Can't we give same input to two nand and then apply nand of the outputs. It will finally give expression like (A'.B')' and It can be written as (A')' + (B') '
sorry sir, i don't know why you drew the necessary gates before the expression. i myself from the expression A'B+AB' got that it's equal ((A'B)'.(AB')')' and for that i needed 5 nand gates...should i memorize the circuit and then find the expression from that????
at 4:30 the expression (AB + A'B')' would always give zero according to complement law ( if we consider AB as q then A'B' is q' and we know that (q+q' =1)) isn't this going wrong?? please help
Sir please do the videos of digital electronics practicals which would be more helpful.. cos your explanation is very good and useful
Your explanation is too good and helpful for all computer science students....Thank you
Wow, this was very helpful
very nice explaination.....helped a lot...i am gonna subscribe
Thanks Neso ♥
After 6 years this is using for me tq so much bro I am from banglore tq
Tussi grt ho sirji!!!
Can't we give same input to two nand and then apply nand of the outputs. It will finally give expression like (A'.B')' and It can be written as (A')' + (B') '
very nicely explained..
Thanks a lot
Why is it called 'Realizing'?🤔
Hello sir how could u draw the circuit before simplifying the expression
Simply the best
Can i get any suggestions for
Online Boolean Expression calculator
Mn to kr rha h college ki fees thi de dun pr abba nhi manege😂
sorry sir, i don't know why you drew the necessary gates before the expression. i myself from the expression A'B+AB' got that it's equal ((A'B)'.(AB')')' and for that i needed 5 nand gates...should i memorize the circuit and then find the expression from that????
Watch the previous lecture in which he implemented EXOR gate using NAND gates
who else thought of applying to intel or AMD after watching all the digital electronics play list :)
hahahahaha i have that feeling bro
may Allah see us through this journey
@@hamzah4792 bhai log abhi kya kar rahe aap sab
what the symbol for the XNOR expression shown at 4:21???
I didn't get it😢
lovely
Sir, half adder can be implemented using XOR gate and AND gates, so why is there a need for NAND gate, if we already have XOR and AND gates?
JUST TO DECREASE THE NUMBER OF GATES FOR IMPLEMENTING HALF ADDER WE USE NAND INSTEAD OF XOR AND "AND" .
link to previous video plz
Sir ji
This is wrong process.
This is diagram of XOR gate by NAND gate only.
means you did not finish watching the video. after 4:45 it is about the Carry
I know distributive law
But
I can't understand this exprseion
[(A'+A.B)(B'+A.B)]'
Change to (A'B'+A.B)'
Could anyone help to solve this?
distributive law(A'+AB=A+B)
[(A'+B)(B'+A)]'
A'B' +A'A +BB' + BA
(A'B' + AB)'
@@vamshishette4148 tnhks
jhakasss explained
"100Thanks"
can we do it in different way?.at first draw sum and carry with AND OR then just replace this gate with NAND gate..check the ansr is same..reply me
I think usse nhi ho payega
@@sanjaypareek9486 it will be...but there will huge amount of nand gate...that will very uncomputable
@@sanjaypareek9486 kyun nhi hoga? You're replacing gates with their nand equivalent
at 4:30 the expression (AB + A'B')' would always give zero according to complement law ( if we consider AB as q then A'B' is q' and we know that (q+q' =1)) isn't this going wrong??
please help
Your assumption is wrong. If AB = q, then A'B' ≠ q'. But, q' = A'+B'.
Yes, if q=AB then q'=(AB)'=(A'+B')
Sir help me
ok
POO IN LOO