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Math problem

A logical circuit was given, I had to create a boolean expression from it; which I did correctly (I guess). I calculated the output of each gate separately, considering the variables.

From the given circuit I got this expression:

('xy + x) + ((x + y) + z) * 'u

I than would need to put this in a Karnaugh's map to simplify the expression (and later on the circuit). But this is the step I got stuck on.

How to put that expression in Karnaugh's map? Even when I perform algebra on the current expression I get terms which aren't present in the map, or single terms like only 'x', but 'only x' ain't present in the map.

I have already converted expressions to Karnaugh map with no problems, but I guess I have difficulty now because it's with 4 variables, and a tough expression.

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  • $\begingroup$ I think if you construct a logic table by using the final expression you get, putting the 1's and 0's to Karnaugh Map could be easier. But in this case, I think you can do it by checking the cases when value of your final expression is $1$. $\endgroup$
    – ArsenBerk
    Commented Dec 14, 2017 at 21:37

2 Answers 2

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HINT

To put a term like $x$ in the map, simply put a $1$ in all the cells where $x$ is true.

So, in your kmap, that would be the top four cells as well as the bottom four cells.

Also, to find the terms to put into the map, it helps to first put it into DNF:

$$x'y+x+((x+y)+z)u'=x'y+x+xu'+yu'+zu'$$

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Sorry, I couldn't send any photo.
Your function is false when $xyzu = {0000, 0001,0011}$.
First and second columns are when $x$ = true (1), another columns are false (0).
Second and third columns are when $y$ = true (1), another columns are false (0).
First and second lines are when $z$ = true (1), another lines are false (0).
Second and third lines are when $u$ = true (1), another lines are false (0).

$$1|1|1|1$$ $$1|1|1|0$$ $$1|1|1|0$$ $$1|1|1|0$$

As a result, $$f = x \lor y \lor \overline uz$$

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