数字逻辑电路与系统设计分习题及解答

更新时间:2023-11-14 03:12:01 阅读量: 教育文库 文档下载

说明:文章内容仅供预览,部分内容可能不全。下载后的文档,内容与下面显示的完全一致。下载之前请确认下面内容是否您想要的,是否完整无缺。

第1章习题及解答

1.1 将下列二进制数转换为等值的十进制数。 (1) (11011)2

(2) (10010111)2 (4) (11111111)2 (6) (0.0111)2

(3) (1101101)2 (5) (0.1001)2 (7) (11.001)2 题1.1 解:

(8) (101011.11001)2

(1) (11011)2 =(27)10

(2) (10010111)2 =(151)10 (4) (11111111)2 =(255)10 (6) (0.0111)2 =(0.4375)10

(3) (1101101)2 =(109)10

(5) (0.1001)2 =(0.5625)10 (7) (11.001)2 =(3.125)10

(8) (101011.11001)2 =(43.78125)10

1.3 将下列二进制数转换为等值的十六进制数和八进制数。 (1) (1010111)2

(2) (110111011)2

(4) (101100.110011)2

(3) (10110.011010)2 题1.3 解:

(1) (1010111)2 =(57)16 =(127)8

(2) (110011010)2 =(19A)16 =(632)8 (3) (10110.111010)2 =(16.E8)16 =(26.72)8 (4) (101100.01100001)2 =(2C.61)16 =(54.302)8 1.5 将下列十进制数表示为8421BCD码。 (1) (43)10 (3) (67.58)10 题1.5 解:

(1) (43)10 =(01000011)8421BCD

(2) (95.12)10 =(10010101.00010010)8421BCD (3) (67.58)10 =(01100111.01011000)8421BCD (4) (932.1)10 =(100100110010.0001)8421BCD

1.7 将下列有符号的十进制数表示成补码形式的有符号二进制数。 (1) +13

(2)?9

(3)+3

(4)?8

(2) (95.12)10

(4) (932.1)10

题1.7解:

(1) +13 =(01101)2 (3) +3 =(00011)2

(2)?9 =(10111)2 (4)?8 =(11000)2

1.9 用真值表证明下列各式相等。 (1) (2) (3) (4)

AB?B?AB?A?B A?B?C???AB???AC?

AB?C?A?BC AB?AC?AB?AC

??题1.9解: (1)

证明AB?B?AB?A?B

A B AB?B?AB A?B 0 0 1 1 (2)

A0 1 0 1 0 1 1 1 0 1 1 1

证明A?B?C???AB???AC?

B C A?B?C? ?AB??AC 0 0 0 0 1 1 1 1 (3)

A0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 证明AB?C??A?B?C

B C AB?C ?A?B?C1 0 1 0 0 0 0 0 0 0 1 1 0 1 0 1 1 0 1 0 1 1 1 1 (4)

A0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 证明AB?AC?AB?AC

B C AB?AC AB?AC 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 1 0 1 1 0 0 1 0 1 0 1 1 0 0 1.11 用逻辑代数公式将下列逻辑函数化成最简与或表达式。 (1)F?AB?AC?BC?ACD (2)F??A?AC??A?CD?D? (3)F?BD?D?D?B?C??AD?B? (4)F?ABC?AD??B?C?D (5)F?AC?BC?B?A?C? (6)F??A?B??B?C? 题1.11解:

(1)F?AB?AC?BC?ACD?A?BC (2)F??A?AC??A?CD?D??A?CD

(3)F?BD?D?D?B?C??AD?B??D?AB?BC (4)F?ABC?AD??B?C?D?ABC?D (5)F?AC?BC?B?A?C??AC?BC

(6)F??A?B??B?C??AB?BC?AC或?AB?BC?AC 1.13 用卡诺图将下列逻辑函数化成最简与或表达式。

(1)F??A?B?CD?ABC?ACD 且AB?CD?0 (2)F?AC?AB 且A,B,C不能同时为0或同时为1 (3)F?A,B,C???m?3,5,6,7???d?2,4?

?m?0,4,6,8,13???d?1,2,3,9,10,11? ?m?0,1,8,10???d?2,3,4,5,11? ?m?3,5,8,9,10,12???d?0,1,2,13?

(4)F?A,B,C,D??(5)F?A,B,C,D??(6)F?A,B,C,D??题1.13解:

(1)F??A?B?CD?ABC?ACD 且AB?CD?0

F?B?AD?AC

(2)F?AC?AB 且A,B,C不能同时为0或同时为1

F?B?C

(3)F?A,B,C??

F?A?B

?m?3,5,6,7???d?2,4?

(4)F?A,B,C,D??

?m?0,4,6,8,13???d?1,2,3,9,10,11?

F?AD?ACD?B

(5)F?A,B,C,D??

?m?0,1,8,10???d?2,3,4,5,11?

F?BD?AB 或 F?BD?AC

(6)F?A,B,C,D??

?m?3,5,8,9,10,12???d?0,1,2,13?

F?BD?AB?CD?AC

1.15将下列逻辑函数化简为或非—或非式。 (1)F?ABC?BC

(2)F??A?C??A?B?C??A?B?C? (3)F??ABC?BC?D?ABD

(4)F(A,B,C,D)?题1.15解:

?m?0,2,3,8,9,10,11,13?

(1)F?ABC?BC

F?B?C?A?C?B?C

或 F?B?C?B?C?A?B

(2)F??A?C??A?B?C??A?B?C?

F?B?C?A?C?A?B?C

(3)F?A,B,C,D??

?m?0,1,8,9,10?

F?B?C?D?A?C

(4)F(A,B,C,D)?

?m?0,2,3,8,9,10,11,13?

F?A?C?D?B?C?B?D

第2章习题及解答

2.1判断图P2.1所示电路中各三极管的工作状态,并求出基极和集电极的电流及电压。

+12V1kΩ30kΩ+6V50kΩβ=50+6V3kΩβ=20(a)图P2.1

(b)

题2.1 解:

(a)三极管为放大状态;设VCES?0.3V有:

IB?6?0.750?0.106mA IC?0.106?50?5.3mA

VB?0.7V VC?6.7V

(b)三极管为饱和状态;

本文来源:https://www.bwwdw.com/article/ca4v.html

Top