hwsol-0402

# hwsol-0402 - ECE 178 Hwk#2 Solutions 1 Problem 2.9(a The...

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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 width-to-height 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, 8-bit 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 4-connected 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.

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4) Problem 2.12 The solution to this problem consists of defining all possible neighborhood shapes to go from a diagonal segment to a corresponding 4-connected segment, as shown in Fig. P2.12.
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hwsol-0402 - ECE 178 Hwk#2 Solutions 1 Problem 2.9(a The...

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