Title sparse image representation via combined transforms English 1.7 MB 210
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SPARSE IMAGE REPRESENTATION

VIA COMBINED TRANSFORMS

a dissertation

submitted to the department of statistics

and the committee on graduate studies

of stanford university

in partial fulfillment of the requirements

for the degree of

doctor of philosophy

Xiaoming Huo

August 1999

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ii

Page 105

3.7. PROOFS 77

the ijth component of matrix Σ is

Σij =
N−1∑
k=0

λkα
2
1(k)α2(i)α2(j)

2
N

cos
(

π(k + δ1)(i + δ2)
N

)
cos
(

π(k + δ1)(j + δ2)
N

)

= α2(i)α2(j)
1
N

N−1∑
k=0

λkα
2
1(k)

[
cos

π(k + δ1)(i − j)
N

+ cos
π(k + δ1)(i + j + 2δ2)

N

]
.

In the above equation, let the first term 1
N

∑N−1
k=0 λkα

2
1(k) cos

π(k+δ1)(i−j)
N

be the ijth

element of matrix Σ1, and let the second term 1N
∑N−1

k=0 λkα
2
1(k) cos

π(k+δ1)(i+j+2δ2)
N

be the

ijth element of matrix Σ2. It is easy to see that the matrix Σ1 is Toeplitz, and the matrix

Σ2 is Hankel. Inserting different values of δ1 and δ2, we establish the Theorem 3.5.

Proof of Theorem 3.6

Consider the z-transform of sequence {h(n) : n ∈ Z} and sequence {g(n) : n ∈ Z}:

H(z) =

n∈Z

h(n)zn,

G(z) =

n∈Z

g(n)zn.

From (3.27), note that all the even terms in H(z)H(z−1) have zero coefficient except the

constant term. Hence we have

H(z)H(z−1) + H(−z)H(−z−1) = 2. (3.38)

Similarly from (3.28) we have

G(z)G(z−1) + G(−z)G(−z−1) = 2. (3.39)

From (3.29), all the even terms in H(z)G(z−1) have zero coefficients, so we have

G(z)H(z−1) + G(−z)H(−z−1) = 0. (3.40)

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78 CHAPTER 3. IMAGE TRANSFORMS AND IMAGE FEATURES

Equations (3.38) (3.39) and (3.40) together are equivalent to the matrix multiplication

(
H(z) H(−z)
G(z) G(−z)

)(
H(z−1) H(−z−1)
G(z−1) G(−z−1)

)
=

(
2 0

0 2

)
.

Taking the determinant of both sides, we have

[H(z)G(−z) − G(z)H(−z)][H(z−1)G(−z−1) − G(z−1)H(−z−1)] = 4. (3.41)

If both sequence {h(n) : n ∈ Z} and sequence {g(n) : n ∈ Z} have finite length (thinking of
the impulse responses of FIR filters), then the polynomial H(z)G(−z) − G(z)H(−z) must
have only one nonzero term. Since if the polynomial H(z)G(−z) − G(z)H(−z) has more
than two terms and simultaneously has finite length, (3.41) can never be equal to a constant.

On the other hand, we have

[H(z−1) + H(−z−1)][H(z)G(−z) − G(z)H(−z)]

= H(z−1)H(z)G(−z) + H(−z−1)H(z)G(−z)

−H(z−1)G(z)H(−z) − H(−z−1)G(z)H(−z)
(3.40)
= H(z−1)H(z)G(−z) − H(z−1)H(z)G(z)

+H(−z−1)G(−z)H(−z) − H(−z−1)G(z)H(−z)
(3.38)
= 2[G(−z) − G(z)].

We know that H(z)G(−z) − G(z)H(−z) only has one nonzero term. The polynomial
H(z−1)+H(−z−1) can only have terms with even exponents and polynomial G(−z)−G(z)
can only have terms with odd exponents, so from the previous equation, there must exist

an integer k, such that

H(z)G(−z) − G(z)H(−z) = 2z2k+1,

and

[H(z−1) + H(−z−1)]z2k+1 = G(−z) − G(z). (3.42)

Page 209

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