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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 Finally, the answer is 17.25 10 = 10001.01 2 1 (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 +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 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 2 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. is the length of the code words....
View Full Document

This note was uploaded on 04/12/2008 for the course ECE 3710 taught by Professor Haris during the Spring '07 term at Georgia Institute of Technology.

Page1 / 12

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online