01-thong tin va bieu dien thong tin (1)
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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. Thng tin v Tin hc
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
1.2. Biu din s trong h m
1.3. Biu din d liu trong my tnh
3
Ni dung
1.1. Thng tin v Tin hc
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
1.2. Biu din s trong h m
1.3. Biu din d liu trong my tnh
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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.
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X l d liu (2) Ni dung
1.1. Thng tin v Tin hc
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
1.2. Biu din s trong h m
1.3. Biu din d liu trong my tnh
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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.
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My tnh in t c mt khp ni
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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.
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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
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OFF ON
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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
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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)
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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)
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S
u
p
e
r
C
o
m
p
u
t
e
r
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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)
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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
...
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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
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c. Cc th h my tnh
S pht trin v cng ngh S pht trin v my tnh
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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
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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
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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)
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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
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Nh hn,
R hn,
Hiu qu hn
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IBM 360
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Thit k trn cng ngh IC
Tc tnh ton: 1000 t php ton trong 1 giy
iv. Th h th t (1974 nay)
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Microprocessor = Central Processing Unit (CPU) thit k trong 1 vi mch n
1971 : Intel 4004
Vi x l (Microprocessor)
1975 Altair 8800
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My tnh c nhn u tin Altair 8800
Vi x l (Microprocessor)
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1981 IBM PC
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Th h my tnh c nhn mi vi kin trc m IBM
1984 Apple Macintos
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1990 - Personal Computers
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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
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Pentium
Th h th t (tip)
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More Pentium
Pro
III
IV
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Itanium
64-bit Intel
Microprocessors
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Th h th t (tip)
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Th h th t (tip)
N
e
t
w
o
r
k
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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
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Ni dung
1.1. Thng tin v Tin hc
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
1.2. Biu din s trong h m
1.3. Biu din d liu trong my tnh
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1.1.3. Tin hc v cc ngnh lin quan
Tin hc (Computer Science/Informatics)
Cng ngh thng tin (Information Technology - IT)
Cng ngh thng tin v truyn thng (Information and Communication Technology ICT).
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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)
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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
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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
1.1. Thng tin v Tin hc
1.2. Biu din s trong h m
1.2.1. H m
1.2.2. Chuyn i c s
1.2.3. i s Boolean
1.3. Biu din d liu trong my tnh
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Ni dung
1.1. Thng tin v Tin hc
1.2. Biu din s trong h m
1.2.1. H m
1.2.2. Chuyn i c s
1.2.3. i s Boolean
1.3. Biu din d liu trong my tnh
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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
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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)
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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
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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)
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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)
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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:
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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
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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)
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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)
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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)
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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
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1 1 0 0
- 1 1 1
--------------------
0 1 0 1
Tnh ton trong h nh phn V d
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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)
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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
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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)
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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
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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)
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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
1.1. Thng tin v Tin hc
1.2. Biu din s trong h m
1.2.1. H m
1.2.2. Chuyn i c s
1.2.3. i s Boolean
1.3. Biu din d liu trong my tnh
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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
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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)
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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)
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b. Chuyn i phn thp phn (2)
V d: Cch chuyn phn thp phn ca s 12.6875(10) sang h nh phn:
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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)
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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
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Ni dung
1.1. Thng tin v Tin hc
1.2. Biu din s trong h m
1.2.1. H m
1.2.2. Chuyn i c s
1.2.3. i s Boolean
1.3. Biu din d liu trong my tnh
83 84
i s Boolean
a NOT a
0 1
1 0
Cc php ton logic vi tng bit nh phn:
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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:
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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
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Php quay
Quay tri khng nh
89
Quay phi khng nh
Quay tri c nh
Quay phi c nh
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
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
1.1. Thng tin v Tin hc
1.2. Biu din s trong h m
1.3. Biu din d liu trong my tnh
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
91 92
1.3.1. Nguyn l chung
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,
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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
1.1. Thng tin v Tin hc
1.2. Biu din s trong h m
1.3. Biu din d liu trong my tnh
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
96
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1.3.2. Biu din s nguyn
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
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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
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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
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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:
v. Tnh ton s hc vi s nguyn (3)
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
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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
v. Tnh ton s hc vi s nguyn (4)
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)
v. Tnh ton s hc vi s nguyn (5)
Cha hai s nguyn khng du
123
v. Tnh ton s hc vi s nguyn (6)
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
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32
v. Tnh ton s hc vi s nguyn (7)
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
1.1. Thng tin v Tin hc
1.2. Biu din s trong h m
1.3. Biu din d liu trong my tnh
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
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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33
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
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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
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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)
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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
1.1. Thng tin v Tin hc
1.2. Biu din s trong h m
1.3. Biu din d liu trong my tnh
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
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)
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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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38
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
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39
Tho lun
153
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