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Unformatted text preview: ) x • y • z + x • y z + x • y
(d) x • y + x • z + y z
(e) (x + y)(x+z)(y + z) (d)xy+xz + yz
= xy+xz + yzl
= xy+xz + yz(x+x)
= xy+xz + yzx + yz x
=x■y+x•z+x■y•z+x•y■z
= xyl + xz + xyz+xyz
= xy1+xyz+xz+xyz
= xy(l +z)+ x z(l +y)
= xy(z+ 1)+ x z(y+ 1)
=x•y•1+x•z■1
= x y+ x ■ z
(e) (x + y)(x +z)( = (x + y)(x + z)
Note that in Example 6.1, functions (a) and (b) are the dual of each other and use dual
expressions in corresponding minimization steps. Function (c) shows the equality of the
functions Fi and F2 of Figure 6.14. Function (d) illustrates the fact that an increase in the
number of literals sometimes leads to a final simpler expression. Observe that function
(e) is the dual of function (d). Hence it is not minimized directly and can be easily
derived from the dual of the steps used to derive function (d).
Complement of a Function
The complement of a Boolean function F is F and is obtained by interchanging O's for 1
's and 1 's for O's in the truth table that defines the function. For example, Figure 6.15
defines the function F = x y + x • z and its
complement F.
X y z F F 0
0
0
0
1
0
0
1
1
0
0
1
0
0
1
0
1
1
1
0
1
0
0
1
0
1
0
1
1
0
1
1
0
0
1
1
1
0
1
Figure 6.15. Truth table for the function F = x • y + X • z and its comolemont
Algebraically, the complement of a function may be derived through De Morgan's
theorems whose generalized forms are as follows:
A1+A2+A3+... + An = A1.A2.A3 ….An
A1 A2 • A3 •... • An = A, + A2 + A3 +.. + An These theorems state that the complement of a function is obtained by interchanging the
OR and the AND operators and complementing each literal. The method is illustrated
below through an example.
Example 6.2.
Find the complement of the following functions:
(a) F, = x • y • z + x • y • z
(b) F2=x(yz+yz)
Solution:
Applying De Morgan's theorems as many times as necessary, the complements are
obtained as follows:
(a) F, = x • y • z + x • y • z
=(...
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 Spring '14

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