bài tập môn điện tử số
TRANSCRIPT
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7/31/2019 Bi tp mn in t s
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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.
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7/31/2019 Bi tp mn in t s
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L c Thanh TunSHSV: 20092978
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
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7/31/2019 Bi tp mn in t s
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L c Thanh TunSHSV: 20092978
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
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7/31/2019 Bi tp mn in t s
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L c Thanh TunSHSV: 20092978
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
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L c Thanh TunSHSV: 20092978
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
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7/31/2019 Bi tp mn in t s
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L c Thanh TunSHSV: 20092978
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
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L c Thanh TunSHSV: 20092978
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)
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7/31/2019 Bi tp mn in t s
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L c Thanh TunSHSV: 20092978
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
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7/31/2019 Bi tp mn in t s
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L c Thanh TunSHSV: 20092978
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
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7/31/2019 Bi tp mn in t s
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L c Thanh TunSHSV: 20092978
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)
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7/31/2019 Bi tp mn in t s
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L c Thanh TunSHSV: 20092978
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
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7/31/2019 Bi tp mn in t s
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L c Thanh TunSHSV: 20092978
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)
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L c Thanh TunSHSV: 20092978
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
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7/31/2019 Bi tp mn in t s
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L c Thanh TunSHSV: 20092978
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
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7/31/2019 Bi tp mn in t s
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L c Thanh TunSHSV: 20092978
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
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L c Thanh TunSHSV: 20092978
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)
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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