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!
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
Thank you for your kind words. Here is a playlist of all my digital logic videos from beginning to end. th-cam.com/play/PLRZOuAYqms1Nf5cfrswAv7BFYuLeRN8pQ.html
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.
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!
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
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.
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
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.
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.
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|
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.
Wow, of all the explanations for this, yours is the clearest, simplest, shortest, most concise. Bravo! Subscribed!
no!he is not
@@loopnation5359 we uko down tu! kama kiatu yako
@@emanueliden6351 english please sir
Thank you, clearest explanation that I have come across so far.
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!
same
Concise and precise.
You just nailed it. Thanks
This video really helped me, other videos are 17 minutes long and I cannot understand completely.
Thank you so much
you explained it wayy better than my lecturers! TYSM!!!!
i struggled so much with topic and i learned more from this 5 minute video than two years of studying computer science
Thanks, for the clear explanation...Also, this exact question came out in one of my assignments so thank you again for the free marks :-)
I feel like I've been enlightened after your video thank you so much!!! :D
May God bless you for this🙏🏽 this is the best explanation ever
I didn't know about this until now big thanks!
Muy bien, excelente explicacion,. Me agrado y lo entendi perfectamente.
Thank you so much for this. It really helped me.
Well explained. Thanks
Thank you for making it so simple
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
Thank you for your kind words. Here is a playlist of all my digital logic videos from beginning to end.
th-cam.com/play/PLRZOuAYqms1Nf5cfrswAv7BFYuLeRN8pQ.html
Wowwww you are amazing 😮🎉 thank you so much 😊
thank yuuuuuuuuuuuuuu for e video. everything is clear now
Excellent explanation......I like it plz.....make more such vedio on various problems related to De Morgan theorem...
Here is a link to the playlist of all my digital logic videos: th-cam.com/play/PLRZOuAYqms1Nf5cfrswAv7BFYuLeRN8pQ.html
can you explain the equation after you said "This is our ultimate goal"? I didnt get the equation method. 4:08
Were did you get another b bar
Thankss this helps me alot
Thanks!
i have a question..if the " + " does't bounded with the longest bar..are we need to change it to multiplication or just leave + ?
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.
Thank you bro ❤
Thank you
Thank you brother
I dont know why so many down votes, this is crazy good
Thanks sir ❤
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!
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
@@dougtougaw7681 Oh yes i see, well explained, Thank you!!!
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.
Im also confused can you explain how he multiplied the distribution part cuz I don't get it
@@deandawiz he took the whole A .B on the right parenthesis and multiplied to each A and B.C separately in the left parenthesis.
@@junh4807 This helped so much thanks!
Well explained 👏
Could have explained the distribution part
why can't we just use double paranthesis and brackets??
What happened to C? HELP I DON'T UNDERSTAND
best one
Nice explain tq
ty bro, you saved my fk life i love you sm.
Thanks
I am more confused now then I ever was. 5:00
What rule did you use to come to this conclusion??
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
hmmm
@@HaloWolf102
Shouldn't A+(B•C) be (A+B)•(A+C)?
10/10
Yes
shloud noty it be : A .(not)B + A.C ?
Because B.(not)B = 1 ?
I was following aong until you got rid of the C in the second equation.
Is B.B(Bar).C.A also 0?
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.
Doug Tougaw thanks! You have helped me in simplifying Boolean algebras. I hope I do well for my test tomorrow
ESE MAQUINA
wow
I don't get what this DeMorgen's theorem is. Isn't this just basic boolean algebra?
A(AUB) = A
4:13 I'm lost
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.
Not really understanding at 4:15....
Same 😓
Oh wait I figured it out
😀
(A+(B)'+C+D)-(A+(B)'+(C)'-(D)')'=Y
Please solve it
I am from Belgium and my professor his explenations are not so clear as this video
Cause there is only one
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.
「動画の音が良くない」、
Keep Change Flip on steroids
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|
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.
Nasty equation 😂
didnt help at all
This was a useless video
@Doug Tougaw
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'
Sorry i had mistake A.B'(1+B.C) =A.B'(1)=A.B'