Hi,

I have a boolean function that im not sure how to solve.

im supposed to write this on disjunctive normal form.

I tried multiplying the and the term

by (z+z') and (y+y') but that makes the whole expression really long.

Any ideas?

Jones

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- Nov 19th 2009, 07:11 AMJonesBoolean algebra help
Hi,

I have a boolean function that im not sure how to solve.

im supposed to write this on disjunctive normal form.

I tried multiplying the and the term

by (z+z') and (y+y') but that makes the whole expression really long.

Any ideas?

Jones - Nov 19th 2009, 09:41 PMqmechA start to rewriting this
I assume (x+z')' means not ( x or (not z)). If so, then doesn't this simplify to

x'z? Then your expression simplifies to:

= x'zy'z + xy' + xz'

= x'y'z + xy' + xz' - Nov 20th 2009, 10:53 AMJones
No, thats incorrect.

The correct answer should be - Nov 20th 2009, 10:59 AMqmechA little more manipulation
Note:

xy' = xy'z + xy'z'

because z + z'=1 - Nov 21st 2009, 01:50 PMJones
So,

Simplifying the first two paranthesis

Should simplify down to just

Applying Morgans law gives:

Using de morgans again gives:

The whole expression now becomes

Is this correct? - Nov 21st 2009, 02:43 PMmosta86
you need my advice it is a 3 literals function so it is easy to build the truth table and see where the function set to 1 then build the karnough map and you will get the most simplified form

- Nov 21st 2009, 02:50 PMmosta86
the simplified form is z'x+zy'

waiting your reply :)

truth table :

X Y Z F

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 0

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 0

K-map :

xy 00 01 11 10

z 0 0 0 (1 1)

1 1) 0 0 (1

=> z'x+zy'