bai tap kts

i hc Bch Khoa TP.HCM Khoa in-in t

L Ch Thng

BI TP K THUT SChng 1: Cc h thng s m1.1 Biu din cc s sau trong h nh phn (binary) a. 23 b. 14 c. 27 d. 34 S 1-2 Biu din cc s sau trong h nh phn (binary) a. 23H b. 14H c. C06AH d. 5DEFH S 1.3 Biu din cc s sau trong h thp phn (decimal) a. 01101001B b. 01111111B c. 10000000B d. 11111111B S 1.4 Biu din cc s sau trong h thp phn (decimal) a. 1FH b. 10H FFH 03H S 1.5 Biu din cc s sau trong h thp lc phn (hex) a. 100 b. 128 c. 127 d. 256 S 1.6 Biu din cc s sau trong h thp lc phn (hex) a. 01111100B b. 10110001B c. 111100101011100000B d. 0110110100110111101B S 1.7 Biu din cc s cho bi 1-1 v 1-3 thnh h thp lc phn (hex). 1.8 Biu din cc s cho bi 1-2 v 1-6 thnh h thp phn (decimal). 1.9 Biu din cc s cho bi 1-4 v 1-5 thnh h nh phn (binary). 1.10 i cc s sau sang h nh phn a. 27,625 b. 12,6875 c. 6,345 d. 7,69 S Bi tp K Thut S Trang 1/22

c. d.

i hc Bch Khoa TP.HCM Khoa in-in t 1.11 i cc s sau sang h bt phn (octal) a. 1023H b. ABCDH c. 5EF,7AH d. C3,BF2H 1.12 i cc gi tr sau thnh byte a. 2KB b. 4MB c. 128MB d. 1GB S 1.13 Ly b 1 cc s sau a. 01111010B b. 11101001B c. 00000000B d. 11111111B S 1.14 Ly b 2 cc s sau a. 10101100B b. 01010100B c. 00000000B d. 11111111B S 1.15 Ly b 9 cc s sau a. 3 b. 14 c. 26 d. 73 S 1.16 Ly b 10 cc s sau a. 7 b. 25 c. 62 d. 38 S 1.17 Biu din cc s sau trong h nh phn c du 4 bit a. 5 b. -5 c. 7 d. -8 S 1.18 Biu din cc s sau trong h nh phn c du 8 bit a. 5 b. -5 c. 34 d. -26 e. -128 f. 64 g. 127 S

L Ch Thng

Bi tp K Thut S Trang 2/22


L Ch Thng

1.19 Cho cc s nh phn c du sau, hy tm gi tr ca chng a. 0111B b. 1000B c. 0000B d. 1111B e. 0011B f. 1100B g. 0111111B h. 00000000B i. 11111111B j. 10000000B S 1.20 Cho cc s nh phn sau, hy xc nh gi tr ca chng nu chng l (i) s nh phn khng du; (ii) s nh phn c du a. 0000B b. 0001B c. 0111B d. 1000B e. 1001B f. 1110B g. 1111B S 1.21 Biu din cc s sau thnh m BCD (cn gi l m BCD 8421 hay m BCD chun) a. 2 b. 9 c. 10 d. 255 S 1.22 Lm li bi 1-21, nhng i thnh m BCD 2421 (cn gi l m 2421) S 1.23 Lm li bi 1-21, nhng i thnh m BCD qu 3 (cn gi l m qu 3 XS3) S 1.24 Cho cc m nh phn sau, hy i sang m Gray a. 0111B b. 1000B c. 01101110B d. 11000101B S 1.25 Cho cc m Gray sau, hy i sang m nh phn a. 0110B b. 1111B c. 11010001B d. 00100111B S 1.26 Cho cc m nh phn sau, hy xc nh gi tr ca chng nu chng l (i) s nh phn khng du; (ii) s nh phn c du; (iii) m BCD; (iv) m 2421; (v) m qu 3; (vi) m Gray a. 1000011B b. 110101B


i hc Bch Khoa TP.HCM Khoa in-in t c. 1101100B d. 01000010B

L Ch Thng

S 1.27 Lm li bi 1-26 vi a. 10000101B b. 0101101B c. 10000000B d. 01111111B S 1.28 Thc hin cc php ton sau trn s nh phn c du 4 bit a. 3+4 b. 4-5 c. -8+2 d. -4-3 1.29 Thc hin cc php ton sau trn s nh phn c du 4 bit, nu kt qu b trn th tm cch khc phc a. 5-7 b. 5+7 c. -2+6 d. -1-8 1.30 Thc hin cc php ton sau trn s nh phn c du 8 bit v cho bit kt qu c b trn hay khng a. 15+109 b. 127-64 c. 64+64 d. -32-96 S 1.31 Thc hin cc php ton sau trn s BCD a. 36+45 b. 47+39 c. 66-41 d. 93-39 e. 47-48 f. 16-40



L Ch Thng

Chng 2: i s Boole2-1 a. b. c. d. e. 2-2

Chng minh cc ng thc sau bng i sAB + A D +BC D =( A +D )( A +C )( B +D ) C D +BC + AB =( A +C )( B +C )( B +D ) D Z + X + X Z = ( X +Z )( Y +Z ) Y

A B = A B AB ( A B C ) = ABC

Cho bng chn tr sau

C B A F1 F2 0 0 0 0 1 0 0 1 0 0 0 1 0 1 0 0 1 1 0 1 1 0 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 0 a. Vit biu thc ca hm F1 v F2 b. Vit biu thc hm F1 di dng tch cc tng (POS) c. Vit biu thc hm F2 di dng tng cc tch (SOP) d. Vit hm F1 di dng v e. Vit hm F2 di dng v 2-3 Cho bng chn tr sau A B C F1 F2 0 0 0 1 1 0 0 1 0 X 0 1 0 X 0 0 1 1 0 1 1 0 0 0 1 1 0 1 1 X 1 1 0 X X 1 1 1 0 0 a. Vit biu thc cc hm F1 v F2 b. Vit dng v cho hm F1 v F2 2-4 Cho cc hm sauF1 ( A, B, C , D ) = ABC D + A BD + A D + A.C C F2 ( A, B, C , D ) = ( B +C + D )( A +C + D )( B + D)

2-5

Hy lp bng chn tr ca F1 v F2 Cho cc hm sau

F1 ( A, B, C , D ) = 0,1,2,4,6,8,12 ) +d (3,13 ,15 ) (

2-6

Hy lp bng chn tr ca F1 v F2 Cho gin xung sau

F2 ( A, B, C , D ) =1,3,4,5,1 ,1 ,14 ,15 ). d (0,6,7,8) ( 1 2



L Ch Thng

A B C D F1 F2 F3a. Vit biu thc cc hm F1, F2 v F3 b. Vit dng v cho hm F1, F2 v F3 2-7

Cho bng chn tr sau A B C D F1 F2 0 0 0 0 1 1 0 0 0 1 0 1 0 0 1 0 0 0 0 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 0 0 0 1 1 1 0 1 1 X X X 1 0 a. Vit biu thc cc hm F1 v F2 b. Vit dng v cho hm F1 v F2 Biu din cc hm cho trong cc bi t 2-2 n 2-7 trn ba Karnaugh Cho s mch sau, hy vit biu thc chun 1 v 2 ca F1 v F2Y F X 1

2-8 2-9

Z F 2

2-10 Cho s mch v gin xung cc tn hiu vo nh sau, hy v dng tn hiu

F.A B C F


i hc Bch Khoa TP.HCM Khoa in-in t A B C

L Ch Thng

2-11 Cho s mch nh sauA Y 0

B

Y

1

E

Y

2

Y D

3

Lp bng chn tr v vit cc hm trong cc trng hp sau a. E=0 v D=0 b. E=0 2-12 Tm dng chun 1 v 2 ca cc hm sauF1 ( X , Y , Z ) = XY +YZ + XZ F2 ( X , Y , Z ) = XY + X Z F3 ( A, B, C ) = A + C + AB F4 ( A, B, C ) = ( A B ) + A BC

2-13 Dng ba Karnaugh rt gn cc hm sau

F1 ( A, B, C , D ) = 0,1,2,4,5,8,1 ,1 ,1 ) ( 0 2 4 F2 ( A, B, C ) =0).d (1,2,3,4,5,6,7) ( F3 ( A, B, C , D ) = A BC D + AB + A(C D ) + A C +C D B

2-14 Dng ba Karnaugh rt gn cc hm sau F1 ( A, B, C , D ) = 1,2,4,7,9,15 ) + d (3,5) (

F4 ( A, B, C , D, E ) =1,3,4,5,6,9,12 ,14 ,20 ,2 ,2 ,2 ,28 ,2 ). d (1 ,1 ,3 ) ( 1 2 5 9 3 6 0

F3 ( A, B, C , D ) = 2,5,7,8,13 ,15 ).d (0,10 ) ( F4 ( A, B, C , D ) = 0,2,4,5,6,8,10 ,12 ,13 ) (

F2 ( A, B, C , D ) = 0,1,2,4,5,8,10 ,11 ,14 ,15 ) (

2-15 Cho hm F(A,B,C,D) biu din trn gin xung nh sau


i hc Bch Khoa TP.HCM Khoa in-in t A B C D Fa. Vit biu thc chun 2 ca hm F b. Biu din hm trn ba Karnaugh c. Rt gn hm F v v mch thc hin ch dng cng NAND 2-16 Rt gn hm sau v thc hin bng cng NAND 2 ng vo 2-17 Rt gn hm sau v thc hin bng cng NOR 2 ng voF ( A, B, C , D ) = 4,6,9,1 ,1 ,1 ) +d (8,1 ,1 ) ( 0 2 4 1 3

L Ch Thng

2-14 Thc hin hm F ( A, B, C , D) =B (C +D) +AC D ch dng cng NAND C 2-15 Thc hin hm F ( A, B, C , D ) =( A +B )( C +B D ) ch dng cng NOR 2-16 Cho cc hm sau F1 ( A, B, C , D) = A B + ( BCD + BCD )C + A B + BD CF2 ( A, B, C , D ) = ( A +C )( C + D ) + A B DF3 ( A, B, C , D ) = A B + A D ( B +C D ) B

F ( A, B, C , D ) =0,2,3,4,6,9,1 ,1 ).d (7,1 ,1 ) ( 0 1 3 5

a. b. c. 2-17 Cho cc hm sau

Hy biu din cc hm trn ba Karnaugh Vit biu thc tch cc tng (POS) cho cc hm Rt gn v v mch thc hin dng ton cng NAND

F1 ( A, B, C , D ) = 0,2,3,4,6,7,8) +d (5,1 ,14 ) ( 2

a. AND-OR b. OR-AND c. d. 2-18 Cho bng chn tr sau G1 0 X 1 1 1 1 1 1 1 1 G2 X 1 0 0 0 0 0 0 0 0 X2 X X 0 0 0 0 1 1 1 1

F2 ( A, B, C , D ) =2,3,8,9,1 ,1 ,1 ,1 ). d (0,11 ,1 ) ( 0 2 4 5 3

Rt gn hm F1 v thc hin F1 dng cu trc cng Rt gn hm F2 v thc hin F2 dng cu trc cng Thc hin F1 dng cu trc ton NAND Thc hin F2 dng cu trc ton NOR X0 X X 0 1 0 1 0 1 0 1 Y0 0 0 1 0 0 0 0 0 0 0 Y1 0 0 0 1 0 0 0 0 0 0 Y2 0 0 0 0 1 0 0 0 0 0 Y3 0 0 0 0 0 1 0 0 0 0 Y4 0 0 0 0 0 0 1 0 0 0 Y5 0 0 0 0 0 0 0 1 0 0 Y6 0 0 0 0 0 0 0 0 1 0 Y7 0 0 0 0 0 0 0 0 0 1

X1 X X 0 0 1 1 0 0 1 1


i hc Bch Khoa TP.HCM Khoa in-in t a. b. Vit biu thc cc hm Y0 n Y7 V s logic ca cc hm trn

L Ch Thng



L Ch Thng

Chng 3: H t hp3-1

Cho mt h t hp hot ng theo bng sau E X1 X0 Y0 Y1 Y2 Y3 1 X X 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 a. Thit k h t hp ny dng cng bt k b. Dng h t hp thit k cu a (v dng s khi) v cc cng logic thc hin hm Thit k mch gii m 2421 thnh thp phn (m 1 trong 10)a. Thc hin bng cng logic b. Thc hin bng mch gii m (decoder) 416 c ng ra tch cc mc 1F ( A, B, C ) = 4,6) (

3-2

Thit k mch cng bn phn (HA) thc hin bng cng logic. Sau , ch dng HA (v dng s khi) thc hin php tnh (x+1)2, bit rng x l s nh phn 2 bit (x = x1x0). 3-4 Mt mch t hp c 5 ng vo A, B, C, D, E v mt ng ra Y. Ng vo l mt t m thuc b m nh sau E D C B A 0 0 0 0 0 0 0 1 1 1 0 1 0 0 0 0 1 1 1 1 1 0 0 0 0 1 0 1 1 1 1 1 0 0 0 1 1 1 1 1 a. Thit k mch t hp dng cng AND-OR sao cho Y=1 khi ng vo l mt t m ng v Y=0 khi ng vo l mt t m sai. b. Thc hin li cu a ch dng ton cng NAND 3-5 Cho mt h t hp hot ng theo bng sau E X1 X0 Y0 Y1 Y2 Y3 1 X X 1 1 1 1 0 0 0 0 1 1 1 0 0 1 1 0 1 1 0 1 0 1 1 0 1 0 1 1 1 1 1 0 a. Thit k h t hp ny dng ton cng NOT v NAND 3 ng vo b. Dng h t hp thit k cu a (v dng s khi) v mt cng AND 2 ng vo thc hin mt h t hp hot ng theo gin xung nh sau (vi U, V, W l cc ng vo; Z l ng ra)3-3


i hc Bch Khoa TP.HCM Khoa in-in t U V W Z

L Ch Thng

Thc hin mch cng ton phn (FA) trn c s mch chn knh (Mux) 41 Lp bng chn tr ca mch chn knh (Mux) 161. Sau , thc hin mch chn knh 161 trn c s mch chn knh 41. 3-8 Cho 4 b m nh sau A=a3a2a1a0 B=b3b2b1b0 C=c3c2c1c0 D=d3d2d1d0 Hy thit k mch chn m (vi Y= y3y2y1y0 l ng ra) trn c s mch chn knh 41 theo bng chn tr sau x1 x0 Y 0 0 A 0 1 B 1 0 C 1 1 D 3-9 Thit k mch chuyn m qu 3 thnh nh phn ch dng vi mch 7483 (mch cng 4 bit ). 3-10 Thit k mch chuyn m BCD 2 decade thnh nh phn ch dng vi mch 7483 (mch cng 4 bit ). 3-11 Thit k mch gii m BCD thnh m LED 7 on anode chung dng cng logic 3-12 Lm li bi trn dng vi mch 74154 (mch gii m 416) v cc cng cn thit 3-13 Thit k mch tr hai s mt bit, trong V l bin iu khin, Ci-1 l s mn ng vo, Ci l s mn ng ra. Khi V=0 th mch thc hin D=A-B, khi V=1 th thc hin D=B-A 3-14 Thit k mch tr hai s 3 bit A v B vi bin iu khin V, da trn c s mch tr hai s mt bit bi trn. 3-15 Thit k mch tr hai s 3 bit A v B sao cho kt qu lun lun dng. 3-16 Thit k mch cng/tr hai s nh phn 4 bit X v Y dng vi mch 7483 (mch cng 4 bit) v cc cng logic (nu cn). Mch c tn hiu iu khin l v, khi v=0 mch thc hin X+Y, khi v=1 mch thc hin X-Y 3-17 Ch s dng mch cng ton phn FA, hy thit k h t hp c bng chn tr sau x1 x0 y0 y1 y2 y3 0 0 0 1 0 0 0 1 1 0 1 0 1 0 1 0 1 0 1 1 0 1 1 1 3-18 Dng vi mch 7483 (mch cng 4 bit) v cc cng logic (nu cn) thit k mch t hp c hot ng nh sau3-6 3-7


i hc Bch Khoa TP.HCM Khoa in-in t x3 x2 x1 x0 C y3 y2 y1 y0

L Ch Thng

Nu C=0 th y3y2y1y0 = x3x2x1x0 Nu C=1 th y3y2y1y0 = b 2 ca x3x2x1x0 3-19 Cho hm F vi 4 bin vo. Hm c tr bng 1 nu s lng bin vo c tr bng 1 nhiu hn hoc bng s lng bin c tr bng 0. Ngc li, hm c tr bng 0. a. Hy biu din hm trn ba Karnaugh b. Rt gn hm v v mch thc hin dng ton cng NAND 3-20 Thit k mch chuyn m nh phn 4 bit sang m BCD ch dng vi mch so snh 4 bit (ng ra tch cc cao) v vi mch cng ton phn FA. 3-21 Thit k mch chuyn m Gray 4 bit sang m nh phn, s dng a. Cc cng logic. b. Mch gii m (decoder) 416. 3-22 Thit k mch chuyn m BCD thnh 7421 s dng decoder 416 c ng ra tch cc mc 0 v khng qu 4 cng NAND.3-23 a.

Thit k mch so snh hai s nh phn mt bit A v B vi cc ng ra tch cc mc 1 s dng cng logic. b. Thit k mch so snh hai s nh phn 4 bit X=x3x2x1x0 v Y=y3y2y1y0 s dng cng logic. Bit rng ng ra F=1 khi X=Y v F=0 khi XY. c. Thc hin mch cu (b) ch dng mch so snh thit k cu (a) v m cng AND. V mch dng s chc nng . 3-24 Mch t hp c chc nng chuyn t m BCD thnh m BCD qu 3. a. Thit k mch s dng cu trc NOR-NOR. b. Thit k mch s dng vi mch 7483 (mch cng 4 bit). 3-25 S dng cc mch chn knh (Mux) 81 v mch chn knh 41 thit k mch chn knh 321. 3-26 Cho F l mt hm 4 bin A, B, C, D. Hm F=1 nu tr thp phn tng ng vi cc bin ca hm chia ht cho 3 hoc 5, ngc li F=0. a. Lp bng chn tr cho hm F. b. Thc hin hm F bng mch chn knh (Mux) 161. c. Thc hin hm F bng mch chn knh (Mux) 81 v cc cng (nu cn). d. Thc hin hm F bng mch chn knh (Mux) 41 v cc cng (nu cn). e. Hy biu din hm F trn ba Karnaugh f. Hy rt gn F v thc hin F ch dng cc mch cng bn phn HA. 3-27 Cho hm F ( A, B, C ) = AB + BC + AC . Hy thit k mch thc hin hm F ch s dng a. Mt vi mch 74138 (decoder 38, ng ra tch cc thp) v mt cng c ti a 4 ng vo. b. Mt vi mch 74153 (mux 41, c ng cho php tch cc thp). c. Hai mch cng bn phn HA v mt cng OR. 3-28 S dng mt decoder 416 khng c ng cho php (enable) thc hin mt decoder 38 c ng cho php. Khng s dng thm cng.



L Ch Thng

3-29 S dng ba mch chn knh (Mux) 21 thc hin mt mch chn knh

41. Khng dng thm cng. 3-30 S dng hai vi mch 74148 (mch m ha 83) thc hin mt mch m ha (encoder) 164.



L Ch Thng

Chng 4: H tun tThit k mch m ni tip mod 16 m ln dng T-FF (xung clock cnh ln, ng Pr v ng Cl tch cc mc thp). 4-2 Thit k mch m ni tip mod 16 m xung dng T-FF (xung clock cnh ln, ng Pr v ng Cl tch cc mc thp). 4-3 Da trn kt qu bi 4-1, thit k mch m ni tip mod 10 m ln 01290 4-4 Da trn kt qu bi 4-2, thit k mch m ni tip mod 10 m xung 151413615 4-5 Da trn kt qu bi 4-2, thit k mch m ni tip mod 10 m xung 98709 4-6 Nu s dng JK-FF hoc D-FF thay cho T-FF trong cc bi 4-1 v 4-2 th thay i th no? 4-7 Thit k mch m ni tip c ni dung thay i theo quy lut ca m 2421, s dng JK-FF (xung clock cnh xung, ng Pr v ng Cl tch cc mc cao) 4-8 Thit k mch m ni tip ln/xung 4 bit dng T-FF (xung clock cnh xung) vi bin iu khin U / D . Khi U / D =1 th mch m ln, khi U / D =0 th mch m xung. 4-9 Thit k mch m song song dng JK-FF (xung clock cnh xung) c dy m nh sau 000010011100110111000 4-10 Lm li bi 4-9 vi yu cu cc trng thi khng s dng trong dy m c a v trng thi 111 xung clock k tip. 4-11 Lm li bi 4-9 dng D-FF. 4-12 Lm li bi 4-9 dng T-FF. 4-13 Lm li bi 4-9 dng SR-FF. 4-14 Thit k mch m song song mod 10 c ni dung thay i theo quy lut ca m 2421 dng T-FF. 4-15 Cho mch m sau4-1

PR

PR

1 C K

T C K

Q

A

1

T C K

Q

B

1

PR

T C K

Q

C

C LR

C LR

Hy v dng sng A, B, C theo CK v cho bit dung lng m ca mch 4-16 Cho mch m sau


C LR

Q

Q

Q


L Ch Thng

1

S C K

Q

A

S C K

Q

B

S C K

Q

C

0

R

Q

R

Q

R

Q

C

K

a. Vit hm kch thch (biu thc cc ng vo) cho mi FF. b. V graph (gin ) trng thi ca b m. c. Cho bit h s m ca b m. d. B m c t kch c khng? Gii thch? 4-17 Cho mch m sauT C K Q Q A T C K Q Q B T C K Q Q

C

K

a. Vit hm kch thch (biu thc cc ng vo) cho mi FF. b. Lp bng trng thi chuyn i ca mch. c. V graph (gin ) trng thi ca b m. d. B m c t kch c khng? Gii thch? 4-18 Cho mch m sau

T C K

Q

A

T C K

Q

B

Q

Q

C

K

a. Vit hm kch thch (biu thc cc ng vo) cho mi FF. b. Lp bng trng thi chuyn i ca mch. c. V graph (gin ) trng thi ca b m v cho bit h s m. d. V gin tn hiu ra, gi s trng thi u l AB=11. e. Mch c cn nh trng thi u hay khng? Gii thch? f. Nu cn xy dng b m c mod 12 th cn ghp ni tip thm bao nhiu FF? C bao nhiu cch ghp v v mch kt ni mi cch ghp. 4-19 Cho mch m sau



L Ch Thng

T C K

Q

A

T C K

Q

B

T C K

Q

C

Q

Q

Q

C

K

a. b. c. d. e.4-20 4-21 4-22 4-23 4-24

Vit hm kch thch (biu thc cc ng vo) cho mi FF. Lp bng trng thi chuyn i ca mch. V graph (gin ) trng thi ca b m v cho bit h s m. B m c t kch c khng? Gii thch? V gin xung ng ra cc FF theo xung CK, bit trng thi u l ABC=011 S dng mt vi mch 7490 thc hin mch m mod 10. S dng mt vi mch 7492 thc hin mch m mod 12. S dng mt vi mch 7493 thc hin mch m mod 16. S dng mt vi mch 7490 thc hin mch m mod 6. S dng hai vi mch 7490 thc hin mch m mod 60.



L Ch Thng

Ph lc A: Cc vi mch cng v FF thng dng1 7 4 L S 2 0 4 3 7 4 L S 4 0 4 5 7 4 L S 6 0 4 9 7 4 L S 8 0 4 1 1 7 4 L S 1 0 0 4 1 3 7 4 L S 1 2 0 4

1 2 7 4 L S 0 8

3

4 5 7 4 L S 0 8

6

9 1 0 7 4 L S 0 8

8

1 1

2 3 7 4 L S 0 8

1

1

1 3 2 7 4 L S 0 0

4 6 5 7 4 L S 0 0 1

9 8 0 7 4 L S 0 0

1 1

2 1 3 7 4 L S 0 0 1

1 2 7 4 L S 3 2 3

4 5 7 4 L S 3 2 6 1

9 0 7 4 L S 3 2 8

1 1

2 3 7 4 L S 3 2 1 1

2 3 7 4 L S 0 2

1

5 6 7 4 L S 0 2

4

8 9 7 4 L S 0 2

1

0

1 1

1 2 7 4 L S 0 2

1

3

1 2 7 4 L S 8 6

3

4 5 7 4 L S 8 6

6 1

9 0 7 4 L S 8 6

8

1 1

2 3 7 4 L S 8 6

1

1

2 3

C LK Q CL 6 74LS 74

11

C LK Q CL 8 74LS 74 13

1

2 4 3

C LK K Q CL 7 74LS 109

12 13

C LK K Q CL 9 74LS 109 15

1

3 1 2

C LK K Q CL 6 74LS 112 15

13 12

C LK K Q CL 7 74LS 112 14


PR

PR

J

Q

5

11

10

4

PR

PR

J

Q

6

14

11

5

PR

PR

D

Q

5

12

10

4

D

Q

9

J

Q

10

J

Q

9


L Ch Thng

Ph lc B: Cc vi mch t hp thng dngMch gii m (decoder) 24, 38, 4162 3 1 A B G 7 4 1 4 1 3 A B 1 5 G 7 4 L S L S Y Y Y Y 1 4 0 5 1 6 2 7 3 3 9 1 2 0 1 1 1 1 0 2 9 3 1 3 9 6 4 5 1 2 3 1 1 1 1 1 1 9 7 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 1 1 1 1 1 1 L S 1 0 1 2 3 4 5 1 2 3 4 5 6 7 8 9 1 1 1 1 1 1 1 5 4

A B C G G G 7 4 1 2 2

Y Y Y Y

A B L S

Y Y Y Y Y Y Y Y 1

0 1 2 3 4 5 6 7 3 8

2 2 2 2

3 2 A 1 B 0 C D

1 8 1 9 G G 7 4

1 2

0 1 3 4 5 6 7

Mch m ha (encoder) c u tin 83, 1041 1 1 1 1 2 3 4 5 0 1 0 2 1 3 2 3 4 5 6 7 E 7 4 I L S A A A G 9 0 7 1 6 2 S 1 4 1 1 1 2 1 1 3 2 3 1 4 2 5 3 6 4 7 5 1 0 8 9 7 4 4 8 9 7 6 1 4

A B C D

E 1

O

1 5

L S

1

4 7

Mch chn knh (mux) 81, 41, 214 3 2 1 1 1 1 1 5 4 3 2 D D D D D D D D 0 1 2 3 4 5 6 7 W Y 6 5 1 1 1 1 6 5 4 3 1 1 1 1 C C C C 0 1 2 3 0 1 2 3 1 Y 7 2 3 5 6 1 1 1 1 1 0 4 3 1 1 2 2 3 3 4 4 A B A B A B A B / B L S 1 5 7 1 Y 2 Y 3 Y 4 Y 4 7 9 1 2

1 1 1 0 A B 9 C 7 G 7 4 L S 1 5 1

0 1 2 C 2 2 C 3 2 C 2 C

2 Y

9

1 4 A 2 B 1 1 5 1 G 2 G 7 4 L S 1 5 3

1 1 5 A G 7 4

Mch phn knh (demux) 141 3 A 3 B 2 1 1 G 1 C 1 1 1 1 2 2 2 2 L S 1 Y Y Y Y Y Y Y Y 7 06 15 24 39 01 0 11 1 21 2 3 5 5

1 4 1 5 2 G 2 C 7 4

Mch cng nh phn 4 bit1 8 3 1 7 4 0 A A A A 1 2 3 4 1 2 3 4 0 4 L S C 4 8 3 1 4 S S S S 9 1 6 2 2 3 1 4 5

1 1 1

1

B B 6 B B 3 C 7

Mch so snh 4 bit, 8 bit


i hc Bch Khoa TP.HCM Khoa in-in t2 4 6 8 1 9 1

L Ch Thng

1 1 1 1 1 1

9 1 2 3 4

0 2 A 3 A 5 A A 1 B 4 B B B A A A 7 4

0 1 2 3 0 1 2 3 < B = B > B L S

1 1 1 1 7 < B 6o = B 5o > B o 5

1 3 5 7

P P P P P P P P Q Q Q Q Q Q Q Q 7 4

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7

P P

= Q > Q

i A i A i A 8

1 1 1 1

3 5 7 9

2 4 6 8

L S

6

8 2

Mch to/kim tra parity8 9 1 1 1 1 1 2 4 0 1 2 3 A B C D E F G H I 7 4 L E O V D E 5 6 D N

S

2 8

0

Mch chuyn m BCD m LED 7 on anode chung7 1 2 6 4 5 3 1 2 4 8 B I / R R B I L T 7 4 L S A B C D O E F G 4 7 1 1 1 1 9 1 1 3 2 1 0 5 4

B

Mch m 8 bit2 4 6 8 1 1 1 1 2 2 2 2 A A A A A A A A 1 2 3 4 1 2 3 4 1 1 1 1 2 2 2 2 Y Y Y Y Y Y Y Y 1 11 21 31 49 17 25 33 4 8 6 4 2 2 3 4 5 6 7 8 9 A A A A A A A A 1 2 3 4 5 6 7 8 I R L S 2 4 5 B B B B B B B B 1 2 3 4 5 6 7 8 1 1 1 1 1 1 1 1 8 7 6 5 4 3 2 1

1 1 1 1

1 3 5 7

1 1 9 1 G 2 G 7 4 L S 2 4 4

1 9 G 1 D 7 4



L Ch Thng

Ph lc C: Cc vi mch tun t thng dngMch m nh phn 4 bit ng b1 2 A C 7 Q Q Q L R Q 4 L S 3 A 4 B 5 C 6 D 3 9 3 4 5 6 7 10 2 9 1 3 A B C D EN P EN T C LK LO A D C LR QA QB QC QD RC O 14 13 12 11 15 1 1 3 2 A C 7 Q Q Q L R Q 4 L S 1 A 1 B 9 C 8 D 3 9 3 1 0

74LS 163

Cac ngo vaoC LR LO D A

Cac ngo ra

Chc nang

ENPx x x L H x

ENTx x L x H x

CLK

QAL D

Q B QC Q DL C L B L A Reset ve 0 Nhap d lieu vao Khong em Khong em em Khong em

L H H H H x

x L H H H x

Khong thay oi Khong thay oi em len Khong thay oi

RCO (Ripple Carry Out) = ENT.QA.QB.QC.QD Mch m ln/xung ng b nh phn 4 bit15 1 10 9 5 4 11 14 A B C D UP DN LO A D C LR 74LS 193 QA QB QC QD C O BO 3 2 6 7 12 13

UP

DNH H

L A OD

CLRL L L L L H

Chc nang em len Khong em em xuong Khong em Nhap d lieu vao Reset ve 0

H H H H

H H x x x x

L x

Mch m mod 10 (mod 2 v mod 5)


i hc Bch Khoa TP.HCM Khoa in-in t1 4 A 1 B 2 3 6 7 R R R R 7 4 0 0 9 9 ( ( ( ( 1 2 A 9 B 8 C 1 1 D

L Ch Thng

Q Q Q Q 1 2 1 2

1 4 2

C C C 7 4

K K L

A B R L S 3

) ) ) ) 9

Q Q Q Q

3 A 5 B 6 C 7 D

1 5 1 2 C C 1 4 C 7 4

K K L

A B R L S 3

Q Q Q Q

1 3 A 1 1 B 1 0 C 9 D

9 0

9 0

L S

0

Mch m mod 12 (mod 2 v mod 6)14 1 6 7 A B R 0 (1 ) R 0 (2 ) 74LS 92 QA QB QC QD 12 11 9 8

Mch m mod 16 (mod 2 v mod 8)14 1 2 3 A B R 0 (1 ) R 0 (2 ) 74LS 93 QA QB QC QD 12 9 8 11

Thanh ghi dch PIPO3 4 6 11 13 14 9 1 D D D D D D 1 2 3 4 5 6 Q Q Q Q Q Q 1 2 3 4 5 6 2 5 7 10 12 15

C LK C LR 74LS 174

Thanh ghi dch SIPO1 2 A B QA QB QC QD QE QF QG QH 3 4 5 6 10 11 12 13

8 9

C LK C LR 74LS 164

Thanh ghi dch PISO10 11 12 13 14 3 4 5 6 2 15 1 SE R A B C D E F G H C LK IN H S H /L D 74LS 165

QH QH

9 7

Thanh ghi dch tri/ phi PIPO


i hc Bch Khoa TP.HCM Khoa in-in t2 3 4 5 6 7 11 9 10 1

L Ch Thng

SR A B C D SL C S S C LK 0 1 LR

QA QB QC QD

15 14 13 12

74LS 194

Mch cht 8 bit3 4 7 8 D D D D D D D D 0 1 2 3 4 5 6 7 C L S 3 7 3 Q Q Q Q Q Q Q Q 0 1 2 3 4 5 6 7 2 5 6 9 1 1 1 1 3 4 7 8 D D D D D D D D 0 1 2 3 4 5 6 7 C L K L S 3 7 4 7 4 L S 5 7 3 Q Q Q Q Q Q Q Q 0 1 2 3 4 5 6 7 2 5 6 9 1 1 1 1 1 1 C 1 O 2 5 6 9 2 3 4 5 6 7 8 9 D D D D D D D D C 1 2 3 4 5 6 7 8 Q Q Q Q Q Q Q Q 1 2 3 4 5 6 7 8 1 1 1 1 1 1 1 1 9 8 7 6 5 4 3 2

1 1 1 1

3 4 7 8

2 5 6 9

1 1 1 1

3 4 7 8

1 1 1 O G 7 4

1 1 1 O C 7 4


bai tap kts

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