Image Processing Basics - Digital Signal Processing: Fundamentals and Applications - page 726

EXAMPLE 14.11
Determine the 2D-DFT coefficients and magnitude spectrum for the following 2 2 image:
" 100 50
100 10

#

Solution:
Since M ¼ N ¼ 2, applying Equation (14.16) leads to
X u; v ¼ p 0; 0 e j 2pu0 þ p 0; 1 e j 2pu0 e j 2pv0 e j 2pv1

2 2 2 2

þp

e j 2pu1
2

þ p

e j 2pu1
2

e j 2pv0
2

e j 2pv1
2

1; 0 1; 1

For u ¼ 0 and v ¼ 0, we have
X 0; 0 ¼ 100e j0 e j0 þ 50e j0 e j0 þ 100e j0 e j0 10e j0 e j0
¼ 100 þ 50 þ 100 10 ¼ 240
For u ¼ 0 and v ¼ 1, we have
X 0; 1 ¼ 100e j0 e j0 þ 50e j0 e jp þ 100e j0 e j0 10e j0 e jp
¼ 100 þ 50 ð1Þþ100 10 ð1Þ¼160
Following similar operations,

X ð1; 0Þ¼60; and X ð1; 1Þ¼60

Thus, we have the following DFT coefficients:

" 240 160
60 60

#

X u; v ¼

Using Equation (14.17) , we can calculate the magnitude spectrum as

" 60 40
15 15

#

A u; v ¼

We can use the MABLAB function fft2() to verify the calculated DFT coefficients:
>> X¼fft2([100 50;100 -10])
X ¼
240 160
60 60

EXAMPLE 14.12
Given the 200 200 grayscale image shown in Figure 14.35A with a white rectangle (11 3 pixels) at its center
and a black background, we can compute its magnitude spectrum (which ranges from 0 to 255). We can display
the spectrum in terms of the grayscale. Figure 14.35B shows the spectrum image.
The displayed spectrum has four quarters. The left upper quarter corresponds to the frequency components,
and the other three quarters are the image counterparts. In the spectrum image, the upper left corner area in the
left upper quarter is white and hence has a highest scale value. Therefore, the image signal has low-frequency