Homework1_sols

Homework1_sols - Problem 1.23 Solution The bit...

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ECE 25 Homework 1 Solutions Problem 1.6 Solution The values are H 11100111 L 2 = 231, H 22120 L 3 = 231, H 3113 L 4 = 215, H 4110 L 5 = 530, H 343 L 8 = 227. As an example 343 8 = 3 8 0 + 4 8 1 + 3 8 2 227 Problem 1.15 Solution The decimal number 694 can be represented as 0110 1001 0100 and that for 835 is 1000 0011 0101. To add them using binary we start with the lowest order digit 0100 4 1001 9 0001 Carry 0101 5 0011 3 0110 6 _____ 110 (Add six) 1000 8 _____ 0110 (add six) 1001 9 1 0010 (12 in BCD) _______ 1 0101 (15) Putting the terms together we have 1 0101 0010 1001 or 1529 in BCD. Problem 1.18 Solution We need a Gray code for 16 digits as opposed to 8 digits. Following the procedure in Section 1-5 we have 0 1 2 3 4 5 6 7 8 9 A B C D E F 0000 0001 0011 0010 0110 0111 0101 0100 1100 1101 1111 1110 1010 1011 1001 1000
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Unformatted text preview: Problem 1.23 Solution The bit configurations are as follows: 365= 101101101 2 365 = 0011 0110 0101 in BCD and 0011 0011 0011 0110 0011 0101 in 8-bit ASCII. Note that this problem emphasizes the fact that the bit configuration can represent different types of information depending on the code. Problem 1.24 Solution The total number for binary is 2 32 = 4294967296. For BCD each decimal digit takes four bits so there are 8 digits that can be represents so the total number of integers is 10 8 which is about a factor of 43 smaller that the binary representation. For ASCII each digit requires 8 bits so the total number of integers is 10 4 , 2...
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This note was uploaded on 05/28/2009 for the course ECE ece 25 taught by Professor Bill lin during the Fall '08 term at UCSD.

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Homework1_sols - Problem 1.23 Solution The bit...

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