ECE 100 05 DeMorgan's Theorem Example

[@funmeister]

Wow, of all the explanations for this, yours is the clearest, simplest, shortest, most concise. Bravo! Subscribed!

[@loopnation5359]

no!he is not

[@emanueliden6351]

@@loopnation5359 we uko down tu! kama kiatu yako

[@loopnation5359]

@@emanueliden6351 english please sir

[@Nicolasgtix]

I have been struggling for long a time trying to understand the explanations of my teacher... you made it all clear in 5 minutes. Thanks a lot. Subscribed!

[@Hdhdhdjqqlsjdndndjwjwjs]

same

[@wonderbuddy3204]

you explained it wayy better than my lecturers! TYSM!!!!

[@ObeyTehPro]

can you explain the equation after you said "This is our ultimate goal"? I didnt get the equation method. 4:08

[@ejsafara456]

didnt help at all

[@againstthemachine86]

Thanks, for the clear explanation...Also, this exact question came out in one of my assignments so thank you again for the free marks :-)

[@kamyabratadebnath9389]

Thank you sir , you projected the theorem very well , All my confusions are gone

[@MiMejorAmigoTablon]

This video really helped me, other videos are 17 minutes long and I cannot understand completely. Thank you so much

[@HaloWolf102]

I am more confused now then I ever was. 5:00 What rule did you use to come to this conclusion??

[@HaloWolf102]

Found the answer. I made a circuit with ABC, and it included 3 AND gates. There was a NOT gate hooked up to B. Made a 3 input binary table, and the table did not produce a single '1'. So that is why B•C•A•B'=0

[@adammustapha4150]

hmmm @@HaloWolf102

[@leleuxfamily8364]

I really like your videos, i am more relieved now after seeing these. I did have a question though could you post a reference to which videos to follow in sequence to what your teaching. I keep having to skip around to get basics before I can watch other videos so if you could tell me like a list of which ones to start and follow with I would greatly appreciate it. Either way though great job on the videos you teach this so much better than my professor and now I feel I have a chance to make a decent grade this semester. Best Wishes Donald

[@dougtougaw7681]

Thank you for your kind words. Here is a playlist of all my digital logic videos from beginning to end. kzbin.info/aero/PLRZOuAYqms1Nf5cfrswAv7BFYuLeRN8pQ

[@Devi--Devi]

What happened to C? HELP I DON'T UNDERSTAND

[@rafainfernal]

I dont know why so many down votes, this is crazy good

[@OccupiedBot]

Not really understanding at 4:15....

[@humanbeing7779]

Same 😓

[@humanbeing7779]

Oh wait I figured it out

[@nathaniell.1886]

I was following aong until you got rid of the C in the second equation.

[@Prof-BOF]

May God bless you for this🙏🏽 this is the best explanation ever

[@pierregiuseppegilles9135]

Shouldn't A+(B•C) be (A+B)•(A+C)?

[@Nathouuuutheone]

This is really great! But since my algebra is a little far, I kinda got lost at the distribution part. I'm gonna have to relearn how to multiply parantheses together cause that totally took me by surprise.

[@deandawiz]

Im also confused can you explain how he multiplied the distribution part cuz I don't get it

[@junh4807]

@@deandawiz he took the whole A .B on the right parenthesis and multiplied to each A and B.C separately in the left parenthesis.

[@coodlevlogs7770]

@@junh4807 This helped so much thanks!

[@kithminijoseph2646]

I feel like I've been enlightened after your video thank you so much!!! :D

[@troymd2009]

Thank you, clearest explanation that I have come across so far.

[@razvypld]

ty bro, you saved my fk life i love you sm.

[@inventing...3717]

4:13 I'm lost

[@dougtougaw7681]

We are multiplying each of the terms in parentheses with the terms in the other parentheses. It's like FOIL: (a+b)(c+d)=ac+ad+bc+bd.

[@TacoDaddy-mr8ig]

This was a useless video

[@kintrix007]

I don't get what this DeMorgen's theorem is. Isn't this just basic boolean algebra?

[@jora1308]

shloud noty it be : A .(not)B + A.C ?

[@jora1308]

Because B.(not)B = 1 ?

[@r3dx936]

ESE MAQUINA

[@庆海孟]

「動画の音が良くない」、

[@jakeday5224]

i struggled so much with topic and i learned more from this 5 minute video than two years of studying computer science

[@mathieubourgoy8581]

I am from Belgium and my professor his explenations are not so clear as this video

[@Mr97gilad]

Wowwww you are amazing 😮🎉 thank you so much 😊

[@uninor1928]

(A+(B)'+C+D)-(A+(B)'+(C)'-(D)')'=Y Please solve it

[@thoughtinfliction8693]

why can't we just use double paranthesis and brackets??

[@jirehreid8213]

Could have explained the distribution part

[@banoubaid450]

Thank you

[@arnaldoferreira632]

Thank you so much for this. It really helped me.

[@thechrisricci]

Thanks!

[@AichatouOmor]

Yes

[@brendamudereri1683]

thank yuuuuuuuuuuuuuu for e video. everything is clear now

[@universoinfinito7125]

Muy bien, excelente explicacion,. Me agrado y lo entendi perfectamente.

[@eggxecution]

I didn't know about this until now big thanks!

[@rachitajain4405]

Thank you for making it so simple

[@vanithanegali5947]

Well explained. Thanks

[@nealmaxwell303]

Keep Change Flip on steroids

[@zahinhusain]

Thankss this helps me alot

[@jyotiaher3003]

Excellent explanation......I like it plz.....make more such vedio on various problems related to De Morgan theorem...

[@dougtougaw7681]

Here is a link to the playlist of all my digital logic videos: kzbin.info/aero/PLRZOuAYqms1Nf5cfrswAv7BFYuLeRN8pQ

[@panashepanaxe629]

Well explained 👏

[@danieldian8708]

Can't we solve it by factoring =>A+B'(1+B.C)which the value of parenthesis is equal to 1 and final result is A+B'

[@danieldian8708]

Sorry i had mistake A.B'(1+B.C) =A.B'(1)=A.B'

[@abrahamabraham7650]

Thank you brother

[@danielakoyleek3757]

Concise and precise. You just nailed it. Thanks

[@imoilman7079]

Thank you bro ❤

[@ahmaddzulhakim4189]

i have a question..if the " + " does't bounded with the longest bar..are we need to change it to multiplication or just leave + ?

[@dougtougaw7681]

If I understand your question correctly, then you are right-if the plus sign is not enclosed under the bar, then it does not change to a multiplication sign.

[@computersciencetutor9678]

You are a legend!!! thank you ever so much!!!!!! But i have one question, on 4:20 , why did you OR it? Since you have and and OR,s in the bracket, how would you know? Thank you!

[@dougtougaw7681]

CST, Thank you for the positive feedback! It's just like distribution in algebra: (X+Y)(WZ)=XWZ+YWZ. The AND operation behaves like multiplication, and the OR operation behaves like addition. Hope that helps, Doug Tougaw

[@computersciencetutor9678]

@@dougtougaw7681 Oh yes i see, well explained, Thank you!!!

[@marthamwansa30]

Were did you get another b bar

[@nayeckmuhammadsafwaan3278]

best one

[@Do-It_With_D]

Thanks

[@Akash-ph2ft]

😀

[@abduljabbar-eh9ko]

wow

[@hanifpetri]

Nice explain tq

[@marthamwansa30]

A(AUB) = A

[@moazelsawaf2000]

Thanks sir ❤

[@marthamwansa30]

Cause there is only one

[@dougtougaw7681]

At 4:14, I distributed the AND operation across the parentheses. This is just like multiplying out (x+y)*z=x*z+y*z. It appears that another z was created, but in reality it was just distributing the z across the parentheses.

[@chaoslegend1000]

10/10

[@frostbitepokin9520]

Is B.B(Bar).C.A also 0?

[@dougtougaw7681]

Yes, Boolean algebra follows the commutative property, which means that a sum or product is independent of the order of the variables being added and multiplied. This means that if there is a variable and its complement anywhere in the product, then it is equal to zero.

[@frostbitepokin9520]

Doug Tougaw thanks! You have helped me in simplifying Boolean algebras. I hope I do well for my test tomorrow

[@furquannaser3549]

Nasty equation 😂

[@merabaap9809]

Sir, i can solve this in 2steps ... Instead of going from outside bar to inside go through inside to outside...(consider | as a bar ) I.e; ||A+(B+C)|+|A.|B|||= ||{A+(B.C)}{A.|B|}||= {A+(B.C)}{A.|B|}= A.|B|

[@dougtougaw7681]

It's true that there are many ways to solve problems in Boolean algebra, and I included additional steps in order to try to make it easier to understand.

[@ObeyTehPro]

@Doug Tougaw