Boolean Algebra 3 - De Morgan’s Theorem
This video follows on from the one about simplifying complex Boolean expressions using the laws of Boolean algebra. In particular this video covers De Morgan’s theorem and how it can be applied, along with the other laws, to simplify complex Boolean expressions. It includes worked examples and some exercises you can try yourself, along with solutions.
Пікірлер: 62
So professionally done. This is what KZread was made for.
@ComputerScienceLessons
4 жыл бұрын
Thanks for the comment. Much appreciated :)KD
Excellent series, what a great help it's been so far! Thanks for publishing them publicly.
This channel deserves a million subs. The videos are very to the point and helpful. I've learnt a lot from these.
I am in a 5week accelerated program for Mechatronics Engineering. Part of Mechatronics is computer science. I am in a Digital Systems 1 class. This is about as hard as it has gotten so far, and I am a junior. Drinking through a waterfall is more like it than a firehose as Iwould normally say. I am definately thankful for these videos you have made. I would be 100% lost without them. Everything is so well explained and put together and clean and concise. My text book for this course is just hard to look at in general. Thank you for these videos. I will definately be revisting them time and time again, I will not be able to master this subject in 5 weeks the way I would like, but I am already scribbling these videos into my future schedule until I can pound these expressions down with quickness and confidence. Thnak you again Computer Science! You are awesome!
@ComputerScienceLessons
Жыл бұрын
That's music to my ears. You are most welcome :)KD
God bless this video series, it helps me to pass Computer Science class. Thanks for the info sharing, have a super day!
@Szoki666
4 жыл бұрын
Really! At first, I did not even understand the whole logic. With these videos I can practise a lot, and I am so happy that I cannot stop it :)
@ComputerScienceLessons
4 жыл бұрын
It's great to hear they have been so useful. :)
Thank you. I am enjoying myself greatly working through these videos.
Many thanks for the video, excellent resource for use with A-level CS students
@ComputerScienceLessons
4 жыл бұрын
You are most welcome. It's nice to hear from another A level CS teacher. If there's anything else you would like me to cover, please let me know :)KD
FINALLY some who explains each step of the solution! I couldn't find any video or tutorial about this in my first language. Each time the teacher jumped through the steps like it was all self explanatory, so eventually i started to think i was too just dumb to understand. With your videos i finally get it!! Thank you for giving me hope again ;) , especially by saying it takes time to develope this skill!!
@ComputerScienceLessons
2 жыл бұрын
You are very welcome. Stick with it :)KD
Thank you so much for all these videos man! Definitely valuable resources for those who didn't really pay attention during class, lol.
This video need more likes for his work.
Thank you very much for these videos, you taught Boolean Algebra very well.
@ComputerScienceLessons
3 жыл бұрын
You are very welcome :)KD
that was a good video along with practice set. i have a question: The old DLD has been replaced by CMOS technology so are these laws (DeMorgans) still used in CMOS or are they obsolete?
Wow, the execution of the series is just perfect
@ComputerScienceLessons
Жыл бұрын
Thank you :)KD
YOU ARE AWESOMEEE !!! MUCH LOVE
This is very important and one must understand
great lecture, thank you!
Hi Kevin, how would you apply De Morgan to simplify this one ~a~b+~bc+ac?
thank you a lot.I spend a whole day looking for a good explamention
@ComputerScienceLessons
3 жыл бұрын
You are most welcome :)KD
Thank you, this series was very helpful.
Thank you for the amazing video!
@ComputerScienceLessons
Жыл бұрын
You are very welcome. :)KD
Do you have anything on consensus theorem? I'm trying to understand why X' + XY = X' + Y
At 11:48 you could remove the brackets which would result in (X V Y) ^ `X ^ Y. Using the absorptive law (X V Y) ^ Y = Y. So `X ^ Y
thank you so much
@ComputerScienceLessons
5 ай бұрын
You're very welcome :)KD
On the second expression at around 6 minutes into the video could you have gone the other way and converted the AND to OR's?
@ComputerScienceLessons
3 жыл бұрын
More than likely - there's more than one way to crack a nut. In my videos I am trying to illustrate different techniques, rather than the shortest route to a solution (well, that's my excuse anyway!) :)KD
Kaway-kaway sa mga gikan sa Moodle. ✋✋✋
Thank you so much
@ComputerScienceLessons
2 жыл бұрын
You're welcome :)KD
Your are excellent Mr professor
@ComputerScienceLessons
4 жыл бұрын
Thank you. :) KD
@repenttoreflect8933
4 жыл бұрын
How many inputs are required for a 6-output decoder? How can one write the logical equations and its corresponding circuit? Help out please Sir
I find: ¬(A ∧ B) = ¬a ∨ ¬b really similar to: -(a + b) = -a - b Since both expressions switch the middle operator, and negates the values of A and B.
@ComputerScienceLessons
3 жыл бұрын
Good observation 💡 Perhaps Augustus De Morgan was thinking along the same lines :)KD
At 2:54 how can you just swap the OR for the AND? Wouldn't that change the meaning of the expression?
@ComputerScienceLessons
5 жыл бұрын
It's OK to do that as long as you do something else as well. Take a look at my video on logic gate combinations, there is a proof of de-morgan's in there. kzread.info/dash/bejne/hoCglqdsebmcipM.html
@codythompson9973
3 жыл бұрын
@@ComputerScienceLessons The link leads to a private video that we cannot view now.
@ComputerScienceLessons
3 жыл бұрын
@@codythompson9973 I do apologise. Please search my channel for Logic Gate Combinations
Thank you for the amazing lecture. I think solution 3 can be reduced more to A+~B*~C
@ComputerScienceLessons
4 жыл бұрын
Quite possibly. Thanks for the comment. :)KD
Great video by the way but l think the answer to the last question is not right if l am not mistaking. In 5th row you have x + y~xy therefore why didn't you erase that y because and operation is more important and you can reduce number of y's and you will end up with x + y~x, and then (x+~x)(x+y) = x+y. Please correct me if l am wrong. Sincere regards sir for the professional video. Everything else was perfect.
🐐🐐🐐 channel
@ComputerScienceLessons
Жыл бұрын
Thank you (I think) :)KD
Help can't seem to solve this problem: x*z+(notx)*y+yz The answer is x*z+(notx) Note that * means and; + means or
@ReZhorw
2 жыл бұрын
That seems to be the consensus theorem, search for it on youtube and you'll find good explanations. Practically it means that you can remove the" + yz " term from xz + x'y I believe the answer should be xz + x'y and not xz +x' as you've written. Swapping places with y and z would be preferred Try multiplying the yz term in xy + x'z + yz with 1 and write the 1 as (x + x')
It’s a good channel, I just don’t understand why few people subscribe it.
Was this re-uploaded?
@ComputerScienceLessons
6 жыл бұрын
Yes. There was a silly error explaining the algorithm which someone kindly pointed out. Fixed now.
@LMinett
6 жыл бұрын
Excellent. Please do keep up the good work. My pupils especially appreciate these videos during their revision.
@ComputerScienceLessons
6 жыл бұрын
It's great to hear my videos are helping other people as well as my own students.
Kmap ka video banayiye sir
@ComputerScienceLessons
4 жыл бұрын
kiya hua :)KD kzread.info/dash/bejne/Zaqfr8mNnczLiqw.html