Ngan Hang Dien Tu So 5367

8/2/2019 Ngan Hang Dien Tu So 5367

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2/121

Ngn hng in t s

Cu 6. i s th p l c phn sau sang d ng nh phn: CB7E

$. 1100 1011 0111 1110

#. 1100 1011 0111 1111

#. 1101 1011 0111 1110

#. 1100 1111 0111 1110

Cu 7. i s nh phn sau sang d ng b 1 t ng ng: 1011 0101 0110

$. 0100 1010 1001

#. 0101 1010 1001

#. 1100 1010 1001

#. 0100 1011 1001

CU H I LO I 2.

Cu 1. Th c hi n php c ng hai s c d u sau theo ph ng php b 1:

0001 11012 + (- 0000 0111)2

$. 0001 0110

#. 0010 0110

#. 0010 0111

#. 0001 1110

Cu 2. Th c hi n php c ng hai s c d u sau theo ph ng php b 2: 0000 11012 + ( 1001 1001)2

$. 0111 0100

#. 0111 0101

#. 0110 0100

#. 0111 0110

Cu 3. Th c hi n php c ng hai s sau: 675 16 + 77316

$. DE816

#. DF816

#. CE816

#. DE716

Cu 4. Th c hi n php tr hai s sau: 84 16 2A16

$. 5A16

#. 6A16

#. 7A16

H C VI N CNG NGH B U CHNH VI N THNG


3/121

Ngn hng in t s

#. 8A16

Cu 5. i s nh phn sau sang d ng b 2 t ng ng: 1011 0101 0110

$. 0100 1010 1010

#. 0100 1010 1001

#. 1100 1010 1001

#. 0100 1011 1001

CU H I LO I 3.

Cu 1. Th c hi n php c ng hai s sau theo b 1: (5) 10 + (-9)10

$. 1111 1011

#. 0000 0100

#. 1111 1010

#. 1000 0100

Cu 2. Th c hi n php c ng hai s sau theo b 2: (5) 10 + (-9)10

$. 1111 1100

#. 0000 0100

#. 1111 1010

#. 1000 0100

Cu 3. Th c hi n php c ng hai s sau: 2 6 (,101)2 + 210 (,101101)2$. 210 (,1011111)2

#. 210 (,1111111)2

#. 210 (,1011011)2

#. 210 (,1011101)2

Cu 4. Th c hi n php chia 2 s sau: 2 7 (,001)2 v 24 (,01)2

$. 22 (,001)2

#. 23 (,001)2

#. 22 (,01)2

#. 22 (,1)2



4/121

Ngn hng in t s

CU H I CH NG 2. I S BOOLE V P 2 BI U DI N HM (6t)

CU H I LO I 1.

Cu 1. A B =$. A B A B+ .

#. A B A B+ .

#. A B A B+ .

#. A B A B+ .

Cu 2. A B =$. A B A B+ .

#. A B A B+ .

#. A B A B+ .

#. A B A B+ .

Cu 3. A 1 =

$. A .

#. A.

#. 1.

#. 0.

Cu 4. A 0 =$. A.

#. A .

#. 1.

#. 0.

Cu 5. A A =$. 0.

#. A .#. A.

#. 1.

Cu 6. A A =$. 1.

#. A .

#. A.

#. 0.



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Ngn hng in t s

Cu 7. A B A B+ =

$. A B A B+ .

#. A B A B+ .

#. A B A B+ .

#. A B A B+ .

Cu 8. A B A B+ =

$. A B A B+ .

#. A B A B+ .

#. A B A B+ .

#. A B A B+ .

Cu 9. A B A B+ =$. A.

#. B.

#. 0.

#. 1.

Cu 10. A + AB =$. A.

#. B.#. 0.

#. 1.

Cu 11. A A B+ =$. A + B.

#. A.

#. B.

#. AB.

Cu 12. A =$. A

#. 1.

#. 0.

#. A .



6/121

Ngn hng in t s

Cu 13. Hai m ch i n hnh 2-1 t ng ng v i nhau v chng :

Hnh 2.1

$. u b ng A #. u b ng B #. u b ng AB #. u b ng A+B

Cu 14. A B C+ + =

$. A.B.C

#. A B C+ +

#. A.B.C#. A + B+ C

Cu 15. A.B.C =

$. A B C+ +

#. A.B.C

#. A.B.C#. A + B+ C

CU H I LO I 2.

Cu 1. Cho m ch i n nh hnh 2-2. Bi u th c hm ra l :

Hnh 2-2

$. A B

#. A B

#. A B#. A B

Cu 2. Cho m ch i n nh hnh 2-3. Bi u th c hm ra l :

Hnh 2-3

$. A B


AB B

A

A

B

F

A

B

F


7/121

Ngn hng in t s

#. A B

#. A B

#. A B

Cu 3. Rt g n: (A + B)(A + C)

$. A + BC#. B + AC#. C + AB#. A + B

Cu 4. Rt g n: A (B C)

$. A.B.C A.B.C+

#. A.B.C A.B.C+

#. A.B.C A.B.C+

#. A.B.C A.B.C+

Cu 5. ng th c sau ng hay sai: A B A B = $. ng.

#. Sai.

Cu 6. ng th c sau ng hay sai: A B A B =

$. Sai.

#. ng.


$. ng.

#. Sai.


$. ng.

#. Sai.


$. Sai.

#. ng.


$. Sai.

#. ng.

CU H I LO I 3.

Cu 1. Rt g n: A (B C)$. AB AC

#. AB A#. AB C

#. A BCH C VI N CNG NGH B U CHNH VI N THNG


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Ngn hng in t s

Cu 2. Rt g n: A BC A BC ABC ABC+ + +

$. AB + AC + BC

#. B + AC + AB#. C + AB + BC

#. A + BC+ ACCu 3. Rt g n : F (A, B, C) = (0, 2, 4, 6,7).

$. AB C+

#. AB C+

#. AB C+

#. AB C+

Cu 4. Rt g n : F (A, B, C, D) = (0, 1, 8, 9, 10)$. BC ABD+

#. BC D+

#. BC ABD+

#. BC ABD+

Cu 5. Rt g n : AB A C BC+ +

$. AB A C+

#. AB C+

#. AB C+

#. AB AC+

CU H I LO I 4.

Cu 1. Rt g n : AB BCD A C BC+ + +

$. AB C+

#. AB C+ + D

#. AB C+

#. AB C+ + DCu 2. Rt g n : C D C D . A C D+ +

$. CD

#. CD

#. CD

#. CD

Cu 3. Rt g n : A BC . A B BC CA+ +

$. AB AC BC+ +#. AB AC+



9/121

Ngn hng in t s

#. AC BC+

#. AB BC+

Cu 4. Rt g n : A C AB BC BCD E+ + +

$. AB C+

#. AB C+ + D

#. AB C+

#. AB C+ + D

CU H I CH NG 3. C NG LOGIC TTL V CMOS (7t)

CU H I LO I 1.

Cu 1. S nguyn l c a c ng AND trong hnh 3-1 l :

Hnh 3-1

$. Hnh (a)

#. Hnh (b)

#. Hnh (c)

#. Hnh (d)

Cu 2. S nguyn l c a c ng NAND trong hnh 3-2 l :

Hnh 3-2

$. Hnh (b)

#. Hnh (a)

#. Hnh (c)

#. Hnh (d)

Cu 3. S nguyn l c a c ng OR trong hnh 3-3 l :



10/121

Ngn hng in t s

Hnh 3-3

$. Hnh (c)

#. Hnh (a)

#. Hnh (b)

#. Hnh (d)

Cu 4. S nguyn l c a c ng NOR trong hnh 3-4 l :

Hnh 3-4

$. Hnh (d)

#. Hnh (a)

#. Hnh (b)

#. Hnh (c)

Cu 5. B ng tr ng thi no xc nh c ng AND?

A B Y A B Y A B Y A B Y

1 1 0 1 1 1 1 1 0 1 1 1

1 0 0 1 0 0 1 0 1 1 0 10 1 0 0 1 0 0 1 1 0 1 1

0 0 1 0 0 0 0 0 1 0 0 0

(a) (b) (c) (d)

$. (b)

#. (a)

#. (c)

#. (d)

Cu 6. B ng tr ng thi no xc nh c ng NAND?



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Ngn hng in t s


1 1 0 1 1 1 1 1 0 1 1 1

1 0 0 1 0 0 1 0 1 1 0 1

0 1 0 0 1 0 0 1 1 0 1 1

0 0 1 0 0 0 0 0 1 0 0 0

(a) (b) (c) (d)

$. (c)

#. (a)

#. (b)

#. (d)

Cu 7. B ng tr ng thi no xc nh c ng OR?


1 1 0 1 1 1 1 1 0 1 1 1

1 0 0 1 0 0 1 0 1 1 0 1

0 1 0 0 1 0 0 1 1 0 1 1

0 0 1 0 0 0 0 0 1 0 0 0

(a) (b) (c) (d)

$. (d)

#. (a)

#. (b)

#. (c)

Cu 8. B ng tr ng thi no xc nh c ng NOR?


1 1 0 1 1 1 1 1 0 1 1 1

1 0 0 1 0 0 1 0 1 1 0 1

0 1 0 0 1 0 0 1 1 0 1 1

0 0 1 0 0 0 0 0 1 0 0 0

(a) (b) (c) (d)

$. (a)

#. (b)

#. (c)

#. (d)

Cu 9. u ra c a c ng AND m c cao:



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Ngn hng in t s

$. Khi t t c l i vo m c cao. #. M i lc.#. Khi c b t k l i vo no m c th p. #. Khi c b t k l i vo no m c cao.

Cu 10. C ng NOT s d ng :

$. o tn hi u u vo c a n.

#. m tn hi u u vo c a n.

#. khu ch i tn hi u u vo c a n.

#. lm tr pha tn hi u u vo c a n.

Cu 11. u ra c a c ng NAND m c th p: $. Khi t t c l i vo m c cao. #. M i lc.#. Khi c b t k l i vo no m c th p. #. Khi c b t k l i vo no m c cao.

Cu 12. u ra c a c ng OR m c cao:

$. Khi u vo b t k m c cao.

#. M i lc

#. Khi u vo b t k m c th p.

#. Khi t t c cc u vo m c th p.

Cu 13. u ra c a c ng NOR m c th p:

$. Khi u vo b t k m c cao.

#. M i lc.

#. Khi u vo b t k m c th p.

#. Khi t t c cc u vo m c th p.

Cu 14. Cc c ng h collector

$. c th n i v i cc u ra v u vo c a cc c ng khc. #. khng th n i v i cc u vo c a c ng khc.

#. ph i n i v i cc u ra c a cc c ng collector khc.

#. s o ng c m c ra c a chng n u n i v i t.

Cu 15. C ng NOT h TTL:

$. dng o m c logic.

#. c th s d ng nh b khu ch i. #. i h i t nh t 1 u vo m c cao.



13/121

Ngn hng in t s

#. i h i t nh t 1 u vo m c th p.

CU H I LO I 2.

Cu 1. Trn hnh 3-5, tr ng thi t ng ng c a cc u ra t A n D l n l t l:

$.Th p-Cao-Th p-Cao. #. Th p-Cao-Th p-Th p. #. Cao-Th p-Th p-Th p. #. Cao-Cao-Th p-Th p.

Cu 2. Trong m ch trn hnh 3-6, tr ng thi t ng ng c a cc u ra t A n D l n l t l:

$. Th p Cao - Th p - Th p. #. Th p-Cao-Th p-Cao. #. Cao-Th p-Th p-Th p. #. Cao-Cao-Th p-Th p.

Cu 3. M ch nh hnh 3-7 s :

$. T o m c u ra th p


Hnh 3-5

U RA B U RA B U RA

A

U RA C

U RA D U VOM C CAO

U RA A U RA C

U RAB

U RAD

U VOM C CAO

Hnh 3-6


14/121

Ngn hng in t s

#. T o m c u ra cao

#. Khng ho t ng v cc u ra c a c ng NAND c n i v i nhau t i c ng NOR

#. Khng ho t ng v cc k t n i ngu n cung c p khng c ch ra

Hnh 3-7

Cu 4. M ch nh hnh 3-8 s :

$. T o m c u ra cao.

#. T o m c u ra th p

#. Khng ho t ng v cc u ra c a c ng NAND c n i v i nhau t i c ng NOR

#. Khng ho t ng v cc k t n i ngu n cung c p khng c ch ra

Hnh 3-8

Cu 5. C ng XOR t o ra u ra v i m c logic cao:

$. V i i u ki n l tr ng thi l i vo khc nhau .

#. V i i u ki n l tr ng thi l i vo gi ng nhau .#. M i lc.

#. Khng lc no c .

Cu 6. C ng XOR t o ra u ra v i m c logic th p:

$. V i i u ki n l tr ng thi l i vo gi ng nhau.

#. V i i u ki n l tr ng thi l i vo khc nhau.

#. M i lc.

#. Khng lc no c .



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Ngn hng in t s

Cu 7. Theo i u ki n m ch trong hnh 3-9 th:

$. n LED t t.

#. n LED sng.

#. m i c ng phn chia dng qua n LED.

#. n bo c kch ho t.

Cu 8. M ch logic DDL c s nh hnh v 3-10 lm ch c nng g: $. AND

#. NOR

#. OR

#. NAND

Cu 9. M ch logic DDL c s nh hnh v 3-11 lm ch c nng g:

$. OR

#. NOR

#. AND

#. NAND


Hnh 3-9

C A TR C

C A SAU

C A S

N BO (LED)

D2B

fD1

+5V

A

Hnh3-10

D2B

fD1

GND

A

Hnh3-11


16/121

Q3

f

D1

R14k

Q1A

+5V

Q2

R21 6k

R31 6k

Hnh 3-13

Ngn hng in t s

Cu 10. M ch logic RTL c s nh hnh v 3-12 lm ch c nng g:

$. NOT

#. OR

#. AND

#. NAND

1 k

1 k

10 k

12 V

-12 V 0 V

5 V

D

5 V

0 V

0 V

5,7 V

Hnh 3-12

Cu 11. M ch logic TTL c s nh hnh v 3-13 lm ch c nng g:

$. NOT collector h

#. NOT

#. AND

#. NAND

Cu 12. M ch logic PMOS c s nh hnh v 3-14 lm ch c nng g:

$. NOT

#. OR

#. AND#. NAND


V

SG

D

Q2

A

V

S

G

D

Q1

f

Hnh 3-14


17/121

Ngn hng i n t s

CU H I LO I 3.

Cu 1. M ch logic RTL c s nh hnh v 3-15 lm ch c nng g :

1 k

10 k

12 V

-12 V 0 V

5 V

D

5 V

0 V

0 V

5,7 V

A

B

Z

5 k

5 k

Hnh 3-15

$. NAND

#. NOR

#. OR

#. AND

Cu 2. M ch logic DTL c s nh hnh v 3-16 lm ch c nng g :

Hnh 3-16

$. NAND

#. NOR

#. OR#. AND

Cu 3. M ch logic PMOS c s nh hnh v 3-17 lm ch c nng g :

$. NOR

#. NAND

#. OR

#. AND


D4B

5k

Q

2k

f

+5V

D3

D1

+5V

D2A

V

SG

D

Q5

A

B

V

S

G

D

Q4

S

G

D

Q3

fHnh 3-17


18/121

Ngn hng in t s

Cu 4. M ch logic NMOS c s nh hnh v 3-18 lm ch c nng g :

$. NAND

#. NOR

#. OR

#. AND

Cu 5. M ch logic CMOS c s nh hnh v 3-19 lm ch c nng g :

$. NAND

#. NOR

#. OR

#. ANDCu 6. C ng collector h s ho t ng bnh th ng nh cc c ng logic bnh th ng n u :

$. L i ra c n i ln ngu n thng qua m t tr gnh

#. L i ra c n i ln ngu n thng qua m t t gnh

#. L i ra n i xu ng t thng qua m t tr

#. L i ra n i xu ng t thng qua m t t


VSS

V

Q1

Q2

Q3

A

B

f

Hnh 3-18

S

G

D

S

G

D

Q4

A

B

VDD

Q2

Q3

Q1

fD

G

SHnh 3-19


19/121

Ngn hng in t s

Cu 7. C cho php u vo c a m ch CMOS h khng? m ch ho t ng bnh th ng th u vo khng dng ph i c m c logic no?

$. Khng c- m ch ho t ng bnh th ng th u vo khng dng ph i n i v i m c logic 1 ho c 0 tu tnh ch t t ng m ch.

#. c- C th coi l m c 1

#. c- Ph i coi l m c 0

#. Khng c- m ch ho t ng bnh th ng th u vo khng dng ph i n i v i m c logic 0

CU H I LO I 4.

Cu 1. Ch c nng c a diode D 3 trong s 3-20 l g?

$. D ch m c i n p lm cho Q 3 v Q4 khng bao gi cng ng ho c cng m . #. Cch ly transistor Q3 v Q4

#. Ch ng nhi u l i ra

#. Cch ly Q4 kh i m ch ngoi n i vo u ra f

Cu 2. M ch i n c bi u di n trong s 3-21 ho t ng nh th no n u nh l i vo E m c th p ?

Hnh 3-21


Hnh 3-20


20/121

Ngn hng in t s

$. M ch tr thnh c ng NAND hai l i vo

#. M ch tr thnh c ng NOR hai l i vo

#. M ch tr thnh c ng AND hai l i vo

#. Tr ng thi l i ra khng theo logic c b n no Cu 3. M ch i n c bi u di n trong s 3-22 ho t ng nh th no n u nh l i vo E m c logic cao ?

Hnh 3-22

$. Tr ng thi l i ra khng theo m c logic c b n no

#. M ch tr thnh c ng NOR hai l i vo #. M ch tr thnh c ng AND hai l i vo

#. M ch tr thnh c ng NAND hai l i vo

Cu 4. Tc d ng c a tr ng thi tr khng l i ra cao trong c ng ba tr ng thi l :

$. Cch ly cc l i ra c a cc c ng logic khi chng cng c n i t i m t l i vo

#. a ra m c logic th 3 l trung bnh c a hai m c cao v th p #. a ra m c logic th p nh ng c gi tr tr khng cao

#. a ra m c logic cao nh ng c gi tr tr khng cao



21/121

Ngn hng in t s

CU H I CH NG 4. M CH LOGIC T H P

CU H I LO I 1.

Cu 1. M ch logic t h p l m ch:

$. C tn hi u u ra ch ph thu c vo tn hi u u vo c a m ch t i th i i m ang xt.

#. Khng nh ng tn hi u u ra ph thu c vo tn hi u u vo m cn ph thu c vo tr ng thi trong c a m ch t i th i i m ang xt.

#. C hai ph ng n trn u ng.

#. Khng c ph ng n no ng.

Cu 2. C m y lo i Hazard?

$. 4.#. 5.

#. 3.

#. 2.

Cu 3. Lo i Hazard no trong m ch logic t h p l lo i nguy hi m nh t?

$. Hazard logic.

#. Hazard tnh.

#. Hazard ng.

#. Hazard hm s .

Cu 4. B m ho u tin l b m ho cho php m ho khi:

$. C hai tn hi u tr ln ng th i tc ng vo.

#. Ch c m t tn hi u tc ng vo.

#. Ch hai tn hi u tc ng vo.

#. C 3 ph ng n trn u ng.

Cu 5. Khi b m ho u tin ti n hnh m ho th cc tr ng thi c u tin th p h n c x l th no?

$. Khng quan tm xem n tr ng thi no.

#. N lun m c logic th p.

#. N lun m c logic cao.

#. C 3 ph ng n trn u ng. Cu 6. B gi i m BCD 8-4-2-1 sang th p phn lm nhi m v bi n i:



22/121

Ngn hng in t s

$. u vo BCD 8-4-2-1 thnh u ra th p phn t ng ng.

#. u vo nh phn thnh u ra th p l c phn (h hexa).

#. u vo th p phn thnh m BCD 8-4-2-1.


Cu 7. D ng c hi n th 7-o n An t chung c :

$. b y An t c a b y thanh LED c u chung v i nhau.

#. b y Kat t c a b y thanh LED c u chung v i nhau.

#. m t An t c a m t thanh LED n bn trong.

#. m t Kat t c a m t thanh LED n bn trong.

Cu 8. D ng c hi n th 7-o n Kat t chung c :

$. b y Kat t c a b y thanh LED c u chung v i nhau.

#. b y An t c a b y thanh LED c u chung v i nhau.

#. m t An t c a m t thanh LED n bn trong.

#. m t Kat t c a m t thanh LED n bn trong.

Cu 9. B h p knh c kh nng:

$. n i m t l i vo trong m t nhm cc l i vo v i m t l i ra.

#. n i m t l i vo m ch v i m t l i ra trong m t nhm cc l i ra. #. n i ng th i m t l i vo m ch v i m t ho c nhi u l i ra .

#. n i ng th i m t ho c nhi u l i vo v i m t l i ra.

Cu 10. B phn knh c kh nng:

$. n i m t l i ra m ch v i m t trong m t nhm cc l i vo.

#. n i ng th i m t l i ra m ch v i m t ho c nhi u l i vo.

#. n i m t l i ra trong m t nhm cc l i ra v i m t l i vo. #. n i ng th i m t ho c nhi u l i vo v i m t l i ra.

Cu 11. N u b t o bit ch n/ l pht ra ch th parity ch n th m u d li u g m:

$. m t s ch n cc bit 1.

#. m t s ch n cc bit 0.

#. m t s l cc bit 1.


Cu 12. N u b t o bit ch n l pht ra ch th parity l th m u d li u g m:



23/121

Ngn hng in t s

$. m t s l cc bit 1.




Cu 13. M t ALU c ch a: $. M t kh i s h c v m t kh i logic.

#. M t kh i s h c.

#. M t kh i logic.

#. M t kh i so snh.

Cu 14. S nh phn A = 1000 v B = 0111, sau khi so snh hai s nh phn thu c k t qu l :

$. A > B.

#. A = B.

#. A < B.

#. B > A.

Cu 15. S nh phn A = 1101 v B = 1110, sau khi so snh hai s nh phn thu c k t qu l :

$. A < B.

#. A > B.

#. A = B.

#. B < A.

Cu 16. D a trn b so snh trn hnh 4-1, cc l i ra :

$. c gi tr 0, 1 v 0.

#. s khng th hi n ch c nng n u khng c cc n LED. #. c gi tr 1, 1 v 1.

#. c gi tr 0, 0 v 0.

Hnh 4-1.



24/121

Ngn hng in t s

Cu 17. A = 1001, B = 1010. B so snh s quy t nh A < B :

$. D a trn c p BIT 1.

#. D a trn c p LSB (c p BIT 0).

#. B i v c hai c p MSB b ng nhau.

#. B i v c hai c p MSB khng b ng nhau.

Cu 18. A = 1001, B = 1000. B so snh s quy t nh A > B :

$. D a trn c p LSB (c p BIT 0).

#. D a trn c p BIT 1.

#. B i v c hai c p MSB b ng nhau.

#. B i v c hai c p MSB khng b ng nhau.

Cu 19. Cho LED 7 o n A chung, mu n hi n th s 3 th nh ng thanh no sng?

$. Thanh a, b, c, d v g.

#. Thanh a, b, c , d v e.

#. Thanh a, b, c v e.

#. Thanh a, b, c, d v f.

Cu 20. Cho LED 7 o n A chung, mu n hi n th s 2 th nh ng thanh no sng?

$. Tt c cc thanh u sng tr thanh c v f. #. Thanh a, b, c v e.

#. Thanh a, b, d v e.

#. Thanh a, b, c v f.

Cu 21. Cho LED 7 o n A chung, mu n thanh no sng th Kat t c a thanh c m c logic g?

$. M c logic 0.

#. M c logic 1.

#. M c 0 v m c 1.

#. Khng m c no c .

Cu 22. Cho LED 7 o n K chung, mu n thanh no sng th An t c a thanh c m c logic g?

$. M c logic 1.

#. M c logic 0.#. M c 0 v m c 1.



25/121

Ngn hng in t s

#. Khng m c no c .

Cu 23. M ch gi i m 7 o n c m y u vo v m y u ra?

$. 4 vo v 7 ra.

#. 3 vo v 7 ra.

#. 2 vo v 7 ra.

#. 4 vo v 5 ra.

Cu 24. M ch h p knh 15 ng d li u c n bao nhiu ng a ch ?

$. 4 ng.

#. 3 ng.

#. 5 ng .

#. 6 ng.

CU H I LO I 2.

Cu 1. B ng tr ng thi no l b ng c a b MUX hai l i vo a ch ?

A B Y A B Y0 Y1 Y2 Y3 A B Y0 Y1 Y2 Y3

0 0 D0 0 0 D0 0 0 0 0 0 1 0 0 0

0 1 D1 0 1 0 D1 0 0 0 1 0 1 0 0

1 0 D2 1 0 0 0 D2 0 1 0 0 0 1 01 1 D3 1 1 0 0 0 D3 1 1 0 0 0 1

(a) (b) (c)

$. (a)

#. (b)

#. (c)

#. (b) v (c)

Cu 2. B ng tr ng thi no l b ng c a b DEMUX hai l i vo a ch ?

A B Y A B Y0 Y1 Y2 Y3 A B Y0 Y1 Y2 Y3

0 0 D0 0 0 D0 0 0 0 0 0 1 0 0 0

0 1 D1 0 1 0 D1 0 0 0 1 0 1 0 0

1 0 D2 1 0 0 0 D2 0 1 0 0 0 1 0

1 1 D3 1 1 0 0 0 D3 1 1 0 0 0 1

(a) (b) (c)



26/121

Ngn hng in t s

$. (b)

#. (a)

#. (c)

#. (a) v (c)

Cu 3. B ng tr ng thi no l b ng c a b gi i m a ch hai l i vo? A B Y A B Y0 Y1 Y2 Y3 A B Y0 Y1 Y2 Y3

0 0 D0 0 0 D0 0 0 0 0 0 1 0 0 0

0 1 D1 0 1 0 D1 0 0 0 1 0 1 0 0

1 0 D2 1 0 0 0 D2 0 1 0 0 0 1 0

1 1 D3 1 1 0 0 0 D3 1 1 0 0 0 1

(a) (b) (c)

$. (c)

#. (a)

#. (b)

#. (a) v (b)

Cu 4. N u ta c l i vo b c ng l Q A = QB = 1 v QC = QD = 0 (QDQCQBQA).D a vo thng tin gi tr u ra b c ng c tnh l :

$. 0100 n u s nh phn A c gi tr l 0001.

#. 1100.

#. 0011 n u s nh phn A c gi tr l 0001. #. Khng c tr ng h p no trn.

Cu 5. N u s nh phn B = 0100 v s A = 1100, th k t qu thu c sau php c ng l (1) 0000. i u ny ng khng ?

$. ng, b i v k t qu ng l 16 10.

#. ng, b i v k t qu ng l 15 10.

#. Khng, b i v c hai bit LSB u b ng 00.

#. Khng, b i v k t qu ng l (1) 1111. Cu 6. D a vo cc thng tin cho trn hnh 4-2. Gi tr u ra c a b c ng l :

$. 1010.

#. 1001.

#. 0101.

#. (1) 1010.



27/121

Ngn hng in t s

Hnh 4-2

Cu 7. Cc u ra c a b gi i m trong hnh 4-3:

$. tch c c m c th p.

#. tch c c m c cao.

#. tch c c ch 3 tr ng thi.

#. t t c u m c th p khi u vo l 0000.

Cu 8. D a trn hnh 4-4, kho ng gi tr u vo xc nh l:

$. 0000 n 1001 .

#. 1111 n 0110.

#. 0001 n 1001.

#. Khng ph i cc tr ng h p k trn.


Hnh 4-4. B gi i m t BCD sang th p phn

Hnh 4-3. B gi i m t BCD sang th p phn


28/121

A

Bf

Hnh 4-6.

Ngn hng in t s

Cu 9. N u t d li u 8-bit c m u bit l 1010 0101, h th ng parity l :

$. yu c u bit ch n/ l m c logic cao.

#. yu c u bit ch n/ l m c logic th p.

#. khng yu c u bit ch n/ l .

#. khng th s d ng t d li u ny. Cu 10. N u t d li u 8-bit c m u bit l 1010 0101, h th ng parity ch n:

$. yu c u bit ch n/ l m c logic th p.

#. yu c u bit ch n/ l m c logic cao.

#. khng yu c u bit ch n l .

#. khng th s d ng t d li u ny. Cu 11. Cho m ch t h p hnh 4-5, hy xc nh hm ra c a m ch:

$. A BC A BC A BC A BC+ + +

#. (A B C)( A B C)( A B C)( A B C)+ + + + + + + +

#. A BC A BC A BC A BC+ + +

#. (A B C)( A B C)( A B C)( A B C)+ + + + + + + +

Cu 12. Cho m ch t h p hnh 4-6, hy xc nh hm ra c a m ch:

$. f AB A AB B=

#. f A+B +A . A+B +B=

#. f A+B +A + A+B +B=

#. f AB A + AB B=


ABCABCABCABC

f

Hnh 4-5


29/121

Ngn hng in t s

Cu 13. Cho m ch m ho hnh 4-7, hy xc nh hm D c a m ch:

$. D 1 . 3 . 5 . 7 . 9=

#. D 1 3 5 7 9= + + + +

#. D 1 . 3 . 5 . 6 . 9=

#. D 1 3 5 8 9= + + + +

Cu 14. Cho m ch m ho hnh 4-8, hy

xc nh hm C c a m ch:

$. C 2 . 3 . 6 . 7=

#. C 2 3 6 7= + + +

#. C 2 . 3 . 4 . 6 . 7=

#. C 2 3 4 6 7= + + + +


R4

R3

R2

R1

+5V

9

8

7

6

5

4

3

2

1

DCBAHnh 4-8 M ch i n c a b m ho dng diode.

R4

R3

R2

R1

+5V

9

8

7

6

5

4

3

2

1

DCBAHnh 4-7 M ch i n c a b m ho dng diode.


30/121

Ngn hng in t s

Cu 15. N u E = 1 th hnh 4-9 l m ch i n c ch c nng g:

$. B ch n a ch nh phn 2 l i vo.

#. B h p knh 2 l i vo.

#. B phn knh 2 l i vo.

#. B m ho 2 l i vo.

Cu 16. N u A l ng a ch , K 1 v K0 l ng d li u th hnh 4-10 l m ch i n c ch c nng g:

$. B h p knh 2 l i vo.

#. B ch n a ch nh phn 2 l i vo

#. B phn knh 2 l i vo.

#. B m ho 2 l i vo.

Cu 17. N u A l ng a ch , K l ng d li u th hnh 4-11 l m ch i n c ch c nng g:

$. B phn knh 2 l i vo.

#. B ch n a ch nh phn 2 l i vo

#. B h p knh 2 l i vo.

#. B m ho 2 l i vo.Cu 18. Hnh 4-12 l m ch i n c ch c nng g:

$. M ch bn t ng .

#. M ch bn hi u .

#. M ch t ng ton ph n.

#. M ch hi u ton ph n .


D0

D1

D2

D3

A

E

A

Hnh 4-9

Hnh 4-10K

AK

K1

K

KA

K

Hnh 4-11

b0

a0

S0

C0

Hnh 4-12


31/121

Ngn hng in t s

CU H I LO I 3.

Cu 1. D a vo s kh i v b ng cho trong hnh 4-13, ta k t lu n:

$. u ra Y m c th p v u Y m c cao khi IC khng c php ho t

ng.

#. u ra Y m c th p v u ra Y m c cao khi IC c php ho t ng.

#. cc u vo ch n lun lun cho php m t u vo ho t ng.

#. t i m t th i i m, c th ch n nhi u h n m t u vo.

Cu 2. D a vo s kh i v b ng cho trong hnh 4-14, ta k t lu n:

$. u ra Y l y m c logic c a u vo c ch n v u ra Y l y m c logic

o c a u vo c ch n.

#. cc m c u ra khng th xc nh.

#. l i vo kh ng ch STRB khng th l d ng xung.

#. cc u ra Y v Y c cng m c logic v i u vo c ch n.


Hnh 4-13

U VO

D LI U U RA

CH N

Hnh 4-14

U VOD LI U U RA

CH N


32/121

Ngn hng in t s

Cu 3. Trn b gi i m 7 o n c minh ho trong hnh 4-15 th:

$. t i m t th i i m ho t ng, c th c nhi u h n m t u ra tr ng thi tch c c.

#. t i m t th i i m ho t ng , ch m t u ra tr ng thi tch c c.

#. t t c cc u ra ph i ng th i tr ng thi khng tch c c.

#. t t c cc u ra ph i ng th i tr ng thi tch c c.

Cu 4. Trong b gi i m trong hnh 4-16, m c u ra tch c c m c logic: $. th p.

#. trung bnh.

#. cao.

#. khng xc nh c, v khng ch r k t n i v i ngu n c p.


Hnh 4-15.

BCD TO 7-SEGMENTDECODER

Hnh 4-16.

BCD TO 7-SEGMENTDECODER


33/121


34/121

Ngn hng in t s

$. CV0 = 0, CR0 n i v i C V1,CR1 n i v i C V2,CR2 n i v i C V3 .

#. CV0 =1, CR0 n i v i C V1,CR1 n i v i C V2,CR2 n i v i C V3

#. CV1 n i v i C R3 , CR0 n i v i C V1,CR1 n i v i C V2,CR2 n i v i C V3

#. Khng tr ng h p no ng. Cu 8. M ch i n hnh 4-19 c ch c nng g?

$. M ch t ng ton ph n.

#. M ch hi u ton ph n.

#. M ch bn hi u.

#. M ch bn t ng.

Cu 9. Trong hnh 4-20, n u u vo 6 m c th p. D n n:

$. u ra BCD c m c logic l 1001. #. khng c u ra no m c logic th p.

#. khng c u ra no m c logic cao.

#. u ra BCD c m c logic l 0110.


C

b

a SCi-1

Hnh 4-19

Hnh 4-20. B m ha u tin BCD


35/121

Ngn hng in t s

Cu 10. Trong hnh 4-21, n LED 2:

$. sng khi m t trong hai u vo c ng OR m c th p.

#. sng khi c hai u vo c ng OR m c cao.

#. khng th c i u khi n sng b i c ng OR.

#. lun lun c i u khi n sng b i c ng OR.

Hnh 4-21

Cu 11. Trong m ch hnh 4-22, chi u dng i n c a o n-g:

$. i vo b i u khi n v IC tr ng thi ht dng.

#. i ra kh i b i u khi n v IC tr ng thi phun dng.

#. i ra kh i b i u khi n v IC tr ng thi ht dng.

#. i vo b i u khi n v IC tr ng thi phun dng.

Hnh 4-22

Cu 12. Trong m ch hnh 4-23, M l u i u khi n, n u M = 0 th m ch c ch c

nng g?$. M ch bn t ng.



36/121

Ngn hng i n t s

#. M ch bn hi u

#. M ch t ng ton ph n


Cu 13. Trong m ch hnh 4-24, M l u i u khi n, n u M = 1 th m ch c ch c nng g:

$. M ch bn hi u

#. M ch bn t ng.

#. M ch t ng ton ph n


CU H I LO I 4 .

Cu 1. Cho b ng tr ng thi c a m ch ton t ng nh sau, bi u th c l i ra l:

$. Si = ai bi Ci-1 v Ci = aibi + (ai bi) Ci - 1

#.i i i i 1

S a b C

= v Ci = aibi + (ai bi) Ci - 1

#. Si = ai bi Ci-1 v i i i i i i - 1C = a b + (a b ) C

#.i i i i 1

S a b C

= vi i i i i i - 1

C = a b + (a b ) C


Ci-1 ai bi Si Ci

0

0

0

0

1

1

1

1

0

0

1

1

0

0

1

1

0

1

0

1

0

1

0

1

0

1

1

0

1

0

0

1

0

0

0

1

0

1

1

1

B ng tr ng thi c a m ch ton t ng.

b0

a0

S0

C0/ B

0

Hnh 4-23M

b0

a0

S0

C0/ B

0

Hnh 4-24M


37/121

Ngn hng in t s

Cu 2. Cho b ng tr ng thi c a m ch hi u ton ph n nh sau, bi u th c l i ra l:

$. Si = ai bi Bi-1 v i i i-1 i i i-1B = a B + b (a B )

#.i i i i 1

S a b B

= vi i i-1 i i i-1

B = a B + b (a B )

#. Si = ai bi Bi-1 v i i i-1 i i i-1B = a B + b (a B )

#.i i i i 1

S a b B

= vi i i-1 i i i-1

B = a B + b (a B )

Cu 3. IC 7483 l b c ng 2 s nh phn 4 bit, m ch hnh 4-25 c ch c nng g? $. M ch c ng 2 s nh phn 4 bit theo b 1.

#. M ch c ng 2 s nh phn 4 bit theo b 2.

#. M ch c ng 2 s nh phn 4 bit.

#. M ch nhn 2 s nh phn 4 bit.


ai bi Bi-1 Si Bi

0

01

1

0

0

1

1

0

10

1

0

1

0

1

0

00

0

1

1

1

1

0

11

0

1

0

0

1

0

10

0

1

1

0

1

B ng tr ng thi c a m ch hi u ton ph n.

A3

A2

A1

A0

B3

B2

B1

B0

Co 7483 Ci

' ' ' '

3 2 1 0B B B B

M

Hnh 4-25


38/121

Ngn hng in t s

Cu 4. IC 7483 l b c ng 2 s nh phn 4 bit, m ch hnh 4-26 c ch c nng g?

$. M ch c ng 2 s nh phn 4 bit theo b 2.

#. M ch c ng 2 s nh phn 4 bit theo b 1.

#. M ch c ng 2 s nh phn 4 bit.

#. M ch nhn 2 s nh phn 4 bit.

Cu 5. IC 7483 l b c ng 2 s nh phn (s A v B) 4 bit, m ch hnh 4-27 l m ch c ng tr 2 s theo b 1, m ch c ch c nng g khi M = 0 v M = 1:

$. M = 0 (A + B); M = 1 (A B).

#. M = 0 (A - B); M = 1 (A + B).

#. C hai tr ng h p trn u ng.

#. Khng th c hi n c php tnh .

Cu 6. IC 7483 l b c ng 2 s nh phn (s A v B) 4 bit, m ch hnh 4-28 l m ch c ng tr 2 s theo b 2, m ch c ch c nng g khi M = 0 v M = 1:

$. M = 0 (A + B); M = 1 (A B).


A3

A2

A1

A0

B3

B2

B1

B0

Co 7483 Ci

' ' ' '

3 2 1 0B B B B

M

Hnh 4-26

A3

A2

A1

A0

B3

B2

B1

B0

Co 7483 Ci

' ' ' '

3 2 1 0B B B B

M

Hnh 4-27


39/121

Ngn hng in t s

#. M = 0 (A - B); M = 1 (A + B).

#. C hai tr ng h p trn u ng.

#. Khng th c hi n c php tnh .

CU H I CH NG 5. M CH LOGIC TU N T

CU H I LO I 1.

Cu 1. M ch logic tu n t l m ch:

$. Khng nh ng tn hi u u ra ph thu c vo tn hi u u vo m cn ph thu c vo tr ng thi trong c a m ch t i th i i m ang xt.

#. C tn hi u u ra ch ph thu c vo tn hi u u vo c a m ch t i th i i m ang xt.

#. C hai ph ng n trn u ng.


Cu 2. Trong cc lo i trig sau, trig no cn t n t i t h p c m:

$. Trig RS.

#. Trig D.

#. Trig T

#. Trig JK.

Cu 3. Trig JK ng b c u t o t c ng NAND ho t ng :

$. S n d ng c a xung nh p.

#. S n m c a xung nh p. #. C hai s n xung.


A3

A2

A1

A0

B3

B2

B1

B0

Co 7483 Ci

' ' ' '

3 2 1 0B B B BM

Hnh 4-28


40/121

Ngn hng in t s

#. C ba ph ng n trn u ng.

Cu 4. Trig JK ng b c u t o t c ng NOR ho t ng :

$. S n m c a xung nh p.

#. S n d ng c a xung nh p.

#. C hai s n xung.


Cu 5. Cc lo i trig MS ho t ng :

$. C hai s n xung.

#. S n m c a xung nh p.

#. S n d ng c a xung nh p.


Cu 6. N u u vo D c a trig thay i th u ra

$. s thay i theo D sau khi c xung nh p clock u vo.

#. thay i tr ng thi c a n m t cch t c th i

#. s thay i sau khi c 2 xung nh p clock u vo .

#. s khng thay khi c xung nh p ti p theo.

Cu 7. M t trig JK ch l t. N u t n s Clock c a n l 2000 hz th t n s

t i l i ra l: $. 1000 hz.

#. 4000 hz.

#. 2000 hz.

#. 500 hz.

Cu 8. Ph ng trnh c tr ng c a trig JK l:

$. kQ J Q K Q= + .

#. kQ J Q K Q= + .

#. kQ J Q K Q= +

#. kQ J Q K Q= +

Cu 9. Ph ng trnh c tr ng c a trig RS l:

$. kQ S R Q= + vSR = 0.

#. kQ S R Q= + vSR = 1.

#. kQ S R Q= + vSR = 0



41/121

Ngn hng in t s

#. kQ S R Q= + vSR = 1

Cu 10. Ph ng trnh c tr ng c a trig D l:

$. kQ D= .

#. kQ D= .

#. kQ DQ= .

#. kQ D Q= +

Cu 11. Ph ng trnh c tr ng c a trig T l:

$. kQ T Q T Q= + .

#. kQ T Q T Q= + .

#. kQ T Q T Q= + .

#. kQ T Q T Q= + .

Cu 12. M hnh Mealy l m hnh:

$. c hm ra ph thu c vo tn hi u vo v tr ng thi trong c a m ch.

#. c hm ra ph thu c vo tn hi u vo.

#. c hm ra ph thu c vo tr ng thi trong c a m ch.

#. khng c ph ng n no ng.

Cu 13. M hnh Moore l m hnh:

$. c hm ra ph thu c vo tr ng thi trong c a m ch.

#. c hm ra ph thu c vo tn hi u vo v tr ng thi trong c a m ch.

#. c hm ra ph thu c vo tn hi u vo.

#. khng c ph ng n no ng.

Cu 14. Ph n t l u gi thng tin c a b ghi d ch l:

$. Trig D

#. Trig T.#. Trig RS.

#. Trig JK.

Cu 15. t o ra c m t Trig Chnh - ph (MS) c n:

$. hai trig cng lo i ng b .

#. hai trig cng lo i.

#. ba trig cng lo i ng b .

#. b n trig cng lo i.



42/121

Ngn hng in t s

Cu 16. M t b m nh phn 5 bit th t n s t i l i ra c a bit c tr ng s l n nh t so v i t n s xung nh p:

$. nh h n 32 l n.

#. nh h n 64 l n.

#. nh h n 16 l n.

#. nh h n 8 l n.

Cu 17. Trong b m ng b , cc l i vo Clock

$. l chung cho m i trig c a b m.

#. ph i c n i v i trig LSB c a b m.

#. ph i c n i v i trig MSB c a b m.

#. ph i l d ng xung c pht theo ki u n b c.

Cu 18. Trong b m khng ng b , tn hi u c n m

$. ph i c n i v i trig LSB c a b m.

#. ph i c n i v i trig MSB c a b m.

#. l chung cho m i trig c a b m.

#. ph i l d ng sng sin.

Cu 19. T n s u vo c a m t b m khng ng b 4 bit l 1MHz. V y t n

s t i u ra t i l i ra c tr ng s l n nh t (MSB) l bao nhiu ? $. 62,5 KHz.

#. 1000 KHz.

#. 500 KHz.

#. 125 KHz.

Cu 20. Khi pht xung clock vo b m khng ng b th xung clock l :

$. Tn hi u i u khi n trig LSB c a b m.

#. Tn hi u i u khi n t t c cc u vo.

#. Tn hi u i u khi n trig MSB c a b m.

#. Tr ng thi tnh.

Cu 21. H s chia t n s cho m t b m khng ng b 5 bit l n l t l :

$. 2, 4, 8, 16 v 32.

#. 1, 2, 4 , 8 v 16.

#. 1, 2, 4, 16 v 32.



43/121

Ngn hng in t s

#. T t c cc tr ng h p trn, ph thu c vo t n s xung clock.

Cu 22. M t b m ng b Mod 10 s :

$. m t 0 9.

#. m t 0 10.

#. Lun l 0 15.

#. Lun l 15.

Cu 23. B ghi d ch dng d ch tri d li u vo n i ti p th lu ng bit d li u chuy n ng t :

$. T ph i qua tri.

#. T tri qua ph i.

#. T ph i qua tri sau t t ri qua ph i .

#. Khng c tr ng h p no trn.

Cu 24. B ghi d ch dng d ch ph i d li u vo n i ti p th lu ng bit d li u chuy n ng t :

$. T tri qua ph i.

#. T ph i qua tri.

#. T ph i qua tri sau t t ri qua ph i .


Cu 25. D li u n p vo b ghi d ch c th :

$. L b t k ki u d li u no.

#. ch l ki u d li u m c cao.

#. ch l ki u d li u m c th p.

#. ch l ki u d li u thay i lun phin.

Cu 26. Khi t n s xung nh p c a b m n i ti p tng th :

$. Ch c nng c a cc u vo xo (CLEAR) v l p (SET) khng b nh h ng g.

#. Cc u vo xo (CLEAR) v l p (SET) s khng i u khi n t t c cc trig c a b m.

#. Tng kh nng m l n nh t c a n.

#. Gi m kh nng m l n nh t c a n. H C VI N CNG NGH B U CHNH VI N THNG


44/121

Ngn hng in t s

Cu 27. N u c p m t xung clock vo b m n i ti p th :

$. Cho php m t b m n i ti p ch y trong ch khng ng b .

#. Xc nh s m l n nh t c a b m n i ti p.

#. Thay i l n l t cc ch ho t ng c a b m n i ti p.

#. Chuy n m t b m n i ti p thnh m t b m song song.

Cu 28. N u nh b m c xo v sau u ra Q (BIT 4) c n i v i l i CLEAR (xo) c a b m th :

$. B m s khng ho t ng .

#. B m s m n s th 8, sau s reset (xo) l i.

#. B m s m n s th 8, sau s preset (l p) l i

#. T t c cc u ra c a b m s trng pha. Cu 29. Chn CLEAR (xo) c a b m ho t ng m c tch c c th p. Khi chn CLEAR c a xu ng m c th p th b m :

$. Ti p nh n xung xo, lc ny t t c cc u ra khng o c t m c th p.

#. Khng ti p nh n xung xo b i v xung CLOCK ch y t do.

#. Ti p nh n xung xo, lc ny t t c cc u ra o c t c nh m c

th p.#. Dao ng gi a gi tr m l n nh t v gi tr nh nh t.

Cu 30. Chn SET (l p) c a b m ho t ng m c tch c c th p. Khi chn SET (l p) c a xu ng m c th p th b m:

$. Ti p nh n xung l p, lc ny t t c cc u ra khng o c t m c cao.

#. Khng ti p nh n xung l p b i v xung CLOCK ch y t do.

#. Ti p nh n xung l p, lc ny t t c cc u ra o c t c nh m c cao.

#. Dao ng gi a gi tr m l n nh t v gi tr nh nh t.

Cu 31. N u kch ho t m t b m n i ti p 4 bit th t i cc l i ra o c a chng s :

$. m t 15 0.

#. m t 0 15.

#. Lun l 0.

#. Lun l 15.



45/121

Ngn hng in t s

Cu 32. C n bao nhiu chu k xung clock u vo pht ra m t chu k xung t i l i ra c tr ng s l n nh t (MSB) c a b m n i ti p 4 bit.

$. 16.

#. 32.

#. 8.

#. 4.

Cu 33. Cc Trig JK s d ng trong b m n i ti p c xy d ng b ng cch:

$. S d ng b t k c u trc no d i y.

#. N i l i vo J v K v i V CC v v hi u ho cc l i vo CLR (xo) v PR (l p).

#. C u trc m ch Trig JK gi ng nh m t m ch Trig T.

#. N i t t c cc l i vo J, K, CLR v PR v i V CC.Cu 34. C n bao nhiu chu k xung clock u vo pht ra m t chu k xung t i l i ra c tr ng s l n nh t (MSB) c a b m song song 4 bit.

$. 16.

#. 32.

#. 8.

#. 4.

Cu 35. N u kch ho t m t b m song song 4 bit th t i cc l i ra o c a chng s :

$. m t 15 0.

#. m t 0 15.

#. Lun l 0.

#. Lun l 15.

Cu 36. Khi t n s xung nh p c a b m song song gi m th :

$. Ch c nng c a cc u vo xo (CLEAR) v l p (SET) khng b nh h ng g.

#. Cc u vo xo (CLEAR) v l p (SET) khng i u khi n t t c cc trig c a b m.

#. Tng kh nng m l n nh t c a n.

#. Gi m kh nng m l n nh t c a n.

Cu 37. Ch c nng n p d li u vo song song c a b ghi d ch s d ng trig D:



46/121

Ngn hng in t s

$. Yu c u s n d ng c a xung clock.

#. L m t l i vo u tin.

#. L l i vo u tin cng chung v i d li u vo n i ti p.

#. Yu c u s n m c a xung clock. Cu 38. Trong m t s chu k xung clock, h ng d ch c a d li u:

$. C th thay i l n l t gi a ph i v tri.

#. Ph i l m t h ng.

#. C th ng th i hai h ng.


Cu 39. Trong b m vng, d li u c d ng:

$. Ch c 1 bit 1 ch y vng trn.

#. Ch c 1 bit 0 ch y vng trn .

#. T t c l bit 1 ch y vng trn .

#. T t c l bit 0 ch y vng trn .

Cu 40. Trong b m vng xo n, d li u c d ng:

$. Tng d n bit 1 sau gi m d n bit 1.

#. Tng d n bit 0 sau gi m d n bit 0.

#. Gi m d n bit 1 sau tng d n bit 1.

#. Tng d n bit 1.

Cu 41. Trong b ghi d ch 4 bit c n bao nhiu xung clock l y d li u ra theo cch song song:

$. 4 xung.

#. 5 xung.

#. 6 xung.

#. 7 xung.

Cu 42. Trong b ghi d ch 4 bit c n bao nhiu xung clock l y d li u ra theo cch n i ti p?

$. 7 xung.



47/121

Ngn hng in t s

#. 5 xung.

#. 6 xung.

#. 8 xung.

Cu 43. Trong b ghi d ch 4 bit, d li u c n n p theo cch n i ti p l D 3D2D1D0,khi th c hi n d ch ph i d li u c n d ch bit no tr c?

$. D0.

#. D1.

#. D2.

#. D3.

Cu 44. Trong b ghi d ch 4 bit, d li u c n n p theo cch n i ti p l D 3D2D1D0,khi th c hi n d ch tri d li u c n d ch bit no tr c?

$. D3.

#. D0.

#. D1.

#. D2.

Cu 45. Trong b ghi d ch 8 bit, c n bao nhiu trig ? $. 8.

#. 7.

#. 6.

#. 5.

Cu 46. B m vng xo n l b m m Johnson?

$. ng.

#. Sai.

Cu 47. B m vng l b m m Johnson?

$. Sai.

#. ng.

Cu 48. Trig JK ng b c th c dng xy d ng b ghi d ch?

$. ng.



48/121

Ngn hng in t s

#. Sai.

CU H I LO I 2.

Cu 1. Cho hnh 5-1. Cho bi t y l Mod m y?

$. Mod 2.

#. Mod 1.

#. Mod 3.

#. Mod 4.

Cu 2. Cho hnh 5-2. Cho bi t d ng sng c a Q 0?

$. Hnh (a).

#. Hnh (b).

#. Hnh (c).

#. Hnh (d).



Clock

'1'

Hnh 5-1

'1' J0

Q0

>

K

Clock

'1'Hnh 5-2

'1' J0

Q0

>

K0

Clock

'1'

Hnh 5-3'1' J0 Q0

>

K

Q

C

(a)

Q

C

Q

C

Q

C

(b)

(c)(d)


49/121

Ngn hng in t s

$. Mod 2.

#. Mod 1.

#. Mod 3.#. Mod 4.


$. Hnh (b).

#. Hnh (a).

#. Hnh (c).

#. Hnh (d).


$. Mod 4.

#. Mod 2.

#. Mod 3.

#. Mod 5.

Cu 6. Cho hnh 5-6. Cho bi t d ng sng c a Q 1?H C VI N CNG NGH B U CHNH VI N THNG

Clock

'1'Hnh 5-4

'1' J0

Q0

>

K

Q

C

(a)

Q

C

Q

C

Q

C

(b)

(c)(d)

Clock

J1

Q1

>

K1

'1'

Hnh 5-5'1'

J0

Q0

>

K0


50/121

Ngn hng in t s

$. Hnh (d).

#. Hnh (a).

#. Hnh (b).

#. Hnh (c).

Cu 7. Cho hnh 5-7. Cho bi t d ng sng c a Q 1 v Q0?


Clock

J1

Q1

>

K1

'1'

Hnh 5-6'1'

J0

Q0

>

K0

(c)

(a)(b)

(d)

Q

C

Q

C

Q

C

Q

C

Clock

J1

Q1

>

K1

'1'

Hnh 5-7'1'

J0

Q0

>

K0

(c)

(a)(b)

(d)

Q

C

Q

Q

C

Q

Q

C

Q

Q

C

Q


51/121

Ngn hng in t s

$. Hnh (c).

#. Hnh (a).

#. Hnh (b).

#. Hnh (d).



52/121

Ngn hng in t s


$. Hnh (b).

#. Hnh (a).

#. Hnh (c).

#. Hnh (d).



(c)

(a)(b)

(d)

C

Q

C

Q

C

Q

C

Q

(c)

(a) (b)

(d)

Q

C

Q

Q

C

Q

Q

C

Q

Q

C

Q

Clock

'1'

Hnh 5-9'1'

J0

Q0

>

K

J1

Q1

>

K1

Clock

'1'

Hnh 5-8

'1'

J0

Q0

>

K

J1

Q1

>

K1


53/121

Ngn hng in t s

$. Hnh (b).

#. Hnh (a).

#. Hnh (c).

#. Hnh (d).


$. Mod 4.

#. Mod 2.

#. Mod 3.

#. Mod 5.Cu 11. Cho hnh 5-11. Cho bi t d ng sng c a Q 1?

$. Hnh (c).

#. Hnh (a).#. Hnh (b).

#. Hnh (d).



Clock

'1'

Hnh 5-10

'1''1'

J0

Q0

>

K0

'1' J1

Q1

>

K1

Clock

'1'

Hnh 5-11

'1''1'

J0

Q0

>

K0

'1' J1

Q1

>

K1

Clock

'1'

Hnh 5-12

'1''1'

J0

Q0

>

K0

'1' J1

Q1

>

K1

(d)

(a)(b)

(c)

C

Q

C

Q

C

Q

C

Q


54/121

Ngn hng in t s

$. Hnh (d).

#. Hnh (a).

#. Hnh (b).#. Hnh (c).


$. Mod 4, m li.#. Mod 4, m ti n.

#. Mod 3, m li.

#. Mod 3, m ti n.



Clock

'1'

Hnh 5-13

'1''1'

J0

Q0

>

K

'1' J1

Q1

>

K1

Clock

'1'

Hnh 5-14'1''1'

J0

Q0

>

K

'1' J1

Q1

>K

1

(c)

(a)(b)

(d)

Q

C

Q

Q

C

Q

Q

C

Q

Q

C

Q


55/121

Ngn hng in t s

$. Hnh (b).

#. Hnh (a).#. Hnh (c).

#. Hnh (d).

Cu 15. Cho hnh 5-15. Cho bi t y l b m Mod m y?

$. Mod 4, m li.

#. Mod 4, m ti n.

#. Mod 5, m li.

#. Mod 5, m ti n.



Clock

'1'

Hnh 5-15

'1''1'

J0

Q0

>

K0

'1' J1

Q1

>

K1

Clock

'1'

Hnh 5-16

'1''1'

J0

Q0

>

K0

'1' J1

Q1

>

K1

(d)

(a)

(c)

(b)

Ck

Q0

Q1

Ck

Q0

Q1

Ck

Q0

Q1

Ck

Q0

Q1


56/121

Ngn hng in t s

$. Hnh (d).

#. Hnh (a).

#. Hnh (b).

#. Hnh (c).

Cu 17. Cho b m n i ti p 4 bit, n u th i gian tr c a 1 trig l th th i gian tr c a b m l bao nhiu (b qua cc th i gian tr khc)?

$. 4 .

#. 3 .

#. 2 .

#. .

Cu 18. Cho b m song song 4 bit, n u th i gian tr c a 1 trig l th th i gian tr c a b m l bao nhiu (b qua cc th i gian tr khc)?

$. .

#. 2 .

#. 3 .

#. 4 .

Cu 19. Cho hnh 5-17. hnh tr ng thi c a m ch l hnh no?


(a)

(c)

(b)

Ck

Q0

Q1

Ck

Q0

Q1

Ck

Q0

Q1

(d)

Ck

Q0

Q1

Clock

J1

Q1

>

K1

'1'

Hnh 5-17'1'

J0

Q0

>

K


57/121


58/121

Ngn hng in t s

#. Hnh (b).

#. Hnh (c).

#. Hnh (d).


$. Hnh (b).

#. Hnh (a).

#. Hnh (c).

#. Hnh (d).


$. Hnh (c).

#. Hnh (a).

#. Hnh (b).

#. Hnh (d).


00 11

1001

00 01

1011

00 10

0111

00 11

0110(a) (b) (c) (d)

Clock

'1'

Hnh 5-19

'1''1'

J0

Q0

>

K

'1' J1

Q1

>

K1

00 11

1001

00 10

0111

00 01

1011

00 11

0110

(a) (b) (c) (d)

Clock

'1'Hnh 5-20

'1'

J0 Q0

>

K

J1 Q1

>

K1


59/121

Ngn hng in t s

Cu 23. Cho hnh 5-21. B ng tr ng thi c a m ch l b ng no?

C Q1 Q0 k1

Q k0

Q C Q1 Q0k

1Q k

0Q C Q1 Q0 k

1Q k

0Q C Q1 Q0 k

1Q k

0Q

0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0

1 0 1 1 0 1 0 1 1 0 1 0 1 1 1 1 0 1 1 1

2 1 0 1 1 2 1 0 0 1 2 1 0 1 0 2 1 0 0 1

3 1 1 0 0 3 1 1 0 0 3 1 1 0 1 3 1 1 0 0(a) (b) (c) (d)

$. B ng (a).

#. B ng (b).

#. B ng (c).

#. B ng (d).


C Q1 Q0 k1

Q k0

Q C Q1 Q0k

1Q k

0Q C Q1 Q0 k

1Q k

0Q C Q1 Q0 k

1Q k

0Q

0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0

1 0 1 1 0 1 0 1 1 0 1 0 1 1 1 1 0 1 1 12 1 0 0 1 2 1 0 1 1 2 1 0 1 0 2 1 0 0 1

3 1 1 0 0 3 1 1 0 0 3 1 1 0 1 3 1 1 0 0

(a) (b) (c) (d)

$. B ng (b).

#. B ng (a).

#. B ng (c).

#. B ng (d).


Clock

J1

Q1

>

K1

'1'

Hnh 5-22'1'

J0

Q0

>

K

Clock

'1'

Hnh 5-21

'1'

J0

Q0

>

K

J1

Q1

>

K1


60/121

Ngn hng in t s


C Q1 Q0 k1

Q k0

Q C Q1 Q0k

1Q k

0Q C Q1 Q0 k

1Q k

0Q C Q1 Q0 k

1Q k

0Q

0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 1 0

1 0 1 1 1 1 0 1 1 0 1 0 1 1 0 1 0 1 1 1

2 1 0 1 0 2 1 0 1 1 2 1 0 0 1 2 1 0 0 1

3 1 1 0 1 3 1 1 0 0 3 1 1 0 0 3 1 1 0 0

(a) (b) (c) (d)

$. B ng (c).

#. B ng (a).

#. B ng (b).

#. B ng (d).


C Q1 Q0 k1

Q k0

Q C Q1 Q0k

1Q k

0Q C Q1 Q0 k

1Q k

0Q C Q1 Q0 k

1Q k

0Q

0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 1 0

1 0 1 1 1 1 0 1 1 0 1 0 1 1 0 1 0 1 1 1

2 1 0 1 0 2 1 0 1 1 2 1 0 0 1 2 1 0 0 1

3 1 1 0 1 3 1 1 0 0 3 1 1 0 0 3 1 1 0 0

(a) (b) (c) (d)

$. B ng (b).

#. B ng (a).

#. B ng (c).

#. B ng (d).Cu 27. Cho hnh 5-25. B ng tr ng thi c a m ch l b ng no?


Clock

'1'

Hnh 5-23

'1''1'

J0

Q0

>

K

'1' J1

Q1

>

K1

Clock'1'

Hnh 5-24

'1''1'

J0 Q0

>

K0

'1' J1 Q1

>

K1


61/121

Ngn hng in t s

C Q1 Q0 k1

Q k0

Q C Q1 Q0k

1Q k

0Q C Q1 Q0 k

1Q k

0Q C Q1 Q0 k

1Q k

0Q

0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 1 0

1 0 1 1 1 1 0 1 1 0 1 0 1 1 0 1 0 1 1 1

2 1 0 1 0 2 1 0 1 1 2 1 0 0 1 2 1 0 0 1

3 1 1 0 1 3 1 1 0 0 3 1 1 0 0 3 1 1 0 0

(a) (b) (c) (d)

$. B ng (c).

#. B ng (a).

#. B ng (b).

#. B ng (d).


C Q1 Q0 k1

Q k0

Q C Q1 Q0k

1Q k

0Q C Q1 Q0 k

1Q k

0Q C Q1 Q0 k

1Q k

0Q

0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1

1 0 1 1 1 1 0 1 1 1 1 0 1 1 0 1 0 1 1 0

2 1 0 1 0 2 1 0 0 1 2 1 0 0 1 2 1 0 1 1

3 1 1 0 1 3 1 1 0 0 3 1 1 0 0 3 1 1 0 0

(a) (b) (c) (d)

$. B ng (d).

#. B ng (a).

#. B ng (b).

#. B ng (c).


Clock

'1'

Hnh 5-25

'1''1'

J0

Q0

>

K0

'1' J1

Q1

>

K1

Clock

J1 Q1

>

K1

'1'Hnh 5-26

'1'

J0 Q0

>

K


62/121

Ngn hng in t s

CU H I LO I 3.

Cu 1. Cho b m hnh 5-27. Cho bi t y l b m Mod m y?

$. Mod 3.

#. Mod 4.

#. Mod 5.

#. Mod 6.


$. Hnh (c).

#. Hnh (a).

#. Hnh (b).H C VI N CNG NGH B U CHNH VI N THNG

Clock

J0

Q0

>

K0

J1

Q1

>K

1'1'

Hnh 5-27'1'

Clock

J0

Q0

>

K0

J1

Q1

>

K1'1'

Hnh 5-28

'1'

Ck

Q0

Q1

Ck

Q0

Q1

Ck

Q0

Q1

Ck

Q0

Q1

(a) (b)

(c) (d)


63/121

Ngn hng in t s

#. Hnh (d).


C Q1 Q0 k1

Q k0

Q C Q1 Q0k

1Q k

0Q C Q1 Q0 k

1Q k

0Q C Q1 Q0 k

1Q k

0Q

0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1

1 0 1 1 1 1 0 1 1 1 1 0 1 1 0 1 0 1 1 0

2 1 0 1 0 2 1 0 0 1 2 1 0 0 1 2 1 0 0 0

3 1 1 0 1 3 1 1 0 0 3 1 1 0 0 3 1 1 0 0

(a) (b) (c) (d)

$. B ng (d).

#. B ng (a).

#. B ng (b).

#. B ng (c).



Clock

J0

Q0

>

K

J1

Q1

>

K1'1'

Hnh 5-29

'1'

00 11

1001

00 01

1011

00 01

1011

00 11

1001

(a) (b) (c) (d)

Clock

J0

Q0

>

K

J1

Q1

>

K1'1'

Hnh 5-30

'1'


64/121

Ngn hng in t s

$. Hnh (c).

#. Hnh (a).

#. Hnh (b).#. Hnh (d).


$. Mod 8.

#. Mod 6.

#. Mod 7.

#. Mod 9 .


$. Hnh (a).

#. Hnh (b).


Clock

J1

Q1

>

K1

'1'

Hnh 5-31'1'

J0

Q0

>

K

J2

Q2

>

K2

Clock

J1

Q1

>

K1

'1'

Hnh 5-32'1'

J0

Q0

>

K0

J2

Q2

>

K2

Ck

Q

Ck

Q

Ck

Q

Ck

Q

1 8 1 8

1 81 8

(a) (b)

(c) (d)


65/121

Ngn hng in t s

#. Hnh (c).

#. Hnh (d).


C Q2 Q1 Q0 k2

Q k1

Q k0

Q C Q2 Q1 Q0 k2

Q k1

Q k0

Q

0 0 0 0 0 0 1 0 0 0 0 1 1 1

1 0 0 1 0 1 0 1 0 0 1 1 1 0

2 0 1 0 0 1 1 2 0 1 0 1 0 1

3 0 1 1 1 0 0 3 0 1 1 1 0 0

4 1 0 0 1 0 1 4 1 0 0 0 1 1

5 1 0 1 1 1 0 5 1 0 1 0 1 0

6 1 1 0 1 1 1 6 1 1 0 0 0 1

7 1 1 1 0 0 0 7 1 1 1 0 0 0

(a) (b)

$. B ng (a).#. B ng (b).


000 001 010 011 111 110 101 100

000 111 110 101 001 010 011 100

(a) (b)

$. Hnh (a).

#. Hnh (b).

Cu 8. Cho hnh 5-35. Cho bi t y l c th m c Mod m y?

H C VI N CNG NGH B U CHNH VI N THNG Clock

'1'

Hnh 5-35'1''1'

J0

Q0

>

K0

'1' J1

Q1

>

K1 '1'

'1' J2

Q2

>

K2

Clock

J1 Q1

>

K1

'1'Hnh 5-33

'1'

J0 Q0

>

K

J2 Q2

>

K2

Clock

J1

Q1

>

K1

'1'

Hnh 5-34'1'

J0

Q0

>

K0

J2

Q2

>

K2


66/121

Ngn hng in t s

$. Mod 8.

#. Mod 5.

#. Mod 6.#. Mod 7.


$. Hnh (c).#. Hnh (a).

#. Hnh (b).

#. Hnh (d).


C Q2 Q1 Q0 k2

Q k1

Q k0

Q C Q2 Q1 Q0 k2

Q k1

Q k0

Q

0 0 0 0 1 1 1 0 0 0 0 0 0 1

1 0 0 1 1 1 0 1 0 0 1 0 1 0

2 0 1 0 1 0 1 2 0 1 0 0 1 1

3 0 1 1 1 0 0 3 0 1 1 1 0 0

4 1 0 0 0 1 1 4 1 0 0 1 0 1

5 1 0 1 0 1 0 5 1 0 1 1 1 0

6 1 1 0 0 0 1 6 1 1 0 1 1 1H C VI N CNG NGH B U CHNH VI N THNG

Clock

'1'

Hnh 5-36'1''1'

J0

Q0

>

K

'1' J1

Q1

>

K1 '1'

'1' J2

Q2

>

K2

Clock

'1'

Hnh 5-37'1''1'

J0

Q0

>

K

'1' J1

Q1

>

K1

'1'

'1' J2

Q2

>

K2

Ck

Q

Ck

Q

Ck

Q

Ck

Q

1 8

1 8

1 8

1 8

(c)

(b)(a)

(d)


67/121

Ngn hng in t s

7 1 1 1 0 0 0 7 1 1 1 0 0 0

(a) (b)

$. B ng (b).

#. B ng (a).


000 111 110 101

001 010 011 100

000 001 010 011

111 110 101 100

(a) (b)

$. Hnh (b).

#. Hnh (a).


$. Mod 8, m li.

#. Mod 8, m ti n.

#. Mod 7, m li.

#. Mod 7, m ti n.

Cu 13. Cho hnh 5-40. Cho bi t d ng sng c a Q 2, Q1 v Q0?


Clock

'1'

Hnh 5-39'1''1'

J0

Q0

>

K0

'1' J1

Q1

>

K1 '1'

'1' J2

Q2

>

K2

Clock

'1'

Hnh 5-38'1''1'

J0

Q0

>

K0

'1' J1

Q1

>

K1 '1'

'1' J2

Q2

>

K2

Clock

'1'

Hnh 5-40'1''1'

J0

Q0

>

K0

'1' J1

Q1

>

K1 '1'

'1' J2

Q2

>

K2


68/121

Ngn hng in t s

$. Hnh (a).

#. Hnh (b).

#. Hnh (c).

#. Hnh (d).


C Q2 Q1 Q0 k2

Q k1

Q k0

Q C Q2 Q1 Q0 k2

Q k1

Q k0

Q

0 0 0 0 1 1 1 0 0 0 0 0 0 1

1 0 0 1 1 1 0 1 0 0 1 0 1 0

2 0 1 0 1 0 1 2 0 1 0 0 1 13 0 1 1 1 0 0 3 0 1 1 1 0 0

4 1 0 0 0 1 1 4 1 0 0 1 0 1

5 1 0 1 0 1 0 5 1 0 1 1 1 0

6 1 1 0 0 0 1 6 1 1 0 1 1 1

7 1 1 1 0 0 0 7 1 1 1 0 0 0

(a) (b)

$. B ng (a).

#. B ng (b).


Clock'1'

Hnh 5-41'1''1'

J0 Q0

>

K0

'1' J1 Q1

>

K1 '1'

'1' J2 Q2

>

K2

Ck

Q

Ck

Q

Ck

Q

Ck

Q

1 8

1 8

1 8

1 8

(c)

(b)(a)

(d)

Q

Q

Q

Q

Q

Q

Q

Q


69/121

Ngn hng in t s


000 111 110 101 001 010 011 100

000 001 010 011 111 110 101 100

(a) (b)

$. Hnh (a).

#. Hnh (b).


$. Mod 8, m ti n.

#. Mod 8, m li.#. Mod 7, m li.

#. Mod 7, m ti n.



Clock

'1'

Hnh 5-43'1''1'

J0

Q0

>

K

'1' J1

Q1

>

K1 '1'

'1' J2

Q2

>

K2

Clock

'1'

Hnh 5-42'1''1'

J0

Q0

>

K0

'1' J1

Q1

>

K1 '1'

'1' J2

Q2

>

K2

Clock

'1'

Hnh 5-44'1''1'

J0

Q0

>

K

'1' J1

Q1

>

K1 '1'

'1' J2

Q2

>

K2

Ck

Q

Ck

Q

Ck

Q

Ck

Q

1 8

1 8

1 8

1 8

(c)

(b)(a)

(d)

Q

Q

Q

Q

Q

Q

Q

Q


70/121

Ngn hng in t s

$. Hnh (c).

#. Hnh (a).

#. Hnh (b).#. Hnh (d).


C Q2 Q1 Q0 k2

Q k1

Q k0

Q C Q2 Q1 Q0 k2

Q k1

Q k0

Q

0 0 0 0 1 1 1 0 0 0 0 0 0 1

1 0 0 1 1 1 0 1 0 0 1 0 1 0

2 0 1 0 1 0 1 2 0 1 0 0 1 1

3 0 1 1 1 0 0 3 0 1 1 1 0 0

4 1 0 0 0 1 1 4 1 0 0 1 0 1

5 1 0 1 0 1 0 5 1 0 1 1 1 0

6 1 1 0 0 0 1 6 1 1 0 1 1 1

7 1 1 1 0 0 0 7 1 1 1 0 0 0

(a) (b)

$. B ng (b).

#. B ng (a).



Clock

'1'

Hnh 5-45'1''1'

J0

Q0

>

K

'1' J1

Q1

>

K1 '1'

'1' J2

Q2

>

K2

Clock

'1'

Hnh 5-46'1''1'

J0

Q0

>

K

'1' J1

Q1

>

K1 '1'

'1' J2

Q2

>

K2


71/121

Ngn hng in t s

000 111 110 101 001 010 011 100

000 001 010 011 111 110 101 100

(a) (b)

$. Hnh (b).

#. Hnh (a).


$. Mod 8, m li.

#. Mod 8, m ti n.

#. Mod 7, m li.

#. Mod 7, m ti n.



Clock'1'

Hnh 5-47'1''1'

J0 Q0

>

K

'1' J1 Q1

>

K1 '1'

'1' J2 Q2

>

K2

Clock

'1'

Hnh 5-48'1''1'

J0

Q0

>

K

'1' J1

Q1

>

K1 '1'

'1' J2

Q2

>

K2

Ck

Q

Ck

Q

Ck

Q

Ck

Q

1 8

1 8

1 8

1 8

(c)

(b)(a)

(d)

Q

Q

Q

Q

Q

Q

Q

Q


72/121

Ngn hng in t s

$. Hnh (a).

#. Hnh (b).

#. Hnh (c).

#. Hnh (d).


C Q2 Q1 Q0 k2

Q k1

Q k0

Q C Q2 Q1 Q0 k2

Q k1

Q k0

Q

0 0 0 0 1 1 1 0 0 0 0 0 0 1

1 0 0 1 1 1 0 1 0 0 1 0 1 0

2 0 1 0 1 0 1 2 0 1 0 0 1 1

3 0 1 1 1 0 0 3 0 1 1 1 0 0

4 1 0 0 0 1 1 4 1 0 0 1 0 1

5 1 0 1 0 1 0 5 1 0 1 1 1 0

6 1 1 0 0 0 1 6 1 1 0 1 1 1

7 1 1 1 0 0 0 7 1 1 1 0 0 0

(a) (b)

$. B ng (a).

#. B ng (b).



Clock

'1'

Hnh 5-49'1''1'

J0

Q0

>

K

'1' J1

Q1

>

K1 '1'

'1' J2

Q2

>

K2

Clock

'1'

Hnh 5-50'1''1'

J0

Q0

>

K

'1' J1

Q1

>

K1

'1'

'1' J2

Q2

>

K2


73/121

Ngn hng in t s

000 111 110 101 001 010 011 100

000 001 010 011 111 110 101 100

(a) (b)

$. Hnh (a).

#. Hnh (b).

CU H I LO I 4.

Cu 1. Cho b m hnh 5-51. Cho bi t y l b m Mod m y?

$. Mod 6.

#. Mod 5.

#. Mod 7.

#. Mod 8.

Cu 2. Cho hnh 5-52. Gi s tr ng thi ban u Q 0Q1Q2 l 100, sau 2 xung Clockth tr ng thi l i ra l bao nhiu?

$. 001.

#. 100.

#. 010.

#. 000.

Cu 3. Cho hnh 5-53. Gi s tr ng thi ban u Q 0Q1Q2 l 000, sau 3 xung Clock

th tr ng thi l i ra l bao nhiu?


Clock

J1

Q1

>

K1

J2

Q2

>

K2'1'

Hnh 5-51

'1'

J0 Q0

>

K0

Clock

D1

Q1

>

D2

Q2

>

Hnh 5-52

D0

Q0

>

Clock

D1

Q1

>

D2

Q2

>

Hnh 5-53

D0

Q0

>


74/121

Ngn hng in t s

$. 111.

#. 110.

#. 011.

#. 001.


$. 111.

#. 110.#. 011.

#. 001.


Clock

J1

Q1

>

K1

J2

Q2

>

K2

Hnh 5-54

J0

Q0

>

K0


75/121

Ngn hng in t s


$. 000.

#. 110.

#. 011.

#. 001.

Cu 6. Cho hnh 5-56. Gi s d li u c n n p vo thanh ghi l 1101, sau 4 xung Clock th tr ng thi l i ra Q 0Q1Q2Q3 l bao nhiu?

$. 1101.

#. 1110.

#. 1011.

#. 1001.

Cu 7. Cho hnh 5-57. Gi s tr ng thi ban u Q 0Q1Q2Q3 l 0000, sau 4 xungClock th tr ng thi l i ra Q 0Q1Q2Q3 l bao nhiu?

$. 0000.

#. 1111.

#. 1011.#. 1101.


Clock

D1

Q1

>

D2

Q2

>

Hnh 5-55

D0

Q0

>

Clock

D1

Q1

>

D2

Q2

>

Hnh 5-56

D0

Q0

>

D3

Q3

>

Clock

D1

Q1

>

D2

Q2

>

Hnh 5-57

D0

Q0

>

D3 Q3

>


76/121

Ngn hng in t s

CU H I CH NG 6. M CH PHT XUNG

CU H I LO I 1.

Cu 1. c i m n i b t nh t c a m ch dao ng a hi dng th ch anh l g? $. T n s tn hi u l i ra n nh

#. Bin tn hi u l i ra n nh

#. Bin l i ra c th i u ch nh c

#. T n s l i ra c th i u ch nh c

Cu 2. c i m quan tr ng nh t c a trig Schmitt l g?

$. Tnh ch ng nhi u cao v n ho t ng nh b so snh hai ng ng

#. T n s ho t ng cao

#. Cng su t tiu th th p

#. L b so snh m t ng ng

Cu 3. M ch a hi i l g?

$. L m ch dao ng a hi c m t tr ng thi n nh v m t tr ng thi t m n nh

#. L m ch pht xung vung

#. L m ch dao ng a hi c chn i u khi n

#. L m ch pht xung i u ho

Cu 4. Trong m ch a hi i ki u vi phn nh hnh 6-1, n u xung i u khi n c r ng l n h n xung a hi i l i ra th :

R

C

V I

Vo1

Vo

ED

Vi2

Hnh 6-1

$. M ch v n ho t ng bnh th ng

#. Tn hi u l i ra lun b ng 0

#. Tn hi u l i ra lun b ng 1

#. Xung l i ra b ng xung l i vo

Cu 5. Cc v tr ng ng c a c ng Schmitt c t o ra b i

$. h i ti p d ng. #. h i ti p m.



77/121

Ngn hng i n t s

#. h i ti p thu n.

#. h i ti p ng c.

Cu 6. K hi u no d i y bi u di n c ng Schmitt o?

$. Hnh b.#. Hnh a.

#. Hnh c.

#. Hnh d.

Cu 7. Trong m t c ng Schmitt, h i ti p d ng (hay tr ) dng

$. Khng c tr ng h p no ng..

#. gi m m c ng ng c a c ng.

#. tng vng khng xc nh c a i n p kch (m c chuy n tr ng thi). #. tng kh nng dng ra.

Cu 8. Trong m ch a hi hnh 6-2, n u khng c i n tr R 1 th:

5553

1

7

48

5

6

CE

CO

V

0,01 F0

CV

2

1R

2R

Hnh 6-2

$. Khng c tn hi u l i ra

#. M ch v n pht xung v t n s l i ra ch ph thu c vo gi tr c a R 2 v C

#. Xung l i ra l xung vung c l p y l 50%

#. M ch v n pht xung nh ng t n s r t cao


b.


78/121

Ngn hng in t s

Cu 9. Trong m ch a hi hnh 6-3, chu k dao ng c a m ch c tnh theo cng th c:

5553

1

7

48

5

6

CE

CO

V

0,01 F0

CV

2

1R

2R

Hnh 6-3

$. T = 0,7 (R1 + 2R2)C

#. T = 0,7 (2R1 + R2)C

#. T = 1,4 (2R1 + R2)C

#. T = 1,1 (2R1 + R2)C

Cu 10. Trong m ch a hi i hnh 6-4, r ng xung c a m ch c tnh theo cng th c:

555

3

1

7

48

5

6

CE

COV

0,01 Fm

2

R

Hnh 6-4

$. T = 1,1 RC

#. T = 0,7 RC

#. T = 1,4 RC

#. T = 2,2 RC

Cu 11. Trong m ch dao ng a hi c b n CMOS hnh 6-5, chu k dao ng c a m ch c tnh theo cng th c:



79/121

Ngn hng in t s

RC

VK

Vi1

Vo

Hnh 6-5

$. T = 1,4 RC

#. T = 0,7 RC

#. T = 1,1 RC

#. T = 2,2 RC

Cu 12. Trong m ch a hi i ki u vi phn dng NOR CMOS hnh 6-6, chu k

dao ng c a m ch c tnh theo cng th c:

R

C

V I

Vo1

Vo

ED

Vi2

Hnh 6-6

$. T = 0,7 (R + R0) C

#. T = 1,4 (R + R0) C#. T = 1,1 (R + R0) C

#. T = 2,2 (R + R0) C

Cu 13. D ng sng ra c a trig Schmitt l:

$. Xung vung

#. sin.

#. tam gic.

#. rng c aCu 14. T n s c a m ch dao ng a hi th ch anh ph thu c vo:

$. Tinh th th ch anh

#. R v C c trong m ch.

#. R c trong m ch.

#. C c trong m ch.



80/121

Ngn hng i n t s

CU H I LO I 2.

Cu 1. Trong m ch a hi i hnh 6-7, cho R = 50k , C = 2,2F tnh r ng xung ra c a m ch:

555

3

1

7

48

5

6

CE

COV

0,01 Fm

2

R

Hnh 6-7$. T = 1,21 ms

#. T = 12,1 ms

#. T = 11,2 ms

#. T = 1,11 ms

Cu 2. Trong m ch a hi i hnh 6-8, cho i n tr u ra c a c ng 1 R 0 =1000, R= 10k, C = 1F , tnh r ng xung ra c a m ch:

R

C

V I

Vo1

Vo

ED

Vi2

Hnh 6-8

$. T = 7,7 s

#. T = 7,7 ms#. T = 7,7 ns

#. T = 7,7 s

Cu 3. Trong m ch a hi hnh 6-9, cho R 1 = R2 = 1k, C = 4,7F tnh t n s dao ng c a m ch:



81/121

Ngn hng in t s

555

3

1

7

48

5

6

CE

CO

V

0,01 Fm

2

R1

R2

Hnh 6-9

$. f = 100 kHz

#. f = 10 kHz

#. f = 1000 kHz

#. f = 1 kHz

Cu 4. Trig Schmitt c s d ng:

$. cho qu trnh chuy n i sng u vo ch m

#. cho i n p vo m t chi u.

#. gi ng nh m t b khu ch i.

#. cho qu trnh chuy n i sng u vo nhanh. Cu 5. Trong m ch a hi i hnh 6-10, n u gi tr c a t C r t b (


82/121

Ngn hng in t s

555

3

1

7

48

5

6

CE

CO

V

0,01 Fm

2

R

R

Hnh 6-11

$. h s l p y b ng (1/2).

#. h s l p y b ng 1.

#. h s l p y b ng 2.

#. h s l p y b ng (1/4).

Cu 7. Trong m ch dao ng a hi c bao nhiu tr ng thi n nh?

$. 2.

#. 1.

#. 3.

#. 4.

Cu 8. M t d ng sng sin c th c bi n i sang d ng sng hnh vung b ng cch s d ng m t:

$. trig Schmitt.

#. b dao ng a hi.

#. b dao ng a hi i.

#. b dao ng a hi dng IC 555.

Cu 9. B dao ng a hi c yu c u xung kch kh i ?

$. Sai.

#. ng.

Cu 10. M ch no c dng bi n i cc tn hi u bi n thin m t cch ch m ch p lm u vo cho cc m ch logic?

$. trig Schmitt.

#. b dao ng a hi.

#. b dao ng a hi i.

#. b dao ng a hi dng IC 555.



83/121

Ngn hng in t s

CU H I LO I 3.

Cu 1. Cho m ch i n trig Schmitt hnh 6-12, n u tn hi u l i vo c d ng tn hi u nh hnh sau, tn hi u l i ra n m

hnh no?

Hnh 6-12$. Hnh c.

#. Hnh a.

#. Hnh b.

#. Hnh d.


+V

-V

-V/2

+V/2

-

+

+

V-

Vi

Vo

R

OpamO

Hnh c

Hnh d

Hnh a.nn

Hnh b


84/121

Ngn hng i n t s

Cu 2. Cho m ch i n trig Schmitt hnh 6-13, n u tn hi u l i vo c d ng tn hi u nh hnh sau, tn hi u l i ra n m hnh no.

$. Hnh (a).

#. Hnh (b).

#. Hnh (c).

#. Hnh (d).

CU H I CH NG 7. B NH BN D N

CU H I LO I 1.

Cu 1. Thng tin trong b nh c l u tr d ng _________.

$. nh phn.

#. Hexa.

#. th p phn.#. Bt phn.


-

+

V+

V-

Vi

Vo

RR

Opam

+V

+V/2

-V/2

-VHnh 6-13

+V

+V/2

-V/2

-VHnh (a)

+V

+V/2

-V/2

-VHnh (b)

+V

+V/2

-V/2

-VHnh (c)

+V

+V/2

-V/2

-VHnh (d)


85/121

Ngn hng in t s

Cu 2. RAM l b nh m:

$. d li u b m t khi m t ngu n nui.

#. d li u khng b m t khi m t ngu n nui.

#. c hai cu trn u ng.

#. Khng c tr ng h p no ng. Cu 3. ROM l b nh m:

$. d li u khng b m t khi m t ngu n nui.

#. d li u b m t khi m t ngu n nui.

#. d li u b m t khi v n cn ngu n nui.

#. Khng c tr ng h p no ng.

Cu 4. DRAM l:

$. RAM ng.

#. RAM tnh.

#. b nh ch c.

#. b nh ch vi t.

Cu 5. SRAM l:

$. RAM tnh.

#. RAM ng.

#. b nh ch c.

#. b nh ch vi t.

Cu 6. DRAM l lo i b nh :

$. c th b m t d li u khi c ngu n nui.

#. khng m t d li u khi c ngu n nui.

#. ch c th c d li u.

#. ch c th vi t d li u. Cu 7. SRAM l lo i b nh :

$. khng m t d li u khi c ngu n nui.

#. c th b m t d li u khi c ngu n nui.

#. ch c th c d li u.

#. ch c th vi t d li u.

Cu 8. C u t o c a m t nh DRAM g m c:

$. 1 transistor tr ng MOS v 1 t i n. #. 1 transistor tr ng MOS v 1 trig .



86/121

Ngn hng in t s

#. 1 transistor tr ng MOS v 1 diode.

#. 1 transistor l ng c c v 1 t i n.

Cu 9. Linh ki n l u gi bit thng tin c a DRAM l:

$. T i n.

#. Transistor.

#. Trig .

#. Diode.

Cu 10. Linh ki n l u gi bit thng tin c a SRAM l:

$. Trig .

#. Transistor.

#. T i n.

#. Diode.

Cu 11. DRAM c ch t o b ng cch s d ng cng ngh :

$. MOS.

#. l ng c c.

#. l ng c c v MOS.


Cu 12. Th i gian truy nh p c a b nh l ng c c so v i b nh MOS l:

$. nhanh h n.

#. lu h n.

#. b ng nhau.

#. Khng c tr ng h p no ng.

Cu 13. PROM l lo i b nh c th s a i d li u c sau khi l p trnh, ng hay sai?

$. Sai.

#. ng.

Cu 14. Trong chip EPROM cho nh sng t ngo i i qua khi c n xo d li u trong b nh th n ph i c ___________.

$. c a s lm b ng thu tinh th ch anh.

#. m t c a s .

#. Hai c a s lm b ng thu tinh th ch anh.

#. Hai c a s .

Cu 15. Th i gian truy nh p c a cc chip ROM hi n nay so v i cc chip RAM l:



87/121

Ngn hng in t s

$. lu h n.

#. nhanh h n.

#. b ng nhau.

#. Khng c p n no ng.

CU H I LO I 2. Cu 1. S cc byte c c i c th c l u tr trong b nh c dung l ng 1024 x 8 l bao nhiu?

$. 1024.

#. 1024 x 2

#. 1024 x 4.

#. 1024 x 8.

Cu 2. S ng a ch c n thi t trong b nh c dung l ng 1024 x 8 l bao nhiu?

$. 10.

#. 11

#. 12.

#. 13.

Cu 3. S ng vo/ ra (I/O) c n thi t trong b nh c dung l ng 1024 x 8 l

bao nhiu?$. 8.

#. 10

#. 12.

#. 14.

Cu 4. S ng a ch c n thi t trong b nh c dung l ng 128KB l bao nhiu?

$. 17.

#. 7

#. 15.

#. 16.

Cu 5. Th i gian truy nh p c a 1 RAM l 10 ns, th i gian t i thi u ph i m t i gi a hai thao tc c l bao nhiu ?

$. 10 ns.

#. 20 ns.

#. 30 ns.



88/121

Ngn hng in t s

#. 40 ns.

Cu 6. M t EPROM c th b xo b i :

$. tia t ngo i.

#. i n.

#. t c u ch. #. b c u ch.

Cu 7. B nh FLASH l b nh :

$. Khng bay h i.

#. Bay h i.

#. C hai p n trn u ng.


Cu 8. B nh FLASH c c u trc gi ng nh b nh :

$. EEPROM.

#. EPROM.

#. SRAM.

#. DRAM.

Cu 9. M c ch s d ng chnh c a b nh FLASH l thay th cho :

$. Cc a m m v a c ng dung l ng nh .

#. CDROM.

#. RAM.

#. ROM.

Cu 10. B nh CACHE c dung l ng so v i b nh chnh c a my tnh l :

$. nh h n.

#. l n h n.

#. b ng nhau.#. tu tr ng h p.

Cu 11. B nh CACHE l b nh ch a cc thng tin m CPU:

$. v a s d ng g n y nh t.

#. lu khng c s d ng.

#. C hai tr ng h p trn.


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Ngan Hang Dien Tu So 5367