0132130084_ism07

# 0132130084_ism07 - CHAPTER 7 Exercises E7.1 (a) For the...

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281 CHAPTER 7 Exercises E7.1 (a) For the whole part, we have: Quotient Remainders 23/2 11 1 11/2 5 1 5/2 2 1 2/2 1 0 1/2 0 1 Reading the remainders in reverse order, we obtain: 23 10 = 10111 2 For the fractional part, we have 2 × 0.75 = 1 + 0.5 2 × 0.50 = 1 + 0 Thus, we have 0.75 10 = 0.110000 2 Finally, the answer is 23.75 10 = 10111.11 2 (b) For the whole part, we have: Quotient Remainders 17/2 8 1 8/2 4 0 4/2 2 0 2/2 1 0 1/2 0 1 Reading the remainders in reverse order, we obtain: 17 10 = 10001 2 For the fractional part, we have 2 × 0.25 = 0 + 0.5 2 × 0.50 = 1 + 0 Thus, we have 0.25 10 = 0.010000 2 Finally, the answer is 17.25 10 = 10001.01 2

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282 (c) For the whole part we have: Quotient Remainders 4/2 2 0 2/2 1 0 1/2 0 1 Reading the remainders in reverse order we obtain: 4 10 = 100 2 For the fractional part, we have 2 × 0.30 = 0 + 0.6 2 × 0.60 = 1 + 0.2 2 × 0.20 = 0 + 0.4 2 × 0.40 = 0 + 0.8 2 × 0.80 = 1 + 0.6 2 × 0.60 = 1 + 0.2 Thus we have 0.30 10 = 0.010011 2 Finally, the answer is 4.3 10 = 100.010011 2 E7.2 (a) 1101.111 2 = 1 × 2 3 + 1 × 2 2 +0 × 2 1 +1 × 2 0 +1 × 2 -1 +1 × 2 -2 +1 × 2 -3 = 13.875 10 (b) 100.001 2 = 1 × 2 2 +0 × 2 1 +0 × 2 0 +0 × 2 -1 +0 × 2 -2 +1 × 2 -3 = 4.125 10 E7.3 (a) Using the procedure of Exercise 7.1, we have 97 10 = 1100001 2 Then adding two leading zeros and forming groups of three bits we have 001 100 001 2 = 141 8 Adding a leading zero and forming groups of four bits we obtain 0110 0001 = 61 16 (b) Similarly 229 10 = 11100101 2 = 345 8 = E5 16 E7.4 (a) 72 8 = 111 010 = 111010 2 (b) FA6 16 = 1111 1010 0110 = 111110100110 2 E7.5 197 10 = 0001 1001 0111 = 000110010111 BCD
283 E7.6 To represent a distance of 20 inches with a resolution of 0.01 inches, we need 20/0.01 = 2000 code words. The number of code words in a Gray code is 2 L in which is the length of the code words. Thus we need L = 11, which produces 2048 code words. E7.7 (a) First we convert to binary 22 10 = 16 + 4 + 2 = 10110 2 Because an eight-bit representation is called for, we append three leading zeros. Thus +22 becomes 00010110 in two’s complement notation. (b) First we convert +30 to binary form 30 10 = 16 + 8 + 4 + 2 = 11110 2 Attaching leading zeros to make an eight-bit result we have 30 10 = 00011110 2 Then we take the ones complement and add 1 to find the two’s complement: one’s complement: 11100001 add 1 +1 11100010 Thus the eight-bit two’s complement representation for -30 10 is 11100010. E7.8 First we convert 19 10 and -4 10 to eight-bit two’s complement form then we add the results. 19 00010011 -4 111111100 15 00001111 Notice that we neglect the carry out of the left-most bit. E7.9 See Tables 7.3 and 7.4 in the book. E7.10 See Table 7.5 in the book.

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## This note was uploaded on 11/20/2011 for the course ELE 220 taught by Professor Anas during the Spring '11 term at American Dubai.

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0132130084_ism07 - CHAPTER 7 Exercises E7.1 (a) For the...

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