(a) Given the image and a filter, calculate the convoluting result.
• i). Given an image of 1 × 8 resolution and a filter of 1 × 3 size as following, show the result of convoluting the image with filter
Image: 𝐼 = 1
Filter: 𝐹 1 0 −1 1
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56 64 79 98
Flip the filter:
115−79126−98132−115133−126 =[23343628177]
(a) Given the image and a filter, calculate the convoluting result.
• ii). Given an image of 4×5 resolution and a filter of 3×3 size as following, show the result of convoluting the image with filter
58 010 𝐼= 98 𝐹=1−11
467 128 775
2992 39556
Answer:𝐼 ∗𝐹 = 22
20 15 21 16 24 19
(b) Name two applications of NMS in image processing.
Answer: Non-Maximum Suppression (NMS) is widely used in image processing tasks such as Canny edge detection, Harris corner detection.
Harris corner detection: after calculating the 𝑓 value, NMS is employed to locate the corner positions.
Canny edge detection: Perform comparisons on the edge strength of the current pixel with the edge strength of the pixel in the positive and negative gradient directions.
Preserve the largest and suppress the others.
(c) Can Harris corner detector be used to detect edges? Explain your answer by describing how Harris matrix will respond to edges.
Answer: Yes, when one eigenvalue has large value and the other has small value.
(e). What is the difference between the 1st derivative of Gaussian filter and Laplacian of Gaussian (LoG) filter ?
𝜕2𝐺 + 𝜕2𝐺 𝜕𝑥2 𝜕𝑦2
1. 1st derivative of Gaussian filter :
2. Laplacian of Gaussian (LoG) filter:
(e). Explain how SIFT can achieve invariances in (i) illumination, (ii) scale, (iii) rotation.
(i). Illumination : the SIFT descriptor is based on gradient magnitude, instead of pixel value. (ii). Scale: SIFT detects the interest points across different scale levels.
(iii). Rotation: the image is firstly rotated to the dominant orientation, thus SIFT is invariant to rotation.
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