ECE 178 Hwk #2 Solutions
1)
Problem 2.9
(a) The total amount of data (including the start and stop bit) in a 256 gray level or
8bit ,1024 x 1024 image, is (1024)
2
x [8 + 2] bits. The total time required to transmit this
image over a At 56K baud link is (1024)
2
x [8 + 2] /56000
= 187:25 sec or about 3.1 min.
(b) At 750K this time goes down to about 14 sec.
2)
Problem 2.10
The widthtoheight ratio is 16/9 and the resolution in the vertical direction is 1125 lines
(1125 pixels in the vertical direction). It is given that the resolution in the horizontal
direction is in the 16/9 proportion, so the resolution in the vertical direction is
(1125)x(16/9) = 2000 pixels per line. The system paints a full 1125x2000, 8bit image
every 1/30 sec for each of the red, green, and blue component images. There are 7200 sec
in two hours, so the total digital data generated in this time interval is
(1125)(2000)(8)(30)(3)(7200) = 1.166 x 10
13
bits, or 1.458 x 10
12
bytes (i.e., about 1.5
terrabytes). These numbers show why image data compression is so important.
Note: Many books where worded to say that there were 8 pixels for each color.
This does
not make any sense and should have said 8 bits of color for each of the three primary
colors.
3)
Problem 2.11
Let p and q be as shown in Fig. P2.11. Then, (a) S1 and S2 are not 4connected because q
is not in the set N
4
(p) (b) S1 and S2 are 8connected because q is in the set N
8
(p) (c) S1
and S2 are
m
connected because (i) q is in N
D
(p), and (ii) the set N
4
(p)
∩
N
4
(q) is empty.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document4)
Problem 2.12
The solution to this problem consists of defining all possible neighborhood shapes to go
from a diagonal segment to a corresponding 4connected segment, as shown in Fig. P2.12.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 MANJUNATH
 Distance, Vertical direction, Halftone, Error diffusion

Click to edit the document details