电子技术基础(数字部分)康华光(第五版)习题解答 下载本文

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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

A?BC?(A?B)(A?C),得

由交换律 (b)

A?AB?(A?A)(A?B)?A?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?E3.1.3 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) (d)

_______ABC(B?C)

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

A?ABC?ABC?CB?CB_____

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

(e) (f)

____________________________AB?AB?AB?AB____________________________

_________AB?AB?AB?AB?A?A?0____________________________________________________________________________

(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) (h)

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

ABC?ABC?ABC?A?BCABC?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?ACABCD?ABD?BCD?ABCD?BC

(l)

__________________________________________________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__________________________________________________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 (b)

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

_________________________________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 (b)

_______________________L?AB?AC_____________

L?D(A?C)

(c)

_____________________________L?D(A?C)?DACL?(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)