Digital Electronics - Boolean Algebra and Logic Simplification - Discussion
Discussion Forum : Boolean Algebra and Logic Simplification - General Questions (Q.No. 19)
19.
When 
are the inputs to a NAND gate, according to De Morgan's theorem, the output expression could be:
are the inputs to a NAND gate, according to De Morgan's theorem, the output expression could be:X = A + B
X = (A)(B)
Discussion:
8 comments Page 1 of 1.
Mayur said:
4 years ago
(A'. B') ' = (A') '+ (B') ' = A+B.
(1)
Sanjeev said:
1 decade ago
If x,y are the inputs for NAND gate, the output is xy bar.
So, if A bar and B bar are inputs, then output is A bar B bar whole bar, which implies A+B.
So, if A bar and B bar are inputs, then output is A bar B bar whole bar, which implies A+B.
Komal said:
1 decade ago
A AND B = A*B
A NAND B = BAR(A*B)
BAR(A) NAND BAR(B) = BAR[BAR(A)* BAR(B)]
According to DeMorgan's Theorem:
BAR[BAR(A)* BAR(B)] = BAR[BAR(A)] + BAR[BAR(B)]
According to Double negation Theorem:
BAR[BAR(A)] + BAR[BAR(B)] = BAR(A) + BAR(B)
A NAND B = BAR(A*B)
BAR(A) NAND BAR(B) = BAR[BAR(A)* BAR(B)]
According to DeMorgan's Theorem:
BAR[BAR(A)* BAR(B)] = BAR[BAR(A)] + BAR[BAR(B)]
According to Double negation Theorem:
BAR[BAR(A)] + BAR[BAR(B)] = BAR(A) + BAR(B)
Chandu said:
1 decade ago
For AND Gate A AND B = A.B.
For NAND Gate A NAND B = BAR(AB).
Inputs are Bar(A) NAND Bar(B) = BAR(BAR(A)+BAR(BAR(B)) = A+B.
For NAND Gate A NAND B = BAR(AB).
Inputs are Bar(A) NAND Bar(B) = BAR(BAR(A)+BAR(BAR(B)) = A+B.
PRATYAKSH said:
10 years ago
A' and B' through NAND gate. (a'*b')' = ((a+b)')' that is equals to a+b.
Amuthan said:
9 years ago
(A.B)' = A' + B'
_Meth_ said:
6 years ago
(A'.B')' = (A')'+(B')' = A+B.
Mayur said:
4 years ago
Bar A Nand Bar B equals to bar (barA . barB).
By Demorgan's theorem;
Bar (barA . barB) OR (A'.B')' = bar ( bar A + bar B) or (A'+B')' = A+B.
By Demorgan's theorem;
Bar (barA . barB) OR (A'.B')' = bar ( bar A + bar B) or (A'+B')' = A+B.
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