fa06-sw-s2 - ECE 2030B 1:00pm 4 problems, 4 pages Problem 1...

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ECE 2030B 1:00pm Computer Engineering Fall 2006 4 problems, 4 pages Exam Two Solutions 25 October 2006 1 Problem 1 (3 parts, 28 points) Representations and Arithmetic Part A (10 points) Convert the following notations: binary notation decimal notation 11011 16 + 8 + 2 + 1 = 27 1010.011 8 + 2 + .25 + .125 = 10.375 2 38 256 x 1K x 1K x 1K = 256 billion hexadecimal notation octal notation 8.1 1000.0001 = 10.04 DA.FA 11011010.11111010 = 332.764 Part B (10 points) For the 24 bit representations below, determine the most negative value, most positive value, and step size (difference between sequential values). All answers should be expressed in decimal notation . Fractions (e.g., 3/16ths) may be used. All signed representations are two’s complement. representation most negative value most positive value step size signed integer (24 bits) . (0 bits) - 8 million 8 million 1 unsigned fixed-point (12 bits) . (12 bits) 0 4 thousand 1/4096 unsigned fixed-point (18 bits) . (6 bits) 0 256 thousand 1/64 signed fixed-point (18 bits) . (6 bits) - 128 thousand
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This note was uploaded on 06/17/2008 for the course ECE 2030 taught by Professor Wolf during the Fall '07 term at Georgia Institute of Technology.

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fa06-sw-s2 - ECE 2030B 1:00pm 4 problems, 4 pages Problem 1...

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