Purdue ECE 27000 - Lecture notes (2 pages)

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Lecture notes

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Purdue University
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Ece 27000 - Introduction to Digital System Design
Introduction to Digital System Design Documents
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2006 Pearson Education Inc Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing by the publisher For exclusive use of adopters of the book Digital Design Principles and Practices Fourth Edition by John F Wakerly ISBN 0 13 186389 4 3e2 1 3e2 3 3e2 5 3e2 6 3e2 18 2 1 2 3 2 5 2 6 a 1101011 2 6B 16 b 174003 8 1111100000000011 2 c 10110111 2 B7 16 d 67 24 8 110111 0101 2 e 10100 1101 2 14 D 16 f F3A5 16 1111001110100101 2 g 11011001 2 331 8 h AB3D 16 1010101100111101 2 i 101111 0111 2 57 34 8 j 15C 38 16 101011100 00111 2 a 1023 16 1000000100011 2 10043 8 b 7E6A 16 1111110011010102 77152 8 c ABCD 16 1010101111001101 2 125715 8 d C350 16 1100001101010000 2 141520 8 e 9E36 7A 16 1001111000110110 01111012 117066 364 8 f DEAD BEEF 16 1101111010101101 1011111011101111 2 157255 575674 8 a 1101011 2 107 10 b 174003 8 63491 10 c 10110111 2 183 10 d 67 24 8 55 3125 10 e 10100 1101 2 20 812510 f F3A5 16 62373 10 g 12010 3 138 10 h AB3D 16 43837 10 i 7156 8 3694 10 j 15C 38 16 348 21875 10 a 125 10 1111101 2 b 3489 10 6641 8 c 209 10 11010001 2 d 9714 10 22762 8 e 132 10 1000100 2 f 23851 10 5D2B 16 g 727 10 10402 5 h 57190 10 DF66 16 i 1435 10 2633 8 j 65113 10 FE59 16 2 20 hj b4j i 2 j i 0 Therefore 4n 1 B n 1 bi 2i i 0 hi 16i i 0 4n 1 B 2 4n i 0 n 1 bi 2i 16 n hi 16 i i 0 Suppose a 3n bit number B is represented by an n digit octal number Q Then the two s complement of B is represented by the 8 s complement of Q 3e2 22 2 24 Starting with the arrow pointing at any number adding a positive number causes overflow if the arrow is advanced through the 7 to 8 transition Adding a negative number to any number causes overflow if the arrow is not advanced through the 7 to 8 transition 2006 Pearson Education Inc Upper Saddle River NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing by the publisher For exclusive use of adopters of the book Digital Design Principles and Practices Fourth Edition by John F Wakerly ISBN 0 13 186389 4 3e2 24 2 26 Let the binary representation of X be x n 1 x n 2 x 1 x 0 Then we can write the binary representation of Y as x m x m 1 x 1x where m n d Note that x m 1 is the sign bit of Y The value of Y is 0 Y 2 m 1 xm 1 xi 2 i i 0 The value of X is n 2 X 2n 1 xn 1 xi 2i i 0 n 2 2n 1 xn 1 Y 2m 1 xm 1 xi 2 i i m 1 n 2 xi 2 i n 2 Case 1 x m 1 0 In this case X Y if and only if 2 n 1 x n 1 x 2i i m i only if all of the discarded bits x x are 0 the same as x 2n 1 xn 1 Y 2 2m 1 m n 1 m 1 Case 2 x m 1 1 In this case X Y if and only if 2 n 1 x n 1 2 2 m 1 is true if and only if all of the discarded bits x m x n 1 are 1 the same as x m 1 3e2 25 3e2 28 0 which is true if and i m x i 2 i n 2 0 which 2 27 If the radix point is considered to be just to the right of the leftmost bit then the largest number is 1 11 1 and the 2 s complement of D is obtained by subtracting it from 2 singular possessive Regardless of the position of the radix point the 1s complement is obtained by subtracting D from the largest number which has all 1s plural 2 30 n B bn 1 2n 1 2 bi 2i i 0 n 2 2B b n 1 2 n bi 2i 1 i 0 Case 1 b n 1 0 First term is 0 summation terms have shifted coefficients as specified Overflow if bn 2 1 4e2 33 Case 2 b n 1 1 Split first term into two halves one half is cancelled by summation term b n 2 2 n 1 if b n 2 1 Remaining half and remaining summation terms have shifted coefficients as specified Overflow if bn 2 0 2 35 001 010 011 100 101 110 111 000 4e2 37 2 38 The manufacturer s code fails every 4th time for a total of 2 n 2 for an n bit encoding disc A standard binary code fails when the LSB changes from 1 to 0 which changes the next bit and possibly others guaranteeing a problem So half the boundaries in a standard binary code are bad a total of 2 n 1 for an n bit encoding disc The manufacturer s code is only half as bad as a standard binary code 3e2 36 2 41 In the string representation each position may have a 0 a 1 or an x a total of three possibilities per position and 3 n combinations in all


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