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Explain the behaviour of the following logic circuit with level input $A$ and output $B$.

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2 Answers

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This is a sequential circuit (whose output depends not only on the present value of its input signals but on the sequence of past inputs) not a combinational one (whose output depends only on the present inputs), therefore solving using just input variable does not yields correct output.

First we need to simplify the circuit.

The two $\textsf{NOT}$ gates at the input end of the $\textsf{NOR}$ gate can be combined with the gate to get: $(A'+B')' = AB$

Now, since we have two variables we will have $4$ combinations $00\; 01\; 10\; 11.$

On analyzing each we will see that for every combination where

  • $A = 0$ we have the stable output of $0$
  • $A=1$ we will have a RACE condition
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There is a NOT after NAND i,e. It is AND.

So it will be A.B

and other side : $\bar{A}$

after NORING $\overline{AB+\overline{A}}$

$\overline{AB} + \overline{\overline{A}}$

$\bar{A} + \bar{B} + A = 1$

So Answer is 1.

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