如何从组合表达式中确定sop和pos部分

Question

(，，)=(′+′)(+′)′+′(′+’)+′

就像这个SOP和POS都可用的组合表达式一样，我怎么能知道哪些是SOP&哪些是pos？

我正在努力寻找sop&pos，但我不能。我们知道(′+′)(+′)这是一个pos函数，也是一个′a+′sop函数。但我不明白如何从组合表达式中找到两个sop-pos都可用的地方。

Answer 1

要将表达式F确定为乘积之和，可以重写它：

F(A, B, C, D) = (D' + AB')(AD + C')B' + BD'(A' + C') + A'
= (ADD' + C'D' + AAB'D + AB'C')B' + (A'BD' + BC'D') + A'
= (AB' + B'C'D' + AB'D + AB'C) + (A'BD' + BC'D') + A'
= C'D' + B'D + A'

结果是乘积的总和。

真相表：

A  B  C  D  |  F
------------+------
0  0  0  0  |  1
0  0  0  1  |  1
0  0  1  0  |  1
0  0  1  1  |  1
0  1  0  0  |  1
0  1  0  1  |  1
0  1  1  0  |  1
0  1  1  1  |  1
1  0  0  0  |  1
1  0  0  1  |  1
1  0  1  0  |  0
1  0  1  1  |  1
1  1  0  0  |  1
1  1  0  1  |  0
1  1  1  0  |  0
1  1  1  1  |  0
------------+-------

具有F=1的每一行对应于minterm，minterm是输入(反转或非反转)的乘积。因此，真值表给出了乘积的总和形式

要获得和的乘积形式，可以取F=0的四行并反转输入(参见De Morgan定律)：

A  B  C  D  |
--------------+------
1  0  1  0  |  0
1  1  0  1  |  0
1  1  1  0  |  0
1  1  1  1  |  0
--------------+------

(A'+B+C'+D)(A'+B'+C+D')(A'+B'+C'+D)(A'+B'+C'+D')

Answer 2

(D'              + AB')(AD + C')B'  + BD'(A' +    C') + A'
D'ADB' + D'C'B' + AB'ADB' + AB'C'B' + BD'A' + BD'C'  + A'  distributive law
D'ADB' + D'C'B' + AB'ADB' + AB'C'B'         + BD'C'  + A'  absorptive law
D'ADB' + D'C'B' + AB'D    + AB'C'           + BD'C'  + A'  idempotent law
0 A B' + D'C'B' + AB'D    + AB'C'           + BD'C'  + A'  complement law
0      + D'C'B' + AB'D    + AB'C'           + BD'C'  + A'  annulment law
D'C'B' + AB'D    + AB'C'           + BD'C'  + A'  identity law
D'C'B'           +  B'D    +  B'C' + BD'C'  + A'  absorptive law  
B'C'D' + BC'D'   +  B'D    +  B'C'          + A'  commutative law 
C'D'(B' + B)   +  B'D    +  B'C'          + A'  distributive law  
C'D'1          +  B'D    +  B'C'          + A'  complement law
C'D'           +  B'D    +  B'C'          + A'  identity law
C'D'           +  B'D                     + A'  kmap reduction
(populating kmap in above order, B'C' is alreay included when we get to it)
kmap
BCD  0 1
00    1 1
01    0 1
11    0 0
10    1 0