Assignment-I (EEL715)
Meera (2011EEZ8311), Ritu (2011EEZ8469) and Adersh (2011EEZ8471)
Answer – 1
Make a BMP image file with a white square in center in paintbrush.
a. Display the magnitude of its 2-D Fourier Transform in the form of the image. Bring
the (0, 0) – frequency coordinate to the center of this display.
b. Change the size of the square and observe the change in the corresponding DFT
display. Comment on the observation.
c. Rotate the image by some angle (say 45°) and observe the change in FT display.
Explain your observation with mathematical equations.
Results:
If we rotate image by 45 deg, as shown in the display of magnitude also rotates by same
angle and changes in phase is also symmetric.
Original Image
Original Image Magnitude
Original Image Phase
Rotated Image
Rotated Square Magnitude
Rotated Square Phase
If we decrease the size of square, the overall dc value shown in the spectrum also reduces. While
displaying the magnitude we used min and max of spectrum. That’s why the display at the center is
not dense as compared to the spectrum of original image without scaling of White Square. Without
using min and max of spectrum, the while area at the center would have been smaller. The converse
results for image with bigger square is visible in the spectrum as shown in the figure.
Assignment-I (EEL715)
Meera (2011EEZ8311), Ritu (2011EEZ8469) and Adersh (2011EEZ8471)
Smaller Square Image
Smaller Square Magnitude
Smaller Square Phase
Bigger Square Image
Bigger Square Magnitude
Bigger Square Phase
Answer – 2
Load an image. Take its FFT. Reconstruct the image using only
a) Magnitude information of FFT
b) Phase information of FFT
Comment on the observation.
Results
Original Image
100
200
300
400
500
100 200 300 400 500
Magnitude
Phase
100
100
200
200
300
300
400
400
500
500
100 200 300 400 500
100 200 300 400 500
Assignment-I (EEL715)
Meera (2011EEZ8311), Ritu (2011EEZ8469) and Adersh (2011EEZ8471)
Figure 2.1 (a) Woman (b) Magnitude (c) Phase Angle
Image A FFT2 Phase
Image A FFT2 Magnitude
Figure 2.2 (a) Woman Reconstructed using only the Magnitude (b) Woman Reconstructed
using only the Phase Angle
Figure 2.2 (a) is obtained using only the spectrum/magnitude and computing the inverse
DFT. This means setting the exponential term to 1, which in turn implies setting the phase
angle to 0. It contains only the intensity information, with the dc term being the most
dominant. There is no shape information.
Figure 2.2 (b) is obtained using only the phase and computing the inverse DFT. The intensity
information has been lost, but the key shape feature in this image is recovered.
Answer – 3
Load an image in MATLAB in gray shades. Change the number of gray levels to 64, 16, 8, 4
and 2 to display the same image. Comment on the observation. Now change the spatial
resolution of this image. Explain the effect of sampling error.
Results
Gray Scale Resolution:
Assignment-I (EEL715)
Meera (2011EEZ8311), Ritu (2011EEZ8469) and Adersh (2011EEZ8471)
Original Image 256
64 Gray Levels
16 Gray Levels
8 Gray Levels
4 Gray Levels
4 Gray Levels
Figure 3.1 (a) 452x374, 256-level image (b)-(f) Image displayed in 64, 16, 8, 4 and 2
intensity levels, while keeping the image size constant.
This experiment is tested on a gray scale image ctskull-256 8-bit image. The resolution of an
image is to be reduced from 256 to 2, each time by a factor of 2. The change of the image
quality and the false contouring effect are to be observed. It is performed while keeping a
constant spatial resolution for the image (256x256). This is because that a higher number of
gray levels would give a smooth transition along the details of the image.
Spatial Resolution:
Assignment-I (EEL715)
Meera (2011EEZ8311), Ritu (2011EEZ8469) and Adersh (2011EEZ8471)
Original Image
Figure 3.1: Original Image Rose 1024x1024
This experiment is tested on a gray scale image rose 1024x1024. There are some small
visual differences between the resized images figs. 3.2 to 3.5, the most noticeable being a
slight distortion in the petals of the flower. There is blocky appearance in the petals of the
flower. It is performed while keeping a constant gray level for the image (256).
Assignment-I (EEL715)
Meera (2011EEZ8311), Ritu (2011EEZ8469) and Adersh (2011EEZ8471)
Image Resized to 1/2
Figure 3.2: Image Scale down to 512x512
Image Resized to 1/4
Figure 3.2: Image Scale down to 256x256
Assignment-I (EEL715)
Meera (2011EEZ8311), Ritu (2011EEZ8469) and Adersh (2011EEZ8471)
Image Resized to 1/8
Image Resized to 1/16
Figure 3.4: Image Scale down to (a) 128x128 (b) 64x64
Image Resized to 1/8
Image Resized to 1/16
Figure 3.5: Image Enlarged of Scale down image (a) 128x128 (b) 64x64
Answer – 4
Take an image having only two point sources. Pass it through a lowpass filter defined
(a) by a mask in spatial domain
(b) in frequency domain
Observe the result. Change the cut-off frequency of the filter and comment
on the result.
Results: Low pass filter blurs the image and cut-off frequency constrains the blurriness. In
case of special domain, the size of filter actually controls the cut-off frequency. Larger the
low pass filter size (e.g. nxn) means larger the cut-off frequency. We used Gaussian filter for
low pass frequency and cut off frequency is represented by D0 and should be some % of
image size. Here, we chose it to be the dimension of filter. For example, filter size in spatial
domain is 5x5, the D0 is taken as 25. The resultant image after low pass filter is shown in the
following figure. Please see the label of figure for operation.
Assignment-I (EEL715)
Meera (2011EEZ8311), Ritu (2011EEZ8469) and Adersh (2011EEZ8471)
Original
Frequency 3x3
Spatial 3x3
Frequency 5x5
Spatial 5x5
Answer – 5
Take an image that has black and white alternating vertical strips.
Pass it through a high pass filter defined in
(a)
(b)
in frequency domain
by a mask in spatial domain.
Observe the result. Change the cut-off frequency of the filter and comment on the result.
Results: High pass filter gives information on the higher intensity values in the image and
cut-off frequency constrains the number neighbour in X and Y directions shown in response
after application of high pass filter on image. It means, if cut off is higher, the thinner edges
would be seen in the response. Please see the label of figure for operation.
Original
Frequency 9x9
Spatial 9x9
Frequency 5x5
Spatial 5x5
Assignment-I (EEL715)
Meera (2011EEZ8311), Ritu (2011EEZ8469) and Adersh (2011EEZ8471)
Answer – 6
Take an image. Carry out edge detection in both directions using
a. Sobel operator
b. Prewitt operator
c. Roberts operator
Results: As shown in the label, first one is original image andother three are images after
applying Sobel, Prewitt and Roberts filters in special domain.
Original Image
After Sobel
After Robert
After Prewitt
Figure 6.1