Solutions Manual
Digital Image Processing 3rd Edition
By
Rafael C. Gonzalez,
Richard E. Woods
( All Chapters Included - 100% Verified Solutions )
1
,Chapter 2
Problem Solutions
Problem 2.1
The diameter, x , of the retinal image corresponding to the dot is obtained from
similar triangles, as shown in Fig. P2.1. That is,
(d /2) (x /2)
=
0.2 0.017
which gives x = 0.085d . From the discussion in Section 2.1.1, and taking some
liberties of interpretation, we can think of the fovea as a square sensor array
having on the order of 337,000 elements, which translates into an array of size
580 × 580 elements. Assuming equal spacing between elements, this gives 580
elements and 579 spaces on a line 1.5 mm long. The size of each element and
each space is then s = [(1.5mm)/1, 159] = 1.3 × 10−6 m. If the size (on the fovea)
of the imaged dot is less than the size of a single resolution element, we assume
that the dot will be invisible to the eye. In other words, the eye will not detect a
dot if its diameter, d , is such that 0.085(d ) < 1.3 × 10−6 m, or d < 15.3 × 10−6 m.
Edge view of dot
Image of the dot x x/2
d/2 on the fovea
d
0.2 m 0.017 m
Figure P2.1
9
2
,10 CHAPTER 2. PROBLEM SOLUTIONS
Problem 2.2
Brightness adaptation.
Problem 2.3
The solution is
λ = c /v
= 2.998 × 108 (m/s)/60(1/s)
= 4.997 × 106 m = 4997 Km.
Problem 2.4
(a) From the discussion on the electromagnetic spectrum in Section 2.2, the
source of the illumination required to see an object must have wavelength the
same size or smaller than the object. Because interest lies only on the boundary
shape and not on other spectral characteristics of the specimens, a single illu-
mination source in the far ultraviolet (wavelength of .001 microns or less) will
be able to detect all objects. A far-ultraviolet camera sensor would be needed to
image the specimens.
(b) No answer is required because the answer to (a) is affirmative.
Problem 2.5
From the geometry of Fig. 2.3, (7 mm)/(35 mm) = (z )/(500 mm), or z = 100 mm.
So the target size is 100 mm on the side. We have a total of 1024 elements per
line, so the resolution of 1 line is 1024/100 = 10 elements/mm. For line pairs we
divide by 2, giving an answer of 5 lp/mm.
Problem 2.6
One possible solution is to equip a monochrome camera with a mechanical de-
vice that sequentially places a red, a green and a blue pass filter in front of the
lens. The strongest camera response determines the color. If all three responses
are approximately equal, the object is white. A faster system would utilize three
different cameras, each equipped with an individual filter. The analysis then
would be based on polling the response of each camera. This system would be
a little more expensive, but it would be faster and more reliable. Note that both
solutions assume that the field of view of the camera(s) is such that it is com-
3
, 11
Intensity
255
0
(x0, y0)
(a)
Intensity
DG
255
Equally
spaced
subdivisions
0
(b)
Figure P2.7
pletely filled by a uniform color [i.e., the camera(s) is (are) focused on a part of
the vehicle where only its color is seen. Otherwise further analysis would be re-
quired to isolate the region of uniform color, which is all that is of interest in
solving this problem].
Problem 2.7
The image in question is given by
f (x , y ) = i (x , y )r (x , y )
2 +(y −y 2]
= 255 e −[(x −x 0 ) 0)
× 1.0
)2 +(y −y )2 ]
= 255 e −[(x −x 0 0
A cross section of the image is shown in Fig. P2.7(a). If the intensity is quan-
tized using m bits, then we have the situation shown in Fig. P2.7(b), where
△G = (255 + 1)/2m . Since an abrupt change of 8 intensity levels is assumed to
be detectable by the eye, it follows that △G = 8 = 256/2m , or m = 5. In other
words, 32, or fewer, intensity levels will produce visible false contouring.
4
Digital Image Processing 3rd Edition
By
Rafael C. Gonzalez,
Richard E. Woods
( All Chapters Included - 100% Verified Solutions )
1
,Chapter 2
Problem Solutions
Problem 2.1
The diameter, x , of the retinal image corresponding to the dot is obtained from
similar triangles, as shown in Fig. P2.1. That is,
(d /2) (x /2)
=
0.2 0.017
which gives x = 0.085d . From the discussion in Section 2.1.1, and taking some
liberties of interpretation, we can think of the fovea as a square sensor array
having on the order of 337,000 elements, which translates into an array of size
580 × 580 elements. Assuming equal spacing between elements, this gives 580
elements and 579 spaces on a line 1.5 mm long. The size of each element and
each space is then s = [(1.5mm)/1, 159] = 1.3 × 10−6 m. If the size (on the fovea)
of the imaged dot is less than the size of a single resolution element, we assume
that the dot will be invisible to the eye. In other words, the eye will not detect a
dot if its diameter, d , is such that 0.085(d ) < 1.3 × 10−6 m, or d < 15.3 × 10−6 m.
Edge view of dot
Image of the dot x x/2
d/2 on the fovea
d
0.2 m 0.017 m
Figure P2.1
9
2
,10 CHAPTER 2. PROBLEM SOLUTIONS
Problem 2.2
Brightness adaptation.
Problem 2.3
The solution is
λ = c /v
= 2.998 × 108 (m/s)/60(1/s)
= 4.997 × 106 m = 4997 Km.
Problem 2.4
(a) From the discussion on the electromagnetic spectrum in Section 2.2, the
source of the illumination required to see an object must have wavelength the
same size or smaller than the object. Because interest lies only on the boundary
shape and not on other spectral characteristics of the specimens, a single illu-
mination source in the far ultraviolet (wavelength of .001 microns or less) will
be able to detect all objects. A far-ultraviolet camera sensor would be needed to
image the specimens.
(b) No answer is required because the answer to (a) is affirmative.
Problem 2.5
From the geometry of Fig. 2.3, (7 mm)/(35 mm) = (z )/(500 mm), or z = 100 mm.
So the target size is 100 mm on the side. We have a total of 1024 elements per
line, so the resolution of 1 line is 1024/100 = 10 elements/mm. For line pairs we
divide by 2, giving an answer of 5 lp/mm.
Problem 2.6
One possible solution is to equip a monochrome camera with a mechanical de-
vice that sequentially places a red, a green and a blue pass filter in front of the
lens. The strongest camera response determines the color. If all three responses
are approximately equal, the object is white. A faster system would utilize three
different cameras, each equipped with an individual filter. The analysis then
would be based on polling the response of each camera. This system would be
a little more expensive, but it would be faster and more reliable. Note that both
solutions assume that the field of view of the camera(s) is such that it is com-
3
, 11
Intensity
255
0
(x0, y0)
(a)
Intensity
DG
255
Equally
spaced
subdivisions
0
(b)
Figure P2.7
pletely filled by a uniform color [i.e., the camera(s) is (are) focused on a part of
the vehicle where only its color is seen. Otherwise further analysis would be re-
quired to isolate the region of uniform color, which is all that is of interest in
solving this problem].
Problem 2.7
The image in question is given by
f (x , y ) = i (x , y )r (x , y )
2 +(y −y 2]
= 255 e −[(x −x 0 ) 0)
× 1.0
)2 +(y −y )2 ]
= 255 e −[(x −x 0 0
A cross section of the image is shown in Fig. P2.7(a). If the intensity is quan-
tized using m bits, then we have the situation shown in Fig. P2.7(b), where
△G = (255 + 1)/2m . Since an abrupt change of 8 intensity levels is assumed to
be detectable by the eye, it follows that △G = 8 = 256/2m , or m = 5. In other
words, 32, or fewer, intensity levels will produce visible false contouring.
4
