Correction at 11:38 The two prime Implicants which I have mentioned ( m1, m5 and m9, m13) are in fact Implicants. And instead of that there will be a prime implicant, which consists of all 4 minterms. ( m1, m5, m9 and m13). That prime implicant is non-essential prime implicant and that's why it was not included in the final minimal solution.
I'm very thankful for your channel and explanations, but it's honestly sad that I have to resort to this since my professor doesn't bother to explain this concept very well, leaving most of us students including myself lost and confused. This video is a lifesaver.
after searching all day on youtube and google,i found this video as crystal clear,easy and more accurate relative to others.best for understanding i,pi,epi.(thank you).
Thanks a lot brother from BD, the way you describe it, take all my confusion although my exam was bad but your video is a gem that I missed, I am just a bit late in here
14:24 , (m12,m13,m8,m9) or (m12,m13,m15,m14) should be EPI. isn't it ? otherwise we cant get minimalized prime implicant ryt ? what is the reason for that ?
14:36, how can this be non-essential prime implicant if two of it's minterms are not included in any other group? Isn't it also an Essential prime implicant?
The other prime implicant m12, m13, m14, m15 is not shown here. But the remaining two minterms are covered by that. That's why it is non-essential prime implicant. I hope, it will clear your doubt.
The condition for calling any prime implicant as the essential prime implicant is that, it should have atleast one minterm, which is not being covered by any other prime implicant. And while selecting that, you should consider all the prime implicants of the function. In this case, that condition is satisfied only for two prime implicants. We just can't ignore any of the prime implicant of the given function while making the decision. I hope, it will clear your doubt.
14:32 is it possible to make an essential prime implicant group of two with (m12 and m13) instead of making the non essential group of four with (m12,m13,m8 and m9) ? If it’s not then why so ?
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?
Sir if we group (m12 , m13 , m15 , m14) and (m12 , m13 , m8 , m9) simultaneously then both will be Non-Esp but if we group only one at a time then both group will be Esp . Suppose we have grouped only (m12 , m13 , m15 , m14) and not (m12 , m13 , m8 , m9) group then m12 and m13 are not covered by any other prime implicant , therefore (m12 , m13 , m15 , m14) group should be Esp. I hope you understood what i want to say.
at 14:21 why the group of [m12, m13, m15, m14] or [m12, m13, m8, m9] is not a essential prime implicant? the minterms m12 and m13 are grouped once either in the group.
Yes, you are right. It is being covered by other other prime implicants. So, it is non essential. But to cover all minterms, it needs to be included during minimization.
😢 What happens when all the groups are separate ie doesn’t overlap with any other implicants eg: minterms(0,2,4,5,10,11,13,15) . Will essential PI will be al the PI or thre is 0 Essential PI?
It is a part of an essential prime implication. If you closely observe, (m0, m2, m8, m10) forms an essential prime implication. And it is also shown in the image.
It is prime implicant. But it’s not essential prime implicant. Because all the 1s are getting covered by other prime implicants. It is prime implicant because all the 1s are not getting covered by other group at a same time. I hope it will clear your doubt. You may watch the video again to understand the concept.
Because here the minimization is being done for the Boolean expression which is in SOP form (Sum of Product Form). For POS form (Product of Sum) form of expression, the 0s in the K-map are combined. I hope, it will clear your doubt.
Because, minterms m12, and m13 are also covered by the other prime implicant m12, m13, m14 and m15. Thats why it is non essential prime implicant. I hope, it will clear your doubt.
Correction at 11:38
The two prime Implicants which I have mentioned ( m1, m5 and m9, m13) are in fact Implicants. And instead of that there will be a prime implicant, which consists of all 4 minterms. ( m1, m5, m9 and m13).
That prime implicant is non-essential prime implicant and that's why it was not included in the final minimal solution.
Thank you sir, this is really helpful.
I'm very thankful for your channel and explanations, but it's honestly sad that I have to resort to this since my professor doesn't bother to explain this concept very well, leaving most of us students including myself lost and confused. This video is a lifesaver.
Same for me bro
after searching all day on youtube and google,i found this video as crystal clear,easy and more accurate relative to others.best for understanding i,pi,epi.(thank you).
Amazing explaination saved my grade!!! My TA struggled to explain the concept. You are the best!
quality of animation and explanation is too good sir. Thank you sir.
Thanks a lot brother from BD, the way you describe it, take all my confusion although my exam was bad but your video is a gem that I missed, I am just a bit late in here
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very nicely explained ,thank you sir
14:24 , (m12,m13,m8,m9) or (m12,m13,m15,m14) should be EPI. isn't it ? otherwise we cant get minimalized prime implicant ryt ? what is the reason for that ?
14:36, how can this be non-essential prime implicant if two of it's minterms are not included in any other group?
Isn't it also an Essential prime implicant?
The other prime implicant m12, m13, m14, m15 is not shown here. But the remaining two minterms are covered by that. That's why it is non-essential prime implicant. I hope, it will clear your doubt.
thank you so much your explanation is so good I finally understand the concept clearly thanks to you
Thank you so much. This helped a lot.
14:45 , we can also take 12,13,14,15 row also as non-essential prime implicant right ?
Sir, at 11:39, does not m1,m5,m9,m13 form a prime implicant instead of two different implicants (m1,m5, and m9,m13)?
Thank you.
Yes, you are absolutely right !! m1, m5,m9, m13 will make a prime implicant. It's a mistake from my side. Thanks for pointing it out 👍🏻
Thank you for doing a video lile this !
14:32
But why would be call that Non-Essential PI? Once we remove one of the 2 Non-Essential, isn't the other one becoming Essential?
The condition for calling any prime implicant as the essential prime implicant is that, it should have atleast one minterm, which is not being covered by any other prime implicant. And while selecting that, you should consider all the prime implicants of the function. In this case, that condition is satisfied only for two prime implicants.
We just can't ignore any of the prime implicant of the given function while making the decision.
I hope, it will clear your doubt.
I think, You are correct, request if anyone can explain if this statement is wrong.
I think it IS also prime implicant in that boolen function. Same as your viewpoint
In 5:26, why we take m9&m13 as prime implicant? It is not clear to me. Sir please reply
because m9 cannot be covered if you dont take it as prime implicant.
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14:32 is it possible to make an essential prime implicant group of two with (m12 and m13) instead of making the non essential group of four with (m12,m13,m8 and m9) ? If it’s not then why so ?
m12 and m13 is not even prime implicant. Because it is being covered by other two groups. (m12, m13, m14, m15) and (m12, m13, m8 and m9)
Thank you sir!
Just superb
nice explanation thanks
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?
so how many total prime implicants and essential prime implicants in last example.?? 14:36
Sir if we group (m12 , m13 , m15 , m14) and (m12 , m13 , m8 , m9) simultaneously then both will be Non-Esp but if we group only one at a time then both group will be Esp . Suppose we have grouped only (m12 , m13 , m15 , m14) and not (m12 , m13 , m8 , m9) group then m12 and m13 are not covered by any other prime implicant , therefore (m12 , m13 , m15 , m14) group should be Esp. I hope you understood what i want to say.
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thank you so much
sir in 11.35 u r saying 2 more PI
Which is wrong because it's 1 "vertical Quad" ...it can't violate priority of higher order group in PI formation
Yes, you are right. I have already mentioned that in the pinned comment.
sir when you add non-epi to the equation does they prevent glitchs ?
at 14:21 why the group of [m12, m13, m15, m14] or [m12, m13, m8, m9] is not a essential prime implicant? the minterms m12 and m13 are grouped once either in the group.
If you consider [ m12, m14, m13, m15] then all the minterms are covered by other prime implicants. m12,m13 is also covered by ( m12, m13, m8 , m9)
sir ,at 11:13 isn't blue circle non essential since all the 1's are covered other prime implicants as well?
Yes, you are right. It is being covered by other other prime implicants. So, it is non essential. But to cover all minterms, it needs to be included during minimization.
Thank you sir
فدوة لقلبك ♥
Thanku sir ....❤️🙌
😢 What happens when all the groups are separate ie doesn’t overlap with any other implicants eg: minterms(0,2,4,5,10,11,13,15) . Will essential PI will be al the PI or thre is 0 Essential PI?
In that case, all those non-overlapping groups are essential prime implicant. It means, you need to include them for minimizing the Boolean function.
Thank you
thanks
non essential PI and SELECTIVE PI are same and the definition is
those PIs which r neither EPI nor RPI is SPI
Sir. At 10:07 why isnt m0 an essential prime implicant?
It is a part of an essential prime implication. If you closely observe, (m0, m2, m8, m10) forms an essential prime implication. And it is also shown in the image.
Understood sir thank you 😊
And thank you for the reply.most educational youtubers wont respond when you ask a question
good video
what program do you use for drawings?
at 5:27 why we use the red one as a prirme implicant bcz all the ones that are covered before so whats the reason of making the red one..
It is prime implicant. But it’s not essential prime implicant. Because all the 1s are getting covered by other prime implicants. It is prime implicant because all the 1s are not getting covered by other group at a same time. I hope it will clear your doubt. You may watch the video again to understand the concept.
Sir what is non prime implicant in kmap
at 10.22 why don t we comsider 0 position as a esp
Because here the minimization is being done for the Boolean expression which is in SOP form (Sum of Product Form).
For POS form (Product of Sum) form of expression, the 0s in the K-map are combined.
I hope, it will clear your doubt.
For more info, also check this example.
ruclips.net/video/ZayoUTi2tsA/видео.html
in 14:35 why m12,m13,m8,m9 doesn 't essential prime implicant ?.thank you
Because, minterms m12, and m13 are also covered by the other prime implicant m12, m13, m14 and m15. Thats why it is non essential prime implicant. I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS But you said either of these m12, m13, m8, m9 or m12, m13, m14, m15 are non essential prime implicant.
in the video at 5.34 only 3 prime implicants only come group 2 is irredundant actually
The second group shown in yellow color is not redundant one. It will be required to cover the minterm m9 and m12.
i don't understand why m0 is not ESPI. Only one prime implicant cover just like m10
Would you please mention where you are referring to in the video ??
@@ALLABOUTELECTRONICS it's at 10:07
@@hiramahiram-d5i m0 is a part of the prime implicant, (m0, m2, m8, m10) And as you can see, it is already ESPI.
At 8:48 how green box is ESP
The minterm m9 is not being covered by any other prime implicit. That’s why it satisfy the condition of ESP.
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need caption pls🙁
Will be updated soon.
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thank you sir