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Unformatted text preview: straightforward conversion to very nice gating networks which are
more desirable from most implementation points of view. In their purest, nicest form they
go into two-level networks, which are networks for which the longest path through which
the signal must pass from input to output is two gates.
Conversion Between Canonical Forms
The complement of a function expressed as the sum-of-minterms equals the sum-ofminterms missing from the original function. This is because the original function is
expressed by those minterms that make the function equal to 1, while its complement is a
1 for those minterms for which the function is a 0. For example, the function
F (A, B, C) = (1,4, 5,6, 7) = nij +m4 +m5+m6 +m7
has a complement that can be expressed asf
F (A, B, C) = (0, 2, 3) = m0 +m2 +m3
Now, if we take the complement of F, by De Morgan's theorem we obtain F back in a
F = m0 +m2 +m3
= m0 • m2 • m3
= II (0, 2, 3)
The last conversion follows from the definition of minterms and maxterms as shown in
Figure 6.16. From the figure, it is clear that the following relation holds true:
That is, the maxterm with subscript/ is a complement of the minterm with the same
subscript/, and vice-versa.
The last example has demonstrated the conversion between a function expressed in sumof-minterms and its equivalent in product-of-maxterms. A similar argument will show
that the conversion between the product-of-maxterms and the sum-of-minterms is similar.
We now state a general conversion procedure: "To convert from one canonical form to another, interchange the symbol and list those
numbers missing from the original form."
For example, the function
F (x, y, z) = TI (0, 2, 4, 5)
is expressed in the product-of-maxterms form. Its conversion to sum-of-minterms is:
F(x,y,z) = I(l,3,6,7)
Note that in order to find the missing terms, one must realize that the total number of
minterms or maxterms is always 2n, where n is the number of binary variables in the
All operations within a co...
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- Spring '14