What to do with inputs i1 and i2 ?

How should I group ? In kmap every adjacent coloum differ by one bit , so should I group only adjacent coloums ?

Im not able to understand this question.

46 votes

Consider a Boolean function $ f(w,x,y,z)$. Suppose that exactly one of its inputs is allowed to change at a time. If the function happens to be true for two input vectors $ i_{1}=\left \langle w_{1}, x_{1}, y_{1},z_{1}\right \rangle $ and $ i_{2}=\left \langle w_{2}, x_{2}, y_{2},z_{2}\right \rangle $ , we would like the function to remain true as the input changes from $ i_{1}$ to $ i_{2}$ ($ i_{1}$ and $ i_{2}$ differ in exactly one bit position) without becoming false momentarily. Let $ f(w,x,y,z)=\sum (5,7,11,12,13,15)$ . Which of the following cube covers of $f$ will ensure that the required property is satisfied?

- $ \overline{w}xz,wx\overline{y},x\overline{y}z,xyz,wyz$
- $ wxy, \overline{w}xz,wyz$
- $ wx\overline{y} \overline{z}, xz, w\overline{x}yz$
- $ wx\overline{y}, wyz, wxz, \overline{w}xz, x\overline{y}z, xyz$

5

Could someone explain me the question plz.. ?

What to do with inputs i1 and i2 ?

How should I group ? In kmap every adjacent coloum differ by one bit , so should I group only adjacent coloums ?

Im not able to understand this question.

What to do with inputs i1 and i2 ?

How should I group ? In kmap every adjacent coloum differ by one bit , so should I group only adjacent coloums ?

Im not able to understand this question.

4

Static Hazard - A static hazard is the situation where, when one input variable changes, the output changes momentarily before stabilizing to the correct value.

- Static-1 Hazard: the output is currently 1 and after the inputs change, the output momentarily changes to 0,1 before settling on 1.
- Static-0 Hazard: the output is currently 0 and after the inputs change, the output momentarily changes to 1,0 before settling on 0.

Dynamic Hazard - A dynamic hazard is the possibility of an output changing more than once as a result of a single input change

How to eliminate static hazard ?

Ans . refer(2) video.

As a rule, dynamic hazards are more complex to resolve, but note that if all static hazards have been eliminated from a circuit, then dynamic hazards cannot occur.

Notice -> Sometimes we can not fix static Hazards. And sometimes they may not impact circuit functionality.

I think $f(w,x,y,z) = xz+wxy' +wyz$ is also free from all kind hazards.

Refer ->

- https://www[dot]youtube.com/watch?v=s8qF3p7bjlI
- https://www[dot]youtube.com/watch?v=fUTCtn_b4qs
- https://www[dot]youtube.com/watch?v=6NaMSJq4SVA
- https://en.wikipedia.org/wiki/Hazard_(logic)

Please notify if anything is not proper.

**ping @****rahul sharma 5**, @**habedo007, **@Orochimaru @**sushmita** @**hem chandra joshi @****gari** and @**Paras Nath**

52 votes

Best answer

The question is indirectly asking for static-1 hazard in the circuit - that is output becoming $0$ momentarily when it is supposed to be $1.$

Here $f(w,x,y,z) = \sum (5,7,11,12,13,15)$

So, $K$-map will be

So, its minimized sum of products expression will be $xz + wxy' + wyz$. Since all the minterms are overlapping, there is no chance of static hazard here.

Now, lets consider the options one by one:

A. $\overline{w}xz,wx\overline{y},x\overline{y}z,xyz,wyz$

Chance of static hazard.

Here, when $y$ changes from $0$ to $1$, the gate for $wyz$ should give $1$ (from earlier $0,$ assuming $w=z=1$) and that of $xy'z$ should give $0$ (from earlier $1$). But there is a possibility of circuit giving $0$ (static $1$ hazard) momentarily due to gate delays ($xy'z$ coming first and $wyz$ coming later). In order to avoid this, we must add a gate with $wxz$ also which ensure that all adjacent blocks in $K$-map are overlapped or a single variable change cannot momentarily change the circuit output.

B. $wxy, \overline{w}xz,wyz$

This is not correct as $wxy$ is not a minterm for the given function

C. $wx\overline{y} \overline{z}, xz, w\overline{x}yz$

Here, also static-1 hazard is possible as the middle $4$ pairs are separated by $1$ bit difference to both $wxy'z'$ as well as $wx'yz$. Could have been avoided by using $wxy'$ instead of $wxy'z'$ and $wyz$ instead of $wx'yz$ which will ensure that all neighboring blocks are overlapped.

D. $wx\overline{y}, wyz, wxz, \overline{w}xz, x\overline{y}z, xyz$

These minterms cover all the minterms of $f$ and also, all the neighboring 1's are overlapped by minterms. So, no chance of hazard here and hence is the required answer.

Correct Answer: $D$

2

IMHO, the question means that when there is a 1-bit change from 0 to 1 or vice-versa, the state of circuit/bit is indeterminate because it has left one of the stage and not reached the other for that *fraction-of-second/moment. *When the circuit does not understand whether it is 1 or 0, it declares it as false.

–1

@Sonam,Is D the right answer?You mentioned that there is no one bit difference?Can you please tell what is the correct answer?And I have seen at some places the first term in D option is not there,and with remaining 5 terms there is no static hazard as all adjacent blocks are overlapped.

What do you mean by comment

"and in option d ) which have 5 correct term but one is not matched ... in fact that is not 1 bit difference ...."

Please help

1

@Rahul, the answer was later edited by me and hence the comments don't make sense now. Everything is explained now in answer- there is no use in saying what is the "correct answer" unless you yourself get to know the reason. (GATE is not ISRO exam where they copy the same question and options again).

The question paper was taken from here:

https://drive.google.com/file/d/0B8_aYGBndW4HbHBnM05WTXZjaDg/view

33 votes

Given that, function to remain true as input changes from i_{1} to i_{2} and i_{1,} i_{2} differ in exactly one bit position.

We know that adjacent cells in kmap differ by exactly 1 bit. So, we have to group cells in function represented by kmap such that all adjacent cells are grouped as below.

all these 6 groups gives :

17 votes

I found a better way to answer this question .

Go according to the Question . i1 and i2 differs in exactly one bit position without becoming false momentarily.

Go to option (a)

**Draw K map and find the binary covered by implicant**

w'xz =5,7 0101 , 0111 here you will see it is one bit different

*w**x**y*'=12,13 1100,1101 here you will see it is also one bit different

*x**y*'*z**=5,13 0101,1101 *here you will see it is one bit different

*xy**z*=7,15 0111, 1111 here you will see it is one bit different

*wyz=11,15 1011,1111 *here you will see it is one bit different

Option (a) misses wxz possible one bit change.So it is false.

(b)

wxy = 15,14 but 14 is not covered by K map so b is wrong option

(c)

wxy'z' = 12 =1100 here there is not bit change so it is wrong option

(d)

*wyz=11,15 1011,1111 *here you will see it is one bit different

*wxz=13,15 1101,1111 *here you will see it is one bit different

w'xz =5,7 0101 , 0111 here you will see it is one bit different

*x**y*'*z**=5,13 0101,1101 *here you will see it is one bit different

*xy**z*=7,15 0111, 1111 here you will see it is one bit different

*w**x**y*'=12,13 1100,1101 here you will see it is also one bit different

So option d satisfy whole f so it is Answer .