布尔逻辑设计-约简 [英] Boolean Logic Design - Reduction

问题描述

我具有以下要简化/简化的功能.

I have the following function to be reduced/simplified.

F(A，B，C，D)= BC +(A + C'D')，其中'表示补数

F(A,B,C,D) = BC + (A + C'D') where ' denotes the complement

Here's my solution:

= BC + (A + C'D')'

= BC + (A + (C+D)

= BC + (A + C + D)

= BC + C + A + D

= C(B + 1) + A + D

= C*1 + A + D

= C + A + D

这是正确的吗?

推荐答案

与传统代数一样，如果对等式的一侧进行某些操作，则必须对另一侧进行处理，包括补数.在这里，我们陈述原始方程式:

As in traditional algebra, if you do something to one side of the equation, you must do it to the other side, including complementing. Here we state the original equation:

F'(A，B，C，D)= BC +(A +(CD)')

F'(A,B,C,D) = BC + (A + (CD)')

由于我们有F'而不是F，所以我的直觉告诉我要对双方进行补充，但首先我将补语分配为(CD)'，从长远来看使生活更轻松:

Since we have F' instead of F, my intuition tells me to complement both sides, but first I distribute the complement in the term (CD)' to make life easier in the long run:

F'= BC +(A +(C'+ D'))

F' = BC + (A + (C'+ D'))

现在，我们可以对等式的两边进行补充:

Now we can complement both sides of the equation:

1:F ='(BC)'(A +(C'+ D'))OR分配补数后变为AND

1: F = '(BC)'(A + (C'+ D')) The OR becomes AND after distributing complement

现在，让我们在内部分配补码，看看我们得到了什么:

Now let's distribute the complements inside just to see what we get:

2:F =(B'+ C')(A'(CD))

2: F = (B'+ C')(A'(CD))

现在，我们可以将正则项(A'(CD))分配给被OR'ed的两个项:

Now we can just distribute the right term (A'(CD)) over the two terms being OR'ed:

3:F = B'(A'(CD)) + C'(A'(CD))

3: F = B' (A'(CD)) + C' (A'(CD))

我们看到正确的术语消失了，因为我们有CC'，因此我们剩下:

We see that the right term goes away since we have a CC' and thus we are left with:

4:F = A'B'CD

4: F = A'B'CD

希望我没有犯错.我知道您已经找到了答案，但是其他阅读此问题的人可能也有类似的问题，因此我这样做是为了避免重复出现的问题.祝你好运！

Hopefully I didn't make a mistake. I know you've found the answer, but others reading this might have a similar question and so I did it out to save duplicate questions from being asked. Good Luck!

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