Image Processing Reference
In-Depth Information
lines and planes embedded in 3D such that f does not change value. Subsequently,
we will show that the problems of describing the direction when isovalues are planes
or lines are mathematically the same. We will first be concerned with the case of
planar loci since these have some notational advantages. We define them directly for
the general N D case, from which the 3D case follows directly.
Definition 12.1. The image f ( r ) , where r
E N , is called linearly symmetric if the
loci of its isogray points constitute hyperplanes that have a common direction, i.e.,
there exists a scalar function of one variable g such that
f ( r )= g ( k T r )
(12.1)
for some vector k
k .
It should be noted that although g ( x ) is a one-dimensional function, g ( k T r ) is a
function defined on N D. The way they are defined assures that the linearly symmet-
ric images have the same gray values at all points r satisfying
E N . The direction of the linear symmetry is
±
k T r = constant
(12.2)
which is the equation of a hyperplane embedded in N D having the normal k . The
hyperplane itself is an ( N − 1)-dimensional surface (which is embedded in N D).
This corresponds to an ordinary plane (surface) in E 3 when N =3.
Example 12.1. A step edge, defined in 3D as:
f ( r )= 1 , if k T
0;
0 , otherwise;
r
0
(12.3)
is equal to χ ( k T
0
r ), where χ is the usual one-dimensional step function:
χ ( x )= 1 , if x
0;
0 , otherwise.
Thus f defined in Eq. (12.3) is linearly symmetric.
WecanFouriertransformalinearlysymmetricimagedefinedin3D,byfollowing
steps analogous to Eqs. (10.102)-(10.103):
( g ( k T r ))(
)= G ( k T
T
T
F (
ω
)=
F
ω
ω
) δ ( u 1
ω
) δ ( u 2
ω
)
(12.4)
In Fig. 12.1, the dashed triangular plane shows a linearly symmetric image in the
3D space having the coordinate axes represented by three black basis vectors. The
red basis vectors show the corresponding axes in the 3D Fourier transform domain,
whereas the vector in black shows k , the axis on which all the energy of the Fourier
transform of the image is concentrated. Note that k is orthogonal to the plane. Natu-
rally, this can be extended to N D, a result which we state as a lemma.
Search WWH ::




Custom Search