SHEET 2
Digital Electronics
1. Prove the following by using Boolean algebra rules:
- xy + x’z + yz = xy + x’z
- x’y’z + yz + xz = z
- (x + y)[x’(y’ + z’)]’ + x’y’ + x’z’ = 1
- w’x + wxz + wx’yz’ + xy = x(w’ + z) + wyz’
2. Using the rules of Boolean algebra, simplify the following Boolean
expressions:
- {[(xy)’x]’[(xy)’y]’}’
- {[x’yw’ + xwz]’ + [(xwz)’ + y’w’z’ + ywz’]’}’
- (x + y)’(x’+ y’)’
- y(wz’ + wz) + xy
- xyz + x’y’z + x’yz + xyz’ + x’y’z’
3. Reduce the following Boolean expressions by using the rules of Boolean
algebra:
- A’C’ + ABC + AC’
- (A’ + C)(A’ + C’)(A + B + C’D)
- A’B(D’ + C’D) + B(A + A’CD)
- ABCD + A’BD + ABC’D
4. Find the complements of the following Boolean expressions and reduce
them to minimum number of literals:
- (xy’ + w’z)(wx’ + yz’)
- wx’ + y’z’
- x’z + w’xy’ + wyz + w’xy
5. Find the complements of the following expressions:
- xy’ + x’y
- (AB’ + C) D’ + E
- (x’ + y + z)(x + y’)(x + z)