chapter 1 digital systems and numbers system digital logic design i 1 dr. wissam hasan mahdi alagele...

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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

العلمي والبحث العالي التعليم وزارةالكوفة الحاسوب- – جامعة علوم قسم التربية كلية

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Outline of Chapter 11.1 Digital Systems1.2 Number System1.2.1 Decimal Numbers 1.2.2 Binary Numbers1.2.3 Octal Numbers 1.2.4 Hexadecimal Numbers1.3 Number-base Conversions

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Analog and Digital SignalAnalog 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

t

X(t)

Analog signal

t

X(t)

Digital signal

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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.t

V(t)

Binary digital signal

Logic 1

Logic 0

undefine

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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 a base (=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

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Decimal Number SystemBase (also called radix) = 10

◦ 10 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }Digit Position

◦ Integer & fractionDigit Weight

◦ Weight = (Base) Position

Magnitude◦ Sum of “Digit x Weight”

Formal Notation

1 0 -12 -2

5 1 2 7 4

10 1 0.1100 0.01

500 10 2 0.7 0.04

d2*B2+d1*B

1+d0*B0+d-1*B

-1+d-2*B-2

(512.74)10

5 *102+1 *101+2 *100+7 *10-1+4 *10-2

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Binary Number SystemBase = 2

◦ 2 digits { 0, 1 }, called binary digits or “bits”Weights

◦ Weight = (Base) Position

Magnitude◦ Sum of “Bit x Weight”

Formal NotationGroups of bits 4 bits = Nibble 8 bits = Byte

1 0 -12 -2

2 1 1/24 1/4

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

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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

1 0 -12 -2

8 1 1/864 1/64

5 1 2 7 4

5 *82+1 *81+2 *80+7 *8-1+4 *8-2

=(330.9375)10

(512.74)8

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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 0 -12 -2

16 1 1/16256 1/256

1 E 5 7 A

1 *162+14 *161+5 *160+7 *16-1+10 *16-2

=(485.4765625)10

(1E5.7A)16

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Number Base Conversions

Decimal(Base 10)

Octal(Base 8)

Binary(Base 2)

Hexadecimal(Base 16)

Evaluate Magnitude

Evaluate Magnitude

Evaluate Magnitude

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Decimal (Integer) to Binary Conversion

Divide the number by the ‘Base’ (=2)Take the remainder (either 0 or 1) as a

coefficientTake the quotient and repeat the division

Example: (13)10

Quotient Remainder Coefficient

Answer: (13)10 = (a3 a2 a1 a0)2 = (1101)2

MSB LSB

13/ 2 = 6 1 a0 = 1 6/ 2 = 3 0 a1 = 0 3/ 2 = 1 1 a2 = 1 1/ 2 = 0 1 a3 = 1

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Decimal (Integer) to Binary Conversion

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Decimal (Fraction) to Binary Conversion

Multiply the number by the ‘Base’ (=2)Take the integer (either 0 or 1) as a coefficientTake the resultant fraction and repeat the

divisionExample: (0.625)10

Integer Fraction Coefficient

Answer: (0.625)10 = (0.a-1 a-2 a-3)2 = (0.101)2

MSB LSB

0.625* 2 = 1 . 250.25* 2 = 0 . 5 a-2 = 00.5 * 2 = 1 . 0 a-3 = 1

a-1 = 1

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Decimal to Binary Conversion

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Table of binary equivalent decimal numbers

Decimal

BinaryDecima

lBinary

Decimal

Binary

1 1 11 1011 21 10101

2 10 12 1100 22 10110

3 11 13 1101 23 10111

4 100 14 1110 24 11000

5 101 15 1111 25 11001

6 110 16 10000 26 11010

7 111 17 10001 27 11011

8 1000 18 10010 28 11100

9 1001 19 10011 29 11101

10 1010 20 10100 30 11110

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Decimal to Octal Conversion

Example: (175)10

Quotient Remainder Coefficient

Answer: (175)10 = (a2 a1 a0)8 = (257)8

175/ 8 = 21 7 a0 = 7 21/ 8 = 2 5 a1 = 5 2 / 8 = 0 2 a2 = 2

Example: (0.3125)10

Integer Fraction Coefficient

Answer: (0.3125)10 = (0.a-1 a-2 a-3)8 = (0.24)8

0.3125* 8 = 2 . 50.5 * 8 = 4 . 0 a-2 = 4

a-1 = 2

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Example: (1983)10

Quotient Remainder Coefficient

Answer: (1983)10 = (a2 a1 a0)16 = (7BF)16

1983/ 16 = 123 15 a0 = F 123/ 16 = 7 11 a1 = B 7 / 16 = 0 7 a2 = 7

Example: (0.5625)10

Integer Fraction Coefficient

Answer: (0.5625)10 = (0.a-1 a-2 a-3)16 = (0.9)16

0.5625*16 = 9 . 0 a-1 = 9

Decimal to Hexadecimal Conversion

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Binary − Octal Conversion

8 = 23

Each group of 3 bits represents an octal digit

Octal Binary

0 0 0 0

1 0 0 1

2 0 1 0

3 0 1 1

4 1 0 0

5 1 0 1

6 1 1 0

7 1 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)

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Binary − Hexadecimal Conversion

16 = 24

Each group of 4 bits represents a hexadecimal digit

Hex Binary0 0 0 0 01 0 0 0 12 0 0 1 03 0 0 1 14 0 1 0 05 0 1 0 16 0 1 1 07 0 1 1 18 1 0 0 09 1 0 0 1A 1 0 1 0B 1 0 1 1C 1 1 0 0D 1 1 0 1E 1 1 1 0F 1 1 1 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)

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Octal − Hexadecimal Conversion

Convert to Binary as an intermediate stepExample:

( 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

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Decimal, Binary, Octal and Hexadecimal

Decimal Binary Octal Hex00 0000 00 001 0001 01 102 0010 02 203 0011 03 304 0100 04 405 0101 05 506 0110 06 607 0111 07 708 1000 10 809 1001 11 910 1010 12 A11 1011 13 B12 1100 14 C13 1101 15 D14 1110 16 E15 1111 17 F

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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]

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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]

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AdditionDecimal Addition

5 5

55+

011

= Ten ≥ Base

Subtract a Base

11 Carry

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Binary AdditionColumn Addition

1 0 1111

1111 0+

0000 1 11

≥ (2)10

111111

= 61

= 23= 84

Decimal

Binary

1 1

2 10

3 11

0 + 0=0 0 + 1=1

1 + 0=1 1 + 1=0 with a carry of 1 or =10

Carry

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Binary SubtractionBorrow a “Base” when needed

0 0 1110

1111 0−

0101 1 10

= (10)22

22 2

1

000

1

= 77

= 23= 54

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Binary MultiplicationBit by bit

01 1 1 1

01 1 0

00 0 0 0

01 1 1 1

01 1 1 1

0 0 000

0110111 0

x

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1.5Complements 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

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Complements1’s Complement (Diminished Radix

Complement)◦All ‘0’s become ‘1’s◦All ‘1’s become ‘0’sExample (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

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ComplementsRadix 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 987602The 10's complement of 246700 is 753300

The 2's complement of 1101100 is 0010100 The 2's complement of 0110111 is 1001001

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Complements2’s Complement (Radix Complement)

◦Take 1’s complement then add 1◦Toggle all bits to the left of the first ‘1’

from the rightExample: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

0 1 01 0 0 0 0

OR

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ComplementsIn digital work, both types of complements of a binary number are used for complemental subtraction :

as follows:

• 1’s Complemental Subtraction1. compute the 1’s complement of the subtrahend by changing all its 1s to 0s and all its 0s to 1s.2. add this complement to the minuend3. perform the end-around carry of the last 1 or 0

Suppose we want to subtract (101)2 from (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.

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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)2 from (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

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Complements

• 2’s Complemental SubtractionIn 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)2 from (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.

← 2’s complement of (1010)2

← The final answer is (0011)2 .

1 1 0 1

+ 0 1 1 0

1 0 0 1 1DROP

0 0 1 1

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ComplementsExample

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.

D Drop final cary

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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.

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

Binary Codes

◦ 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.

• BCD CodeIt 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 ).

Binary Code Example:

◦ Consider decimal 185 and its corresponding value in BCD and binary:

Other Decimal Codes

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.

000 001

010

100

110 111

101011

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.