in Digital Logic
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61 votes
61 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? 

  1. $ \overline{w}xz,wx\overline{y},x\overline{y}z,xyz,wyz$
  2. $ wxy, \overline{w}xz,wyz$
  3. $ wx\overline{y} \overline{z}, xz, w\overline{x}yz$
  4. $ wx\overline{y}, wyz, wxz, \overline{w}xz, x\overline{y}z, xyz$
in Digital Logic
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4 Comments

Before going through best answer , read this link

https://www.geeksforgeeks.org/digital-logic-static-hazards/

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This Geeksforgeeks link gives the most concise explanation. For those wondering, this topic is also present in the book by Morris Mano 3rd edition (it seems to have been removed from the 5th edition), under the chapter Async Sequential Circuits - Hazards in Combinational Circuits.

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Isn’t it asking for monotonic property, as in this question https://gateoverflow.in/8294/gate2015-1-39

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4 Answers

74 votes
74 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 $1.$

$\text{Static 1 Hazard:}$ Output going to $0$ when it should remain $1$

  • Happens in a $2$ level $\text{SOP}$ implementation.
  • Suppose only one $\text{AND}$ gate (minterm) is $1$ in an $\text{SOP}$ implementation.
  • A variable in that minterm changes causing the output of that $\text{AND}$ gate to become $0$ and another AND gate to be $1.$
  • Depending on the propagation delay of the gates, the output can momentarily become $0$ before finalizing on $1.$ For example, consider $f = a’b’ + ac$ and initially $a = 0,b = 0, c = 1,$ so that output is $1.$ Now, if $a$ changes to $1,$ again output should be $1$ due to $ac$ term. But if $a’b’$ turns to $0$ before $ac$ turns to $1,$ output can be momentarily $0$ causing a static- $1$ hazard.
  • Can be detected in a K-map if there are any adjacent $1$'s not covered by an implicant. i.e., to avoid static hazard, all adjacent $1$’s in a K-map must be covered by some implicant. In the below K-map, the implicant shown in green ensures no static-$1$ hazard.   

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

So, $K$-map will be 

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

Now, let's 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.

edited by
by

4 Comments

Done (y)
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arjun sir now it is clear ,nice edit sir
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what is the significance of two inputs in this question . I did not understand . Can any one please explain me .
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39 votes
39 votes

Given that,  function to remain true as input changes from i1 to i2 and i1, i2 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 : 

3 Comments

Nice Explanation!@Orochimaru 

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But, if the option had been xz+wxy′+wyz, then it also would have been correct, right?
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19 votes
19 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

wxy'=12,13   1100,1101   here you will see it is also one bit different

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

xyz=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

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

xyz=7,15 0111, 1111 here you will see it is one bit different

wxy'=12,13   1100,1101   here you will see it is also one bit different

So option d satisfy whole f so it is Answer .

edited by

1 comment

Nice explanation :)
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4 votes
4 votes

 

According to the question we have to find all the pairs out of the given minterms where only 1 bit changes. 

So first of all we have to put all the given minterms i.e 5,7,11,12,13,15 in K- map. It will be look like this $:$ 

 

 

 

Option B.   wxy, w'xz, wyz 

wxy →  1110 (14) and 1111 (15) 

w'xz →  0101 (5) and 0111 (7) 

wyz →  1011 (11) and 1111 (15) 

But in the given minterms 14 is not there, so this option is Wrong. 

 

Option C.   wx(yz)', xz, wx'yz 

wx(yz)' →  wx(y' + z') → wxy’ + wxz’

wxy' → 1100 (12) and 1101 (13) 

wxz' →  1100 (12) and 1110 (14) 

As again the minterm is 14 coming. So no need to check further. this option is also Wrong 

 

Option A. w'xz, wxy', xy'z, xyz, wyz

If we put this into k map we will find only one pair (wxz) is missing. So this is also not covering all the pairs. So wrong. 

Here is the K map for this: 

 

 

Option D.   w'xz, wxy', xy'z, xyz, wyz, wxz

Similarly put all this into k map, we will find all the pairs are covering.  

Here is the K map for this: 

 

So, Option D is the correct ans. 

edited by
Answer: