01-Thong Tin Va Bieu Dien Thong Tin (1)

1

TRNG I HC BCH KHOA H NIVIN CNG NGH THNG TIN V TRUYN THNG

TIN HC I CNGBi 1: Thng tin v biu din thng tin

Ni dung

1.1. Thng tin v Tin hc

1.2. Biu din s trong h m

1.3. Biu din d liu trong my tnh

2

Ni dung


1.1.1. Thng tin v x l thng tin

1.1.2. My tnh in t (MTT)

1.1.3. Tin hc v cc ngnh lin quan



3

Ni dung







4

2

D liu, Tn hiu, Thng tin

Thng tin(ngha rng): s phn nh s vt, s vic, hin tng ca th gii khch quan. Mang li nhn thc cho con ngi v th gii khch quan

D liu: nhng gi tr nh tnh v nh lng ca s vt, hin tng c xc nh thng qua cc php o c Cha ng thng tin

Khng c nng lng

Tn hiu: s vt (hoc thuc tnh vt cht, hin tng) phn nh, kch thch vo mt s vt, hin tng khc. Cha ng thng tin

C nng lng

Truyn ti thng tin t vt ny sang vt khc

5

X l d liu (Data processing)

Thng tin nm trong d liu Cn phi x l d liu thu c thng tin cn thit, hu ch phc v cho con ngi

Qu trnh x l d liu

NHP(INPUT)

X L(PROCESSING)

XUT (OUTPUT)

LU TR (STORAGE)

Khi d liu t, c thlm th cng

Khi d liu nhiu ln,cc cng vic lp ilp li ???

S dng my tnhin t h tr chovic lu tr, chn lcv x l d liu.

7

X l d liu (2) Ni dung







8

3

1.1.2. My tnh in t

My tnh in t (Computer):

Lm vic khng bit chn

Tit kim rt nhiu thi gian, cng sc

Tng chnh xc trong vic t ng ha mt phn hay ton phn ca qu trnh x l d liu.

9

My tnh in t c mt khp ni

10

a. Biu din thng tin trong MTT

Trong my tnh mi thng tin u c biu din bng s nh phn

a d liu vo cho my tnh, cn phi m ho n v dng nh phn.

Vi cc kiu d liu khc nhau cn c cch m ho khc nhau.

11

a. Biu din thng tin trong MTT (2)

n v nh nht biu din thng tin gi l bit.

BIT l ch vit tt ca BInary digiT.

Mt bit c 2 trng thi: 0 hoc 1

0 = OFF ; 1 = ON

12

OFF ON

4

a. Biu din thng tin trong MTT (3)

Tn gi K hiu Gi tr

Byte

KiloByte

MegaByte

GigaByte

TeraByte

Petabyte

Exabyte

B

KB

MB

GB

TB

PB

EB

8 bit

210 B = 1024 Byte

220 B = 1024 KB

230 B = 1024 MB

240 B = 1024 GB

250 B = 1024 TB

260 B = 1024 PB

13

Cc n v biu din thng tin ln hn:

b. Phn loi MTT

Theo kh nng s dng chung:

My tnh ln/Siu my tnh (Mainframe/SuperComputer)

My tnh tm trung (Mini Computer)

My vi tnh ( Micro Computer)

14

i. My tnh ln/Siu my tnh

Phc tp, c tc rt nhanh

S dng trong cc cng ty ln/vin nghin cu

Gii quyt cc cng vic ln, phc tp

Rt t (hng trm ngn ~ hng triu USD).

Nhiu ngi dng ng thi (100 500)

15

S

u

p

e

r

C

o

m

p

u

t

e

r

5

ii. My tnh tm trung (Mini computer)

Cng ging nh cc my Mainframe

S khc bit chnh:

H tr t ngi dng hn (10 100)

Nh hn v r hn (vi chc nghn USD)

17

iii. My vi tnh (Micro computer)

S dng vi x l

Nh, r, hiu nng cao,

Ph hp cho nhiu i tng ngi dng, s dng nhiu trong cng nghip v gii tr: My tnh c nhn Personal Computer (PC)

My tnh nhng Embedded Computer

Cc thit b cm tay nh in thoi di dng, my tnh b ti

...

18

My tnh c nhn (Personal Computer PC)

My tnh bn Desktop Computer

My tnh di ng Portable Computer

My tnh xch tay (Laptop Computer)

My tnh b ti (PDA - Personal Digital Assistant)

My tnh bng Tablet Computer

My tnh bn

Laptop

My tnh bng

PDA

My tnh nhng (Embedded computer)

L my tnh chuyn dng (special-purpose computer)

Gn trong cc thit b gia dng, my cng nghip

Gip con ngi dng s dng thit b hiu qu hn

20

6

c. Cc th h my tnh

S pht trin v cng ngh S pht trin v my tnh

21

i. Th h u (1950 1958)

1930s: Bng n c s dng lm cc bng mch tn hiu iu khin (electric circuits or switches)

iu khin bng tay, kch thc rt ln22

Bng n chn khng (vacumm tube)

ENIAC

My tnh in t u tin vi cng ngh bng chn khng:

Kch thc: di 10m, rng 3m, cao 3m

Trong 1 giy thc hin c 3 php ton23

ENIAC -Electronic Numerical Integrator and Calculator

UNIVAC 1

L my tnh thng mi u tin

Thc hin 30000 php ton / 1 giy24

UNIVAC I -UNIVersal Automatic Computer

7

ii. Th h th hai (1958 1964)

1947: Bng bn dn c pht minh ti Bell Laboratories

Bng bn dn c s dng thay bng n chn khng25

Cng ngh bn dn (diodes, transistors)

TRADIC

26

My tnh u tin s dng hon ton bng bn dn:

8000 transistors

Nhanh hn

Nh hn

R hn.

TRADIC - TRAnsistorized Airborne DIgital Computer

iii. Th h th ba (1965 1974)

27

1959 thit k ra vi mch u tin da trn cng ngh silicon (silicon chip or microchip)

Trn 1 vi mch tch hp hng triu transitor

Cng ngh mch tch hp (IC integrated circuit)

Vi mch Integrated Circuit

28

Nh hn,

R hn,

Hiu qu hn

8

IBM 360

29

Thit k trn cng ngh IC

Tc tnh ton: 1000 t php ton trong 1 giy

iv. Th h th t (1974 nay)

30

Microprocessor = Central Processing Unit (CPU) thit k trong 1 vi mch n

1971 : Intel 4004

Vi x l (Microprocessor)

1975 Altair 8800

31

My tnh c nhn u tin Altair 8800

Vi x l (Microprocessor)

32

9

1981 IBM PC

33

Th h my tnh c nhn mi vi kin trc m IBM

1984 Apple Macintos

34

1990 - Personal Computers

35

Tc vi x l tng nhanh:

CPU 1 li,

CPU a li

Kin trc t thay i

Th h th t (tip)

L

a

p

t

o

p My

tnh

bn

Pocket

10

Pentium

Th h th t (tip)

37

More Pentium

Pro

III

IV

38

Itanium

64-bit Intel

Microprocessors

39

Th h th t (tip)

39

Th h th t (tip)

N

e

t

w

o

r

k

11

e. Th h 5 (1990 - nay)

Artificial Intelligence (AI)

Cng ngh vi in t vi tc tnh ton cao v x l song song.

M phng cc hot ng ca no b v hnh vi con ngi

C tr khn nhn to vi kh nng t suy din pht trin cc tnh hung nhn c

Xu hng ngy nay

Nhanh hn

Nh hn

R hn

D s dng hn

42

Ni dung







43


Tin hc (Computer Science/Informatics)

Cng ngh thng tin (Information Technology - IT)

Cng ngh thng tin v truyn thng (Information and Communication Technology ICT).

12

1957, Karl Steinbuch ngi c xng trong 1 bi bo c thut ng "Informatik "

1962, Philippe Dreyfus ngi Php gi l informatique "

Phn ln cc nc Ty u, tr Anh u chp nhn. Anh ngi ta s dng thut ng computer science, hay computing science,

1966, Nga cng s dng tn informatika45

a. Tin hc (Informatics) a. Tin hc (2)

Tin hc c xem l ngnh khoa hc nghin cu cc phng php, cng ngh v k thut x l thng tin mt cch t ng.

Cng c ch yu s dng trong tin hc l my tnh in t v mt s thit b truyn tin khc.

Ni dung nghin cu ca tin hc ch yu gm 2 phn:

K thut phn cng (Hardware engineering)

K thut phn mm (Software engineering)

46

Xut hin Vit nam vo nhng nm 90 ca th k 20.

CNTT x l vi cc my tnh in t v cc phn mm my tnh nhm chuyn i, lu tr, bo v, truyn tin v trch rt thng tin mt cch an ton.

(Information Technology Association of America)

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b. Cng ngh thng tin

Mt ngnh s dng h thng cc thit b v my tnh, bao gm phn cng v phn mm cung cp mt gii php x l thng tin cho cc c nhn, t chc c yu cu

C nh hng v c ng dng trong nhiu ngnh ngh khc nhau ca x hi

Cc ng dng ngy nay ca IT:

Qun tr d liu

Qun l h thng thng tin

Thit k sn phm

ng dng khoa hc

48

b. Cng ngh thng tin (2)

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Information and Communication Technology

Truyn thng my tnh l s kt ni mt s lng my tnh vi nhau

L thut ng mi, nhn mnh s khng th tch ri hin nay ca CNTT vi cng ngh truyn thng trong thi i tt c u ni mng

Internet - Mng my tnh ton cu

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c. Cng ngh thng tin v truyn thng (ICT) Ni dung



1.2.1. H m

1.2.2. Chuyn i c s

1.2.3. i s Boolean


50

Ni dung



1.2.1. H m

1.2.2. Chuyn i c s

1.2.3. i s Boolean


51 52

L tp hp cc k hiu v qui tc biu din v xc nh gi tr cc s.

Mi h m c mt s k t/s (k s) hu hn. Tng s k s ca mi h m c gi l c s (base hay radix), k hiu l b.

V d: Trong h m c s 10, dng 10 k t l: cc ch s t 0 n 9.

1.2.1. H m

14

53

V mt ton hc, ta c th biu din 1 s theo h m c s bt k.

Khi nghin cu v my tnh, ta quan tm n cc h m sau y:

H thp phn (Decimal System) con ngi s dng

H nh phn (Binary System) my tnh s dng

H m bt phn (Octal System), h mi su (Hexadecimal System) dng vit gn s nh phn

1.2.1. H m (2)

54

H m thp phn hay h m c s 10 bao gm 10 k s theo k hiu sau:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Dng n ch s thp phn c th biu din c 10n gi tr khc nhau:

00...000 = 0

....

99...999 = 10n-1

a. H m thp phn

55

Gi s mt s A c biu din di dng:

A = an an-1 a1 a0 . a-1 a-2 a-m Gi tr ca A c hiu nh sau:

1 1 0 11 1 0 110 10 ... 10 10 10 ... 10

10

n n mn n m

ni

ii m

A a a a a a a

A a

a. H m thp phn (2)

56

V d: S 5246 c gi tr c tnh nh sau:

5246 = 5 x 103 + 2 x 102 + 4 x 101 + 6 x 100

V d: S 254.68 c gi tr c tnh nh sau:

254.68 = 2 x 102 + 5 x 101 + 4 x 100 + 6 x 10-1 + 8 x 10-2

a. H m thp phn (3)

15

57

C b k t th hin gi tr s. K s nh nht l 0 v ln nht l b-1.

S N(b) trong h m c s (b) c biu din bi:

N(b)=anan-1an-2a1a0.a-1a-2a-m

b. H m c s b (vi b 2, nguyn) b. H m c s b (2)

Trong biu din trn, s N(b) c n+1 k s biu din cho phn nguyn v m k s l biu din cho phn l, v c gi tr l:

58

59

S dng 2 ch s: 0,1

Ch s nh phn gi l bit (binary digit)

V d: Bit 0, bit 1

Bit l n v thng tin nh nht

c. H m nh phn

60

Dng n bit c th biu din c 2n gi tr khc nhau:

00...000 (2) = 0 (trong h thp phn)

...

11...111 (2) = 2n - 1 (trong h thp phn)

VD: Dng 3 bit th biu din c cc s t 0 n 7 (trong h thp phn)

c. H m nh phn (2)

16

61

Gi s c s A c biu din theo h nh phn nh sau:

A = an an-1 a1 a0 . a-1 a-2 a-m

Vi ai l cc ch s nh phn, khi gi tr ca A l:

1 1 0 1 21 1 0 1 22 2 ... 2 2 2 2 ... 2

2

n n mn n m

ni

ii m

A a a a a a a a

A a

c. H m nh phn (3)

62

V d:

S nh phn 1101001.1011 c gi tr:

1101001.1011(2) = 26 + 25 + 23 + 20 + 2-1

+ 2-3 + 2-4

= 64 + 32 + 8 + 1 + 0.5 + 0.125 + 0.0625 = 105.6875(10)

c. H m nh phn (4)

63

Php cng:

1+0=0+1=1;

0+0=0;

1+1=10;

Php tr:

0-1=1; (vay 1)

1-1=0;

0-0=0;

1-0=1

Tnh ton trong h nh phn

64

1 0 1

+ 1 1 1

---------

1 1 0 0

Tnh ton trong h nh phn V d

17

65

1 1 0 0

- 1 1 1

--------------------

0 1 0 1

Tnh ton trong h nh phn V d

66

S dng cc ch s: 0,1,2,3,4,5,6,7

Dng n ch s c th biu din c 8n

gi tr khc nhau:

00...000 = 0 (trong h thp phn)

...

77...777 = 8n -1 (trong h thp phn)

d. H m bt phn (Octal System b=8)

67

Gi s c s A c biu din theo h nh phn nh sau:

A = an an-1 a1 a0 . a-1 a-2 a-m Vi ai l cc ch s trong h bt phn, khi

gi tr ca A l:

d. H m bt phn (2)

1 1 0 1 21 1 0 1 28 8 ... 8 8 8 8 ... 8

8

n n mn n m

ni

ii m

A a a a a a a a

A a

68

V d:

235 . 64 (8) c gi tr nh sau:

235 . 64 (8) = 2x82 + 3x81 + 5x80 + 6x8-1

+ 4x8-2

= 157. 8125 (10)

d. H m bt phn (3)

18

69

S dng 16 k s: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

Cc ch in:

A, B, C, D, E, F

biu din cc gi tr s tng ng (trong h 10) l 10, 11, 12, 13, 14, 15

e. H m 16, Hexadecimal, b=16

70

Gi s c s A c biu din theo h thp lc phn nh sau:

A = an an-1 a1 a0 . a-1 a-2 a-mVi ai l cc ch s trong h thp lc phn, khi gi tr ca A l:

1 1 0 1 21 1 0 1 216 16 ... 16 16 16 16 ... 16

16

n n mn n m

ni

ii m

A a a a a a a a

A a

e. H m 16 (2)

71

V d: 34F5C.12D(16) c gi tr nh sau:

34F5C.12D(16)= 3x164 + 4x163 + 15x162 + 5x161

+ 12x160 +?

= 216294(10) + ?

e. H m 16 (3) Ni dung



1.2.1. H m

1.2.2. Chuyn i c s

1.2.3. i s Boolean


72

19

73

1.2.2. Chuyn i c s

Trng hp tng qut, mt s N trong h thp phn (N(10)) gm phn nguyn v phn thp phn.

Chuyn 1 s t h thp phn sang 1 s h c s b bt k gm 2 bc:

i phn nguyn (ca s ) t h thp phnsang h b

i phn thp phn (ca s ) t h thpphn sang h c s b

74

a. Chuyn i phn nguyn

Bc 1:Ly phn nguyn ca N(10) chia cho b, ta c thng l T1 s d d1.

Bc 2: Nu T1 khc 0, Ly T1 chia tip cho b, ta c thng s l T2 , s d l d2

(C lm nh vy cho ti bc th n, khi ta c Tn =0)

Bc n: Nu Tn-1 khc 0, ly Tn-1 chia cho b, ta c thng s l Tn =0, s d l dn

Kt qu ta c s N(b) l s to bi cc s d (c vit theo th t ngc li) trong cc bc trn

Phn nguyn ca N(10) = dndn-1d1 (b)

75

a. Chuyn i phn nguyn (2)

V d: Cch chuyn phn nguyn ca s 12.6875(10) sang s trong h nh phn:

Dng php chia cho 2 lin tip, ta c mt lot cc s d nh sau

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b. Chuyn i phn thp phn

Bc1: Ly phn thp phn ca N(10)nhn vi b, ta c mt s c dng x1.y1(x l phn nguyn, y l phn thp phn)

Bc 2: Nu y1 khc 0, tip tc ly 0.y1nhn vi b, ta c mt s c dng x2.y2

(c lm nh vy cho n khi yn=0)

Bc n: Nu yn-1 khc 0, nhn 0.yn-1 vi b, ta c xn.0

Kt qu ta c s sau khi chuyn i l:

Phn thp phn ca N(10) = 0.x1x2xn (b)

20

77

b. Chuyn i phn thp phn (2)

V d: Cch chuyn phn thp phn ca s 12.6875(10) sang h nh phn:

78

V d: Chuyn t thp phn sang nh phn

12.6875(10) = 1100.1011 (2)

69.25(10) = ?(2)

Cch 2: Tnh nhm

Phn tch s thnh tng cc ly thaca 2, sau da vo cc s m xcnh dng biu din nh phn

Nhanh hn.

V d: 69.25(10) = 64 + 4 + 1+

= 26 + 22 + 20 + 2-2

= 1000101.01(2)

79 80

Mt s v d

Nh phn Hexa: 11 1011 1110 0110(2) = ?

11 1011 1110 0110(2) = 3BE6(16) Hexa Nh phn: AB7(16) = ?

AB7(16) = 1010 1011 0111(2) Hexa Thp phn: 3A8C ?

3A8C (16) = 3 x 163 + 10 x 162 + 8 x 161

+12 x 160

= 12288 + 2560 + 128 + 12

= 14988(10)

21

81

Mt s v d (tip)

Thp phn Hexa: 14988 ?

14988 : 16 = 936 d 12 tc l C

936 : 16 = 58 d 8

58 : 16 = 3 d 10 tc l A

3 : 16 = 0 d 3

Nh vy, ta c: 14988(10) = 3A8C(16)

Bi tp

Chuyn sang h nh phn

124.75

65.125

82

Ni dung



1.2.1. H m

1.2.2. Chuyn i c s

1.2.3. i s Boolean


83 84

i s Boolean

a NOT a

0 1

1 0

Cc php ton logic vi tng bit nh phn:

22

85

i s Boolean (tip)

a b a AND b a OR b a XOR b

0 0 0 0 0

0 1 0 1 1

1 0 0 1 1

1 1 1 1 0

Cc php ton logic vi cp bit nh phn:

86

i s Boolean (tip)

Thc hin cc php ton logic vi 2 s nh phn:

Kt qu l 1 s nh phn khi thc hin cc php ton logic vi tng cp bit ca 2 s nh phn

Cc php ton ny ch tc ng ln tng cp bit m khng nh hng n bit khc.

87

V d

VD: A = 1010 1010 v B = 0000 1111

AND OR XOR NOT

1010 1010 01010101

0000 1111 11110000

00001010 10101111 10100101

Php dch

Dch tri logic

88

Dch phi logic

Dch phi s hc

23

Php quay

Quay tri khng nh

89

Quay phi khng nh

Quay tri c nh

Quay phi c nh

Ni dung




1.3.1. Nguyn l chung

1.3.2. Biu din s nguyn

1.3.3. Biu din s thc

1.3.4. Biu din k t

90

Ni dung







1.3.4. Biu din k t

91 92


Mi d liu khi a vo my tnh u phic m ha thnh s nh phn

Cc loi d liu:

D liu nhn to: Do con ngi quy c

D liu t nhin:

Tn ti khch quan vi con ngi.

Ph bin l cc tn hiu vt l nh m thanh, hnhnh,

24

93

a. Nguyn tc m ha d liu

M ha d liu nhn to:

D liu s: M ha theo cc chun quy c

D liu k t: M ha theo b m k t

M ha d liu t nhin:

Cc d liu cn phi s ha trc khi avo my tnh

Theo s m ha v ti to tn hiu vt l

94

S m ha v ti to tn hiu vt l

V d: MODEM: MOdulator and DEModulator

(iu ch v Gii iu ch)

95

b. Cc loi d liu trong my tnh

D liu c bn

S nguyn: M nh phn thng thng(khng du) v m b hai (c du)

S thc: S du chm ng

K t: B m k t

D liu c cu trc

L tp hp cc loi d liu c bn c cuthnh theo mt cch no .

V d: Kiu d liu mng, xu k t, tp hp,bn ghi,

Ni dung







1.3.4. Biu din k t

96

25

97


Dng 1 chui bit biu din.

i vi s nguyn c du, ngi ta sdng bit u tin(Most significant bit) biu din du - v bit ny gi l bit du.

* di t d liu:

L s bit c s dng m ha loi d liutng ng

Trong thc t, di t d liu thng l bis ca 8.

98

a. S nguyn khng du

Dng tng qut: gi s dng n bit biu din cho mt s nguyn khng du A:

an-1an-2...a3a2a1a0 Gi tr ca A c tnh nh sau:

Di biu din ca A:

T 0 n 2n-1

1 2 1 01 2 1 0

1

0

2 2 ... 2 2

2

n nn n

ni

ii

A a a a a

A a

V d 1

Biu din cc s nguyn khng du sau y bng 8 bit:

A = 45 B = 156

Gii:

A = 45 = 32 + 8 + 4 + 1 = 25 + 23 + 22 + 20

A = 0010 1101(2)

B = 156 = 128 + 16 + 8 + 4 = 27 + 24 + 23 + 22

B = 1001 1100 (2)

99

V d 2

Cho cc s nguyn khng du X, Y c biu din bng 8 bit nh sau:

X = 0010 1011

Y = 1001 0110

Gii:

X = 0010 1011 = 25 + 23 + 21 + 20

= 32 + 8 + 2 + 1 = 43

Y = 1001 0110 = 27 + 24 + 22 + 21

= 128 + 16 + 4 + 2 = 150100

26

101

Trng hp c th: vi n = 8 bit

Di biu din l [0, 255]

0000 0000 = 0

0000 0001 = 1

0000 0010 = 2

0000 0011 = 3

.....

1111 1111 = 255

Trc s hc:

Trc s hc my tnh:

Vi n = 8 bit

123 + 164 =?

Ch trng hp php tnh vt qu di biu din

1111 1111

+ 0000 0001

1 0000 0000

KQ sai: 255 + 1 = 0 ?(do php cng b nh ra ngoi)

102

Vi n = 16 bit, 32 bit, 64 bit

n = 16 bit:

Di biu din l [0, 65535]

n = 32 bit:

Di biu din l [0, 232-1]

n = 64 bit:

Di biu din l [0, 264-1]

103 104

b. Biu din s nguyn c du

S dng bit u tin biu din du - v bit ny gi l bit du

S dng s b hai biu din

27

i. Phn b l g?

105

U: Universal Set (Tp ton th)

A U

Ac = U \ A

106

ii. S b chn v s b mi (h thp phn)

Gi s c 1 s nguyn thp phn A c biu din bi n ch s thp phn. Ta c: S b chn ca A = (10n 1) A

S b mi ca A = 10n A

NX: S b mi = S b chn + 1

V d: Xt n = 4 ch s, A = 2874

S b chn ca A = (104 1) 2874 = 7125

S b mi ca A = 104 2874 = 7126

iii. S b mt v s b hai (h nh phn)

Gi s c 1 s nguyn nh phn c biu din bi n bit. Ta c:

S b mt ca A = (2n - 1) A

S b hai ca A = 2n A

NX: S b hai = S b mt + 1

V d

Xt n = 4 bit, A = 0110

S b mt ca A = (24 - 1) - 0110 = 1001

S b hai ca A = 24 - 0110 = 1010

107 108

Nhn xt

V d (c)

Xt n = 4 bit, A = 0110

S b mt ca A = (24 - 1) - 0110 = 1001

S b hai ca A = 24 - 0110 = 1010

C th tm s b mt ca A bng cch o ngc tt c cc bit

S b hai = S b mt + 1

A + S b hai ca A = 0 nu b qua bit nh ra khi bit cao nht

28

109

iv. Biu din s nguyn c du

Biu din s nguyn c du bng s b hai

Dng n bit biu din s nguyn c du A

Biu din s b 2 ca A (s dng n bit)

V d: Biu din s nguyn c du sau y bng 8 bit: A = - 70(10)Biu din 70 = 0100 0110

B 1: 1011 1001 (nghch o cc bit)

+ 1

B 2: 1011 1010

Vy: A = 1011 1010(2)

110

iv. Biu din s nguyn c du (2)

Dng tng qut ca s nguyn c du A:

an-1an-2...a2a1a0 Gi tr ca A c xc nh nh sau:

Di biu din: [-2n-1, 2n-1-1]

10000000

.

01111111

Nhn xt: Vi s dng, s m?

21

10

2 2n

n in i

i

A a a

111

V d

Xc nh gi tr ca cc s nguyn c du 8 bit sau y:

A = 0101 0110

B = 1101 0010

Gii:

A = 26 + 24 + 22 + 21 = 64 + 16 + 4 + 2 = +86

B = -27 + 26 + 24 + 21 =

= -128 + 64 + 16 + 2 = -46

Trng hp c th: vi n = 8 bit

Di biu din l [-128, 127]

0000 0000 = 0

0000 0001 = +1

0000 0010 = +2

.......

01111111 = +127

10000000 = -128

10000001 = -127

.......

1111 1110 = -2

1111 1111 = -1

Trc s hc my tnh

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113

v. Tnh ton s hc vi s nguyn

Cng/ tr s nguyn khng du:

Tin hnh cng/tr ln lt tng bt t phi qua tri.

Khi cng/tr hai s nguyn khng du n bit ta thu c mt s nguyn khng du n bit.

Nu tng ca hai s ln hn 2n-1 th khi s trn s v kt qu s l sai.

Tr s khng du th ta ch tr c s ln cho s nh. Trng hp ngc li s sai

114

V d: Cng tr s nguyn khng du

Dng 8 bit biu din s nguyn khng du

Trng hp khng xy ra trn s (carry-out): X = 1001 0110 = 150 Y = 0001 0011 = 19 S = 1010 1001 = 169 Cout = 0

Trng hp c xy ra trn s (carry-out): X = 1100 0101 = 197 Y = 0100 0110 = 70 S = 0000 1011 267 Cout = 1 carry-out (KQ sai = 23 + 21 + 20 = 11)

115

v. Tnh ton s hc vi s nguyn (2)

Cng s nguyn c du

Cng ln lt cc cp bit t phi qua tri, b qua bit nh (nu c).

Cng hai s khc du: kt qu lun ng

Cng hai s cng du: Nu tng nhn c cng du vi 2 s hng th

kt qu l ng

Nu tng nhn c khc du vi 2 s hng th xy ra hin tng trn s hc (overflow) v kt qu nhn c l sai

116

V d: Cng/tr s nguyn c du

VD: khng trn s

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117

V d: Cng/tr s nguyn c du

C xy ra trn s:


Tr s nguyn c du

tr hai s nguyn c du X v Y, cn ly b hai ca Y tc Y, sau cng X vi Y tc l: X Y = X + (-Y).

Cng ln lt cc cp bit t phi qua tri, b qua bit nh (nu c).

V d:

118

V d

Cho A = 0x3D, B = 0x50. Mnh no sauy l sai?

a. not A = 0xC2

b. A and B = 0x10

c. A or B = 0x7E

d. A xor B = 0x6D

119

V d

Cho A = -25(10), B = +58(10) l 2 s nguyn, c m ha di dng s nguyn 8 bit. Mnh no sau y l ng ?

a. A = 1010 0111

b. B = 0011 0110

c. A + B = 0010 0001

d. A B = 1010 1101

120

31

A v B c m ha di dng s nguyn co du8 bit, A = 1010 1010, B = 0011 1100. Mnh no sau y l ng?

a. not A = +85

b. not B = -61

c. A xor B = -22

d. A or B = -66

121 122


Nhn/chia s nguyn khng du Cc bc thc hin nh trng h 10 VD: Php nhn

1011 (11 c s 10)x

1101 (13 c s 10) -------------

10110000

1011 1011--------------

10001111 (143 c s 10)


Cha hai s nguyn khng du

123


Nhn s nguyn c du: Bc 1: Chuyn i s nhn v s b nhn

thnh s dng tng ng

Bc 2: Nhn 2 s bng thut gii nhn s nguyn khng du c tch 2 s dng

Bc 3: Hiu chnh du ca tch:

Nu 2 tha s ban u cng du Kt qu l tch thu c trong bc 2.

Nu khc du Kt qu l s b 2 ca tch thu c trong bc 2.

124

32


Chia s nguyn c du:

Bc 1: Chuyn i s chia v s b chia thnh s dng tng ng

Bc 2: Chia 2 s bng thut gii chia s nguyn khng du Thu c thng v d u dng

Bc 3: Hiu chnh du ca kt qu theo quy tc sau:

125

Ni dung







1.3.4. Biu din k t

126

127

a. Nguyn tc chung

biu din s thc, trong my tnh ngi ta thng dng k php du chm ng (Floating Point Number)

V d: 12.3 = 12.3 * 100

= 123 * 101

= 1.23 * 10-1

128

a. Nguyn tc chung (2)

Mt s thc X c biu din theo kiu s du chm ng nh sau:

X = M * RE

Trong :

M l phn nh tr (Mantissa)

R l c s (Radix) thng l 2 hoc 10.

E l phn m (Exponent)

Vi R c nh th lu tr X ta ch cn lu tr M v E (di dng s nguyn)

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129

V d - Biu din s thc

Vi c s R = 10, gi s 2 s thc N1 v N2 c lu tr theo phn nh tr v s m nh sau: M1 = -15 v E1 = +12

M2 = +314 v E2 = -9

C ngha l

N1 = M1 x 10 E1 = -15x1012

= -15 000 000 000 000

v

N2 = M2 x 10 E2 = 314 x 10-9

= 0.000 000 314

130

b. Php ton vi s thc

Khi thc hin php ton vi s du chm ng s c tin hnh trn c s cc gi tr ca phn nh tr v phn m.

131

c. Php ton vi s thc (2)

Gi s c 2 s du phy ng sau:

N1 = M1 x RE1 v N2 = M2 x RE2

Khi , vic thc hin cc php ton s hc s c tin hnh:

N1 N2 = (M1 x R E1-E2 M2) x RE2 ,

(gi thit E1 E2)

N1 x N2 = (M1x M2) x R E1+E2

N1 /N2 = (M1 / M2) x R E1-E2

132

c. Chun IEEE 754/85

L chun m ha s du chm ng

C s R = 2

C cc dng c bn:

Dng c chnh xc n, 32-bit

Dng c chnh xc kp, 64-bit

Dng c chnh xc kp m rng, 80-bit

34

133

c. Chun IEEE 754/85 (2)

Khun dng m ha:

S e m

S e m

S e m

31 30 23 22 0

63 62 52 51 0

79 78 64 63 0

134

c. Chun IEEE 754/85 (3)

S l bit du, S=0 l s dng, S=1 l s m.

e l m lch (excess) ca phn m E, tc l: E = e b

Trong b l lch (bias):

Dng 32-bit : b = 127, hay E = e - 127

Dng 64-bit : b = 1023, hay E = e - 1023

Dng 80-bit : b = 16383, hay E = e - 16383

135

c. Chun IEEE 754/85 (4)

m l cc bit phn l ca phn nh tr M, phn nh tr c ngm nh nh sau:

M = 1.m

Cng thc xc nh gi tr ca s thc tng ng l:

X = (-1)S x 1.m x 2e-b

S e m

136

V d 1

V d 1: C mt s thc X c dng biu din nh phn theo chun IEEE 754 dng 32 bit nh sau:

1100 0001 0101 0110 0000 0000 0000 0000

Xc nh gi tr thp phn ca s thc .

Gii:

S = 1 X l s m

e = 1000 0010 = 130

m = 10101100...00

Vy X = (-1)1 x 1.10101100...00 x 2130-127

= -1.101011 x 23 = -1101.011 = -13.375

35

137

V d 2

Xc nh gi tr thp phn ca s thc X c dng biu din theo chun IEEE 754 dng 32 bit nh sau:

0011 1111 1000 0000 0000 0000 0000 0000

Gii:

S = 0 X l s dng

e = 0111 1111= 127

m = 000000...00

Vy X = (-1)0 x 1.0000...00 x 2127-127

= 1.0 x 20 = 1

138

V d 3

Biu din s thc X = 9.6875 v dng s du chm ng theo chun IEEE 754 dng 32 bit

Gii:

X = 9.6875(10) = 1001.1011(2) = 1.0011011 x 23

Ta c:

S = 0 v y l s dng

E = e 127 nn e = 127 + 3 = 130(10) = 1000 0010(2)

m = 001101100...00 (23 bit)

X = 0100 0001 0001 1011 0000 0000 0000 0000

V d

1. Biu din cc s thc sau di dng chunIEEE 754 32 bit

a. X = 0.75

b. Y = -27.0625

2. Xc nh gi tr ca cc s thc c biu dindi dng IEEE 754 32 bit

a. X = 1111 0000 1110 0100 0000 0000 0000 0000

b. Y = 0000 1111 0001 1000 0100 0000 0000 0000

139 140

Cc quy c c bit

Nu tt c cc bit ca e u bng 0, cc bit ca m u bng 0, th X = 0

Nu tt c cc bit ca e u bng 1, cc bit ca m u bng 0, th X =

Nu tt c cc bit ca e u bng 1, m c t nht mt bit bng 1, th X khng phi l s (not a number - NaN)

36

141

Trc s biu din

Dng 32 bit: a = 2-127 10-38 b = 2+127 10+38

Dng 64 bit: a = 2-1023 10-308 b = 2+1023 10+308

Dng 80 bit: a = 2-16383 10-4932 b = 2+16383 10+4932

Ni dung







1.3.4. Biu din k t

142

143

a. Nguyn tc chung

Cc k t cng cn c chuyn i thnh chui bit nh phn gi l m k t.

S bit dng cho mi k t theo cc m khc nhau l khc nhau.

VD: B m ASCII dng 8 bit cho 1 k t.

B m Unicode dng 16 bit.

144

a. B m ASCII

Do ANSI (American National Standard Institute) thit k

ASCII l b m c dng trao i thng tin chun ca M. Lc u ch dng 7 bit (128 k t) sau m rng cho 8 bit v c th biu din 256 k t khc nhau trong my tnh

B m 8 bit m ha c cho 28 = 256 k t, c m t 00(16) FF(16), bao gm: 128 k t chun c m t 00(16) 7F(16) 128 k t m rng c m t 80(16) FF(16)

37

145

i. K t chun B m ASCII

95 k t hin th c: C m t 20(16) 7E(16) 26 ch ci hoa Latin 'A' 'Z' c m t 41(16) 5A(16) 26 ch ci thng Latin 'a' 'z' c m t 61(16)

7A(16) 10 ch s thp phn '0' '9' c m t 30(16) 39(16) Cc du cu: . , ? ! : ;

Cc du php ton: + - * /

Mt s k t thng dng: #, $, &, @, ...

Du cch (m l 20(16))

33 m iu khin: m t 0016 1F16 v 7F16dng m ha cho cc chc nng iu khin

146

147

K t iu khin nh dng

BS Backspace - Li li mt v tr: K t iu khin con tr li li mt v tr.

HT Horizontal Tab - Tab ngang: K t iu khin con tr dch tip mt khong nh trc.

LF Line Feed - Xung mt dng: K t iu khin con tr chuyn xung dng di.

VT Vertical Tab - Tab ng: K t iu khin con tr chuyn qua mt s dng nh trc.

FF Form Feed - y sang u trang: K t iu khin con tr di chuyn xung u trang tip theo.

CR Carriage Return - V u dng: K t iu khin con tr di chuyn v u dng hin hnh.

148

K t iu khin truyn s liu

SOH Start of Heading - Bt u tiu : K t nh du bt u phn thng tin tiu .

STX Start of Text - Bt u vn bn: K t nh du bt u khi d liu vn bn v cng chnh l kt thc phn thng tin tiu .

ETX End of Text - Kt thc vn bn: K t nh du kt thc khi d liu vn bn c bt u bng STX.

EOT End of Transmission - Kt thc truyn: Ch ra cho bn thu bit kt thc truyn.

ENQ Enquiry - Hi: Tn hiu yu cu p ng t mt my xa.

ACK Acknowledge - Bo nhn: K t c pht ra t pha thu bo cho pha pht bit rng d liu c nhn thnh cng.

NAK Negative Aknowledge - Bo ph nhn: K t c pht ra t pha thu bo cho pha pht bit rng vic nhn d liu khng thnh cng.

SYN Synchronous / Idle - ng b ha: c s dng bi h thng truyn ng b ng b ho qu trnh truyn d liu.

ETB End of Transmission Block - Kt thc khi truyn: Ch ra kt thc khi d liu c truyn.

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149

K t iu khin phn cch thng tin

FS File Separator - K hiu phn cch tp tin: nh du ranh gii gia cc tp tin.

GS Group Separator - K hiu phn cch nhm: nh du ranh gii gia cc nhm tin (tp hp cc bn ghi).

RS Record Separator - K hiu phn cch bn ghi: nh du ranh gii gia cc bn ghi.

US Unit Separator - K hiu phn cch n v: nh du ranh gii gia cc phn ca bn ghi.

150

Cc k t iu khin khc

NUL Null - K t rng: c s dng in khong trng khi khng c d liu.

BEL Bell - Chung: c s dng pht ra ting bp khi cn gi s ch ca con ngi.

SO Shift Out - Dch ra: Ch ra rng cc m tip theo s nm ngoi tp k t chun cho n khi gp k t SI.

SI Shift In - Dch vo: Ch ra rng cc m tip theo s nm trong tp k t chun.

DLE Data Link Escape - Thot lin kt d liu: K t s thay i ngha ca mt hoc nhiu k t lin tip sau .

DC1

DC4

Device Control - iu khin thit b : Cc k t dng iu khin cc thit b ph tr.

CAN Cancel - Hy b: Ch ra rng mt s k t nm trc n cn phi b qua.

EM End of Medium - Kt thc phng tin: Ch ra k t ngay trc n l k t cui cng c tc dng vi phng tin vt l.

SUB Substitute - Thay th: c thay th cho k t no c xc nh l b li.

ESC Escape - Thot: K t c dng cung cp cc m m rng bng cch kt hp vi k t sau .

DEL Delete - Xa: Dng xa cc k t khng mong mun.

151

b. K t m rng B m ASCII

c nh ngha bi:

Nh ch to my tnh

Ngi pht trin phn mm

V d:

B m k t m rng ca IBM: c dng trn my tnh IBM-PC.

B m k t m rng ca Apple: c dng trn my tnh Macintosh.

Cc nh pht trin phn mm ting Vit cng thay i phn ny m ho cho cc k t ring ca ch Vit, v d nh b m TCVN 5712.

152

c. B m Unicode

Do cc hng my tnh hng u thit k

L b m 16-bit, Vy s k t c th biu din (m ho) l 216

c thit k cho a ngn ng, trong c ting Vit

39

Tho lun

153

01-Thong Tin Va Bieu Dien Thong Tin (1)

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