@@aviralkhanduja5834 because the blocks(in the complemented form) were not parallel to the non-complimented form...you must add those two blocks individually to get the correct answer.
At the end of k-mapping (09:01) , Those two cells are not in the same parallel. Which mean A is changing from zero to one, therefor it should be + ABCE + A'CE ?
***** The answer should be C'E' + BD + A'DE + A'B'CE + ABCE. It is because we cannot combine the way you did it. so, we have to group it in twos( in each map).
on the last part of the function A is changing because you took group of 4 ( 2 from the first side and 2 from the other side) C is not changing also E is not changing so it should be C'E'+BD+A'DE+CE
Sorry, incorrect. those can't be grouped. So, therefore, two more literals will be added each for the groups in both sides. If you calculate further, you will find C'E'+BD+A'DE+A'B'CE+ABCE to be the answer
so in short , we can say that for 5 variables , the k map is a 3d object , again like the lower dimensions we can only connect the 1s along the axes ( since the minterms only differ by 1 term ) , and this 3D map is connected at the opposite edges , i.e. like the borders of a 2d k map were connected to a closed surface , so the boundaries of this would fold to connect in 4d . Understood clearly sir , thnx
Last two pairings are really confusing. Let me write what I understand. If we remove A for only last particular case. Can over lap the two 4 variable k'maps for BCDE in that case C will remain constant and E will remain constant. So final value of F should be F = C'E' + BD + A'DE + CE.
Number of cells filled with 1s are equal to no. Of implicants ,if we consider don't care as 1 and grouped with output , then our number of implicants changes we should consider them as 1s
Hello Your work is really good,helps me alot but can do me a small favour by uploading a video on quine-mcclusky method in digital electronics as soon as possible because i have an exam in 3 days.and your tutorial really helps me alot
for those who find this 3 dimentional map irritating, you can use what we call the shannon theorem: f(Xn,.....Xi,......X0)=Xi.f(Xn,.....,1,......X0)+/Xi.f(Xn,.....,0,......X0)
grouping of last two pairs into quad should be like this .... A is changing from 0 to 1 so we will neglect A B is changing from 0 to 1 so we will neglect B C is same i.e 1 so we will write C as it is D is chnaging from 0 to 1 so we will neglect D E is same i.e 1 so we will write E as it is Answer = CE We can also check this if we add both pairs individually for first pair it is A'B'CE for second pair it is ABCE NOW if we add both A'B'CE + ABCE = CE(A'B'+AB) CE(1) = CE
Last grouping of 2 pairs into quad is incorrect. Add them both separately, it comes out A'.B'.C.E + A.B.C.E
Please check
I can't understand last 2 groupings
Yes true I was confuse though nice video sir
you added A video annotation, mr curry
Parth Jinwala thanks
Thank God..
The fact that we got the error shows how well you taught us 😇
I know you people caught the error but it's very tough to teach this 5 variable so that happening just ignore it.
He made us to understand
i realized it to be wrong but just by instinct can you please tell me the reason for it to be wrong
@@aviralkhanduja5834 because the blocks(in the complemented form) were not parallel to the non-complimented form...you must add those two blocks individually to get the correct answer.
last grouping of two pair is incorrect. we add both separately...final ans is like.. C'.E'+B.D+A.'D.E+A'.B'.C.E+A.B.C.E
he didn't apply D-Morgan law so better to get it you can do further part and get the right ans .......
It can be minimize bro
@@naveenkorlepara how?
@@anshulyadav1060 lol he did missed the part where that's my problem.
Thank you Neso! Even with the error I still understood you, difficult concept but is well explained here.
thank you from the bottom of my heart
i want to thak u to neso academy for such soperb approach towards video learning ...... thank u alll
Sir i have a doubt last step A'B'CE+ ABCE is that correct sir
awesome video sir i just learned and caught your points very quickly sir
thank you sir
At the end of k-mapping (09:01) , Those two cells are not in the same parallel. Which mean A is changing from zero to one, therefor it should be + ABCE + A'CE ?
Oh.. yes, I missed typing it.
Anyway. You're welcome :)
***** You forgot about BD, didn't you?
***** The answer should be C'E' + BD + A'DE + A'B'CE + ABCE. It is because we cannot combine the way you did it. so, we have to group it in twos( in each map).
@@SomasundharamSampath i agree with you on this one well done and thanks for the answer
yes
is the answer not equal to C'E' + BD + A'DE + CE. we must have grouped as quad
Yea you are right
Yesssssssssss
No it cannot be grouped as a quad. They will be in diagonal
@@kirtilodha5242 yes they are in the 2 and 3rd row respectively. we cannot group them
I think it's wrong.bcz for quad there will be three variables in expression
You did a great job, with that 3 D drawing and I understood every single thing you said. That you very much from the core of my heart :)
please cover videos on integrated logic gates including cmos and ttl..as it is in our course
Hello ,what do u do currently ,
I guess MBA
on the last part of the function A is changing because you took group of 4 ( 2 from the first side and 2 from the other side) C is not changing also E is not changing so it should be C'E'+BD+A'DE+CE
Sorry, incorrect. those can't be grouped. So, therefore, two more literals will be added each for the groups in both sides. If you calculate further, you will find C'E'+BD+A'DE+A'B'CE+ABCE to be the answer
@@abdullahelkafi6280 after 6 years .. 😐
Well the last 2 groupings are incorrect and many here rectified it.
so in short , we can say that for 5 variables , the k map is a 3d object , again like the lower dimensions we can only connect the 1s along the axes ( since the minterms only differ by 1 term ) , and this 3D map is connected at the opposite edges , i.e. like the borders of a 2d k map were connected to a closed surface , so the boundaries of this would fold to connect in 4d . Understood clearly sir , thnx
Legends say, he intentionally did the wrong part in order to help us gain confidence in K'map by checking the comment section.
:)
Underrated 😃
😂
But sir last 2 cells done at 9.1 doesnt overlap each other then how can it be grpef
the last 2 duo pairs' A doesn't change from 0 to 1, hence should be included
which making A'B'CE + ABCE
The minimal for the given expression will be C'D'+ BD+A'DE+A'B'C+ABCE. As for last two prime implicants A is varied .
👍🏿
Thanks, understand everything 2 days before my exam 👌
Last two pairings are really confusing.
Let me write what I understand.
If we remove A for only last particular case. Can over lap the two 4 variable k'maps for BCDE in that case C will remain constant and E will remain constant. So final value of F should be F = C'E' + BD + A'DE + CE.
thank you neso...you people just saved me from a carry over
He is a genius making an error on purpose so people would come and comment.
I think the answer is wrong sir , bcos those doublets are not parallel to each other
Sir , the last quad is not possible which is shown at the end of the video
Sir thank you so much. Can you make lecture on solving by VEM. PLZ
Thank you for this. Best explanation I've seen yet!
Number of cells filled with 1s are equal to no. Of implicants ,if we consider don't care as 1 and grouped with output , then our number of implicants changes we should consider them as 1s
you are doing tremendous job. I request you to plzzz complete all the subject of ECE for GATE and ESE2018 .I will be very thankful to you.
Thnks sir.You helped me a lot in digital electronics.
Great tutorial!
Thx for u support my friend
sir..thanks for all lecture..
I have a request...that please upload " variable entered mapping method "
to solve higher order k-map problem...
why the third group is A'DE? i THOUGHT ONLY B is changing from 0 to 1 and C does not change? So it should be A'C'DE?
Thankyou neso
Thank you so much sir for all the tutorials.
Helped me a lot in my preparation
AMAZING. THIS COURSE IS GREAT . THANK YOU!
Thanks a lot sir
Hello
Your work is really good,helps me alot but can do me a small favour by uploading a video on quine-mcclusky method in digital electronics as soon as possible because i have an exam in 3 days.and your tutorial really helps me alot
example of 5 variable in k__ map
thanku...soo much sir
if i am not wrong...sir in 4th minterm isn't we get CE bcoz B is changing from 0 to 1.
Thanq so much
Sir please take another example to solve 5 variable k- Map 🙏🙏🙏🙏🙏
Last quad grouping is incorrect. On the left face its B'C and on the right face it is BC
for those who find this 3 dimentional map irritating, you can use what we call the shannon theorem: f(Xn,.....Xi,......X0)=Xi.f(Xn,.....,1,......X0)+/Xi.f(Xn,.....,0,......X0)
Thanks Sir☺️☺️
good explanation
I think the Last grouping is your visual error
Didn't understand
How we get BCE at 9:10
Please clarify this sirr
He made a mistake on it.
He has made a mistake
The answer for that grouping is EC
It must be CE
sir ,can u please tell me how you have filled the map with 0's and 1's. didn't you considered the table
pasham shyamsundhar reddy it's like already filled question
there is a mistake at 8 min 43 sec, 1s group is not overlapping
Thank you! I found the 3D picture very helpful in picturing this concept
Last solution is incorrect
It should be F= C'E' + BD + A'DE + CE
Last 2 grouping is wrong.
It's not quad, you make group of 2 pair 2 times.
how the last pair or rather quad of ones has been grouped....????
can u tell me ,I can't understand....
5 var map k is so complicated that it even confused our sensei
thanks! :)
last pair is not possible please check it.
If it were not for your videos, I would not be able to do my EE project.. Thanks man!
Can we solve it by using 1 table only for 5 variables
I think the last minterm should be CE , not BCE.
Yes right 👍
grouping of third pair is also incorrect ... c is not changing for group 3 and c =0 so according to me it answer should be A'C'DE
Appreciate it .. Really helped understanding K-Map..
Thank You So much !
its very helpful sir thanks alot!
Is this me or there should be 5 implicants in the last one right??
Sir Iam having two doubts ,can dontcares were grouped with ones twice, as we do for output 1's when they r in need only
Don't can be considered as 1 when need and can treated as same
THANK YOU VERY MUCH..SAVED MY DE EXAM
You have to learn kmap first.. i have watched this before going to exam.lol
Cant we simply take combination of 3 variable same as we took combination of 2 vars
There is mistake. . . .F=C'E'+BD+A'DE+ABCE+A'B'C'E is the right answer. . .
You also make a mistake
yes at 9 min the pair is not parallel
How did u consider 1 s for mapping
please do the video of variable entered map
Sir..can u give another example for this
C'E' + BD + A'DE + ABCE + A'B'CE
It helped me a lot Thank u so much
Sir Plzz make a video on 6 variable it will means a lot to me...... Plzzz 😖😖
6 variable K map wont come in exam...so chill man!!!
muchas gracias . muy bien explicado.
In last at the place of BCD should be CE.......??
Good 👍
A will not change in the last one
Matao'=Puto Amo=Neso Academy
Aapka sara video achha h lekin jo padhne ke baad feeling aana chahiye wo nahi aata h. Yahi lagata h ku mai ratta mar raha hu.
good morning sir
sir mujhe BCE last ans me samjh nahi aai,
Pls follow notes to explain
A'.B'.C.E + A.B.C.E=CE
Can't understand anything
IS THERE ANY METHOD TO SOLVE 5 VARIABLE TRUTH TABLE CONSIDERING 3 VARIABLE OF IT AS ONE AND SOLVING AS THREE VARIABLE
Please correct the Solution
Wrong grouping at the end of the video
grouping of last two pairs into quad should be like this ....
A is changing from 0 to 1 so we will neglect A
B is changing from 0 to 1 so we will neglect B
C is same i.e 1 so we will write C as it is
D is chnaging from 0 to 1 so we will neglect D
E is same i.e 1 so we will write E as it is
Answer = CE
We can also check this if we add both pairs individually
for first pair it is A'B'CE
for second pair it is ABCE
NOW if we add both A'B'CE + ABCE = CE(A'B'+AB)
CE(1) = CE
Thanks for the good explanation. Can you support us for the Quine Mecluskey method of simplification. ?
make ys clear about last group u made .... u just forgot about group in first k map :v
the last answer of k-map is wrong
sir i didn't understand.....take another eg
that one is super just a mistake in 2ones group
How to group 2nd row in A'and third row in A...???
make a video on quine-mcCLUESKY method
yo it's already there bro
thanks
No clarity.....
i didnt got how you fill the ones.
just a random example