Digital Image Processing HW#1 Solution
02/05/2004
TA: Jessie Hsu yh2117@columbia.edu
1. Design of an image acquisition system:
(a) Image acquisition system
(b) Derivation of pixel size
(c) Minimum coverage of image of dark blob
From (b), every imaging element is a square measuring 8mx8m, with spaces
between two 2m, therefore we consider each pixel to be a square of size
10m10m = 0.01mm0.01mm.
From (c), we can see that, for the image of the dark blob to fully cover 22 pixels, we
need it to cover 33 pixels such that even when it shifts a little bit, there are 22
pixels guaranteed to be covered.
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Accordingly, it is the square inside the dark blob that should be projected onto 33
pixels, rather than the dark blob itself. So we have the square measuring
D
D .

2
2
(d) From the diameter of dark blob to the dimension of the square inside it
Specifications:
L = 80 mm
D = 0.8 mm ( the diameter of the dark blob )
f = 25k mm ( k = 1, 2, …, 8 )
35k mm ( k = 1, 2, …, 5 )
w = 51210 m = 5120.01 mm = 5.12 mm
102410 m = 10240.01 mm = 10.24 mm
204810 m = 20480.01 mm = 20.48 mm
v = 310 m = 30.01 mm = 0.03 mm
This is an open problem, and there is not a single answer to it. Here are two ways that
you can approach the problem.
Approach 1.
Geometry analysis:
Observing the imaging geometry in (a), we have 2 equations
D
2 v
a
f
L w

a f
(1) ( from the geometry )
(2) ( for the viewing screen to be entirely projected onto CCD array )
The task is to determine a, f, w such that w gives us the lowest cost.
2
(2)
2
2
w
vL 
 0.03  80  3 2  4.24mm
(1)
D
0.8
So we choose w = 5.12 mm ( the lowest cost )
f can be arbitrarily chosen, as long as the corresponding a is determined as
D f 40
a
 
2f
3
2 v
We want a to be small, so that there will be less noise in the imaging process. One
possible choice is
f = 35mm
a = 659.97 mm or 0.66 m
Approach 2.
Rather than setting the = sign in (1), we relax it to
D
2 v
a
f
L w

a f
(3) ( from the geometry )
(4) ( for the viewing screen to be entirely projected onto CCD array )
Then we have a lower bound and an upper bound for the ratio R 
L
D
R
w
2v
R  18.86  Rmax
a
f
(5)
(6)
by analysis of the geometry again, we can see that the minimum resolution of the
L
CCD camera has to be
Rmax  0.01

80
 429.65 (7)
18.86  0.01
which means the camera must have resolution greater than 430430. In our
specifications, 512512 will do the job, so we fix w to be 5.12 mm.
Then the lower bound of R becomes
L
80

 15.62
w 5.12
The parameters can therefore be determined by choosing a ratio between the lower
bound and the upper bound followed by choosing a focal length not to small ( lower
cost on lens ) and choosing an a not too large ( for less noise ). One possible choice is
R = 16
f = 35 mm
a = 560 mm = 0.56 m
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2. You can calculate the results either by a simple MATLAB script or by hand
calculation. Here is an example MATLAB script:
% HW 1, Prob 2
clear;
clr=zeros(3,6);
% Colors in NTSC RGB
clr(:,1)=[1 0 0]';
% Red
clr(:,2)=[1 1 0]';
% Yellow
clr(:,3)=[0 1 0]';
% Green
clr(:,4)=[0 1 1]';
% Cyan
clr(:,5)=[0 0 1]';
% Blue
clr(:,6)=[1 0 1]';
% Magenta
% Convert to CIE RGB
A=[1.167 -0.146 -0.151
0.114 0.753 0.159
-0.001 0.059 1.128];
cclr=A*clr;
% Convert to I,V1,V2
B=[1/3 1/3 1/3
-1/sqrt(6) -1/sqrt(6) 2/sqrt(6)
-1/sqrt(6) 1/sqrt(6) 0];
IV=B*cclr;
% Convert to HSI
hclr=zeros(3,6);
% Hue, Saturation, Intensity
hclr(3,:)=IV(1,:);
% Intensity
for k=1:6
v1=0;
v1=IV(2,k);
v2=0;
v2=IV(3,k);
hclr(1,k)=atan(v2/v1);
% Hue
hclr(2,k)=sqrt(v1^2+v2^2); % Saturation
end
Starting with NTSC RGB colors
Red
Yellow
R
1.0
1.0
G
0
1.0
B
0
0
Green
0
1.0
Cyan
0
1.0
Blue
0
0
Magenta
1.0
0
0
1.0
1.0
1.0
Blue
-0.1510
0.1590
1.1280
Magenta
1.0160
0.2730
1.1270
We will get the CIE RGB color representations
Red
Yellow
Green
Cyan
R
1.1670
1.0210
-0.1460
-0.2970
G
0.1140
0.8670
0.7530
0.9120
B -0.0010
0.0580
0.0590
1.1870
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Performing the RGB to HSI conversion, we obtain
H
S
I
Red
0.6873
0.6776
0.4267
Yellow
0.0867
0.7261
0.6487
Green
-1.0726
0.4178
0.2220
Cyan
0.6022
0.8714
0.6007
Blue
0.1370
0.9264
0.3787
Magenta
-0.6561
0.4972
0.8053
Therefore
(a) Magenta has the highest intensity.
(b) Blue has the highest saturation.
[Bonus]
MATLAB code for displaying the colors:
% Display the colors
M=64;
img=zeros(M,M*6,3);
for k=1:6
img(:,(k-1)*M+1:k*M,1)=clr(1,k);
img(:,(k-1)*M+1:k*M,2)=clr(2,k);
img(:,(k-1)*M+1:k*M,3)=clr(3,k);
end
figure
imshow(img)
set(gcf,'color',[1,1,1]);
title({'Display of the 6 colors in NTSC RGB','Red, Yellow, Green, Cyan, Blue, Magenta'});
imgcie=zeros(M,M*6,3);
for k=1:6
if cclr(1,k)<0
imgcie(:,(k-1)*M+1:k*M,1)=0;
elseif cclr(1,k)>1
imgcie(:,(k-1)*M+1:k*M,1)=1;
else
imgcie(:,(k-1)*M+1:k*M,1)=cclr(1,k);
end
if cclr(2,k)<0
imgcie(:,(k-1)*M+1:k*M,2)=0;
elseif cclr(2,k)>1
imgcie(:,(k-1)*M+1:k*M,2)=1;
else
imgcie(:,(k-1)*M+1:k*M,2)=cclr(2,k);
end
if cclr(3,k)<0
imgcie(:,(k-1)*M+1:k*M,3)=0;
elseif cclr(3,k)>1
imgcie(:,(k-1)*M+1:k*M,3)=1;
else
imgcie(:,(k-1)*M+1:k*M,3)=cclr(3,k);
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end
end
figure
imshow(imgcie)
set(gcf,'color',[1,1,1]);
title({'Display of the 6 colors in CIE RGB','Red, Yellow, Green, Cyan, Blue, Magenta'});
Observations and remarks:
1. In displaying the colors in CIE RGB, the values are clipped to 0~1 so that imshow
can be used correctly.
2. The colors represented in CIE RGB have values less than zero or greater than one.
This means these colors cannot be reproduced in the real world because they are
outside the RGB color cube.
3. Observing CIE RGB, perceptually, magenta exhibits the maximum intensity,
followed by cyan. The most saturated colors are blue and red (from CIE RGB).
These facts roughly match the experimental results we get. Possible mismatch can
be due to the clipping of RGB values in CIE RGB, which results in changes in
intensity, hue, and saturation values.
4. NTSC RGB is used in televisions, while CIE RGB is used in computers.
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