Homework 4
Due: Thursday, April 30
This homework serves as the take-home final exam.
It will count as two homework grades if this improves your grade.
The following exercises are for both 5750 and 6880.
0) Verify that z-transform is linear and that the z-transform of a convolution
is a product of z-transforms.
The biorthogonal transform CDF(2,2) is the transform we went through as
an example where the predicition would perfectly predict linear signals. Here it
is with the normalization steps. (CDF stands for Cohen, Daubechies, Feauveau.)
1
(1)
P : dj+1 [n] = aj [2n + 1] − (aj [2n] + aj [2n + 2]);
2
1 (1)
(1)
(1)
U : aj+1 [n] = aj [2n] + (dj+1 [n − 1] + dj+1 [n]);
4
1 (1)
N1 : dj+1 [n] = √ dj+1 [n];
2
√ (1)
N2 : aj+1 [n] = 2aj+1 [n].
1) Verify that CDF(2,2) preserves the first moment, i.e.
X
n
1 X
naj+1 [n] = √
naj [n].
2 2 n
2) In the z-transform formulation, we found T (z) for CDF(2,2) in class. Determine the entire matrix factorization in the z-transform for one lifting step of
CDF(2,2). Now find the low pass filter h0 (simply h in our filter bank notation)
from the identity H0 (z) = H00 (z 2 ) + zH01 (z 2 ).
3) Send the signal
s = [1 1 2 3 4 5 4 4]
1
through the CDF(2,2) filter twice using a lifting implementation. (This can
easily be done by hand.) Then invert the transform to recover s.
4) Suppose Lift(s) performs the lifting implementation for some transform on
a one dimensional signal, s = {s[n] : n ∈ Z}. Describe how to perform a two
dimensional, separable transform on an image, im = {im[n1 , n2 ] : n1 , n2 ∈ Z}.
(Pseudo code is appropriate if that is easier for you.)
The following exercise is for students registered for 6880.
G5) Determine the z-transform factorization of the form

 K
H(z) = 
0

m
Y

 1


0
i=1

0
K −1

Si (z)  
1

1
−Ti (z)

0 

1
for the biorthogonal transfrom CDF(3,3):
1
(1)
U1 : aj+1 [n] = aj [2n] − aj [2n − 1]
3
1
(1)
(1)
(1)
P1 : dj+1 [n] = aj [2n + 1] − (9aj+1 [n] + 3aj+1 [n + 1]);
8
1
(2)
(1)
(1)
(1)
(1)
U2 : aj+1 [n] = aj+1 [n] + (3dj+1 [n − 1] + 16dj+1 [n] − 3dj+1 [n + 1]);
36
√
2 (1)
N1 : dj+1 [n] =
d [n];
3 j+1
3 (2)
N2 : aj+1 [n] = √ aj+1 [n].
2
2