082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

5.4 Boolean algebra 189
However, we know from Engineering application 5.1 that the much simpler expression
(A·B) +C has the same truth table and if a circuit were to be built corresponding
tothisitwouldrequirefewercomponents.Clearlywhatweneedisatechniqueforfinding
the simplest expression which does the desired job since the d.n.f. is not in general
the simplest. It is not obvious what is meant by ‘simplest expression’. In what follows
we shall be concerned with finding the simplest d.n.f. It is nevertheless possible that a
logically equivalent statement exists which would give a simpler circuit. Simplification
canbeachievedusingthelawsofBooleanalgebraasweshallseeinExample5.11and
Engineering application 5.3.
Engineeringapplication5.2
TheexclusiveORgate
WehavealreadylookedattheORgateinSection5.3.Thefullnameforthistypeof
ORgateistheinclusiveORgate.Itissocalledbecauseitgivesanoutputof1when
either or both inputs are 1. The exclusive OR gate only gives an output of 1 when
eitherbutnotbothinputsare1.ThetruthtableforthisgateisgiveninTable5.15and
its symbolisshown inFigure 5.15. Usingthe truthtable, the d.n.f. forthe gateis
F=A·B +A ·B
The exclusive OR often arises in the design of digital logic circuits. In fact, it is
possible tobuy integrated circuits thatcontain exclusive ORgates as basic units.
Table5.15
The truthtable foran
exclusive OR gate.
A B F
1 1 0
1 0 1
0 1 1
0 0 0
A
B
F = A . B + A . B
Figure5.15
Symbol foran exclusive OR gate.
Example5.11 UsethelawsofBooleanalgebragiveninTable5.8tosimplifythefollowingexpressions.
(a) A·B+A·B
(b) A+A ·B
(c) A+A ·B·C
(d) A·B·C+A·B·C+A·B·C+A·B ·C +A ·B·C
Solution (a) Using the distributive lawwe can write
A·B+A·B=A·(B+B)
Using the complement law,B +B = 1 and hence
A·B+A·B=A·1
=A using the identity law