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?
Sir Sir I am a beginner now. And may want to become electronic engineer. Where do I start and what is my path? What is my path for electronic engineer. I were saw your playlist but I am very confused where to start. Technology playlist?
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.
No, that's not necessary. The don't care terms can be combined even for making the group of two 1s or two 0s, provided by making them it is possible to reduce the Boolean expression further. If by making the group using don't care term does not reduce theBoolean expression, then there is no need to add that group.
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.
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.
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.
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.
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.
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.
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.
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This single video covered all my doubts about sop , pos and k map which I was searching every where bit by bit from whole day, Thanks alot
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Actually I have no words to say This video helped me a lot🥲 I cleared my DLD examination with good Mark's...thank U so much❤....
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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.
Please sir
Thank you Sir, The exam is 3hrs from now and it helped a lot
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In 11:31 wasn’t it will be BCD'?
Great video, you're a good educator.
Thank you sir watched all the videos given to me as per the handout,and understand all the concepts really well.
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?
In 9:22 s why we don't use M4?
1:01 We only use don’t care when the fonction is unspecified for certain values?
Sir Sir I am a beginner now. And may want to become electronic engineer. Where do I start and what is my path? What is my path for electronic engineer. I were saw your playlist but I am very confused where to start. Technology playlist?
Great explanation thank you sir
At 11:26 ,will 0010 will not be added ,this single min term as pair will not counted at final expression of answer?
7:38 In POS, Complements are taken as 1 and variables are take as 0, right?... then how it is C'D'
Thank you man this work is so good!!!!
Sir at 7:43 there should be one more group of 0x00 in first row of k-map?
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.
at 5:52 why you not grouped (5,7,13,15)?
So the don't care combinations are involved if they form 4 square variable. If it's 2 variables don't group with 0/1?
No, that's not necessary. The don't care terms can be combined even for making the group of two 1s or two 0s, provided by making them it is possible to reduce the Boolean expression further. If by making the group using don't care term does not reduce theBoolean expression, then there is no need to add that group.
for the example 2, why when we do PO we don't include the A'B'?
Because just making the two groups, we are able to cover all the maxterms (0s). So, there is no need to cover A'B'. I hope, it will clear your doubt.
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that random indian always helping
Thank you sir ☺️
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Nice sir 👍
thanks a lot it was really helpful.
Thank u sir ❤️🔥
Thanks a lot
9:41 In POS form we represent 0 by A and 1 by A' but you have represented in SOP form
In the SOP form, I have represented F3'. If you invert it then you will get F3 in POS form. (10:01)
5:50, why you have not considered B in the 4 cell group while it's not changing? Please help me😢
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.
thank you! was helpful
11.30 , while making the groups , we should use 2^n cells right?
But u are used 10 cells in this question
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.
Thanks really helpful
10:12
Shouldn't the small group be Abar*Bbar*Dbar, instead of just Bbar*Dbar?
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 4.40 when we make group of 4 then why we considered only b complement not any term of a
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.
thank you
What would happen for single term?
Would you please elaborate, what you mean to say ??
At 8:03 m2 and m3 shouldn't be A'B'C?
Can u explain how why it's B'C?
It is the combination of 4 terms. The pink group consists of M2, M3, M10 and M11. That is why it’s B’C. I hope it will clear your doubt.
sir in this dont care we should consider all x?
No, not necessarily all x. You just only need to consider those x, which can be used to minimise the Boolean expression further.
Thanks a lot❤❤❤❤❤❤❤❤❤❤❤
Thanks. Can you pleasse the example of 5 veriables with don't care method
will try to cover one such example.
thank you sir
Good video
awasome thanks
Sir why Bbar C ?7:40
Not A barB barC?
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.
sir how to draw circuit diagram for this output
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.
If f(A,B,C) =π(5,10, 11,12)
Then what is F’ ???
Please reply I'm so confused😵😵
It will be equal to Σ ( 5, 10, 11, 12)
For more info please check the last part of SOP and POS form video. Here is the link : ruclips.net/video/YmKmS9bpMqM/видео.html
Best ....
thanks
At 9:46 i think there should be A'B'D' instead of only B'D'.
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.
Okay i got u ... Thanks for clearing the 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.
I think ..there is some mistake in POS form SIMPLIFICATION USING K MAP
Nope, that’s alright. In case, if you have any doubt, let me know here. Happy to help you.
@@ALLABOUTELECTRONICSIn example 3 ..the Answer given is AB‘+B‘D‘+CD
How this expression is in POS form..? It is in SOP form..ri8.?
Sir Questions POS ke and answer SOP ke
Ye bhed bhav kyun😂
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At 9:12 why x at 4 is not grouped in the kmap
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.
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Thank You sir❤