More On Linear Predictive Analysis 主講人：虞台文. Contents Linear Prediction Error Computation of the Gain Frequency Domain Interpretation of LPC Representations

More On Linear Predictive Analysis

主講人：虞台文

Contents Linear Prediction Error Computation of the Gain Frequency Domain Interpretation of LPC Representations of LPC Coefficients

– Direct Representation– Roots of Predictor Polynomials– PARCO Coefficients– Log Area Ratio Coefficients– Line Spectrum Pair


Linear Prediction Error

LPC Error

1 ),()()( 00

αnGuknsαnep

kk

1 ),()()( 00

αnGuknsαnep

kk

Examples

PremphasizedSpeech Signals

Could be used for Pitch Detection

Normalized Mean-Squared Error

1

0

2

1

0

2

)(

)(

N

mn

pN

mn

n

ms

meV

1

0

2

1

0

2

)(

)(

N

mn

pN

mn

n

ms

meV

m

mn

ms

meV

)(

)(

2

2

m

mn

ms

meV

)(

)(

2

2

AutocorrelationMethod

1

0

2

1

0

2

)(

)(

N

mn

N

mn

n

ms

meV

1

0

2

1

0

2

)(

)(

N

mn

N

mn

n

ms

meVCovariance

Method

General Form

Normalized Mean-Squared Error

p

i

p

jj

ijin α

c

cαV

0 0 00

p

i

p

jj

ijin α

c

cαV

0 0 00

p

i

iin r

rαV

0 0

p

i

iin r

rαV

0 0

n

iin kV

1

2 )1(

n

iin kV

1

2 )1(

1

0

2

1

0

2

)(

)(

N

mn

pN

mn

n

ms

meV

1

0

2

1

0

2

)(

)(

N

mn

pN

mn

n

ms

meV

AutocorrelationMethod

1

0

2

1

0

2

)(

)(

N

mn

N

mn

n

ms

meV

1

0

2

1

0

2

)(

)(

N

mn

N

mn

n

ms

meVCovariance

Method

Experimental Evaluation on LPC Parameters

Frame Width N

Filter Order p

Conditions:

1. Covariance method and autocorrelation method

2. Synthetic vowel and Nature speech

3. Pitch synchronous and pitch asynchronous analysis

Pitch Synchronous Analysis

/i/

zero: the same order as the synthesizer.

The frame was beginning at the beginning of a pitch period.

Covariance method is more suitable for pitch synchronous analysis.

Covariance method is more suitable for pitch synchronous analysis.

Why the error increases?

Pitch Asynchronous Analysis

/i/

Both covariance and autocorrelation methods exhibit similar performance.

Both covariance and autocorrelation methods exhibit similar performance.

Monotonically decreasing

Frame Width Variation

/i/

Why the errors jump high when the frame size nears the multiples of pitch period?

Why the errors jump high when the frame size nears the multiples of pitch period?

The errors resulted by covariance and autocorrelation methods are compatible when N > 2P.

The errors resulted by covariance and autocorrelation methods are compatible when N > 2P.

Pitch Synchronous Analysis

Both for synthetic and nature speeches, covariance method is more suitable for pitch synchronous analysis.

Both for synthetic and nature speeches, covariance method is more suitable for pitch synchronous analysis.

Pitch Asynchronous Analysis

Both for synthetic and nature speeches, two methods are compatible.

Both for synthetic and nature speeches, two methods are compatible.

Frame Width Variation

Both for synthetic and nature speeches, the errors resulted by covariance and autocorrelation methods are compatible when N > 2P.

Both for synthetic and nature speeches, the errors resulted by covariance and autocorrelation methods are compatible when N > 2P.


Computation of the Gain

Speech Production Model (Review)

Impulse Train Generator


Random NoiseGenerator


Time-VaryingDigital Filter


Vocal TractParameters

G

u(n)

s(n)

H(z)

Speech Production Model (Review)







Vocal TractParameters

G

u(n)

s(n)

p

k

kk za

1

1

1

p

k

kk za

1

1

1

p

k

kk za

G

zU

zSzH

1

1)(

)()(

p

k

kk za

G

zU

zSzH

1

1)(

)()(

)()()(1

nGuknsansp

kk

)()()(1

nGuknsansp

kk

Linear Prediction Model (Review)

)()()(1

nGuknsansp

kk

)()()(1

nGuknsansp

kk

p

kk knsαns

1

)()(ˆLinear Prediction:

)()(ˆ)( nensns Error compensation:

)()()(1

neknsαnsp

kk

)()()(1

neknsαnsp

kk

Speech Production vs. Linear Prediction

)()()(1

nGuknsansp

kk

)()()(1

nGuknsansp

kk

)()()(1

neknsαnsp

kk

)()()(1

neknsαnsp

kk

Speech production:

Linear Prediction:

Vocal Tract Excitation

Linear Predictor Error

ak = k

)()( nGune )()( nGune

Speech Production vs. Linear Prediction

)()()(1

nGuknsansp

kk

)()()(1

nGuknsansp

kk

)()()(1

neknsαnsp

kk

)()()(1

neknsαnsp

kk

Speech production:

Linear Prediction:

p

kk knsansnGu

1

)()()(

p

kk knsαnsne

1

)()()(

)()( nGune )()( nGune

The Gain

p

kk knsansnGu

1

)()()(

p

kk knsαnsne

1

)()()(

Generally, it is not possible to solve for G in a reliable way directly from the error signal itself.

Instead, we assume

1

0

221

0

2 )()(N

m

N

mn muGmeE

Energy of Error Energy of Excitation

1/A(z)

Assumptions about u(n)

1

0

221

0

2 )()(N

m

N

mn muGmeE

1

0

221

0

2 )()(N

m

N

mn muGmeE

Voiced Speech

Unvoiced Speech

)()( nδnu )()( nδnu

)()]()([ mδmnunuE )()]()([ mδmnunuE

This requires that both glottal pulse shape and lip radiation are lumped into the vocal tract model.

G(z)G(z) V(z)V(z) R(z)R(z)

G

u(n)=(n)h(n)

1/A(z)

Gain Estimation for Voiced Speech

1

0

221

0

2 )()(N

m

N

mn muGmeE

1

0

221

0

2 )()(N

m

N

mn muGmeE

Voiced Speech )()( nδnu )()( nδnu

This requires that both glottal pulse shape and lip radiation are lumped into the vocal tract model.


G

u(n)=(n)h(n)

This requires that p is sufficiently large.This requires that p is sufficiently large.

1/A(z)

Gain Estimation for Voiced Speech


G

u(n)=(n)h(n)

)()(

zA

GzH

)()(

zA

GzH (n) h(n)

p

k

kk zazA

1

1)(

p

k

kk zazA

1

1)(

Correlation Matching

)()(

zA

GzH

)()(

zA

GzH (n) h(n)

p

k

kk zazA

1

1)(

p

k

kk zazA

1

1)(

Define

n

ij jnhinhρ )()(

0

|)|()(n

jinhnh Assumed causal.

||̂ jir Autocorrelation function of the impulse response.

0

)()(ˆn

m nmhnhr

0

)()(ˆn

m nmhnhr


)()(

zA

GzH

)()(

zA

GzH (n) h(n)

p

k

kk zazA

1

1)(

p

k

kk zazA

1

1)(

0

)()(ˆn

m nmhnhr

0

)()(ˆn

m nmhnhr

Autocorrelation function of the speech signal

0

)()(n

m nmsnsr

0

)()(n

m nmsnsrIf H(z) correctly model the speech production system, we should have

mm rr ˆ mm rr ˆ


)()(

zA

GzH

)()(

zA

GzH (n) h(n)

p

k

kk zazA

1

1)(

p

k

kk zazA

1

1)(

0

)()(ˆn

m nmhnhr

0

)()(ˆn

m nmhnhr

0

)()(n

m nmsnsr

0

)()(n

m nmsnsr

GzAzH )()( )()(0

nGδinhap

ii

nn

p

ii jnhnGδjnhinha )()()()(

0

)()()(0

jGhjnhinhap

i ni


0

)()(ˆn

m nmhnhr

0

)()(ˆn

m nmhnhr

0

)()(n

m nmsnsr

0

)()(n

m nmsnsr

)()()(0

jGhjnhinhap

i ni

)(ˆ0

|| jGhrap

ijii

Assumed causal.

mm rr ˆ mm rr ˆ

2

0

Grap

iii

0j

00

||

p

ijiira pj ,,2,1


2

0

Grap

iii

0j

00

||

p

ijiira pj ,,2,1

The same formulation as autocorrelation method.

The Gain for voice speech

2

0

Grap

iii

0j

E n

nEG 2nEG 2

More on Autocorrelation

H(z)H(z)x(n) y(n))()()( knhkxny

k

Define )()(),( mnxnxmnxx

The stationary assumption implies

)]()([)],([)( mnxnxEmnEm xxxx )]()([)],([)( mnxnxEmnEm xxxx

AssumedStationary

Properties of LTI Systems

H(z)H(z)x(n) y(n)( ) ( ) ( )

k

y n h k x n k

Define )()(),( mnynymnyy


lk

lmnxlhknxkh )()()()(

lk

lmnxknxlhkh )()()()(



])()([)()()],([

lk

yy lmnxknxElhkhmnE

][)()(

l

xxk

lkmlhkh

Define )()(),( mnynymnyy

lk

lmnxlhknxkh )()()()(

lk

lmnxknxlhkh )()()()(



])()([)()()],([

lk

yy lmnxknxElhkhmnE

][)()(

l

xxk

lkmlhkh

Independent on n

)]()([)],([)( mnynyEmnEm yyyy )]()([)],([)( mnynyEmnEm yyyy

y(n) is also stationary.



][)()(][

l

xxk

yy lkmlhkhm

][)()(

k l

xx lkmlhkh ll+k

][)()(

k l

xx lmklhkh

l k

xx klhkhlm )()(][


l k

xxyy klhkhlmm )()(][][

k

klhkhlc )()()(

k

klhkhlc )()()(Define

l

xxyy lclmm )(][][

Estimatedfrom input

Estimatedfrom output

FilterDesign

The Gain for Unvoiced Speech

)()]()([ mδmnunuE )()]()([ mδmnunuE

)()(

zA

GzH

)()(

zA

GzH u(n) s(n)

p

k

kk zazA

1

1)(

p

k

kk zazA

1

1)(

)()()(1

nGuknsansp

kk

)()()()]()([1

mnsnGuknsaEmnsnsEp

kk

p

kkm mnsnuGEmnsknsEar

1

)]()([)]()([ˆ

p


1

)]()([)]()([ˆ


p


1

)]()([)]()([ˆ

p


1

)]()([)]()([ˆ

p

kkmkm mnsnuGErar

1|| )]()([ˆˆ

=?


p

kkmkm mnsnuGErar

1|| )]()([ˆˆ

=?

0 ,0)]()([ mmnsnuE

Why?

:0m

)()()()]()([1

nGuknsanuEnsnuEp

kk

GnunuGE )()(


p

kkmkm mnsnuGErar

1|| )]()([ˆˆ

0 ,0)]()([ mmnsnuE

0 ,)]()([ mGmnsnuE

Estimated using rm

pmrap

kkmk ,2,1 ,0

0||

0 ,0

2

mGrap

kkk


pmrap

kkmk ,2,1 ,0

0||

0 ,0

2

mGrap

kkk

Once again, we have the same formulation as autocorrelation method.

Furthermore, nEG 2nEG 2


Frequency Domain Interpretation of LPC

Spectral Representation of Vocal Tract

)(1

)(

1

zA

G

za

GzH p

k

kk

)(1

)(

1

jp

k

kjk

j

eA

G

ea

GeH

Spectra

Frequency Domain Interpretation of Mean-Squared Prediction Error

)(1

0

2 meEpN

mnn

)()()( zAzSzE nn

Parseval’s Theorem

dωeAeSπ

E jωπ

π

jωnn

22 |)(||)(|2

1


dωeAeSπ

E jωπ

π

jωnn

22 |)(||)(|2

1

)()(

jωjω

eA

GeH

)()(

jωjω

eH

GeA

dωeH

eS

πE

π

π jω

jωn

n 2

2

|)(|

|)(|

2

1 dωeH

eS

πE

π

π jω

jωn

n 2

2

|)(|

|)(|

2

1


dωeH

eS

πE

π

π jω

jωn

n 2

2

|)(|

|)(|

2

1 dωeH

eS

πE

π

π jω

jωn

n 2

2

|)(|

|)(|

2

1

|Sn(ej)| > |H(ej)| contributes more to the total error than |Sn(ej)| < |H(ej)|.

Hence, the LPC spectral error criterion favors a good fit near the spectral peak.

Spectra


Representations of LPC Coefficients ---

Direct Representation

Direct Representation

pp zazazazA 2

21

11)(

Coding ai’s directly.

z1

z1

z1

a1

a2

ap

uG[n] uL[n]G

G/A(z)G/A(z)

Disadvantages

The dynamic ranges of ai’s is relatively large.

Quantatization possibly causes instability problems.



Roots of Predictor Polynomials

Roots of the Predictor Polynomial

pp zazazazA 2

21

11)(

p

kk zz

1

1)1(

Coding p/2 zk’s.

kjkk erz

Dynamic range of rk’s?

Dynamic range of k’s?

10 kr

k0

The Application

2/

1

11 )1)(1()(p

k

jk

jk zerzerzA kk

2/

1

221cos21p

kkkk zrzr

kbk er Let kk rb ln

2/

1

221cos21p

k

bk

b zeze kk

Formant Analysis Application.Formant Analysis Application.

Implementation

Each Stage represents one formant frequency and its corresponding bandwidth.

11 cos2 r

21r

22 cos2 r

22r

2/2/ cos2 ppr

22/pr

][nuG ][nuLG

z1

z1

z1

z1

z1

z1

G/A(z)G/A(z)



PARCO Coefficients

PARCO Coefficients

1)(0 na

1,,2,1 ,)1()1()(

niakaa ninn

ni

ni

nn

n ka )(

Step-Up Procedure:

piaa pii ,,2,0 )( piaa p

ii ,,2,0 )(

PARCO Coefficients

1,,2,1 ,1 2

)()()1(

nik

akaa

n

ninn

nin

i

)(nnn ak

Step-Down Procedure:

where n goes from p to p1, down to 1 and initially we set:

piaa ip

i ,,2,1 ,)(

Dynamic range of ki’s? 11 ik



Log Area Ratio Coefficients

Log Area Ratio Coefficients

ki’s: Reflection Coefficients

ii

iii AA

AAk

1

1

1/

1/

1

1

ii

ii

AA

AA

i

i

i

i

k

k

A

A

1

11

gi’s: Log Area Ratios

1

1logi

ii A

Ag

i

i

k

k

1

1log

1

1

i

i

g

i g

ek

e



Line Spectrum Pair

LPC Coefficients

mmm zazazazA 2

21

11)(

where m is the order of the inverse filter.

Line Spectrum Pair (LSP) is an alternative LPC spectral representation.

If the system is stable, all zeros of the inverse filter are inside the unit circle.

Line Spectrum Pair LSP contains two polynomials. The zeros of the the two polynomials have the f

ollowing properties:– Lie on unit circle– Interlaced– Through quantization, the minimum phase property

of the filter is kept.

Useful for vocoder application.

Recursive Relation of the inverse filter

)()()( 1)1(11

zAzkzAzA m

mmmm

where km+1 is the reflection coefficient of the m+1th tube.

Recall that

ii

iii AA

AAk

1

1

ii

iii AA

AAk

1

1ki =1:

ki =1:

Special cases:

Ai+1

Ai+1=0

LSP Polynomials

)()()( 1)1( zAzzAzP mm

m

)()()( 1)1( zAzzAzQ mm

m

)()()( 21 zQzPzAm )()()( 21 zQzPzAm

mmm zazazazA 2

21

11)(

Properties of LSP Polynomials


Show thatThe zeros of P(z) and Q(z) are on the unit circle and interlaced.

Proof )()()( 1)1( zAzzAzP m

mm


m


m)()()( 1)1( zAzzAzQ m

mm

)(

)(1)()(

1)1(

zA

zAzzAzP

m

mmm

)(

)(1)()(

1)1(

zA

zAzzAzQ

m

mmm

)(

)()(

1)1(

zA

zAzzH

m

mm

)(

)()(

1)1(

zA

zAzzH

m

mm

)(1)( zHzAm

)(1)( zHzAm

Proof

)(

)()(

1)1(

zA

zAzzH

m

mm

)(

)()(

1)1(

zA

zAzzH

m

mm

)(1)()( zHzAzP m )(1)()( zHzAzP m

)(1)()( zHzAzQ m )(1)()( zHzAzQ m

m

iim zzzA

1

1)1()(

m

iim zzzA

1

1 )1()(

m

iim

m zzzzAz1

111)1( )()(

m

i i

i

zz

zzz

11

11

)1(

)(

m

i i

i

zz

zzzzH

1

1

)(

)1()(

m

i i

i

zz

zzzzH

1

1

)(

)1()(

1 , ijω

ii rerz i 1 , ijω

ii rerz i

Proof

m

i i

i

zz

zzzzH

1

1

)(

)1()(

m

i i

i

zz

zzzzH

1

1

)(

)1()(

1 , ijω

ii rerz i 1 , ijω

ii rerz i

)||1)(||1(|||1| 2222iii zzzzzz )||1)(||1(|||1| 2222

iii zzzzzz

)1)(1(|1| **2 zzzzzz iii **22 ||||1 zzzzzz iii

))((|| **2iii zzzzzz

zzzzzz iii**22 ||||



Proof )||1)(||1(|||1| 2222

iii zzzzzz )||1)(||1(|||1| 2222iii zzzzzz

m

i i

i

zz

zzzzH

1

1

)(

)1()(

m

i i

i

zz

zzzzH

1

1

)(

)1()(

1 , ijω

ii rerz i 1 , ijω

ii rerz i

>0

1||1

1||1

1||1

|)(|

z

z

z

zH



Proof )||1)(||1(|||1| 2222

iii zzzzzz )||1)(||1(|||1| 2222iii zzzzzz

m

i i

i

zz

zzzzH

1

1

)(

)1()(

m

i i

i

zz

zzzzH

1

1

)(

)1()(

1 , ijω

ii rerz i 1 , ijω

ii rerz i

1||1

1||1

1||1

|)(|

z

z

z

zH



P(z)=0Q(z)=0

iff H(z) = 1.

This concludes that the zeros of P(z) and Q(z) are on the unit circle.

Proof (interlaced zeros)

m

i i

i

zz

zzzzH

1

1

)(

)1()(

m

i i

i

zz

zzzzH

1

1

)(

)1()( 1 , i

jωii rerz i 1 , i

jωii rerz i

Fact: H(z) is an all-pass filter.

i

iap zz

zzzH

i

*1

)(

)cos(1

)sin(tan2)( 1

ii

iii ωωr

ωωrωωφ

)()( ωjφjω eeH )()( ωjφjω eeH

)()( ωφjωap

i

ieeH

m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()(

m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()( Phase

One can verify that (0) = 0 and (2) = 2(m+1)


)()( ωjφjω eeH )()( ωjφjω eeH

m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()(

m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()( Phase

πnωφ

nπωφeH jω

)12()(1

2)(1)(



Therefore, z=1 is a zero of Q(z).

zeros of P(z)

zeros of Q(z)

One can verify that (0) = 0 and (2) = 2(m+1)


m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()(

m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()(

(0) = 0

(2) = 2(m+1)

0 2

()

2(m+1)

Is this possible?Is this possible?


m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()(

m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()(

0 2

()

2(m+1)

Is this possible?Is this possible?

(0) = 0

(2) = 2(m+1)


m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()(

m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()(

Group Delay

ω

ωφωτ

)(

)(

m

i iii

i

ωωrr

r

12

2

)cos(21

11 > 0

() is monotonically decreasing.

(0) = 0

(2) = 2(m+1)


m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()(

m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()(

0 2

()

2(m+1)

2345 ...

Typical shape of ()

Typical shape of ()

(0) = 0

(2) = 2(m+1)

m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()(

m

i ii

ii

ωωr

ωωrωmωφ

1

1

)cos(1

)sin(tan2)1()(

0 2

()

2(m+1)

2345 ...

Typical shape of ()

Typical shape of ()




πnωφ

nπωφeH jω

)12()(1

2)(1)(

πnωφ

nπωφeH jω

)12()(1

2)(1)(

Q(ej)=0Q(ej)=0

P(ej)=0P(ej)=0

Q(ej)=0Q(ej)=0

P(ej)=0P(ej)=0

Q(ej)=0Q(ej)=0

P(ej)=0P(ej)=0

Proof (interlaced zeros) 0 2

()

2(m+1)

2345 ...

Typical shape of ()

Typical shape of ()

Q(ej)=0Q(ej)=0

P(ej)=0P(ej)=0

Q(ej)=0Q(ej)=0

P(ej)=0P(ej)=0

Q(ej)=0Q(ej)=0

P(ej)=0P(ej)=0

There are 2(m+1) cross points from 0 , these constitute the 2(m+1) interlaced zeros of P(z) and Q(z).

Quantization of LSP Zeros


m

i

j zezP i

2

1

1)1()( 2

m

i

j zezQ i

2

1

1)1()( 12

For effective transmission, we quantize i’s into several levels, e.g., using 5 bits.

Is such a quantization detrimental? Is such a quantization detrimental?

Minimum Phase Preserving Property

Show that in quantizing the LSP frequencies, the reconstructed all-pole filter preserves its minimum phase property as long as the zeros has the properties shown in the left figure.

Find the Roots of P(z) and Q(z)

mmm zazazazA 2

21

11)(


m)()()( 1)1( zAzzAzP m

mm


m)()()( 1)1( zAzzAzQ m

mm

mmm zazazazA 2

21

11 1)(

)1(1

21

11)1( )(

mmmmm

m zzazazazAz

)1(

11)(1)(

m

m

i

iimi zzaazP

)1(

11)(1)(

m

m

i

iimi zzaazQ

Symmetric

Anti-symmetric


)1(

11)(1)(

m

m

i

iimi zzaazP

)1(

11)(1)(

m

m

i

iimi zzaazP )1(

11)(1)(

m

m

i

iimi zzaazQ

)1(

11)(1)(

m

m

i

iimi zzaazQ

)1)(1()( 22

11

1 mm zpzpzpzzP

11 1 paa m

2112 ppaa m

3223 ppaa m

1)( 11 maap

1122 )( paap m

2233 )( paap m

/ 2 / 2 1 / 2 1 / 2m m m ma a p p / 2 / 2 / 2 1 / 2 1( )m m m mp a a p ...

...

/ 2 1 / 2 1m mp p ...

11 ppm

1mp

...


)1(

11)(1)(

m

m

i

iimi zzaazP

)1(

11)(1)(

m

m

i

iimi zzaazP )1(

11)(1)(

m

m

i

iimi zzaazQ

)1(

11)(1)(

m

m

i

iimi zzaazQ

1

)(

0

11

p

paap iimii

1

)(

0

11

p

paap iimii

We only need compute the values on 1 i m/2.

)1)(1()( 22

11

1 mm zpzpzpzzP

1)( 11 maap

1122 )( paap m

2233 )( paap m

/ 2 / 2 / 2 1 / 2 1( )m m m mp a a p ...

/ 2 1 / 2 1m mp p ...

11 ppm

1mp


)1(

11)(1)(

m

m

i

iimi zzaazP

)1(

11)(1)(

m

m

i

iimi zzaazP )1(

11)(1)(

m

m

i

iimi zzaazQ

)1(

11)(1)(

m

m

i

iimi zzaazQ

)1)(1()( 22

11

1 mm zqzqzqzzQ

11 1 qaa m

2112 qqaa m

3223 qqaa m

1)( 11 maaq

1122 )( qaaq m

2233 )( qaaq m

/ 2 / 2 1 / 2 1 / 2m m m ma a q q / 2 / 2 / 2 1 / 2 1( )m m m mq a a q ...

...

/ 2 1 / 2 1m mq q ...

11 qqm

1mq

...


)1(

11)(1)(

m

m

i

iimi zzaazP

)1(

11)(1)(

m

m

i

iimi zzaazP )1(

11)(1)(

m

m

i

iimi zzaazQ

)1(

11)(1)(

m

m

i

iimi zzaazQ

1

)(

0

11

q

qaaq iimii

1

)(

0

11

q

qaaq iimii

We only need compute the values on 1 i m/2.

)1)(1()( 22

11

1 mm zqzqzqzzQ

1)( 11 maaq

1122 )( qaaq m

2233 )( qaaq m

/ 2 / 2 / 2 1 / 2 1( )m m m mq a a q ...

/ 2 1 / 2 1m mq q ...

11 qqm

1mq


)1)(1()( )1(1

22

11

1 mm zzpzpzpzzP

)1)(1()( )1(1

22

11

1 mm zzqzqzqzzQ

zeroon 1

zeroon +1

P’(z)

Q’(z)

Both P’(z) and Q’(z) are symmetric. Both P’(z) and Q’(z) are symmetric.


mm zzpzpzpzP )1(1

22

111)(

jmmjjjj eepepepeP )1(1

2211)(

12/

12/

)()( 22222

m

im

ijiji

jjj peepeeemmmmm

2

/ 2 11

/ 22 2 21

2 cos( ) cos( )m

mj m m

i mi

e p i p

)(~ P To find its zeros.


/ 2 11

/ 22 2 21

( ) cos( ) cos( )m

m mi m

i

P p i p

/ 2 1

1/ 22 2 2

1

( ) cos( ) cos( )m

m mi m

i

P p i p

)cos()cos(coscos2 )cos()cos(coscos2

cos3cos43cos 3

)2(2cos4cos 1)2(cos2 2

cos5cos2cos3cos2

cos2cos3cos25cos

1cos22cos 2


Define kck cos xc cos1

odd is 2

even is 12

2/)1(2/)1(

22/

kxcc

kcc

kk

kk

odd is 2

even is 12

2/)1(2/)1(

22/

kxcc

kcc

kk

kk

)cos()cos(coscos2 )cos()cos(coscos2

/ 2 11

/ 22 2 21

( ) cos( ) cos( )m

m mi m

i

P p i p

/ 2 1

1/ 22 2 2

1

( ) cos( ) cos( )m

m mi m

i

P p i p


xc 1

odd is 2

even is 12

2/)1(2/)1(

22/

kxcc

kcc

kk

kk

odd is 2

even is 12

2/)1(2/)1(

22/

kxcc

kcc

kk

kk

12 22 xc

xxxccc 342 3123

18812 24224 xxcc

xxxxccc 520162 35235 ...

/ 2 11

/ 22 2 21

( ) cos( ) cos( )m

m mi m

i

P p i p

/ 2 1

1/ 22 2 2

1

( ) cos( ) cos( )m

m mi m

i

P p i p


xc 1

12 22 xc

xxxccc 342 3123

18812 24224 xxcc

xxxxccc 520162 35235

Consider m=10.

54321 cos2cos3cos4cos5cos)(~

pppppP 54321 cos2cos3cos4cos5cos)(~

pppppP

/ 2 11

/ 22 2 21

( ) cos( ) cos( )m

m mi m

i

P p i p

/ 2 1

1/ 22 2 2

1

( ) cos( ) cos( )m

m mi m

i

P p i p


xc 1xc 1

12 22 xc 12 2

2 xc

xxc 34 33 xxc 34 3

3

188 244 xxc 188 24

4 xxc

xxxc 52016 355 xxxc 52016 35

5

11 2 3 4 52( ) cos5 cos 4 cos3 cos 2 cosP p p p p p 1

1 2 3 4 52( ) cos5 cos 4 cos3 cos 2 cosP p p p p p

5 3 4 21

3 2 12 3 4 52

( ) (16 20 5 ) (8 8 1)

(4 3 ) (2 1)

P x x x p x x

p x x p x p x p

215 3 1 4 2 3 12

3 4 52 1

( ) ( ) ( 3 5) (2 8 )

(4 20) 8 16

P p p p p p x p p x

p x p x x

215 3 1 4 2 3 12

3 4 52 1

( ) ( ) ( 3 5) (2 8 )

(4 20) 8 16

P p p p p p x p p x

p x p x x


1)( 1011 aap

1922 )( paap

2833 )( paap

3744 )( paap

4655 )( paap

215 3 1 4 2 3 12

3 4 52 1

( ) ( ) ( 3 5) (2 8 )

(4 20) 8 16

P p p p p p x p p x

p x p x x

215 3 1 4 2 3 12

3 4 52 1

( ) ( ) ( 3 5) (2 8 )

(4 20) 8 16

P p p p p p x p p x

p x p x x

4655 )( qaaq


215 3 1 4 2 3 12

3 4 52 1

( ) ( ) ( 3 5) (2 8 )

(4 20) 8 16

Q q q q q q x q q x

q x q x x

215 3 1 4 2 3 12

3 4 52 1

( ) ( ) ( 3 5) (2 8 )

(4 20) 8 16

Q q q q q q x q q x

q x q x x

1)( 1011 aaq

1922 )( qaaq

2833 )( qaaq

3744 )( qaaq

Find the Roots of P(z) and Q(z) 21

5 3 1 4 2 3 12

3 4 52 1

( ) ( ) ( 3 5) (2 8 )

(4 20) 8 16

P p p p p p x p p x

p x p x x

215 3 1 4 2 3 12

3 4 52 1

( ) ( ) ( 3 5) (2 8 )

(4 20) 8 16

P p p p p p x p p x

p x p x x

215 3 1 4 2 3 12

3 4 52 1

( ) ( ) ( 3 5) (2 8 )

(4 20) 8 16

Q q q q q q x q q x

q x q x x

215 3 1 4 2 3 12

3 4 52 1

( ) ( ) ( 3 5) (2 8 )

(4 20) 8 16

Q q q q q q x q q x

q x q x x

cosxWe want to find i’s such that

0)(~ iP 0)(~ iP

0)(~

iQ 0)(~

iQ


Algorithm:

We only need to find zeros for this half.

0

0

)(~

iQ


2 3

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