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
Tussi grt ho sirji!!!
Thanks Neso ♥
very nice explaination.....helped a lot...i am gonna subscribe
After 6 years this is using for me tq so much bro I am from banglore tq
Thanks a lot
very nicely explained..
Mn to kr rha h college ki fees thi de dun pr abba nhi manege😂
Simply the best
Hello sir how could u draw the circuit before simplifying the expression
Can i get any suggestions for
Online Boolean Expression calculator
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') '
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
Why is it called 'Realizing'?🤔
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
I didn't get it😢
lovely
link to previous video plz
what the symbol for the XNOR expression shown at 4:21???
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" .
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
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
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')
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
"100Thanks"
Sir help me
ok
POO IN LOO