GATE IT 2007 | Question: 45

Question

The line $T$ in the following figure is permanently connected to the ground.

Which of the following inputs $(X_1 X_2 X_3 X_4)$ will detect the fault ?

• A. $0000$
• B. $0111$
• C. $1111$
• D. None of these

Comments

@bikram sir

what is meant by fault here ?

Answer 1

To detect the fault, we should get an unexpected output. The final gate here is a NOR gate which produces output 0 if either of its input is 1 and else 1. i.e., the output will be 0 for inputs $(0,1), (1,0)$ and $(1,1)$ and output will be 1 for (0,0).

By grounding $T$ is at 0. So, we can ignore the inputs $(1,0)$ and $(0,0)$ to the final NOR gate as they won't be detecting faults. Now, expected $(1,1)$ input will become $(1,0)$ due to grounding of $T$ but produces same output $0$ as for $(1,1)$. Hence this also cannot detect the defect. So, to detect the defect, the input to the final gate must be $(0,1)$ which is expected to produce a 0 but will produce a 1 due to grounding of $T$.

Now, for $(0,1)$ input for the final gate, we must have,

$X_3 = X_4 = 1$

But if $X_4=1,$the OR gate makes 1 output and we won't get $(0,1)$ input for the final gate. This means, no input sequence can detect the fault of the circuit.

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Alternatively, we can write equation for the circuit as

$(((x_1.x_2.x_3)' + x_4) + x_3.x_4)' \\= ((x_1.x_2.x_3)' + x_4)' . (x_3.x_4)' \\=x_1.x_2.x_3 . x_4' . (x_3'+x_4') \\= x_1.x_2.x_3 . x_4'$

For the faulty circuit output will be

$((x_1.x_2.x_3)' + x_4)'  = x_1.x_2.x_3 . x_4'$

So, there is no effect of $T$ being grounded here. Answer is D option.

Comments

Arjun sir:-

You mentioned  Now, expected (1,1)(1,1) input will become (1,0)(1,0) due to grounding of TT but produces same output 00 as for (1,1)(1,1).

1,1 should be changed to 1,0 also and output wil be 1,how can it will give 0?

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Arjun Sir:-

You mentioned :-

Now, expected (1,1) input will become (1,0) due to grounding of T but produces same output 0 as for (1,1).

1,1 will become 1,0.How can this produce 1 output?

(0,1),will also get changed to 0,0 ,so how can we use this to detect fault?

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@Arjun sir

Can you please explain the last line

For the faulty circuit output will be

((x1.x2.x3)′+x4)′=x1.x2.x3.x′4

?

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Let the upper input terminal of the last NOR gate be A and the lower be B.
Since T is connected permanently to the ground so the NOR gate will receive (A and 0) and will output (A+0)’=A’.
But if it was not connected to the ground i.e. if the connections were not faulty then Nor gate would have given (A+B)’.
To detect the fault we need to see when the difference between A’ and (A+B)’ appears.

• if A=0 and B= 0

           A’=1 and (A+B)’=1 …so can’t detect the diff.

• if A=1 and B=0

          A’=0 and (A+B)’=0 …again can’t detect the diff

• if A=1 and B=1

          A’=0 and (A+B)’=0 …again no

• if A=0 and B=1

          A’=1 and (A+B)’=0 …YES!!

So we need to create a condition where A=0 and B=1.
Now what is B? B=x3x4 so for B to be 1, x3=1 and x4=1
A= (x1x2x3)’+x4. Since x4=1 so A will always be 1. But we need A=0 to detect the fault.
So it can’t be detected. Hence it should be D.

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great explanation @Arjun sir

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thank you .. really it's brief and to the point , great help

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final output=$x_{1}.x_{2}.x_{3}.x'_{4}$ it will be one only when x3=1 and x4=0 but there is no such any combination so answer is (d) is my argument correct????

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Thanku

Answer 2

I'm going to answer question according to image in my question. ( I suspect image might be wrong, in that case answer will change. )

Stuck at 0 -> Stuck at 0 means that line is always one, no matter what input is given to it.

Here T is stuck at 0 , it means that  whatever input given to x3 & x4 value of T does not change.

T = x3 AND x4.

Now we will give 1 to x3 & 1 to x4. In that scene T should be 1, as we are suspecting T might be stuck at 0, we are trying to make it 1 with input.

Now we have NOR GATE at the gate, which have only one input, T. NOR Gate act likes as inverter here.

T	Output
0	1
1	0

So we have given input to x3 & x4 so that T should be 1. So if line is not stuck at 0, T =1 & we get 0 as output.

If line is stuck at 0, then we will gate T = 0, & 1 as output.

Now output is independent of x1 & x2.

SO we have total 4 test cases. Only x3 & x4 should be set to 1.

So here B) & C) are both answers.

(Here I've solved the question according to current diagram available, in case diagram is changed , then answer might be different but method will be same)

Comments

Output of OR gate is connected to NOR in actual question! Anyways! Using this method I got answer! Thanks :)

Answer 3

The final gate is a NOR gate. 

NOR gate works as follows-
1. The output of a NOR gate is 0 if any one of its input is 1. 
2. The output of the NOR gate is 1 only if both the inputs are 0.

In order to detect the fault in line $T$, we would provide such an input combination to $X_{1}, X_{2}, X_{3}, X_{4}$ which makes the effect of the top input line to NOR gate negligible. This is possible only if the top input line has  an input 0, i.e.  $(X_{1}.X_{2}.X_{3})' + X_{4} = 0$, i.e. $X_{1} = 1$ and $X_{2} = 1$ and $X_{3} = 1$ and $X_{4} = 0$.

Now, the final output of the NOR gate only depends on its bottom input line or line $T$.
If we apply an input 1 to the bottom input line, the final output of the NOR gate should ideally be 0. On the other hand, if we apply an input 0 to the bottom input line, the final output of the NOR gate should be 1. However, as $T$ is permanently connected to the ground, the input to this (bottom input) line will always be 0. 
Therefore, we will be able to detect the error by providing an input 1 to this line, i.e. $X_{3}.X_{4} = 1$, i.e. $X_{3} = 1$ and $X_{4} = 1$. But this is not possible as we already made $X_{4} = 0$ in order to make the effect of top input line negligible.

Therefore, no input combination will be possible to detect this error. 

Hence, the correct option is Opt (D).    

Comments

Can you please explain me what we consider fault as here? I am not getting what is fault here?

Answer 4

ans d)

Comments

How?

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Thanks