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Minimum-Phase & All-Pass Filters PDF

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Minimum-Phase & All-Pass Filters Peter Kabal Departmentof Electrical & ComputerEngineering McGill University Montreal, Canada Version 2.1March 2011 (cid:13)c 2011PeterKabal 2011/03/11 Youarefree: toShare–tocopy,distributeandtransmitthiswork toRemix–toadaptthiswork Subjecttoconditionsoutlinedinthelicense. ThisworkislicensedundertheCreative CommonsAttribution 3.0Unported License. To view a copy of this license, visit http://creativecommons. org/licenses/by/3.0/orsendalettertoCreativeCommons,171Second Street,Suite300,SanFrancisco,California, 94105, USA. RevisionHistory: • 2011-03v2.1:CreativeCommonslicence,minorupdates(simplifications) • 2010-01v2.0:Newmaterialonmin-phase&all-passfilters • 2007-11v1.6:Initialrelease Minimum-Phase&All-PassFilters 1 1 Introduction This report examines the properties of minimum-phase and all-pass filters. The analysis deals with the case of complex filter coefficients. This strategy simplifies the analysis since we will be able to express general filters as the cascade of first order filters sections. We will be interested in the amplitude, phase, and group delay responsesof thesefilters. Of particular interestwill be filterswhichhavearesponsedescribedbyarationalz-transform, B(z) H(z) = A(z) M ∏(1−z z−1) k = Gz−n0 k=1 N ∏(1−pkz−1) (1) k=1 M ∏(1−r ejθkz−1) k = Gz−n0 k=1 , N ∏(1−ρ ejφkz−1) k k=1 whereinthelastpartoftheequationthezeroshavebeenexpressedaszk = rkejθk andthepolesas pk = ρkejφk. 1.1 AmplitudeResponse Themagnitude-squaredresponseofafilter H(z)canbeevaluatedfromthez-transformas |H(ω)|2 = H(ω)H∗(ω) (2) = H(z)H∗(1/z∗)| . z=ejω Thez-transformrelationshipforthesecondtermcanbeseentoresultin ∞ H∗(1/z∗) = ∑ h∗zn. (3) n n=−∞ Thedoubleconjugation(onceonzandagainonH(·))hastheeffectofleavingthezn termsuncon- jugated. Onlythecoefficientsareconjugated. TheresponseH∗(1/z∗)hasanexpansioninpositive powersofz,i.e. thetimeresponseisflippedaboutn = 0. Minimum-Phase&All-PassFilters 2 FortherationaltransformofEq.(1),themagnitude-squaredresponseis M ∏(1−z z−1)(1−z∗z) k k |H(ω)|2 = |G|2 k=1 (cid:12)(cid:12) N (cid:12) ∏(1− pkz−1)(1− p∗kz)(cid:12)(cid:12) (cid:12) k=1 (cid:12)z=ejω (cid:12) M (cid:12) ∏(1−2Re[z e−jω]+|z(cid:12)|2) k k = |G|2 k=1 (4) N ∏(1−2Re[p e−jω]+|p |2) k k k=1 M ∏(1−2r cos(ω−θ )+r2) k k k = |G|2 k=1 . N ∏(1−2ρ cos(ω−φ )+ρ2) k k k k=1 The frequency dependent terms in the magnitude-squared response are sinω and cosω. For a system with real coefficients, the magnitude-squared response is symmetrical about ω = 0 and canbeexpressedasaratioofpolynomialsincosω. 1.2 PhaseResponse ThephaseresponseofafilterwithfrequencyresponseH(ω)is1 H (ω) arg[H(ω)] = tan−1 I , (5) H (ω) (cid:18) R (cid:19) where H (ω)and H (ω)aretherealandimaginarypartsofthefrequencyresponse,respectively. R I FortherationalresponsegivenbyEq.(1),thephaseresponseis M −Im[z e−jω] N −Im[p e−jω] arg[H(ω)] = arg[G]−n ω+ ∑ tan−1 k − ∑ tan−1 k 0 k=1 1−Re[zke−jω]! k=1 1−Re[pke−jω]! M r sin(ω−θ ) N ρ sin(ω−φ ) = arg[G]−n ω+ ∑ tan−1 k k − ∑ tan−1 k k . 0 1−r cos(ω−θ ) 1−ρ cos(ω−φ ) k=1 (cid:18) k k (cid:19) k=1 (cid:18) k k (cid:19) (6) Forasystemwithrealcoefficients,thephaseresponseisanti-symmetricalaboutω = 0. 1Someauthorsdefinethephasewithanegativesign.Here,wewillrefertothenegativeofthephaseasthephaselag. Minimum-Phase&All-PassFilters 3 1.3 GroupDelayResponse Thegroupdelayresponseisthenegativederivativeofthephaseresponse2 darg[H(ω)] τ (ω) = − g dω dH (ω) dH (ω) (7) H (ω) R −H (ω) I I R dω dω = . H2(ω)+H2(ω) R I Amorecompactformofthesameresultis dH(ω)/dω τ (ω) = −Im . (8) g H(ω) (cid:20) (cid:21) Sincethegroupdelayisthederivativeofthephaselag,theareaunderthegroupdelaycurveover afrequencyintervalof2π isequaltothechangeinthephaselagoverthesameinterval. Consider the group delay for just one of the terms in the numerator of the rational response giveninEq.(1),3 |z |2−Re[z e−jω] τ(zk)(ω) = k k g 1−2Re[z e−jω]+|z |2 k k (9) r2−r cos(ω−θ ) = k k k . 1−2r cos(ω−θ )+r2 k k k Theoverallgroupdelayfortherationaltransferfunctionis M |z |2−Re[z e−jω] N |p |2−Re[p e−jω] τ (ω) = n + ∑ k k − ∑ k k g o 1−2Re[z e−jω]+|z |2 1−2Re[p e−jω]+|p |2 k=1 k k k=1 k k (10) M r2−r cos(ω−θ ) N ρ2−ρ cos(ω−φ ) = n + ∑ k k k − ∑ k k k . o 1−2r cos(ω−θ )+r2 1−2ρ cos(ω−φ )+ρ2 k=1 k k k k=1 k k k Thegroupdelayresponsehastermsinsinωandcosω. Forasystemwithrealcoefficients,the groupdelayresponseissymmetricalaboutω = 0andcanbeexpressedasaratioofpolynomialsin cosω. Asaratioofpolynomials,thegroupdelayisoftenbetterbehavedthanthephaseresponse – the phase response can have phase jumps and ambiguities of multiples of 2π. Thus it may be preferable to characterize a system or filter by the group delay response rather than the phase 2The derivationof the expressionfor the groupdelay responseusesthe identity dtan−1(y/x)/du = (xdy/du− ydx/du)/(x2+y2). 3TheresultusestherelationsdRe[z e−jω]/dω=Im[z e−jω]anddIm[z e−jω]/dω=−Re[z e−jω]. k k k k Minimum-Phase&All-PassFilters 4 response. 1.4 Example As an example, consider the filter H(z) = 1−z−1. This is just the case in which there is a single zeroinEq.(1), H(ω) = 1−e−jω (11) = 2je−jω/2sin(ω/2). The amplitude response is 2|sin(ω/2)|, which has a discontinuous derivative at ω = 0. From Eq.(4),themagnitude-squaredresponseis |H(ω)|2 = 2(1−cosω). (12) Thisisapolynomialincosω. Thephaseresponseofthisfilteris(fromEq.(6)) sinω arg[H(ω)] = tan−1 . (13) 1−cosω (cid:18) (cid:19) Thephaseresponseisπ/2forω = 0+ andis−π/2forω = 0−. Thereisaphasejumpofsizeπ at ω = 0. DirectlyfromEq.(11),wecanseethatthephasecanalsobewrittenas arg[H(ω)] = π/2−ω/2+arg[sin(ω/2)]. (14) Inthisform,thelineartrendinthephaseisclear. Thelasttermgivesaphasejumpof±πatω = 0. Forthissameexample,thegroupdelayevaluatesto 1 1−cosω τ (ω) = . (15) g 2 1−cosω The group delay is well behaved and is equal to 1/2 sample everywhere, including at ω = 0 (as shown by invoking L’Hoˆpital’s rule twice). The area under the group delay curve (over 2π) is equaltoπ. 2 CausalStable Filters For the sequel, we consider causal stable filters with rational z-transforms. The poles of these filtersmustlieinsidetheunitcircle. Thepositionsofthezeros,however,can lieinsideoroutside theunitcircle. Minimum-Phase&All-PassFilters 5 Wewillworkwithasimplifiedversionoftherationalfunction H(z)inEq.(1), B(z) H(z) = A(z) M (16) b ∏(1−z z−1) 0 k = k=1 . A(z) Wehavesetthedelayfactorn whichappearedinEq.(1)tozero.4 0 2.1 SystemswiththeSameMagnitudeResponse Considerthezeroatz correspondingtoarootfactor(1−z z−1)intheexpansionofthenumerator k k polynomial B(z). Then H∗(1/z∗) which appears in the expression Eq. (2) for the magnitude- squaredresponsehas a rootfactor (1−z∗z) in its numeratorcorrespondingtoa zero at 1/z∗, i.e. k k atthereciprocalradius. Ifweexpresszk asrkejθk,wegettherootsymmetriesshowninFig.1. Ifzk isinsidetheunitcircle(asshown),then1/z∗ isoutsidetheunitcircle. Similarlyifz isoutsidethe k k unitcircle,then1/z∗ isinsidetheunitcircle. k 1 ej k r ej k rk k Fig.1 Zerosymmetries: reciprocalzerosatz and1/z∗. k k Consider therootsof theproduct H(z)H∗(1/z∗) usedto calculate themagnitude responseof H(z). We can randomly assign one of each pair of roots (z ,1/z∗) to a new response G(z). The k k new responsewill have the same magnitude response as H(z) since one of roots will be in G(z) andtheotherwillbeinG∗(1/z∗). Thisshowsthatingeneraltherearemanyresponseswhichhave thesamemagnituderesponse. However,thephaseresponsesofthesesystemswillbedifferent. 4Havingn <0wouldviolatecausality.Havingn >0givesazeroatinfinitywhichwillpreventtheresponsefrom 0 0 beingminimumphase. Minimum-Phase&All-PassFilters 6 3 Minimum-PhaseSystems Acausalstablefilterissaidtominimum-phaseifallofitszerosareinsidetheunitcircle. Amaximum- phase systemhas all of its zeros outside the unit circle. A systemwith a zero on the unit circle is notstrictlyminimum-phase. ConsiderarootfactorofthenumeratorinEq.(16). Thefrequencyresponseofthatfactoris B(k)(ω) = 1−r ej(θk−ω). (17) k We can see that this is the equation for a circle of radius r centred at (1,0). Figure 2 shows two k < cases plottedin the complex plane. The first is minimum-phase (r 1). As ω changes, the end k ofthedashedlinefollowsthecircle. Thephaseoftheresultistheanglethatthedashedlinemake with the real axis. The phase varies up and down, with maximum excursions of less than ±π/2 > andreturnstoitsinitialvalue. Thesecondcaseshownismaximumphase(r 1). Thephasehas k atotalexcursionof2π sincetheendofthedashedlinenowencirclestheorigin. 1(cid:16)r ej((cid:84)k(cid:16)(cid:90)) 1(cid:16)rkej((cid:84)k(cid:16)(cid:90)) k (1,0) (1,0) (a)Phase for a minimum-phase (b)Phaseforamaximum-phaseroot rootfactor factor Fig.2 Illustratingthephaseofarootfactorinthecomplexplane. 3.1 Minimum-Phase/All-PassDecomposition Thenumeratorpolynomial B(z)oftheexpansionof H(z)inEq.(16)canbesplitintothreeparts. B(z) = B (z)B (z)B (z), (18) min uc max where B (z)isminimum-phase, B (z) hasunitcirclezeros,and B (z)ismaximum-phase. A min uc max minimum-phasesystemhasonlythefirstfactor. Minimum-Phase&All-PassFilters 7 Letthemaximum-phasetermbewrittenas K B (z) = β ∏(1−β z−1). (19) max 0 k k=1 Sincethistermismaximumphase,|β | > 1fork = 1,...,K. Nowwrite B (z)as k max B (z) B (z) = z−KB∗ (1/z∗) max max max z−KBm∗ax(1/z∗) (20) = G (z)G (z). min all ThefirsttermG (z) isminimum-phase. Thedelayfactorz−K shiftsthistermtomakethecorre- min sponding time responsecausal. The second term G (z) is all-pass. It has a constant magnitude all responseequaltounity. ThiscanbeseenfromEq.(2)byevaluating G (z)G∗ (1/z∗) = 1 forallz. (21) all all Theall-passfiltercanbewrittenas β K 1−β z−1 G (z) = 0 ∏ k . (22) all β∗ z−1(1−β∗z) 0 k=1 k ThisisastablecausalIIRfilter–thedenominatorhasrootsinsidetheunitcircle. Theall-passfilter ismaximumphase. Based on the decomposition of Eq. (18), we can express any causal stable filter H(z) as the productofaminimum-phasefilter,afilterwithunitcirclezeros,andanall-passfilter, H(z) = H (z)B (z)G (z). (23) min uc all Bythisconstruction H (z)canbewrittenas min B (z)z−KB∗ (1/z∗) H (z) = min max . (24) min A(z) 4 All-PassFilters An all-pass filter is a filter for which the magnitude of the frequency response is constant. All- passfilters,exceptfortrivialfilters,areIIR.Forthesequel,wesetthemagnitudeofthefrequency responsetounity. Startwiththesimplestnon-trivial all-pass filter,which canformonesectionof Minimum-Phase&All-PassFilters 8 acausal,stableall-passfilter, z−1−α∗ H(k)(z) = k . (25) all 1−α z−1 k Herewehaveexpressedtheall-passfilterintermsofitspolelocationα . Inthisform,|α | < 1. k k Ageneralall-passfiltercanbeformedasthecascadeofall-passsections K z−1−α∗ H (z) = ∏ k . (26) all 1−α z−1 k=1 k Iftheall-passfilterhasrealcoefficients,thenforeachcomplexrootα ,theremustbeacorrespond- k ingconjugaterootα∗. k FromEq.(26)wecanseethatanall-passfilterischaracterizedbyhavinganumeratorpolyno- mialwhichhasthecoefficientsofthedenominatorpolynomial,conjugatedandinreverseorder, z−KD∗(1/z∗) H (z) = . (27) all D(z) ThenfromEq.(2),weseethatanyfilterofthisformhasconstantmagnitude |H (ω)|2 = 1. (28) all 4.1 PhaseResponseofanAll-PassFilter Ifαk isexpressedasrkejθk,thephaseresponseforanall-passsectionis arg[H(k)(ω)] = arg[e−jω −α∗]−arg[1−α e−jω)] all k k = arg[e−jω]+arg[1−α∗ejω]−arg[1−α e−jω] k k (29) r sin(ω−θ ) = −ω−2tan−1 k k 1−r cos(ω−θ ) (cid:18) k k (cid:19) The denominator inside the tan−1(·) function is positive for r < 1 (corresponding to a stable, k causal all-pass filter). The numerator can change sign, and so the phase contribution due to the tan−1(·) function is limited to ±π/2 (and 2tan−1(·) is limited to ±π). The tan−1(·) function contributes an oscillation around the linear phase term – the phase is monotonic downward (as showninthediscussionofthegroupdelayresponsewhichfollows). Thephaseresponseforeach all-passsectiondecreasesby2π asω increasesby2π. The phase response for the overall filter is the sum of the phases for each section. Thus the

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in the amplitude, phase, and group delay responses of these filters. The circle of Apollonius is closely related to the property that the magnitude
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