data types and representations - iuma - ulpgcnunez/clases-fdc/irvine-gajski... · 2006-10-11 ·...

Principles OfPrinciples Of

Digital DesignDigital DesignChapter 2Chapter 2

Data Types and Representations

Decimal, Binary, Octal, and Hexadecimal Number System Complement Number SystemFixed Point and Floating Point NumbersCharacter Codes, Cube Representation

2Copyright © 2004-2005 by Daniel D. Gajski Slides by Philip Pham, University of California, Irvine

Each number is represented by a string of digits, in which the position of each digit has an associated weight

1234.56 = 1 1234.56 = 1 ·· 1000 + 2 1000 + 2 ·· 100 + 3 100 + 3 ·· 10 + 4 10 + 4 ·· 1 + 5 1 + 5 ·· 0.1 + 6 0.1 + 6 ·· 0.010.01

In general, any decimal number D of the form

has the value

Positional Number SystemPositional Number System

ddm m –– 11 ddm m –– 22 ……dd11 dd00 .d.d––11 dd––22 ……dd––nn

Least significant digitRadix pointMost significant digit

DD = = ddm m –– 11 ·· 1010m m –– 11 ++ ……+ + dd00 ·· 101000 + + dd––11 ·· 1010––11 + + ……+ + dd––nn ·· 1010––nn == ∑−

−=

⋅1m

ni

ii rd∑

−

−=

⋅1m

ni

ii rd

Radix


General form of a binary number:

Its value is equivalent to

Examples

Binary Number SystemBinary Number System

bbmm –– 11 bbmm –– 22 ……bb11 bb00 bb––11 bb––22 ……bb––nn

Least significant bit (LSB)Binary pointMost significant bit (MSB)

B =B = ∑−

−=

⋅1

2m

ni

iib∑

−

−=

⋅1

2m

ni

iib

Radix 2

101011010122 = 1 ·· 16 + 0 16 + 0 ·· 8 + 1 8 + 1 ·· 4 + 0 4 + 0 ·· 2 + 1 2 + 1 ·· 1 = 211 = 211010

11010111010122 = 1 = 1 ·· 32 + 1 32 + 1 ·· 16 + 0 16 + 0 ·· 8 + 1 8 + 1 ·· 4 + 0 4 + 0 ·· 2 + 1 2 + 1 ·· 1 = 531 = 531010

10.10110.10122 = 1 = 1 ·· 2 + 0 2 + 0 ·· 1 + 1 1 + 1 ·· 0.5 + 0 0.5 + 0 ·· 0.25 + 1 0.25 + 1 ·· 0.125 = 2.6250.125 = 2.6251010

0.11110.111122 = 1 = 1 ·· 0.5 + 1 0.5 + 1 ·· 0.25 + 1 0.25 + 1 ·· 0.125 + 1 0.125 + 1 ·· 0.0625 = 0.93750.0625 = 0.93751010


Octal and Hexadecimal NumbersOctal and Hexadecimal Numbers

16Hexadecimal

10Decimal

8Octal

2Binary

RadixNumbers

F15171111E14161110

777111666110555101444100333112221011110000

B11131011A101210109911100188101000

1000110000

11011100

Binary

2120

1514

Octal

11171016

D13C12

HexadecimalDecimal

Different representation of numbers 0 through 21Back to Conversion of Numbers (Part II)Back to Conversion of Numbers (Part II)


Conversion of NumbersConversion of Numbers(Binary, Octal, Hexadecimal)(Binary, Octal, Hexadecimal)

Binary to Octal ConversionStart at the binary point and work left. Separate the bits intoStart at the binary point and work left. Separate the bits into groups groups of three, replace each group with corresponding octal digit.of three, replace each group with corresponding octal digit.

Binary to Hexadecimal ConversionStart at the binary point and work left. Separate the bits intoStart at the binary point and work left. Separate the bits into groups groups of four, replace each group with corresponding hexadecimal digitof four, replace each group with corresponding hexadecimal digit..

1010011100101001110022 = 001 010 011 100 = 1234= 001 010 011 100 = 123488

1010011100101001110022 = 0010 1001 1100 = 29C= 0010 1001 1100 = 29C1616


Conversion of FractionsStart at the binary point. Group the binary digits on the rightStart at the binary point. Group the binary digits on the right into into groups of three (Octal) or four (Hexadecimal).groups of three (Octal) or four (Hexadecimal).

Conversion to Binary NumbersReplace each octal or hexadecimal digit with corresponding 3Replace each octal or hexadecimal digit with corresponding 3--bit or bit or 44--bit binary string from the bit binary string from the Octal and Hexadecimal TableOctal and Hexadecimal Table..

Conversion of NumbersConversion of Numbers(Binary, Octal, Hexadecimal)(Binary, Octal, Hexadecimal)

0.101110.1011122 = 0.101 110 = 0.101 110 = 0.56= 0.5688= 0.1011 1000 = 0.B8= 0.1011 1000 = 0.B81616

76576588 == 11111010111111010122765.432765.43288 == 111110101.100011010111110101.10001101022

FEDFED1616 = = 11111110110111111110110122FED.CBAFED.CBA1616 = = 111111101101.110010111010111111101101.11001011101022


Conversion to Decimal NumbersConversion to Decimal Numbers

DD = = = ((= ((……((((ddm m –– 11))r +r + ddm m –– 22))rr ++ ……))rr + + dd11))rr + + dd00∑−

=

⋅1

0

m

i

ii rd∑

−

=

⋅1

0

m

i

ii rd S = S ·· r + di

i = m – 1

S = 0

Start

i = 0

i = i – 1

Done

no

yes


Dividing the top equation by r, we obtain the quotient Q and remainder R

Conversion from Decimal NumbersConversion from Decimal Numbers

DD = = = ((= ((……((((ddm m –– 11))r +r + ddm m –– 22))rr ++ ……))rr + + dd11))rr + + dd00∑−

=

⋅1

0

m

i

ii rd∑

−

=

⋅1

0

m

i

ii rd

QQ = (= (……((((ddm m –– 11))r +r + ddm m –– 22))rr ++ ……))rr + + dd11

RR = = dd00

Divide S by rS = quotientdi = remainder

i = 0

S = D

Start

i = m – 1

i = i + 1

Done

no

yes


Conversion of Decimal NumbersConversion of Decimal NumbersExamplesExamples

Problem: Convert (a) 179 to binary, (b) 467 to octal, and (c) 3417 to hexadecimal.

Solution:

Therefore,Therefore,101100111011001122==1791791010

remainder 1 (MSD)remainder 1 (MSD)00==22÷÷11

5511224499

1122

11224488

==

====

========

remainder 1remainder 1÷÷ 221111remainder 1remainder 122÷÷55

remainder 0remainder 022÷÷2222remainder 0remainder 022÷÷4444

22

9999

8877

÷÷

÷÷÷÷

remainder 0remainder 022

remainder 1remainder 122remainder 1 (LSD)remainder 1 (LSD)2211

Therefore,Therefore,72372388==4674671010

88

007755

======

remainder 7 (MSD)remainder 7 (MSD)88÷÷778877

5566

÷÷÷÷


Therefore,Therefore,D59D591616==341734171010

88

007755

======

remainder 13 (MSD)remainder 13 (MSD)1616÷÷33111111

2244

÷÷÷÷

3377


(a)(a)(b)(b)

(c)(c)


Addition of Binary NumbersAddition of Binary Numbers

ci+1, si = xi + yi + ci

c0 = 0

i = 0

Start

i = m

i = i + 1

Done

no

yes

11110000

xi

11001100

yi

10101010

ci sici+1

0010

110101100110

s7

0

1

1

128

s8

0

1

1

256

s9

0

1

1

512

s10

1

1

s4

1

1

1

1

16

s5

0

1

1

0

32

s6

1

1

1

1

64

x + y (1110)

Carries

y (123)

x (987)

s0s1s2s3

0110

110

1101

1101

8 4 12

0111x + y

111Carries

321y

789x

Addition of Binary DigitsDecimal Addition

Binary Addition


11110000xi

11001100yi

10101010

ci dibi+1

0011

110000100111

Subtraction of Binary NumbersSubtraction of Binary Numbers

bi+1, di = xi – yi – bi

b0 = 0

i = 0

Start

i = m

i = i + 1

Done

no

yes

d7

0

1

1

128

d8

1

0

1

256

d9

1

0

1

512

d4

0

0

1

1

16

d5

1

0

1

0

32

d6

1

1

1

1

64

x – y (864)

Borrows

y (123)

x (987)

d0d1d2d3

0000

000

1101

1101

8 4 12

468x – y

000Barrows

321y

789x

Subtraction of Binary DigitsDecimal Subtraction

Binary Subtraction

Back to TwoBack to Two’’ss--complement Subtractioncomplement Subtraction


Sign Magnitude RepresentationSign Magnitude RepresentationStart

additionStart

subtraction

s2 = s′2

s1 = s2

mr = m1 + m2sr = s1

m1 > m2

m1 = m2

mr = m1 – m2sr = s1

mr = 0sr = 0

mr = m2 – m1sr = s2

Done

yesno

yesno

yesno

A sign magnitude number <s, m>, consists of two parts:

Sign and MagnitudeSign and MagnitudeThe sign is either + or –The magnitude is an integer between 0 and the largest representable value

Examples:011110110111101122 = +123= +1231010111110111111101122 = = ––1231231010

D1 = < s1, m1 > and D2 = < s2, m2 > yields the result Dr = < sr, mr >


Radix-complement of a number DD == is equal to:

If digit complement dd′′ = = ((r r –– 1) 1) –– dd then

Therefore,is a negative number of D D , since

Complement Number SystemComplement Number System

∑−

=

⋅1

0

m

i

ii rd∑

−

=

⋅1

0

m

i

ii rd

1))1(( +−−=−= DrDrD mm 1))1(( +−−=−= DrDrD mm

∑−

=

1

0'

m

iid∑

−

=

1

0'

m

iid

((rrmm –– 1)1) –– DD = ( (= ( (rr –– 1) (1) (rr –– 1) 1) …… ((rr –– 1) 1) –– ((ddm m –– 11 ddm m –– 2 2 …… dd0 0 ) )) )= ( (= ( (rr –– 1) 1) –– ddm m –– 11)) ( (( (rr –– 1) 1) –– ddm m –– 2 2 ) ) …… ( (( (rr –– 1) 1) –– dd0 0 ))

= = dd′′mm –– 1 1 dd′′mm –– 2 2 …… dd′′00 = = = D= D′′

1'+= DD 1'+= DD

DD 0=+ DD 0=+ DD


Complement Number SystemComplement Number System

60--971--8820-7931-6A42-5B53-4

FEDCBA

3210

Digit

2---3---4---5---

C64-D75-E860F971

0---1---

Binary Octal Hexa-decimalDecimal

011101117

011001106

010101015

010001004

001100113

001000102

000100011

1000 or 000000000

10011111-1

10101110-2

10111101-3

11001100-4

11011011-5

11101010-6

11111001-7

-1000-8

Sign-Magnitude

Two’s ComplementDecimal

Digit Complements Two’s Complement and Sign-Magnitude Representations


TwoTwo’’ss--complement Additioncomplement Addition

Adding two positive numbers generates a correct result

Adding two negative numbers generates a correct result if carry is ignored

000000

(+6)(+6)111100(+4)(+4)001100++(+2)(+2)110000

Ignored carry = Ignored carry = 11 000000

((––6)6)110011((––4)4)001111++((––2)2)111111


TwoTwo’’ss--complement Additioncomplement AdditionAdding a positive and a negative number generates a correct result

Adding large numbers may generate an incorrect result because of an overflow

000000

((––2)2)111111((––4)4)001111++(+2)(+2)110000

111100

((––7)7)000011(+5)(+5)001100++(+4)(+4)001100

Ignored carry = Ignored carry = 11 111100

(+7)(+7)111100((––5)5)110011++((––4)4)001111

Overflow rule: an overflow occurs when the sign of the sum is different than the signs of both operands


Subtraction using two’s complementDirect subtraction using subtraction

standard procedure

TwoTwo’’ss--complement Subtractioncomplement Subtraction

Startaddition

Startsubtraction

B2 = B2‘ + 1

Br = B1 + B2

Done

–

1

0

0

1

011(–2) ignore barrow =1

011Barrows

010(+4)

000(+2)

+

1

0

0

1

011(–2)

00Carries

011Two’s complement of (+4)

000(+2)

+

0

0

0

0

010(+2) ignore carry =1

00Carries

001Two’s complement of (–8)

011(–4)


Decimal Multiplication

Binary Multiplication

Binary MultiplicationBinary Multiplication

××

productproduct22881111 ×× multiplicandmultiplicand441133 ×× multiplicandmultiplicand2244multipliermultiplier3311multiplicandmultiplicand4411

0011111100000000

11 111111

000011

1111

11

00

1111

11

0011

××

product product (182)(182)001100111111

multiplier multiplier (13)(13)11multiplicand multiplicand (14)(14)00


ShiftShift--andand--add Multiplicationadd MultiplicationStart

PP = 0i = 0

bi = 1

Product = PP

Done

yesno

PP = PP + (2i × MD)

i = i + 1

i = m

MR = multiplier bm–1 …b1 b0 (m-bit)MD = multiplicand (n-bit)PP = partial product

Example of shift-and-add multiplication

shifted multiplicandshifted multiplicand00111111fourth partial productfourth partial product00111100000011

shifted zerosshifted zeros00000000

first partial productfirst partial product00000000shifted multiplicandshifted multiplicand00111111second partial productsecond partial product00111111

11 1111

00

0000

11

1111

11

11

1111

11

11

0011

××

product product (182)(182)001100shifted multiplicandshifted multiplicand1111

third partial productthird partial product00

multiplier multiplier (13)(13)11multiplicand multiplicand (14)(14)00

no

yes


TwoTwo’’ss--complement Multiplicationcomplement MultiplicationUse multiplication procedure for unsigned numbers

Negate multiplicand if multiplier sign is negative

ignore carryignore carry

extended partial productextended partial product00111100110011111111extended, shifted, and negated extended, shifted, and negated muliplicandmuliplicand

001111110000

11

00110011

00

001100111111

all zerosall zeros00000000extended partial productextended partial product00111100110011

extended shifted multiplicandextended shifted multiplicand0011000011

extended partial productextended partial product000000000000extended multiplicandextended multiplicand001100001111extended partial productextended partial product001100001111

00 11

00

11

1111

00

00

00

0000

11

11

0000

11

11

1111

××

product product (182)(182)001100

all zerosall zeros0000

extended partial productextended partial product0000

multiplier multiplier ((––13)13)11multiplicand multiplicand ((––14)14)00

Note:•Carry out of MSB in the third PP is ignored before sign extension

•Carry out of MSB is ignored in the final product

•Sign truncation is needed in the final product


Decimal Division

Binary Division

Binary DivisionBinary Division

shifted shifted ((divisor x divisor x 3)3)2244

)1414

remainderremainder44

reduced dividendreduced dividend6644shifted shifted ((divisor x divisor x 11))4411dividenddividend668811quotientquotient3311

11110011

110000001111001111

11

1111001111

11001111

reduced dividendreduced dividend001100shifted divisorshifted divisor000000

shifted divisorshifted divisor00reduced dividendreduced dividend001100

shifted divisorshifted divisor001111

)11101110

remainder remainder (14)(14)000011

reduced dividendreduced dividend001100

shifted divisorshifted divisordividend dividend (186)(186)001100quotient quotient (13)(13)110011


Floating-point numbers have the formmantissamantissa ×× ((radixradix) ) exponentexponent

Since radix is implicit, only mantissa and exponent must be represented explicitlyFloating-point numbers are fixed-point numbers given by the mantissa, whose radix point is specified by the exponentExponent is represented in the excess-code format called characteristic, obtained by adding a bias to the exponent:

bias =bias = radixradixss –– 11where ss is equal to the number of bits in the exponent field

FloatingFloating--point Numberspoint Numbers

2121

Mantissamagnitude

Signed exponent

Mantissasign

NormalizedFraction

Excess-127 characteristicSign

NormalizedFraction

Excess-1032 characteristicSign

32-bit standard

0 1 9 31

0 1 12 63

Implied binary point

Implied binary point64-bit standard

General format


FixedFixed--point vs. Floatingpoint vs. Floating--pointpoint

ExponentMantissaInteger

11 numbersbetween 1000 and 2000

1001 numbersbetween 1000 and 2000Example

~ 1×~ 100×Precision

~ 10101~ 104Range

0 – 99 × 100 – 9999Representable numbers

4-digit floating-point number4-digit fixed number

The range is the interval of numbers from the largest to the smallest representable numberThe precision is the amount of numbers in a number interval


Binary Codes for Decimal NumbersBinary Codes for Decimal Numbers

Binary-coded Decimals (BCD)

2421 Code

Excess-3 Code

Biquinary Code

9

8

7

6

5

4

3

2

1

0

Decimal digit

1000100101011010111

1000010100111000110

1000001100010110101

0110000011101000100

0101000011000110011

0100100010100100010

0100010010000010001

0100001001100000000

1010000110011111001

1001000101111101000

BCD (8421) 2421 BiquinaryExcess-3


Character CodesCharacter Codesb6b5b4

O

N

M

L

K

J

I

H

G

F

E

D

C

B

A

@

100

_

^

]

\

[

Z

Y

X

W

V

U

T

S

R

Q

P

101

o

n

m

l

k

j

i

h

g

f

e

d

c

b

a

‘

110

?

>

=

<

;

:

9

8

7

6

5

4

3

2

1

0

011

z*SUBLF1010

{+ESCVT1011

|,FSFF1100

}-GSCR1101

~.RSSO1110

y)EMHT1001

1111

1000

0111

0110

0101

0100

0011

0010

0001

0000

b3b2b1b0

w’ETBBEL

v&SYNACK

u%NAKENQ

t$DC4EOT

s#DC3ETX

r“DC2STX

q!DC1SOH

pSPDLENUL

DEL/USSI

x(CANBS

000 001 111010

American Standard Code for Information Interchange (ASCII)

NUL NullSOH Start of headingSTX Start of textETX End of textEOT End of transmissionENQ EnquiryACK AcknowledgeBEL BellBS BackspaceHT Horizontal tabLF Line feedVT Vertical tabFF Form feedCR Carriage returnSO Shift outSI Shift inSP SpaceDLE Data link escapeDC1 Device control 1DC2 Device control 2DC3 Device control 3DC4 Device control 4NAK Negative acknowledgementSYN SynchronizeETB End transmission blockCAN CancelEM End of mediumSUB SubstituteESC EscapeFS File separatorGS Group separatorRS Record separatorUS Unit separatorDEL Delete or rubout


Any n-bit string can be represented by an n-cube with 2n vertices, each corresponding to a particular string of n bits

Any m-subcube has the samen – m bits and m bits that take all 2n possible combinations of 0s and 1s

Distance between two vertices is equal to 1 + the number of vertices on the shortest path between two vertices

Cube RepresentationCube Representation0 1

10 11

00 01

101 111

001 011

000 010

100 110

1110 1111

0000 0001

1010 0111

1000 1001

1100 1101

0110 0111

0010 0011

0100 0101

(a)

(b)

(c)

(d)

n-cubes for n = 1, 2, 3, and 4

data types and representations - iuma - ulpgcnunez/clases-fdc/irvine-gajski... · 2006-10-11 ·...

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