PHN II

K THUT XUNG - s

Chng 4

TM TT L THUYT

1 . 'ranzito ch chuyn mch (ch kha) c in p ra chi mt trong hai trng thi phn bit.

a) Trng thi in p thp khi tranzito m bo ha (viBi - T l khi c hai it ca n u m) vi gi tr 0 <

^ khi tha mn iu kinb) Trng thi in p cao khi tranzito kha dng (vi Bi-T

thng c gi tr Ujj^ gp 0,1E, 0,3E vi E l mcngun ni.

2 . IC ch khta chi 1 trong hai trng thi in p ra phn lit : hoc mc in p cao l hoc mc in

p thp l (gi l 2 mc bo ha ca IC, nu c nuibng ngun i xng E n c gi tr thp hn ngun t IV n 3V)

a) B so snh l 1 IC khda cd trng thi ra c thit lp nh hai in p t ti hai li vo p v N ca IC. Mt in p c chn lm mc ngng c nh (nu H Up tacd b so snh o, cn nu s Uj^ ta c D so sanh

118

thun), in p kia l in p tn hiu cn so snh nhn bit trng thi gi tr ca nd ang hn hay km ngng, th hin kt qu mc ra ang hay (ty loi so snhang s dng l thun hay o).

b) Nu s dng hai IC khda kiu mt thun mt o vi 2 ngng c nh khc nhau t ti chng v cng lm vic v mt in p tn hiu cn so snh, ta nhn c kiu bso snh ca s (so snh 2 ngng) cho php ta nhn bit cd nm trong (hay nm ngoi) khong ngng ny nh trng thi ra 1 trong hai tr bo ha tng ng. ,

3. B so snh 2 ngng c tr (Trig Smit) l b to dng xung vung gc cng tn s t mt tn hiu tun hon cd dng bt k. y l dng 1 b so snh 2 ngng ch dng mt IC v cc gi tr in p ngng c ly t cc mc ra bo ha^max ^max thng qua 1 mch hi tip dng. Khi in pcn so snh t ti li p ta c Smit kiu thun, ngcli, khi = Uj^ ta C Smit kiu o. Cc gi tr ngngc xc nh theo thig s ca mch hi tip dng bi ce h thc (3.9) n (3.13) SGK.

4. a) B a hi i dng to dng xung vung gc cd rng ty chn (theo tham s ca s ), vi chu k xung bng chu k in p kch thch li vo. Thi im xut hin in p kch thch (cng l lc bt u xut hin xung vung gc li ra) mang ngha l 1 mc thi gian nh du lc bt u hay kt thc mt thao tc no trong mt h c iu khin (ch ng c ch i). H thc xc nh tham s xung l (3.19) (3.21)

b) B a hi t dao ng dng to xung vung gc c chu k v rng t chn (theo tham s ca s , xem cc cng thc (3*23), (3.26) (3.27) v (3.28). Cc xung vung do a hi to ra cd n nh tn s cao (nh vo bin php k thut c bit) c dng lm dy xung nhp o thi gian v iu khin trt t lm vic ca mt h thng xung s.

5 . B to xung tam gic da trn nguyn l mch tch phn to dng in p bin i tuyn tnh theo thi gian. in

119

p tam gic o Coi nh 1 dng tn hiu chun theo hai bc t do (theo ln v theo khong thi gian) c th thc hin c php bin 'gia hai i lng ny 1 cch n tr (trong nguyn l ADC).

a) Cd th s dng qu trnh phng in hay np in chm cho 1 t in bng 1 dng in n nh t 1 ngun n dng to xung in p dng tam gic. Cht lng xung tam gic do n nh ca ngun dng quyt nh.

b) C th kt hp 1 b to xung vung gdc v 1 b to xung tam gic (ni tip pha sau) thc hin trong 1 vng hi tip ng thi to ra 2 dng tn hiu trn (h.3.30 SGK), in p ra ca b ny dng lm in p vo iu khin ca b kia khng cn dng kch thch ngoi.

6 . i s logic l cng c ton hc phn tch v tng hp trng thi ca cc mch s. Quan h logic (hm logic) gia cc bin trng thi (gi l bin logic) c thc hin nh ba php ton logic c bn : php ph nh logic, php cng logic (hoc) v php nhn logic (v) kt hp vi cc nh lut c bn : lut hon v, liit phn phi v lut kt hp gia cc php cng v nhn logic v hai hng s 1 v hng s 0 .

^ sl) Lut hon v i vi php cng v php nhn logic : nu k hiu cc bin logic l X, y, z, php cng (du +), php nhn (du .) th :

Vi php cng logic : x + y + z = y + x + z = z + x + y = ... Vi php nhn logic : x . y . z = y . x . z = z . x , y = . . .

b) Lut phn phi gia php cng v php nhn logic :

x(y + z) = xy + xz.

c) Lut kt hp gia 2 php cn^ v nhn logic :

X y H- z = (x + y) + z = z + (y + z) = ...

: X, y . z = (x . y ) . z = X . (y . z) = ...

7. Cn ghi nh 10 tin (quy tc) quan trng ca i slogic i vi cc php tnh logic nu :

120

a) Quy tc vi php ph nh logic : (2 = X(X ) = X.

b) Quy tc vi php cng logic x + x = x ; x + l = 1 x + 0 = x ; x 4 * x = 1c) Quy tc vi php nhn logic x . x = x ; x . 0 = 0 x . l = x ; x . x = 0 .d) Trong s cc nh l suy ra t h tin trn, nh l

lp hm ph nh ca 1 hm bt k cho (nh l Demorgan l quan trng nht :

F (x , , z , ...) . , + , = F (x , y, Z, ...) + , . ,

nh l Demorgan cho php xy dng cc cu trc logic c tnh ng nht cao, tnh i ln cao v nh ti u v tnh kinh t k thut cng nh cng ngh thc hin n giB r tin hn. Ch rng cc quy tc v lut nu trn cng ng cho trng hp cc k hiu X, y, z i din cho 1 t hp phc tp ca cc bin logic.

8 . Vi 1 hm logic bt k cho trc cch biu din quan h hm d dng mt biu thc k hiu hm, bin v cc php ton logic gia chng l ph bin nht, trong c 2 dng c bn :

a) Biu thc c dng l 1 tng ca cc tch cc bin logic. Mi s hng ca tng c th mt cc bin (dng y ) hay khng mt cc bin (dng khuyt) :

V d : Fj = xy + xy ; F2 = xy + xy

F 3 = x y z + x y z + x z + x y i ( d n g y )

F4 = xy + z (dng khuyt).b) Biu thc c dng l 1 tch cc tng cc bin, cng c

th dng y hoc dng khng y (khuyt).V d : F2 = G3 = ( x + y + z ) ( x+ y + z ) ( x+ y + z) X

X (x + y* + z)F4 = G4 = (X + ) Z-

9. Hm logic bt k cn c th biu din tng ng hai dng thng dng khc.

121

a) Hm c biu din di dng 1 bng lit k mi trng thi gi tr c th ca cc bin v gi tr tng ng ca hm tng trng thi k (gi l bng chn l).

V d vi cc hm nu trn ;

Fj = xy + x ; F2 = x +

hay F3 = xyz + xyz + xyz + xyz.

ta c cc bng chn l tng ng sau :

0 0 0 1 1 0 1 1

1001

X y0 00 11 0 1 1

01

1 0

X y z F30 0 0 00 0 1 00 I 0 0

0 1 1 11 0 0 01 0 1 11 1 0 11 1 1 1

b) Hm c cho di dng mt hnh cc vung (ba Cacno) sao cho mi tng ng vi 1 kh nng (1 trng thi)c th ca cc tr cc bin logic v 2 k nhau (tnh k nhauxt vi c bin gii gia cc hng v cc ct mp ba) ch c php ct 1 bin logic khc tr s nhau, cc bin cn li ca chng phi ng tr. Nh vy mi cng tng ng vi mt s hng ca tng trong cch biu din bng biu thc hay 1 dng trong cch biu din bng bng.

V d vi cc hm Fj, F2 v F3 nu trn, ta c (ch : Trng thi no t hm nhn tr 1 ti tng ng sc gn s 1, cc ng vi tr F = 0 s trng hoc ghis 0 )

122

X .

0 F ,v0 1 Fxy

yi 01

11

0

1

1

1

01

r------00 01 V 10

11 1 1

10. Cn nm vng cc phng php biu din hm logic nu trn v cch thc chuyn i t dng biu din ny sang dng khc, khi chuyn cch biu din, cn lu cc nhn xt sau :

a) Cc cch biu din bng bng hay ba Car no ch tng ng vi dng biu thc y ( mt tt c cc bin trong tt G cc s hng). Khi gp dng rt gn, trc khi chuyn sang biu din bng bng hay ba, phi a biu thc hm v d n g y n h cc q u y tc_thch hp (v d X + X = 1 ;X + X = X . . . , x . l = x , x . x = 0

b) Dng biu thc l tng cc tch (y ) tng ng vi cc dng (hay cc ba) hm logic nhn tr 1. Ngc li dng biu thc l tch cc tng cc bin s tng ng vi biu din ca hm o (ca hm cho dng tng cc tch) v do vy s tng ng vi dng hay hm nhn tr 0 .

V d : ta ly cch biu din bng hay ba ca Fj hay F2 cho :

X y Fi F20 0 1 00 1 0 11 0 0 11 1 1 0

X X

0

1

0 1 ^2 >1

1 y

Nu vit Fj vi dng 2 v dng th 3 (tr Fj == 0) tac

Fj = ( + y)(x + y) (2)

Nu vit Fj theo cc dng 1 v dng 4 ta ct :Fj = (x + y)(x + ) (3)

Nu khai trin (2) hoc (3) ta s a c Fj v ng nhtvi dng (1) ; v d :

T (2) : = XX + x + yx + y (p dng lut phn phi)

= 0 + x + y .x + 0 (p d n g t i n XX = 0).

= x xy (p dng lut hon v)hoc t (3) lp :

= Fj (theo tin 2 ln ph nh) = (k + y) . (x + )

= X . + X . (theo nh l Demorgan)= x + xy (tin 2 ln ph nh).11. Ti thiu hda hm logic bi ton a hm v dng

rt gn theo cc ngha : S lng cc php ton logic (hay cc phn t logic thc

hin php ton tng ng) ng thc hin hm logic cho l t nhat/

S loi phn t (loi dng php ton logic) thc hin hm l ti thiu.

Khi s dng quy tc Cacno ti thiu ha hm logic (dn ) cn ch cc nhn xt quan trng sau :

a) Quy tc pht biu l nu cd 2" c tr 1 nm k nhau hp thnh 1 khi vung hay ch nht th c th thay 2 nh ny bng ch 1 ln vi s ng bin gim i n.

b) S nh c gom li trong 1 ln phi hnh thnh 1 khi yung hay ch nht v l ti a t mc c th, tha mn iu kin 2 (vi n l 1 s nguyn n = 1, 2, 3...)

c) S cc n (nhm) c lp (khng cha nhau) sau khigoxn li l lng t nht. f

124

d) S cc nm ti mp ba theo nh ngha cng l cc nm k nhau (l chi c 1 bin khc tr nhau).

e) 1 nh c tr 1 c th tham gia ng thi vo nhiunhm ( ln) khc nhau do h qu ca tin X + X = X.

f) Nu trong biu din ba Cacno ca 1 hm no c cc m y hm khng xc nh (cc t hp trng thi khng dng n) th c th s dng chng cho mc ch ti thiuha bng cch gn cho ny tr 1.

g) Nu s lng cc trng (c tr 0) t hn th c th ti thiu hda hm ph nh logic ca hm cho bng cch dn cc tr 0 ging nh quy tc lm i vi cc tr 1 nu trn.

12. a) Cc hm logic c bn bao gm :Hm ph nh logic (khng) Fj^Q = XHm nhn logic (v) = Xj . %2Hm cng logic (hoc) Fqj = Xj +Hm v -khng = x^Tx ^ = Xj +X2

Hm hoc -khng Fj^ Qj^ = Xj~ -t- = Xj . X2

b) thc hin 5 hm logic c bn, ngi ta xy dng 5 phn t logic c bn (bng cc mch in t thch hp), chng cd tn gi tng ng v c k hiu l :

^NO X

F.nand = Xj -X fN f r n f i = X jtX i

Hnh 4.1

125

,/c) Cc phn t v -khng, hoc -khng c tnh tng thch k thut cao, tnh vn nng th hin c im l cc phn t logic c bn cn li u ct th c xy dng ch t 1 vi phn t v -khng hay 1 vi phn t hoc -khng :

V d : t v -khng ta ct th nhn c cc phn t cn li bng cch sau :

X- d w > -

Hnh 4.2

13. Cc hm logic thng dng thng gp bao gm :

Hm khc du (hay cng mun nh phn) v k hiu phn t logic tng ng = XjX2 + XjX2 = X2

Hm cng du (hay hm tng ng) v k hiu phn t tng ng :

^t = * 1 * 2 + * 1 X 2 = X j e X2

Hm a s :

^ s = * 1 ^ * 2 * * 3 = * 1 * 2 + * 2 * 3 + * 1 * 3

= + jX2X3 + XjjX3 +, XjX2X3

s = X j Hm na tng T^ _ip = XjX2 Ff

126

Hm tng y : S|J = [X|j Yi^ ] P|J_J

Pk = *kyk + f*k yic Pk-IBng trng thi cc hm trn :

XjX2 Fkd na tngs p

0 0 0 1 0 00 1 1 0 1 01 0 1 0 1 01 1 0 1 0 1

X1X2X3 ^a s *kykPk-1 t^ngPk

0 0 0 0 0 0 0 0 00 0 1 0 0 0 1 1 00 1 0 0 0 1 0 1 00 1 1 1 0 1 1 0 11 0 0 0 1 0 0 1 01 0 1 1 1 0 1 0 11 1 0 1 1 1 0 0 11 1 1 1 1 1 1 1 1

Tt c cc hm trn u c th xy dng t cu trc hnhp cc phn t c bn (khng, v, hoc logic) hay t cu trcthun nht ch gm cc phn t v -khng hoc ch gm cc phn t hoc -khng. thc hin c cu trc thun nht thng phi bin i biu thic ca hm v dng thch hp nh nh l Demorgau.

14. Trig s c xy dng t 1 cu to gm 2 phn tNAND hoc hai phn t NOR bao nhau nh 2 vng hi tipdng kn ;

a) Bng trng thi tng ng :

127

Sn Rnn n Qn+i :0 0 0 1 1 0 1 1

cm10

. ' Qn

Sn Rn Qn+10 0 Qn0 1 01 0 1 .....1 1 cm

mb) Lu nhm c cu to t NAND chi chuyn bin vi

sn m (i xung "1 -0") ca xung vo, cn nhm vi cu to t NOR ch chuyii trng thi ra vi sn dng (i ln

0 - 1) ca xung vo :

Q

Hnh 4.3

c) Cc loi Trig s phc tp hn (D, T, MS, JK) u c xy dng trn c s t hai cu trc c bn nu trn.

d) Trig JK c tnh cht vn nng, tc l t n c kh nng xy dng tt c cc loi RS, T... cn li. Hai bng trng thi ca Trig m T v Trig vn nng JK c dng :

Tn0

1

Qnur

n+ 1 'T* 'QQ

J n K n Q n+1

0 0 Q n0 1 01 0 11 1 Q n

Tc l Trig T lt sau mi xung vo ca m T.

Trig JK ct 3 kh nng hot ng :

128

Khi J = K = 1 n lm vic nh Trig m T. Khi J K trng thi ra c tr ging trng thi gi tr

li vo J Khi J = K = 0 trng thi ra ca Trig c bo ton

(gi nguyn nh trc )..

Chng 5

BI TP PHXn II C LI GII

1 P'

t

u

R.

Hnh 5.1

Bi tp 5. . Cho mch in hnh 5.1Bit rng E = 15V.

;:,ax= + 1 2 V ; - 1 2 VRj = 10 kQ ; Rj = 30 kQ

Uj(t) cd dng in p hnh tam gic i xng qua gc ta vi bin = 6Vv chu k Tj = 20ms.

a) Hy v dng c tuyntruyn t in p ca mch 2 (Uj) trong hai trng hp :

1) IC l l tng (vi tc chuyn mch gia 2 mc bo ha l v cng n - thi gian tr chuyn mch bng 0)

2) IC thc t c tc tng in p l 0,5 s /vb) Xc nh dng 2 (t) v cc tham s : chu k, bin v

thi gian tr pha u ca U jt) so vi j(t) bit khi coi IC l l tng.

c) nhn c gi tr bin trong gii hn :- 0,6V 2 ^ +5V vi I2 = lOmA

cn b sung 1 mch hn ch bin li ra, xc nh gi tr in tr v v dng mch ny.

9-BTKTT-A 1 2 9

Bi gi :a) Ta tm dng c tuyii 2 (Uj) trong trng hp l tung.

Mch cho cd dng l 1 b so snh c tr kiu o (Trig Smit o) vi 2 mc ngng t ti li vo p l :

Khi 2 mc bo ha dng : 2 = = + 12V, quamch hi tip dng Rj v R2 vi h s hi tip :

R, 1 0 kQR j+ R 2 10kQ +30kQ

ngt

= 0,25

ta nhn c = 0,25 . 12 = + 3V.

Khi 2 mc bo ha m 2 = = - 12V

ta nhn c ngng th hai ca s :

u ; = = 0,25 (-12V) = - 3V

Vy c tuyn. truyn t l tng c dng hnh 5.2a.c tuyn truyn t thc t vi tc thay i in p i

ra l 0,5 fis/v, chuyn t mc bo ha dng sang mc bo ha m hoc ngc li cn tn 1 khong thi gian chuyn mch sau khi J t ti ngng l :

I( tr thuyn

mch)

Ung Ung-v 0 3^

~2^

u,

b )

Hnh 5,2

130 0-BTKTT-B

t^r 0,5 sfv ( u ;^ + I u ; , j )max'= 0,5 . sV . 24V = 12 s.

b) Xc nh cc tham s v dng ca 2 (t) trong trng hp l tng, vi Uj(t) dng tam gic cho trc.

Biu din U jit) theo Uj(t) ta nhn c th hnh (5.3). Dng 2 (t) l 1 xung vung gdc, cng chu k Uj(t) ; .

T2 = Tj = 20ms

Bin 2 (t) d mc bo ha v ca IC quyt

nh nn = + 12V (= v

U z ^ i n = - u ^ ; 3 , = - 1 2 V .

Thi gian chm pha ca 2 so vi Uj c xc nh bi thi gian tng ca Uj(t) t lc t = 0 n lc t = l lc Uj t tai n ga.g = + 3V tri d dng xc nht quy tc taitt gic ng dng OAB v OAB c cc cnh tdng ng :

1 3 1 '

OA OB ABOA " OB AB OB =

O B . AB AB

^T,. 5{m s), 3(V)Suy ra = OB = ------------------------------------- ------ = 2,5 ms.

c) Mch hn bin b sung vo c v nh hnh 5.4Chn mc 2 = 5y ta cd :khi 2 = - 12V =

(trong khong < t < tj)

Dj, m theo chiu thuu nh 1 it thng thng v U3 = - Uj3 = - 0,6V.

Cn trong khong thi gian tj < t < t2,

Hnh 5.4 u, = u ;;,, = + 1 2 V.2 maxlm vic ch i nh thng zener, do :

3 = = + 5V. Vy mch hn bin thc hin c iukin hn ch 2 pha - 0,6V + 5V.

Vi dng I2 = lOmA, st p trn in tr l+12V - 5V = + do vy trong khong tj < t < 2 tr s

c xc nh bi1 2 V -5V

hc = - 3 = 700Q' 10.l (^A)Trong khong < t < 2, st p trn Rjjj, l

-12V + 0,6V = - 11,4V do = l,l4k Q .

Kt qu chn gi tr l tr trung bnh ca hai gi tr tnh trong 2 na chu k ca U2 (t) :

700 + 1140- 2 2 = 9 2 0

1 3 2

Lu rng, nu ti tnh cht q (tr) ca vi mch thc th dng U 2 (t) v do d u^t) c khc i th hin trn hnh 5.5

Hn/ 5. 5Ta nhn thy ngay rng do tc dng ca khu mch hn

ch, thi gian tr ca in p li ra gim i ng k so vi trc y :

Khi cha c mch hn ch, ta tnh c thi gian cn thit 2 chuyn t mc n hay ngc li l :

= 0,5 fisW . [+ 12V - (- 12V)] = 12/is.Sau khi c mch hn ch, thi gian ny l :

= 0,5 /XSW . [+5V - (- 0,6V)] = 2,8 fist^r c khong trn 75% so vi gi tr trc.

(Ch rng thi gian tr tnh c ph hp vi loi IC khng chuyn dng, trohg trng hp dng IC so snh chuyn dng, thi gian ny s gim i hai hay ba cp).

133

Bi tp 5.2. Cho mch in hnh 5.6a) Nu cc nhim v ca

R O)

/V

cp i >

-1--E

w )

A

4--_L

mch cho.

b) V cc dng in p bin i theo thi gian ti cc im N, p, A ca mch (gi thit bit

c) Bil R = lOkQ ;VR = lOkQ

Hnh 5,6= 9,lkQ

c = 0,ljuF ; E = 15V

Xc nh gi tr tn s ca in p ti im A tng ng vi hai v tr gii hn (1) v (2) ca VR.

d) Xc nh dng in p ti A v tr s in tr khikhng ti, bit rng it Zener c = +5V ; = lOmA v

- + 3,6V.

Bi gii :

a) Mch cho c dng 1 b a hi t dao ng dng IC ch khda, c bao vng hi tip dng dng R , R2 kthp vi mt khu mch hn ch bin in p ra dng D^ .Vy mch c hai nhim v :

1 ) T to xung vung gc c tn s thay i c (do IC,R, R2 , R, VR v c m nhim).

t *

2) Hn ch bin xung vung to ra c hai pha trnv pha di (do R ,^ v Dz thc hin).

b) IC lm vic ch kha nn bin in p li ra ch mt trong hai trng thi bo ha ca vi mch l

== (khi c bo ha dng) haybo ha m), c dng l 1 xung vung gdc, qua mch hi tip dng R, R2 , ti/li vo p c mt trong hai in p ngng t ti l

134

" Rj + Rj + R2 ^ max

= ^ < 3 x 5 = - ^ u ; 3,

R, y ta k hiu = -p p l h s hi tip dng.

xrC J ^^ 2^

Cc gi tr hay U g^ qua mch R, VR np (hay phng)

cho t c cho ti khi t ti ngng (hay ths lt sang trng thi bo ha kia. vic phn tch trn,cc in p Up, U ^ = V v c dng nh hnh 5.7.

c) Chu k ca xung vung gc u^(t) c xc nh theo h thc :

2RT = 2(R + VR). c . ln( 1 +

khi Rj = R2 ta c

T = 2 , 2 (R + VR) . cTi ( 1 ) VR = 0

ta c = 2,2 RC ; thay gi tr R v c cho c := 2,2.10.10^ . 0 ,1.10 '^ = 2,2.10 '^s

ti (2) c VR = IDkQ, thay cc gi tr cho ca R v c vo, t c :

T(2 ) = 4,4 . 10.10^ . 0,1.10'^ = 4,4.10'^s.

Vy tn s ca u^(t) thay i trong gii hn :

F ,..='1

1 1'^ (2 ) 4,4.10'^

227 Hz.

135

Ur

u* \ imax

O

max

tr t;

d) Khu mch gm Rq, Eq, Dz nh nu trn cd nhim v hn ch bin xung vung gdc ^(t) c 2 mc trn v di. Cc ngng hn ch ny c th xc nh c t cc mc boha u

= 12V vmax= - 12V. V itmax

Zener lm vic 2 ch nn :

1) Khi =^ax = 12V (trongkhong 0 < t < tj) th Dz ch n p vi = + 5V dovy = , + E= + 5V + 3,6V = 8 ,6 V

Ngng hn ch pha trn ca S l + 8 ,6 V.

2) Khi =(trong khong

thi gian tj < t

h - C R p ,.ln (l

Tnh vi 1 gi tr c nh ca VR ^ (gi s c = 650Hz)min

16 t A 1 f>3T = 0,01.10", 10.10^ In 1,4.

2 = O.Ol.lO * 90.10^.1n 1,4 t y :

1 = 5

Bi tp 5.5. Cho cc bin logic Xj(t), X2 (t) vi th thi gian bit trc dng hnh 5.12. Hy xy dng th thi gian ca cc hm logic c bn ca hai bin cho.

a) th cc hm ph nh ca Xj(t) v X2 (t) c xy dng t biu thc v bng trng thi

^NOl ^NOl X2 Fn 0 2^N0 2 ^ 0 0 1

1 0 1 0

N^O.2

0

0

a

0 0

0

I

\0

I I

1 "T I >

I .1;

1 1;

i 1. 1

0 0 0

Hnh 5.2 a)

p dng bng nh ngha xt trong tng khong thi gian khc nhau khi cho bin thi gian tng dn t tr t = 0 ta nhn c th tng ng ca Fj^ jQi v ca Fj^ Q2

b) T biu thc nh ngha v bng chn l ca cc hm v (nhn logic) v hm hoc (cng logic) ta c :

143

^AND ~ ^ 1 ^ 2 Xi X2 F A N D ^ORFor = X, + X2 0 0 0 0

0 1 0 1

1 0 0 1

1 1 1 1

Tng dn t t gi tr t = 0, xt trong tng khong thi gian ( Xj v nhn nhng gi tr c nh cho theo gi thit), p dng kt qu gi tr hm trong bng chn l vit, ta nhn c th hnh .1 2 b.

FNOR

1 i- p _ ,_

01 1 1

1 ' 1 0

0

J i 0

I I

0 ! 0

Hnh S.12b)

144

c) Vi cc hm v ph nh v hoc ph nh,' ta c biu thc nh ngha : ^NAND = Xj . X2 v

F n o r = + X2Bng trng thi tng ng ca chng l

Xi X2 ^NAND ^NOR0 0 1 1

* 0 1 1 0

1 0 1 0

1 1 0 0

p dng bng trng thi trong tng khong thi gian theo gi thit Xj v X2 nhn 1 gi tr xc nh v khng thy i, ta c kt qu tng ng ca cc hm hoc Fj^ Qjtheo tng dng thch hp ca bng trng thi v nhn c th kt qu trn hnh 5 .12b.

Ch l c kt qu th cc hm v tac th dng biu thc nh ngha chng l hm o ca v Fqj, do vy c th nhn trc tip kt qu th ca hai hm ny bng cch nghch o (ly ph nh) cc kt qu thu c ca v Fqj. Phng php tm dng th nu trnct th m rng cho 3 bin Xj(t), X2 (t), Xj(t) hay nhiu bin hn da trn cc nh ngha c bn i vi hm nhiu bin v bng trng thi tng ng ca chng.

Bi tp 5.6. Cho cc mch logic c cu trc hnh 5.13. a) v b) vi 2 u vo c cc bin logic X j v X2 tc ng, 1 ura nhn c cc hm logic ln lt l F v F2 -

a) Hy tm biu thc ca Fj v F2 dng y

b) Bin i cc biu thc tm c cu a) v dng tigin theo hai cch : dng tng cc tch v dng tch cc tng cc bin qua chng minh rng Fj = F2 -

O-BTKTT-A 145

x,p-D-

D B

Cn'B' -5

3 ) b)Hnh 5.13

c ) Tim C u t r c tng ng vi c u t r c (5.13) t r o n g ch s dng 1 loi phn t NAND (hoc ch 1 loi phn t NOR).

Bi gii : a) K hiu thm cc hm logic trung gian A, B, c , D trong hnh 5.13 a) v b) ln lt vit cc quan h hm t u vo ti u ra ca tng cu trc, ta c :

Vi cu trc (5.13a), theo nh ngha v k hiu cc hmlogic c bn, c cc quan h :

c = Xj ; A = Xj + X2D = 5^; B = c + D = x, +X2

T t c Fj = ,B = (Xj + X2 )(Xj + X2)

Vi cu trc (5.13b) nhn c ;C = ^ ; A = . C = X j^

D = Xj ; B = X2.D = Xj Xj

T c F2 = A + B

= X,X2 + X2X,

Vy dng y ca hai hm logic cn tim l

= (Xj + X2)Ci + X2) v F j = X,X2 + XjXjb) Bin i Fj v F2 v dng thu gn theo nh l Demorgan

Fj = (Xj + X2)Cj + X2) (1)

146 10-BTKTt-B

p dng tin 2 ln ph nh (x) = X c kt qu :Fj = Xj . X2 + Xj . X / (2)

Dng (1) ca Fj l cch biu din tch cc tng cc bin.Dng (2) ca Fj l cch biu din tng cc tch cc bn.Vi hm F2 , p dng nh l Demorgan c

F2 = XjX2 + X^X

= . X^Xj

= (Xi + X )^(X2 + X,)

p ng tin (X) = X v lut Qn b gia php cng v php nhn, sau ch tin XX = 0, ta nhn c :

F2 = (Xj + X2)(X2 + X) (3)= Xj . 5^ + X j . Xj + X2X2 + X2X1

= XjXj + X2Xj, do lut hon v vi php nhn :

= XjX2 + XjX2 (4)Dng (4) biu din kiu tng cc tch v dng (3) l biu

din kiu tch cc tng cc bin ca F2 .So snh dng cc biu thc (2) r_(4) ta nhn c iu

phi chng minh : Fj = F2 = F = XjX2 + XjXj ' (4)

c) Thc hin ph nh lin tip 2 n v phi ca (4) c

F = + X1X2

p dng nh l Demorgan cho du ph nh bn trong ca F c :

= (X, , x^) + (X, . X2 )

F = X1X2 . XjXj = A.B (5) y ta k hiu A = v B = x jx (6 )Cc hm (5) v (6 ) c thc hin vi 5 phn t NAND

loi c 2 u vo. Do vy cu trc tng ng cn tm c dng sau (h.5.14> :

147-

o B ^ x^.x^

c > -

A -F = A . B

- Y - , \ + x ^ x

Hnh 5.14.

Nu t cc biu thc (5) v (6 ) tip tc p dng nh l DeMorgan, c

F = + B (7)A = Xj + X2 A = Xj + X2

B = X j + ^ B = (Xj +X 2 )

thay cc biu thc (8 ) vo (7) cF = X T ^ + X, +X 2 (9)

T (9) ta nhn c cu trc loi thun nht dng 6 phn t NOR (h.5.15K

/O-

B

Hnh 5.15

Bi tp 5.7. Mt hm logic 3 bin F(Xj, X2 , X3 ) gm c 6 S hng, dng y c biu thc sau :

F(Xp Xj, X3) = Xj X2 X3 + Xj Xj X3 + Xj X2 X3 +

+ XjX^Xj + XjXjXj + XjXjX,

a) Hy thit lp bng trng thi v vit- ba Cacno ca F.

148

b) Tm biu thc ti thiu ca F nh quy tc Cacno

c) Xy dng cu trc thc hin F t cc phn t NOR c2 u vo.

Bi gii :

a) Bng trng thi c thit lp nh lit k tt c cc trng thi t hp c th c ca cc tr cc bin vo v gi tr tng ng ca hm nhn c vi mi trng thi k. Vi quy c rng khi bin logic Xj nhn tr 0 ta vit l X, cn khiX nhn tr 1 ta vit l X, nh vy biu thc ca F c dng :F = ^ 0 + nij + m2 + m^ + m +

Vi = X, X2 X3 = 0 0 0 m4 = X ,X 2 ^ = 1 0 0

mi = X2 X3 = 0 0 1 5 = = 1 0 1

m2 = X2 X3 = 0 1 0 ; 6 = 1 1 0

Vit di dng bng trng thi v bia Cacno ta nhn c hnh 5.16

Hl F

*"0 0 0 0 1m. 0 0 1. 1m2 0 1 0 1m3 0 1 1 0r r . 1 0 0 1m 1 0 1 1- 6 1 1 0 1n i j 1 1 1 0

0

Nhm

00-

01 n /

/ / hm c

Hnh 5.

Cc t hp bin i c gi l cc mintec F = 1 khi c t nht 1 mintec Iti nhn gi tr 1 .

b) T hnh Cacno thit lp, theo quy tc Cacno vi cc ch km theo c th thit lp c 2 ahm ln i vi cc (cc mintec) c tr 1 nh sau :

149

Nhm A bao gm 2 ^ = 4 hng trn ng v tr X3 = 0.

^Nhdm B bao 2 ^ = 4 nm ti 4 gdc ba ng vi 2ct XjX2 = 00 v XjX2 = 10.

Mi kh nng kt hp to cc nhm khc vi 2 nhm A v B nn trn u cha t hn hoc b cha trong A hocB (tc l vi phm cc iu ch trong khi thc hin quy tcCacno v do vy l khng ti gin).

Vy c kt qu ti thiuF = + nij + m2 + = A + BA v B l hai s hng mi ca F c s bin gim i 2 . Trong mi nhm, khi chuyn t 1 nh ny sang nh

k vi n, bin no c gi tr thay i (o tr) th s khng cn trong kt qu. v d trong nhdm A, t hng 1 ct 1 n cui hng 1 ct 4 cd X v X2 n n lt o tr nn kt qu ta nhn c A = X3 (X3 nhn tr 0 ).

Tng t vi nhdm B cd B = X2 (X2 nhn tr 0).

T F = 5^ + X2 = X ^ T lq

y l dng ti gin ca F. Ta cng c th ti gin F bng cc quy tc v nh l

ca i s logic, v d :A = + X,X2X3 +

= + m2 + + 1114thc hin nhtm nig vi v m2 vi rig c :

A = (Xj + X,)X2X3 + + Xi)X2X3.

p dng quy tc X + X = 1 v X . 1 = X, ta c ;A = X2 X3 + X2X3

= 0 ^ 2 ^ 2 ) ^ 3= X3

150

Tng t vi B = + rtij += X,X.X3 + X,X2X3 + X.X^Xj + X1X2X3

B = (X, + Xi)X2X3 + (Xj + X|)X2X3

X.X, + X-,. X,= X2X3 + X2 . = X2

B i tp 5.8. Cho hai hm logic 3 bin cd dng sau := X iX 3 -f*X3 X2 + X 2 Xj ( 1 )

- XjX 3 +X 3 X2 +X 2 Xi (2 )

a) Chng minh rng v F2 l 2 dng rt gn ca cng1 hm F. Tm biu thc y ca F v lp bng trng thi,ba Cacno ca F.

b) Xy dng cu trc thc hin hm F bng cc phn t NAND t 1 trong hai dng rt gn cho.

c) Tm cu trc thc hin F nh cc phn t NOR cd 2 u vo.

Bi gi :a) Trong cc biu thc (1) v (2) cho cc s hng u

vng 1 bin, trc tin cn a chng v dng y ( bin) bng cch thm cc tha s (X + X) = 1 i vi bin vngmt i vo, p dng lut phn phi v hon v, t (1 ) cd :Fi = X iX 3 (X2 +X 2 ) + X 3 X2 ( X , + X , ) + X 2 Xi(X3 + X 3)

= Xj X 2 X 3 + X ,X 2 X 3 + X ,X 2 X 3 + X i X2X3+Xj } ^ X 3 + X j X2X3

Vi cch quy c ni ti bi tp 5.7, ta c := 0 0 0 = XiX 2 ^ = 1 0 0 =

X1 X2 X3 = 0 0 1 = XJX2 X3 = 1 0 1 =

X1 X2 X3 = 0 1 0 = m2 X1 X2 X3 = 1 1 0 =

X1 X2 X3 = 0 1 1 = m3 XjXjXj = 1 1 1 =

ta nhn c biu thc thu gn ca Fj d dng :Fj = + nij + + m2 + (3)

Tng t vi F2 c : F2 = X jX 3 (X2 +X 2) + X 3 X2 ( X j + X i ) + X 2 X i ( X 3 +X 3)

= X1 X2 X3 + X 1 X2 X3 + X 1 X2 X3 + X 1 X2 X3 + X 1 X2 X3

+ X 1 X2 X31 5 2

v biu thc thu gn ca F2 :

So snh 2 biu thc (3) v (4) vi ch p dng lut hon v i vi php cng logic nhn c kt qu F v F2 chnh l 2 dng thu gn ca cng 1 hm F vi F CQ dng y l :

F = Iij + m2 + + ini ^ + m .^

T y, cd bng trng thi v ba Carno ca F nh hnh 5.18.

F2 = + mj + II3 + m2 (4)

m X1X2X3 F

0 0 0 0

mi 0 0 1 1m2 0 1 0 1m3 0 1 1 1- 1 0 0

1 0 1 1

1 1 0 11 1 1 0

0

X-

r - .7 /

^ ( l

Hnh 5. 8

b) Xy dng cu trc thc hin F t phn t NAND c 2 u vo. Xut pht t 2 biu thc (1) v (2) ca Fj v F2 :

Fj = XjX2 +X 2 X3 +X 3 XJ

F2 = XJX2 +X 2 X3 +X 3

Ly ph nh 2 ln cc biu thc trn sau p dng nh l DeMorgan vi Fj ta c :

(5)

trong - Xj X2 ; = X2 X3 v Cj = X3 Xj

Tng t vi F2 ta c :

153

F2 = XJX2 + X 2 X3 + X 3 X1 = XiX 2 .X2 X3 .X3 X^

A2 .B 2 .C2 A2 B C2 (6)

T cc biu thc (5) v (6 ) thu c c th xy dng cc cu trc sau (hnh 5.19a, v b) :

LLP

, c ,

j_

Br

F, --

B,c,

a)

0^ - 0 ------

JO- c >

G

!Hnh 5.19

154

c) xy dng cu trc Fj v F2 t cc phn t NOR hai u vo, ta vit li cc biu thc (1) v (2) di dng khc sau khi p dng tin hai ln ph nh :

Fj - X1X2 + X2X3 + X3XJ = X1X2 + X2X3 + X3XJ

F2 = + X2X3 + X3XJ = + X2X3 + X3Xj

hay dng d nhn hn :

F2 = XjX ^ + X2X3 + X3XJ = + B2 + C2

T cc biu thc (7) v (8) vi

(7)

(8 )

X, + X2

X2 + X3

X3 + x .

X, + X2

X2 + X3

X3 + X,

xy dng c cc cu trc hnh 5 .20a v hinh 5 .20b.

A

,

c.

F,

B,fC,

Hnh 5.20 a )

155

X,

X,

&

0-C ^

C:

F^

Bp t c

B tC

Hnh 5.20 b)

Bi tp 5.9. Mch in hnh 5.21 l 1 trig RS c cc tham s sau ;

E = 5V ; = R, = 2,7 kQ ; = Rj = 15 kQ

= R2 = 27kQ

Tj, T2 loi 2N3904 c yS = 70

R t ,

+

Q

I

rs V

L

R U

h ,

a) Tnh cc gi tr in p baz v colect ca Tj v T2qua v ic p hn tch hot n g c a m ch.

Xc nh in p trn cc t chuyn mch (t nh) Cj vC2 khi s 1 trong hai trng thi n nh bn.

b) V dng th thi gian (l tng) ca in p v

khi ti lc t = c 1 xung in p cc tnh m t ti

cc i vo s (hoc R).

Bi gii :a) Mch hnh 5.21 ch c 2 trng thi phn bit : Tj dng

T2 ngt (khi d = 0 ; Q2 = 1 ) v trng thi Tj ngt T2 ng (Q = 1 , Q2 = 0 ) ; khi ch ng, dng colect ti thiu tha mn I < l^.

Gi thit rng Tj ng T2 ngt, dng q qua Rj bng 0 ;, Rj v R2 to thnh mt b chia p nh thin cho baz T .

m bo ch ng (Ug = 4*0,7V) nh ngun E. Qj

trng thi 0 (in p bo ha) qua phn p RjR ^ to thin p m < 0 lm T2 ngt. Trng thi ny c duy tr lu

ty . Khi tc ng ngoi lm Qj = 1 (Tj chuyn ln trngthi ngt v d nh xung s cc tnh m qua j kha Tj). S chuyn sang trng thi th 2 ; T2 ngj Q2 = 0. Hnh 5.22 s thu gn ca 5.21 tnh cc in p : trn cc cc ca tranzito :

tEx>

uCE

Lr

^

I.

ub

Hnh 5.22

Gi s T ng : U = 0,7V. (viTil

loi tranzito Si).

157

in p trn R2 v dng qua n := U - (-E ) = 0,7 V + 5V = 5,7V.

I2 = U^ /R2 = 5,7V/27k 211 A.

it p ri trn (Rj + Rj) l ;

R,^ + = E - = 5 - 0,7 = 4,3V

dng qua chng :Uri 4 3Y

^2 + Ri 2,7kQ+ l^ Q

Suy ra Ig = 243 h - 211 = 32 fK.T 1 = /3Ig = 70.32 n k = 2240 fiA = 2,24mA. Lc c

bo ha ca Tj, dng thc t l :

E-UcEj,h 5 V - 0 .2 V , ^C^(thc) - - 2,7k 1,78 mA

,

B6)

r >

Hnh- 6.14

Bi tp 6.16. Trn hnh 6.15 l th thi gian ca cc bin logic Xj(t), X2 (t), Xjit).

a) Hy thit lp th thi gian ca cc hm logic c bn c xy dng t 3 php tnh logic i vi cc bin Xj, Xj, X3

= X, ; = X, ; F^o, = XNO, 3I "2 " '3^and = X..X2 .X3 ; For = X, + X2 + X3

Fnani) = X, . X2 . X3

13-BTKTT-A 173

TnOR ^ 1 ^ 2

X.

0

t1 1

0 0 0

1

0 0

1

0 ...... . 0 .

0

Hnh 6.15

b) Hy thit lp bng chn l ca cc hm trn.c) Xy dng cu trc cc hm Fqpj, Fj^ or t

cc phn t NAND v NOR loi chi cd hai ca vo.d) Theo nh ngha hm cng mun 3 bin :

F = Xj e X3Vit biu thi gian v bng trng thi ng vi Xp X2 v

X3 cho hnh 6.15.

B i tp 6.17. Cho cc hm logic 3 bin c biu thc dng sau :

1) FjiXj, X2, X3) = XjXjXj + XjXj + X2X3

2) F2 (Xj, X2, X3) = X1X3 + X2X3 + X1X2X3

3) F3 (Xj, Xj, Xj) = XjX2 + X2X3 + X3X1 + X1X2X34) F , (Xj, X2 , X3 ) = X1 X2 + X2 X3 + XjXj + X1 X2 X3

-5) F5 (Xj, X2 , X3 ) = XjX 2 + X2 X3 + X1 X3

6) (Xj, X2, X3) = XjXj + X2X3 + XjX3 + Xj X2X3

a) Hy thit lp bng trng thi ca Fj -- tng ng vi biu thc cho, t xy dng ba Cacno ca chng.

174 13-BTKTT-B

b) Nu bit trc th thi gian ca Xj(t) X2 (t) v Xgt) (dng t chn), v dng F|(t) theo gi thit chn.

c) Tm cu trc thc hin cc hm Fj (i = 1 - 6 ) bng ch cc phn t NAND c hai li vo.

Bi tp 6.18. Cho mch in hnh 6.16 vi hai li vo bin Xj v X2 , hai li ra nhn c cc hm Fj v F2 .

a) Vit cc biu thc iogic ca Fj v ca F2 v a chng v dng ti thiu.

b) Lp cc bng trng thi tong ng ca Fj v F2 .c) Vi dng Xj(t) v X2 (t) bit trc (h. 6.17). V dng

th thi gian ca Fj(t) v Fjt) ph hp vi X, X2 cho.

0 0t

0 0 H

Hnh 6.17

Bi tp 6.19. Cho cc hm logic 3 bin c gi tr c xc nh bi cc bng trng thi di y :

175

X 1 X2 X3 Fl 80 0 0 o o 1 o 1 o0 1 11 o o 1 o 1 1 1 o 1 1 1

1010101 o

01010101

11o010

o1

o01

1

1

01 o

01

1 1

01

1

1

1

1

o01

1 1

01 1 101 1 o

01 10101 1

a) Vit ba cacno cho cc hm Fj F cho.b) Ti thiu ha cc hm trn bng phng php Cacno.c) Xy dng cu trc cc hm Fj -- F ch dng thun nht

1 loi NAND 2 ca hay 1 loi NOR 2 ca vo.

Bi tp 6 .2 0 . Hm khc du (cng modu nh phn) 3 bin c nh ngha F = Xj X2 *

a) Hy lp biu thc y v bng trng thi ca F.b) Xy dng cu trc F t cc phn t logic c bn NO

AND, OR

c) Nu trong biu thc nh ngha thay X3 bng X3 ^F = Xj X2 Xj th cc kt qu ca cu a) v b) c g thay i.

Bi tp 6.21. Da vo cc nh lut v quy tc (tin ) ca i s logic, hy chng minh mt s nh l sau :

a) (X, + X2) (X + X3) = X, + X2 X3 X, X2 + Xj X2 = Xj

+ Xj X2 = Xj + X2

b) Xj (Xi + X2 ) = Xj X2

Xj (X, + X2) = X,

+ X1X2 = Xj

1 7 6

c) Chng minh cc tnh cht sau ca php cng modun ; x 0 = x ; x i = x x x = 0 ; x x = 1

n u Xj X2 = X3 th X3 = X2 v X2 X3 = X

Bi tp 6.22. Cho cc hm logic 3 bin sau :

Fj = X 1X2 + X3 ; Gi = (Xj + X2 )PCj -I- X3)

F2, = XjX^ + XjX3 + X2 X3 ;G2 = (X 4- X2)(Xj + X3 )(X2 + X3)

a) Chng minh rng Fj v F2 cng biu din 1 hm F v Gj v G2 cng biu din 1 hm G,

b) Vit bng trng thi v ba Cacno ca F v ca GCt nhn xt g khi ng thi c X3 = X2 = 1 i vi cc

hm v F2 , trong trng hp ny dng hm Fj hay F2 thunli hn nu ti tnh cht qu khi chuyn trng thi0 ^ 1 ca bt k 1 bin hay hm logic no o ? Tng tvi Gj v G2 khi ng thi Xo = X3 = 0

c) Xy dng cu trc thc hin F2 v cu trc thc hin G2 t cc phn t NAND cd 2 ca vo.

Bi tp 6.23. Cho 2 hm logic 3 bin cd biu thc sau :

Fj = xy + yz + 2X F2 = x + z + z x

a) Tm mi lin h gia 2 hm Fj v F2b) Thit lp bng trng thi v ba cacno ca Fj v ca F2c) Xy dng c trc Fj t cc phn t NAND 2 ca vo

v cu trc thc hin j t cc phn t NOR 2 ca vo.

Bi tp 6.24. Cho hai hm logic 3 bin sau ;

G| = (x + y)( + z)(z + x)G2 = xy + yz + zx

177

a) Tm quan h logic gia Gj v G2b) Vit ba cacno v bng trng thi ca Gj v Gj ^c) Xy dng cu trc thc hin Gj t cc phn t i NAN

2 ca vo v cu trc thc hin G2 t cc phn t NOH 2 cvo.

Bi tp 6.25. Hnh 6.18 v 6.19 biu din hai cu c thc hin cc hm Fj v F2 tng ng t cc bin vo A, B, c, D v E.

B

cDE

Hnh 6.19Hnh 6.18

a) Tm biu thc v F2 ' dng ti thiu.b) dng y , tm

mi lin h logic gia Fj v F2 lp bng trng thi ca chng.

Bi tp 6.26. Cho ccmch in t hinh 6 . 2 0 v 6 .2 1 .

Gi thit u vo cc bin logic Xj, Xj, X3 l cc xung in p dng 2 mc ov v 4V, ti ni ti u ra c gi tr ln (R >

R = R2 R3 R4 =

lv2 - XV

5s,6 kQ,

^2

^Hnh 6,20

l+5f

R .

X.

' ^

1 7 S

o

X.

RX

R

a) Xc nh trng thi logic ca F| 2 theo tt c cc tng thi logic ca cc bin vo Xj, X2, X3 (vi quy c Xj = 0 khi OVvX = 1 khi+5V).'

b) Vi ni tr ngun ' in p X b

I ;

m cn knh n l loi khta thng ng lin quan ti 2 ch ngho v ch . giu tng ng ca chng)

b) Minh ha kt qu cu a) trn cc th thi gian thng hng F(t), F2(t) theo cc x^ (t) v X2 (t) cho trc (vi mi kh nng c th). Lp bng trng thi ca Fj v F2 .^

c) Hy suy rng cu trc v cc kt qu cho

+

o

t; ( p)o-

2 ( P )

o

f (n) n (n )Hnh 6.23

3 bin Xj, X2 , X3 ; (hnh v cu to v bng trng thi). Bi tp 6.28. Cho cc mch in t hnh 6.24 v 6.25

o + E (5y) ,

h

R, (ikS) (4-k)

/. R .O l )

Ta1

i-io----- ---- - - - - - - -3^ [Vk)

o-------------------

-ok

Hnh 6.24

Cc Tranzito Bi-T v MOSFET lm vic ch chuyn mch khi cd xung dng XjX2 v Xj tc ng li yo (vi MOSFET loi knh p l kha thng m v ch ngho cn MOSFET knh n l loi kha thng ngt v lm vic ch giu).

180

X,o-

o

O-

o

-o

r;

nh 6.25

a) Xc nh tt c cc trng thi gi tr cd th ca Fj v ca F2 theo cc kh nng ca bin vo (khi phn tch trng thi ca cc tranzito trong s ).

Qua lp bng trng thi ca Fj v ca F2 theo X.

b) Biu din Fj(t) v F2 (t) theo cc gi tr X(t) t chn (cha mi kh nng cd th).

Bi tp 6.29. Chocc hm logic 4 bin

(Xp X-), X3 , X )^ c cc bng trng thi sau :t

m, X2 X3 X4 F2 F3 F4 m X, X2 X3 X4 Fi Fz F3 F.i

nio 0 0 0 0 0 0 0 0 1 0 0 0 i 1 0 0m 0 0 0 1 0 0 0 1 my 1 0 0 1 1 1 0 1m2 0 0 1 0 0 0 1 1 mio 1 0 1 0 1 1 1 1m3 0 0 1 1 0 0 1 0 mn 1 0 1 1 1 1 1 0m4 0 1 0 0 0 1 1 0 mi2 1 1 0 0 1 0 1 0ni5 0 1 0 1 0 1 1 1 ni3 1 1 0 1 1 0 1 1m6 0 1 1 0 0 1 0 1 mi4 1 1 1 0 1 0 0 1m? 0 1 1 0 1 0 0 mi5 1 1 1 1 1 0 0 0

a) Thit lp ba Cacno cho h cc hm ra F| (i = 1, 2, 3,4 ) tng ng vi bng trng thi cho.

b) Thc hin tm hm ti thiu ca F| theo phng php Cacno.

c) Xy dng cu trc logic thc hin cc hm ti thiu t cc phn t logic c bn ;

1 ) T cc phn t hn hp NO, AND v OR.2) T ch cc phn t NAND (hoc NOR)

181

3) T cc phn t thng dng (nu c th)Bi tp 6.30. Cho cc hm 4 bin G j (Xp X2 , X3 , x^) (j =

1, 2 , 3, 4) c cc b n g tr n g th i sa u :

a) Thit lp ba Cacno cho cc hm Gj (j = 1 , 2 , 3, 4) ph hp vi bng trng thi cho.

b) Ti thiu ha cc hm Gj theo quy tc Cacno.c) Xy dng cu trc cc hm ti thiu t :

mi Xi X 2 X3,X4 Gi G 2 G 3 G, nii Xi X 2 X 3 X 4 Gi G 2 G 3 G 4

mo 0 0 0 0 0 0 0 0 ni8 1 0 0 0 1 1 0 0mi 0 0 0 1 0 1 0 0 m 1 0 0 1 1 0 0 0m2 0 0 1 0 0 1 0 1 mio 1 0 1 0 1 0 0 1m3 0 0 1 1 0 1 1 1 mu 1 0 1 1 1 0 1 1x r i A 0 1 0 0 0 1 1 0 mi2 1 1 0 0 0 1 0ms 0 1 0 1 1 1 1 0 mi3 1 1 0 1 0 0 1 01X16 0 1 1 0 1 1 1 1 mi4 1 1 1. 0 0 0 1 1m? 0 1 1 1 1% 1 0 1 mi5 1 1 1 1 0 0 0 1

1) Cc hm NAND hoc NOR2) Cc hm F tng ng hay F khc du (nu c th)Bi tp 6.31. Cho cc hm 4 bin H^ (Xp X2 , X3 , x^) (k

= 1, 2, 3, 4) c cc bng trng thi sau

li Xi X 2 X 3 X 4 Hi H3 H 4 nii XI X 2 X 3 X 4 H2 H3 H 4

mo 0 0 0 0 0 0 0 0 ni8 1 0 0 0 1 0 0 0mi 0 0 0 1 1 1 1 1 m9 1 0 0 1 0 1 1 1ni2 0 0 1 0 1 1 1 0 mio 1 0 1 0 0 1 1 0m3 0 0 1 1 1 1 0 1 mn 1 0 1 1 0 1 0 1ni4 - 0 1 0 0 1 1 0 0 mu 1 1 0 0 0 1 0 0ms 0 1 0 1 1 0 1 1 mi3 1 1 0 1 0 0 1 1m6 0 1 1 0 1 0 1 0 1 1 1 0 0 0 1 0m? 0 1 1 1 1 0 0 1 mi5 1 1 1 1 0 0 0 1

a) Vit ba Cacno cho cc hm H|. tng ng vi cc bng trng thi cho.

182

b) Tm cc hm Hj. sau khi ti thiu ha bng phng php Cacno.

c) Xy dng cu trc logic thc hin h hm Hy. ti thiu. (Cn trii dng cxi trc gn nht c th khi dng cc phn t logic c bn hay phn t logic thng dng)

Bi tp 6.32. Cho cc hm 4 bin A, Bj, C^ vi cc bin vo l X3 X2 Xj biu th cc bng trng thi sau :

ni X3 X2 X, Xo A3 A , A, Ao B3 B2 B, Bo C3 C2 Co0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1

mj 0 0 0 1 0 0 0 1 0 0 0 1 0 1 0 0m2 0 0 1 0 0 0 1 0 0 0 1 0 0 1 0 1m3 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 0ni4 0 1 0 0 1 0 0 0 0 1 0 0 0 1 1 1^ 5 0 1 0 1 0 1 1 1 1 0 1 1 1 0 0 0

0 1 1 0 1 1 0 0 1 1 0 0 1 0 0 1m - 0 1 1 1 1 1 0 1 1 1 0 1 1 0 1 0iix 1 0 0 0 1 1 1 0 1 1 1 0 1 0 1 1ni 1 0 0 1 1 1 1 1 1 1 1 1 1 1 0 0m)

a) Thit lp ba Cacno cho cc hm ra A (Bj hoc c^)b) Ti thiu cc h hm ra A (hoc B hoc Cy) theo quy

tc Cacno (lu rng cc t hp khng ang ti 4 * I1 5 c th gn cho tr 1 ti thiu cc hm cho).

c) Xy dng cu trc logic thc hin cc hm A, B, C^ sau khi ti thiu.

Bi tp 6.33. Xut pht t bng trng thi thu gn ca Trig vn nng JK :

a) Hy thit lp bng trng thi y ca hm ra ph thuc 3 binlogic J, K v

b) Vit ba Cacno cho hm ra

J K Qn+10 0 Qn

' 0 1 01 0

1 1 Qn

183

R s Qn.i0 0 Qn0 1 0

1 0 1

1 1 X

c) Ti thiu ha Qp+P a n v dng ti gin v qua o vit phng trnh c tnh ca Trig JK.

Bi tp 6.34. Xut pht t bng trng thi rt gn ca Trig RS, trong trng thi nh du X l trng thi cm.

a) Lp bng trng thi y cahm ra ph thuc 3 bin vo R, sv Q.

b) Vit h phng trnh hm ra (mtphng trnh cho v 1 phng trnhcho iu kin cm).

c) Rt gn h phng trnh thit lp a v dng phngtrnh c tnh ca RS-Trig (nh cc quy tc ca i s logic).

Bi tp 6.35. Cho 1 hm 4 bin cd dng F = (X2 + X3 )+ X3X,

a) Vit biu thc ca F dng y v qua thit lp bng trng thi ca F.

b) Xy dng cu trc logic thc hin F vi cc phn t thun nht NAND hoc NOR 2 ca vo.

c) Vi dng X|(t), X2 (t), X3 (t) v x^t) cho trc (t chn) lp th F(t) tng ng (ch phi bao gm mi t hp trng thi c th ca X(t)).

Bi tp 6.36. Mt mng cu trc t hp kt hp vi 1 trig RC lm vic vi 3 bin logic li vo (theo trt t X2 , Xp X ng vi cc cp nh phn 2 ,^ 2^ v 2 ^ tng ng). Hm ra c tnh cht sau :

1) Nu X2XjX^ ^ th li ra trng thi 1.

- 2) Nu X2 XjX^ < m2 th li ra trng thi 0.

3) Nu X2 X^X ^ < ni5 v X2XjX ,^ > m 2

y = X2 XjX ^ , I2 = X2 XjX^ ^ th li ra gi trng thi trc n cd.

a) Thit lp bng trng thi ca mch t hp tha mn cc iu kin nu trn.

iW

b) Ti thiu hda hm ra ca mng t hp trn.c) Hy xy ng s ton b bng cch ni tip h trn

vi 1 RS Trig v gii thch hot ng qua bng trng thi RS ca nd.

Sfs = X,(X_2 + X3)

(X2 + X3)

Bi tp 6.37. Xy dng cc cu trc thun nht (NAND hoc NOR) thc hin cc hm ogic c biu thc sau :

1) (A + B)(C + D)E2) (A + B) C (D + E)3) AB + C + DE4) A (B + C) D5) A + BC + D

6 ) AB + CD + E7) (A + B) (C + D) 8 ) AB + CD + E9) ( + B) (C + D)1 0 ) (A + I) (C + D)

Tm cch vit bng trng thi ca cc hm trn 1 cch thu gn (nh t thm cc bin ph).

( y A, B, c, D, E l cc bin logic li vo).Bi tp 6.38. Cho cu trc logic hnh 6.26.

Hnh 6.26

a) Vit biu thc h hm logic F|, F2 , F3 theo cc bin logic u vo X.

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b) Thit lp bng chn l ca cc hm Fq, Fj, F2 ; F3 (vit chung trong 1 bng 4 ct bin vo, 4 ct hm ra).

c) Vit ba Cacno cho cc hm Fq, Fp Fj, v Fj v tm cc biu thc ti thiu ca chng.

Bi tp 6.39. Cho cu trc logic hnh 6.27.

A,

t X.

o-

X

F

Hnh 6.27

'^ a) Tm biu thc logic ca hm F theo cc bin vD

b) C nhn xt g v tnh cht ca mch (nu coi cc bin vo X l cha thng tin cn cc bin Aj iu khin).

c) M rng cho trng hp nhm Aj gm 3 bin X gm 2^ bin v trng hp Aj gm 4 bin X gm 2* bin.

d) Tm cu trc tng ng vi cu trc hnh 6.27 thc hin bng cc phn t NAND {hay bng cc phn t NOR).

Bi tp 6.40. TVong cc bi tp 6.29 'n 6.32, vi cc bng trng thi cho, i ln vai tr u vo v u ra (tc l li ra s l cc hm X, cn li vo s l cc bin Aj (hoc Bj... Fj, Gj, hoc Hj) v d vi bi 6.29 c bng trng thi khi vit n ^ c li l :

186A,-

Vo Ra Vo R aFi F2 F3 F4 X2 X 3 X 4 F2 F3 F4 X, Xz X 3 X 40 0 0 0 0 0 0 0 1 1 0 0 1 0 0 00 0 0 1 0 0 0 1 1 1 0 1 1 0 0 10 0 1 1 0 0 1 0 1 1 1 1 1 0 1 00 0 1 0 0 0 1 1 1 1 1 0 1 0 1 10 1 1 0 0 1 0 0 1 0 1 0 1 1 0 00 1 1 1 0 1 0 1 1 0 1 1 1 1 0 10 1 *0 1 0 1 1 0 1 0 0 1 1 1 1 00 1 0 0 0 1 1 1 1 0 0 0 1 1 1 1

Cc cu hi tng t vi cc bng thit lp trn :a) Lp ba Cacno cho h cc hm ra X ng vi bng trng

thi cho.b) Ti thiu ha cc hm X theo quy tc Cacno.

- c) Xy dng cu trc logic ca cc b bin i m loi ny

MC LC

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