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1 3 / 1 / 16 / John M. Yarbrough: Digital Logic Applications and Design % 70%

2 1 CHAPTER 1 Digital Concepts and Number Systems

4 1.1 Digital and Analog: Basic Concepts P1

5 1.1

6 1.1 V c (0)=0V, Initial voltage at t 0 is 0V V c =V a (1-e -t/rc ) P1

7 1.1 Discrete voltage values for time interval (a) Analog representation (b) Discrete representation P2

8 1.1

9 1.1 Analog volt meter Digital volt meter P2/3

10 1.2 Some History of Digital Systems 1642 Blaise Pascal 1820 Charles Babbage P4

11 George Boole The Mathematical Analysis of Logic 0 1

12 Walter Brattain John Bardeen William Shockley Jack Kilby(Texas Instruments) Robert Noyce(Fairchild Semiconductor) ) (IC )

13 Noyce , i486 1,200,

15 1.3 Impact of Digital Technology on Society 4

16 G MID PMP / LED / LED / (HEV)

17 1.4 Defining the Problem, an Introduction to Algorithms Algorithm P5

19 1.5 Digital Systems Overview P6

20 1.5 SSI (small-scale integration) contains 0-9 gates MSI (medium-scale integration) contains 10-99gates LSI (large-scale integration) contains 100 or more gates VLSI (very large-scale integration ) contains more than 1000 gates P6

21 1.6 Introduction to Number Systems P7

22 1.7 Positional Number Systems P7

23 Decimal Numbers =( )+( )+( )+( )+( )+( ) Radix or Base Weight P7

24 r ( r) r N = A n-1 r n-1 + A n-2 r n A 1 r +A 0 + A -1 r -1 + A -2 r A -m r -m Most Significant Bit (MSB) Least Significant Bit (LSB)

25 Let r = radix or base of a number system; Let c = character from the character set of the radix; Let N = number to be represented in radix; Let n = the number of digits in the integer portion of N Let m = the number of digits in the fractional portion of N ( ) m m n n n n r r c r c r c r c c r r c r c N = ( ) = = 1 n m i i i r r c N P7 P7

26 1 e.g. let r = 6 (312.4) 6 = = (116.66) 10 Conversion from r-radix (any system with radix r) to decimal follows similar process as above

27 2 Most common number systems for computers: Binary (r = 2) Octal (r = 8) Hexadecimal (r = 16)

28 Binary Numbers Computers represent all data as strings of bits, each bit being either 0 or 1 base 2, with 2 digits: 0 and 1 e.g. ( ) 2 = (in decimal) = ½ + 0 = (45.5) 10 P8

29 Powers of two Memorize at least through 2 16

30 Octal Numbers base 8 with 8 digits: 0..7 Eg = (in decimal) = P8

31 Hexadecimal Numbers r = 16 Digits (convention): 0..9, A, B, C, D, E, F A=10, B=11,, F = 15 P8

32 e.g1. (3FB) 16 = (in decimal) = = (1019) 10 e.g2. A59C.3A 16 = (A 16 3 )+( )+( )+(C 16 0 )+( )+(A 16-2 )

33 1.8 Number System Conversion Decimal binary octal and hexadecimal number P9

34 A B C D E F P10

35 Binary Octal ( ) 2 ( ) 8 E.g. Convert the binary number to octal Solution 010,011,110, ,101,010 2 =

36 Binary Hex ( ) 2 ( 6 A 8. F 5 C ) 16 E.g. Convert the binary number to hexadecimal Solution 0100,1111, ,0101, =4F7.D50 16

37 hexadecimal and octal to Binary Ex 1.9/1.11 Convert the numbers F37A.B2 16, to binary Solution F37A.B2=1111,0011,0111, , =111,011, Practice problem: convert convert to binary then hexadecimal Solution =011,110, ,011, ,110, ,011,110 2 =0,1111, ,1111,0 2 =F7.4F 16 P10/11

38 Binary to decimal Conversion =(1 2 4 )+(1 2 3 )+(0 2 2 )+(0 2 1 )+(1 2 0 ) +(0 2-1 )+(1 2-2 )+(1 2-3 ) = = P11

39 Decimal to Binary Conversion 2 0 E.g. Convert to binary LSB Solution = MSB P12

40 D = B b b b b b b 5 0

41 (215) D =(? ) B 1=b 0 1=b 1 1=b =b 3 1=b 4 0=b 5 1=b 6 1=b 7 ( 215 ) D =( ) B (215) D = 1 2 = ( ) 6 B

42 Decimal to Binary Conversion 2 Ex 1.13 Convert to binary Solution Multiply.75 by 2 (.75)2=1.5 1(MSB) Multiply.75 by 2 (.5)2=1.0 1 Multiply.75 by 2 (0)2=0.0 0(LSB) (.75) 1 =.110 = D 2 P12/13

43 1.14 Convert to binary Solution First, convert the integer part of the decimal number by successive division 1.Devide 95 by 2 5.Devide 5 by 2 (LSB) 2.Devide 47 by 2 6.Devide 2 by 2 3.Devide 23 by 2 7.Devide 1 by 2 4.Devide 11 by 2 (MSB) = = P13

44 1.14 Convert to binary Solution Second, convert the fraction = (.0625) 2 = (MSB) 2. (.125) 2 = (.25) 2 = (.5) 2 = (LSB) = = P13

45 Decimal to Any Radix Conversion The conversion of decimal numbers to any other radix applies the successive division and successive multiplication algorithms Ex 1.15 Convert to octal (base 8) Solution Convert the integer portion by successive division 1.Divide 23 by 8 (LSB) Convert te fraction portion by successive multiplication (.625) 8 = (MSB) 2.Divide 2 by =27 8 (MSB) = = P14

46 Any Radix to Decimal Conversion e.g. Convert to decimal Solution =3(25)+2(5)+4(1)+2(1/5) = /5 = P15

47 Any Radix Conversion Practice problem: 1. Convert to decimal 2. Convert to base 5 Solution:

48 P33/34

49 1.9 Binary Codes n 2 n N 2 n N P16

50 1.9.1 Natural Binary Coded Decimal 4 0~9 e.g. Convert into BCD Solution: 9=1001 2=0010 7=0111 5=0101 6= =1001,0010,0111, in BCD P16

51 1.9.2 Binary Codes (Weighted) P17

52 ( ) 8421BCD =(90) 3 : , 1 8,

53 BCD BCD 0111 = = ( 7) D [ ] 8421BCD [ 1101 ] 2421BCD = = () 7 D

54 BCD BCD ( 463.5) ( 863.2) = = BCD 2421BCD

55 FF H = B = O = D = 8421BCD B = 9DF1 H

56 1.9.3 BCD BCD Self-Complementing Codes P17

57 1.9.4 Unit Distance Code P18

58 P18

59 1.9.5 Alphanumeric Codes ASCII (American standard code for information interchange) 128 EBCDIC (extended BCD interchange code) P19

60 b 3 b 2 b 1 b 0 b 6 b 5 b NUL DLE SP P p 0001 SOH DC1! 1 A Q a q 0010 STX DC2 2 B R b r 0011 ETX DC3 # 3 C S c s 0100 EOT DC4 $ 4 D T d t 0101 ENQ NAK % 5 E U e u 0110 ACK SYN & 6 F V f v 0111 BEL ETB 7 G W g w 1000 BS CAN ( 8 H X h x 1001 HT EM ) 9 I Y i y 1010 LF SUB * : J Z j z 1011 VT ESC + K [ k { 1100 FF FS L \ l 1101 CR GS = M ] m } 1110 SO RS N n 1111 SI US / O _ o DEL

61 1.9.6 Signed Number Binary +11 ( ) B 11 ( ) B P20/21

62 1.9.7 Signed Magnitude Codes P20/21

63 1.9.8 Complement Codes Sign magnitude 2s complement 1s complement D

64 12 12 (mod) = =12+5=5 9-4=(9 + 8)mod

65 s 9s 8s 8s 2s s P21

66 [0,M) M M a,b [0, M) f(a-b)==f(a+c) c= M-b b f 0<=x< M, f(x)=x; x>= M,f(x)=x % M; x<0,f(x)=f(m +x). % ( ) f ab==a+c -b c n M=2 n c=2 n -b 2 n -b 2 n 1 n 0 b 2 n -1 n 1 b 1 2 n -b 1

67 Ex: Find 2 s complement of Complement the bits + 1 Add [N] 10 =10 n -(N) 10 [N] 2 = 2 n -(N) 2 [N] 2 (N) 2 [N] 2 = (101001) 2 = ( ) 2 - (101001) 2 = (010111) 2 (N) 2 + [N] 2 = +0

68 Radix Complement Number Systems Quzi Find the 2s complement of the binary number Find the 1s complement code for Find the 2s complement code for Answer:

69 1.10 Arithmetic Arithmetic = = = P22

70 = = = =

71 P24

72 P24

73 1 1 = 1 + ( 1) = 0 ( ) 8bits = = ? = ~0~ = = !

74 X>Y X<Y X=Y (+X)+(+Y) (+X)+(- Y) (- X)+(+Y) (- X)+(- Y) (+X) (+Y) (+X) (- Y) (- X) (+Y) (- X) (- Y) +(X+Y) - (X+Y) +(X+Y) - (X+Y) +(X-Y) - (X-Y) +(X-Y) - (X-Y) - (X-Y) +(X-Y) - (X-Y) +(X-Y) +(X-Y) +(X-Y) +(X-Y) +(X-Y) P25

75 = 12 0^ ^ ^ = 6 0^ ^ ^ = -6 0^ ^ ^ (-9) + (-3) = -12 1^ ^ ^ (-3) = 12 0^ ^ ^

76 /31 0^ ^ (+X )+(+Y ) 0^ ^ ^ ^ ^ ^ The error occurred due to insufficient bit positions to hold the answer P25

77 Add X and Y. Let X= 8 10 and Y= 10 10, both in 2s complement. 1^ ^ ^1110 1^ ^ ^ ( +( )= 18 10, 0^1110 is not and 1^01110 is the Complement of P26

78 Subtract Y from X. Let X=+8 10 and Y=-6 10, X-Y=X+(-Y) =0^1000 The 2s complement of (1^1010) is 0^0110 0^1000 = ^0110 =(complement of -6 ) 0^1110 = P27

79 Subtract Y from X. Let X= = 1^0110 and Y= 5 10 = 1^1011 The radix complement of 1^1011(-5 10 ) is 0^0101 1^0110 = ^0101 =(complement of ) 1^1011 =-5 10 P27

80 perform the following operations by finding the radix complement of the subtrahend( ) and adding the result to the minuend( ):

81 P35/36

Microsoft Word - 最新正文.doc

Microsoft Word - 最新正文.doc 9 21 1.1.1 1.1.2 1 2 2 Windows 7+Office 2010 3 4 5 6 4 7 1.1.3 5 1.1.4 1 3 2 NII 1993 3 CNNIC 2014 1 16 33 1 2013 12 6.18 5358 45.8% 2012 3.7 2 2013 12 5 19.1% 2012 74.5% 81.0% 2013 3G 2013 12 2.47 2012