Page 175 - FUNDAMENTALS OF COMPUTER

NPP Number System, Boolean Algebra and Logic Circuits 175

Problem 3.25 àíZ 3.25
Simplify the following Boolean {ZåZ ~y{b`Z ì`§OH$m| H$mo gab H$amo…
expressions:

(a) F = A B. ( A + AB + ABC ) (b) Y= (A+ B )( B.A. + B . A )
Solution: hc:

(a) F = AB. ( A + A B + . ABC. ) (b) Y= A . A B . + A . A B . + . B A B+ A . B B .
F = A B A + .. A BA B + .. . A BA BC.. . Y = 0 + A . B + B . A + 0
F = 0 + 0 + AB.C Y = A B +. BA.
F = A.B.C

Problem 3.26 àíZ 3.26
Simplify the following Boolean {ZåZ ~y{b`Z ì`§OH$m| H$mo gab H$amo:
Expressions:

(a) F = ( A AB+ NPP . ) (b) F = A B +. AB (c) Y = A B . + A B .
) ( B B A+ .
.
Solution: hc:

(a) F = ( A A B+ ) ( B BA+ . + . ) F = A 1 . + B
F = A + B
( F = )A + ( B + A )B +
(c) Y = A B . + A B .
F = A + B

F = AB. Y = A B AB. . .

(b) F = A B +. AB. Y =    A + B . AB

F = AB + .   A +  B   Y = (A B+ ) AB.
Y = A B B +.. BAB
F = AB +. A + B Y = AB + AB

F = AB + ( ) + 1 B Y = AB

Problem 3.27
Simplify the following Boolean expr-
ession :
C (b) ( ) ( )
(a) F = A B C +.. A B C+. + AB + F = A + AB + C . A B + . C