Boolean Algebra Function

Question

Implement the following Boolean function with Decoder and external logic gates as necessary. Draw the logic diagram and label all input and output lines.

F = XYZ + (X' + Z')

I am having trouble converting the Function above to K-maps. I believe once i understand that, I will be able to finish my problem on my own.

score 0 · Accepted Answer · answered Oct 15 '14 at 07:11

The parenthesis are not necessary, since 'or' is associative, so we have:

F = XYZ + X' + Z'

So the K-map would look something like:

    XY  XY' X'Y X'Y'
   +---+---+---+---+
Z  |   |   |   |   |
   +---+---+---+---+
Z' |   |   |   |   |
   +---+---+---+---+

We start out with the XYZ term, which occupies only one square since it fixes the value of all three variables:

    XY  XY' X'Y X'Y'
   +---+---+---+---+
Z  | T |   |   |   |
   +---+---+---+---+
Z' |   |   |   |   |
   +---+---+---+---+

Then X' touches one half of the table:

    XY  XY' X'Y X'Y'
   +---+---+---+---+
Z  | T |   | T | T |
   +---+---+---+---+
Z' |   |   | T | T |
   +---+---+---+---+

And finally Z' touches the bottom row, and only XY'Z is left as false:

    XY  XY' X'Y X'Y'
   +---+---+---+---+
Z  | T | F | T | T |
   +---+---+---+---+
Z' | T | T | T | T |
   +---+---+---+---+

Boolean Algebra Function

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