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第一章
1.8
(1) 42.5 (2) 25.25 (3) 2.3125 (4) 0.78125 (5)4.125
1.11
(100111001) 8421 (10001011)B (213)
O (8B)H (10001101100)EX-3 (100111100)5421
1. 13
(1) [87]DEC= 1010111 BIN = 127 OCT = 57 HEX
(2) [94] DEC= 1011110 BIN = 136 OCT = 5E HEX
(3) [108] DEC=1101100 BIN = 154 OCT = 6C HEX
(4) [175] DEC=10101111 BIN = 257 OCT = AF HEX
1.14
(1) [C5]HEX= 11000101 BIN= 305 OCT= 197 DEC
(2) [F4] HEX= 11110100 BIN= 364 OCT= 244 DEC
(3) [AB] HEX= 10101011 BIN= 253 OCT= 171 DEC
(4) [97] HEX= 10010111 BIN= 227 OCT= 151 DEC
(5) [3C8] HEX= 1111001000 BIN= 1710 OCT= 968 DEC
1.15
(1) 91 11000100
(2) 587 10001011
(3) 10111010 11101101
(4) 10011100 10011100
1.17
(1) 010011 010011
B
(2) 101110 101111
(3) 11001110 10110010
(4) 010111011 010111011
第二章
2.4
(1)∑m(1,4,5,6)
(2)∑m(0,1,7)
(3)∑m(4,5,10,11,13,14)
(4)∑m(0,1,4,5,6,7,8,9,10,11,12,13,14,15)
2.5
(1)∏M(0,2,4,6,7)
(2)∏M(1,2,6,7)
(3)∏M(10)
(4)∏M(4,9,7,9)
2.6
(1)
B +
AC
(2)
Z +
XY
(3)
DBDB +
(4)
ZYWZXYX
+
+
2.7
(1)
(
CBBA
)(
+
+
)
(2)
(
YZX
)(
+
+
Z
)
(3)
(
DBDB
)(
+
+
)
YXXWZ
)(
)(
+
+
)
(4)
Y
(
+
2.8
(1)
BABA +
(2)
ZW +
ZWX
(3)
YZXYX +
+
DAC
(4) ①
+
*注:卡诺图化简所得的结果并不是唯一的
BD
BA
AB
②
C
+
+
+
+
DBADA
2.24
(2)有一式错误
2.34
F=Σm(2,4,5,6) F=ПM(0,1,3,7) F =Σm(0,1,3,7) F =ПM(2,4,5,6)
*
F
*
=Σm(0,4,6,7) F
=ΠM(1,2,3,5)
2.36
CEBP
=
+
EDBA
+
ACDE
+
EDCAEDCB
+
+
EDCBAECAB
+
P
E
ABC
DE
C
000 001
011
00
01
11
10
0
1
3
2
D
4
5
7
6
12
13
15
14
010 100
16
8
9
11
10
17
19
18
B
图P2-9
A
C
101
20
21
23
22
111
110
24
25
27
26
28
29
31
30
B
3.1
υO(t)
VIm
REFV
(b)
3.2
υO(t)
VIm
O
VREF
(b)
3.3
υI(t)
υO(t)
第三章
t
t
t
t
O
O
(b)
3.8
(a),(b)都为
BAF
⊕=
3.9
P
1
=
X
1
+
X
2
+
YY
21
P
2
=
Y
1
+
Y
2
+
XX
21
P
3
=
3.10
EDCBA
+
•
(
)
三态输出高阻时,后一级门的输入端悬空,对 TTL 来说悬空为逻辑‘1’
P
1
=
ABF
(
+
C
)
+
ABF
(
+
EDC
⊕+
)
DJFE
+
+
)
+
(
AB
+
DCJFE
++
+
)
P
2
=
(
AB
+
3.11
=⊕
F
AB
⊕
F
C=0
C=1
P
1
P =1
P
= 2
Z
=2
ABC
+
EDC
P
4.1
(a) =P
A ⊙ B C⊕ (b)
第四章
CBAP
+
=
(c)
P
=
CBA
( +
)
PQ =
CB
⊕+
)
A
(f)
( ⊕=
CBA
)
Q
=
(
(d)
P
=
CBA
)
( +
(e)
P
4.4
3, 4, 16
BAP
+
=
Q =
AB
4.6
C
B
A
1
1
2
4
&
EN
0
1
2
3
4
5
6
7
4.10
C
B
1
A
1
(1)
&
&
P2
P1
MUX
G 0
3
0
1
EN
0
1
2
3
EN
0
1
2
3
C
B
1
A
1
F
(2)
MUX
G 0
3
0
1
EN
0
1
2
3
EN
0
1
2
3
F
MUX
G 0
3
0
1
EN
0
1
2
3
EN
0
1
2
3
C
B
A
1
1
(3)
4.11
C
B
1
A
1
F
(4)
MUX
G
0
7
F
EN
0
2
0
1
2
3
4
5
6
7
Z
1
1
(2)
0
D
B
A
1
C
1
(1)
4.17
0
1
EN
0
1
2
3
EN
0
1
2
3
0
Y
X
W
MUX
G 0
3
F
F
MUX
G
0
7
EN
0
2
0
1
2
3
4
5
6
7
F
1
∑=
m
)15,14,10,5,1,0(
,F2 的 MUX 实现与 F1 完全一样
A
B
C
1
1
D
1
MUX
G
0
3
0
2
EN
0
1
2
3
4
5
6
7
F2
4.20
至少要用 8 个异或门
d0
d1
d2
d3
d4
d5
d6
d7
=1
=1
=1
=1
A
=1
或
4.24
(1)
(2)
⊕=
BAF
⊕⊕=
CBAF
B
=1
=1
F
=1
=1
=1
=1
=1
d0
d1
d2
d3
d4
d5
d6
d7
=1
=1
=1
1
F
=1
=1
1
A
B
C
=1
=1
4.25
C
B
A
1
MUX
G 0
3
0
1
EN
0
1
2
3
EN
0
1
2
3
F
F
)14,13,11,8,17,4,2,1(m
(1) ∑=
注(1)也可以用三个异或门来实现即
1
D
DCBAF
⊕⊕⊕=