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@chyavanphadke48134 жыл бұрын
The final answer was : (A+B) . (A+B') . (A+B) can't we remove the common terms as we did in SSOP ? So the answer will be : (A+B) . (A+B')
@ishaanbhola24283 жыл бұрын
Same doubt , whats the answer ?
@madsoul93042 жыл бұрын
Ur answer is right bro
@harekrishna7063 Жыл бұрын
Using idempotent law
@techguru68515 жыл бұрын
Bro u teach in very short duration and clear our concept... Keep it up... Helpful.
@SimpleSnippets5 жыл бұрын
Thank you so much bro, I have a request, please do share the videos and our channel with your friends as well. That will be the biggest help 😇
@techguru68515 жыл бұрын
Already did bro.. And will share with more frnds
@SimpleSnippets5 жыл бұрын
Thank you so much 🤟😁
@colonelmuska82165 жыл бұрын
Hi, At 2:13, isn't the expression you provided already in SOP form? If it's not in SOP form, why not? If it is in SOP form, why would someone want to convert it to POS form? Thanks so much for these great videos!
@yinkak3921 Жыл бұрын
I don't know if ur still active but I have a question. Why did (A+B) . (A+BB') evaluate to (A+B) . (A+B).(A+B'). shouldn't it be (A+B).((A+B).(A+B')) because we evaluated A = (A+B).(A+B') and the "." operation was performed on A so shouldn't the "." operation be performed on ((A+B).(A+B')) as a whole?
@anirudhsengupta9638Ай бұрын
Bro thats a product take an example x.(y.z) = x.y.z
@LIFE_OF_COPYWRITER6 жыл бұрын
Thanx it really helpful nd clear my all doubt. Tomorrow is my internal. Let c.....✌
@SimpleSnippets6 жыл бұрын
Hi Nikita, I'm happy that you've cleared your doubts 😇✌️ All the best for your exams 😇✌️
@LIFE_OF_COPYWRITER6 жыл бұрын
Simple Snippets thankuh tanmay😊
@SimpleSnippets6 жыл бұрын
Do share it with your classmates 😉 Maybe even they'll find it helpful 😋✌️
@vivekprasad_official_music22674 жыл бұрын
Can you help me in this 3 topic 1.Other logic operating 2.Digital logic operation 3.integrated circuits
@aksharsalunke61764 жыл бұрын
Please Sir at for next playlist use white board and thanks for this all knowledge to gave us.
@WinifredKokosu7 ай бұрын
Thanks a lot I want to know if the question given to you will state that you should leave your answer in the standard product of sum form And also should the expression be simplified to the lowest level
@Funland-v5d3 ай бұрын
Very nice and easy lecture
@raunitjaiswal3629 Жыл бұрын
hello bhaiya your channel content is just too good .
@vishalm27666 жыл бұрын
i think you can still modify that answer as [(A+B).(A+B')]. because A.A=A
@SimpleSnippets6 жыл бұрын
correct ✌
@ekeshwarlalsahu57476 жыл бұрын
(A+B) is not equal to (A+B')
@geemochi99384 жыл бұрын
@Lifes Lessons what
@SAMRATsofficial6 жыл бұрын
In last Ans why don't you use Idempotent Law for (A+B)(A+B)
@vpgsalty6 жыл бұрын
hey dude. what would I need to do if more than one of the variables is missing from a term and I want to convert the expression to standard form? I understood how to do that if one is missing. like (A+B) where only C is missing. but what if B or A were missing too?
@SimpleSnippets6 жыл бұрын
You introduce that variable the same way you introduce the first missing variable ✌️
@yashkadostt6 жыл бұрын
Concept cleared bro✌️😎
@SimpleSnippets6 жыл бұрын
Thank you so much Yash. Really happy to hear this from you 😇✌️
@himanshu71034 жыл бұрын
thanks sir , beautiful
@AsmaaDiis Жыл бұрын
thank u❤
@lubienaali27635 жыл бұрын
Thanks sir it's really helpful for me
@SimpleSnippets5 жыл бұрын
Most welcome my friend. Please do share the video with your friends too. That'll be the biggest help and support ✌️☺️
@chinmayeegouda1515 жыл бұрын
your video helped me thx
@SimpleSnippets5 жыл бұрын
Most welcome Chinmayee 😊 glad to know this. Please do subscribe on this channel and share the videos with your friends too. That's the biggest help and support ✌️
@beastwasil23372 ай бұрын
5:46 Can't we to using 3 variables ?
@siddumair23025 жыл бұрын
Thank you 🙏💕
@SimpleSnippets5 жыл бұрын
Most welcome Umair 😇 please share the video with your friends too ✌️
@wilfredalmeida61096 жыл бұрын
How did A+(B×Bbar) become (A+B)×(A+Bbar)???
@SimpleSnippets6 жыл бұрын
That's the distributive law ✌️
@rinnivatsa45695 жыл бұрын
Hey i have doubt
@kathanshah330511 ай бұрын
can't we dominace law here.?
@chandaoswal53967 жыл бұрын
Why we really need POS and SOP? And why we really need to convert Sum to Product & Product to Sum as already we have AND gate & OR gate?
@seeni20056 жыл бұрын
dude i feel its easier than logic gates i think algebra screwed you
@LegitGamer23453 жыл бұрын
to get marks , its simple .
@chandaoswal53967 жыл бұрын
Can POS contain single complementary variables. Like for ex: A'+A.B+B'+A'.B?
@SimpleSnippets7 жыл бұрын
+Chanda Oswal yes but then A'+A ultimately is 1 right? So i in your above mentioned example the final answer after simplifying this will be 1.1.B
@chandaoswal53967 жыл бұрын
+Simple Snippets :-In our classes it was really confusing me.Thank you so much sir for such a good explanation.Now i think its one of the easiest topics ever. Its so easy to catch, Your videos should be viraled all over so that everyone can get this topic in 5 mins.
@mirandoo3006 жыл бұрын
✌
@SimpleSnippets6 жыл бұрын
Thank you 😇 please do share the videos with your friends and contacts 😇
@akshayawasthy59897 жыл бұрын
A+B . A+B . A+B' isn't this exp suppose to be like this A+B . A+B' as A.A=A
@SimpleSnippets7 жыл бұрын
A+B . A+B' here the 2 sums ie (A+B) & (A+B') are different because in one term we have B and in other we have B' so we cant apply the law that you're mentioning.
@haythama-n-m20717 жыл бұрын
No, he is talking about (A+B)(A+B) (A+B`)=(A+B)(A+B`)
@kasimkarachiwala13917 жыл бұрын
??
@Amitkrdas178 жыл бұрын
apply idempotent law dude..
@SimpleSnippets8 жыл бұрын
These laws have been discussed in other tutorials.Here we simply explain what POS means, why and how to use them.
@justastudent49655 жыл бұрын
how does this come about? At kzbin.info/www/bejne/noTKlqyCiNKfick Because i dont see how you jumped from A+(B.B')