PERE Jean-Baptiste COEN 312

ID number: 9436413

Assignment #2

Chapter 3: Gate-level minimization

3.1) Simplification of the following Boolean function, using three-variable maps:

d) F(x, y, z) = ∑(3, 5, 6, 7)

x y z / 00 / 01 / 11 / 10
0 / / /
1 /
1 /
1 /
1 / 1

F(x, y, z) = xy + yz + xz

3.3) Simplification of the following Boolean expression, using three-variable maps:

b) x’y’ + yz + x’yz’

x y
z / 00 / 01 / 11 / 10
0 / 1 / 1 / /
1 / 1 /
1 / 1

x’y’ + yz + x’yz’ = x’ + yz

3.5) Simplification of the following Boolean function, using four-variable maps:

d) F(A, B, C, D) = ∑ (0, 2, 4, 5, 6, 7, 8, 10, 13, 15)

A B C D / 00 / 01 / 11 / 10
00 / 1 /
1 / / 1
01 / /
1 / 1 /
11 / / 1 / 1 /
10 / 1 / 1 / / 1

F(A, B, C, D) = A’B + B’D’ + BD = A’B + (B xor D)’

3.6) Simplification of the following Boolean expression, using four-variable maps:

b) x’z + w’xy’ + w(x’y + xy’)

w x y z / 00 / 01 / 11 / 10
00 / /
1 / 1 /
01 /
1 / 1 / 1 /
1
11 / 1 / / /
1
10 / / / / 1

x’z + w’xy’ + w(x’y + xy’) = xy’ + x’z + wx’y

3.8) Minterms of the following Boolean expression by first plotting each function in a map:

b) C’D + ABC’ + ABD’ + A’B’D

A B C D / 00 / 01 / 11 / 10
00 / / 1 /
01 / 1 /
1 / 1 / 1

11 / 1 /
10 / / / 1

C’D + ABC’ + ABD’ + A’B’D = ∑(1, 3, 5, 9, 12, 13, 13, 14)

3.9) All the prime implicants (the essential or not) for the following Boolean function:

b) F(A, B, C, D) = ∑(0, 2, 3, 5, 7, 8, 10, 11, 14, 15)

A B C D / 00 / 01 / 11 / 10
00 / 1 /
1
01 / /
1 / /

11 /
1 / 1 /
1 /
1
10 / 1 / / 1 / 1

Essential Prime implicants:

∑(0, 2, 8, 10) = B’D’

∑(3, 7, 11, 15) = CD or ∑(2, 3, 10, 11) = B’C

∑(10, 11, 14, 15) = AC

∑(5, 7) = A’BD

3.12) Simplification of all the following Boolean function to product-of-sums form:

b) F(A, B, C, D) = ∏(1, 3, 5, 7, 13, 15)

A B C D / 00 / 01 / 11 / 10
00 /
01 /
0 /
0 /
0 /

11 / 0 / 0 / 0
10 / / / /

F(A, B, C, D) =(A + D’)(B’ + D’)

3.15) Simplification of the following function F, together with the don’t care conditions d, and then expression of the simplified function in sum-of-minterms form:

d) F(A, B, C, D) = ∑(1, 3, 8, 10, 15) d(A, B, C, D) = ∑(0, 2, 9)

A B C D / 00 / 01 / 11 / 10
00 /
X / 1
01 / 1 /
/
/
X
11 / 1 / /
1 /
10 / X / 1

F(A, B, C, D) = B’D’ +A’B’ + ABCD = ∑(0, 1, 2, 3, 8, 10, 15)

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