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Chapter 1 Digital Systems and Numbers System

وزارة التعليم العالي والبحث العلمي جامعة الكوفة - كلية التربية – قسم علوم الحاسوب. Chapter 1 Digital Systems and Numbers System. Digital Logic Design I. Dr. Wissam Hasan Mahdi Alagele. e-mail:wisam.alageeli@uokufa.edu.iq. http :// edu-clg.kufauniv.com/staff/Mr.Wesam.

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Chapter 1 Digital Systems and Numbers System

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  1. وزارة التعليم العالي والبحث العلمي جامعة الكوفة - كلية التربية – قسم علوم الحاسوب Chapter 1 Digital Systems and Numbers System Digital Logic Design I Dr. WissamHasan Mahdi Alagele e-mail:wisam.alageeli@uokufa.edu.iq http://edu-clg.kufauniv.com/staff/Mr.Wesam

  2. Outline of Chapter 1 • 1.1 Digital Systems • 1.2 Number System • 1.2.1 Decimal Numbers • 1.2.2 Binary Numbers • 1.2.3 Octal Numbers • 1.2.4 Hexadecimal Numbers • 1.3 Number-base Conversions

  3. Analog and Digital Signal • 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. • Greater accuracy X(t) X(t) t t Analog signal Digital signal

  4. Binary Digital Signal • An information variable represented by physical quantity. • For digital systems, the variable takes on discrete values. • Two level, or binary values are the most prevalent 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 On and Off • Binary values are represented by values or ranges of values of physical quantities. V(t) Logic 1 undefine Logic 0 t Binary digital signal

  5. Number System • Integers are normally written using positional number system, in which each digit represents the coefficient in a power series Where is the number of digit, is the radix or base and is the coefficient Ex. There are four systems of arithmetic which are often used in digital circuits. These systems are: • decimal: it has a base (=10) and coefficients are in the range 0 to 9 • binary: it has a base (=2) and coefficients are all either 0 or 1 • octal : it has abase (=8) and coefficients are in the range 0 to 7 • Hexadecimal: it has a base (=16) and coefficients are in the range { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F } 97142=9*104+7*103+1*102+4*101+2*100

  6. 2 1 0 -1 -2 100 10 1 0.1 0.01 500 10 2 0.7 0.04 Decimal Number System • 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 5 1 2 7 4 d2*B2+d1*B1+d0*B0+d-1*B-1+d-2*B-2 5*102+1*101+2*100+7*10-1+4*10-2 (512.74)10

  7. 4 2 1 1/2 1/4 2 1 0 -1 -2 Binary Number System • Base = 2 • 2 digits { 0, 1 }, called binary digits or “bits” • Weights • Weight = (Base) Position • Magnitude • Sum of “Bit x Weight” • Formal Notation • Groups of bits 4 bits = Nibble 8 bits = Byte 1 0 1 0 1 1 *22+0 *21+1 *20+0 *2-1+1 *2-2 =(5.25)10 (101.01)2 1 0 1 1 1 1 0 0 0 1 0 1

  8. 64 8 1 1/8 1/64 2 1 0 -1 -2 Octal Number System • Base = 8 • 8 digits { 0, 1, 2, 3, 4, 5, 6, 7 } • Weights • Weight = (Base) Position • Magnitude • Sum of “Digit x Weight” • Formal Notation 5 1 2 7 4 5 *82+1 *81+2 *80+7 *8-1+4 *8-2 =(330.9375)10 (512.74)8

  9. 256 16 1 1/16 1/256 2 1 0 -1 -2 Hexadecimal Number System • 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 1 E 5 7 A 1 *162+14 *161+5 *160+7 *16-1+10 *16-2 =(485.4765625)10 (1E5.7A)16

  10. Number Base Conversions Evaluate Magnitude Octal (Base 8) Evaluate Magnitude Decimal (Base 10) Binary (Base 2) Hexadecimal (Base 16) Evaluate Magnitude

  11. Decimal (Integer) to Binary Conversion • 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 Coefficient Quotient Remainder 13 / 2 =6 1a0 =1 6 / 2 =3 0a1 =0 3 / 2 =1 1a2 =1 1 / 2 =0 1a3 =1 Answer: (13)10= (a3 a2 a1 a0)2= (1101)2 MSB LSB

  12. Decimal (Integer) to Binary Conversion

  13. Decimal (Fraction) to Binary Conversion • Multiply the number by the ‘Base’ (=2) • Take the integer (either 0 or 1) as a coefficient • Take the resultant fraction and repeat the division Example: (0.625)10 Coefficient Integer Fraction a-1 =1 0.625 * 2 =1 . 25 0.25 * 2 =0 . 5a-2 =0 0.5 * 2 =1 . 0a-3 =1 Answer: (0.625)10= (0.a-1 a-2 a-3)2= (0.101)2 MSB LSB

  14. Decimal to Binary Conversion

  15. Table of binary equivalent decimal numbers

  16. Decimal to Octal Conversion Example: (175)10 Coefficient Quotient Remainder 175 / 8 =21 7a0 =7 21 / 8 =2 5a1 =5 2 / 8 =0 2a2 =2 Answer: (175)10= (a2 a1 a0)8= (257)8 Example: (0.3125)10 Coefficient Integer Fraction a-1 =2 0.3125 * 8 =2 . 5 0.5 * 8 =4 . 0a-2 =4 Answer: (0.3125)10= (0.a-1 a-2 a-3)8= (0.24)8

  17. Decimal to Hexadecimal Conversion Example: (1983)10 Coefficient Quotient Remainder 1983 / 16 =123 15a0 =F 123 / 16 =7 11a1 =B 7 / 16 =0 7a2 =7 Answer: (1983)10= (a2 a1 a0)16= (7BF)16 Example: (0.5625)10 Coefficient Integer Fraction a-1 =9 0.5625 *16 =9. 0 Answer: (0.5625)10= (0.a-1 a-2 a-3)16= (0.9)16

  18. Binary − Octal Conversion • 8 = 23 • Each group of 3 bits represents an octal digit Example: Assume Zeros ( 1 0 1 1 0 . 0 1 )2 ( 2 6 . 2 )8 Works both ways (Binary to Octal & Octal to Binary)

  19. Binary − Hexadecimal Conversion • 16 = 24 • Each group of 4 bits represents a hexadecimal digit Example: Assume Zeros ( 1 0 1 1 0 . 0 1 )2 ( 1 6 . 4 )16 Works both ways (Binary to Hex & Hex to Binary)

  20. Octal − Hexadecimal Conversion • Convert to Binary as an intermediate step Example: ( 2 6 . 2 )8 Assume Zeros Assume Zeros ( 01 0 1 1 0 . 0 1 0 )2 ( 1 6 . 4 )16 Works both ways (Octal to Hex & Hex to Octal)

  21. Decimal, Binary, Octal and Hexadecimal

  22. Tutorial Problems • Find the decimal equivalents of the following binary numbers: (a) (101)2(b) (1001)2 (c) (10.011)2 [(a) (5)10 (b) (9)10 (c) (3.0375)10] • What are the decimal equivalents of the following binary numbers ? (a) (1111)2(b) (10100)2(c) (11011001)2(d) (10011001)2 [(a) (15)10(b) (20)10(c) (109)10(d) (153)10] • Express the following binary numbers into their equivalent decimal numbers : (a) (11.01)2(b) (101.11)2(c) (110.01)2 [(a) (3.25)10(b) (5.75)10(c) (6.25)10] • Convert the following decimal numbers into their binary equivalents: (a) (25)10(b) (125)10(c) (0.85)10 [ (a) (11001)2(b) (1111101)2(c) (0.110110)2] • What are the binary equivalents of the following decimal numbers ? (a) (27)10(b) (92)10(c) (64)10 [(a) (11011)2(b) (1011100)2(c) (1000000)2] • Convert the following real numbers to the binary numbers: (i) (12.0)10(ii) (25.0)10 (iii) (0.125)10 [(i) (1100)2 (ii) (11001)2 (iii) (0.001)2]

  23. Tutorial Problems • Convert the following numbers : (a) (35)8 to decimal (b) (6421)8 to decimal (c) (1359)10 to octal (d) (7777)10 to octal [(a) (239)10(b) (3345)10(c) (2517)8(d) (17141)8]

  24. Addition • Decimal Addition 1 1 Carry 5 5 + 5 5 1 1 0 = Ten ≥Base  Subtract a Base

  25. Binary Addition • Column Addition 1 1 1 1 1 1 Carry = 61 = 23 1 1 1 1 0 1 + 1 0 1 1 1 = 84 1 0 1 0 1 0 0 ≥ (2)10 0 + 0=0 0 + 1=1 1 + 0=1 1 + 1=0 with a carry of 1 or =10

  26. Binary Subtraction • Borrow a “Base” when needed 1 2 = (10)2 2 0 2 0 0 2 1 = 77 = 23 0 0 1 1 0 1 − 1 0 1 1 1 0 1 1 0 1 1 0 = 54

  27. Binary Multiplication • Bit by bit 1 0 1 1 1 x 1 0 1 0 0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 1 1 0 0 1 1 0

  28. 1.5 Complements • 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: (rn–1) – N • Example for 6-digit decimal numbers: • 9’s complement is (rn–1)–N = (106–1)–N = 999999–N • 9’s complement of 546700 is 999999–546700 = 453299 • Example for 7-digit binary numbers: • 1’s complement is (rn–1) – N = (27–1)–N = 1111111–N • 1’s complement of 1011000 is 1111111–1011000 = 0100111 • Observation: • Subtraction from (rn–1) will never require a borrow • Diminished radix complement can be computed digit-by-digit • For binary: 1 – 0 = 1 and 1 – 1 = 0

  29. Complements • 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 1 1 1 1

  30. Complements • Radix Complement • Example: Base-10 • Example: Base-2 The r's complement of an n-digit number N in base r is defined as rn – N for N ≠ 0 and as 0 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 rn – N = [(rn 1) – N] + 1. The 10's complement of 012398 is 987602 The 10's complement of 246700 is 753300 The 2's complement of 1101100 is 0010100 The 2's complement of 0110111 is 1001001

  31. Complements • 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.: OR 1 0 1 1 0 0 0 0 1 0 1 1 0 0 0 0 0 1 0 0 1 1 1 1 + 1 OR 0 1 01 0 0 0 0 0101 0 0 0 0

  32. Complements • In digital work, both types of complements of a binary number are used for complemental subtraction : • as follows: • 1’s ComplementalSubtraction • compute the 1’s complement of the subtrahend by changing all its 1s to 0s and all its 0s to 1s. • add this complement to the minuend • perform the end-around carry of the last 1 or 0 • Suppose we want to subtract (101)2from (111)2 . 1 1 1 0 1 0 1 0 0 1 + 1 010 ← 1’s complement of subtrahend (101)2 ← end-around carry As seen, we have removed from the addition sum the 1 carry in the last position and added it onto the remainder. It is called end-around carry.

  33. Complements 4. if there is no end-around carry (i.e. 0 carry), then the answer must be re-complemented and a negative sign attached to it. Example. Subtract (1101)2from (1010)2 . 1 1 0 1 0 + 0 0 1 0 1 1 0 0 - 0 0 1 1 ← 1’s complement of subtrahend (1101)2 ← no end-around carry re-complementing with negative sign

  34. Complements • 2’s Complemental Subtraction In this case, the procedure is as under : 1.find the 2’s complement of the subtrahend, 2.add this complement to the minuend, 3.drop the final carry, 4.if the carry is 1, the answer is positive and needs no re-complementing, 5.if there is no carry, re-complement the answer and attach minus sign. Example.Using 2’s complement, subtract: (A) (1010)2from (1101)2, (B) (1101)2 from (1010)2. Solution. The 1’s complement of 1010 is 0101. The 2’s complement is 0101 + 1 = 0110. We will add it to 1101. 1 1 0 1 + 0 1 1 0 1 0 0 1 1 DROP 0 0 1 1 ← 2’s complement of (1010)2 ← The final answer is (0011)2 .

  35. Complements • Example 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. D Drop final cary There is no end carry. Therefore, the answer is Y – X =  (2's complement of 1101111) =  0010001.

  36. Complements • Example • Repeat Example, but this time using 1's complement. There is no end carry, Therefore, the answer is Y – X =  (1's complement of 1101110) =  0010001.

  37. Digital Coding • In digital logic circuits, each number or piece of information is defined by an equivalent combination of binary digits. A complete group of these combinations which represents numbers, letters or symbols is called a digital code • Codes have been used for security reasons so that others may not be able to read the message even if it is intercepted. • The choice of a code depends on the function or purpose it has to serve. • There are many types of coding • Binary codes • BCD code • Gray code • ASCII code

  38. Binary Codes • BCD Code • It is a binary code in which each decimal digit is represented by a group of four bits. it is also called an 8421 code. The 8 4 2 1 indicates the • binary weights of the four bits (2 3 , 2 2 , 2 1 , 2 0 ). • A number with N decimal digits will require 4N 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. • A decimal number in BCD is the same as its equivalent binary number only when the number is between 0 and 9.

  39. Binary Code • Example: • Consider decimal 185 and its corresponding value in BCD and binary: Other Decimal Codes

  40. 000 001 010 011 101 100 110 111 Binary Codes) • Gray Code • The advantage is that only bit in the code group changes in going from one number to the next. • Error detection. • Representation of analog data. • Low power design.

  41. ASCII Character Codes • A popular code used to represent information sent as character-based data. • The ASCII code is a seven-bit code, and so it has 27 (=128) possible code groups. This is more than enough to represent all of the standard keyboard characters.

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