ECE240-Set 2

Synopsys University Courseware – ECE 240Lecture Set 2 – Numbers

ECE 240 -Introduction to Logic Design

Synopsys University Courseware – ECE 240Lecture Set 2 – 2

Number Systems and Codes

Positional number systemsPositional number systems Positional number system conversionsPositional number system conversions Binary system operationsBinary system operations Encoding techniquesEncoding techniques

Synopsys University Courseware – ECE 240Lecture Set 2 – 3

Positional Number System

General form of a number:

dp-1dp-2…d0.d-1d-2..d-n

The value of the number:

Radix Radix PointPoint

Most significant digit (MSB)Most significant digit (MSB) Least significant digit Least significant digit (LSB)(LSB)

ip

ni

i rdD

1

Radix or BaseRadix or Base

Synopsys University Courseware – ECE 240Lecture Set 2 – 4

Decimal System

d : 0, 1, 2, . . . , 9 r = 10

dd33

dd22 dd11 dd00

dd-1-1

dd-2-2

1x101x1033

4x104x1022

5x105x1011 3x103x1000

0x100x10-1-1

5x105x10--

22

1x101x1033 + 4x10 + 4x1022 + 5x10 + 5x1011 + 3x10 + 3x1000 + 0x10 + 0x10-1-1 + 5x10 + 5x10-2 -2 = =

1453.051453.05

p p = 4 , = 4 , n n = 2= 2 1453.01453.055

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b : 0 , 1 r = 2

bb33

bb22 bb11 bb00

bb-1-1

bb-2-2

1x21x233

1x21x222 0x20x211

1x21x200

0x20x2--

11

1x21x2-2-2

1x21x233 + 1x2 + 1x222 + 0x2 + 0x211 + 1x2 + 1x200 + 0x2 + 0x2-1-1 + 1x2 + 1x2-2-2 = =

13.2513.25

p p = 4 , = 4 , n n = 2= 2 1101.01101.011

Binary System

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Exercise

Calculate the equivalent decimal numbers:

1122

101022

111122

10010022

1011101122

1x1x2200 = 1 = 1

1x1x2211 + + 0x0x220 0 = 2 + 0 = 2= 2 + 0 = 2

1x1x2211 + + 1x1x2200 = 2 + 1 = 3 = 2 + 1 = 3

1x1x2222 + + 0x0x2211 + + 0x0x2200 = 4 + 0 + 0 = 4 = 4 + 0 + 0 = 4

1x1x2233 + + 0x0x222 2 + + 1x1x2211 + + 1x1x2200 = 8 + 0 + 2 + 1 = 11 = 8 + 0 + 2 + 1 = 11

Synopsys University Courseware – ECE 240Lecture Set 2 – 7

Octal and Hexadecimal Numbers

Octal number System:

r = 8 d : 0, 1, 2, … , 7

Hexadecimal number System:

r = 16 d : 0, 1, 2, …, 9, A, B, C, D, E, F

Used for representations of long binary numbers

Synopsys University Courseware – ECE 240Lecture Set 2 – 8

Binary, Decimal, Octal and Hex #sDecimal Binary Octal Hexadecimal0 0000 00 01 0001 01 12 0010 02 23 0011 03 34 0100 04 45 0101 05 56 0110 06 67 0111 07 78 1000 10 89 1001 11 910 1010 12 A11 1011 13 B12 1100 14 C13 1101 15 D14 1110 16 E15 1111 17 F

Synopsys University Courseware – ECE 240Lecture Set 2 – 9

Binary – Octal Conversion

Starting from the decimal point: Separate the bits into groups of three Replace each group with its corresponding Octal digit

100100011011110110 .. 101101010122 ==

100 100 011011 110110 . . 101101 01001022 = =

436 . 52436 . 5288

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110001000111101110 .. 101010101122 = =

00010001 00010001 11101110 .. 10101010 1000100022 = =

11E . A811E . A81616

Starting from the decimal point: Separate the bits into groups of four Replace each group with its corresponding

Hexadecimal digit

Binary - Hexadecimal Conversion

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Octal - Binary Conversion

Convert each Octal digit into its corresponding three bit string

436 . 52436 . 528 8 ==

100100 011011 110110 . 101. 101 01001022

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Convert each Hexadecimal digit into its corresponding four bit string

A5E . C8A5E . C816 16 ==

10101010 01010101 11101110 . 1100. 1100 1000100022

Hexadecimal - Binary Conversion

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Radix-r to Decimal Conversion

ip

ni

i rdD

1

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Decimal to Radix-r Conversion

Successive division of D by r

The remainder of the long division will give the digits starting from the least significant digit

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Decimal to Binary Conversion

4510

45/2 = 22 The remainder is 1 LSB22/2 = 11 0

11/2 = 5 15/2 = 2 1

2/2 = 1 01/2 = 0 1 MSB

MSBMSBLSBLSB

3232 1616 88 44 22 11

11 00 11 11 00 11

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Exercises

Convert 5 into binary:Convert 5 into binary: 1 0 11 0 1

Convert 17 into binary:Convert 17 into binary: 1 0 0 0 11 0 0 0 1

Convert 34 into binary:Convert 34 into binary: 1 0 0 0 1 01 0 0 0 1 0

2222

hexadecimal:hexadecimal: 55

hexadecimal:hexadecimal: 1111

hexadecimal:hexadecimal:

Synopsys University Courseware – ECE 240Lecture Set 2 – 17

Binary Addition

Binary addition table:

carry in X Y X+Y carry out 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1

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Example:

1 0 1 0 1

0 1 1 0 0

1 0 0 0 0 1

2121

1212

3333

16 8 4 2 116 8 4 2 1

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Binary Subtraction

Binary subtraction table:

barrow in X Y X-Y barrow out

0 0 0 0 0 0 0 1 1 1 0 1 0 1 0 0 1 1 0 0 1 0 0 1 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1 1

Synopsys University Courseware – ECE 240Lecture Set 2 – 20

Example:

1 1 0 1 0

0 1 1 1 1

0 1 0 1 1

2626

1515

1111

16 8 4 2 116 8 4 2 1

Synopsys University Courseware – ECE 240Lecture Set 2 – 21

Representation of Negative Numbers in Binary Systems

Signed-magnitude Representation.

Two’s-Complement Representation.

One’s-Complement Representation.

Synopsys University Courseware – ECE 240Lecture Set 2 – 22

Signed Magnitude Representation

The MSB represents the sign bit (0 = positive, 1 = negative)

The range for n-bit is from - 2n-1 – 1 to +2n-1 – 1.

Example: n=5, Range from -15 to 15 00000 = 0 , 10000 = -0 10011 = -3 , 01100 = +12

Disadvantages: Complicated digital adders Two possible representations of zero

Synopsys University Courseware – ECE 240Lecture Set 2 – 23

Two’s Complement Representation The MSB represents the sign bit (0 = positive, 1 =

negative)

To calculate the negative number:

1. Complement all bits of the positive number

2. Add 1

For n-bit number the decimal value =

B b bi

i

ni

nn

0

2

112 2

The range for n-bit is: The range for n-bit is: from -2from -2n-1 n-1 to +2to +2n-1 n-1 – 1.– 1.

AdvantagesAdvantages: Addition/subtraction performed directly and only : Addition/subtraction performed directly and only one zeroone zero

DisadvantageDisadvantage: One extra negative number (not symmetric): One extra negative number (not symmetric)

Synopsys University Courseware – ECE 240Lecture Set 2 – 24

Two’s Complement Example

N = 8: from -128(10000000) to 127 (01111111)N = 8: from -128(10000000) to 127 (01111111)

001010 = = 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0

1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 step step 11

+ 1+ 1 step step 22

1 0 0 0 0 0 0 0 0 = 01 0 0 0 0 0 0 0 0 = 01010

0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 = 0 = 01010

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1 0 0 1 1 0 1 11 0 0 1 1 0 1 1

+ 1+ 1

+100+1001010 = = 0 1 1 0 0 1 0 00 1 1 0 0 1 0 0

1 0 0 1 1 1 0 0 = -1001 0 0 1 1 1 0 0 = -1001010

64 32 16 8 4 2 164 32 16 8 4 2 1

step 1step 1step 2step 2

Two’s Complement Example

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One’s Complement Representation The MSB represents the sign bit (0 = positive, 1 =

negative) To calculate the negative number, complement all

bits of the positive number For n-bit number the decimal value =

B b bi

i

ni

nn

0

2

112 2 1( )

The range for n-bit is: from 2The range for n-bit is: from 2n-1 n-1 – 1 to +2– 1 to +2n-1 n-1 – 1. – 1. AdvantagesAdvantages: Symmetry, ease of : Symmetry, ease of complementation.complementation.

DisadvantagesDisadvantages: : Two possible representations of zero. Two possible representations of zero. Complicated digital adders. Complicated digital adders.

Synopsys University Courseware – ECE 240Lecture Set 2 – 27

One’s-Complement Example

N = 8: form -127(10000000) to 127 (01111111)

+10010 = 0 1 1 0 0 1 0 0

-10010 = 1 0 0 1 1 0 1 1

010 = 0 0 0 0 0 0 0 0

-010 = 1 1 1 1 1 1 1 1

Synopsys University Courseware – ECE 240Lecture Set 2 – 28

Comparison (4-bit) Decimal Signed One’s Two’s

Magnitude Compl.Compl.-8 - - 1000-7 1111 1000 1001-6 1110 1001 1010-5 1101 1010 1011-4 1100 1011 1100 -3 1011 1100 1101 -2 1010 1101 1110 -1 1001 1110 11110 0000 or 1000 0000 or 1111 0000 1 0001 0001 0001

2 0010 0010 0010 3 0011 0011 0011 4 0100 0100 0100 5 0101 0101 0101 6 0110 0110 0110 7 0111 0111 0111

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Exercise

What is the representation of What is the representation of +11+11, , -11-11 in: in:

5-bit 5-bit signedsigned magnitude representation magnitude representation

0 1 0 1 10 1 0 1 1+11 = +11 =

5-bit 5-bit one’sone’s complement representation complement representation

0 1 0 1 10 1 0 1 1+11 = +11 =

1 1 0 1 11 1 0 1 1-11 = -11 =

1 0 1 0 01 0 1 0 0-11 = -11 =

5-bit 5-bit two’stwo’s complement representation complement representation

0 1 0 1 10 1 0 1 1+11 = +11 = 1 0 1 0 11 0 1 0 1-11 = -11 =

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Two’s Complement Addition (A+B)

1. Use binary addition rules

2. Ignore any carry beyond the sign bit If the range is not exceeded, addition result

will be correct including the sign bit.

Examples:

(-6) 1 1 0 1 0(-6) 1 1 0 1 0 (0) 1 0 0 0 0 (0) 1 0 0 0 0

(-2) 1 1 1 0(-2) 1 1 1 0

+ (-4) 1 1 0 0+ (-4) 1 1 0 0

(-3) 1 1 0 1(-3) 1 1 0 1

+ (+3) 0 0 1 1+ (+3) 0 0 1 1

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Two’s Complement Subtraction: Method 1 (a-b)

1. Use binary Subtraction rules

2. Ignore any borrow beyond the sign bit

Example:

(-2) 1 1 1 0(-2) 1 1 1 0

(2) 0 0 1 0 (2) 0 0 1 0

- (4) 0 1 0 0- (4) 0 1 0 0

Synopsys University Courseware – ECE 240Lecture Set 2 – 32

Two’s Complement Subtraction Method 2 (a+(-b)) Add A to the Two’s complement of B:

1. Take the One’s complement of B

2. Add it to A with initial carry-in i.e. 1

Example: 2-4

(-2) 1 1 1 0(-2) 1 1 1 0

11

0 0 1 0 0 0 1 0

+ 1 0 1 1+ 1 0 1 1

One’s Complement of One’s Complement of 44

22

Initial Carry Initial Carry inin

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Coding

Coding: Representing a set of objects by a set of strings.

Code: The set of bit strings. Code Word: A particular bit string in the Code.

Examples: Data Objects: Decimal Numbers, Characters. Nondata Objects: Machine states, Control Actions.

Synopsys University Courseware – ECE 240Lecture Set 2 – 34

Binary Codes For Decimal Numbers

To represent the 10 decimal digits, we need 4 bits.

Examples: 4 bits Codes

1. BCD (8421) - Binary Coded Decimal

2. 2421

3. Excess-3

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Decimal CodesDecimal BCD BCD Excess-3

(8421) 2421

0 0000 0000 0011

1 0001 0001 0100

2 0010 0010 0101

3 0011 0011 0110

4 0100 0100 0111

5 0101 1011 1000

6 0110 1100 1001

7 0111 1101 1010

8 1000 1110 1011

9 1001 1111 1100

Synopsys University Courseware – ECE 240Lecture Set 2 – 36

Exercise

Codes for 517

BCD: BCD:

2421:2421:

Excess-3:Excess-3:

0101 0001 01110101 0001 0111

1011 0001 11011011 0001 1101

1000 0100 10101000 0100 1010

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Gray Code

Decimal Binary Code Gray Code0 000 0001 001 0012 010 0113 011 0104 100 1105 101 1116 110 1017 111 100

ApplicationApplication: Electromechanical applications.: Electromechanical applications.

The Mechanical Encoding disk. Altimeters. - The Mechanical Encoding disk. Altimeters. - is an instrument used to measure the altitude of an object above a fixed level

One bit changes between two successive code One bit changes between two successive code wordswords

Binary Code and Gray Code (n = 3):Binary Code and Gray Code (n = 3):

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Binary Code to Gray Code Conversion Convert n bit Binary code into n bit Gray Code:

Bit i of the Gray Code is obtained by comparing

bits i and i+1 of the Binary Code: Add extra 0 to the left of the Binary code

If bits i and i+1 in the Binary code are the same then bit i

of the Gray code is 0, else bit i is 1.

Example:

1001110011 010011010011BinaryBinary GrayGray11 11 0011 00

Synopsys University Courseware – ECE 240Lecture Set 2 – 39

Exercise

1110111101 011101011101BinaryBinary 1 0 0 1 11 0 0 1 1 GrayGray

i and i+1:i and i+1: same same 0 0;; else, else, 11

1 0 0 1 11 0 0 1 1 0 1 1 1 0 10 1 1 1 0 1GrayGray BinaryBinary

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Codes for Nondata Objects

n objects encoded with b=log2n=log10n/log102 binary bits.

Example: 124 objects, b=7

Application: The control unit runs one device at a time.

ControControl Unitl Unit

Enable?Enable?

DEVICEDEVICE

Enable?Enable?

DEVICEDEVICE

Enable?Enable?

DEVICEDEVICE

Binary-coded device selectBinary-coded device select

Synopsys University Courseware – ECE 240Lecture Set 2 – 41

Coding for Error Control

Bit error occurs due to noise, etc., if

0 sent is received as 1 or

1 sent is received as 0

EDC (error detecting codes) to detect bit errors

ECC (error correcting codes) to correct bit errors

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Detects any one bit error in a 2-bit block, but will miss two

11001111001100111100

10000011011000001101

To send:To send: 1011010110

send:send:

If we receiveIf we receive

we detect two errors but miss onewe detect two errors but miss one

EDC: Repetition-2 Code

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ECC: Repetition-3 Code

111000111111000111000111111000

110000100001001110000100001001

To send:To send: 1011010110

send:send:

If we receiveIf we receive

we correct two and miss two.we correct two and miss two.

Detects any one bit error in a 3-bit block, but will miss any two or three

Synopsys University Courseware – ECE 240Lecture Set 2 – 44

Parity-bit EDCs

To sendTo send mm bits,bits, bb11, b, b22, …, b, …, bmm..

SendSend m+1m+1 bits,bits, bb11, b, b22, …, b, …, bmm,,

pp,, wherewhere pp = = bb1 1 b b2 2 … … b bmm

(b(b11 b b2 2 … … b bm m pp = = 00))

SenderSender ReceiverReceiver

bb11, b, b22, …, b, …, bmm,, pp

When When rr11, r, r22, …, r, …, rmm, r, rm+1m+1 is received is received

computecompute s = s = rr11 r r2 2 … … r rm m r rm+1m+1

if if s=1s=1, then , then 11 (or 3 or 5 or …) (or 3 or 5 or …) bit errors occurred.bit errors occurred.

ifif s=0s=0,, then then 00 (or 2 or 4 or …) (or 2 or 4 or …) bit errors occurred, no error.bit errors occurred, no error.

P(2 or 4 or … errors)=P(2 or 4 or … errors)=P (undetected P (undetected

errors)errors)small enough for independent bit small enough for independent bit errorserrors

Synopsys University Courseware – ECE 240Lecture Set 2 – 45

Error-Detection

Error detection: use a dictionary of code words that is a subset of the 2n possible code words

A failure occurs when an invalid code word is sent

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Summary

Binary forms used in digital systems Positive/Negative numbers, addition/subtraction Arithmetic operations performed directly on negative numbers

represented in Two’s complement.

Hexadecimal numbers are used for shorthand representation of binary numbers.

Decimal numbers, characters, and actions encoded into binary strings.

Error-detecting and error-correcting codes use extra bits.