Yes..... totally agreed..

☠️

hello styrish

@@870affan8 hi bro.. All the best for tomorrow's electronics exam 🤘❤️

Here the goal is to cover all the 1s in the k-map by making the minimum groups. Of course, you can make the group of 5 and 13, but If you make the group of 5 and 13, then to cover 9, you need to make another group with 13. So, group (5, 13) is redundant over here, More technically if I say, then these three groups (shown in the video) are essential prime implicants. The essential prime implicants will always be the part of the final minimized expression. The (5, 13) is prime implicant. For more info, you can check this video : kzfaq.info/get/bejne/nNNxqqag2Ky3gqc.htmlsi=3XL4fZlaaXm40y4l

Yes, all the 1s needs to be covered in the k-map.

Thanks again God richly bless u

Please l need the laws of boolean algebra and l also want to know if a boolean expression must be simplified to the lowest level

After making two groups of 4 1s, only two mimterms m13 and m9 are remaining. And it can be covered by making the group of 2 1s. If there is one more mimterms m11, then we would have made a group of 4 1s. I hope, it will clear your doubt.

You can if you want. But here the idea is to cover all the minterms (with '1') by making minimum number of groups. Here, just by making three groups {(4,5,6,7) (13,9) and (14,15)}, we are able to cover all the 1s in the K-map. If you make a group of 4 1s with (5,7,13, and 15) then to cover remaining 1s, you need to make another three groups. So, that is why, here (5,7,13 and 15) are not grouped, and instead 4,5,6 and 7 is grouped. I hope, it will clear your doubt.

No, all the minterms for which output is 1 needs to be covered. If that 1 ( minterm) cannot be grouped with any other minterm then you just need to make a group of only that 1.

8 and 9 can be grouped but to cover the minterm 1, one needs to make the group of 1 and 9. Similarly, to cover 10, one needs to make group of 1 and 10. So, yes we can make the group of 8 and 9, but during the minimization or for finding the minimal expression, that group will be redundant. Because just by making these two groups, we are able to cover all the 1s in the K-map. In the next video, I am going to cover it in detail. (The concept of Prime implicant and the essential prime implicant) After that, I think your all doubts related to K-map will get clear.

you can make the group of 2. For more info, you can check this video, where couple of questions related to K-map have been solved. kzfaq.info/get/bejne/mbqCd62j1pvdg4E.html

It will be covered seperately. ( In the seperate video)

@@ALLABOUTELECTRONICS okay 😉❤️

You can make those groups. But those are redundant groups. Because to cover 9 and 14, you need to make another groups. Just by making these three groups, we are able to cover all the 1s in the K-map. (basically the two groups you were talking are the non-essential prime implicants) For more info, please check this video. Your doubts will get clear. kzfaq.info/get/bejne/nNNxqqag2Ky3gqc.html

On the vertical side of the map, we have two variables A and B. And on the horizontal side we have C and D. Now, in this group if you see, on the vertical side the variable B is not changing. ( 01 and 11, the second variable B value is 1). Likewise, on the horizontal side, the variable D is not changing. ( 01 and 11, the second digit is representing variable D, and its value is 1). Therefore, we can say that this group of 4 represents B.D. I hope, it will clear your doubt.

@@ALLABOUTELECTRONICS ok yes it cleared my doubts thank you and fabulous explanation 👍🏻👍🏻👍🏻

we can represent this only by D since it is not changing

Which group ? I can only see 1 group of 8. That is C'. The other groups are groups of 4. The group should be in the power of 2. And it has to be symmetrical. It can be either square or rectangle. As per that, there are only three groups.

@@ALLABOUTELECTRONICS oh, I mistakenly assumed that second group can have 8 elements but I am now clear that it will be actually group of six which is not possible as you told.Thanks and sorry.

No, you cannot group them. You can make the group of four 1s in the corner, but you cannot group the two 1s which are in the opposite corner. You can combine these groups of 1s. (0,2) (0,8) (8,10) (2,10). But you cannot make the group of (0,10) and (2,8)

Thank you@@ALLABOUTELECTRONICS

I will cover the state diagram very soon.

Lol😂

Like what

You should thank him for his efforts rather than this. There is an Arabic proverb that says : "They did not find any flaw in the gold, and they said its sparkle blinds the eyes."

Is your keyboard doesn't have spacecap

As far as I can see, there seems no mistake in example 4. But still, if you feel there is a mistake, please mention the timestamp where you feel, there is a mistake.