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

Hello, 😃

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@js__ How is the fault detection condition known to be 1?

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Answer : D

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

Excellent explanation @MiNiPanda mam...really helpful

Thanks

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There is no such combination

To detect fault you must need to do alter the output before and after grounding

Before T is grounded

output of OR	output of AND	output of NOR
0	0	1
0	1	0
1	0	0
1	1	0

 

After grounding T

output of OR	output after AND	output of NOR
0	0	1
0	0	1
1	0	0
1	0	0

 

See, we have only single combination of inputs for which we can see this alteration and that is 01(after grounding) but we can't make these input combination to NOR gate with the given circuit.

if You try to make input 01 to NOR, inputs to circuit should be $x_1x_2x_3x_4 = d d 1 1$ to get output 1 after AND and becuse $x_3$ is 1 you will definitely get output 1 after OR gate.

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thank you for such a clear and well stated answer to both @MiNiPanda mam and @Arjun sir

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

Correct Circuit:

$f = [\{(x_1x_2x_3)’ + x_4\}’ + x_3x_4]’ = [x_1’ + x_2’ + x_3’ + x_4 + x_3x_4]’ = [x_1’ + x_2’ + x_3’ + x_4]’ = x_1x_2x_3x_4’$

Faulty Circuit:

$f = [\{(x_1x_2x_3)’ + x_4\}’ + \mathbb{T}]’ = [\{(x_1x_2x_3)’ + x_4\}’ + 0]’ = x_1x_2x_3x_4’$

which is the same as the correct circuit.

$\therefore$ There is no way to detect the fault.

ANS: (D.)

Answer 4

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 5

ans d)

Comments

How?

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Thanks

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No explanation direct correct option nice..

Answer 6

We consider circuit in two different configurations to check the inputs. And then apply the three available inputs to get output values. On doing so, we observe that all inputs give same outputs in both the configurations. Thus None of these is the answer.