








Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
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
1 / 14
This page cannot be seen from the preview
Don't miss anything!
2
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
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
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
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]
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.
???
because Hp-1(z) must be a polynomial.
15
16
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
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
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
2
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.