Richard Brown III D Richard Brown III 1 6 brPage 2br DSP MinimumPhase AllPass Decomposition MinimumPhase AllPass Decomposition Suppose we have a causal stable rational transfer function with one or more zeros outside the unit circle We denote the z ID: 22139
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DSP:Minimum-PhaseAll-PassDecomposition DigitalSignalProcessingMinimum-PhaseAll-PassDecomposition D.RichardBrownIII D.RichardBrownIII 1/6 DSP:Minimum-PhaseAll-PassDecomposition Minimum-PhaseAll-PassDecomposition SupposewehaveacausalstablerationaltransferfunctionH(z)withoneormorezerosoutsidetheunitcircle.Wedenotethezerosoutsidetheunitcircleasfc1;:::;cMg.Wecanformaminimumphasesystemwiththesamemagnituderesponsebyre\rectingthesepolestotheirconjugatesymmetriclocationsinsidetheunitcircle,i.e.,Hmin(z)=H(z)z 1 c1 1 c1z 1z 1 cM 1 cMz 1| {z }unit-magnitudeall-pass lterItshouldbeclearthatHmin(z)andH(z)havethesamemagnituderesponse.Moreover,wehavethedecompositionH(z)=Hmin(z)z 1 c1 1 c1z 1z 1 cM 1 cMz 1 1=Hmin(z)Hap(z)whereHap(z)isanall-pass lter. D.RichardBrownIII 2/6 DSP:Minimum-PhaseAll-PassDecomposition Example1(part1of2) SupposeH(z)=1 2z 1 1+1 3z 1Thissystemisclearlynotminimumphasesinceithasazeroatz=2.Wecanre\rectthiszeroinsidetheunitcirclewithanall-pass ltertowriteHmin(z)=H(z)z 1 2 1 2z 1=z 1 2 1+1 3z 1henceH(z)=z 1 2 1+1 3z 1| {z }Hmin(z)1 2z 1 z 1 2| {z }Hap(z) D.RichardBrownIII 3/6 DSP:Minimum-PhaseAll-PassDecomposition Example1(part2of2) Notethat,justrequiringthezerostobeinsidetheunitcircledoesnotuniquelyspecifyHmin(z).Forexample,wecouldalsowriteH(z)=2 z 1 1+1 3z 1| {z }Gmin(z)2z 1 1 z 1 2| {z }Gap(z)whereG= Hforbothsystems.Doesitmatterwhichonewepick?Tosatisfytheminimumphasedelayproperty,recallthatwerequire\Hmin(ej0)=0.NotethatHmin(ej0)=1 2 1+1 30andGmin(ej0)=2 1 1+1 3�0hence\Hmin(ej0)= and\Gmin(ej0)=0.ThecorrectansweristochoosetheGmin(z)Gap(z)decomposition. D.RichardBrownIII 4/6 DSP:Minimum-PhaseAll-PassDecomposition Example2 SupposeH(z)=(1+3z 1)(1 1 2z 1) z 1(1+1 3z 1)Thisisalsoclearlynotminimumphaseduetothezeroatz= 3.Wecanre\rectthiszeroinsidetheunitcircletowriteHmin(z)=H(z)z 1+3 1+3z 1=(z 1+3)(1 1 2z 1) z 1(1+1 3z 1)=3(1 1 2z 1) z 1Recallthatminimumphase ltersmustbecausal.SinceHmin(z)=3z 1 2isclearlynotcausal,wecanfactoroutthez 1denominatorterm(puttingitintheall-pass ltersinceitdoesnota ectthemagnituderesponse)toarriveatH(z)=31 1 2z 1| {z }Hmin(z)z 1+3 z 1(1+3z 1)| {z }Hap(z) D.RichardBrownIII 5/6 DSP:Minimum-PhaseAll-PassDecomposition EqualizationofNonminimumPhaseChannel SupposeH1(z)=(z 4)(z+5) (z+0:5)(z 0:3)withROCjzj�0:5.WeformtheinversesystemH2(z)=(z+0:5)(z 0:3) (z 4)(z+5).Notethatthereisnocausalstableinversehere.OneapproachinthiscaseistofactorH1(z)intoacausalstableminimumphase lterandacausalstableallpass lter,i.e.H1(z)=Hmin(z)Hap(z)=(4z 1)(5z+1) (z+05)(z 03)| {z }minimumphase(z 4)(z+5) (4z 1)(5z+1)| {z }allpassROC:z�05andinvertjusttheminimumphasecomponent,i.e.H2(z)=(z+0:5)(z 0:3) (4z 1)(5z+1).ThenH1(z)H2(z)=1butratherH1(z)H2(z)=Hap(z).Hence,theequalizerH2(z)correctsthemagnitudedistortion,butleavessomeresidualphasedistortion. D.RichardBrownIII 6/6