solving for N cosh−1 (λ/�)N ≥ cosh−1 (Ωr/Ωc) (6) The characteristic equation of the power transfer function is 1 + �2TN 2 � jΩ s � = 0 or TN � jΩ s � = ± j�c c Now TN (x) = cos(N cos−1(x)), so that � � s �� j cos N cos−1 = ± (7)jΩc � If we write cos−1 (s/jΩc) = γ + jα, then s = Ωc (j cos (γ + jα)) = Ωc (sinh α sin γ + j cosh α cos γ) (8) which defines an ellipse of width 2Ωc sinh(α) and height 2Ωc cosh(α) in the s-plane. The poles will lie on this ellipse. Substituting into Eq. (16) � s �TN = cos (N (γ + jα))jΩc = cos Nγ cosh Nα − j sin Nγ sinh Nα, the characteristic equation becomes j cos Nγ cosh Nα − j sin Nγ sinh Nα = ± . (9) Equating the real and imaginary parts in Eq. (21), (1) since cosh x = 0 for real x we require cos Nγ = 0, or γn = (2n − 1)πn = 1, . . . , 2N (10)2N 7–4and, (2) since at these values of γ, sin Nγ = ±1 we have 1 1 α = ±N sinh−1 � (11) As in the Butterworth design procedure, we select the left half-plane poles as the poles of the filter frequency response. Design Procedure: 1. Determine the filter order cosh−1 (λ/�)N ≥ cosh−1 (Ωr/Ωc) 2. Determine α α = ±N 1 sinh−1 1 � 3. Determine γn, n = 1 . . . N (2n − 1)π γn = n = 1, . . . , N 2N 4. Determine the N left half-plane poles pn = Ωc (sinh α sin γn + j cosh α cos γn) n = 1, . . . , N 5. Form the transfer function (a) If N is odd −p1p2 . . . pN H(s) = (s − p1)(s − p2) . . . (s − pN ) (b) If N is even 1 p1p2 . . . pN H(s) = 1 + �2 (s − p1)(s − p2) . . . (s − pN ) The difference in the gain constants in the two cases arises because of 2the ripple in the pass-band. When N is odd, the response H(j0) = 1, 2 | |whereas if N is even the value of H(j0) = 1/(1 + �2).| | 7–5Example 1 Repeat the previous Butterworth design example using a Chebyshev Type 1 design. From the previous example we have Ωc = 10 rad/s., Ωr = 20 rad/s., � = 0.3333, λ = 4.358. The required order is cosh−1 (λ/�) cosh−1 13.07 N ≥ cosh−1 (Ωr/Ωc) = cosh−1 2 = 2.47 Therefore take N = 3. Determine α: 1 �1� 1 α = sinh−1 = sinh−1(3) = 0.6061 N � 3 and sinh α = 0.6438, and cosh α = 1.189. Also, γn = (2n − 1)π/6 for n = 1 . . . 6 as follows: n: 1 2 3 4 5 6 γn: π/6 π/2 5π/6 7π/6 3π/2 11π/6 sin γn: 1/2 1 1/2 −1/2 -1 −1/2 cos γn: √3/20 −√3/2 −√3/2 0 √3/2 Then the poles are pn = Ωc (sinh α sin γn + j cosh α cos γn)� 1 √3� p1 = 10 0.6438 + j1.189 …


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