Identify each of these logic gates by name, and complete their respective truth tables:
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In order to familiarize students with the standard logic gate types, I like to given them practice with identification and truth tables each day. Students need to be able to recognize these logic gate types at a glance, or else they will have difficulty analyzing circuits that use them.
Identify each of these relay logic functions by name (AND, OR, NOR, etc.) and complete their respective truth tables:
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In order to familiarize students with standard switch contact configurations, I like to given them practice with identification and truth tables each day. Students need to be able to recognize these ladder logic sub-circuits at a glance, or else they will have difficulty analyzing more complex relay circuits that use them.
AKarnaugh地图无非就是一种特殊的真实表，可用于将逻辑函数减少为最小的布尔表达式。
这是特定三输入逻辑电路的真实表：
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Complete the following Karnaugh map, according to the values found in the above truth table:
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Showing students that Karnaugh maps are really nothing more than truth tables in disguise helps them to more readily learn this powerful new tool.
AKarnaugh地图无非就是一种特殊的真实表，可用于将逻辑函数减少为最小的布尔表达式。
Here is a truth table for a specific four-input逻辑电路:
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Complete the following Karnaugh map, according to the values found in the above truth table:
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Showing students that Karnaugh maps are really nothing more than truth tables in disguise helps them to more readily learn this powerful new tool.
Here is a truth table for a four-input logic circuit:
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如果我们将此真实表转化为Karnaugh地图，我们将获得以下结果：
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Note how the only 1’s in the map are clustered together in a group of four:
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If you look at the input variables (A, B, C, and D), you should notice that only two of them actually change within this cluster of four 1’s. The other two variables hold the same value for each of these conditions where the output is a “1”. Identify which variables change, and which stay the same, for this cluster.
对于这个四个1的集群，变量A和C是仅有的两个变化的输入。对于四个“高”输出中的每一个，变量B和D保持相同（B = 1和D = 1）。
这个问题向学生介绍了karnaugh的探测原则矛盾的变量在分组中。
Here is a truth table for a four-input logic circuit:
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如果我们将此真实表转化为Karnaugh地图，我们将获得以下结果：
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Note how the only 1’s in the map all exist on the same row:
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如果您查看输入变量（A，B，C和D），则应注意，Karnaugh地图上每个“ 1”条件中只有两个是恒定的。识别这些变量，并记住它们。
Now, write an SOP (Sum-of-Products) expression for the truth table, and use Boolean algebra to reduce that raw expression to its simplest form. What do you notice about the simplified SOP expression, in relation to the common variables noted on the Karnaugh map?
对于这个四个1的集群，变量A和B是唯一在Karnaugh地图中显示的四个“ 1”条件保持恒定的两个输入。真实表的简化布尔表达是AB。在这里看到图案？
这个问题向学生强烈暗示，Karnaugh地图是一种图形方法，可以为真实表降低形式的SOP表达。一旦学生意识到Karnaugh的映射是逃避艰苦的布尔代数简化的关键，他们的兴趣将受到兴趣！
One of the essential characteristics of Karnaugh maps is that the input variable sequences are always arranged in Gray code sequence. That is, you never see a Karnaugh map with the input combinations arranged in binary order:
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当我们考虑使用karnaugh地图检测输出集中的共同变量时，原因很明显。例如，在这里，我们在中心有一个Karnaugh地图，其中有四个1的集群：
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Arranged in this order, it is apparent that two of the input variables have the same values for each of the four “high” output conditions. Re-draw this Karnaugh map with the input variables sequenced in binary order, and comment on what happens. Can you still tell which input variables remain the same for all four output conditions?
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看着这个，我们仍然可以说所有四个“高”输出条件的b = 1和d = 1，但这是not显然是像以前一样的接近度。
您可以简单地告诉学生，必须根据灰色代码对输入变量进行测序，以使Karnaugh映射以简化工具工作，但这不会向学生解释为什么它需要这样。这个问题向学生展示了Karnaugh地图中灰色代码测序的目的，通过向他们展示替代性（二进制测序），并允许他们查看寻求非界限变量的任务如何使其变得复杂。
检查这个真相表和对应的Karnaugh地图：
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尽管从初次出现中可能并不明显，但Karnaugh地图中的四个“高”条件实际上属于同一组。为了更明显，我将绘制一个新的（超大）Karnaugh地图模板，沿每个轴重复了两次灰色代码序列：
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用真实表中的0和1值填写该地图，然后查看四个“高”条件的分组是否变得显而易见。
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后续问题：这个问题告诉我们有关分组的问题？换句话说，我们如何识别“高”国家的群体而不必制作超大的卡诺地图？
延伸超过卡诺地图边界的位群体的概念往往会使学生感到困惑。实际上，这是唯一使学生讨厌Karnaugh地图的事情！简单地告诉他们将地图的边界分组并不是真正教导他们为什么the technique is valid. Here, they should see with little difficulty why the technique works.
而且，如果出于某种原因无法想象一下Karnaugh地图的边界的位群体，他们就知道他们可以绘制超大的地图，这将变得显而易见！
A student is asked to use Karnaugh mapping to generate a minimal SOP expression for the following truth table:
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按照显示的真实表，学生绘制了这张卡诺地图：
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“这很容易，”学生对自己说。“所有'1'条件都属于同一组！”然后，学生突出显示了一个单一组的三胞胎：
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Looking at this cluster of 1’s, the student identifies C as remaining constant (1) for all three conditions in the group. Therefore, the student concludes, the minimal expression for this truth table must simply be C.
但是，第二个学生决定在此问题上使用布尔代数，而不是karnaugh映射。从原始真相表开始，并为其生成产品总和（SOP）表达式，简化如下：
$$\overline{A}BC+A\overline{B}C+ABC$$
$$BC(\overline{A}+A)+A\overline{B}C$$
$$ bc+a \ overline {b} c $$
$$ C（b+a \ overline {b}）$$
$$ C（b+a）$$
$$ AC+BC $$
显然，第二名学生的布尔减少（AC BC）给出的答案与第一学生的Karnaugh地图分析（C）给出的答案不符。
Perplexed by the disagreement between these two methods, and failing to see a mistake in the Boolean algebra used by the second student, the first student decides to check his Karnaugh mapping again. Upon reflection, it becomes apparent that if the answer really were C, the Karnaugh map would look different. Instead of having three cells with 1’s in them, there would be four cells with 1’s in them (the output of the function being “1”任何time C = 1:
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Somewhere, there must have been a mistake made in the first student’s grouping of 1’s in the Karnaugh map, because the map shown above is the only one proper for an answer of C, and it is not the same as the real map for the given truth table. Explain where the mistake was made, and what the proper grouping of 1’s should be.
Proper grouping of 1’s in the Karnaugh map:
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The purpose of this question is to illustrate how it is incorrect to identify clusters of arbitrary size in a Karnaugh map. A cluster of three, as seen in this scenario, leads to an incorrect conclusion. Of course, one could easily quote a textbook as to the proper numbers and patterns of 1’s to identify in a Karnaugh map, but it is so much more informative (in my opinion) to illustrate by example. Posing a dilemma such as this makes studentsthink关于为什么答案是错误的，而不是要求他们记住看似任意的规则。
State the rules for properly identifying common groups in a Karnaugh map.
Any good introductory digital textbook will give the rules you need to do Karnaugh mapping. I leave you to research these rules for yourself!
The answer speaks for itself here - let your students research these rules, and ask them exactly where they found them (including the page numbers in their textbook(s)!).
A七个部分解码器是一个数字电路，旨在驱动非常常见的数字显示设备类型：一组LED（或LCD）段，该段以四位代码的命令呈现数字0到9：
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The behavior of the display driver IC may be represented by a truth table with seven outputs: one for each segment of the seven-segment display (a through g). In the following table, a “1” output represents an active display segment, while a “0” output represents an inactive segment:
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这样的真实示例为Karnaugh映射等技术提供了出色的展示。让我们以输出为例，显示它没有真实表中的所有其他输出：
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Plotting a Karnaugh map for output a, we get this result:
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Identify adjacent groups of 1’s in this Karnaugh map, and generate a minimal SOP expression from those groupings.
注意,6个细胞是空白的,因为truth table does not list all the possible input combinations with four variables (A, B, C, and D). With these large gaps in the Karnaugh map, it is difficult to form large groupings of 1’s, and thus the resulting “minimal” SOP expression has several terms.
However, if we do not care about output a’s state in the six non-specified truth table rows, we can fill in the remaining cells of the Karnaugh map with “don’t care” symbols (usually the letter X) and use those cells as “wildcards” in determining groupings:
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用这个new Karnaugh map, identify adjacent groups of 1’s, and generate a minimal SOP expression from those groupings.
Karnaugh地图groupings with strict “1” groups:
$$\overline{D}B+\overline{D}CA+D \ \overline{C} \ \overline{B}+\overline{C} \ \overline{B} \ \overline{A}$$
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卡诺（Karnaugh）与“不在乎”通配符的地图分组：
$突
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后续问题：这个问题和回答仅集中在BCD到7段解码器电路的A输出上。想象一下，如果我们要在这两个时尚中处理解码器电路的所有七个输出，则首先使用严格的“ 1”输出分组开发SOP表达式，然后使用“不关心”通配符。您认为这两种方法中的哪种方法总体上会产生最简单的门电路？两种不同的解决方案对解码器电路的行为有什么影响，对于六个未指定的输入组合1010、1011、1100、1100、1101、1111和1111的影响？
One of the points of this question is for students to realize that bigger groups are better, in that they yield simpler SOP terms. Also, students should realize that the ability to use “don’t care” states as “wildcard” placeholders in the Karnaugh map cells increases the chances of creating bigger groups.
众所周知，我选择了一个非常糟糕的例子来尝试从中表达SOP的表达，因为十个中只有两个非零输出条件！制定POS表达会更容易，但这是另一个问题的主题！
When designing a circuit to emulate a truth table such as this where nearly all the input conditions result in “1” output states, it is easier to use Product-of-Sums (POS) expressions rather than Sum-of-Products (SOP) expressions:
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是否可以使用Karnaugh地图为此真实表生成适当的POS表达式，还是Karnaugh Maps仅限于SOP表达式？解释您的答案，以及您如何获得答案。
是的，您可以使用Karnaugh地图来生成POS表达式，而不仅仅是SOP表达式！
I am more interested in seeing students’ approach to this problem than acknowledgment of the answer (that Karnaugh maps may be used to generate SOP and POS expressions alike). Setting up a Karnaugh map to see if a POS expression may be obtained for this truth table is not difficult to do, but many students are so unfamiliar/uncomfortable with “experimenting” in this manner than they tend to freeze when presented with a problem like this. Without specific instructions on what to do, the obvious steps of “try it and see” elude them.
作为他们的教练，您负责鼓励学生的实验性思维方式。不要简单地告诉他们如何自行“发现”答案，因为如果您这样做，您会抢劫他们的真实发现体验。
使用karnaugh映射为此真实表生成简单的布尔表达式，并绘制相当于该表达式的继电器逻辑电路：
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简单的表达和继电器电路：
$$ b \ overline {c} $$
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One of the things you may want to have your students share in front of the class is their Karnaugh maps, and how they grouped common output states to arrive at Boolean expression terms. I have found that an overhead (acetate) or computer-projected image of a blank Karnaugh map on a whiteboard serves well to present Karnaugh maps on. This way, cell entries may be easily erased and re-drawn without having to re-draw the map (grid lines) itself.
Ask your students to compare using a Karnaugh map versus using standard SOP/Boolean simplifications to arrive at the simplest expression for this truth table. Which technique would they prefer to use, and why?
使用karnaugh映射为此真实表生成简单的布尔表达式，并绘制与该表达式相等的栅极电路：
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简单的表达式和门电路：
$$AC+BC\overline{D}$$
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挑战问题：使用布尔代数技术将表的原始SOP表达式简化为最小的形式，而无需使用Karnaugh地图。
One of the things you may want to have your students share in front of the class is their Karnaugh maps, and how they grouped common output states to arrive at Boolean expression terms. I have found that an overhead (acetate) or computer-projected image of a blank Karnaugh map on a whiteboard serves well to present Karnaugh maps on. This way, cell entries may be easily erased and re-drawn without having to re-draw the map (grid lines) itself.
使用karnaugh映射为此真实表生成简单的布尔表达式，并绘制与该表达式相等的栅极电路：
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简单的表达式和门电路：
$$ ac+a \ overline {b} \ \ edline {d} $$
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挑战问题：使用布尔代数技术将表的原始SOP表达式简化为最小的形式，而无需使用Karnaugh地图。
One of the things you may want to have your students share in front of the class is their Karnaugh maps, and how they grouped common output states to arrive at Boolean expression terms. I have found that an overhead (acetate) or computer-projected image of a blank Karnaugh map on a whiteboard serves well to present Karnaugh maps on. This way, cell entries may be easily erased and re-drawn without having to re-draw the map (grid lines) itself.
This is one of those situations where an important group “wraps around” the edge of the Karnaugh map, and thus is likely to be overlooked by students.
Use a Karnaugh map to generate a simple Boolean expression for this truth table, and draw a relay circuit equivalent to that expression:
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简单的表达和继电器电路：
$$ b \ overline {d}+\ edline {c} \ \ edline {d} $$
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后续问题：尽管上面显示的继电器电路确实满足了最小的SOP布尔表达，但有一种方法可以使其更简单。提示：做得正确，您可以消除电路中的一个联系人！
挑战问题：使用布尔代数技术将表的原始SOP表达式简化为最小的形式，而无需使用Karnaugh地图。
One of the things you may want to have your students share in front of the class is their Karnaugh maps, and how they grouped common output states to arrive at Boolean expression terms. I have found that an overhead (acetate) or computer-projected image of a blank Karnaugh map on a whiteboard serves well to present Karnaugh maps on. This way, cell entries may be easily erased and re-drawn without having to re-draw the map (grid lines) itself.
This is one of those situations where an important group “wraps around” the edge of the Karnaugh map, and thus is likely to be overlooked by students.
Use a Karnaugh map to generate a simple Boolean expression for this truth table, and draw a relay circuit equivalent to that expression:
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简单的表达和继电器电路：
$$ b +\ edline {d} $$
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鉴于在此真实表中拥有一定数量的1，尝试开发POS表达而不是SOP表达是合理的。但是，您的学生可能会发现Karnaugh映射的优雅使这两种方式都很容易！这是一个问题，您绝对希望让学生在班上解释他们的解决方案方法，并且您绝对希望他们看到如何使用Karnaugh地图这两种方式（SOP和POS）。
我发现白板上的空白karnaugh地图的开销（乙酸）或计算机项目的图像很好地呈现Karnaugh地图。这样，可以轻松地删除和重新绘制单元格条目，而不必重新绘制地图（网格线）本身。
I am studying for my midterms, and this helped me a lot. The author is trying to share the information through pure comprehension, explaining the origin and target of each and every rule, which most teachers tend not to do… Thank you so much !!