JEE MainPhysicsSemiconductors
In a digital logic circuit, two inputs A and B are connected to an OR gate and a NAND gate. The outputs of these two gates are then fed into an AND gate to produce the final output X . The inputs A and B are provided as time-varying square waveforms. Over four consecutive time intervals, the sequence of logic states for A is 0, 1, 1, 0 and for B is 0, 0, 1, 1 . The corresponding sequence of logic states for the outpu
Options
- A0, 1, 0, 1
- B0, 1, 1, 1
- C0, 0, 1, 0
- D1, 0, 1, 0
Correct answer
A. 0, 1, 0, 1
Step-by-step solution
Let the output of the OR gate be Y₁ and the output of the NAND gate be Y₂ . Y₁ = A + B Y₂ = A B These outputs are fed into an AND gate, so the final output X is: X = Y₁ Y₂ = (A + B) ( A B ) Applying De Morgan's laws and distributive property: X = (A + B) ( A + B ) = A A + A B + B A + B B = A B + A B This is the Boolean expression for an XOR gate. The output X is 1 if and only if A and B are different. Let us evaluate X for each time interval: Interval 1: A = 0, B = 0 X = 0 Interval 2: A = 1, B = 0 X = 1 Interval 3: