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

5.4 Boolean algebra 191
S0
Stage
0
A0 B0
C0
S1
Stage
1
A1 B1
C1
S2
Stage
2
A2 B2
C2
S3
Stage
3
A3 B3
Figure5.16
Four full-addersconnected to allow two 4-bitbinarywordsto
be added.
C3
Table5.16
Truthtable forafull-adder.
C1 A2 B2 S2 C2
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0
0 1 1 0 1
1 0 0 1 0
1 0 1 0 1
1 1 0 0 1
1 1 1 1 1
Notice that there are three inputs to the circuit,C1, A2 and B2. There are also
twooutputsfromthecircuit,S2andC2.Writingexpressionsfortheoutputsind.n.f.
yields
S2 =C1 ·A2·B2+C1·A2·B2+C1·A2 ·B2+C1·A2·B2
C2 =C1·A2·B2+C1·A2·B2+C1·A2·B2+C1·A2·B2
Itisimportanttoreducetheseexpressionstoassimpleaformaspossibleinorderto
minimize the number of electronic gates needed to implement the expressions. So,
startingwithS2,
Let
S2 =C1 ·A2·B2+C1·A2·B2+C1·A2 ·B2+C1·A2·B2
=C1·(A2·B2+A2 ·B2) +C1·(A2·B2 +A2·B2)
by the distributive law (5.2)
X=A2·B2 +A2·B2 (5.3)
Notice this isanequation for an exclusive OR gate with inputsA2 andB2.
UsingEquation (5.3) wehave
X =A2·B2 +A2·B2
= (A2·B2)·(A2·B2) by DeMorgan’s law
= (A2 +B2)·(A2 +B2) by DeMorgan’s law
= (A2+B2)·(A2+B2) by the complement law
=A2·A2+A2 ·B2+B2·A2+B2·B2
by the distributive law
= 0 +A2 ·B2 +B2·A2 +0 by the complement law
=A2 ·B2 +B2·A2 by the identity law
=A2·B2+A2 ·B2
Itisnow possible towrite Equation (5.2) as
by the commutative law
S2=C1·X +C1·X
➔