edited by
12,059 views
38 votes
38 votes

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 ?

  1. $0000$
  2. $0111$
  3. $1111$
  4. None of these
edited by

8 Answers

Best answer
89 votes
89 votes

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.


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.

selected by
11 votes
11 votes

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)

2 votes
2 votes

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).    

Answer:

Related questions

23 votes
23 votes
6 answers
1
Ishrat Jahan asked Oct 29, 2014
6,380 views
What is the final value stored in the linear feedback shift register if the input is $101101$?$0110$$1011$$1101$$1111$
32 votes
32 votes
3 answers
3
30 votes
30 votes
3 answers
4
Ishrat Jahan asked Oct 30, 2014
4,656 views
Consider the following expression$a\bar d + \bar a\bar c + b\bar cd$Which of the following Karnaugh Maps correctly represents the expression?