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MINIMUM-PHASE FIR FILTER DESIGN the Lifting Procedure MINIMUM-PHASE FIR FILTER DESIGN The Lifting Procedure 1. INTRODUCTION 2. PROBLEM FORMULATION 3. THE SQUARE MAGNITUDE 4. A TRANSFORMATION 5. MEETING SPECIFICATIONS 6. SUMMARY OF THE LIFTING PROCEDURE 7. MEETING SPECIFICATIONS | EXAMPLE 8. PLOTTING THE PHASE 9. PHASE UNWRAPPING 10. GROUP DELAY I. Selesnick EL 713 Lecture Notes 1 INTRODUCTION FIR filters that have symmetric or antisymmetric impulse responses have no phase distortion (their phase is linear). However, the prob- lem with linear-phase filters is that the delay can be too large. The delay of a linear-phase filter is equal to (N − 1)=2, where N is the length of the filter. To obtain a linear-phase FIR filter with a narrow transition-band and high stop-band attenuation requires making the filter long. Therefore, linear-phase FIR filters satisfying demanding specifications will have a large delay. This large delay could be a major drawback. (1) If the filter is used inside a feedback loop in a control system it could cause instability. (2) If the filter is used in a communication system it could cause delays that are longer than is acceptable. For applications where it is important to minimize the delay caused by a filter, a minimum-phase filter can be a good choice. Minimum- phase filters have all their zeros inside or on the unit circle. A minimum-phase filter can be obtained from a linear-phase filter by reflecting all of the zeros that are outside the unit circle to inside the unit circle. In other words, those zeros located at z = r ejθ are 1 jθ moved to z = r e . This modification results in a minimum-phase filter that has the same frequency response magnitude jH(ej!)j as the linear-phase filter. However, the minimum-phase filter obtained in this way will not be optimal in general. There will generally be a minimum-phase filter that is superior than the one obtained by simply reflecting the zeros in this way. I. Selesnick EL 713 Lecture Notes 2 PROBLEM FORMULATION In this section we consider the design of low-pass filters. The design problem can be stated as follows. Problem 1: Find the FIR filter h(n) of minimal length (not nec- essarily with linear-phase) such that j! jjH(e )j − 1j ≤ ∆p for 0 ≤ ! ≤ !p (1) j! jH(e )j ≤ ∆s for !s ≤ ! ≤ π: (2) The given design parameters are the band-edges !p, !s and the maximum error in the pass-band and stop-band ∆p, ∆s. This design problem is more difficult than the design of a linear- phase FIR filter, because the frequency response magnitude jH(ej!)j is a nonlinear function of the filter coefficients h(n). N−1 j! X j!n jH(e )j = h(n) e : (3) n=0 This section describes how to design minimum-phase FIR filters so that the error between jH(ej!)j and a desired D(!) is minimized. I. Selesnick EL 713 Lecture Notes 3 EXAMPLE 0.6 0.6 0.4 0.4 0.2 h(n) 0.2 h(n) 0 0 −0.2 −0.2 0 5 10 0 5 10 n n 1.2 1.2 1 1 0.8 0.8 )| )| ω ω j 0.6 j 0.6 |H(e |H(e 0.4 0.4 0.2 0.2 0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 ω/π ω/π 1 1 0.5 0.5 12 12 0 0 −0.5 −0.5 Zeros of H(z) Zeros of H(z) −1 −1 −1 0 1 −1 0 1 The filter on the left is a linear-phase FIR filter of length 13 designed so that the Chebyshev error is minimized. The maximum value of the error in the pass-band and stop-band is 0.137. The filter on the right is a minimum-phase FIR filter of the same length, however, the maximum value of the error in the pass-band is 0.0882, and in the stop-band it is 0.1099. By not restricting the impulse response to be symmetric, the frequency response magnitude jH(ej!)j can be improved. The remainder of this section describes how to design a non-symmetric low-pass filter h(n) so that jH(ej!)j has the smallest error. I. Selesnick EL 713 Lecture Notes 4 THE SQUARE MAGNITUDE Because jH(ej!)j is not a linear function of the coefficients h(n), it turns out to be more convenient to work with the square magnitude jH(ej!)j2. Assuming the filter coefficients h(n) are real, we can write: jH(ej!)j2 = H(ej!) · H(ej!) (4) = H(ej!) · H(e−j!) (5) = H(z) · H(1=z)jz=ej! : (6) Suppose h(n) is a length N FIR impulse response. If we define R(z) := H(z) · H(1=z) (7) then r(n) = h(n) ∗ h(−n) (8) and R(ej!) = jH(ej!)j2: (9) Note that r(n) will be symmetric with r(n) = r(−n) and the length of r(n) will be 2N − 1. It turns out to be more straight-forward to design the filter R(z), and then to obtain H(z) from R(z). Herrmann and Sch¨usslerproposed a method for designing minimum- phase filters for which the Chebyshev error is minimized [1]. Their method is based on the transformation of a linear-phase FIR Cheby- shev filter in to a minimum-phase FIR filter. I. Selesnick EL 713 Lecture Notes 5 Transformation of Type I FIR equi-ripple linear-phase filter into an equi-ripple minimum-phase filter The method consists of 4 steps. 1. Design an equi-ripple Type I linear-phase filter h1(n) of length 2N − 1. 2. Obtain a second Type I linear-phase filter h2(n) by adjusting h1(n) so that A2(!) ≥ 0. 3. Spectrally factor h2(n) to obtain a minimum-phase FIR filter h3(n) of length N. 4. Scale h3(n) by an appropriate constant to obtain a minimum- phase FIR filter h(n) that is equi-ripple. This procedure is best illustrated by an example. I. Selesnick EL 713 Lecture Notes 6 STEP ONE Step One: Design an equi-ripple Type I FIR filter of length 2N −1. For example, to design a length 11 minimum-phase FIR filter, we will first design a length 21 Type I FIR filter. 1.1931 1 0.8069 ) ω ( 1 A 0.0966 0 −0.0966 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 ω/π 0.6 2 0.5 1.5 0.4 1 (z) 0.3 1 0.5 20 (n) 1 0.2 0 h 0.1 −0.5 Zeros of H 0 −1 −0.1 −1.5 −0.2 −2 0 5 10 15 20 −2 −1 0 1 2 n Define δ1 and δ2 to be the maximum value of the error in the pass- band and stop-band. δ1 := max jA1(!) − 1j for 0 ≤ ! ≤ !p (10) δ2 := max jA1(!)j for !s ≤ ! ≤ π (11) We can then write: 1 − δ1 ≤ A1(!) ≤ 1 + δ1 for 0 ≤ ! ≤ !p (12) −δ2 ≤ A1(!) ≤ δ2 for !s ≤ ! ≤ π: (13) I. Selesnick EL 713 Lecture Notes 7 N = 11; Kp = 1; Ks = 2; wp = 0.50*pi; ws = 0.56*pi; wo = (wp+ws)/2; L = 1000; w = [0:L]*pi/L; D = (w<=wo); W = Kp*(w<=wp) + Ks*(w>=ws); [h1,del] = fircheb(2*N-1,D,W); d1 = del/Kp; d2 = del/Ks; In this example, δ1 = 0:1931, δ2 = 0:0966. I. Selesnick EL 713 Lecture Notes 8 STEP TWO Step Two: Step two is to adjust the filter h1(n) so that the filter can be spectrally factored. This can be done by adding δ2 to the middle coefficient of h1(n). 1.2897 1 0.9034 ) ω ( 2 A 0.1931 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 ω/π 2 0.6 1.5 1 2 0.4 2 (z) 2 0.5 2 20 (n) 2 0 h 0.2 −0.5 2 Zeros of H 2 0 −1 −1.5 −0.2 −2 0 5 10 15 20 −2 −1 0 1 2 n h2(n) = h1(n) + δ2 · δ(n − N + 1) (14) In Matlab this can be done with the command h2 = h1; h2(N) = h2(N) + d2; This has the effect of `lifting' the amplitude response A2(!) = A1(!) + δ2: (15) The frequency response amplitude of h2(n) then satisfies the fol- I. Selesnick EL 713 Lecture Notes 9 lowing bounds, 1 − δ1 + δ2 ≤ A2(!) ≤ 1 + δ1 + δ2 for 0 ≤ ! ≤ !p (16) 0 ≤ A2(!) ≤ 2 δ2 for !s ≤ ! ≤ π: (17) Notice that the zeros of H2(z) on the unit circle are now of even multiplicity. Then the filter h2(n) can be spectrally factored be- cause A2(!) ≥ 0 (18) and because h2(n) is symmetric and of odd length. I. Selesnick EL 713 Lecture Notes 10 STEP THREE Step Three: We can obtain a minimum-phase filter by spectrally factoring the filter h2(n). If h3(n) is a spectral factor of h2(n), then we have j! 2 jH3(e )j = A2(!): (19) or j! p jH3(e )j = A2(!): (20) 1.1356 1 0.9505 )| ω j (e 3 |H 0.4395 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 ω/π 2 0.6 1.5 0.5 1 0.4 (z)| 0.5 3 0.3 10 (n)| 0 3 h 0.2 −0.5 0.1 Zeros of H 0 −1 −0.1 −1.5 −0.2 −2 0 5 10 15 20 −2 −1 0 1 2 n This can be done with the Matlab command, h3 = sfact(h2); where the command sfact is described in the notes about spectral I.
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