# GATE1989-4-ix

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Provide short answers to the following questions:

Explain the behaviour of the following logic circuit (Fig.4) with level input A and output B

9

1st NAND gate = $\overline{(A\cdot B)}$
then NOT = $A\cdot B$

2nd NOT = $\overline{A}$

then NOR = $\overline{(A\cdot B)+\overline{A}} = \overline{B+\overline{A}} = A \cdot \overline{B}$

this shows
when A = 0  output is 0
when A = 1 output is toggled i.e,
$\overline{B}$

correct me if i m wrong??

1
You are treating it as a combinational ckt. U didnt account for the feedback of the output .
0
Please, someone, explain this question in brief.

This is a sequential Ckt not a combinational one, therefore solving using just input var doesnt yeilds correct output.

First we need to simplify the ckt.

The two not gates at the input end of the nor gate can be combined with the gate to get : (A'+B')' = AB

Now, since we have two Vars we will have 4 combinations 00 01 10 11

On analysinh 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.

There is a NOT after NAND i,e. It is AND.

So it will be A.B

and other side : $\bar{A}$

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

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

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

edited
1
Same mistake as above.

Treating a sequential ckt as combinational. Your answer is correct if it had been a combinational ckt just dependent on inputs for all cycles.

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