Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Wavelets and Filter Banks: Understanding Modulation and Polyphase Matrices, Slides of Banking and Finance

This document delves into the concepts of wavelets and filter banks, focusing on modulation and polyphase matrices. It covers noble identities, block toeplitz matrices, and block z-transforms, providing examples and explanations of the modulation matrix and synthesis modulation matrix. The document also discusses the relationship between modulation and polyphase matrices and the importance of perfect reconstruction in the polyphase domain.

Typology: Slides

2012/2013

Uploaded on 07/29/2013

sathyanna
sathyanna 🇮🇳

4.4

(8)

103 documents

1 / 14

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Course 18.327 and 1.130
Wavelets and Filter Banks
Modulation and Polyphase
Representations:
Noble Identities;
Block Toeplitz Matrices
and Block z-transforms;
Polyphase Examples
2
Modulation Matrix
Matrix form of PR conditions:
[F0 (z) F1 (z)] H0(z) H0(-z) = [ 2z œ?
??
? 0 ]
H1(z) H1(-z)
So
[ F0(z) F1(z)] = [2z œ?
??
? 0] H œ1(z)
H
m
œ1(z) = 1
&'(
&'(&'(
&'(
Modulation matrix, H
m(z)
H0(z) H1(-z) - H0 (-z) H1 (z) (must be non-zero)
H1(-z) -H0(-z)
-H1(z) H0(z)
1
Docsity.com
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe

Partial preview of the text

Download Wavelets and Filter Banks: Understanding Modulation and Polyphase Matrices and more Slides Banking and Finance in PDF only on Docsity!

Course 18.327 and 1.

Wavelets and Filter Banks

Modulation and Polyphase

Representations:

Noble Identities;

Block Toeplitz Matrices

and Block z-transforms;

Polyphase Examples

2

Modulation Matrix

Matrix form of PR conditions:

[F 0 (z) F 1 (z)] H 0 (z) H 0 (-z) = [ 2z œ????^ 0 ]

H 1 (z) H 1 (-z)

So

[ F 0 (z) F 1 (z)] = [2z œ????^ 0] H œ1^ (z)

H (^) mœ1^ (z) = 1

&'(&'(^ &'(&'( Modulation matrix, H (^) m(z)

H 0 (z) H 1 (-z) - H 0 (-z) H 1 (z) (must be non-zero)

H 1 (-z) -H 0 (-z) -H 1 (z) H 0 (z)

???

+,+,*+,

???

Ω F 0 (z) = 2z-?^ H 1 (-z)

Require these F 1 (z) = 2z-?^ H 0 (-z)^ to be FIR

Suppose we choose = 2z -^? Then

F 0 (z) = H 1 (-z)

F 1 (z) = -H 0 (-z)

3

1 *+,

4

Synthesis modulation matrix:

Complete the second row of matrix PR conditions by replacing z with œz:

F 0 (z) F 1 (z) H 0 (z) H 0 (-z) z-????^0

F 0 (-z) F 1 (-z) H 1 (z) H 1 (-z) 0 (-z) -^ ????

Synthesis modulation matrix, Fm(z)

&'(&'(&'(&'(

Note the transpose convention in Fm(z).

7

Derivation of Polyphase Form

  1. Filtering and downsampling: x[n] y[n] H(z) éééé 2

H(z) = Heven(z 2 ) + z-1^ Hodd(z 2 ); heven[n] = h[2n] hodd[n] = h[2n+1]

x[n] �

Heven(z 2 )

Hodd(z 2 )

éééé 2

y[n]

z-

8

x[n]

z -

Heven(z 2 )

Hodd(z 2 )

éééé 2

éééé 2

y[n]

z -

Heven(z)

Hodd(z)

éééé 2

éééé 2

y[n]

x[n]

Polyphase Form

xeven[n]

xodd[n-1]

9

  1. Upsampling and filtering

x[n] y[n]

F(z) = Feven (z 2 ) + z-1^ Fodd (z 2 )

åååå 2 F(z)

x[n] (^) +

Feven(z 2 )

Fodd(z 2 )

åååå 2

y[n]

z-

10

Feven(z 2 )

Fodd(z 2 )

åååå 2

åååå 2

y[n]

x[n]

Feven(z)

Fodd(z)

åååå 2

åååå 2

y[n]

x[n]

yeven[n]

yodd[n]

z-

Polyphase Form

z-

13

Similarly, for the synthesis filter bank:

Fb = f 0 [0] f 1 [0] f 0 [1] f 1 [1]

f 0 [2] f 1 [2] f 0 [3] f 1 [3]

f 0 [0] f 1 [0] f 0 [1] f 1 [1]

f 0 [2] f 1 [2] f 0 [3] f 1 [3]

14

Note transpose convention for synthesis polyphase matrix

f 0 [0] f 1 [0] f 0 [1] f 1 [1]

f 0 [2] f 1 [2] f 0 [3] f 1 [3]

Fp(z) = + z-

F0,even [z] F1,even [z] F0, odd [z] F1, odd [z]

  • Perfect reconstruction condition in polyphase domain:

Fp(z) Hp(z) = I (centered form)

This means that Hp(z) must be invertible for all z on the unit circle, i.e. det Hp(eiwwww) òòòò 0 for all frequencies wwww.

???

  • Given that the analysis filters are FIR, the requirement for the synthesis filters to be also FIR is: det Hp(z) = z-?^ (simple delay)

because Hp-1(z) must be a polynomial.

  • Condition for orthogonality: Fp(z) is the transpose of Hp(z), i.e. HpT(z-1) Hp(z) = I i.e. Hp(z) should be paraunitary.

15

16

Relationship between Modulation

and Polyphase Matrices

h0,even[n] = h 0 [2n] H 0 (z) = H0,even(z 2 ) + z-1^ H0,odd(z 2 ) ; h0,odd[n] = h 0 [2n+1] H 1 (z) = H1,even(z 2 ) + z-1^ H1,odd(z 2 ) Two more equations by replacing z with -z. So in matrix form: H 0 (z) H 0 (-z) H0,even(z 2 ) H0,odd(z 2 ) 1 1 H 1 (z) H 1 (-z) H1,even(z 2 ) H1,odd(z 2 ) z-1^ -z -

H (^) m(z) Modulation matrix

Hp(z 2 ) Polyphase matrix

19

Polyphase Matrix

Example: Daubechies 4-tap filter

1+μμμμ 3 3 + μμμμ 3 3 -μμμμ 3 1- μμμμ 3 4 μμμμ 2 4 μμμμ 2 4 μμμμ 2 4 μμμμ 2

{(1 + μμμμ3 ) + (3 + μμμμ3 ) z-1^ + (3 - μμμμ 3) z-2^ + (1 - μμμμ3) z-3^ }

h 0 [0] =

H 0 (z) =

H 1 (z) =

4 μμμμ 2

{(1 - μμμμ3) œ (3 - μμμμ3) z-1^ + (3 + μμμμ3)z -2^ œ (1 + μμμμ3)z -3^ }

4 μμμμ 2

h 0 [1] = h 0 [2] = h 0 [3] =

20

Time domain: h 0 [0] 2 + h 0 [1] 2 + h 0 [2] 2 + h 0 [3] 2 = 1 {(4 + 2μμμμ3) + (12 + 6 μμμμ3) + (12 œ 6 μμμμ3) + (4 œ 2 μμμμ3)} = 1 h 0 [0] h 0 [2] + h 0 [1] h 0 [3] = 1 {(2μμμμ3) + (-2μμμμ3)}

i.e. filter is orthogonal to its double shifts

32

32

21

Polyphase Domain: {(1 + μμμμ3) + (3 - μμμμ3) z-1^ }

{(3 + μμμμ3) + (1 - μμμμ3) z-1^ }

{(1 - μμμμ3) + (3 + μμμμ3) z-1^ }

(^1) { - (3 - μμμμ3) œ (1 + μμμμ3) z-1 (^) } 4 μμμμ 2

4 μμμμ 2

4 μμμμ 2

(^4) μμμμ 2

H0,even(z)

H0,odd(z)

H1,even(z)

H1,odd(z)

1 + μμμμ 3 3 + μμμμ 3 3 - μμμμ 3 1 - μμμμ 3

1 - μμμμ 3 -(3 - μμμμ3) 3 + μμμμ 3 -(1 + μμμμ 3

Hp(z) =

4 μμμμ 2

4 μμμμ 2

&'(&'(&'(&'( &'(&'(&'(&'( A^ B

z -

22

Hp(z) = A + B z-

HpT(z -1^ ) Hp(z) = (AT^ + BT^ z)(A + Bz -1^ ) = (ATA + BTB) + ATBz -1^ + BTAz

AT^ A =

4 μμμμ 2

1 + μμμμ 3 3 + μμμμ 3

1 - μμμμ 3

  • (3 - μμμμ3)

4 μμμμ 2

1 + μμμμ 3 3 + μμμμ 3 1 - (^) μμμμ 3 -(3-μμμμ3)

(4 + 2μμμμ3) + (4 - 2μμμμ3) (6 + 4μμμμ3) - (6 - 4μμμμ3) (6 + 4μμμμ3) - (6 - 4μμμμ3) (12 + 6μμμμ3) + (12 - 6μμμμ3)

³ μμμμ3/ μμ μμ3/4 ½

www w + pw + pw + p

Modulation domain:

H 0 (z) H 0 (z-1) = P(z) =

(-z^3 + 9z + 16 + 9z-1^ œ z-3) 1 (z^3 œ 9z + 16 œ 9z-1^ + z-3) 16

H 0 (-z) H 0 (-z-1) = P(-z) =

So H 0 (z) H 0 (z-1) + H 0 (-z) H 0 (-z-1) = 2

i.e. |H 0 (w)|^2^ + |H 0 (w + p)|^2^ = 2

25

26

(^0) -1 -0. 8 -0. 6 -0.4 -0.2 0 0.2 0.4 0 .6 0 .8 1

  1. 5

1

  1. 5

2

  1. 5

Ma g nitud e re s po ns e o f Da u be c h ie s 4 -ta p filte r.

Fre

que

nc y re

s pons e pha

s e

Ang ula r fre que nc y (no rm a liz e d b y (^) p)

Magnitude Response of Daubechies 4-tap filter.

27

-4 -1 -0. 8 -0. 6 -0. 4 -0. 2 0 0. 2 0.4 0.6 0.8 1

0

1

2

3

4

P ha s e re s pons e of Da ube c hie s 4-ta p filte r.

Fre

que

ncy re

s pons e phas e

Angula r fre qu e nc y (norm a liz e d by (^) p)

Phase response of Daubechies 4-tap filter.