The SOP Karnaugh map (map of 1's) for a Boolean function can be described as . Note that for the set of variables in the Boolean expression, and can never be 1 at the same time. Plot the 1 's on the K-map below, use the extra information, and develop the simplified SOP expression, and draw only the simplified circuit.

Sol2:

To plot the 1's on the K-map, we start by creating a table with the inputs of the Boolean function f and its output (1's):

x	y	z	f
0	1	0	1
1	0	1	1
1	1	0	1
1	1	1	1
0	1	1	1

Next, we will plot the 1's on the K-map, using the grouping rule:

	00	01	11	10
0
1	X	X	X	X

We can see that all 1's are grouped in a single group that covers the cells 01, 11, 10, and 00. The grouping rule is satisfied because we can draw a rectangle that encloses all 1's in the group, and each cell in the group is adjacent to at least one other cell in the group.

The simplified SOP expression can be obtained by reading the product terms corresponding to the cells in the group, which are:

f = y'z + x'y

To draw the simplified circuit, we first need to apply De Morgan's theorem to obtain the simplified POS expression:

f' = (y+z')(x+y')

Then we can draw the circuit using AND, OR, and NOT gates:

_____

x ----| |

| AND |-- y'

z ----|_____|

|

|

__|__

| |

f' --| OR |-- f

|____|

|

|

___

_| |

y' --| NOT|

|_____|

Finally, we can obtain the simplified circuit for f by complementing the output of the simplified circuit for f':

_____

x ----| |

| AND |-- y'

z ----|_____|

|

|

___

_| |

y' --| NOT|

|_____|

|

|

__|__

| |

f ---| OR |

|_____|