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in Digital Logic by Junior (805 points)
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Let us analyze NOR Gate.

Let A and B be two input lines to NOR gate.

We observe that if any input in NOR Gate is 1 then the output is 0 directly.

and Similarly for AND Gate if any input is 0 then the output is 0 directly.

Now let us analyze each option.

Option A : A = 0 and B = 0

When B = 0 is passed from AND Gate then Confirm output is 0 but due to NOT Gate in front of AND gate,  Q = 1 but Q = 0  will directly pass to one of inputs at NOR gate(see diagram in question) and when B=0 and Q=0 is input to NOR gate Confirm output P = 1

Option B : A = 1 and B = 1

When A = 1 is passed from NOR Gate then Confirm output at P = 0 and that output will directly pass to one of inputs at AND gate and when P=0 is input to AND gate -> Confirm output 0 but there is a NOT gate in front of AND gate which makes Q = 1.  

Option D: A =1 and B = 0

When A = 1 is passed from NOR Gate then Confirm output at P = 0 and that output will directly pass to one of inputs at AND gate and when P=0 is input to AND gate  ->Confirm output 0 but there is a NOT gate in front of AND gate which makes Q = 1.  

Option C: A = 0 and B = 1

When A = 0 is passed as input to NOR Gate then output is undecidable. It depends on the other input which actually is an output from AND Gate .

Similarly when B = 1 is passed from AND gate then output is undecidable, it depends on the other input which actually is an output from NOR Gate.

So P and Q in this case depends on what was the previous state P or Q value and is therefore undecidable.

by Active (1.8k points)
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classical explanation buddy
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@kamal pratap, they are asking for unreliable here, i didn't get it, how undecidable state is unreliable?
0
I meant to say that we cant decide the values unless we know the previous values of P or Q. Therefore, if you pass A=0 and B=1 then you cant say what might be the output. They are comparing with other options and it seems by choosing rest of the options one can decide what will be the output without knowing previous value of  P or Q state . its not the Turing machine decidability I was talking. It is in normal context.
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