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Unformatted text preview: x-y-z)-(x-y-z)
= (x + y + z) ■ (x + y + z)
(b) F2=x-(y-z + y
x + (y-z)-(y-z)
A simpler procedure for deriving the complement of a function is to take the dual of the
function and then complement each literal. This method follows from the generalized De
Morgan's theorems. Remember that the dual of a function is obtained by interchanging
OR and AND operators and O's and l's. The method is illustrated below with the help of
Find the complement of the functions Fj and F 2 of Example 6.2 by taking their dual and
complementing each literal.
(a) F, = xyz + xy-z
The dual of F1 is: (x + y + z;) ■ (x + y + z)
Complementing each literal we get
F2 = (x + y + z)-(x
(b) F2 = x-(y-z + y
Thedualof F2 is: x + (y + z)-(y + z)
Complementing each literal we get + z)-(y.(y+z)
Canonical Forms for Boolean Functions
Minterms and Maxterms
A binary variable may appear either in its normal form (x) or in its complement form (x).
Now consider two binary variables x and y combined with an AND operator. Since each
variable may appear in either form, there are four possible combinations:
x-y Each of these four AND terms is called a minterm or a standard product.
In a similar manner, n variables can be combined to form 2 n minterms. The 2n different
minterms may be determined by a method similar to the one shown in Figure 6.16 for
three variables. The binary numbers from 0 to 2 n - 1 are listed under the n variables. Each
minterm is obtained from an AND term of the n variables, with each variable being
primed if the corresponding bit of the binary number is 0 and unprimed if it is a 1.
Variables Minterms Maxterms X y z Term Designation Term Designation 0 0 0 x-y-z m0 x+y+z Mo 0
M2 0 1 1
x-y-z m3 - x+y+z M3 1
M5 1 1 0 x-y-z m x+y+z M6 1 1 1 x-y-z x+y+z M7 6
m7 Figure 6.16. Minterms and Maxterms for three variables.
A symbol for each minterm is also shown in the figure and is of the form nij, where j
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- Spring '14