Can Karnaugh Maps Simplify Digital Logic Expressions Without Static Hazards?

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The discussion focuses on simplifying the digital logic expression f = w' z' + w' xy + wx' z + wxyz using Karnaugh Maps (K-Maps). The initial simplification attempts resulted in 14 operations, but further analysis revealed that the expression could be simplified to f = w' z' + xyz + wx' z + w' xy + wyz without static hazards. It was emphasized that to avoid static hazards, adjacent but disjointed regions on the K-Map should not be circled. Additionally, the complexities of static hazard theory in practical applications were highlighted, noting variations in gate delays and the challenges in achieving hazard-free designs. Ultimately, the conversation underscores the importance of careful K-Map analysis to ensure both simplification and hazard elimination.
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Homework Statement


Draw the schematic circuit diagram that implements the following expression using as few basic gates as possible (AND, OR, NOT, XOR, NAND, NOR).

The prime denotes the complement:

f = w^\prime z^\prime + w^\prime xy + wx^\prime z + wxyz

The Attempt at a Solution



From the truth table, I drew a Karnaugh Map (attached and linked to below. I'm just wondering whether I did it right. From the K-Map, the resulting simplification is:

f = w^\prime z^\prime + w^\prime xy + wx^\prime z + wyz

Is there no further simplification?

http://img340.imageshack.us/img340/9333/hw14bkmapwm3.th.png
 

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f = w^\prime z^\prime + w^\prime xy + wx^\prime z + wxyz

The original equation has 15 operations, removing that last x gives you 14. the following gives you 11. There might be other things you could do, but I don't see any right off.

s = xy

f = w^\prime ( z^\prime + s) + w (x^\prime z + sz)
 
cepheid said:
From the truth table, I drew a Karnaugh Map (attached and linked to below. I'm just wondering whether I did it right. From the K-Map, the resulting simplification is:

f = w^\prime z^\prime + w^\prime xy + wx^\prime z + wyz

Is there no further simplification?

If you circled the regions on the map like this:
http://img522.imageshack.us/img522/3419/3termswhazardqs5.th.jpg
you would have obtained a simpler expression:
f = w^\prime z^\prime + xyz + wx^\prime z

BUT this expression, as well as yours, has static hazard.

To eliminate the static hazard there must not be any pair of adjacent but disjointed regions circled on the map. So you must circle the regions like this:
http://img20.imageshack.us/img20/5679/3termswohazardtv7.th.jpg
which will give you the simplest expression without hazards:
f = w^\prime z^\prime + xyz + wx^\prime z + w^\prime xy + wyz
 
Last edited by a moderator:
antonantal said:
If you circled the regions on the map like this:
http://img522.imageshack.us/img522/3419/3termswhazardqs5.th.jpg
you would have obtained a simpler expression:
f = w^\prime z^\prime + xyz + wx^\prime z

BUT this expression, as well as yours, has static hazard.

To eliminate the static hazard there must not be any pair of adjacent but disjointed regions circled on the map. So you must circle the regions like this:
http://img20.imageshack.us/img20/5679/3termswohazardtv7.th.jpg
which will give you the simplest expression without hazards:
f = w^\prime z^\prime + xyz + wx^\prime z + w^\prime xy + wyz

Just to expand on this a bit, the thing about static hazard theory is that it is difficult to match with practice. This is because each type of gate has a different delay, each type's delay varies differently with temperature, and runs have different delays. Also, the above definition of static hazard either assumes NOT operations are infinitely fast, or that you buffer all signals with either an inverter a non-inverter but never both and assume their delays are equal. In practice, edge sensitive logic is either made synchronous (clocked), run lengths are adjusted accordingly and fingers are crossed, or off-the-shelf programmable ICs are used which are hopefully fairly hazard-less (and fingers are still crossed).
 
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