For the Boolean function: F = xy’z
+x’y’z + w’xy + wx’y + wxy
a)determine the k-map
b)write the new reduced equation
c) write the corresponding truth table
a) Given F = xy’z+x’y’z+w’xy+wx’z+wxy

b)

The Reduced SOP is F = wz+y’z+xy
c) Truth Table:
|
w |
x |
y |
z |
y’ |
y’z |
xy |
wz |
F =wz+y’z+xy |
|
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
|
0 |
0 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
|
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
|
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
|
0 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
|
0 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
|
0 |
1 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
|
0 |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
|
1 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
|
1 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |
|
1 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
|
1 |
0 |
1 |
1 |
0 |
0 |
0 |
1 |
1 |
|
1 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
|
1 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |
|
1 |
1 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
|
1 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
For the Boolean function: F = xy’z +x’y’z + w’xy + wx’y + wxy a)determine the...
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F =
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