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EECS 598-
Lecture thirteen: Subdivision, refinability,
z-transform
Igor Guskov
February 21, 2002
1 B-spline refinability
For B-splines we have
Np(v) =
∫ (^) v
v− 1
Np− 1 (u)du = ν? Np− 1 (v).
where ν = N 0 is the box function and the convolution of two functions is defined as
(f? g)(x) :=
−∞
f (y)g(x − y)dy.
Then every B-spline basis function is simply a multiple convolution of the box function Np = ν? Np− 1 = ν?...? ν︸ ︷︷ ︸ p+
1.1 Refinability
Zeroth degree B-splines are obviously refinable:
N 0 (t) = N 0 (2t) + N 0 (2t − 1).
Linear (degree one) B-splines are also refinable (see Figure 1):
N 1 (t) =
N 1 (2t) + N 1 (2t − 1) +
N 1 (2t − 2)
Later we will see how to get refinability for B-spline basis functions of all orders.
-1 0 1 2 3
-1 0 1 2 3
t
t
N 1 (t+1) N 1 (t) N 1 (t-1)
N 1 (2t) N 1 (2t-1) N 1 (2t-2)
Figure 1: Linear b-splines are refinable: N 1 (t) = 12 N 1 (2t) + N 1 (2t − 1) + 1 2 N^1 (2t^ −^ 2)
2 Stationary subdivision
2.1 Notational
A stationary subdivision scheme S with subdivision mask σ is defined via:
fj+1,l = (Sfj )l =
k
σl− 2 kfj,k
Subdivision mask forms columns of the subdivision matrix. It will be convenient to define linear spline corresponding to the sequence on level j as (`j g)(t) :=
k
gkN 1 (2j^ t − k + 1).
We will say that a multilevel sequence (fj,k)j≥ 0 produces a function F (t) when F (t) = limj→∞(`j fj )(t) at every point t. When (fj,k) comes from a
Fact A fundamental solution of a subdivision scheme satisfies the corre- sponding refinement relation.
Theorem Let φ and ξ be the fundamental solutions of subdivision schemes with masks σφ^ and σξ^ correspondingly. Then φ? ξ is the fundamental solution for the subdivision scheme with the mask σφ^? σξ/2. Here the discrete convolution for sequences is defined as (σφ^? σξ)q =
k σ
φ q−kσ
ξ k.
Example See that Chaikin rules are convolutions of linear rules with box rules. Now to make it easier we introduce z-tranforms.
2.3 z-transforms
We define∑ z-transform for a sequence as a polynomial in z via relation f (z) =
k fkz k. Note how heavily we abuse our notation. Here’s the good news: discrete convolution of sequences is equivalent to multiplication of the corresponding z-transforms. Indeed,
(a? b)(z) =
q
k
aq−kbkzq^ =
q
k
aq−kzq−kbkzk^ = ∑
s
k
aszsbkzk^ =
s
aszs^
k
bkzk^ = a(z)b(z).
Examples For a box splines we have subdivision mask σ[0]^ = (... , 0 , σ 0 [0] =
1 , σ 1 [0] = 1, 0 ,.. .). This notation is clearly not good. From now on we’ll write it as σ0 =... + 0z−^1 + 1z^0 + 1z^1 + 0z^2 + · · · = 1 + z.
So the subdivision mask for boxes is 1 + z.
Exercise The theorem above then states that the subdivision mask for linear splines is (1 + z)(1 + z)/2 = (1 + 2z + z^2 )/2 which matches with the previous example. Now do it up to cubics etc.
Smoothness fact If f ∈ Cn−^1 then f? N 0 ∈ Cn.
Subdivision and z-transform For a subdivision scheme the two conse- quent levels are related via
fj+1(z) =
l
k
σl− 2 kfj,kzl
Introduce q = l − 2 k and get
fj+1(z) = σ(z)fj (z^2 ). For B-splines the smoothness is easy. What about other schemes?
Four-point scheme 4pt scheme is an interpolating scheme with the mask
−
z−^3 +
z−^1 + 1 +
z^1 −
z^3.
What can we say about its smoothness?
Figure 2: The fundamental solution for the 4pt scheme after two subdivision steps.
A subdivision scheme is affine if applied to sequence of all ones it pro- duces a sequence of all ones on the next level, that is S 1 = 1. This actually means that
k σ^2 k^ = 1 and^
k σ^2 k+1^ = 1. Then its z-transform has a root at −1, so that σ(z) = (1 + z)τ (z) where τ is another polynomial. Let’s look at difference sequences: (∆g)k := gk−gk− 1. We have (∆g)(z) = (1 − z)g(z). Suppose we have an affine scheme S. Then let’s look for the difference sequence at the level j + 1:
(∆fj+1)(z) = (1 − z)fj+1(z) = (1 − z)σ(z)fj (z^2 ) = (1 − z)(1 + z)τ (z)f(z^2 ) = τ (z)(1 − z^2 )fj (z^2 ) = τ (z)(∆fj )(z^2 ).
