Bài tập môn điện tử số

7/31/2019 Bi tp mn in t s

1/19

L c Thanh TunSHSV: 20092978

Bi tp mn in ts.

Htn: L c Thanh TunSHSV: 20092978Lp: H thng thng tin & truyn thng. Ks cht lng cao K54

Chng 1: Kin thc c sca k thut s.

1.1.

+Bi ton logic l mt bi ton m nhng thng tin cho trc( d liu vo) vcc p ng ca bi ton, u ch c thmt trong hai trng thi i khng

nhau: ng/sai, nng/lnh

+Bin logic l cc bin trong i s Boole. N ch c hai gi tr, k hiu 1/0,c trng cho hai trng thi i khng ca mt hin tng.

+Mc logic l min in p trong cc mch logic in. Trong in p mangthng tin v hai gi tr ca bin logic, v n ch c th nm hai min gi trhon ton phn bit nhau. C hai mc logic l mc caoHv mc thp L.

1.2.+Mch logic l mch in gm nhng linh kin, ch yu l cc kha ng/m,ghp ni vi nhau; nhm thc hin nhng quan hlogic cho trc.

+ Nu trong mch logic, ta quy c mc thp L c trng cho gi tr 0 logic,mc cao H c trng cho gi tr 1 logic, th mch gi l mch logic dng. Nuquy c ngc li( H0, L1), mch s gi l mch logic m.

1.3.

Xc nh p ng dng song ca cc biu thc logic theo dng song ca ccbin A, B, C.


2/19


A

B

AB

AB

A +

A. B. C

A + B + C

A

B

C

A+ B

AA

B+ B

A.1

B+ 1

A.B.C

A+ B+ C


3/19


4/19


Hnh 1.24.ba)Biu thc logic: Q= AC+ BCBng chn l:

A B C AC BC Q0 0 00 0 1

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

00

001010

00

010001

00

011011

b)Biu thc logic: Q=( A+ B+ C)( A+ B+ D)(A+ B+ D)Bng chn l:

3 29

128

9128

9128

91283 2

A

B

C

D

Q

A B C D A+ B+ C A+ B+ D A+ B+ D Q0 0 0 00 0 0 10 0 1 00 0 1 10 1 0 00 1 0 10 1 1 0

0 1 1 11 0 0 0

1111111

10

1111111

11

0101111

11

0101111

10


5/19


1.9.

a) Chuyn mch hnh 1.24a sang mch tng ng, ch dng phn t NAND.

Q= AC+ BC= Q= AC + BC =AC . BC

b)Chuyn mch hnh 1.24b sang mch tng ng, ch dung cc phn t NOR.

Q=( A+ B+ C)( A+ B+ D)(A+ B+ D)= A + B + C A + B + DA + B + D

= A + B + C + A + B + D + A + B + D

C

A

B

1

Q

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

1 1 0 11 1 1 01 1 1 1

0111

111

0101

111

1111

111

0101

111


6/19


1.10.

+ S dng phn tNAND hai u vo v Q1:

Q1(A,B,C)= AC+ AB+ BC= AC+ B(A+ C)=AC + BA + C =AC .BA + C

=AC . B + A + C =AC . B + AC . =AC . B

+ S dng phn tNOR hai u vo v Q2:

Q2(A,B,C)= (A+ B)(B+ C)(A+ C)=(A+ BC)( B+ C)= A + BC B + C

=A+ B C + B + A. B + + B + =A + B +

2

31

2

3

1

912

8

9128

9128

9128

A

B

C

D

0

0

Q

A

C

B

1

Q1


7/19


1.11.

+ Xy dng mch dng cc cng V, HOC, O:

Q=A.k+B.

+ V mch ch dng phn t NAND:

Q= A.k+B. = A. + B. = A. .B.

A

B

C

0

0

Q2

A

dk

B

Q

A

dk

B

Q

1


8/19


1.15.

a) i sang s thp phn:1110.10b= 14,5d100101011101b= 2397d

46AEh= 18094dFA2Ch= 64044d

b) i sang s nh phn:97,75d= 1100001.11b625,7d=

c) i sang s Hexa;921d= 399h6120d= 17F8h

d) i cc s :1001011b= 75h1001010101111101b= 057Dh2ACh= 1010101100bB34Dh= 1011001101001101b27,45d= 100111.01000101bcd11101000110.01bcd= 746,4d10100111b= gray

15d= gray10010110gray= b

1.16. Thc hin cc php cng.100101b+ 10111b= 111100b10011111001b+ 100001111101b= 110101110110bB23CDh+ 17912h= C9CDFhAFEFFEh+ 2FBCADh= DFACABh

1.17.-120= 1,1111000 (s nh phn c du)-120= 1,0001000 (m b 2)


9/19


Chng 3: Tng hp mch logic t hp

3.1.

Q(A, B, C, D)= ; 7,10 ;

1 1 0 1

1 x 0 0

0 0 0 0

1 0 x 1

Biu thc logic dng:+ Tng cc tch: Q(A, B, C, D)= A.C + A.B.D + B.C.D + Tch cc tng: Q(A, B, C, D)= (A+B).(B+C).(C+D).(A+D)

V mch dng phn t NAND :Q(A, B, C, D)= A.C +A.B.D + B.C.D=A.C+A.B.D+B.C.D=A.C.A.B.D.B.C.D

A

C

B

D

1

1

1

1

Q


10/19


3.2.

Bng logic:

E D P C

0 0 1 1

0 1 1 1

1 0 1 0

1 1 0 1

Theo dng hi chun ta c:P= E D C= E D Mch logic :

D

E

P

C


11/19


Chng 5: Mch logic t hp

5.1. Mch cng song song gm cc b cng FA thc hin cng 2 s nh phn:A=00111, B=10101 A+B=11100

1110010101

S1

S2

S3

S4

S5

(D0)


12/19


5.2.

Dy bt ra ca vi mch cng 74LS83A vi cc dy bt vo:A1 =10010110, B1=11111000 S1=10001110

A2 =11101000, B2=11001100 S2=10110100A3 =00001010, B3=10101010 S3=10110100A4 =10111010, B4=00100100 S4=11011110


13/19


5.7. Thc hin hm logic t hp bng MUX 8

F1= ;

F2= ;

X04

X13

X22

X31

X415

X514

X613

X712

A11

B10

C9

E7

Y5

Y6

X04

X13

X22

X31

X415

X514

X613

X712

A11

B10

C9

E7

Y5

Y6

U1

OR_2

(X0)

0

1

0

0

(X0)

(X2)

X04

X13

X22

X31

X415

X514

X613

X712

A11

B10

C

9

E7

Y5

Y6

X04

X13

X22

X31

X415

X514

X613

X712

A11

B10

C9

E7

Y5

Y6

U1

OR_2

(X2)

1

0

1

1

(X1)

(X0)


14/19


5.8.Dng 2 vi mch MUX 4 v 1 vi mch MUX 2 thnh lp mt mch MUX 8

1X06

1Y7

1X15

1X24

1X33

2X010

2Y9

2X111

2X212

2X313

A14

B2

1E1

2E15

1A2

1Y4

1B3

2A5

2Y7

2B6

3A11

3Y9

3B10

4A14

4Y12

4B13

A/B1

E15

0

0

0

Q


15/19


16/19


6.2.

Dng sng ca tn hiu u ra Q2 theo dng sng cc tn hiu vo A,B. Khiu, Q1=Q2 = 0 logic.

6.3.

J

CLK

Q

Q

K

D

CLK

Q

Q+5V

A

B

Q2Q1


17/19


Dng sng ca tn hiu u ra S theo dng sng cc tn hiu vo C, E. Khi uQ0=Q1=L.

C

E

Q0

Q1

S


18/19


6.7.

Mch m thun dng 74192. m t 0 ti 48 th dng:

A7

QA13

B1

QB12

C2

QC11

D6

QD10

BI/RBO4

QE9

RBI5

QF15

LT3

QG14

A7

QA13

B1

QB12

C2

QC11

D6

QD10

BI/RBO4

QE9

RBI5

QF15

LT3

QG14

+5V

D015

Q03

D11

Q12

D210

Q26

D39

Q37

UP5

TCU12

DN4

TCD13

PL11

MR14

D015

Q03

D11

Q12

D210

Q26

D39

Q37

UP5

TCU12

DN4

TCD13

PL11

MR14

+0V

0

(D1)


19/19

L c Thanh Tun

6.9.

Dng vi mch 4017 v cng OR xy dng mch m vng chu trnh m 7xung. Cc trng thi DCBA ng vi:

Khi u 0000 Xung 4 1011Xung 1 0011 Xung 5 0111

Xung 2 0110 Xung 6 0010

Xung 3 1100 Xung 7 v khi u

CLK14

E13

MR15

CO12

Q03

Q12

Q24

Q3 7Q4

10

Q51

Q65

Q76

Q89

Q911

(CLK)

A

B

C

D

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