如何从组合表达式中确定sop和pos部分？

Question

(，，，)=(‘+’)(+‘)’+‘(’+‘)+’

像这样的组合表达式，SOP和POS都可用，我怎么能知道哪一个是sop &哪个是pos？

我们知道(‘+’)(+‘)是一个pos函数，它是一个’A+‘ sop函数。但我不明白如何从两个sop pos都可用的组合表达式中找到。

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'

结果是一批产品.

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
------------+-------

带有F=1的每一行对应一个腹地，一个输入的乘积(倒的或非倒的)。所以，真值表给出了一个乘积之和形式。

要获得求和乘积形式，您可以将F=0的四行转换为输入(cf )。德摩根定律)：

  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
BC\D  0 1
00    1 1
01    0 1
11    0 0
10    1 0