(完整word版)电子技术基础(数字部分)康华光(第五版)习题解答 联系客服

发布时间 : 星期日 文章(完整word版)电子技术基础(数字部分)康华光(第五版)习题解答更新完毕开始阅读90e2cafaf321dd36a32d7375a417866fb94ac05a

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0 LA?A?A?B1?B2?A?B1?B2

LB?LA?A?B1?B2?A?B1?B2?

____________________3.1.2 用逻辑代数证明下列不等式

(a) A?AB?A?B

由交换律 A?BC?(A?B)(A?C),得

A?AB?(A?A)(A?B)?A?B

(b) ABC?ABC?ABC?AB?AC

ABC?ABC?ABC?A(BC?BC?BC)?A(C?BC)?A(C?B)?AB?AC

(c) A?ABC?ACD??C?D?E?A?CD?E

_____A?ABC?ACD??C?D?E?A?ACD?(C?D)E?A?CD?CDE?A?CD?E

3.1.3 用代数法化简下列等式

(a) AB(BC?A)

AB(BC?A)?ABC?AB?AB

(b) (A?B)(AB)

(A?B)(AB)?AB

(c) ABC(B?C)

______________

ABC(B?C)?(A?B?C)(B?C)?AB?BC?AC?BC?C?AB?C_____

(d) A?ABC?ABC?CB?CB

A?ABC?ABC?CB?CB?A?C

_____

(e) AB?AB?AB?AB

AB?AB?AB?AB?A?A?0

_____________________________________________________________________________________________________________________________________________(f) (A?B)?(A?B)?(AB)?(AB)

___________________________________________________________________________________________________________(A?B)?(A?B)?(AB)?(AB)?(A?B)?(A?B)?(AB)?(AB)

?(AB?BA?B)(AB?AB)?B(AB?AB)?AB

(g) (A?B?C)(A?B?C)

(A?B?C)(A?B?C)?A?B

(h) ABC?ABC?ABC?A?BC

ABC?ABC?ABC?A?BC?A?ABC?BC?A?BC?BC?A?C

(i) AB?(A?B)

AB?(A?B)?AB?(A?B)?(A?B)(A?B)?A?B

_______________________________________________________________ (j) B?ABC?AC?AB

(k)

B?ABC?AC?AB?B?ABC?AC?B?AC?AC ABCD?ABD?BCD?ABCD?BC

ABCD?ABD?BCD?ABCD?BC?ABC?ABD?B(CD?C)?ABC?ABD?B(C?D)?ABC?ABD?BC?BD?B(AC?AD?C?D)?B(A?C?A?D)?AB?BC?BD__________________________________________________(l) AC?ABC?BC?ABC

____________________________________________________________________________________________________AC?ABC?BC?ABC?(AC?ABC)?(B?C)?(A?B?C)?(ABC?ABC)(A?B?C)?BC(A?B?C)?ABC?BC?BC (m) AB?ABC?A(B?AB)

____________________________________________________________________________________________________AB?ABC?A(B?AB)?A(B?BC)?AB?AB________________________________________________________________

?A(B?C)?A?A?B?C?A?03.1.4 将下列各式转换成与 – 或形式

(a) A?B?C?D

(1)当A?B?0,C?D?1时,真值为1。于是 AB=01,CD=00或CD=11时,真值为1; AB=10,CD=00或CD=11时,真值为1。

____________________________________则有四个最小项不为0,即ABCD、ABCD、ABCD、ABCD

(2)当A?B?1,C?D?0时,真值为1。 AB=00,CD=10或CD=01时,真值为1;

AB=11,CD=10或CD=01时,真值为1。

__________________则有四个最小项不为0,即ABCD、ABCD、ABCD、ABCD

________A?B?C?D??m(1,2,4,7,8,11,13,14)

____________________________________________________________________________(b) A?B?C?D?C?D?A?D

A?B?C?D?C?D?A?D?(A?B)(C?D)?(C?D)(A?D) ?(C?D)(A?B?D)?AC?AD?BC?BD?CD?D?AC?BC?D________________________________________________________________________________________________________________________________________ (c) AC?BD?BC?AB

__________________________________________________________________________________________AC?BD?BC?AB?AC?BD?BC?AB?(A?C)(B?D)?(B?C)(A?B)?AB?BC?AD?CD?AB?AC?B?BC?B?AD?CD?AC

3.1.7 利用与非门实现下列函数

(a) L=AB+AC

L?AB?AC

____________________________________(b) L?D(A?C)

L?D(A?C)?DAC

_________________________________________________(c) L?(A?B)(C?D)

L?(A?B)(C?D)?ABCD

__________________________________________3.2.2 用卡诺图法化简下列各式

(a) AC?ABC?BC?ABC

_______________________________________________________________AC?ABC?BC?ABC?AC?BC?BC?ABC?AC?C?ABC?C?ABC?C___________

(b) ABCD?ABCD?AB?AD?ABC

ABCD?ABCD?AB?AD?ABC?AB?ABCD?AD?A(B?BCD)?AD?AB?ACD?AD?AB?A(D?DC) ?AB?AD?AC__________________(c) (AB?BD)C?BD(AC)?D(A?B)

__________________(AB?BD)C?BD(AC)?D(A?B)

?ABC?BCD?BD(A?C)?DAB ?ABC?BCD?ABD?BCD?ABD?ABC?BCD?AB?BCD__________ (d) ABCD?D(BCD)?(A?C)BD?A(B?C)

__________ABCD?D(BCD)?(A?C)BD?A(B?C)

?ABCD?BCD?ABD?BCD?ABC ?m11?m1?m9?m12?m14?m6?m14?m4?m5??m(1,4,5,6,9,11,12,14)