K Maps

Example: K MAPS FACTS

Q- Simplify the following Boolean function in (a) sum of products form (SOP)

(b) Product of sums form (POS)

F(x, y, z, w) = ∑(0, 1, 2, 5, 8, 9, 10)

Ans: We mark 1s in the squares corresponding to the terms present in the function and 0s for the terms missing from the function as follow:

1s marked represent min terms and 0s represent max terms

  1. Combine 1s and make groups of adjacent squares.

This is the wrong way of grouping as we always have to make largest groups.

So the correct way of grouping is as follow:

All the corner squares form one group of 4

Also rightmost two squares of top and bottom row form one group of 4

And third we have a group of 2

From this map table we get the simplified expression as F= y’w’+y’z’+x’z’w and so we get SOP form

  1. Combine 0s and try to make groups of maximum size

This is the wrong way of grouping as we always have to make largest groups.

So the correct way of grouping is as

From this map, we get the simplified expression of compliment function as  F’= xy+zw+yw’ in SOP form

Now take compliment of the function to get POS form

F= (xy+zw+yw’)’ = (x’ + y’) (z’ + w’) (y’ + w)

This is the required POS form.

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