Image Processing Reference
In-Depth Information
1
0.8
0.6
0.4
0.2
0
−0.2
−60
−40
−20
0
20
40
60
Fig. 10.3. The graph represents the sinc function, Eq. (10.4)
Example 10.1. The function
ω
=
2
π
g
(
t
)=sin(
ωt
)
,
with
12
,
(10.2)
rendered in Fig. 10.1, oscillates around zero. This sinusoid is used to construct the
linearlysymmetric2Dfunction,showninFig.10.2(top,left).Torenderthenegative
values of
g
we added the constant 0.5 and rescaled the gray values to the effect
that white represents 1 and black represents
1 by using 256 gray tones and linear
mapping.Thedashedgreenlinethroughoriginshowsanexampleisocurve.Werecall
that an isocurve is the curve that joins all points having a certain gray value. The
example isocurve in the constructed 2D image is clearly a line, along which the
gray value of the image is the same. In fact, every other isocurve is also a line and
all isocurves are parallel, just as they should be, because of the way the image is
constructed.
By construction, the gray value of the image cannot change unless the argument
of
g
changes.Foranygivenimagelocation
r
,theargumentof
g
willequalaconstant
value
C
−
k
T
r
=
C
(10.3)
totheeffectthatitispossibletorestrictthechangesof
r
toacurvesothat
C
isinvari-
ant. By virtue of Eq. (10.3), this path is a line, and the equation represents parallel
lines when
k
is fixed. We can move in the image, i.e., change
r
, along a line which
is perpendicular to the line shown as the dashed diagonal, i.e along
k
, and obtain
changesinthevalueof
C
,whichinturnchangesthevaluesof
g
.Inthispath,theob-
tained function or gray values are identical to the values of the original 1D function,
g
(
t
)=cos(
ωt
). The solid green arrow illustrates the vector
k
=(cos(
π
4
)
,
sin(
π
4
))
T
usedtobuildtheimage.Therepresentationisnotuniquebecause
k
(dashed)would
have generated the same image. In other words, the
k
used to construct the linearly
symmetricimageisuniqueonlyuptoasignfactor
−
±
1.Thecolorsurfacein3Dshows
