Chng 3Biu din d liu v s hc my tnh3.1 Cc h m c bn 3.2 M ho v lu
tr d liu trong my tnh 3.3 Biu din s nguyn 3.4 S hc nh phn 3.5 Biu
din s du chm ng 3.6 Biu din k t
GV: inh ng Lng
3.1 Cc h m c bn H thp phn (Decimal System): con ngi s dng
H nh phn (Binary System): my tnh s dng H thp lc phn (Hexadecimal
System): dng biu din rt ngn s hc nh phn Cch chuyn i gia cc h m.
GV: inh ng Lng
H thp phn (decimal)B k t c s gm 10 s: 09 Dng tng qut:
an-1an-2an-3a1a0,a-1 a-2a-mA=V d: 123,45 Phn nguyn : 123 : 10 = 12
d 3 12 : 10 = 1 d 2 1 Phn phn : : 10 = 0 d 1 45 123 =123,45i= m
a *10i
n 1
i
Trong (ai = 09).
0,45*10 = 4,5 0,5 *10 = 5
GV: inh ng Lng
H nh phn (Binary) H thp lc phn (Hexadecimal)H nh phn(Binary) B k
t c s gm 2 s: 0,1 Dng tng qut: an-1an-2an-3a1a0,a-1 a-2a-m
A=
i= m
a i * 2i
n 1
(ai = 0,1)
V d: 11011,0112 = 24+23+21+20+2-2+2-3 =27,375 Thp lc phn
(hexadecimal) B k t c s: 09,AF Dng tngnqut: an-1an-2an-3a1a0,a-1
a-2a-m 1
A=
V d: 89ABH = 1000 1001 1010 1011B.GV: inh ng Lng
i= m
ai *16i
(ai = 0..9, A..F )
3.2 M ho v lu tr trong my tnhNguyn tc chung v m ho d liu Mi d
liu c a vo my tnh c m ho 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 M ho d
liu nhn to D liu s nguyn: m ho theo mt s chun qui c D liu s thc: m
ho bng s du chm ng D liu phi s (k t): m ho theo cc b m k t hin hnh
nh : ASCII, Unicode,GV: inh ng Lng
M hnh m ho v ti to tn hiu vt lB cm bin tn hiu (Sensor) B chuyn i
tng t => s (ADC)
T/h vl
My tnh
T/h vl
B ti to tn hiu
B chuyn i s=> tng t (ADC
Cc d liu vt l thng dng m thanh
Hnh nhGV: inh ng Lng
Th t lu tr cc byte d liu MTB nh chnh t chc lu tr d liu theo n v
byte di t d liu c th chim t 1 n 4 byte. V vy cn phi bit th t chng
lu tr trong b nh chnh i cc d liu nhiu byte. C hai cch lu tr c a ra
Little Endian (u nh): Byte c ngha thp hn c lu tr trong b nh v tr c
a ch nh hn. Big Endian (u to): Byte c ngha thp hn c lu tr trong b
nh v tr c a ch ln hn.
GV: inh ng Lng
Th t lu tr cc byte d liu MTV d: lu tr mt t 32bit 0001 1010 0010
1011 0011 1100 0100 1101B 1 A 2 B 3 C 4 DH Biu din trong ngn nh
theo 2 cch300 301 302 303 4D 3C 2B 1A 300 301 302 303 1A 2B 3C
4D
Little EndianGV: inh ng Lng
Big Endian
Th t lu tr cc byte d liu MT Lu tr ca cc b vi x l in hnh Loi my
Intel: 80x86, Petium -> little endian Motorola 680x0 v cc b x l
RISC -> big endian Power PC & Itanium: tch hp c hai cch
trn
GV: inh ng Lng
3.3. Biu din s nguynMy tnh biu din s nguyn chia thnh 2 loi Biu
din s nguyn khng du (unsign integer) Biu din s nguyn c du (sign
integer) S nguyn khng du: Gi s dng n bit biu din s nguyn khng
du-> di m n bit biu din c t 0 -> 2n-1. Gi tr ca s nguyn c
tnh: n 1 i
ai= 0
i
*2
GV: inh ng Lng
Di min tr ca s nguyn khng du c biu bng hnh trn Gi tr nh nht bng
0 Gi tr ln nht bng 2n-1
V d: S nguyn khng du V d: n=8 n=16 n=32 028-1 (255) 0 216-1
(65535). 0232-1
GV: inh ng Lng
S nguyn c du S b mt v s b hai N: Cho mt s nh phn N c biu din bi
n bit. Ta c S b mt ca N bng (2n-1)-N S b hai ca N bng 2n-N V d: Cho
s N = 0001 00012 c biu din bi n=8bit. Xc nh s b 1 v b 2 ca N. Ap
dng cng thc 1111 1111 (2n-1) 0001 0001 N s b mt ca N 1110 1110GV:
inh ng Lng
S nguyn c du Nhn xt: s b mt ca mt s N c xc nh bng cch o cc bit
trong N Ap dng cng thc 1 0000 0000 (2n) 0001 0001 N s b hai ca N
1110 1111 Nhn xt: s b hai ca mt s N c xc nh bng cch ly s b mt ca N
cng thm 1 S b 2 ca N =(s b 1 ca N)+1
GV: inh ng Lng
S nguyn c duGi s dng n bit biu din s nguyn c du-> di m n bit
biu din c t (- 2n-1 ..-1,0 .. 2n-1-1). Gi tr ca s nguyn c tnh theo
2 phn ring bit: Phn gi tr dng (0 -> 2n-1-1). Phn gi tr m (-
2n-1-1). Di min tr ca s nguyn c du c biu bng hnh trn Gi tr nh nht
bng - 2n-1 Gi tr ln nht bng +2n-1-1GV: inh ng Lng
S nguyn c duTrong : Bt c trng s cao nht (hay bit ngoi cng bn tri
ca dy nh c my tnh s dng biu din du ca gi tr) nu: = 0 : th s nh phn
cn tnh gi tr l s dng. Dng tng qut l: 0an-2an-3a0 = 1 : th s nh phn
cn tnh gi tr l s m. Dng tng qut l: 1an-2an-3a0 2 n 1 +GV: inh ng
Lng
Gi tr tnh
ai * 2i i =0
n 2
V d 1V d 1: Cho
s nguyn c du biu din n=8bit sau: A=B5H v B=6AH
Hy xc nh gi tr ca hai s nguyn c du A v B di dng h s ngi s dng Bi
gii Biu din s nguyn A di dng nh phn A=B5H = 1011 01012 =>A= -128
+ 53 = - 75 Biu din s nguyn B di dng nh phn B=6AH = 0110 10102
=> B = 64+32+8+2 = 106GV: inh ng Lng
V d 2V d 2: Biu din s nguyn c du sau y A=+97 v B=-101 theo hai
dng kiu n=8bit v n=16bit trong my tnh. Li gii Biu din s A dng s
nguyn c du trong my tnh A = 0110 00012 (n=8bit) Biu din s B dng s
nguyn c du trong my tnh Biu din s +101 = 0110 01012 Ly b 2GV: inh
ng Lng
1001 10112
=> B = - 101 = 1001 10112
V d 2 Biu din s A dng s nguyn c du trong my tnh A = 0000 0000
0110 00012 (n=16bit) Biu din s B dng s nguyn c du trong my tnh Biu
din s +101 =0000 0000 0110 01012 Ly b 2 1111 1111 1001 10112 =>
B =-101 = 1111 1111 1001 10112
GV: inh ng Lng
Biu din s nguyn theo m BCD(Binary Coded Decimal Code)
Dng nhm 4 bit m ho mi s nh phn 0 -> 0000 6 -> 0110 1 ->
0001 7 -> 0111 2 -> 0010 8 -> 1000 3 -> 0011 9 ->
1001 4 -> 0100 5 -> 0101 C 6 t hp khng c s dng l: 1010, 1011,
1100, 1101, 1110, 1111.GV: inh ng Lng
Biu din s nguyn theo m BCDBiu din s38 = 0011 1000BCD 61 = 0110
0001BCD
Thc hin php cng:
3861 99 275 816 1091GV: inh ng Lng
0011 1000BCD0110 0001BCD 1001 1001BCD 0010 0111 0101BCD 1000
0001 0110BCD 1010 1000 1011BCD => sai
Biu din s nguyn theo m BCD Chnh li cho ng: Cng thm 6(0110) vo cc
ct c gi tr ln hn 9 c nh. Php ton trn ta cng thm vo ct 1 v ct 3 tnh
li ta c 001BCD Ta uc 1091 275 816 1091 0010 0111 0101BCD 1000 0001
0110BCD 1010 1000 1011BCD => sai 0110 0000 0110 1 0000 1001 0001
=> ngGV: inh ng Lng
3.4 S hc nh phn Php cng s nguyn khng du Khi cng hai s nguyn khng
du n bit Nu khng c nh ra khi bit cao nht th kt qu nhn c lun ng. Nu
c nh ra khi bit cao nht th kt qu nhn c l sai. Hay ni khc hn php cng
b trn(Cout =1) Trn c nh (Carry Out) xy ra khi kt qu php ton nhn c
> 2n-1Y(n-bit) X(n-bit)
cin
B cng n bitS(n-bit)
cout
GV: inh ng Lng
3.4 S hc nh phnCng s nguyn c du. Khi cng hai s nguyn c du n-bit
-> Kt qu nhn c ta khng quan tm n c mang Cout. M ta ch nhn kt qu
tnh c trong n-bit. V vy cng hai s khc du th cho kt qu lun lun ng.
Nu cng hai s cng du nu kt qu tr v cng du ton hng thi kt qu ng. Ngc
li kt qu khc du ton hng th xy ra hin tng trn (Overflow) kt qu
sai.
GV: inh ng Lng
3.4 S hc nh phn Php tr Php tr s nguyn chnh l php cng vi s o du.
V nh X-Y = X+(-Y) Tm s o chnh l ta tm b hai ca s . Nguyn tc php tr:
Ly b hai ca Y ra thnh Y ri cng vi XGV: inh ng Lng
Y (n-bit)
X(n-bit)
B hai-Y
B cng n-bit
S=X-Y
3.4 S hc nh phn Nhn s nguyn khng du 1011 S b nhn 1001 S nhn 1011
0000 Tch ring phn 0000 1011 1100011 (99) -> Tch
GV: inh ng Lng
3.4 S hc nh phn Nu bit ca s nhn l 0 => Tch ring phn bng 0 Nu
bit ca s nhn l 1 => Tch ring phn l gi tr s b nhn Tch ring phn
tip theo c dch tri 1 bit so vi tch ring phn pha trc . Tch bng tng
cc tch ring phn. Nhn hai s nguyn n bit. Tch c di lun chun b 2n bit.
V vy php nhn khng c khi nim trn.
GV: inh ng Lng
3.4 S hc nh phn B nhn s nguyn khng duMn-1 Mn-2 * * * M1 M0
S b nhn
B cng n bit
K cng
B iu khin cng v dch
K dch phi
C
An-1 An-2
*
* *
A1
A0
Qn-1 Qn-2
*
* *
Q1
Q0
S nhn
GV: inh ng Lng
3.4 S hc nh phnSTART
C,A
