computer system architectureartoa.hanbat.ac.kr/lecture_data/computer_architecture/com_arch03.pdf ·...

Hanbat National University Prof. Lee Jaeheung

Computer SystemArchitecture

Computer SystemArchitecture

2Hanbat National University Prof. Lee Jaeheung

데 이 터 표 현

데이터종류

보완

고정소숫점표현

유동소숫점표현

다른 2진표현

에러검출코드


자 료 표 현

Computer가다루는정보

* 데이터-수치데이터숫자 (정수, 실수)

-수가아닌데이터문자, 기호

* 데이터성분사이의구조-데이터구조

Linear Lists, Trees, Rings, etc

* Program(Instruction)

Data Types


수치 데이터 표현

R = 10 Decimal number system, R = 2 BinaryR = 8 Octal, R = 16 Hexadecimal

Radix point(.) 완전수의자리와기수의자리를나눈다.

데이타수치자료 -수(정수, 실수)기호자료 -기호, 문자

진법Nonpositional number system

-로마식 표기Positional number system ( 진수 )

-각자리에따라서 값을가지고있다.- Decimal(10진), Octal(8진), Hexadecimal(16진), Binary(2진)

Base (or radix) R number- R은각자리의의미를정확하게부여한다.- Example AR = an-1 an-2 ... a1 a0 .a-1…a-m

- V(AR ) = ∑−

−=

1n

mi

iiRa

Data Types


왜 디지털 컴퓨터에서는 진수를 사용할까?

중요점은비용과시간이다.-하드웨어설계에드는비용.

Arithmetic and Logic Unit, CPU, Communications-연산에필요한시간.

산술 –수치연산 - 테이블연산* Non-positional Number System

-테이블연산을통해무한히연산가능하다.-->설계가불가능하다.

(설계시매우비싸다.)

* Positional Number System-테이블연산에한계가있다.--> 물리적으로실현할수있다.

-비용때문에작은테이블일수록비용이적게든다.--> 2진수가 10진으로변하기쉽다.

0 10 0 1

1 1 10

0 1 2 3 4 5 6 7 8 90 0 1 2 3 4 5 6 7 8 91 1 2 3 4 5 6 7 8 9 102 2 3 4 5 6 7 8 9 10113 3 4 5 6 7 8 9 1011124 4 5 6 7 8 9 101112135 5 6 7 8 9 10111213146 6 7 8 9 1011121314157 7 8 9 101112131415168 8 9 10111213141516179 9 101112131415161718

Binary Addition Table

Decimal Addition Table

Data Types


REPRESENTATION OF NUMBERS - POSITIONAL NUMBERS

Decimal Binary Octal Hexadecimal00 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

Binary, octal, and hexadecimal conversion

1 0 1 0 1 1 1 1 0 1 1 0 0 0 1 11 2 7 5 4 3

A F 6 3

OctalBinaryHexa

Data Types


진 수 변 환

10진수를변환하기- Separate the number into its integer and fraction parts and convert

each part separately.- Convert integer part into the base R number→ R로연달아나눈후나머지값을정리한다.

- Convert fraction part into the base R number→정수자리를연달아증가시키면서정리(축척) 한다.

십진수의표시

V(A) = Σ ak Rk

A = an-1 an-2 an-3 … a0 . a-1 … a-m

(736.4)8 = 7 x 82 + 3 x 81 + 6 x 80 + 4 x 8-1

= 7 x 64 + 3 x 8 + 6 x 1 + 4/8 = (478.5)10(110110)2 = ... = (54)10(110.111)2 = ... = (6.785)10(F3)16 = ... = (243)10(0.325)6 = ... = (0.578703703 .................)10

Data Types


EXAMPLE

Convert 41.687510 to base 2.

Integer = 414120 110 05 02 11 00 1

Fraction = 0.68750.6875x 21.3750x 20.7500x 21.5000x 21.0000

(41)10 = (101001)2 (0.6875)10 = (0.1011)2

(41.6875)10 = (101001.1011)2

Convert (63)10 to base 5: (223)5Convert (1863)10 to base 8: (3507)8Convert (0.63671875)10 to hexadecimal: (0.A3)16

Exercise

Data Types


보 수

R 진법에는두가지방법이있다. - R 보수와 (R-1)보수

(R-1)보수각자리수를 (R-1) .

Example - 9보수 83510는 16410이다.- 1의보수 10102는 01012 이다. (bit by bit complement operation)

R의보수(R-1) 1을더한다.

Example- 10의보수 83510는 16410 + 1 = 16510

- 2의보수 10102는 01012 + 1 = 01102

Complements


고정 소숫점

Numbers :고정소숫점과유동소숫점.

Fixed Point Representations

2진고정소숫점의표현

X = xnxn-1xn-2 ... x1x0. x-1x-2 ... x-m

Sign Bit(xn): 0 for positive - 1 for negative

Remaining Bits(xn-1xn-2 ... x1x0. x-1x-2 ... x-m)


부호화된 수부호화된 수

부호절대값표현부호화된 1의보수표현부호화된 2의보수표현

Example+9 와 -9 를 7 bit-binary number로표현시

+9의표현 ==> 0 0010013가지방법에의한 -9 표현

부호절대표현에의해 : 1 001001부호화된 1의보수표현에의해 : 1 110110부호화된 2의보수표현에의해 : 1 110111

일반적으로 컴퓨터에서고정된소숫점은 표현은오직 integer part 로표현되거나 fractional part 로표현된다.

양수음수두자기를모두표현한다.- 3가지표현에따라서


CHARACTERISTICS OF 3 DIFFERENT REPRESENTATIONS

보완

부호절대값 : 오직기호비트만보수를취한다.부호화된 1의보수 : 모든비트를보수를취한다.부호화된 2의보수 : 모든비트에보수를취한후하위1비트를더한다.


최대값과최소값는숫자와 0으로나타내어진다.X = xn xn-1 ... x0 . x-1 ... x-m

Signed MagnitudeMax: 2n - 2-m 011 ... 11.11 ... 1Min: -(2n - 2-m) 111 ... 11.11 ... 1Zero: +0 000 ... 00.00 ... 0

-0 100 ... 00.00 ... 0

Signed 1’s ComplementMax: 2n - 2-m 011 ... 11.11 ... 1Min: -(2n - 2-m) 100 ... 00.00 ... 0Zero: +0 000 ... 00.00 ... 0

-0 111 ... 11.11 ... 1

Signed 2’s ComplementMax: 2n - 2-m 011 ... 11.11 ... 1Min: -2n 100 ... 00.00 ... 0Zero: 0 000 ... 00.00 ... 0


2의 보수의 중요성2의 보수의 중요성

Signed 2’s complement representation follows a “weight”scheme similar to that of unsigned numbers

비트표현은 음수를 가지고 있다.다른 비트는 기존 값을 가진다.

X = xn xn-1 ... x0

V(X) = - xn × 2n + xi × 2i

i = 0Σn-1


덧 셈 연 산절 대 값 표 현

[1] 부호를비교한다.[2] 만약, 두수의부호가같으면두수를더한다.

-overflow 를주의한다.[3] 같지않다면, 두수의크기를비교하고큰수로부터작은수를뺀다.

--> 덧셈연산을위해필요하다.[4] Determine the sign of the result

6 0110+) 9 100115 1111 -> 01111

9 1001- ) 6 0110

3 0011 -> 00011

9 1001-) 6 0110- 3 0011 -> 10011

6 0110+) 9 1001-15 1111 -> 11111

6 + 9 -6 + 9

6 + (- 9) -6 + (-9)

Overflow 9 + 9 or (-9) + (-9)9 1001

+) 9 1001(1)0010overflow



덧 셈 연 산2의 보수

Example 6 0 01109 0 1001

15 0 1111

-6 1 10109 0 10013 0 0011

6 0 0110-9 1 0111-3 1 1101

-9 1 0111-9 1 0111

-18 (1)0 1110

두수를더할때부호비트까지연산하여케리비트를가지고부호를결정한다.– overflow를주의한다.

overflow9 0 10019 0 1001+)

+) +)

18 1 0010

+) +)

2 operands have the same signand the result sign changesxn-1yn-1s’n-1 + x’n-1y’n-1sn-1 = cn-1⊕ cn

x’n-1y’n-1sn-1(cn-1 ⊕ cn)

xn-1yn s’n-1(cn-1 ⊕ cn)



덧셈 연산1의 보수 사용

두수의덧셈에부호비트도같이연산한다.-최상의비트를넘는오버플로우비트발생시, 그들의결과값에케리된수를더한다.

6 0 0110-9 1 0110-3 1 1100

-6 1 10019 0 1001

(1) 0(1)00101

3 0 0011

+) +)

+)

end-around carry

-9 1 0110-9 1 0110

(1)0 11001

0 1101

+)

+)

9 0 10019 0 1001

1 (1)0010+)

overflow

Example

not overflow (cn-1 ⊕ cn) = 0

(cn-1 ⊕ cn)



COMPARISON OF REPRESENTATIONS

음수변환수월함

S + M > 1’s Complement > 2’s Complement

Hardware

- S+M: Needs an adder and a subtractor for Addition- 1’s and 2’s Complement: Need only an adder

연산속도

2’s Complement > 1’s Complement(end-around C)

Zero의인식

2’s Complement is fast



뺄 셈 연 산

뺄셈을보완한다. (Sign bit포함)피감수 sign bits를포함하여덧셈연산을한다.

( ± A ) - ( - B ) = ( ± A ) + B ( ± A ) - B = ( ± A ) + ( - B )


2의보수를이용한뺄셈연산


유 동 소 수 점 표 현

The location of the fractional point is not fixed to a certain locationThe range of the representable numbers is wide

F = EM

mn ekek-1 ... e0 mn-1mn-2 … m0 . m-1 … m-m

부호 지수 가수

가수점에의해표시된다. (정수혹은분수에의해)

지수기수에의해서위치를가리키다.

십진표현

V(F) = V(M) * RV(E) M: MantissaE: ExponentR: Radix

Floating Point Representation


유 동 소 수 점

0 .1234567 0 04sign sign

가수 지수

==> +.1234567 x 10+04

Example

Note:유동소숫점표현에서는, 오직가수(Mantissa(M))와지수Exponent(E)중심으로표현된다.. ( 기수와소숫점은 생략된다. )

Example

2진수 +1001.11는 16-bit 유동소숫점으로밑의표현으로나타내어진다.(6-bit exponent and 10-bit fractional mantissa)

0 0 00100 100111000

0 0 00101 010011100

Exponent MantissaSignor



유 동 소 수 점 의 특 성 표 현

표준

-같은수일지라도유종소숫점의표현은배우다르다.→ Need for a unified representation in a given computer

-가수의 non-zero digit 점의자리표현은매우중요하다.

Zero의표현

- Zero가수(Mantissa) = 0

- Real ZeroMantissa = 0표현

= 최소화된수표현은00 ... 0← hardware에의해쉽게확인할수있다.



INTERNAL REPRESENTATION AND EXTERNAL REPRESENTATION

CPUMemory

InternalRepresentation Human

Device

AnotherComputer

ExternalRepresentation




EXTERNAL REPRESENTATIONNumbersMost of numbers stored in the computer are eventually changed

by some kinds of calculations→ Internal Representation for calculation efficiency→ Final results need to be converted to as External Representation

for presentability

Alphabets, Symbols, and some NumbersElements of these information do not change in the course of processing→ No needs for Internal Representation since they are not used

for calculations→ External Representation for processing and presentability

ExampleDecimal Number: 4-bit Binary Code

BCD(Binary Coded Decimal)

Decimal BCD Code0 00001 00012 00103 00114 01005 01016 01107 01118 10009 1001

External Representations


OTHER DECIMAL CODES

Decimal BCD(8421) 2421 84-2-1 Excess-30 0000 0000 0000 00111 0001 0001 0111 01002 0010 0010 0110 01013 0011 0011 0101 01104 0100 0100 0100 01115 0101 1011 1011 10006 0110 1100 1010 10017 0111 1101 1001 10108 1000 1110 1000 10119 1001 1111 1111 1100 d3 d2 d1 d0: symbol in the codes

BCD: d3 x 8 + d2 x 4 + d1 x 2 + d0 x 1⇒ 8421 code.

2421: d3 x 2 + d2 x 4 + d1 x 2 + d0 x 184-2-1: d3 x 8 + d2 x 4 + d1 x (-2) + d0 x (-1)Excess-3: BCD + 3

Note: 8,4,2,-2,1,-1 in this table is the weight associated with each bit position.

BCD: It is difficult to obtain the 9's complement.However, it is easily obtained with the other codes listed above.

→ Self-complementing codes

External Representations


GRAY CODE

Characterized by having their representations of the binary integers differ in only one digit between consecutive integers

Useful in some applications

Decimalnumber

Gray Binaryg3 g2 g1 g0 b3 b2 b1 b0

0 0 0 0 0 0 0 0 01 0 0 0 1 0 0 0 12 0 0 1 1 0 0 1 03 0 0 1 0 0 0 1 14 0 1 1 0 0 1 0 05 0 1 1 1 0 1 0 16 0 1 0 1 0 1 1 0 7 0 1 0 0 0 1 1 18 1 1 0 0 1 0 0 09 1 1 0 1 1 0 0 1

10 1 1 1 1 1 0 1 011 1 1 1 0 1 0 1 112 1 0 1 0 1 1 0 013 1 0 1 1 1 1 0 114 1 0 0 1 1 1 1 015 1 0 0 0 1 1 1 1

4-bit Gray codes

Other Binary codes


그레이 코드 - 분석

Letting gngn-1 ... g1 g0 be the (n+1)-bit Gray code for the binary number bnbn-1 ... b1b0

gi = bi ⊕ bi+1 , 0 ≤ i ≤ n-1gn = bn

andbn-i = gn ⊕ gn-1 ⊕ . . . ⊕ gn-ibn = gn

0 0 0 0 00 0 0001 0 1 0 01 0 001

1 1 0 11 0 0111 0 0 10 0 010

1 10 0 1101 11 0 1111 01 0 1011 00 0 100

1 1001 1011 1111 0101 0111 0011 1011 000

그레이코드는 연산되어진결과를반영한다.-연산없이결과테이블을예상하기쉽지다.- for any n: reflect case n-1 about a

mirror at its bottom and prefix 0 and 1to top and bottom halves, respectively

Reflection of Gray codes

Note:

ε

Other Binary codes


CHARACTER REPRESENTATION ASCII

ASCII (American Standard Code for Information Interchange) Code

Other Binary codes

0123456789ABCDEF

NULSOHSTXETXEOTENQACKBELBSHTLFVTFFCRSOSI

SP!“#$%&‘()*+,-./

0123456789:;<=>?

@ABCDEFGHIJKLMNO

PQRSTUVWXYZ[\]mn

‘abcdefghIjklmno

Pqrstuvwxyz{|}~DEL

0 1 2 3 4 5 6 7DLEDC1DC2DC3DC4NAKSYNETBCANEMSUBESCFSGSRSUS

LSB(4 bits)

MSB (3 bits)


CONTROL CHARACTER REPRESENTAION (ACSII)

NUL NullSOH Start of Heading (CC)STX Start of Text (CC)ETX End of Text (CC)EOT End of Transmission (CC)ENQ Enquiry (CC)ACK Acknowledge (CC)BEL BellBS Backspace (FE)HT Horizontal Tab. (FE)LF Line Feed (FE)VT Vertical Tab. (FE)FF Form Feed (FE)CR Carriage Return (FE)SO Shift OutSI Shift InDLE Data Link Escape (CC)

(CC) Communication Control(FE) Format Effector(IS) Information Separator

Other Binary codes

DC1 Device Control 1DC2 Device Control 2DC3 Device Control 3DC4 Device Control 4NAK Negative Acknowledge (CC)SYN Synchronous Idle (CC)ETB End of Transmission Block (CC)CAN CancelEM End of MediumSUB SubstituteESC EscapeFS File Separator (IS)GS Group Separator (IS)RS Record Separator (IS)US Unit Separator (IS)DEL Delete


에러 검출 코드

Parity System

-에러검색을위해간소화한다. - One parity 비트를이용하여검사한다. -짝수검사기와홀수검사기가있다.

짝수검사기- One bit is attached to the information so that

the total number of 1 bits is an even number

1011001 01010010 1

홀수검사기- One bit is attached to the information so that

the total number of 1 bits is an odd number

1011001 11010010 0

Error Detecting codes


Parity Bit Generation

For b6b5... b0(7-bit information); even parity bit beven

beven = b6 ⊕ b5 ⊕ ... ⊕ b0

For odd parity bit

bodd = beven ⊕ 1 = beven

PARITY BIT GENERATION


PARITY GENERATOR AND PARITY CHECKER

Parity Generator Circuit (even parity)

b6b5b4b3b2b1

b0

beven

Parity Checker

b6b5b4b3b2b1

b0

beven

Even Parity error indicator

Error Detecting codes

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