2. 1 CHAPTER 1 DIGITAL SYSTEMS AND BINARY NUMBERS

3. 2 O UTLINE OF C HAPTER 1 1.1 Digital Systems 1.2 Binary Numbers 1.3 Number-base Conversions 1.4 Octal and Hexadecimal Numbers 1.5 Complements 1.6 Signed Binary Numbers 1.7 Binary Codes 1.8 Binary Storage and Registers 1.9 Binary Logic 2

4. 3 DIGITAL SYSTEMS AND BINARY NUMBERS Digital computers General purposes Many scientific, industrial and commercial applications Digital systems Digital Telephone Digital camera Electronic calculators Digital TV Discrete information-processing systems Manipulate discrete elements of information For example, {1, 2, 3, …} and {A, B, C, …}… 3

5. 4 A NALOG AND D IGITAL S IGNAL Analog system The physical quantities or signals may vary continuously over a specified range. Digital system The physical quantities or signals can assume only discrete values. t X(t) t Analog signalDigital signal 4

6. 5 B INARY D IGITAL S IGNAL For digital systems, the variable takes on discrete values. Two level, or binary values. Binary values are represented abstractly by: Digits 0 and 1 Words (symbols) False (F) and True (T) Words (symbols) Low (L) and High (H) And words Off and On

7. 6 D ECIMAL N UMBER S YSTEM Base (also called radix) = 10 10 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } Digit Position Integer & fraction Digit Weight Weight = ( Base) Position Magnitude Sum of “ Digit x Weight ” Formal Notation 102-2 51274 1010.11000.01 5001020.70.04 d 2 *B 2 +d 1 *B 1 +d 0 *B 0 +d -1 *B -1 +d -2 *B -2 (512.74) 10 6

8. 7 O CTAL N UMBER S YSTEM Base = 8 8 digits { 0, 1, 2, 3, 4, 5, 6, 7 } Weights Weight = ( Base) Position Magnitude Sum of “ Digit x Weight ” Formal Notation 102-2 811/8641/64 51274 5 *8 2 +1 *8 1 +2 *8 0 +7 *8 -1 +4 *8 -2 =(330.9375) 10 (512.74) 8 7

9. 8 B INARY N UMBER S YSTEM Base = 2 2 digits { 0, 1 }, called b inary dig its or “ bits ” Weights Weight = ( Base) Position Magnitude Sum of “ Bit x Weight ” Formal Notation Groups of bits 8 bits = Byte 102-2 211/241/4 10101 1 *2 2 +0 *2 1 +1 *2 0 +0 *2 -1 +1 *2 -2 =(5.25) 10 (101.01) 2 1 1 0 0 0 1 0 1 8

10. 9 H EXADECIMAL N UMBER S YSTEM Base = 16 16 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F } Weights Weight = ( Base) Position Magnitude Sum of “ Digit x Weight ” Formal Notation 102-2 1611/162561/256 1E57A 1 *16 2 +14 *16 1 +5 *16 0 +7 *16 -1 +10 *16 -2 =(485.4765625) 10 (1E5.7A) 16 9

11. 10 T HE P OWER OF 2 n2n2n 02 0 =1 12 1 =2 22 2 =4 32 3 =8 42 4 =16 52 5 =32 62 6 =64 72 7 =128 n2n2n 82 8 =256 92 9 =512 102 10 =1024 112 11 =2048 122 12 =4096 202 20 =1M 302 30 =1G 402 40 =1T Mega Giga Tera Kilo 10

12. 11 A DDITION Decimal Addition 55 55 + 011 = Ten ≥ Base  Subtract a Base 11Carry 11

13. 12 B INARY A DDITION Column Addition 101111 11110 + 0000111 ≥ (2) 10 111111 = 61 = 23 = 84 12

14. 13 B INARY S UBTRACTION Borrow a “Base” when needed 001110 11110 − 0101110 = (10) 2 2 2 2 2 1 000 1 = 77 = 23 = 54 13

15. 14 B INARY M ULTIPLICATION Bit by bit 01111 0110 00000 01111 01111 0 0000 01101110 x 14

16. 15 N UMBER B ASE C ONVERSIONS Decimal (Base 10) Octal (Base 8) Binary (Base 2) Hexadecimal (Base 16) Evaluate Magnitude 15

17. 16 D ECIMAL ( I NTEGER ) TO B INARY C ONVERSION Divide the number by the ‘Base’ (=2) Take the remainder (either 0 or 1) as a coefficient Take the quotient and repeat the division Example: ( 13 ) 10 QuotientRemainder Coefficient Answer: (13) 10 = (a 3 a 2 a 1 a 0 ) 2 = (1101) 2 MSB LSB 13 / 2 = 61 a 0 = 1 6 / 2 = 30 a 1 = 0 3 / 2 = 11 a 2 = 1 1 / 2 = 01 a 3 = 1 16

18. 17 D ECIMAL ( F RACTION ) TO B INARY C ONVERSION Multiply the number by the ‘Base’ (=2) Take the integer (either 0 or 1) as a coefficient Take the resultant fraction and repeat the multiplication Example: ( 0.625 ) 10 IntegerFraction Coefficient Answer: (0.625) 10 = (0.a -1 a -2 a -3 ) 2 = (0.101) 2 MSB LSB 0.625 * 2 = 1. 25 0.25 * 2 = 0. 5 a -2 = 0 0.5 * 2 = 1. 0 a -3 = 1 a -1 = 1 17

19. 18 D ECIMAL TO O CTAL C ONVERSION Example: ( 175 ) 10 QuotientRemainder Coefficient Answer: (175) 10 = (a 2 a 1 a 0 ) 8 = (257) 8 175 / 8 = 217 a 0 = 7 21 / 8 = 25 a 1 = 5 2 / 8 = 02 a 2 = 2 Example: ( 0.3125 ) 10 IntegerFraction Coefficient Answer: (0.3125) 10 = (0.a -1 a -2 a -3 ) 8 = (0.24) 8 0.3125 * 8 = 2. 5 0.5 * 8 = 4. 0 a -2 = 4 a -1 = 2 18

20. 19 B INARY − O CTAL C ONVERSION 8 = 2 3 Each group of 3 bits represents an octal digit OctalBinary 00 0 0 10 0 1 20 1 0 30 1 1 41 0 0 51 0 1 61 1 0 71 1 1 Example: ( 1 0 1 1 0. 0 1 ) 2 ( 2 6. 2 ) 8 Assume Zeros Works both ways (Binary to Octal & Octal to Binary) 19

21. 20 B INARY − H EXADECIMAL C ONVERSION 16 = 2 4 Each group of 4 bits represents a hexadecimal digit HexBinary 00 0 10 0 0 1 20 0 1 0 30 0 1 1 40 1 0 0 50 1 60 1 1 0 70 1 1 1 81 0 0 0 91 0 0 1 A1 0 B1 0 1 1 C1 1 0 0 D1 1 0 1 E1 1 1 0 F1 1 Example: ( 1 0 1 1 0. 0 1 ) 2 ( 1 6. 4 ) 16 Assume Zeros Works both ways (Binary to Hex & Hex to Binary) 20

22. 21 O CTAL − H EXADECIMAL C ONVERSION Convert to Binary as an intermediate step Example: ( 0 1 0 1 1 0. 0 1 0 ) 2 ( 1 6. 4 ) 16 Assume Zeros Works both ways (Octal to Hex & Hex to Octal) ( 2 6. 2 ) 8 Assume Zeros 21

23. 22 D ECIMAL, B INARY, O CTAL AND H EXADECIMAL DecimalBinaryOctalHex 000000000 010001011 020010022 030011033 040100044 050101055 060110066 070111077 081000108 091001119 10101012A 11101113B 12110014C 13110115D 14111016E 15111117F 22

24. 23 C OMPLEMENTS There are two types of complements for each base- r system: the radix complement and diminished radix complement. Diminished Radix Complement (r-1)’s Complement Given a number N in base r having n digits, the ( r–1 )’s complement of N is defined as: (r n –1) – N Example for 4-digit decimal numbers : 9’s complement is (r n – 1)–N = (10 4 –1)– N = 9999– N 9’s complement of 4670 is 9999–4670 = 5329 Example for 4-digit binary numbers: 1’s complement is (r n – 1) – N = (2 4 –1)– N = 1111– N 1’s complement of 1100 is 1111–1100 = 0011 Observation: Subtraction from ( r n – 1) will never require a borrow For binary: 1 – 0 = 1 and 1 – 1 = 0 23

25. 24 C OMPLEMENTS 1’s Complement ( Diminished Radix Complement) All ‘0’s become ‘1’s All ‘1’s become ‘0’s Example (10110000) 2  (01001111) 2 If you add a number and its 1’s complement … 1 0 1 1 0 0 0 0 + 0 1 0 0 1 1 1 1 1 1 1 1 24

26. 25 C OMPLEMENTS Radix Complement Example: Base-10 Example: Base-2 The r's complement of an n-digit number N in base r is defined as r n – N for N ≠ 0. Comparing with the (r  1) 's complement, we note that the r's complement is obtained by adding 1 to the (r  1) 's complement, since r n – N = [(r n  1) – N] + 1. The 10's complement of 012398 is 987602 The 10's complement of 4670 is 5330 The 2's complement of 1101100 is 0010100 The 2's complement of 0110111 is 1001001 25

27. 26 C OMPLEMENTS 2’s Complement ( Radix Complement) Take 1’s complement then add 1 Toggle all bits to the left of the first ‘1’ from the right Example : Number: 1’s Comp.: 0 1 0 1 0 0 0 0 1 0 1 1 0 0 0 0 0 1 0 0 1 1 1 1 + 1 OR 1 0 1 1 0 0 0 0 00001010 26

28. 27 C OMPLEMENTS Subtraction with Complements The subtraction of two n -digit unsigned numbers M – N in base r can be done as follows: 27

29. 28 C OMPLEMENTS Example 1.5 Using 10's complement, subtract 72532 – 3250. Example 1.6 Using 10's complement, subtract 3250 – 72532. There is no end carry. Therefore, the answer is – (10's complement of 30718) =  69282. 28

30. 29 C OMPLEMENTS Example 1.7 Given the two binary numbers X = 1010100 and Y = 1000011, perform the subtraction (a) X – Y ; and (b) Y  X, by using 2's complement. There is no end carry. Therefore, the answer is Y – X =  (2's complement of 1101111) =  0010001. 29

31. 30 C OMPLEMENTS Subtraction of unsigned numbers can also be done by means of the ( r  1)'s complement. Remember that the ( r  1) 's complement is one less then the r 's complement. Example 1.8 Repeat Example 1.7, but this time using 1's complement. There is no end carry, Therefore, the answer is Y – X =  (1's complement of 1101110) =  0010001. + 30

32. 31 S IGNED B INARY N UMBERS To represent negative integers, we need a notation for negative values. Signed-magnitude represents the sign with a bit placed in the leftmost position of the number and the rest of the bits represent the number. The convention is to make the sign bit 0 for positive and 1 for negative. Example: +9 is represented only by 00001001 three different ways to represent -9 31

33. 32  Table 1.3 lists all possible four-bit signed binary numbers in the three representations. 2 n-1 -1 -(2 n-1 -1)

34. 33 S IGNED B INARY N UMBERS Arithmetic addition The addition of two numbers in the signed-magnitude system follows the rules of ordinary arithmetic. If the signs are the same, we add the two magnitudes and give the sum the common sign. If the signs are different, we subtract the smaller magnitude from the larger and give the difference the sign of the larger magnitude. The addition of two signed binary numbers with negative numbers represented in signed-2's-complement form is obtained from the addition of the two numbers, including their sign bits. A carry out of the sign-bit position is discarded. Example: 33

35. 34 The main problem with signed-magnitude system is that it doesn’t support binary arithmetic (which is what the computer would naturally do). That is, if you add 10 and - 10 binary you won’t get 0 as a result. 00001010 (decimal 10) (signed-magnitude) + 10001010 (decimal -10) -------------------------------- 10010100 (decimal -20) (wrong answer)

36. 35 00001100 (decimal 12) + 11110011 (decimal -12) (1’s complement) -------------------------------- 11111111 (decimal -0) 00000011 (decimal 3) + 11111101 (decimal -2) (1’s complement) ------------------------------- 100000000 (decimal 256) If we have only 8 bits to represent the numbers, the leftmost 1 will be discarded, and the result would be 00000000 (decimal +0).

37. 36 S IGNED B INARY N UMBERS Arithmetic Subtraction In 2’s-complement form: Example: 1.Take the 2’s complement of the subtrahend (including the sign bit) and add it to the minuend (including sign bit). 2.A carry out of sign-bit position is discarded. (  6)  (  13)(11111010  11110011) (11111010 + 00001101) 00000111 (+ 7) 36

38. 37 The problems of a part of chapter one tha will be solved this week are : 1.2, 1.3, 1.4, 1.5, 1.7, 1.8, 1.9, 1.12, 1.13, 1.14(a,c), 1.15(a,c), 1.16, 1.17(a,b), 1.18(a,c)

39. 38 A decimal number in BCD is the same as its equivalent binary number only when the number is between 0 and 9. The binary combinations 1010 through 1111 are not used and have no meaning in BCD. BCD numbers are decimal numbers and not binary numbers, although they use bits in their representation. The only difference between a decimal number and BCD is that decimals are written with the symbols 0, 1, 2,..., 9 and BCD numbers use the binary code 0000, 0001, 0010..... 1001. When the binary sum is equal to or less than 1001 (without a carry), the corresponding BCD digit is correct. However, when the binary sum is greater than or equal to 1010, the result is an invalid BCD digit. B INARY C ODES

40. 39 BCD Code (Decimal computers) A number with k decimal digits will require 4k bits in BCD. Decimal 396 is represented in BCD with 12bits as 0011 1001 0110, with each group of 4 bits representing one decimal digit. BCD is very common in electronic systems where a numeric value is to be displayed, especially in systems consisting only of digital logic, and not containing a microprocessor. 39

41. 40 B INARY C ODE Example: Consider decimal 185 and its corresponding value in BCD (it is decimal number) and binary: BCD addition 40

42. 41 B INARY C ODE Example: Consider the addition of 184 + 576 = 760 in BCD: Decimal Arithmetic: (+375) + (-240) = +135 41

43. 42 B INARY C ODES Other Decimal Codes 42

44. 43 BCD is named also 8421 code. BCD and 2421 code are examples of weighted codes. Some digits can be coded in two possible ways in the 2421 code. (4 can be assigned to 0100 or 1010). The excess-3 code is an unweighted code. The 2421 and the excess-3 codes are examples of self-complementing codes. 9’s complement of a decimal number is obtained directly by complementing each bit. 604 ----- ( 1001 0011 0111 ) 999 – 604 = 395 ----- ( 0110 1100 1000 ) 43

45. 44 The 8, 4, -2, -1 code is an example of assigning both positive and negative weights to a decimal code. When you add two XS-3 numbers together, the result is not an XS-3 number. For instance, when you add 1 (0100) and 0 (0011) in XS-3 the answer seems to be 4 (0111) instead of 1 (0100). In order to correct this problem, when you are finished adding each digit, you have to subtract 3 (binary 0011) if the digit is less than decimal 10 and add three if the number is greater than or equal to decimal 10. 1 + 9 = 10 0100 + 1100 = 10000 (there is carry, this means answer > 9). The true answer is (0100 0011) You need to add 0011 to each digit.

46. 45 B INARY C ODES Gray Code The advantage is that only bit in the code group changes in going from one number to the next. Error detection. (it prevents data errors which can occur with natural binary during state changes) Representation of analog data. 000001 010 100 110111 101 011 45

47. 46 G RAY C ODE 1010 (natural binary number) ---- 1111  Most left bit is the same.  Add each two bits in the binary number starting from the left.  0 + 0 = 0 1 + 0 =1 1 + 1 = 0 (discard carry) 1111 (gray code) ---- 1010  Most left bit is the same.  Add the generated bit in the natural binary to the next gray code bit (to the right). One of the application is the rotary encoder on the motor shaft to control the angular position by a controller such as microcontroller) (in robot arm). 46

48. 47 ASCII C HARACTER C ODES American Standard Code for Information Interchange (Refer to Table 1.7) A popular code used to represent information sent as character-based data. It uses 7-bits to represent: 94 Graphic printing characters. 34 Non-printing characters (Control Characters). Some non-printing characters are used for text format (e.g. BS = Backspace, CR = carriage return). (Format effectors) Other non-printing characters are Communication CC (e.g. STX and ETX start and end text areas). Information separators are used to separate the data into divisions such as paragraphs and pages. 47

49. 48 ASCII C HARACTER C ODE American Standard Code for Information Interchange (ASCII) Character Code 48

50. 49 ASCII Character Code 49

51. 50 ASCII P ROPERTIES The seven bits of the code are designated by b1 through b7. with b7 the most significant bit. The letter A. for example is represented in ASCII as 1000001 (column 100, row 0001). ASCII has some interesting properties: Digits 0 to 9 span Hexadecimal values 30 16 to 39 16 Upper case A-Z span 41 16 to 5A 16 Lower case a-z span 61 16 to 7A 16 Lower to upper case translation (and vice versa) occurs by flipping bit 6. 50

52. 51  ASCII is a seven-bit code, but most computers manipulate an eight-bit quantity as a single unit called a byte. Therefore, ASCII characters often are stored one per byte.  Extended ASCII (8 bits) adds the Greek alphabets.

53. 52 B INARY C ODES Error-Detecting Code To detect errors in data communication and processing, an eighth bit is sometimes added to the ASCII character to indicate its parity. A parity bit is an extra bit included with a message to make the total number of 1's either even or odd. Example: Consider the following two characters and their even and odd parity: 52

54. 53 B INARY C ODES Error-Detecting Code Parity can detect all single-bit errors. A code word has even parity if the number of 1’s in the code word is even. A code word has odd parity if the number of 1’s in the code word is odd. Example: If the receiver detects a parity error, it sends back ASCII NAK. If no error is detected, the receiver sends back ACK. 10001001 1 0 (odd parity) Message B: Message A: (even parity) 53

55. 54 B INARY S TORAGE AND R EGISTERS Registers A binary cell is a device that possesses two stable states and is capable of storing one of the two states. A register is a group of binary cells. A register with n cells can store any discrete quantity of information that contains n bits. A binary cell Two stable state Store one bit of information Examples: flip-flop circuits A register A group of binary cells AX in x86 CPU Register Transfer A transfer of the information stored in one register to another. One of the major operations in digital system. n cells 2 n possible states (1100001111001001) 50,121 CI (ASCII with even parity) Ex-3: 9096 BCD ??? (meaningless) 54

56. 55 T RANSFER OF INFORMATION Figure 1.1 Transfer of information among register 55

57. 56 T RANSFER OF INFORMATION Circuit elements to manipulate individual bits of information Load-store machine LDR1; LDR2; ADD R3, R2, R1; SDR3; Figure 1.2 Example of binary information processing 56

58. 57 B INARY L OGIC (B OOLEAN ALGEBRA ) Definition of Binary Logic Binary logic consists of binary variables and a set of logical operations. The variables are designated by letters of the alphabet, such as A, B, C, x, y, z, etc, with each variable having two and only two distinct possible values: 1 and 0, Three basic logical operations: AND, OR, and NOT. 57

59. 58 BINARY LOGIC Truth Tables, Boolean Expressions, and Logic Gates xyz 000 010 100 111 xyz 000 011 101 111 xz 01 10 ANDORNOT z = x y = x yz = x + yz = x = x’

60. 59 S WITCHING C IRCUITS ANDOR 59

61. 60 L OGIC GATES Logic gates are electronic circuits that operate on one or more input signals to produce an output signal. 0 1 2 3 Logic 1 Logic 0 Un-define Figure 1.3 Example of binary signals 60

62. 61 L OGIC GATES Graphic Symbols and Input-Output Signals for Logic gates: Fig. 1.4 Symbols for digital logic circuits Fig. 1.5 Input-Output signals for gates 61

63. 62 Logic gates Graphic Symbols and Input-Output Signals for Logic gates with multiple inputs: 62

64. 63 The problems of chapter one that will be solved are: 1.19, 1.20, 1.21, 1.22, 1.23, 1.26, 1.28, 1.30, 1.32, 1.33, 1.35