In OR gate we need atleast one bit to be equal to 1 to generate the output as 1 because OR means any of the condition out of two is equal to 1 which means if atleast one input is 1 then it shows output as 1 . Number of 1's in input may be more than one but the output will always be 1 in OR gate. So the answer is 'C'.
2: The time required for a gate or inverter to change its state is called
Y=A+B. This is the equation of OR gate. We require 3 NAND gates to create OR gate. We can also write
After 1st NAND operation
Y = (A AND B)'
Y= A' + B' (Demorgan's Law)
After 2nd NAND operation
Y= ( A' + B')'
Y= A . B (Demorgan's Law)
After 3rd NAND operation
Y= ( A . B )'
Y= A' + B ' (Demorgan's Law)
Design and synthesis of combinational and sequential circuits
Number representation and computer arithmetic (fixed and floating point)
Propositional (Boolean) Logic
Satisfiability and Tautology
Logic Families: TTL, ECL and C-MOS gates.
Boolean algebra and Minimization of Boolean functions
Flip-flops-types, race condition and comparison.
Design of combinational and sequential Circuits
Representation of Integers: Octal, Hex. Decimal and Binary.
2's complement and 1 's complement arithmetic
Floating point representation.
This Section covers Digital Logic Questions Answers .
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