Page 180 - FUNDAMENTALS OF COMPUTER

NPP

180 Fundamentals of Computers NPP

Solution: hc:
The output of OR Gate is: OR

( X + Y )
The output of AND Gate is: AND

(X + Y ) Z.
Therefore F = ( X YZ+ ) . F = ( X YZ+ ) .

Problem 3.34 àíZ 3.34
Find the expression for Y: Y hoVw ì`§OH$ {ZH$mbmo…

A

B
Y
C

Solution: hc:

Y = A ⊕ B⊕ C = S ⊕ C

Let, [S = A ⊕ ] B
But we know that

A ⊕ B = A B +. AB.
Therefore Y = S C +. SC.

[ Y = ⊕ ]AB ( C +. ⊕ )AB C.

Y = [ AB + . A B. ] . C + ( AB + . A B. ) C.

Y= [ B.A . B . A ] C. + ( B.A + B . A ) C.

Y = ( [A + B ) (A. + B )] C. + ( B.A + B . A ) C.
Y = ( A B+ ) ( A B+ . ) C + . A BC + . A B C..

Y = ( A A + . AB + B A + . BB ) C + . ABC + A BC.

Y = ( AB + A B ) C + AB C + . AB C

Y = ABC + A BC + ABC + A BC.