Simplify the following Boolean function F using K-map. F = A' * B' * C' + B' * C * D' + A' * BC * D' + A * B' * C' , d = A' * B * C' * D + ACD + A * B' * D' where "d" indicates do not care conditions.
Thank you for the tutorial but what about the real-life examples that can make us get the concept of don't care and when they are actually used while creating the truth table on our own without a set condition?
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Just by making the two groups, we are able to cover all the 0s in the K-map for POS minimization. So, there is no need to make any other group. That additional group will be redundant one.
If you closely observe then here three groups have been made. In each group, the total no of covered cells are in the power of 2. In one group we have 4 cells, in second group 2 cells and in the third group again 4 cells. So, all are in the power of 2. You do need to add them all. Just need to ensure that, in each group, the total no of cells are in the power of 2. I hope, it will clear your doubt.
The small group which you are referring to is made of four minterms. Minterms m0, m2, m8 and m10 (The four corners of the K-map). And that corresponds to B'D'. I hope, it will clear your doubt.
At 5:50, there are two groups. In one group variable A and B are changing. That group is C'D. While in the second group variable B and C is not changing. That group corresponds to BC. If you are talking about the second group, then B is already considered in that.
There are two group of 4. And I guess you are asking about B' ( The one in orange color). Well, in that case, if you see that group, then as you move from one minterm to another minterm vertically, ( 0 to 4 or 1 to 5) then variable A is changing from 0 to 1. similarly if you move from one minterm to another minter horizontally ( 0 to 1 or 4 to 5) then value of variable is changing from 0 to 1. Only non-changing variable is B. And in the group, the value of variable B is 0. Therefore, that group represents B' . Similarly, if you see the second group of 4 in the yellow color, then it represents A'. Because, in that group, as you move horizontally, then variable B and C both are changing. Only non-changing variable is A. And its value is 0. Therefore, it represents A'. I hope, it will clear your doubt.
9:55 here why are 0 and 4 not paired? Also,many k map simplifications include not pairing some visible pairs.why that?and how would we understand when not to pair them?
4 is a don't care term. During the minimization, you need to ensure that all the 1s gets covered (during SOP minimization) and all 0s gets covered (during POS minimization). It doesn't matter, If all the don't care terms are not covered. By including some or all the don't care terms, if you are able to minimize the function further, then you should include them. But its not compulsory to include all of them.
For more info please check the last part of SOP and POS form video. Here is the link : ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-YmKmS9bpMqM.html
If you closely observe then it is group of four 1s ( the group in pink colour). The others two 1s are at the bottom corners. So, this group of four 1s represents B’D’. I hope, it will clear your doubt.
Here we are making a group of four 1s (not two 1s). The minterms m11 and m10 are also part of that group. Considering the group of four 1s, it is B'.C. I hope it will clear your doubt.
Would you please mention the timestamp in a video, where you are referring. In general, once you have a minimized Boolean expression then as per the expression it is easy to draw a circuit diagram. If you will mention the timestamp, I can tell you, how to draw a circuit diagram for that specific expression.
It is a don't care term. It is not necessary to cover all the don't care terms in the K-map. But you need to cover all the minterms in the K-map for which the output function is 1. You can use some or all the don't care terms to minimize the boolean expression. But to include all of them during the minimization is not compulsory. I hope, it will clear your doubt.
