CHAP. 15] BOOLEAN ALGEBRA 399
(b) Herexy′z=00000100,xy=00000011, andz′=01010101. ThenE=xy′z+xy+z′=01010111. Thus
T( 00001111 , 00110011 , 01010101 )= 01010111
15.32. Find the truth tableT=T(E)for the Boolean expressionE=E(x, y, z)where:
(a) E=xyz′+x′yz; (b) E=xyz+xy′z+x′y′z.
HereEis a complete sum-of-products expression which is the sum of minterms. Example 15.13 gives the truth
tables for the minterms (using the special sequences). Each minterm contains a single 1 in its truth table; hence the
truth table ofEwill have 1’s in the same positions as the 1’s in the minterms inE. Thus:
(a) T (E)= 00001010 ; (b) T (E)= 01000101
15.33. Find the truth tableT=T(E)for the Boolean expression
E=E(x, y, z)=(x′y)′yz′+x′(yz+z′)
First expressEas a sum-of-products:
E=(x+y′)yz′+x′yz+x′z′=xyz′+y′yz′+x′yz+x′z′
=xyz′+x′yz+x′z′
Now expressEas a complete sum-of-products:
E=xyz′+x′yz+x′z′(y+y′)
=xyz′+x′yz+x′yz′+x′y′z′
As in Problem 15.32, use the truth tables for the minterms appearing in Example 15.13 to obtainT(E)=10101010.
15.34. Find the Boolean expressionE=E(x, y, z)corresponding to the truth table:
(a) T (E)= 01001001 ; (b) T (E)=00010001.
Each 1 inT(E)corresponds to the minterm with the 1 in the same position (using the truth tables for the minterms
appearing in Example 15.13). For example, the 1 in the second position corresponds tox′y′zwhose truth table has
a single 1 in the second position. ThenEis the sum of these minterms. Thus:
(a) E=x′y′z+x′yz+xyz′; (b) E=xy′z′+xyz
(Again we assume the input consists of the special sequences.)
KARNAUGH MAPS
15.35. Find the fundamental productPrepresented by each basic rectangle in the Karnaugh map in Fig. 15-31.
In each case find those literals which appear in all the squares of the basic rectangle; thenPis the product of
such literals.
(a) x′andz′appear in both squares; henceP=x′z′.
(b) xandzappear in both squares; henceP=xz.
(c) Onlyzappears in all four squares; henceP=z.
Fig. 15-31