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
- 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
- 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
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
- 5
1
- 5
2
- 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.
