GATE CSE 1999 | Question: 2.8

Question

Consider the circuit shown below. In a certain steady state, the line $Y$ is at $'1'$. What are the possible values of $A, B$ and $C$ in this state?

• A. $A=0, B=0, C=1$

• B. $A=0, B=1, C=1$

• C. $A=1, B=0, C=1$

• D. $A=1, B=1, C=1$

Comments

What is feedback on Y indicating...?Do we need to take into account...since it is a combinational circuit

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Feedback on Y(Next state) is represented as y(present state) as input. and This is  asynchronous sequential logic circuit.

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Answer could be either (B) or (D)  

Since what i got AssuMing as NOT gate

B'C=0    { B=0,C=0 OR  B=1,C=0 OR B=1,C=1}

A Could be anything bcoz we got 1+A'

Answer 1

The figure is not clear- I assume there is a NOT gate just before taking Y making the final AND gate a NAND gate.

We have a steady state- meaning output is not changing. Y is 1 and remains 1 in the next state(s). So, we can write

$Y = \overline { \overline{(\overline{(AY)}. B )} . C} $

$ 1 = \overline {A} . B  + \overline{ C}$

So, $C = 0$ or  $\overline{A} . B = 1$

So, option B is TRUE.

Comments

No @Pranavpurkar The answer here is correct. It will be only Option B, because there is a NOT GATE before Y.

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There is NO need to waste time thinking if or if not to consider “that dot” before Y as NOT gate or not. Those days, GATE was subjective exam, & we could have written our assumption while writing the answer.

Just understand when the answer will be B & when it will be A,C,D.

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@Deepak Poonia

okay sir ! Thankyou.

Answer 2

Y = ((AY)'B)'C = ((AY)'' + B')C = (AY + B')C
Y is 1 given.. then
C should be 1and either A is 1 or B is 0.
except B all are true.

Comments

I don't think option A is correct.

As C should be 1 , the above circuit will become SR Latch with NAND Gate

Y = S'+ Ry    [Qn+1 = S'+ RQn] where S'R' should be 0 .

Answer 3

the simplified expression is :-  ((AY')B)'C = Y'

therefore,C=1 ,Y=1 can be deduct from the problem .

now the more simplified expression becomes :-

((A'+Y)B)' = 0

or,((A'+Y)'+B') = 0

or, AY'+B' = 0

therefore B should be 0 to get a 1 in the last or operation as Y is 1.therefore options (a) & (C) are correct.