sparse image representation via combined transforms - Convex ...

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)
)(
H(z −1 ) H(−z −1 )
)
(
2 0
)
G(z)
G(−z)
G(z −1 ) G(−z −1 )
=
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) =2z 2k+1 ,
and
[H(z −1 )+H(−z −1 )]z 2k+1 = G(−z) − G(z). (3.42)