Explanation for number of self-dual expressions - The number of self dual expressions is the number of truth tables when "dualed" that give the same expression. 1) Number of min-terms for a given number of variables n = 2^(n) because each variable can go to 0 or 1 2) Number of truth tables for a given number of min-terms = 2^(number of min-terms) because each min term can go to 0 or 1 3) Consider what makes a truth table "dual-able", it is basically what gives a "mirror image" on the top and bottom half of the truth table. Here we define mirror image as 0s going to 1s and 1s going to 0s 4) Therefore if we decide top (or bottom) half of a dualable truth table, the bottom (or top) half of the truth table is known. So we can choose values of only top (or bottom) half of truth table and the other side is defined by the duality. So the number of such truth tables where each of the top min terms can go to 0 or 1 = 2^(min-terms/2) Therefore total number of truth tables = 2^(min-terms/2) = 2^((2^2)/2) = 2^(2^(n-1)) --------- Example of a mirrorable truth table: F = A, with variables A and B - A B | Y 0 0 | 0 0 1 | 0 ------------- 1 0 | 1 1 1 | 1 F(dual) = A B | Y 1 1 | 1 1 0 | 1 ------------- 0 1 | 0 0 0 | 0 which is the same as the first table re-arranged
any truth table where if you make all 1s as 0s and 0s as 1s, then it gives back itself is self dual. To get a dual of any Boolean Expression, replace- OR with AND i.e. + with . AND with OR i.e. . with + 1 with 0 0 with 1 Such tables can be obtained by fixing any values for the outputs in the top half of a truth table, then fixing the complement of those values as mirror image for the bottom half of the truth table. ie. if first output in the table is 1 in the top half, then bottom value is 0, if second value is 0 then second to last value is 1, and so on.
Thanks for deriving the formula! I think the "mirrorable" quality is the self-dual. In other words, if F is mirrorable then F is self dual. In truth tables, the ones, or positive logic, corresponds to sum of products. While the zeros, or negative logic, corresponds to the product of sums. Because of the mirror quality we obtain the same expression for either sum of products or product of sums. In the example you provided there are two functions, each of which is self-dual. But they aren't dual of each other. It might be said they are complements of each other. You only need one function for it to be self dual. The example in the video and the homework assignment are each self dual b/c they are mirrorable. But they are also complements Example: Y=AB+BC+AC = (0 0 0 1 0 1 1 1) Homework: Y=A'B'+B'C'+A'C' = (1 1 1 0 1 0 0 0) They aren't dual to each other though, as we have worked them out. Hope that is helpful.
For self dual homework problem: n = number of variables but you give only 3 variables such as A,B,C and the second problem is find number of self dual expressions take n=4. There is no matching for first (Y=A'.B'+B'.C'+A'.C')and second problem like first problem n=3 and n=4.
Answer for H.W. Problems at 08:43 *Problem 1 Solution* Y = A'.B' + B'.C' + A'.C' (Given) For Self Dual, Take Dual One Time. Y' = (A'+B').(B'+C').(C'+A') Y' = (B'+A'.C').(A'+C') {Distributive Law} Y' = B'.A' + B'.C' + A'.A'.C' + A'.C'.C' {Distributive Law} Y' = A'.B' + B'.C' + A'.C' + A'.C' {X + X = X} Y' = A'.B' + B'.C' + A'.C' {X + X = X} Y' = Y Taking Dual One Time gives Same Expression. Hence, It is Self Dual. *Problem 2 Solution* Number of Self Duals = 2^2^n-1 (n=4, which is given) Number of Self Duals = 2^2^4-1 Number of Self Duals = 2^2^3 Number of Self Duals = 2^8 Number of Self Duals = 256
Sir in the last lecture you say that, we don't have to complement the variable we keep variable as it is, so why you in 3:15 example no. 1 write A' as it is why you. Not write A'=A
Selfdual:
Y=(a'+b').(b'+c').(a'+c')
=(B'+a'c').(a'+c')
=a'b'+a'a'c'+c'b'+a'c'c'
=a'b'+a'c'+b'c'+a'c'
=a'b'+b'c'+a'c'
Thus it's self dual
2)for n=4....
2 power 2 power (4-1)
Ans:256
Explanation for number of self-dual expressions - The number of self dual expressions is the number of truth tables when "dualed" that give the same expression.
1) Number of min-terms for a given number of variables n = 2^(n) because each variable can go to 0 or 1
2) Number of truth tables for a given number of min-terms = 2^(number of min-terms) because each min term can go to 0 or 1
3) Consider what makes a truth table "dual-able", it is basically what gives a "mirror image" on the top and bottom half of the truth table. Here we define mirror image as 0s going to 1s and 1s going to 0s
4) Therefore if we decide top (or bottom) half of a dualable truth table, the bottom (or top) half of the truth table is known. So we can choose values of only top (or bottom) half of truth table and the other side is defined by the duality. So the number of such truth tables where each of the top min terms can go to 0 or 1 = 2^(min-terms/2)
Therefore total number of truth tables = 2^(min-terms/2) = 2^((2^2)/2) = 2^(2^(n-1))
---------
Example of a mirrorable truth table: F = A, with variables A and B -
A B | Y
0 0 | 0
0 1 | 0
-------------
1 0 | 1
1 1 | 1
F(dual) =
A B | Y
1 1 | 1
1 0 | 1
-------------
0 1 | 0
0 0 | 0
which is the same as the first table re-arranged
How it is self dual
any truth table where if you make all 1s as 0s and 0s as 1s, then it gives back itself is self dual.
To get a dual of any Boolean Expression, replace-
OR with AND i.e. + with .
AND with OR i.e. . with +
1 with 0
0 with 1
Such tables can be obtained by fixing any values for the outputs in the top half of a truth table, then fixing the complement of those values as mirror image for the bottom half of the truth table. ie. if first output in the table is 1 in the top half, then bottom value is 0, if second value is 0 then second to last value is 1, and so on.
Nice explanation
Thanks for deriving the formula!
I think the "mirrorable" quality is the self-dual. In other words, if F is mirrorable then F is self dual. In truth tables, the ones, or positive logic, corresponds to sum of products. While the zeros, or negative logic, corresponds to the product of sums. Because of the mirror quality we obtain the same expression for either sum of products or product of sums.
In the example you provided there are two functions, each of which is self-dual. But they aren't dual of each other. It might be said they are complements of each other. You only need one function for it to be self dual.
The example in the video and the homework assignment are each self dual b/c they are mirrorable. But they are also complements
Example: Y=AB+BC+AC = (0 0 0 1 0 1 1 1)
Homework: Y=A'B'+B'C'+A'C' = (1 1 1 0 1 0 0 0)
They aren't dual to each other though, as we have worked them out.
Hope that is helpful.
thank you sir...i have prepared this as a notes sir
1. Y=A'B'+B'C'+A'C' is a self dual
2.When n=4, no. of self dual =256
Thanks
If n=4 no of sd = 64
2^2^4-1 = 64 use calculator
@@satishreddy9766 No brother you used wrong order of operation.
@@satishreddy9766 🤣🤣🤣🤣☺
1.Self Dual
>>>> After expansion of [(A'+B') . (B'+C') . (A'+C')] We are again getting the same expression.
2. 2 ^ (2 ^ 4-1 )=256
For self dual homework problem: n = number of variables but you give only 3 variables such as A,B,C and the second problem is find number of self dual expressions take n=4. There is no matching for first (Y=A'.B'+B'.C'+A'.C')and second problem like first problem n=3 and n=4.
Sir, What is the significance of Self Dual ? Where do we use these functions?
Thank you
In digital ic design this helps while deciding nmos and PMOs stack
Thanks a lot for the H.W. problems. Helps in consolidating the concepts.
Thanks a lot for homework problems , helps in consolidating the concepts
SIR I HAVE COMPLETED LECTURE 22 OUT OF 202. TREMENDOUS LECTURE AND THANKYOU FOR YOUR GUIDANCE !!
thank you so much sir for these lectures. helped a lot. can't describe in words. very very much greatful.
ok i got it dat (1)self dual (2) 256
but can give examples of self duals......i can get 8 self duals wid single variable...wat about d others
+Mohit Verma I agree with you on number 1, I also got that it was a self dual expression, and for number 2; I also got 256! :)
please explain for the self dual.
Right
One self dual expression is the one he already solved
Homework questions solution ->
1) Y = A'B' + B'C' + A'C' is a self dual
2) No. of self dual if n = 4 are 256.
Y = A'B' + B'C' + A'C' is Self dual ✅
sir i have doubt on 1st example can please say which boolean rules as to use after converting it into positive logic
I Thank You Very Very much for giving a free quality education specially for those students which is comes from poor faimly
Sir your explanation is soo good .I love neso academy ❤️❤️
I Love Neso Academy ❤️
the general rule for dual is F(A, B, C, ...) ----> F(A', B', C', ...)' so a self dual must meet the condition of F(A, B, C, ...) = F(A', B', C', ...)'
1) Y = A'B' + B'C' + A'C' is self dual
2) if n=4 then number of self dual expressions is 256
1)example answer= expression is self dual
2)example answer= 64
🔴H/W Problem: 1
Given: Y = A'B' + B'C' + A'C'
Solution: Y' = (A'+B').(B'+C').(A'+C')
= (B'+A'C').(A'+C')
= A'B'+B'C'+A'.A'C'+A'C'.C'
= A'B' + B'C' + A'C'
= Y
✔️ So, the expression Y is a Self Dual expression. (ans)
..........................................................
🔴H/W Problem: 2
Given: No. of variables, n = 4
Solution: No. of self dual expressions = 2^(2^(4-1))
So, n = 2^8 = 256
✔️No. of self dual expressions = 256
Thanq bhaii
@@ganeshmula4508 u r wlcm
1. Self Dual
2. 256
Yes it's a self dual expression
256
Thank you so much sir g.You are the my best teacher.
Answer for H.W. Problems at 08:43
*Problem 1 Solution*
Y = A'.B' + B'.C' + A'.C' (Given)
For Self Dual, Take Dual One Time.
Y' = (A'+B').(B'+C').(C'+A')
Y' = (B'+A'.C').(A'+C') {Distributive Law}
Y' = B'.A' + B'.C' + A'.A'.C' + A'.C'.C' {Distributive Law}
Y' = A'.B' + B'.C' + A'.C' + A'.C' {X + X = X}
Y' = A'.B' + B'.C' + A'.C' {X + X = X}
Y' = Y
Taking Dual One Time gives Same Expression. Hence, It is Self Dual.
*Problem 2 Solution*
Number of Self Duals = 2^2^n-1 (n=4, which is given)
Number of Self Duals = 2^2^4-1
Number of Self Duals = 2^2^3
Number of Self Duals = 2^8
Number of Self Duals = 256
1- the function is self dual
2- 256
Can u plz explain how do we identify an expression as self dual or not??
Can u plz explain how can we find that an expression is self dual or not??
thanks a lots sir for this lecture
1 . Self dual expression
2. 256
256 no of self dual exp
And given example is self dual of itself
Sir, would you tell us the best website to solve more problems on digital electronic, so that we can solve different types of problems
did you find any?
If n=3 what are the self duals?....as n=2 the self duals are a and a' b and b' c and c'
Same question. If you look at his solved problem, you will find anothee self-dual expression of 3 variables.
1) self dual
2) 2^2^n-1 = 2^2^3 = 2^8 = 256
Thanks to neso team ❤️
Sir in the last lecture you say that, we don't have to complement the variable we keep variable as it is, so why you in 3:15 example no. 1 write A' as it is why you. Not write A'=A
1)Self Dual
2) 2^8=256
no. of self dual expression=256
problem 1 is self dual
self dual
Glad 2 have u sir 😍
Y is self dual
256 self duals are possible for 4 variables
Self Dual & 256
Problem 1 is a self-dual expression.
good job sir
problem 1:) i think its not self dual
problem 2:)256
can u tell is it correct or not???.....
1)is self dual bro (its a self dual)
and second one is 256
Why we are not appling Redundancy Theorm in AB+AC+BC+AC ? (In 7th minute)
PLEASE Explain...
Thank you so much!
Thank you sir
Y=A'.B'+B'.C'+A'.C' is self dual
When n=4 we have 256 no of dual
1) Yes it is self dual
2) 2^2^3= 2^8=256
i) yes its a dual.
ii) 256
1)Expression is Self Dual
2)256
1) Self dual
2) 256
Self dual and 2^8
Wait. So A is the same as A*1. When converted to negative logic we get A+1 which is 1, therefore A dual is 1. What am I doing wrong??
not wrong it is right
Dual of A will be A only
But if you Taking Dual of A.1, it will be A+1
Thank you
The first expression is Self Dual
n= 3 , self duals = 256
Y is self dual
1.) It's self dual form
2.) 256 self dual expressions..
Yes it's self dual and 256
Why AND = 0 and OR = 1 you said in 0:50
Can u plz explain me that how can we find whether an expression is self dual or not??
I got it😅
Sir I am not understanding...that what actually self dual is??.....how you say that this is self dual or not... please answered......
This problem is once again explain
1. given expression is self dual
2. 256
you are really awesome
1) It is a self dual .
2)256 .
thanks
1. It is self dual
2. 256
1.self dual form
2^8=256
yes it is self dual
256
Sir which codes are called self complementing codes. It was asked in my exam paper plz tell me
BCD
2421 CODES , 84-2-2 CODES, GREY CODES
I don't know how to solve First problem. Can you please help me😞😞 am getting confused , Although r video is very very helpful for me . Thanks ☺☺
Y' = (A'+B').(B'+C').(A'+C')
= (B' + A'.C') . (A' + C') //DISTRIBUTIVE LAW
= B'.A' + B'.C' + A'.A'.C' + A'.C'.C' //DISTRIBUTIVE LAW
= B'.A' + B'.C' + A'.C' + A'.C' //IDEMPOTENT LAW
= B'.A' + B'.C' + A'.C' == Y
Hope this would help
@@elainhasheli8793 thanks you Elain
@@melthabah1158 My pleasure ❤️
1) self dual and
2) 256
self dual & 256
Sir i dint understand at number of self duals formula
H/W
1.Self dual
2.256
values of c'+c'=?
As, we know (A + A) = A
So, (C' + C') should also equal to C'
Isn't it..?
1. Self duel
2.256
Yep self dual 🙂
1. yes it is self dual
Self dual
And 256
Self dual
256.
1.ya it's self dual2.256
ans
1) self dual
2) 256
1.it is self dual
2.256
How it is self dual
yes it is self dual
Home work
1.self dual
2.256
What's A'.A' = ????
A'.A'=A'
this is because A' can be 0 or 1
and 0.0=0
1.1=1
2. 2^8=256 self dual form
Self dual & 256 expressions
Hi, when we convert the and expression to OR expression where negative logic handled ?
Or converted expression considered as negative expression
Little bit of confusion for 2) if its 4 then 4-1 is 3 so its 2 ^ 4 ^ 3 which is 4096?
2^(2^(4-1))
= 2^(2^3)
= 2^8
= 256
It is 2^(2^n), not 2^(n^n).
@@zaphyr1371 **2^(2^(n-1))
is it possible to simplify two times dual example...and get back the same function??
Can any one prove the result of self dual expression (2^2^(n-1)) ??
See the upper comments
1.self dual
2.256
Give more hw 😉
2. 256
for 4 variables I got only 8 self duals, like for variables A,B,C,D:- A',B',C',D'. Can u give some other examples
it is self dual
256
Self dual.
256
TAKE CARE FOR HW 2
(2^2)^3 = 2^6!! you multiply the expontenials. so the result is 64!!
just saw that many got it wrong in the comments.