Spring 2012
PROCESSING
CSCE 763: DIGITAL IMAGE
Homework #1
Due time: Tuesday, Jan 24th, before class starts
1.
Using the background information provided in Section 2.1, and thinking purely in geometric
terms, estimate the diameter of the smallest printed dot that the eye can discern if the page on
which the dot is printed is 0.3 m away from the eyes. Assume for simplicity that the visual
system ceases to detect the dot when the image of the dot on the fovea becomes smaller than
the diameter of one receptor (cone) in that area of the retina. Assume further that the fovea
can be modeled as a square array of dimensions 1.5 mm*1.5 mm, and that the cones and
spaces between the cones are distributed uniformly throughout this array.
Solution:
Assume the focal length of the eye is 0.014m since the distance between the object and the eye is
much less than 3m.
Figure 1
The diameter, x, of the retinal image corresponding to the dot is obtained from similar triangles,
as shown in below. That is,
𝑑
𝑥
=
0.3 0.014
which gives x = 0.047d . 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
0.047(d ) <1.3×10−6 m, or d < 27.66×10−6 m.
2. Alternating current certainly is part of the electromagnetic spectrum. Commercial alternating
current in the United States has a frequency of 60 Hz. What is the wavelength in kilometers
of this component of the spectrum?
Solution:
λ = c /v
= 2.998×108(m/s)/60(1/s)
= 4.997×106m=4997Km.
3. A common measure of transmission for digital data is the baud rate, defined as the number of
bits transmitted per second. Generally, transmission is accomplished in packets consisting of
a start bit, a byte (8 bits) of information, and a stop bit. Using these facts, answer the
following:
(a) How many minutes would it take to transmit a 1024*1024 image with 256 intensity levels
using a 56K baud modem?
(b) What would the time be at 3000K baud, a representative speed of a phone DSL (digital
subscriber line) connection?
Solution:
(a) The total amount of data (including the start and stop bit) in an 8-bit, 1024×1024 image, is
(1024)2×[8+2] bits. The total time required to transmit this image over a 56K baud link is
(1024)2 ×[8+2]/56000 = 187.25 sec or about 3.1 min.
(b) At 3000K this time goes down to about 3.5 sec.
4. Consider the two image subsets, S1 and S2, shown in the following figure. For V={1},
determine whether these two subsets are (a) 4-adjacent, (b) 8-adjacent, or (c) m-adjacent.
Solution:
Let p and q be the two pixels in S1 and S2, as shown below. Then, (a) S1 and S2 are not 4connected because q is not in the set N4(p); (b) S1 and S2 are 8-connected because q is in the set
N8(p); (c) S1 and S2 are m-connected because (i) q is in ND(p), and (ii) the set N4(p) ∩ N4(q) is
empty.
Figure2
5. Consider the image segment shown.
(a) Let V={0, 1} and compute the lengths of the shortest 4-, 8-, and m-path between p and q.
If a particular path does not exist between these two points, explain why.
(b) Repeat for V={1, 2}.
Solution:
(a) When V = {0,1}, 4-path does not exist between p and q because it is impossible to get from p
to q by traveling along points that are both 4-adjacent and also have values from V . Figure 3(a)
shows this condition; it is not possible to get to q. The shortest 8-path is shown in Figure 3(b); its
length is 4. The length of the shortest m- path (shown dashed in Figure 3b) is 5. Both of these
shortest paths are unique in this case.
(b) One possibility for the shortest 4-pathwhen V ={1,2} is shown in Figure 3c; its length is 6. It is
easily verified that another 4-path of the same length exists between p and q. One possibility for
the shortest 8-path (it is not unique) is shown in Figure 3d; its length is 4. The length of a shortest
m-path (shown dashed in Figure 3d) is 6. This path is not unique.
Figure 3