Minimum-Phase & All-Pass Filters

30 Pages·2011·0.28 MB·English

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
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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