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ICE_Chapter_07

# ICE_Chapter_07 - CHAPTER 7 Solutions for Exercises E7.1(a...

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CHAPTER 7 Solutions for 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 1

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Finally, the answer is 17.25 10 = 10001.01 2 (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 2
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 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 L is the length of the code words. Thus we need L = 11, which produces 2048 code words.

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