Written Assignment 1 - 16 with 1 bit excess 1B90F 16 + C44E...

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Shayan Rizvi CSE 2300 9-26- 11 Written Assignment 1 Section 15 Problem 1 25 10 = 0001 1001 2 Compliment bits = 1110 0110 2 + 1 2’s compliment = 1110 0111 2 120 10 = 0111 1000 2 111 Compliment bits = 1000 0111 2 + 1 2’s compliment = 1000 1000 2 -42 10 = 1010 1010 1 Compliment bits = 0101 0101 2 + 1 2’s compliment = 0101 0110 2 -6 10 = 1000 0110 2 1 Compliment bits = 0111 1001 2 + 1 2’s compliment = 0111 1010 2 -111 10 = 1110 1111 2 Compliment bits = 0001 0000 2 + 1 2’s compliment = 0001 0001 2 Problem 2 12648430 10 = ? 16 12648430 mod 16 = 14 = E 790526 mod 16 = 14 = E 49407 mod 16 = 15 = F 3087 mod 16 = 15 = F 192 mod 16 = 0 = 0 12 mod 16 = 12 = C 12648430 10 = C0FFEE 16
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Problem 3 F35B 16 + 27E6 16 F35B 16 = 1111 0011 0101 1011 2 27E6 16 = 0010 0111 1110 0110 2 11 1111 1111 11 1111 0011 0101 1011 2 +0010 0111 1110 0110 2 0011 1011 0100 0001 2 with an excess of 1 bit = 3B41
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Unformatted text preview: 16 with 1 bit excess 1B90F 16 + C44E 16 1B90F 16 = 0001 1011 1001 0000 1111 2 C44E 16 = 1100 0100 0100 1110 2 11 1 11 0001 1011 1001 0000 1111 2 + 1100 0100 0100 1110 2 0010 0111 1101 0101 1110 2 = 27D5D 16 Problem 4 The 2’s compliment in subtraction is very useful. Subtracting two binary numbers can get very tedious and is prone for mistakes. When the 2’s compliment is used in subtraction of binary numbers, subtraction is not done directly. The subtrahend is negated by taking its 2’s compliment and then it is added to the minuend. Now the normal addition rules can be applied. By using the 2’s compliment a difficult subtraction problem turns into an easier addition problem....
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This note was uploaded on 02/14/2012 for the course CSE 2300 taught by Professor Stevechu during the Fall '11 term at Central Connecticut State University.

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Written Assignment 1 - 16 with 1 bit excess 1B90F 16 + C44E...

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