giao trinh ky thuat so

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

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

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

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

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

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

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

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 :

(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

. . . . .

- 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

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

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.

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

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 ++=

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 : +, -, *, / =, >,

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)

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)

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

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) :

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

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 :

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 :

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

* 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.

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

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

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.

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.

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

Bng gi tr (Truth table)

A X

0 0

1 1

2.4.2.Cng o (NOT gate )

K hiu

A


A X= A

0 1

1 0

2.4.3.Cng AND (AND gate)

A

B


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


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


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


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


A B X= AB

0 0 1

0 1 1

1 0 1

1 1 0

Cng NAND 3 ng vo :

CBA


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


A B X= BA + 0 0 1 0 1 0 1 0 0 1 1 0

Hm NOR 3 ng vo:

CBA


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


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

Ghi ch cng EX-OR khng c nhiu hn hai ng vo.

2.4.8.Cng EXNOR (EX_NOR gate)

A

B


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

* 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.

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

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

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

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

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

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).

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.

= )).().(( 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

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

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+

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

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

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

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

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

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:

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:

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

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

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

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

*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

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

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

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

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)

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

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

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.

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

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)

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

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.

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:

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

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


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



Tn Qn+1

0 Qn

1 nQ

T

CK

Q


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



Q

CK

K

J


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

[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=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=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

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

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:

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

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.

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

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

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

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

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

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 .

..

*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.

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

giao trinh ky thuat so

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