giao trinh ky thuat so
TRANSCRIPT
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MC LC H THNG S M.......................................................................................................3 1.1 H thng s : ................................................................................................................3 1.1.1.H thng s m thp phn (Decimal) : ................................................................4 1.1.2. H thng s nh phn (Binary) :.............................................................................4 1.1.3. H thng s bt phn (Octal) : ...............................................................................5 1.1.4. H thng s thp lc phn (Hexa-decimal) : ........................................................5 1.2. Chuyn i c s : ......................................................................................................6 1.2.1. Chuyn i c s nh phn, bt phn, thp lc phn sang c s thp phn: ....6 1.2.2. Chuyn i c s t nh phn sang thp lc phn : .............................................6 1.2.3. Chuyn i c s t nh phn sang bt phn : .....................................................7 1.2.4. Chuyn i c s t bt phn sang nh phn : ....................................................8 1.2.5. Chuyn i c s t thp lc phn sang nh phn : .............................................8 1.2.6. Chuyn i c s thp phn sang c s nh phn, bt phn, thp lc phn : ...8 1.2.3. Cc php ton nh phn :......................................................................................10 1.2.3.1. Php ton cng : .................................................................................................10 1.2.3.2. Php ton tr : ....................................................................................................11 1.2.3.3. Php ton nhn :.................................................................................................11 1.2.3.4. Php ton chia : ..................................................................................................12 1.3. M nh phn :............................................................................................................12 1.3.1. M BCD (Binary Coded Decimal) :....................................................................13 1.3.2. M qu 3 : ..............................................................................................................13 1.3.3. M qu 6 : ..............................................................................................................13 1.3.4. M Gray (m vng) :.............................................................................................14 1.3.5. M k t (alphanumeric) : ...................................................................................14 1.3.6.Bng ca mt s m : .............................................................................................15 1.4. S c du _ B 1 _ B 2 : .........................................................................................16 1.4.1. S c du (signed number) : .................................................................................16 1.4.2.1. H thng s nh phn (b 1) : ............................................................................16 1.4.2.2. H thng s bt phn (b 7) : ............................................................................17 1.4.2.3. H thng s thp phn (b 9) :..........................................................................17 1.4.2.4. H thng s thp lc phn (b 15) :..................................................................17 1.4.3. S b 2: ...................................................................................................................17 BI TP CHNG 1....................................................................................................18 I S BOOLE V CNG LOGIC ...........................................................................19 2.1. i s Boole :.............................................................................................................19 2.1.1. nh ngha :............................................................................................................19 2.1.2. Cc php ton (gm c ba php ton c bn) : ..................................................19 2.1.3. Cc cng thc v nh l : ....................................................................................20 2.1.4. Chng minh cc cng thc :.................................................................................22 2.1.5. Ba quy tc v ng thc : .....................................................................................24 2.2. Hm Boole :...............................................................................................................25 2.2.1.Bng gi tr (hay cn gi l bng s tht,bng chn l-Truth table) ................25 2.2.2. Biu thc hm s : .................................................................................................26 2.2.3. Ba Karnaugh : ......................................................................................................26 2.2.4. S mch logic :..................................................................................................26
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2.3. Cc dng chun ca hm Boole : ............................................................................27 2.3.1. Dng chun 1 : (tng cc Mintern - tch chun)....................................................27 2.4. Cc cng logic : .........................................................................................................29 2.4.1.Cng khng o:.....................................................................................................29 2.4.3.Cng AND (AND gate)...........................................................................................30 2.4.4.Cng OR (OR gate) ................................................................................................31 2.4.5.Cng NAND (AND +NOT) ....................................................................................32 2.4.6.Cng NOR (OR +NOT) .........................................................................................33 2.4.7.Cng EXOR (EX-OR gate)...................................................................................34 2.4.8.Cng EXNOR (EX_NOR gate) ............................................................................35 2.5.1.Ba K 2 bin.............................................................................................................37 2.5.2.Ba K 3 bin.............................................................................................................38 2.5.3.Ba K 4 bin: ...........................................................................................................38 2.5.4.Ba K 5 bin: ...........................................................................................................39 2.6.n gin ha hm Boole:..........................................................................................39 2.6.1.Phng php i s :S dng cc cng thc ,cc tin v nh l rt gn....39 2.6.1.Phng php rt gn bng ba K ........................................................................39 2.7.Ty nh (dont care).Thng k hiu l d(v tr ca ) ........................................42 2.7.1.n gin ha theo dng chun 2...........................................................................42 2.7.2.Mt s phng php thc hin hm Boole bng cc cu trc cho trc .........43 BI TP CHNG 2.....................................................................................................45 MCH T HP ..............................................................................................................48 3.1.Gii thiu: ..................................................................................................................48 3.1.1.Mch gii m (Decoder).........................................................................................48 3.1.2.Mch m ha (Encoder): .......................................................................................54 3.1.3.Mch dn knh (Multiplexer)-MUX ....................................................................55 3.1.4.Mch phn knh (DeMultiplexer)-DEMUX ........................................................61 3.1.5.Mch kim tra chn l: ..........................................................................................62 3.2.Phng php thit k mt s mch t hp .............................................................64 BI TP CHNG 3.....................................................................................................69 H TUN T ..................................................................................................................70 4.1.Cc mch cht v FF: ...............................................................................................70 1.Cht:(Latch)..................................................................................................................70 4.1.2.Flip-Flop (FF) .........................................................................................................71 4.2. B m: ( Counter )..................................................................................................76 4.2.1.Phn loi:.................................................................................................................76 4.2.2.Mch m ni tip (khng ng b):....................................................................76 4.2.3.Mch m song song (ng b):............................................................................79 4.2.4.Mch chia tn dng FF ..........................................................................................83 4.3.Nhp data vo FF ......................................................................................................84 4.3.1.Nhp khng ng b ..............................................................................................84 4.3.2.Nhp ng b ..........................................................................................................85 4.4.H ghi dch (Shift regiter) .........................................................................................85 BI TP CHNG 4.....................................................................................................88
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Chng 1
H THNG S M
1.1 H thng s :
H thng s thng s dng l h thng s c v tr. Trong mt h thng nh vy
mt s biu din bng mt chui cc k t s (digit); mi v tr ca k t s
s c mt trng s nht nh.
Trng s y chnh l c ly tha v tr ca k t s trong chui.
C s chnh l s k t s c dng biu din trong mt h thng.
Cc h thng s thng gp l h thng s thp phn (Decimal system), h thng
s nh phn (Binary system), h thng s bt phn (Octal system), h thng
s thp lc phn (Hexa-decimal) v.vGi tr thp phn ca mt s c tnh theo
cng thc sau :
Trong :
- G : l gi tr.
- t : v tr ca k t s ng trc du ngn cch thp phn (0, 1, 2, 3, ).
- n : s k t s ng trc du ngn cch thp phn ca s tr i 1.
- C : c s.
- A : k t s.
- t : v tr ca k t s ng sau du ngn cch thp phn ( -1, -2, -3, ).
- m : s k t s ng sau du ngn cch thp phn ca1 s.
Trong cc h thng s ngi ta thng quan tm n s c ngha cao nht (s c
trng s ln nht) k hiu l MSB (Most Significant Bit) v s c ngha thp
nht (s c trng s nh nht) k hiu l LSB (Less Significant Bit)
t
m
t
tt
n
t
t ACACG =
=
+= 10
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V d :
10010[2] MSD : Most Significant Digit
MSB LSB
1998[10] LSD : Less Significant Digit
1.1.1.H thng s m thp phn (Decimal) :
K t s : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
C s : 10
V d : V tr
3 2 1 0
1 9 9 9[10] = 1.103 + 9.10
2 + 9.101 + 9.100
= 1000 + 900 + 90 + 9
0 -1 -2
1 , 2 5[10] = 1.100 + 2.10-1 + 5.10-2
= 1,00 + 0,2 + 0,05
1.1.2. H thng s nh phn (Binary) :
K t s : 0, 1
C s : 2
Mi con s trong s nh phn (0 hoc 1) c gi l mt bit (vit tt ca binary digit). Cc n v khc :
Tn gi Vit tt Gi tr
Byte B 8 bit
Kilo Byte KB 1024 byte = 210 B
Mega Byte MB 1024 KB = 220 B
Giga GB 1024 MB = 230 B
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V d : V tr
4 3 2 1 0
1 0 1 0 1[2] = 1.24 + 0.23 + 1.22 + 0.21 + 1.20
= 16 + 0 + 4 + 0 + 1 = 21[10]
(S nh phn trn c 5 bit)
1 0 -1 -2 -3
1 1 , 1 0 1[2] = 1.21 + 1.20 + 1.2-1 + 0.2-2 + 1.2-3
= 2 + 1 + 0,5 + 0 + 0,125 = 3,625[10]
(S nh phn trn c 5 bit)
Nhn xt : - Nu bit cui cng l 0 s nh phn l s chn. - Nu bit cui cng l 1 s nh phn l s l.
1.1.3. H thng s bt phn (Octal) :
K t s : 0, 1, 2, 3, 4, 5, 6, 7
C s : 8
V d : V tr
1 0
4 6[8] = 4.81 + 6.80 = 32 + 6 = 38[10]
0 -1 -2
2 , 3 7[ 8] = 2.80 + 3.8-1 + 7.8-2
= 2 + 3.0,125 + 7.0,02 = 2,515[10]
1.1.4. H thng s thp lc phn (Hexa-decimal) :
K t s : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F
C s : 16
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V d :
V tr 1 0 2 E[16] = 2.16
1 + 14.160 = 46[10] 3 2 1 0 -1
0 1 2 C , D[16] = 0.163 + 1.162 + 2.161 + 12.160 + 13.16-1
= 0 + 256 + 32 + 12 + 0,0625
= 300,0625[10]
Ghi ch : Nu s haxa-decimal bt u bng ch th khi vit phi thm s 0 vo
trc (Vd : EF 0EF).
1.2. Chuyn i c s :
1.2.1. Chuyn i c s nh phn, bt phn, thp lc phn sang c s thp
phn:
Nguyn tc : ly mi s hng trong chui s nhn vi c s ly tha v tr ca n
sau ly tng tt c kt qu (cc v d trn).
1.2.2. Chuyn i c s t nh phn sang thp lc phn :
Nguyn tc : Nhm t phi qua tri bn s (bn bit); nhm cui cng nu thiu
th ta c thm cc s 0 vo. Thay th cc nhm 4 bit thnh cc m thp lc phn
tng ng.
V d :
0010 1101[2] = 2D[16] 1100 1010 1111 1110 1101 1010[2] = 0CAFEDA[16]
2 D C A F E D A
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Bng m thp lc phn :
Thp phn Nh phn Thp lc phn
0 0000 0
1 0001 1
2 0010 2
3 0011 3
4 0100 4
5 0101 5
6 0110 6
7 0111 7
8 1000 8
9 1001 9
10 1010 A
11 1011 B
12 1100 C
13 1101 D
14 1110 E
15 1111 F
1.2.3. Chuyn i c s t nh phn sang bt phn :
Nguyn tc : Nhm t phi qua tri ba s (ba bit); nhm cui cng nu thiu th ta c thm cc s 0 vo. Thay th cc nhm ba bit thnh cc m thp lc phn tng ng. V d :
100 111 010[2] = 472[8] 001 000[2] = 10[8] 4 7 2 1 0
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1.2.4. Chuyn i c s t bt phn sang nh phn :
Nguyn tc : Thay th mt k t s bng mt s nh phn ba bit tng ng theo bng sau.
Bt
phn
0 1 2 3 4 5 6 7
Nh
phn
000 001 010 011 100 101 110 111
V d :
3 4 5[8] = 11100101[2] 1 3 7[8] = 1011111[2]
011 100 101 001 011 111
1.2.5. Chuyn i c s t thp lc phn sang nh phn :
Nguyn tc : Thay th mt k t s bng mt s nh phn bn bit tng ng.
V d :
2 F E (H) = 1011111110[2] 0010 1111 1110
1.2.6. Chuyn i c s thp phn sang c s nh phn, bt phn, thp lc
phn :
Chia lm hai phn : phn nguyn (phn N) v phn thp phn (phn L).
* Phn nguyn N :
- Ly N chia cho c s (2 hoc 8 hoc 16), thng s l N0, s d l n0.
- Ly N0 chia cho c s (2 hoc 8 hoc 16), thng s l N1, s d l n1.
- Ly N1 chia cho c s (2 hoc 8 hoc 16), thng s l N2, s d l n2.
. . . . .
- Tip tc chia cho n khi thng s Ni = 0, s d l ni .Khi s N biu din
dng nh phn l :
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(Cc s d c ly theo th t t di ln)
V d 1 :
64[10] = ?[2] 35[10] = ?[2]
64 2 35 2 0 32 2 1 17 2 0 16 2 1 8 2 0 8 2 0 4 2 0 4 2 0 2 2 0 2 2 0 1 2 0 1 2 1 0 1 0 = 1000000[2] = 100011[2]
V du 2 :
1997[10] 16 = 7CD[16] 423[10] 16 = 1A7[16]
13 124 16 7 26 16 12 7 16 10 1 16 7 0 1 0
V d 3 :
266[10] 8 = 412[8] 1999[10] 8 = 3717[8] 2 33 8 7 249 8
1 4 8 1 31 8 4 0 7 3 8 3 0
* Phn thp phn L :
- Ly phn L nhn c s thnh l L c phn nguyn l d1, phn thp phn l
L1.
- Ly phn L1 nhn c s thnh l L1 c phn nguyn l d2, phn thp phn l
L2.
- Ly phn L2 nhn c s thnh l L2 c phn nguyn l d3, phn thp phn l
L3.
N[2] = ni ni-1 n2 n1 n0
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. . . . . .
- Tip tc cho n khi phn thp phn ca tch s bng 0 hay t c s l cn
thit.
Khi phn l s l :
V d 1 : L[10] = 0.6875 L[2]
_ 0.6875 x 2 = 1.3750 (L) d1 = 1; L1 = 0.3750
_ 0.3750 x 2 = 0.750 (L1) d2 = 0; L2 = 0.750
_ 0.750 x 2 = 1.50 (L2) d3 = 1; L3 = 0.50
_ 0.50 x 2 = 1.0 (L3) d4 = 1; L4 = 0
L[2] = 0.1011
V d 2 : L[10] = 0.6875 L[8]
_ 0.6875 x 8 = 5.5 (L) d1 = 5; L1 = 0.5
_ 0.5 x 8 = 4.0(L1) d2 = 4; L2 = 0
L[8] = 0.54
V d 3 : L[10] = 0.6875 L[16]
_ 0.6875 x 16 = 11 (L) d1 = B; L1 = 0
L[16] = 0.B
1.2.3. Cc php ton nh phn :
Cng nh s hc thp phn, s hc nh phn cng c bn php tnh c bn l :
Cng (+), Tr (-), Nhn (*), Chia (/) .
1.2.3.1. Php ton cng :
Nguyn tc : 0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0 nh 1 (carry)
L[2] = d1 d2 d3 d4 dk
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V d :
100110 1010110 1001010 + 001 + 1000101 + 1010010
100111 10011011 10011100 1.2.3.2. Php ton tr :
Nguyn tc : 0 0 = 0
0 1 = 1 mn 1 (borrow)
1 0 = 1
1 1 = 0
V d :
1111 1000
- 0110 - 0011 1001 0101
1.2.3.3. Php ton nhn :
Nguyn tc : 0 x 0 = 0
0 x 1 = 0
1 x 0 = 0
1 x 1 = 1
V d :
1 0 1 0 1 0 0 0 1 x 1 0 1 x 1 0 0 0 1 0 1 0 1 0 0 0 1 0 0 0
0 0 0 0 1 0 1 0 1 1 0 0 1 0
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1.2.3.4. Php ton chia :
V d : 101000 [2] / 11 = ?; 1010[2] / 101 = ?; 111111[2] / 110 = ?
1 0 1 0 0 0 11 - 1 1 1101
0 1 0 0 - 1 1 0 0 1 0 Thng s - 0 0 1 0 0 - 1 1 0 0 1 S d
1010 101 111111 110 - 101 10 - 110000 1010
00 001110 - 1100 11
Thng thng tnh ton khng b nhm ln ta c th chuyn sang s thp
phn tnh ton ,sau chuyn kt qu sang s nh phn.Tuy nhin trong k
thut in t cng nh trong my tnh vic tnh ton ny hon ton c thc
hin rt n gin ta khng cn phi chuyn i.
V d:1000[2] (8) 0011[2] (3) = 0101[2] (5)
1.3. M nh phn :
M nh phn l mt m s dng h thng nh phn v c sp xp theo mt
cu trc no .
Trong cc my tnh hoc cc mch s lun lm vic h thng nh phn; Cc
thit b xut hay nhp ( hin th) thng lm vic h thng thp phn .V th
cc gi tr thp phn phi c m ha bng cc gi tr nh phn.
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1.3.1. M BCD (Binary Coded Decimal) :
M s BCD l s thp phn m ha theo nh phn. M s ny dng nhm bn
bit biu th s thp phn t 0 n 9.
V d : 1 2 0 (D) 1 9 9 9 (D) 0 0 0 1 0 0 1 0 0 0 0 0(BCD) 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1(BCD)
Lu : M BCD ch c gi tr t 0 cho n 9 nn khi ta chuyn i t m
BCD sang gi tr thp phn cn ch trng hp cm ( khng tn ti m BCD).
V d :
0 0 1 0 1 0 0 0 0 1 1 1 0 1 0 1 1 1 0 0 0 1 0 1
2 8 7 5 (cm) 5
1.3.2. M qu 3 :
M qu 3 (tha 3, d 3) l m c c khi tng 3 n v t Binary.Tc l cng
thm 011[2].
V d :
1.3.3. M qu 6 :
M qu 6 (tha 6, d 6) l m c c khi tng 6 n v t Binary.Tc
cng thm 0110[2]
V d:
10000101* (Qua3)
11001001* (Qua3)
10110101* 6) (Qua10110101*
6) (Qua
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1.3.4. M Gray (m vng) :
M Gray hay cn gi l m vng suy ra t m nh phn. Gi s cho m nh nh
phn c bn bit B3 B2 B1 B0, m Gray tng ng l G3 G2 G1 G0 th c th tnh
theo cng thc sau :
n gin khi i t nh phn sang Gray ,ta cn c t s nh phn theo qui
lut sau : sau bit 0 th gi nguyn, sau bit 1 th i 1 thnh 0 v 0 thnh 1
V d :
1.3.5. M k t (alphanumeric) :
L m biu din cc k t (vd: k t bn phm).
M ASCII : l m m hu ht cc my tnh u dng (M chun ca M
American Standard Code for Information Interchange). Mi k t (ch ci, ch
s , du, k hiu t bit ) tng ng vi mt m 8 bit (l dy lin tip cc
ch s 0 v 1)
V d :
K t M ASCII K t M ASCII
. . . . . . . . . . . . . . . . . . . .
0 00110000 A 01000001
1 00110001 B 01000010
2 00110010 Y 01011001
3 00110011 Z 01011010
4 00110100 . . . . . . . . . .
5 00110101 a 01100001
6 00110100 b 01100010
. . . . . . . . . . . . . . . . . . . .
01100100* (Gray)
iB1iBiG ++=
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B m ASCII c 128 k hiu c m ha :
- 26 ch ci Latin in hoa : A Z.
- 26 ch ci Latin in thng : a z.
- 10 ch s thp phn.
- Cc k t ton hc thng thng : +, -, *, / =, >,
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8 1 0 0 0 1 0 0 0 1 0 1 1 1 1 1 0 1 1 0 0
9 1 0 0 1 1 0 0 1 1 1 0 0 1 1 1 1 1 1 0 1
10 1 0 1 0 1 1 0 1 0 0 0 0 1 1 1 1
11 1 0 1 1 1 1 1 0 0 0 0 1 1 1 1 0
12 1 1 0 0 1 1 1 1 0 0 1 0 1 0 1 0
13 1 1 0 1 0 0 0 0 0 0 1 1 1 0 1 1
14 1 1 1 0 0 0 0 1 0 1 0 0 1 0 0 1
15 1 1 1 1 0 0 1 0 0 1 0 1 1 0 0 0
1.4. S c du _ B 1 _ B 2 :
1.4.1. S c du (signed number) :
Khi biu din s c du thng thng s dng thm 1 bit gi l bit du (thng
t v tr s c trng s cao nht MSB) : bit ny l khng (0) ch s
dng;bit ny l mt (1) ch s m.
V d :
1. 0101 = - 5
Bit du 0. 0101 = + 5
1.4.2. S b 1
S b 1 c nh ngha cho mt s N c n s s bng : r n -1 N (vi r l c
s).
1.4.2.1. H thng s nh phn (b 1) :
Cho s N = 1010 [2] (r = 2; n = 4) rn = 2 4 = 10000.
rn 1 - N = 10000 1 1010 = 1111 1010 = 0101.
Lu : Ta c th tm b 1 ca mt s nh phn n gin bng cch thay 0 1;
1 0.
01011010 ( Bu1)
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1.4.2.2. H thng s bt phn (b 7) :
Cho s N = 234 [8] (r = 8; n = 3) rn = 8 3 = 512=1000[8]
rn 1 - N = 1000 1 234 = 543[8]=355
1.4.2.3. H thng s thp phn (b 9) :
Cho s N = 15249[10] (r = 10; n = 5) rn = 10 5 = 100000.
rn 1 - N = 100000 1 15249 = 99999 15249 = 84750.
1.4.2.4. H thng s thp lc phn (b 15) :
Cho s N = 45[16] (r = 16; n = 2) rn = 16 2 = 256=100[16]
rn 1 - N = 100 1 45 = 0FF 45 = 0BA[16]=186
1.4.3. S b 2:
S b 2 c nh ngha cho mt s N c n s s bng : r n N (vi r l c s).
T nh ngha trn ta c s b 2 chnh l s b 1 cng 1.
V d :
8475015249 ( Bu9)
18645 ( Bu15)
011001011010:phan Nh* (Bu2)(Bu1)[2]
356355234:phan Bat* (Bu8)(Bu7)[8]
18718645:phan luc Thap * (Bu16)(Bu15)[16]
847518475015249:phan Thap* (Bu10)(Bu9)[10]
355234 ( Bu7)
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BI TP CHNG 1
1.Chuyn i t s Binary sang Decimal
a.10110 b.10001101 c.100100001001
d.1111010111 e.10111111 f.101010101010
2. Chuyn i t s Decimal sang Binary
a.37 b.14 c.189 d.205 e.2313
3.Mt s nh phn 8 bit c gi tr thp phn tng ng ln nht l bao nhiu?
4.Chuyn i t s Octal sang decimal
a.743 b.36 c.3777 d.257 e.1204
5. Chuyn i t s Decimal sang Octal
a.59 b.372 c.919 d.65536 e.255
6. Chuyn i sang s Binary cc s t bi 2 n bi 4
7.Chuyn i t s Hex sang Decimal
a.92 b.1A6 c.37FD d.2CO e.7FF
8.Chuyn i t Decimal sang Hex
a.75 b.314 c.2048 d.25619
9.M ha nhng s decimal sau sang m BCD
a.47 b.962 c.187 d.1204
10.Chuyn i t m BCD sang Decimal
a.1001011101010010 b.000110000100
c.0111011101110101 d.010010010010
11.Dch sang m ASCII cc k t sau: CDDTK5=2005
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Chng 2
I S BOOLE V CNG LOGIC
2.1. i s Boole :
2.1.1. nh ngha :
i s Boole (hay cn gi l i s logic do George Boole, nh ton hc ngi
Anh, sng to vo th k XIX) l mt cu trc i s c xy dng trn tp
cc phn t nh phn (Binary) cng vi 2 php ton cng v nhn tha cc iu
kin sau :
a) Kn vi cc php ton cng (+) v nhn (*).Tc l A,B X th:
A+B X v A.B X.
b) i- i vi php cng s c phn t trung ha 0 (ng nht) : x + 0 = x.
ii- i vi php ton nhn s c phn t trung ha 1 ( ng nht) : x *
1 = x.
c) Giao hon :
i- x + y = y + x.
ii- x . y = y . x.
d) Phn b v kt hp :
i- a . (b + c) = (a . b) + (a . c)
ii- a + (b . c) = (a + b) .(a + c)
e) Lun lun tn ti mt phn t nghch (b) sao cho :
i- x + x= 1
ii- x. x = 0
+ Ghi ch: Trong chng ny v cc chng sau cc k hiu 0 v 1 l k
hiu cho 2 mc Logic 0 v 1 ch khng phi l k hiu ca s nh phn.Do
cc php ton phi tun th theo nguyn tt ring ca n.
2.1.2. Cc php ton (gm c ba php ton c bn) :
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a) Php cng (OR) :
a b a + b
0
0
1
1
0
1
0
1
0
1
1
1
b) Php nhn (AND) :
a b a . b
0
0
1
1
0
1
0
1
0
0
0
1
c) Php b (NOT) :
a a
0 1
1 0
2.1.3. Cc cng thc v nh l :
a) Quan h gia cc hng s :
Nhng quan h di y gia hai hng s ( 0, 1) lm tin ca i s Boole.
l cc quy tc php ton c bn i vi t duy logic.
Cng thc 1-1: 0 . 0= 0
Cng thc 1-2 1 + 1= 1
Cng thc 2-1: 0 . 1= 0
Cng thc 2-2: 1+ 0 = 1
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Trang 21
Cng thc 3-1: 0+ 0= 0
Cng thc 3-2: 1. 1= 1
Cng thc 4-1: 0 = 1
Cng thc 4-2: 1 = 0
b) Quan h gia bin s v hng s : Cng thc 5-1: x . 1= x
Cng thc 5-2: x + 0 = x
Cng thc 6-1: x . 0 = 0
Cng thc 6-2: x + 1= 1
Cng thc 7-1: 0. =xx
Cng thc 7-2: 1. =+ xx
Bin s y t l x, hai hng s Logic l 0 v 1.
c) Lut giao hon :
Cng thc 8-1: x + y = y+ x
Cng thc 8-2: x . y = y. x
d) Lut kt hp :
Cng thc 9-1: (x . y).z = x.(y. z)
Cng thc 9-2: (x + y) + z = x + (y+ z)
Cng thc 10-1: x . (y + z) = x.y+x.z
Cng thc 10-2: x + y . z = (x+y) . (x +z)
e) Lut phn phi :
Cng thc 11-1: x + x = x
Cng thc 11-2: x . x = x
f) Lut ng nht :
Cng thc 12: xx = g) nh l De_Morgan :
Cng thc 13-1: yxyx .=+
Cng thc 13-2: yxyx +=.
h) nh l hp thu :
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Trang 22
Cng thc 14-1: x + x.y = x
Cng thc 14-2: x . (x+y) = x
Cng thc 15: x+ x .y=x+y
2.1.4. Chng minh cc cng thc :
Mnh i ngu:
Trong cu trc i s Boole ,mt mnh c gi l i ngu vi mnh
khc nu ta thay th 0 thnh 1 v 1 thnh 0,du cng (+) thnh du nhn(.) v
ngc li.
Khi chng minh mt mnh l ng th mnh i ngu ca n cng
ng.
VD: 2 mnh A+1=1 v
A.0 = 0 l 2 mnh i ngu.
Phng php chng minh cc cng thc trn l lp bng tt c cc gi tr c th
c ca cc bin v tnh tng ng vi v phi, v tri ring r. Nu ng thc
tn ti vi tt c cc gi tr th cng thc ng. Sau y s l v d :
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Trang 23
V d 1 : Chng minh cng thc 10-2 x + y . z = (x + y) . (x + z).
x y z y . z x + y . z
(V tri)
x + y x + z (x + y) . (x + z)
(V phi)
0 0 0 0 0 0 0 0
0 0 1 0 0 0 1 0
0 1 0 0 0 1 0 0
0 1 1 1 1 1 1 1
1 0 0 0 1 1 1 1
1 0 1 0 1 1 1 1
1 1 0 0 1 1 1 1
1 1 1 1 1 1 1 1
Tt c cc gi tr ca ba bin x, y, z to thnh 8 t hp. Gi tr ca v tri x + y .
z trng vi gi tr ca v phi (x + y).(x + z). Suy ra ta c x + y . z = (x + y).(x +
z). Vy cng thc 10-2 c chng minh.
Cng thc 13-1: yxyx .=+
Cng thc 13-2: yxyx +=.
V d 2 : Chng minh nh l De_Morgan
Gii :
* Cng thc 13-1:
x y x . y yx. x y x+ y
0 0 0 1 1 1 1
0 1 0 1 1 0 1
1 0 0 1 0 1 1
1 1 1 0 0 0 0
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Trang 24
* Cng thc 13-2 :
x y x + y yx + x y yx.
0 0 0 1 1 1 1
0 1 1 0 1 0 0
1 0 1 0 0 1 0
1 1 1 0 0 0 0
L lun nh v d 1 suy ra nh l De_Morgan c chng minh.
Tng t nh vy ta c th chng minh tt c cc cng thc trn bng phng
php ny hoc dng cng thc ny suy ra cng thc kia.
2.1.5. Ba quy tc v ng thc :
a) Quy tc thay th :
Trong bt k ng thc no, nu thay th mt bin no bng mt hm s
(nhiu bin) th ng thc vn thit lp.
Quy tc ny c ng dng rt nhiu trong vic bin i cc cng thc bit
cho ra mt cng thc mi hay rt gn mt hm Boole no .
V d :
Cho mt hm Boole F1 = (A + B) . C
V : xx=
Thay th (A + B) . C = x ta c
B).CACB)(A +=+ (.
b) Quy tc tm o ca mt hm s :
Z l o ca hm s Z s c bng cch i du . thnh du +; +
thnh du .; 0 thnh 1; 1 thnh 0; bin s thnh o ca bin s ;
o bin s thnh nguyn bin s.
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Trang 25
c) Quy tc i ngu :
Hm Z v Z c gi l i ngu khi cc du cng + v du . ; cc
gi tr 0 v 1 i ch cho nhau mt cch tng ng.
V d :
Z=(A+B).C th hm Y=A.B+C l i ngu ca Z
2.2. Hm Boole :
Mt bin nh phn (x, y, z, ) c th ly gi tr 0 hoc 1. Hm Boole l mt
biu thc to bi cc bin nh phn, cc php ton cng +; nhn .; php b
(o); cc du bng =; du ngoc ( ).
Mt hm Boole c th c biu din bng cc phng php khc nhau ty
theo c im ca tng hm. Thng dng bn phng php. l:
2.2.1.Bng gi tr (hay cn gi l bng s tht,bng chn l-Truth table)
Bng gi tr l bng miu t quan h gia cc gi tr ca hm s tng ng
vi mi gi tr c th c ca cc bin s.
Khi lp bng ta cho bin s gi tr 0 v 1 to thnh cc t hp bin (khng
trng nhau) ri tnh gi tr hm. c im ca phng php ny tng i r
rng, trc quan nhng s rc ri nu bin s nhiu, khng p dng c cc
cng thc v nh l logic tnh ton.
V d :
a b c F
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 1
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Trang 26
2.2.2. Biu thc hm s :
Biu thc hm s dng i s logic dng cc php ton nhn (AND), cng
(OR), b (NOT) biu th quan h gia cc bin trong hm.
C hai dng biu din hm s, l dng chun 1 (tng cc tch hay tch
chun - Minterm) k hiu l m v dng chun 2 (tch cc tng chun hay tng
chun Maxterm) k hiu l M
V d :
521 mmmCBACBACBA(1,2,5)C)B,f(A, ++=++== (Dng chun 1)
530 .M.MM)CBA()CB(AC)B(A(0,3,7)C)B,f(A, =++++++== (Dng chun 2) 2.2.3. Ba Karnaugh :
Ba Karnaugh l phng php hnh v biu th hm logic (s ni k phn sau).
V d :
F xy z 00 01 11 10 0 1 1 1 1 1
2.2.4. S mch logic :
S logic c c khi ta dng cc k hiu logic (k hiu cc cng logic)
biu th hm s.
FA
B
C
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Trang 27
2.3. Cc dng chun ca hm Boole :
2.3.1. Dng chun 1 : (tng cc Mintern - tch chun)
a) Khi nim Minterm :
Cc mintern c c l khi ta kt hp n bin bng php
ton AND.
Nu c n bin ta s c 2n t hp bin c 2n mintern.
Nu bin c gi tr 1 ta s dng dng nguyn bin s, ngc
li, nu bin c gi tr 0 ta s dng dng b bin s.
K hiu ca mintern l mi ; vi i l gi tr thp phn ca t hp
cc bin.
b) Dng chun 1 :
Dng chun 1 l biu thc i s dng php ton cng (OR) cng tt c cc
minterm lm cho hm s logic bng 1.
c) V d :
x y z mi F1 F2
0 0 0 zyx = m0 0 0
0 0 1 zyx = m1 1 1
0 1 0 zyx = m2 1 0
0 1 1 zyx = m3 1 0
1 0 0 zyx = m4 0 1
1 0 1 zyx = m5 0 1
1 1 0 zyx = m6 0 0
1 1 1 x y z= m7 0 1
* Lu :
-Cc bin x, y, z c du b hoc khng b l ty thuc vo gi tr
0 hoc 1.
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Trang 28
-Gi tr ca F1 hoc F2 l gi tr t cho v ta c th chn gi tr
khc.
Cn c vo bng trn ta c dng chun 1 (c ba cch vit u c) ca hai hm
F1 v F2.
F1 = zyx + zyx + yzx
= m1 + m2 + m3
= (1, 2, 3)
F2 = zyx + zyx + zyx + xyz
= m1 + m4 + m5 + m7
= (1, 4, 5, 7)
2.3.2. Dng chun 2 : (tch cc Maxtern tng chun)
a) Khi nim Maxterm :
Cc maxtern c c l khi ta kt hp n bin bng php
ton OR.
Nu c n bin ta s c 2n t hp bin c 2n maxtern.
Nu bin c gi tr 1 ta s dng dng b bin s, ngc li,
nu bin c gi tr 0 ta s dng dng nguyn bin s.
K hiu ca maxtern l Mi ; vi i l gi tr thp phn ca t
hp cc bin.
b) Dng chun 2 :
Dng chun 2 l biu thc i s dng php ton nhn (AND) nhn tt c cc
maxterm lm cho hm s logic bng 0.
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Trang 29
c) V d :
x y z Mi F1 F2 0 0 0 x+y+z = M0 0 0
0 0 1 x + y + z = M1 1 1
0 1 0 x+ y+z = M2 1 0
0 1 1 x+ y+ z = M3 1 0
1 0 0 x + y+z = M4 0 1
1 0 1 x+y+ z = M5 0 1
1 1 0 x+ y+z = M6 0 0
1 1 1 x+ y+ z= M7 0 1
* Lu :
_ Cc bin x, y, z c du b hoc khng b l ty thuc vo gi tr 1 hoc 0.
_ Gi tr ca F1 hoc F2 l gi tr t cho v ta c th chn gi tr khc.
Cn c vo bng trn ta c dng chun 2 (c ba cch vit u c) ca hai hm
F1 v F2.
F1 = ( x + y + z ) ( x + y + z ). ( x + y + z ) . ( x + y + z ) . ( x + y + z )
= M0 . M4 . M5 . M6 . M7
= (0, 4, 5, 6, 7).
F2 = ( x + y + z ) . ( x + y + z ) . ( x+ y + z ) . ( x+ y + z )
= M0 . M2 . M3 . M6
= (0, 2, 3, 6).
2.4. Cc cng logic :
2.4.1.Cng khng o:
K hiu
A X=A
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Trang 30
Bng gi tr (Truth table)
A X
0 0
1 1
2.4.2.Cng o (NOT gate )
K hiu
A
Bng gi tr (Truth table)
A X= A
0 1
1 0
2.4.3.Cng AND (AND gate)
A
B
Bng gi tr (Truth table)
A B X
0 0 0
0 1 0 X=0 khi 1 ng vo =0
1 0 0 X=1 khi ng vo =1
1 1 1
Cng AND 3 ng vo:
CX=A.B.CB
A
X=A.B
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Trang 31
Bng gi tr (Truth table)
A B C X
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 1
2.4.4.Cng OR (OR gate)
Cng OR 2 ng vo:
A
BX=A+B
Bng gi tr (Truth table)
A B X
0 0 0
0 1 1 X=1 khi 1 ng vo =1
1 0 1 X=0 khi ng vo =0
1 1 1
Cng OR 3 ng vo:
BC
X=A+B+CA
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Trang 32
Bng gi tr (Truth table)
A B C X
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1
2.4.5.Cng NAND (AND +NOT)
Cng NAND 2 ng vo :
A
B
Bng gi tr (Truth table)
A B X= AB
0 0 1
0 1 1
1 0 1
1 1 0
Cng NAND 3 ng vo :
CBA
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Trang 33
Bng gi tr (Truth table)
A B C X= ABC
0 0 0 1
0 0 1 1
0 1 0 1
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 0
2.4.6.Cng NOR (OR +NOT)
Hm NOR 2 ng vo
B
A
Bng gi tr (Truth table)
A B X= BA + 0 0 1 0 1 0 1 0 0 1 1 0
Hm NOR 3 ng vo:
CBA
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Trang 34
Bng gi tr (Truth table)
A B C X= CBA ++
0 0 0 1
0 0 1 0
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 0
2.4.7.Cng EXOR (EX-OR gate)
A
B
Bng gi tr (Truth table)
A B X= BABABA =+
0 0 0
0 1 1
1 0 1
1 1 0
Hay s c tng ng nh sau:
B
A
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Trang 35
Ghi ch cng EX-OR khng c nhiu hn hai ng vo.
2.4.8.Cng EXNOR (EX_NOR gate)
A
B
Bng gi tr (Truth table)
A B X= BABAAB =+
0 0 1
0 1 0
1 0 0
1 1 1
Hay s c tng ng nh sau:
B
A
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Trang 36
* Quan h gia cc cng Logic
C mt s k hiu sau c dng trong mt s sch khc nhau
2.5.Ba Karnaugh (Ba K)
Ba Karnaugh l ba c s bng 2n ,vi n l s bin ca hm Boole, mt s
tng ng vi mt t hp ca cc bin cho.
Hai c gi l lin tip nhau(k cn nhau)khi n ch khc nhau 1 bin.
Cc bin phi c sp xp vi nhau sau cho 2 k cn nhau ch khc nhau 1
bit. Nu khng tun theo nguyn tt ny th khng cn l ba karnaugh na.
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Trang 37
2.5.1.Ba K 2 bin
S cn biu din hm l 22= 4 (C n bin s cn 2n )
F A
B 0 1 0 0 2
1 1 3
S th t = gi tr thp phn ca t hp nh phn tng ng.
VD: theo hnh trn th khi A=1;B=0 th t hp nh phn l 10[2]=2[D].Do
ny c s th t l 2
VD:Biu din hm sau bng Ba K :F(A,B)=(0,2).y l dng chun 1 .Nu
biu din di dng bng gi tr th ta c nh sau:
Hng A B F
0 0 0 1
1 0 1 0
2 1 0 1
3 1 1 0
T bng gi tr trn ta thy S th t =S th t hng.Lc ta biu din hm
Boole bng ba K nh sau:
F A B 0 1
0 1 1 1 0 0
+ s 0 v s 2 c gi tr l 1.Cc cn li c gi tr l 0
Tuy nhin ta c th biu din hm trn nh sau:
F A F A B 0 1 B 0 1 0 1 1 0 1 1 0 0
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Trang 38
2.5.2.Ba K 3 bin
S = 23= 8
F AB C 00 01 11 10
0 0 2 6 4 1 1 3 7 5
S th t = gi tr thp phn ca t hp nh phn tng ng.
VD: Cho hm Boole F(A,B,C)=(1,2,4,7).Ta biu din dng ba K nh sau:
F AB F AB C 00 01 11 10 C 00 01 11 10 0 1 1 0 0 0 1 1 1 1 0 0
2.5.3.Ba K 4 bin:
F AB
CD 00 01 11 10 00 0 4 12 8 01 1 5 13 9 11 3 7 15 11 10 2 6 14 10 S th t = gi tr thp phn ca t hp nh phn tng ng.
VD : Cho hm Boole F(A,B,C,D)=(0,1,2,4,7,10,14,15).Biu din bng ba K
F AB
F AB
CD 00 01 11 10 CD 00 01 11 10 00 1 1 00 0 0 01 1 01 0 0 0 11 1 1 11 0 0 10 1 1 1 10 0
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Trang 39
2.5.4.Ba K 5 bin:
F ABC 0 1
DE 00 01 11 10 10 11 01 00 00 0 4 12 8 24 28 20 16 01 1 5 13 9 25 29 21 17 11 3 7 15 11 27 31 23 19 10 2 6 14 10 26 30 22 18
2.6.n gin ha hm Boole:
2.6.1.Phng php i s :S dng cc cng thc ,cc tin v nh l rt
gn
VD: Rt gn hm sau
+F(A,B,C)=ABC+AB+C =AB(C+1) + C = AB+C
+F(x,y) = x(x+y) = xx+xy = x+xy=x
+F(x,y,z) = xyz+xyz+xy = xyz+xy=xy
+F(x,y,z) = xy+xz+yz (khng rt gn c na)
Phng php i s rt gn hm Boole bt buc ta phi nh cc cng thc,cc
quy tt,cc nh l Kt qu cui cng ta cng khng bit l ti u cha.Ta
c mt phng php khc c th khc phc c nhng vn trn l phng
php rt gn bng ba K
2.6.1.Phng php rt gn bng ba K
Nguyn tc: Khi gom 2 lin tip vi nhau th ta s loi i c 1 bin. Bin b
loi chnh l bin khc nhau trong 2 lin tip.Ta c th gom cng lc 2 ,4 ,8
,16 tc l gom 2n k cn nhau.Khi gom 2n k cn nhau ta loi b c n
bin. V tr cc k cn cho php nh sau:
AB 00 01 11 10 AB 00 01 11 10 C 0 1 0 0 0 C 0 1 1 0 0 1 1 0 0 0 1 0 0 0 0
Loi c bin C Loi c bin B
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Trang 40
AB 00 01 11 10 AB 00 01 11 10 C 0 1 0 0 1 C 0 0 1 1 0 1 0 0 0 0 1 0 0 0 0
Loi c bin A Loi c bin A AB 00 01 11 10 AB 00 01 11 10 C 0 1 1 1 1 C 0 0 1 1 0 1 0 0 0 0 1 0 1 1 0
Loi c bin AB Loi c bin AC
AB 00 01 11 10 AB 00 01 11 10 CD 00 1 0 0 1 CD 00 0 0 0 0 01 0 0 0 0 01 0 1 1 0 11 0 0 0 0 11 1 1 1 1 10 1 1 0 0 10 1 0 0 1
Loi c bin A
Loi c bin B
Loi c bin AC
Loi c bin AD
AB 00 01 11 10 AB 00 01 11 10 CD 00 1 0 0 0 CD 00 0 0 0 0 01 1 0 0 0 01 0 0 0 0 11 1 0 0 0 11 0 0 0 0 10 1 0 0 0 10 1 1 1 1
Loi b bin CD Loi bin AB
AB 00 01 11 10 AB 00 01 11 10 CD 00 1 0 0 1 CD 00 1 0 0 1 01 0 0 0 0 01 1 0 0 1 11 0 0 0 0 11 1 0 0 1 10 1 0 0 1 10 1 0 0 1
Loi b bin AC Loi b bin ACD
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Trang 41
AB 00 01 11 10 AB 00 01 11 10 CD 00 1 1 0 0 CD 00 0 0 0 0 01 1 1 0 0 01 1 1 1 1 11 1 1 0 0 11 1 1 1 1 10 1 1 0 0 10 0 0 0 0
Loi b bin BCD Loi b bin ABC Khi gom cc k cn nhau ta loi b nhng bin khc nhau,ch gi li nhng
bin ging nhau.Khi ta gom nhng k cn c gi tr l 1 th bin gi li l
chnh n nu bin mang gi tr l 1 v s c gi tr b nu bin l 0
VD:C 2 t hp c gom c gi tr l
T hp 1:ABC 010
T hp 2:ABC 011
Khi gom 2 ny ta loi b bin C v gi li bin AB.V A c gi tr l 0 v B
c gi tr l 1 nn t hp ny s c biu din l AB
VD 1: Cho hm Boole c bng gi tr nh sau.Rt gn bng ba K
AB 00 01 11 10 AB 00 01 11 10 C 0 1 0 0 0 C 0 0 0 1 0 1 1 0 0 0 1 1 1 1 0
A B F(A,B,C)= A B AC AB F(A,B,C)= AC+AB VD 2: Cho hm Boole c bng gi tr nh sau.Rt gn bng ba K
AB 00 01 11 10 AB 00 01 11 10 CD 00 1 0 0 0 CD 00 1 1 0 0 01 1 0 1 1 01 1 1 0 0 11 1 0 0 0 11 0 0 0 0 10 1 0 0 0 10 1 1 1 1
A B ACD A C CD F(A,B,C,D)= A B + ACD F(A,B,C,D)= A C+CD
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VD3:Rt gn hm Boole F1(A,B,C)= (1, 2, 5,6), F2(A,B,C)= (0,1, 2, 7)
AB 00 01 11 10 AB 00 01 11 10 C 0 0 1 1 0 C 0 1 1 0 0 1 1 0 0 1 1 1 0 1 0
C BC A C ABC F1(A,B,C)=C + BC
A B F2(A,B,C)= A C + A B +ABC
2.7.Ty nh (dont care).Thng k hiu l d(v tr ca )
VD:Cho hm Boole F(A,B,C)=(0,1,4,5,6)+d2 hoc l
F(A,B,C)=)=(0,1,4,5,6)+d(2)
C ngha l khi biu din bng ba K ta c th cho th 2 l 0 hoc 1 ty sao
cho c li nht khi rt gn.
Trong ba K ta c th dng du x cho ty nh.Nhn vo ba K ta thy nu
chn ty nh l 1 th rt gn s ti u.
AB 00 01 11 10 AB 00 01 11 10 C 0 1 x 1 1 C 0 1 1 1 1 1 1 0 0 1 1 1 0 0 1
C B
F2(A,B,C)= C + B 2.7.1.n gin ha theo dng chun 2
Phng php:Vn thc hin tng t nh dng chun 1 nhng khi gom cc
k cn ta gom nhng c k hiu l 0 .Mi s hng l mt tng.Kt qu cui
cng l tch ca cc tng .Khi lin kt th ta ch cc bin c gi tr l 0 th l
chnh n v c gi tr l 1 th s ly b (o).
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Trang 43
VD:Rt gn dng chun 2 hm F(A,B,C)= (0, 2, 3, 6).
AB 00 01 11 10 C 0 1 1 0 1 1 0 1 0 0
(B+C ) ( A+B )
F(A,B,C)= (B+C ).( A+B ) 2.7.2.Mt s phng php thc hin hm Boole bng cc cu trc cho
trc
T 2
T 1 AND OR NAND NOR
AND
x
Chun 1
x
-Chun 2 -Ly b hm F 2 ln. p dng Demorgan
OR
Chun 2
x
-Chun 1 -Ly b F 2 ln. p dng D
x
NAND
-Chun 2 -Ly b cc
thnh phn. p dng D
x
-Chun 1 -Ly b F 2 ln. p dng D
x
NOR
x
-Chun 1 -Ly b cc
thnh phn. p dng D
x
-Chun 2 -Ly b hm F 2 ln. p dng Demorgan
Da vo bng trn ta p dng s gii quyt c cc bi ton.Cc nh du x
s khng thc hin c cu trc dng .
VD:Cho hm F(A,B,C)= ABC+ABC+ A B .Dng cu trc OR -NAND thc
hin hm trn
Thc hin:
Bc 1: a v dng chun 1 (bi ton cho sn).
Bc 2:Ly b 2 ln
F(A,B,C)= ABC+ABC+ A B= BACABABC ++ = BACABABC.
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Trang 44
= )).().(( BACBACBA +++++
Bc 3:V s thc hin mch bng cc cng Logic
C
A
F(A,B,C)
B
Cng cho hm nh trn .Thc hin bng cu trc NAND-NAND
F (A,B,C)= ABC+ABC+ A B= BACABABC ++ = BACABABC.
n y ta thy xut hin cu trc mong mun nn khng p dng tip nh
l Demorgan.
S thc hin mch
F(A,B,C)
C
B
A
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Trang 45
BI TP CHNG 2
1.Rt gn hm Boole bng phng php i s
a.y= DBADBA +
b.z= ))(( BABA ++
c.y= BCDAACD +
d.y= CABCA +
e.y= DCBADCBA +
f. y= ))(( DBCA ++
2.Cho Z= CBA ++ Dng cng NAND v cng o biu din hm trn
3.Cho Z= ABC Dng cng NOR v cng o biu din hm trn
4.Xc nh biu thc ng ra ca cc mch sau
a.
A
B
C
D
X
b.
X
B
D
A
C
d.
D
B
X
A
C
5.V dng sng ng ra X theo tn hiu vo A,B,C
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Trang 46
CB
XC
A
AB
6.Xc nh ng ra X
X
C
B
A
7.Vi nhng gi tr no ca A,B,C th LED sng
C
LED
A
B
8.Rt gn hm sau bng ba K
a.F= BAABCABC ++
b.F= ABBDAABCABCD +++
c.F(A,B,C,D)= )15,10,7,5,4,1(
d.F(A,B,C,D)= )15,10,7,5,4,1( +d0
e.F(A,B,C,D)= )11,10,7,5,4,2,1,0(
f.F(A,B,C)= )7,4,3,0(
g.F(A,B,C,D)= 5)13,10,7,4,3,0( d+
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Trang 47
h. F(A,B,C,D= )())()(( CADCADCBADCBA +++++++++
9.Cho hm Boole F(A,B,C,D)= )15,10,7,5,4,1(
a.Thc hin hm dng cu trc NAND-NOR
b.Thc hin hm dng cu trc NOR-NOR
c.Thc hin hm dng cu trc NAND-NAND
d.Thc hin hm dng cu trc OR-NOR
e.Thc hin hm dng cu trc AND-OR
f. Thc hin hm dng cu trc AND-NOR
g.Thc hin hm dng ton cng NAND
h.Thc hin hm dng ton cng NOR
i.Thc hin hm dng ton cng NAND 2 ng vo
j.Thc hin hm dng ton cng NOR 2 ng vo
-
Trang 48
Chng 3
MCH T HP
3.1.Gii thiu:
Mch t hp l mt mch c xy dng t cc cng Logic c bn thc hin
nhng chc nng m ngi s dng yu cu.Mt s tnh nng thng c s
dng th ngi ta tch hp cc cng thnh nhng IC chuyn s dng cho cc
mc ch . Trong chng ny ta s tm hiu cu trc bn trong cng nh tnh
nng ca mt s IC thng dng.
3.1.1.Mch gii m (Decoder)
Decoder l mch chuyn i N bit u vo thnh M ng ra .Mi ng ra c
chn ( tch cc) tng ng vi mt t hp u vo.
Nu c N ng vo tc c 2N t hp.ng vi mi t hp u vo s c mt ng
ra mc Logic cao cn tt c cc ng ra khc s mc Logic thp
Tuy nhin c nhng Decoder c thit k ngc li tc ng no tch cc th
ng c mc logic thp cn cc ng cn li mc cao.
a.Decoder 2 4 c ng ra tch cc mc cao
Bng gi tr
2-->4X 0
Y 3X 1
Y 0Y 1Y 2
T bng gi tr trn ta c
Y0= 10 XX ; Y1= 10 XX ; Y2= 10 XX ; Y0= 10 XX ;
S tng ng
X1 X0 Y3 Y2 Y1 Y0
0 0 0 0 0 1
0 1 0 0 1 0
1 0 0 1 0 0
1 1 1 0 0 0
-
Trang 49
Y1
Y3
X0
Y0
X1
Y2
Decoder 2 4 c ng ra tch cc mc cao
b.Decoder 2 4 c ng ra tch cc mc thp
2-->4X 0
Y 3X 1
Y 0Y 1Y 2
2-->4
X0 Y0
X1
Y1
Y2
Y3
T bng gi tr trn ta c
Y0=X0+X1;Y1=X0+ 1X ;Y2= 0X +X1;Y3= 0X + 1X
S tng ng
X1 X0 Y3 Y2 Y1 Y0
0 0 1 1 1 0
0 1 1 1 0 1
1 0 1 0 1 1
1 1 0 1 1 1
-
Trang 50
X1
Y0
Y3
Y2
X0
Y1
Decoder 2 4 c ng ra tch cc mc thp
Nhn vo bng gi tr ca Decoder ta thy c mt im bt li l ti mt thi
im phi c mt ng ra c chn.Nu nh ta khng mun chn ng no th
Decoder loi ny khng thc hin c.Xut pht t nguyn nhn ngi ta
cn tch hp mt loi Decoder c ng vo cho php gi l Enable (E).
Xt Decoder 2 4 c ng ra tch cc mc thp c ng vo Enable tch cc mc
cao
Bng gi tr
2-->4X 0
Y 3
X 1Y 0Y 1Y 2
E
E X1 X0 Y3 Y2 Y1 Y0
1 0 0 1 1 1 0
1 0 1 1 1 0 1
1 1 0 1 0 1 1
1 1 1 0 1 1 1
0 x x 1 1 1 1
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Trang 51
Tng t nh cc cch lm trc ta cng s tm c s tng ng ca
Decoder trong trng hp c Enable.
*Nhn xt:ti mt thi im c mt ng ra tch cc .Ng ra tng ng vi
s thp phn ca t hp nh phn
V d:Nu ng vo c t hp l 10 th ng ra c chn l Y2 v 10[B]=2[D].
Xt Decoder 3 8 c ng ra tch cc mc thp
3-->8
X 0
Y 3X 1
Y 0Y 1Y 2
X2Y 4Y 5Y 6Y 7
c.Ghp cc Decoder vi nhau
ghp cc Decoder vi nhau th cc decoder phi c ng vo E.
Khi ghp 2 Decoder 2 4 ta s c 1 Decoder 3 8
Khi ghp 2 Decoder 3 8 ta s c 1 Decoder 4 16
Khi ghp 2 Decoder n 2n ta s c 1 Decoder n+1 2n+1
V d:ghp 2 Decoder 2 4 thnh 1 Decoder 3 8
X2 X1 X0 Y7 Y6 Y5 Y4 Y3 Y2 Y1
Y0
0 0 0 1 1 1 1 1 1 1 0
0 0 1 1 1 1 1 1 1 0 1
0 1 0 1 1 1 1 1 0 1 1
0 1 1 1 1 1 1 0 1 1 1
1 0 0 1 1 1 0 1 1 1 1
1 0 1 1 1 0 1 1 1 1 1
1 1 0 1 0 1 1 1 1 1 1
1 1 1 0 1 1 1 1 1 1 1
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Trang 52
Y7
X0
X1
Y0Y1Y2Y3
E
X0 Y1
Y6Y5
Y2
D 2
X1
Y3
D 1
X2
X0
X1
Y0Y1Y2Y3
E
Y4
Y0
Bng gi tr khi ghp 2 Decoder
X2 X1 X0 Y7 Y6 Y5 Y4 Y3 Y2 Y1 Y0
0 0 0 0 0 0 0 0 0 0 1
0 0 1 0 0 0 0 0 0 1 0 D 1 hot ng
0 1 0 0 0 0 0 0 1 0 0 D 2 b cm
0 1 1 0 0 0 0 1 0 0 0
1 0 0 0 0 0 1 0 0 0 0
1 0 1 0 0 1 0 0 0 0 0 D 2 hot ng
1 1 0 0 1 0 0 0 0 0 0 D 1 b cm
1 1 1 1 0 0 0 0 0 0 0
Nhn vo bng gi tr trn ta thy nu ghp nh trn th 2 Decoder ny hon
ton tng ng nh 1 decoder 3 8
d.Dng Decoder thc hm Boole
V d: Cho hm Boole c F(A,B,C)=(0,1,3,6).Dng Decoder thc hin hm
trn thay cho cc cng Logic khc.
Ta thy Decoder thc s l cc cng Logic c tch hp thnh.thc hin hm
Boole dng Decoder thc cht l ta tn dng cc cng Logic c sn trong
Decoder thc hin hm Boole .
thc hin hm Boole dng Decoder ta ln lt theo cc bc sau:
-
Trang 53
- V c 3 bin nn ta dng Decoder 3 8.Tng qut nu hm c n bin th dng
Decoder n 2n
- Lp bng gi tr hm Boole.Theo nh v d trn ta c bng gi tr nh sau:
A B C Y
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 0
Ta c th biu din hm Boole trn bng dng chun 1 nh sau:
F(A,B,C)= (0,1,3,6)= CABBCACBACBA +++
Bng gi tr ca decoder 3 8 nh sau:
X2 X1 X0 Y7 Y6 Y5 Y4 Y3 Y2 Y1 Y0
0 0 0 0 0 0 0 0 0 0 1
0 0 1 0 0 0 0 0 0 1 0
0 1 0 0 0 0 0 0 1 0 0
0 1 1 0 0 0 0 1 0 0 0
1 0 0 0 0 0 1 0 0 0 0
1 0 1 0 0 1 0 0 0 0 0
1 1 0 0 1 0 0 0 0 0 0
1 1 1 1 0 0 0 0 0 0 0
Ta c:
Y0= CBA ;Y1= CBA ;Y2= CBA ;Y3= BCA ;Y4= CBA ;Y5= CBA ;Y6= CAB ;Y7=
ABC
So snh 2 kt qu ny ta thy c th biu din hm Boole nh sau:
-
Trang 54
F(A,B,C)= (0,1,3,6)= CABBCACBACBA +++ =Y0+Y1+Y3+Y6 (ca
Decoder).
S thc hin dng Decoder nh sau:
C
B
3-->8
X 0
Y 3X 1
Y 0Y 1Y 2
X2Y 4Y 5Y 6Y 7
A
F(A,B,C)
Nu s dng Deoder c ng ra tch cc mc thp th cng cch lm tng t
nh trn ta s c mch nh sau:
3-->8
X 0
Y 3X 1
Y 0Y 1Y 2
X2Y 4Y 5Y 6Y 7
F(A,B,C)
A
B
C
Ta c th c nhiu cch thc hin hm Boole dng Decoder nhng tt c u
cho cng kt qu.
3.1.2.Mch m ha (Encoder):
L mch c 2n ng vo v n ng ra
Xt Encoder 4 2
X0
X2
Y0
Y1
X3
X1
T bng gi tr trn ta thy nu a tn hiu t mch gii m n mch m ha
th d liu s ng nh trng thi ban u.T ngi ta c th m ha d liu
trc khi a ln ng tn hiu.
S tng ng ca Encoder 4 2
X3 X2 X1 X0 Y1 Y0
0 0 0 1 0 0
0 0 1 0 0 1
0 1 0 0 1 0
1 0 0 0 1 1
-
Trang 55
T bng gi tr trn ta c:Y0= ( )1302 XXXX ;Y1= ( )2301 XXXX
Y1
Y0
X2
X0
X3
X1
Cng tng t nh Decoder ta cng c Encoder c ng vo,ng ra tch cc mc
cao hoc thp.Ty tng trng hp c th m ta chn cho ph hp.
Mch m ha c u tin:
Khi c 2 hay nhiu ng vo cng mc tch cc th ng ra c th c th s c
2 gi tr khc nhau , chnh v vy ngi ta qui nh b m ha c u tin .
Khi nu c 2 tn hiu cng mc tch cc th ch c ng vo c u tin
cao hn mi cho tc ng ti ng ra
Cho Encoder nh hnh v
X0
X2
Y0
Y1
X3
X1
u tin gim dn t X0 n X3.Lc bng gi tr ca Encoder nh sau:
3.1.3.Mch dn knh (Multiplexer)-MUX
Mch dn knh hay cn gi l MUX l mch c 2n d liu (Data),n ng vo
iu khin (Selects) v c mt ng ra
a.Xt MUX 4 1 (22 1)
X3 X2 X1
X0
Y1 Y0
x x x 1 0 0
x x 1 0 0 1
x 1 0 0 1 0
1 0 0 0 1 1
-
Trang 56
4-->1
X0
X1
YX2
X3
AB
MUX c 4 ng vo d liu l X0,X1,X2,X3 v c 2 ng vo iu khin l A,B.
T bng gi tr ta c Y= ABXBAXBAXBAX 3210 +++
S mch tng ng
X2X0 X3
Y
B
A
X1
A B Y
0 0 X0
0 1 X1
1 0 X2
1 1 X3
-
Trang 57
b.Xt MUX 8 1 (23 1)
8-->1
X0
X1
Y
X2
X3
A
C
X4
X5
X6
X7
B
Tng t nh trn ta c th v s tng ng ca MUX 8 1
c.Ghp cc MUX
ghp 2 MUX vi nhau th 2 MUX phi c Enable
Ghp 2 MUX 4 1 thnh 1 MUX 8 1.Thc hin ghp nh sau
A
Y
4-->1
X0
X1
Y
X2
X3
A
B
E
X2
X6
X0
X4
X3
MUX 1
X1
X7
4-->1
X0
X1
Y
X2
X3
A
B
E
MUX 2
X5
B
C
A B
C
Y
0 0 0 X0
0 0 1 X1
0 1 0 X2
0 1 1 X3
1 0 0 X4
1 0 1 X5
1 1 0 X6
1 1 1 X7
-
Trang 58
*Ch :Khi ghp nh trn ta phi ch rng A l MSB v B l LSB
Bng gi tr
A C
B
Y
0 0 0 X0
0 0 0 X1 MUX 1 hot ng
0 0 0 X2 MUX 2 b cm
0 0 0 X3
0 0 0 X4 MUX 2 hot ng
0 0 0 X5 MUX 1 b cm
0 0 0 X6
0 0 0 X7
Bng gi tr ny hon ton tng ng nh bng gi tr ca MUX 8 1
d.Dng MUX thc hin hm Boole
+ Dng MUX 2n 1 thc hin hm Boole n bin
Cho hm Boole F(A,B,C)= (1,2,5,7).Dng MUX thc hin hm trn
Hm c 3 bin nn ta dng MUX 8 1(c n bin th dng MUX 2n 1)
Bng gi tr ca hm Boole Bng gi tr ca MUX
A B
C
Y
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 0
1 0 1 1
1 1 0 0
1 1 1 1
A B
C
Y
0 0 0 X0
0 0 1 X1
0 1 0 X2
0 1 1 X3
1 0 0 X4
1 0 1 X5
1 1 0 X6
1 1 1 X7
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Trang 59
Nhn vo 2 bng gi tr trn ta thy nu cho X0=X3=X4=X6=0 v
X1=X2=X3=X7=1 th ng ra ca MUX chnh l hm F(A,B,C).S
mch thc hin nh sau
BA
F(A,B,C)
[1]
8-->1
X0
X1
Y
X2
X3
A
C
X4
X5
X6
X7
BC
+ Dng MUX 2n 1 thc hin hm Boole n+1 bin
Thc hin hm F(A,B,C)= (2,5,6,7) dng MUX 4 1
Trc ht ta phi nh bng sau:
2 gi tr lin tip ca hm Boole Bit LSB
0 0 0
0 1 N0
1 0 0N
1 1 1
Sau lp bng gi tr ca hm Boole ri da vo bng trn ta ghi li thnh
bng nh bn di.Trong X0,X1,X2,X3 l cc ng data ca MUX 41.Cc
bin cao a vo 2 ng iu khin cn bin l LSB th thc hin t bng gi tr
-
Trang 60
Bng gi tr
S mch thc hin
[1]
A
4-->1
X0
X1
Y
X2
X3
A
B
F(A,B,C)
C
B
Xt mt VD khc:Dng MUX 81 thc hin hm Boole F(A,B,C,D)=
(1,2,5,6,7,9,11,15)
A B C
D
F(A,B,C,D)
0 0 0 0 0
0 0 0 1 1
D
X0
0 0 1 0 1
0 0 1 1 0
D
X1
0 1 0 0 0
0 1 0 1 1
D
X2
A B C Y
0 0 0 0
0 0 1 0
0
X0
0 1 0 1
0 1 1 0
C
X1
1 0 0 0
1 0 1 1
C
X2
1 1 0 1
1 1 1 1
1
X3
-
Trang 61
0 1 1 0 1
0 1 1 1 1
1
X3
1 0 0 0 0
1 0 0 1 1
D
X4
1 0 1 0 0
1 0 1 1 1
D
X5
1 1 0 0 0
1 1 0 1 0
0
X6
1 1 1 0 0
1 1 1 1 1
D
X7
D
8-->1
X0
X1
Y
X2
X3
A
C
X4
X5
X6
X7
B
[1]
F(A,B,C,D)
C
B
A
3.1.4.Mch phn knh (DeMultiplexer)-DEMUX
DEMUX l mch t hp c 1 ng vo n ng iu khin v 2n ng ra.Nu d liu
t MUX a n DEMUX th d liu s c phc hi ng trng thi ban u.
Xt DEMUX 1 4
X
A
Y0Y1Y2Y3
B
T bng trn ta c XABYBAXYBXAYBAXY ==== 3;2;1;0
A B Y3 Y2 Y1 Y0
0 0 0 0 0 X
0 1 0 0 X 0
1 0 0 X 0 0
1 1 X 0 0 0
-
Trang 62
S mch
Y2
Y0
B
Y3
A
X
Y1
3.1.5.Mch kim tra chn l:
C 2 dng :
-Even parity (parrity chn)
-Old parity (parity l)
c dng pht hin sai lch trn ng truyn
+Even parity:bit parity c to ra sao cho tng s bit 1 l chn.Nu tng s bit
1 chn th bit P=0.Tng s bit 1 l th P=1
+Old parity: bit parity c to ra sao cho tng s bit 1 l l.Nu tng s bit 1
chn th bit P=1.Tng s bit 1 l th P=0
VD:Thit k h kim tra chn 4 bit (event parity)
-
Trang 63
A B C D P
0 0 0 0 0
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0
0 1 0 0 1
0 1 0 1 0
0 1 1 0 0
0 1 1 1 1
1 0 0 0 1
1 0 0 1 0
1 0 1 0 0
1 0 1 1 1
1 1 0 0 0
1 1 0 1 1
1 1 1 0 1
1 1 1 1 0
Rt gn ta c )()())(( BADCBADCP +=
S mch
C
A
D
B P
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Trang 64
3.2.Phng php thit k mt s mch t hp
a.Phng php
+Xc nh s ng vo-ra t bi ton yu cu
+Lp bng gi tr theo bi ton t ra
+Rt gn cc hm Boole
+Dng cng Logic hoc cc mch t hp thc hin hm Boole
b.Mt s VD v thit k mch t hp
VD1:Thit k mt mch t hp thc hin vic cng 2 s nh phn 1 bit c
nh.Cn gi l mch cng bn phn=Haft-Adder=HA
-Trc ht ta xc nh s ng vo-ra ca mch t hp:Theo bi th mch c 2
ng vo l 2 s nh phn v 2 ng ra l tng 2 s nh phn v s nh
-Bng gi tr
A B S C
0 0 0 0
0 1 1 0
1 0 1 0
1 1 0 1
Trong S(sum) l tng 2 s nh phn v C(carry) l s nh
-Rt gn ta c:S= BA ;C=AB
-Thc hin mch
B
C
AS
Mch cng 1 bit c thnh phn nh trc gi l mch cng ton phn hay
-
Trang 65
Full-Adder=FA
An Bn
Cn-1
Sn
Cn
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0
0 1 1 0 1
1 0 0 1 0
1 0 1 0 1
1 1 0 0 1
1 1 1 1 1
111 );( ++== nnnnnnnnnnn CACBBACBACS S mch
C n-1C n
B n
A n
S n
VD 2:Thit k mch t hp thc hin vic nhn 2 s nh phn 2 bit
-Mch c 4 ng vo 4 ng ra v tch ca 2 s nh phn khi nhn nhau ln nht l
1001.
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Trang 66
- Bng gi tr
A1 B1 A2
B2
Y3 Y2 Y1
Y0
0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0
0 0 1 0 0 0 0 0
0 0 1 1 0 0 0 0
0 1 0 0 0 0 0 0
0 1 0 1 0 0 0 1
0 1 1 0 0 0 1 0
0 1 1 1 0 0 1 1
1 0 0 0 0 0 0 0
1 0 0 1 0 0 1 0
1 0 1 0 0 1 0 0
1 0 1 1 0 1 1 0
1 1 0 0 0 0 0 0
1 1 0 1 0 0 1 1
1 1 1 0 0 1 1 0
1 1 1 1 1 0 0 1
Rt gn ta c
Y3=A1B1A2B2;Y2=A1A2( 21 BB + );Y1= 221221211211 BABBAABBAABA +++
Y0=B1B2.
-Thc hin mch
-
Trang 67
B2
Y3
B1
A1Y0
Y2
Y1
A2
VD3:So snh 2 s nh phn 2 bit
-S ng vo l 4 v ng ra l 3 ;Y0 (2 s bng nhau);Y1 (s th 1 ln hn);Y2
(s th nht nh hn)
-
Trang 68
-Bng gi tr
A1 B1 A2
B2
Y2 Y1 Y0
0 0 0 0 0 0 1
0 0 0 1 1 0 0
0 0 1 0 1 0 0
0 0 1 1 1 0 0
0 1 0 0 0 1 0
0 1 0 1 0 0 1
0 1 1 0 1 0 0
0 1 1 1 1 0 0
1 0 0 0 0 1 0
1 0 0 1 0 1 0
1 0 1 0 0 0 1
1 0 1 1 1 0 0
1 1 0 0 0 1 0
1 1 0 1 0 1 0
1 1 1 0 1 0 0
1 1 1 1 0 0 1
-Rt gn:T bng gi tr trn ta c
Y0=(0,5,10,15);Y1=(4,8,9,12,13);Y2=(1,2,3,6,7,11,14)
-Thc hin mch.C nhiu cch thc hin y ta s dng Decoder thc hin
Y1
Y2
Y0
A
B
4-->16
X 0 Y 3
X 1
Y 0Y 1Y 2
X2
Y 4Y 5Y 6Y 7
X3Y 8Y 9
Y 10Y11
Y13Y12
Y14Y15
C
D
-
Trang 69
BI TP CHNG 3
1.Thc hin ghp 2 Decoder 38 thnh 416
2.Thc hin ghp 4 Decoder 24 thnh 416
3.Cho hm F(A,B,C,D)= )15,14,13,7,4,3,0(
a.Thc hin hm Boole bng cng Logic
b.Thc hin hm Boole bng Decoder
c.Thc hin hm Boole bng MUX
d.Thc hin hm Boole bng MUX 81
4.Cho hm F(A,B,C,D)= )15,10,9,6,4,3,0(
a.Thc hin hm Boole bng cng Logic
b.Thc hin hm Boole bng Decoder
c.Thc hin hm Boole bng MUX
d.Thc hin hm Boole bng MUX 81
5.Thit k mch t hp sao cho m vo l BCD v m ra l LED 7 on
6.Thit k mch t hp sao cho khi cho vo m nh phn 4 bit .Ng ra s mc
0 khi s thp phn tng ng ca ng vo nh hn 10
7.Thit k mch so snh 2 s nh phn 3 bit.Dng LED bo hiu.
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Trang 70
Chng 4
H TUN T Mch tun t nhn thng tin t cc ng vo v kt hp v cc phn t nh
xc nh trng thi ca ng ra. Nh vy mch tun t c cc u ra khng ch
ph thuc vo u vo hin hnh m cn ph thuc vo mt chui cc gi tr
u vo qu kh.
4.1.Cc mch cht v FF:
1.Cht:(Latch)
Mch cht l mch tun t s xt cc ng vo mt cch lin tc v thay i cc
ng ra ca n bt c lc no.
a.Cht cng NOR:
SET/Q
QCLEAR
Bng gi tr
SET CLEAR OUTPUT
0 0 Khng
i
0 1 Q=0
1 0 Q=1
1 1 Cm
*Cm : l trng thi Q=/Q=0 (Trng thi ny khng dng)
a.Cht cng NAND:
-
Trang 71
SETQ
/QCLEAR
Bng gi tr
SET CLEAR OUTPUT
0 0 Cm
0 1 Q=1
1 0 Q=0
1 1 Khng
i
4.1.2.Flip-Flop (FF)
Flip-Flop l mt mch tun t thng thng ly mu cc ng vo v thay i
cc ng ra ca chng nhng thi im xc nh bi tn hiu ng h (xung
clock).
Ngi ta phn loi FF da vo tn hiu iu khin :FF iu khin (tch cc)
cnh ln v FF iu khin (tch cc) cnh xung.
a.DFF
K hiu
D
CLK
Q
/Q
D
CLK
Q
/Q
(FF tch cc cnh ln) (FF tch cc cnh xung)
Bng gi tr :
Dn Qn+1
0 0
1 1
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Trang 72
Dn l trng thi ca tn hiu vo xung th n
Qn+1 l trng thi ca tn hiu ra xung th n+1
Mt cch tng qut: Qn+1=Dn
Biu din dng tn hiu ng ra theo tn hiu vo v xung Clock.Gi s ban u
Q0=0.FF tch cc cnh xung.
D
Q
CK
Biu din dng tn hiu ng ra theo tn hiu vo v xung Clock.Gi s ban u
Q0=0.FF tch cc cnh ln
D
Q
CK
b.TFF
T
CLK
Q
/Q
T
CLK
Q
/Q
(FF tch cc cnh ln) (FF tch cc cnh xung)
Bng gi tr :
Tn l trng thi ca tn hiu vo xung th n
Qn+1 l trng thi ca tn hiu ra xung th n+1
Biu din dng tn hiu ng ra theo tn hiu vo v xung Clock.Gi s ban u
Q0=0.FF tch cc cnh xung.
Tn Qn+1
0 Qn
1 nQ
-
Trang 73
T
CK
Q
Biu din dng tn hiu ng ra theo tn hiu vo v xung Clock.Gi s ban u
Q0=1.FF tch cc cnh ln.
Q
T
CK
c.J_K FF
J
CK
Q
QK
J
CK
Q
QK
Bng gi tr
Jn Kn Qn+1
0 0 Qn
0 1 0
1 0 1
1 1 nQ
Biu din dng tn hiu ng ra theo tn hiu vo v xung Clock.Gi s ban u
Q0=0.FF tch cc cnh xung.
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Trang 74
Q
CK
K
J
Biu din dng tn hiu ng ra theo tn hiu vo v xung Clock.Gi s ban u
Q0=0.FF tch cc cnh ln.
Q
CK
K
J
d.S_R FF
S
CK
Q
QR
S
CK
Q
QR
Bng gi tr
Sn Rn Qn+1
0 0 Qn
0 1 0
1 0 1
1 1 x
x:Trng thi cm khng dng
J_K FF s tng ng T- FF khi hai ng vo J= K = 1
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Trang 75
[1]
J
CK
Q
QK
J_K FF s tng ng D-FF khi hai ng vo J=v K ni vi nhau qua cng
o.
CK
DJ
CK
Q
QK
Ngoi ra FF cn c cc ng vo iu khin l Preset(Pr) v Clear (Cl).K hiu
thng nh sau
J
CK
Q
QK
Pr
Cl
J
CK
Q
QK
Pr
Cl
(FF c Pr v Cl tch cc mc cao) (FF c Pr v Cl tch cc mc thp)
Khi Pr v Cl tc ng th ng ra s mc 0 hoc 1 m khng ph thuc vo tn
hiu ng vo
+Trng hp Pr v Cl tch cc mc cao
-Khi Pr=Cl=0 th FF c ng ra ph thuc vo ng vo v xung CK
-Khi Pr=1 v Cl=0 th FF c ng ra Q=1
-Khi Pr=0 v Cl=1 th FF c ng ra Q=0
-Khi Pr=Cl=1 th FF c ng ra khng xc nh c
+Trng hp Pr v Cl tch cc mc thp
-Khi Pr=Cl=1 th FF c ng ra ph thuc vo ng vo v xung CK
-Khi Pr=0 v Cl=1 th FF c ng ra Q=1
-Khi Pr=1 v Cl=0 th FF c ng ra Q=0
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Trang 76
-Khi Pr=Cl=1 th FF c ng ra khng xc nh c
4.2. B m: ( Counter )
4.2.1.Phn loi:
Ngi ta phn loi mch m theo nhiu kiu khc nhau nh mch m ni tip
(khng ng b)m song song (ng b),mch m ln,mch m
xung,mch m nhp c data,mch m khng nhp c datatrong
chng ny ta s tm hiu mt s mch m khc nhau.
Trong chng ny c mt khi nim gi l Modulo(MOD): l s trng thi
khc nhau trong mt chu k m ( khng lp li gi tr trong mt chu k m ).
4.2.2.Mch m ni tip (khng ng b):
a.m MOD chn 2n
Phng php thit k
-MOD 2n s dng n JKFF
-m ni tip th xung CK ca FF sau c ly t ng ra Q ca FF trc
-Cc ng J&K ni chung v ni vi mc Logic 1.S m c ly t cc ng ra
Q ca cc JKFF (dng BCD).
V d1: Thit k mch m c MOD=4 dng JKFF tch cc cnh xung.Gi s
ban u cc ng ra bng 0.
-MOD=4=22 do dng 2 JKFF
-S thc hin nh hnh v
Q1Q0
J
CK
Q
QK
[1][1]
CK
J
CK
Q
QK
Tn hiu ra c biu din nh sau
1
0 1
10
0
0
Q0
0
1
Q1
CK
1
0
0
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Trang 77
Nhn vo kt qu trn ta thy d liu ra l :00,01,10,11,00,01Nu ta ta s
dng mch gii m chuyn i t BCD sang LED 7 on ta s thy s hin th
ln lt l 0,1,2,3,0,1
V d2: Thit k mch m c MOD=8 dng JKFF tch cc cnh xung.Gi s
ban u cc ng ra bng 0.
-MOD=8=23 do dng 3 JKFF
-S thc hin nh hnh v
Q1
J
CK
Q
QK
Q2Q0
J
CK
Q
QK
[1][1]
CK
[1]
J
CK
Q
QK
Tn hiu ra c biu din nh sau
0
1
0
1
11
1
0 1
1
0
0
1
0
0
0
Q0
0
1
0
0
0
Q1
CK
Q20
1
0
0
0
1
1
0
1
Trng hp mch m ni tip nhng m gim dn th cch thit k tng t
nh trn nhng xung CK ca FF sau c ly t ng Q ca FF trc
Tm li
Qn=CKn+1 th mch m l m ni tip v m ln
nQ =CKn+1 th mch m l m ni tip v m xung
a.m MOD bt k 2n-1< MOD < 2n
thit k mch m loi ny bt but ta phi dng loi JKFF c Pr v Cl
V d1: Thit k mch m m ln dng JKFF tch cc cnh ln c Pr v Cl
tch cc mc cao m chui s sau:0,1,2,3,4,5,0,1,2,.. v c th lp li
Ta thy rng trong mt chu k m xut hin 6 trng thi khc nhau.V vy
mch m c MOD=6 tc l MOD l.Do 22
-
Trang 78
chu k mi.Trong trng hp ny ta li khng mun iu xy ra m ta mun
khi m n 5 mch t ng quay v 0.Trong JKFF c mt ng lm cho FF tr
v 0 lp tc l chn Cl.Nh vy thay v khi m n 5 mch s tip tc m
ln 6 ,ti thi im ny ta s tc ng vo chn Cl mch t ng tr v
0.Phng php tt nht l ta dng cng AND hoc cng NAND
T ta c th a ra s mch cho v d trn nh sau:
J
CK
Q
QK
Pr
Cl
[1][1]
CK
[1]
J
CK
Q
QK
Pr
Cl
J
CK
Q
QK
Pr
Cl
Q1 Q2Q0
* Chn Pr ta khng tc ng n nn khi thc hin mch ta nn ni cc chn
ny xung mc logic 0 (GND)
Tuy nhin ta c th a ra s khc thc hin mch trn nh sau
J
CK
Q
QK
Pr
Cl
[1][1]
CK
[1]
J
CK
Q
QK
Pr
Cl
J
CK
Q
QK
Pr
Cl
Q1 Q2Q0
Ta c c s ny v lc ng ra Q0 =0 nn ta cng khng cn tc ng
vo chn Cl ca FF u tin
V d 2: Thit k mch m m ln dng JKFF tch cc cnh xung c Pr v
Cl tch cc mc thp m chui s sau:0,1,2,3,4,5,6,0,1,2,.. v c th lp li
Tng t nh trn ta c s thc mch nh sau:
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Trang 79
Q0
J
CK
Q
QK
Pr
Cl
CK
Q1
[1] [1]
J
CK
Q
QK
Pr
Cl
J
CK
Q
QK
Pr
Cl
Q2
[1]
4.2.3.Mch m song song (ng b):
Mch m song song thng thit k phc tp hn mch m ni tip rt
nhiu.tuy nhin c mt c im m mch m ni tip khng th p ng c
l m chui m bt k khng theo trnh t no c.
V d nh ta mun m chui s 0,2,4,6,8 th trong trng hp ny ta khng
th s dng mch m ni tip c m phi s dng mch m song song.Mt
chui m no m dng phng php ni tip c th bao gi ta cng c
th dng phng php song song.Tuy nhin iu ny khng nn v thit k theo
kiu ni tip th n gin hn nhiu.
Xt mt v d sau:Thit k mch m MOD=4 dng phng php m song
song
Trc ht ta phi nh bng tra sau:
Qn Qn+1 Jn Kn
0 0 0 X
0 1 1 X
1 0 X 1
1 1 X 0
X: ty nh
Bng ny c c l do ta suy ra t bng gi tr ca JKFF
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Trang 80
Jn Kn Qn+1
0 0 Qn
0 1 0
1 0 1
1 1 nQ
Tip theo ta lp bng gi tr
Xung Q1 Q0
(hin ti)
Q1 Q0
(k tip)
J0 K0
J1 K1
1 0 0 0 1 1 X 0 X
2 0 1 1 0 X 1 1 X
3 1 0 1 1 1 X X 0
4 1 1 0 0 X 1 X 1
Dng ba K rt gn cc hm J0,K0, J1, K1 theo bin Q0,Q1(hin ti) ta c
J0=K0=1;J1=K1=Q0
Mch m song song th cc xung CK ca cc FF c ni chung vi nhau,cn
cc chn J,K th c ni theo kt qu rt gn ca ba K
Trong bi v d ny ta s c s thc hin mch nh sau:
[1]
CK
J
CK
Q
QK
Q1Q0
J
CK
Q
QK
C 2 phng php u cho ta cng kt qu tuy kt ni theo 2 kiu khc
nhau
Ta xt mt v d khc phc tp hn l thit k mch m m 0,2,4,6,0Gi s
rng nu s m u tin xut hin khi cp in cho mch khng nm trong
chui s trn th s k tip s l 0.
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Trang 81
Bng gi tr:
Xung Q2 Q1 Q0
(hin ti)
Q2 Q1 Q0
(k tip)
J0 K0
J1 K1
J2 K2
1 0 0 0 0 1 0 0 X 1 X 0 X
2 0 0 1 0 0 0 X 1 0 X 0 X
3 0 1 0 1 0 0 0 X X 1 1 X
4 0 1 1 0 0 0 X 1 X 1 0 X
5 1 0 0 1 1 0 0 X 1 X X 0
6 1 0 1 0 0 0 X 1 0 X X 1
7 1 1 0 0 0 0 0 X X 1 X 1
8 1 1 1 0 0 0 X 1 X 1 X 1
Dng ba K rt gn ta c J0=K0=Q0;J1=K1=Q1+ 0Q ;J2=K2=Q0Q2+Q1 0Q
S mch
Q0
CK
Q1 Q2
J
CK
Q
QK
J
CK
Q
QK
J
CK
Q
QK
Ty theo cch rt gn ta c th a ra mt s khc.y ch l 1 cch
gii.Nhng kt qu cui cng chui s cho ra l ging nhau.
V d:xt mch m bt k nh s sau:
000 010 101 110 001
100 011 111
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Trang 82
S JKFF cn l 3
Bng gi tr:
Xung Q2 Q1 Q0
(hin ti)
Q2 Q1 Q0
(k tip)
J0 K0
J1 K1
J2 K2
1 0 0 0 0 1 0 0 X 1 X 0 X
2 0 0 1 1 1 1 X 0 1 X 1 X
3 0 1 0 1 0 1 1 X X 1 1 X
4 0 1 1 1 0 0 X 1 X 1 1 X
5 1 0 0 0 0 0 0 X 0 X X 1
6 1 0 1 1 1 0 X 1 1 X X 0
7 1 1 0 0 0 1 1 X X 1 X 1
8 1 1 1 0 0 0 X 0 X 1 X 1
Sau khi rt gn ta c
J0=Q1;K0= 2Q Q1+Q2 1Q =Q2Q1;J1=Q0+ 2Q ;K1=1;J2=Q0+Q2;K2= 0Q +Q1
S mch thc hin
J
CK
Q
QK
Q1
CK
Q2
J
CK
Q
QK
[1]
J
CK
Q
QK
Q1
Q1
Q0
Q2/Q0
Mch m song song m ln-xung
Xt mch m m ln-xung MOD 8 c chui s m l
0,1,2,3,4,5,6,7,0,1 thit k mch ny ta kt hp phng php thit k mch
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Trang 83
m ln v xung kiu song song.S dng cc linh kin mt cch ti u ta c
s mch nh sau:
[1]
J
CK
Q
QK
Counter Up/Q0
/Q0
J
CK
Q
QK
Q0
CK
Counter Down
Q0J
CK
Q
QK
Khi Counter Up tc ng (mc cao) th mch m ln
Khi Counter Down tc ng (mc cao) th mch m xung
Chui s m ly t ng Q0,Q1,Q2.
4.2.4.Mch chia tn dng FF
Mt ng dng rt quan trng ca FF l lm mch chia tn .Ta c th dng DFF
hoc dng JKFF
Mch chia tn s c tn s bng tn s ca xung CK chia cho s MOD ca
mch
Vi d :Cho mt mch nh sau:
D
CLK
Q
Q
D
CLK
Q
Q
CK
Q0
D
CLK
Q
Q
Q1 Q2
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Trang 84
Nu ly t ng ra Q0 th tn s s c chia i.Nu ly t Q1 th tn s s bng
na Q0 tc bng mt phn t CK.Nu ly t Q2 th tn s bng mt phn tm
CK.
Dng tn hiu c biu din nh sau:
Q0
Q1
CK
Q2
S mch nh sau:
CK
Q1
[1]
Q2
J
CK
Q
QK
[1]
Q0
J
CK
Q
QK
[1]
J
CK
Q
QK
Mch ny hon ton ging nh mch trn nhng dng JKFF
4.3.Nhp data vo FF
C 2 cch nhp data vo FF:nhp ng b v nhp khng ng b.
4.3.1.Nhp khng ng b
ALoad
CK
Q
QK
Pr
Cl
J
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Trang 85
Khi ta tc ng vo Pr v Cl
nhp A vo FF
-Xa FF bng cch cho xung Cl tc ng.Trong khi tc ng Cl th cho xung
Load=0,lc Pr=1.Ng ra Q=0,tn hiu A cch li vi FF
-Khi tc ng xung Load (Load=1)khng tc ng Clear th Pr= A
+Nu A=0 th Pr=1 lc ny FF c ng ra Q=0
Do Q=A hay ni cch khc A c nhp vo FF.
4.3.2.Nhp ng b
A
Load 1
B
3
1
Load 2
CKCK
Q
QS
R
2
Khi tc ng vo xung CK nhp A th cng 1 m v cng 2 ng
Load 1=1;Load 2=0,ng ra (3)= A ;R= A v S=A
+Nu A=0 th R=1 v S=0 ;khi c xung CK th Q=0
+Nu A=1 th R=0 v S=1 ;khi c xung CK th Q=1
Do A c nhp vo FF khi c xung CK
4.4.H ghi dch (Shift regiter)
Xt h ghi dch 4 bit dch phi
-Load:Xung nhp data
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Trang 86
-Clear:xung xa
-DSR:Data Shift Right
-Cc SR c ni thnh DFF
CK
Q
QR
S
Pr
Cl
Load
Q2Q1
BC
CK
Q
QR
S
Pr
Cl
Q0Q3
DSRCK
Q
QR
S
Pr
Cl
A
D
CKCK
Q
QR
S
Pr
Cl
Clear
+Khi tc ng vo Cl th Q3Q2Q1Q0=0000
+Khi tc ng vo Load th Q3Q2Q1Q0=DCBA
+Thc hin di phi
Xung Q3Q2Q1Q0
(hin ti)
Q3Q2Q1Q0
(k tip)
1 CDBA DSR1CDB
2 DSR1CDB DSR2DSR1CD
3 DSR2DSR1CD DSR3DSR2DSR1C
4 .
..
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Trang 87
*Xt trng hp Q3 ni vi DSR
Xung Q3Q2Q1Q0
(hin ti)
Q3Q2Q1Q0
(k tip)
1 CDBA ACDB
2 ACDB BACD
3 BACD DBAC
4 DBAC CDBA
5 CDBA ACDB
Trong trng hp ny ta c mch m vng
*Xt trng hp 3Q ni vi DSR
Gi l mch m vng xon.Gi s DCBA=0000
Xung Q3Q2Q1Q0
1 0000
2 1000
3 1100
4 1110
5 1111
6 0111
7 0011
8 0001
9 0000
Ta c th dng mch ny cho h thng n nhp nhy thng thy bn
ngoi.Tc nhanh chm ph thuc vo xung CK.
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Trang 88
BI TP CHNG 4
1.Thit k mch m, m ln MOD 13 theo 2 phng php song song v ni
tip.
2.Thit k mch ng h s hin th gi pht giy.
3.Thit k mch sao cho tn s vo l 32.768Hz v tn s ra l 1Hz.
4.Thit k mch o chu k tn s .
5.Thit k mch c chui s m nh sau:1,3,5,7,9,1,3Nu s ban u khng
nm trong chui s m th s tip theo bng 0.
6.Thit k mch m, m xung MOD 6 theo 2 phng php song song v ni
tip.
7.Cho mch nh hnh v.Xc nh mch m loi g,MOD l bao nhiu
J
CK
Q
QK
Pr
Cl
[1] [1]
CK
Q0 Q1 Q2
J
CK
Q
QK
Pr
Cl
[1]
J
CK
Q
QK
Pr
Cl