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Till now, the Boolean expressions which have been discussed by us were completely specified, i.e., for each combination of input variable we have specified a minterm by representing them as 1 in the K-Map. But, there may arise a case when for a given combination of input we may not have a specified output or the input combination may be invalid. The combinations for which we don't have any output expression specified are called don't care combination.

到現在為止，已經完全指定了我們已經討論過的布爾表達式，即，對于輸入變量的每種組合，我們通過在K-Map中將它們表示為1來指定一個最小項。 但是，可能會出現以下情況：對于給定的輸入組合，我們可能沒有指定的輸出，或者輸入組合可能無效。 沒有為其指定任何輸出表達式的組合稱為“ 無關組合” 。

For Example, in 8421 code, input states 1001, 1010, 1011, 1100, 1101, 1110 and 1111 are invalid and the corresponding output is the don't care. Similarly, in Excess-3 code, binary input states 0000, 0001, 0010, 1101, 1110 and 1111 are unspecified and are also represented by don't care.

例如，在8421代碼中，輸入狀態1001、1010、1011、1100、1101、1110和1111無效，而相應的輸出則無關緊要 。 類似地，在Excess-3代碼中，未指定二進制輸入狀態0000、0001、0010、1101、1110和1111，也由“ 無關”表示。

These don't care combinations in the K-Map are denoted by an X (cross) symbol.

K映射中的這些無關組合用X(叉)符號表示 。

The X is called "Don't care conditions".

X稱為“無關條件” 。

General rules to be followed while minimizing the expressions using K-Map which include don't care conditions are as follows,

使用K-Map最小化表達式時應遵循的一般規則如下，其中包括無關緊要的條件 ，

After forming the K-Map, fill 1's at the specified positions corresponding to the given minterms. Fill X at the positions where don't care combinations are present.

形成K-Map后，在與給定的最小項相對應的指定位置填充1。 在不需要組合的位置填充X。

Now, Encircle the groups in the K-Map. One thing to be kept in mind is, now we can treat Don't Care conditions (X) as 1s if these help in forming the largest groups. No such group can be encircled whose all the elements are X.

現在，在“ K-Map”中將組圈起來。 要記住的一件事是，如果這些條件有助于形成最大的群體，那么我們可以將“無關條件”(X)視為1 。 不能包圍所有元素都是X的此類組。

If still there are 1s left which doesn't get encircled in any of the groups, then these isolated 1s are encircled individually.

如果仍然有1個沒有被任何組包圍的1 ，則這些孤立的1將被單獨包圍。

Now, recheck all the encircled groups, and remove any redundancy if present.

現在，重新檢查所有包圍的組，并刪除所有冗余(如果存在)。

Write the Boolean expression for each encircled group.

為每個包圍的組編寫布爾表達式。

The final minimal expression can be obtained by ORing each Boolean expressions that were obtained from each group.

可以通過對從每個組獲得的每個布爾表達式進行“或”運算來獲得最終的最小表達式。

Point to remember:

要記住的一點：

While designing K-Map using SOP form, don't care conditions (X) are considered as 1, if it helps form the largest group, otherwise it is considered as 0 and are left during encircling. On the contrary, while designing a K-Map using POS form, don't care conditions (X) are considered as a 0, if it helps form the largest group, otherwise it is considered as 1 and are left during encircling.

在使用SOP格式設計K-Map時，如果條件( X )有助于形成最大的組，則將條件( X )視為1，否則將其視為0并在包圍過程中保留。 相反，在使用POS表單設計K-Map時，如果條件( X )有助于形成最大的組，則將條件( X )視為0，否則將其視為1并在包圍過程中保留。

Example 1: Minimize the given Boolean Expression by using the four-variable K-Map.
F (A, B, C, D) = Σ m (1, 5, 6, 12, 13, 14) + d (2, 4).

示例1：使用四變量K映射最小化給定的布爾表達式。
F(A，B，C，D)=Σm(1、5、6、12、13、14)+ d(2、4)。

Solution:

解：

We will fill 1s at the appropriate minterm positions and also the don't care positions will be filled with (X). Now, we can encircle these 1s and X using the rules we have discussed earlier. We can observe that we can encircle them into three groups. First Group (encircled by black ink) contains four 1s and the Boolean expression given by it is B.C. Second group is the rolling group (encircled by blue ink) which also contains four 1s gives B.D. Third group is the pair of 1s which gives A.C.D

我們將在適當的最小期限位置填充1s ，并且不在乎位置將以( X )填充。 現在，我們可以使用前面討論的規則將這些1和X圈起來。 我們可以觀察到我們可以將它們分為三類。 第一組(用黑色墨水包圍)包含四個1，并且它給出的布爾表達式是B. C。 第二組是所述滾動組(由藍色墨水環繞)，其中還包含四個1S給出B. d。 第三組是一對1 ，它們給出A。 ?.D

Hence, the resultant simplified Boolean Expression will be:

因此，結果簡化后的布爾表達式將為：

F (A, B, C, D) = B.C + B.D + A.C.D

Example 2: Minimize the given Boolean Expression by using the four-variable K-Map.
F (A, B, C, D) = Σ m (1, 5, 6, 12, 13, 14) + d (2, 4).

示例2：使用四變量K映射最小化給定的布爾表達式。
F(A，B，C，D)=Σm(1、5、6、12、13、14)+ d(2、4)。

Solution:

解：

Solution:

解：

As we have already discussed in our last example, we will fill 1's at the appropriate minterm positions and X at don't care positions, encircling the groups using the rules we have discussed above results into the formation of five groups as seen in the image. The Boolean Expressions given by the groups are A, C.D, B.C. D, A.B.D and B.C.D respectively.

正如我們在上一個示例中已經討論的那樣，我們將在適當的最小項位置填充1 ，在無關位填充X ，使用上面討論的規則將組圈起來，如圖所示，形成五個組。 由基團給出的布爾表達式為A，C. d，B .C。 D， A B． d和B.?.D分別 。

ORing these individual Boolean expressions to get the simplified Boolean Expression as:

對這些單個布爾表達式進行“或”運算得到簡化的布爾表達式，如下所示：

F (A, B, C, D) = A + C. D + B.C. D + A.B. D + B. C. D

翻譯自: https://www.includehelp.com/basics/karnaugh-maps-with-dont-care-conditions.aspx

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