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Digital Logic Design

by: Haskell Heaney DVM

Digital Logic Design CPEN 214

Haskell Heaney DVM
Christopher Newport University
GPA 3.74

Costa Gerousis

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Costa Gerousis
Class Notes
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This 10 page Class Notes was uploaded by Haskell Heaney DVM on Monday October 5, 2015. The Class Notes belongs to CPEN 214 at Christopher Newport University taught by Costa Gerousis in Fall. Since its upload, it has received 54 views. For similar materials see /class/219470/cpen-214-christopher-newport-university in Computer Engineering at Christopher Newport University.

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Date Created: 10/05/15
CPEN 214 Digital Logic Design Binary Systems C Gerousis Digital Design 3rd Ed Mano Prentice Hall Digital vs Analog 0 An analog system has continuous range of values 7 A mercury thermometer 7 Vinyl records 7 Human eye 0 A digital system has a set of discrete values 7 Digital Thermometer 7 Compact Disc CD 7 Digital camera Benefits of using digital Analog signal Digital signal 39 Advantages of using Digital 39 Cheap electronic circuits 39 Easier to calibrate and ad39ust 39 Resistance to noise Clearer picture and sound Binary System 39 Discrete elements of information are represented With bits called binary codes Example 09m 10012 15111112 Quem on Why are commercial products made With digital circuits as opposed to analog Most digital devices are programmable By changing the program in the device the same underlying hardware can be used for many different applications Decimal Code Review the decimal number system Base Radix is 10 symbols 01 9 Digits For Numbers gt 9 add more signi cant digits in position to the left eg 19gt9 Each position carries a weight WWI 102 101 I100 H10 h0 210 1LSD MSD If we were to write 193625 using a power series eXpansion and base 10 arithmetic 1gtlt1039gtlt1023gtlt1016gtlt1002gtlt10715gtlt1072 Binary number system The binary number system Base is 2 symbols 01 Binary Digits Bits For Numbers gt 1 add more signi cant digits in position to the left eg 10gt1 Each position carries a weight using decimal MSD mm 22 I 21 I20 u212722 LSD If we write 1011101 using a decimal power series we convert from binary to decimal 1gtlt24 0gtlt231gtlt221gtlt211gtlt20 0gtlt2711gtlt272 1gtlt160gtlt81gtlt41gtlt21gtlt10gtlt051gtlt0252325 Binary number system El 11000001112 10 ANS484375 gquot In computer work 210 1024 is referred as K kilo 220 1048576 is referred as M mega 230 240 D What is the exact number of bytes in a 16 Gbyte memory module OctalHex number systems The octal number system from Greek OKTQ 7 Its base is 8 9 eight digits 0 1 2 3 4 5 6 7 23648 10 2gtlt823gtlt816gtlt804gtlt8 1 1585 The hexadecimal number system from Greek AEKAEEI Its base is 16 9 rst 10 digits are borrowed from the decimal system and the letters A B C D E F are used for the digits 101112 13 1415 D63FA1610 13x164 6gtlt1633gtlt162 15gtlt16110gtlt160 877562 Conversion from Decimal to Binary 5 quot Conversion from decimal to binary Let each bit of a binary number be represented by a variable whose subscript bit positions ie 1102 a2a1a02 Its decimal equivalent is 1x221x210x2010 a2 x22 11le aox2010 It is necessary to separate the number into an integer part and a fraction Repeatedly divide the decimal number by 2 Conversion from Decimal to Binary Find the binary equivalent of 37 2 1805 218 9 0 29 4 05 1 LSB 0 1 24 2 0 0 0 1 3710 21001012 22 10 21 0 05 MSB 05310 2 ANS531021101012 Conversion from Decimal to Binary 5 quot Conversion from decimal fraction to bina same method used for integers except multiplication is used instead of division Convert 0854210 to binary give answer to 6 digits 08542 X2 l 07084 a11 M53 07084 X 2 1 04168 a2 1 04168X2 0 08336 a3 0 08336 X 2 1 06672 a4 1 06675 X 2 1 03344 a5 1 03344 X 2 0 06688 a6 0 L53 08542100a71a a a a a 01101102 72 73 74 75 752 El 53854210 2 Conversion from Decimal to Octal 5 quot Conversion from decimal to w The decimal number is rst divided by 8 The remainder is the LSB The quotient is then divide by 8 and the remainder is the next signi cant bit and so on Convert1122tooctal 81122 140O25 R2 lt LSB 8140 1705 R4 817 20125 R1 8E 0025 R2 MSB 112210 21428 Decimal Hex Binary Octal 0 0 0000 00 1 1 0001 m Table 12 2 2 W 02 8 3 3 0011 03 Page 4 4 0100 04 5 5 0101 J 6 e 0110 J 7 7 0111 07 s s 1000 10 9 9 1001 11 10 A 1010 12 11 B 1011 13 12 c 1100 14 13 D 1101 15 14 E 1110 15 15 F 1111 17 Conversion using Table 5 quot Conversion from and to binary octa and hexadecimal plays and important part in digital computers since 23 8 and 24 16 each octal digit corresponds to 3 binary digits and each hexa digit corresponds to 4 binary digits 010 111 100 001 011 0002 2741308 0110 1111 11010001001101002 6FD13416 1 s and 2 s Complements 1 s complement ofN 2quot 1 7N N is a binary 1 s complement can be formed by changing 1 s to 0 s and 0 s to 1 s 2 s complement of a number is obtained by leaving a11 least signi cant 0 s and the rst 1 unchanged and replacing 1 s with 0 s and 0 s with 1 in all higher signi cant digits The 1 s complement of 1101011 0010100 The 2 s complement of0110111 1001001 CI Find the 1 s and 2 scomplement of 10000000 Answer 01111111 and 10000000 Subtraction Using Complements Subtraction with digital hardware using complements Subtraction of two 71 digit unsigned numbers M 7N base r Adtho the r s complement ofNM rquot 7N IfM S N the sum will produce an end carry and is equal to rquot that can be discarded The result is then M 7 N 3 IfM 2 N the sum will not produce an end carry and is equal to rquot 7 N iAI N Binary Subtraction using complements 5 quot Binary subtraction is done using the same procedure Subtract 1010100 7 1000011 using 2 s complement A 1010100 2 s complementofB 0111101 Sum 10010001 Discard end carry 7 10000000 end carry Answer 0010001 CI Subtract 1000011 7 1010100 using 2 s complement Answer 7 0010001 Binary Subtraction using complements Subtract 1010100 7 1000011 using 1 s complement A 1010100 1 s complement ofB 0111100 Sum 10010000 Endaround carry 1 Answer 0010001 CI Subtract 1000011 7 1010100 using 1 s complement Answer 7 0010001 Arithmetic addition Negative numbers must be initially in 2 s complement form and if the obtained sum is negative it is in 2 s complement form 6 00000110 76 11111010 13 00001101 13 00001101 19 00010011 7 00000111 Cl Add 76 and 713 Answer 11101101 Transfer of Information with Registers x 0 me H N u mnmnmmnm i l i i mnmnm mm 1 in P vuc sm Rrgrmr 1w r 1 Wm Fig 11 Transfer or inrormanan with registers


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