02 Bieu Dien So Nguyen

KIN TRC MY TNH &

HP NG

02 Biu din s nguyn

1

H c s q tng qut2

Tng qut s nguyn c n ch s thuc h c s q bt k c biu din:

(mi ch s xi ly t tp X c q phn t)

V d:

H c s 10: A = 123 = 100 + 20 + 3 = 1.102 + 2.101 + 3.100

q = 2, X = {0, 1}: h nh phn (binary)

q = 8, X = {0, 1, 2,, 7}: h bt phn (octal)

q = 10, X = {0, 1, 2,, 9}: h thp phn (decimal)

q = 16, X = {0, 1, 2,,9, A, B,, F}: h thp lc phn (hexadecimal)

Chuyn i: A = 123 d = 01111011 b = 173 o = 7B h

H c s thng c bin din trong my tnh l h c s 2

0

0

1

1

1

1011 ......... qxqxqxxxxn

nn

Chuyn i gia cc h c s

3

c im

Con ngi s dng h thp phn

My tnh s dng h nh phn, bt phn, thp lc phn

Nhu cu

Chuyn i qua li gia cc h m ?

H khc sang h thp phn (... dec)

H thp phn sang h khc (dec ...)

H nh phn sang h khc v ngc li (bin )

Chuyn i gia cc h c s[1] Decimal (10) Binary (2)

4

Ly s c s 10 chia cho 2

S d a vo kt qu

S nguyn em chia tip cho 2

Qu trnh lp li cho n khi s nguyn = 0

V d: A = 123

123 : 2 = 61 d 1

61 : 2 = 30 d 1

30 : 2 = 15 d 0

15 : 2 = 7 d 1

7 : 2 = 3 d 1

3 : 2 = 1 d 1

1 : 2 = 0 d 1

Kt qu: 1111011, v 123 l s dng,

thm 1 bit hin du vo u l 0 vo

Kt qu cui cng: 01111011

Chuyn i gia cc h c s[2] Decimal (10) Hexadecimal (16)

5

Ly s c s 10 chia cho 16

S d a vo kt qu

S nguyn em chia tip cho 16

Qu trnh lp li cho n khi s nguyn = 0

V d: A = 123

123 : 16 = 7 d 12 (B)

7 : 16 = 0 d 7 Kt qu cui cng: 7B

Chuyn i gia cc h c s[3] Binary (2) Decimal (10)

6

Khai trin biu din v tnh gi tr biu thc

V d:

10112 = 1.23 + 0.22 + 1.21 + 1.20 = 1110

0

0

1

1

1

1011 2.2....2.... xxxxxxn

nn

Chuyn i gia cc h c s[4] Binary (2) Hexadecimal (16)

7

Nhm tng b 4 bit trong biu din nh phn ri chuyn

sang k s tng ng trong h thp lc phn (0000

0,, 1111 F)

V d

10010112 = 0100 1011 = 4B16

HEX BIN HEX BIN HEX BIN HEX BIN

0 0000 4 0100 8 1000 C` 1100

1 0001 5 0101 9 1001 D 1101

2 0010 6 0110 A 1010 E 1110

3 0011 7 0111 B 1011 F 1111

Chuyn i gia cc h c s[5] Hexadecimal (16) Binary (2)

8

S dng bng di y chuyn i:

V d:

4B16 = 10010112

HEX BIN HEX BIN HEX BIN HEX BIN

0 0000 4 0100 8 1000 C` 1100

1 0001 5 0101 9 1001 D 1101

2 0010 6 0110 A 1010 E 1110

3 0011 7 0111 B 1011 F 1111

Chuyn i gia cc h c s[6] Hexadecimal (16) Decimal (10)

9

Khai trin biu din v tnh gi tr biu thc

V d:

7B16 = 7.161 + 12 (B).160 = 12310

0

0

1

1

1

1011 16.16....16.... xxxxxxn

nn

H nh phn

10

c dng nhiu trong my tnh biu din cc gi tr lu trong

cc thanh ghi hoc trong cc nh. Thanh ghi hoc nh c kch

thc 1 byte (8 bit) hoc 1 word (16 bit).

n c gi l chiu di bit ca s

Bit tri nht xn-1 l bit c gi tr (nng) nht MSB (Most Significant

Bit)

Bit phi nht x0 l bit t gi tr (nh) nht LSB (Less Significant Bit)

0

0

1

1

1

1011 2.2....2.... xxxxxxn

nn

tng nh phn11

S nh phn c th dng biu din bt k vic g m bn mun!

Mt s v d:

Gi tr logic: 0 False; 1 True

K t:

26 k t (A Z): 5 bits (25 = 32)

Tnh c trng hp vit hoa/thng + k t l 7 bits (ASCII)

Tt c cc k t ngn ng trn th gii 8, 16, 32 bits (Unicode)

Mu sc: Red (00), Green (01), Blue (11)

V tr / a ch: (0, 0, 1)

B nh: N bits Lu c ti a 2N i tng

S nguyn khng du12

c im

Biu din cc i lng lun dng

V d: chiu cao, cn nng, m ASCII

Tt c bit u c s dng biu din gi tr (khng quan tm n

du m, dng)

S nguyn khng du 1 byte ln nht l 1111 11112 = 28 1 = 25510

S nguyn khng du 1 word ln nht l 1111 1111 1111 11112 = 216

1 = 6553510

Ty nhu cu c th s dng s 2, 3 word.

LSB = 1 th s l s l s l

S nguyn c du13

Lu cc s dng hoc m (s c du)

C 4 cch ph bin:

[1] Du lng

[2] B 1

[3] B 2

[4] S qu (tha) K

S c du trong my tnh c biu din dng s b 2

S nguyn c du[1] Du lng

14

Bit tri nht (MSB): bit nh du m / dng

0: s dng

1: s m

Cc bit cn li: biu din ln ca s (hay gi tr tuyt

i ca s)

V d:

Mt byte 8 bit: s c 7 bit (tr i bit du) dng biu din

gi tr tuyt i cho cc s c gi tr t 0000000 (010) n

1111111 (12710)

Ta c th biu din cc s t 12710 n +12710

-N v N ch khc gi tr bit MSB (bit du), phn ln (gi

tr tuyt i) hon ton ging nhau

S nguyn c du[2] B 1

15

Tng t nh phng php [1], bit MSB dng lm bit du

0: S dng

1: S m

Cc bit cn li (*) dng lm ln

S m: Thc hin php o bit tt c cc bit ca (*)

V d:

Dng b 1 ca 00101011 (43) l 11010100 (43)

Mt byte 8 bit: biu din t 12710 n +12710

B 1 c hai dng biu din cho s 0, bao gm: 00000000 (+0) v

11111111 (0) (mu 8 bit, ging phng php [1])

Khi thc hin php cng, cng thc hin theo quy tc cng nh phn thng

thng, tuy nhin, nu cn pht sinh bit nh th phi tip tc cng bit nh

ny vo kt qu va thu c

S nguyn c du[3] B 2

16

Biu din ging nh s b 1 + ta phi cng thm s 1 vo kt

qu (dng nh phn)

S b 2 ra i khi ngi ta gp vn vi hai phng php

du lng [1] v b 1 [2], l:

C hai cch biu din cho s 0 (+0 v -0) khng ng nht

Bit nh pht sinh sau khi thc hin php tnh phi c cng

tip vo kt qu d gy nhm ln

Phng php s b 2 khc phc hon ton 2 vn

V d:

Mt byte 8 bit: biu din t 12810 n +12710 (c li 1

s v ch c 1 cch biu din s 0)

S b 1 v S b 2

0 0 0 0 0 1 0 1S 5 (8 bit)

1 1 1 1 1 0 1 0S b 1 ca 5

1 1 1 1 1 0 1 1S b 2 ca 5

1+

0 0 0 0 0 1 0 1+ S 5

0 0 0 0 0 0 0 01Kt qu

17

Nhn xt s b 218

(S b 2 ca x) + x = mt dy ton bit 0 (khng tnh bit 1 cao nht

do vt qu phm vi lu tr)

Do s b 2 ca x chnh l gi tr m ca x hay x (Cn gi l

php ly i)

i s thp phn m 5 sang nh phn?

i 5 sang nh phn ri ly s b 2 ca n

Thc hin php ton a b?

a b = a + (b) Cng vi s b 2 ca b.

S nguyn c du[4] S qu (tha) K

19

Cn gi l biu din s dch (biased representation)

Chn mt s nguyn dng K cho trc lm gi tr dch

Biu din s N:

+N (dng): c c bng cch ly K + N, vi K c chn sao cho tng ca K v mt s

m bt k trong min gi tr lun lun dng

-N (m): c c bng cck ly K - N (hay ly b hai ca s va xc nh)

V d:

Dng 1 Byte (8 bit): biu din t -12810 n +12710

Trong h 8 bit, biu din N = 25, chn s tha k = 128, :

+2510 = 100110012

-2510 = 011001112

Ch c mt gi tr 0: +0 = 100000002, -0 = 100000002

Nhn xt20

S b 2 [3] lu tr s c du v cc php tnh ca chng trn

my tnh (thng dng nht)

Khng cn thut ton c bit no cho cc php tnh cng v tnh tr

Gip pht hin d dng cc trng hp b trn.

Du lng [1] / s b 1 [2] dng cc thut ton phc tp v

bt li v lun c hai cch biu din ca s 0 (+0 v -0)

Du lng [1] php nhn ca s c du chm ng

S tha K [4] dng cho s m ca cc s c du chm ng

Biu din s m (s b 2)21

-2n-1 2n-2 23 22 21 20

0

0

1

1

2

2

1

1011 2.2....2.)2.(... xxxxxxxn

n

n

nn

N bits

Phm vi lu tr: [-2n-1, 2n-1 - 1]

V d:

1101 01102 = -27 + 26 + 24 + 23 + 22 + 21

= -128 + 64 + 16 + 4 + 2 =

= -4210

V d (s b 2)22

Tnh gi tr khng du v c du23

Tnh gi tr khng du v c du ca 1 s?

V d s word (16 bit): 1100 1100 1111 0000

S nguyn khng du ?

Tt c 16 bit lu gi tr gi tr l 52464

S nguyn c du ?

Bit MSB = 1 do s ny l s m

p dng cng thc gi tr l 13072

Tnh gi tr khng du v c du24

Nhn xt

Bit MSB = 0 th gi tr c du bng gi tr khng du.

Bit MSB = 1 th gi tr c du bng gi tr khng du tr i 256

(28 nu tnh theo byte) hay 65536 (216 nu tnh theo word).

Tnh gi tr khng du v c du ca 1 s?

V d s word (16 bit): 1100 1100 1111 0000

Gi tr khng du = 52464

Gi tr c du: v bit MSB = 1 nn gi tr c du = 52464

65536 = 13072

Php dch bit v php xoay25

Shift left (SHL): 1100 1010 1001 0100

Chuyn tt c cc bit sang tri, b bit tri nht, thm 0 bit phi nht

Shift right (SHR): 1001 0101 0100 1010

Chuyn tt c cc bit sang phi, b bit phi nht, thm 0 bit tri nht

Rotate left (ROL): 1100 1010 1001 0101

Chuyn tt c cc bit sang tri, bit tri nht thnh bit phi nht

Rotate right (ROR): 1001 0101 1100 1010

Chuyn tt c cc bit sang phi, bit phi nht thnh bit tri nht

Php ton LogicAND, OR, NOT, XOR

26

AND 0 1

0 0 0

1 0 1

OR 0 1

0 0 1

1 1 1

XOR 0 1

0 0 1

1 1 0

NOT 0 1

1 0

Php nhn Php cng Php so snh khc

Php ph nh

V d27

X = 0000 1000b = 8d

X shl 2 = 0010 0000b = 32d = 8 . 22

(X shl 2) or X = 0010 1000b = 40d = 32 + 8

Y = 0100 1010b = 74d

((Y and 0Fh) shl 4) = 1010 0000

OR OR

((Y and F0h) shr 4) = 0000 0100

= 1010 0100 = 164d (khng du)

= (164 28) = -92d (c du)

Mt s nhn xt28

i vi s nguyn khng du

x SHL y = x . 2y

x SHR y = x / 2y

i vi s nguyn bt k

AND dng tt bit (AND vi 0 lun = 0)

OR dng bt bit (OR vi 1 lun = 1)

XOR, NOT dng o bit (XOR vi 1 = o bit )

x AND 0 = 0

x XOR x = 0

M rng:

Ly gi tr ti bit th i ca x: (x SHR i) AND 1

Gn gi tr 1 ti bit th i ca x: (1 SHL i) OR x

Gn gi tr 0 ti bit th i ca x: NOT(1 SHL i) AND x

o bit th i ca x: (1 SHL i) XOR x

Cc php ton t29

Php Cng (+)

Php Tr (-)

Php Nhn (*)

Php Chia (/)

Php cng30

Nguyn tc c bn:

V d:

0 1

0

1

0 1

1 10

1 1 1 0

1 0 0 0

1

1110

1

1

01

1 1 1 0

1 0 0 0

1

1

Php cng31

Php tr32

Nguyn tc c bn: a v php cng

A B = A + (-B) = A + (s b 2 ca B)

V d: 11101 10011 = 11101 + 01101

1 1 1 0

1 0 0 0

1

1010

1

1

01

1 1 1 0

0 1 1 0

1

1

Php tr33

Php nhn34

Nguyn tc c bn:

0 1

0

1

0 0

0 1

Php nhn35

0 1

1 0

011

1

1 0 0

0 00 0 0

0 11 0 0

0 11 0 0

0011

0 1

1 01

1 0 0

Php nhn36

Thut ton nhn37

Gi s ta mun thc hin php nhn M x Q vi

Q c n bit

Ta nh ngha cc bin:

C (1 bit): ng vai tr bit nh

A (n bit): ng vai tr 1 phn kt qu nhn ([C, A, Q]: kt qu nhn)

[C, A] (n + 1 bit) ; [C, A, Q] (2n + 1 bit): coi nh cc thanh ghi ghp

Thut ton:Khi to: [C, A] = 0; k = nLp khi k > 0{

Nu bit cui ca Q = 1 thLy (A + M) [C, A]

Shift right [C, A, Q]k = k 1

}

Thut ton nhn ci tin (s khng/c du)

39

Khi to: A = 0; k = n; Q-1 = 0 (thm 1 bit = 0 vo cui Q)

Lp khi k > 0

{

Nu 2 bit cui ca Q0Q-1{

= 10 th A M A

= 01 th A + M A

= 00, 11 th A khng thay i

}

Shift right [A, Q, Q-1]

k = k 1

}

Kt qu: [A, Q]

V dM = 7, Q = -3, n = 4

40

Php chia41

Gi s ta mun thc hin Q / M viKhi to: A = n bit 0 nu Q > 0; A = n bit 1 nu Q < 0; k = n

Lp khi k > 0

{

Shift left (SHL) [A, Q]

A M A

# Nu A < 0: Q0 = 0 v A + M A

# Ngc li: Q0 = 1

k = k 1

}

Kt qu: Q l thng, A l s d

V d php chia42

Prefix in byte (Chun IEC)43

International Electrotechnical Commission (IEC)

Prefix in byte (Chun SI)44

International System of Units (SI)

Ch : khi ni kilobyte chng ta ngh l 1024 byte nhng thc ran l 1000 bytes theo chun SI, 1024 bytes l kibibyte (IEC)

Hin nay ch c cc nh sn xut a cng v vin thng mi dngchun SI

30 GB 30 * 109 ~ 28 * 230 bytes

1 Mbit/s 106 b/s

Homework45

c chng 9, sch ca W.Stalling

c trc slide bi ging s thc

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02 Bieu Dien So Nguyen