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NPP

NPP Number System, Boolean Algebra and Logic Circuits 229

A B C D 00 01 11 10

00 0 1 0 1

01 1 0 0 0

11 0 1 0 1

10 1 1 1 1

The solution contains one quad, four pairs EH$ ŠdmS>, Mma noAa VWm EH$ qgJb H$s ghm`Vm
and one single. The simplified expression can
be written as: go {ZåZ ì`§OH$ àmßV hmoJm:
Y = D . C . B . A + B . A + D . C . A + D . C . A + D . C . B + D . C . B
(c) The given expression is: (c) {X`m J`m ì`§OH$:

F = A C . B . + A C . B . + A C . B . + A B . + A
This is not in a canonical SOP Form because Ho$Zmo{ZH$b ñdê$n _| Zht h¡Ÿ& Bg_| .A B VWm A
B . A and A have less than three terms. There- Eogo na h¢ {OZ_| VrZm| Ma Zht h¢Ÿ& gd©àW_ h_ A B . H$mo
fore we have to convert the terms into three vari- g_w{MV ñdê$n _| ~Xb boVo h¢Ÿ& My±{H$ Bg_| ‘C’ Zht h¡
able terms. Consider .A B which has ‘C’ as the
missing variable. Therefore multiply this term AV… (C+ C ) go JwUm H$aZo na:
by (C+ C ).

B . A = .A . B (C+ C ) = .A C . B + C . B . A
Now, A is to be multiplied by both (C+ C ) A~ A H$mo (C+ C )VWm (B+ B ) XmoZm| go JwUm H$aZo
and (B+ B ). na:

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

Now, put the values of .A B and A in the B . A VWm A Ho$ _mZ aIZo na {ZåZ ì`§OH$ àmßV
given expression: hmoJm:

F = + A.B.C + A.B.C + A.B.C + A.B.C A.B.C + + A.B.C A.B.C + + A.B.C A.B.C
The terms .A C . B and .A C . B occur twice C . B . A VWm .A C . B XmoZm| nX Xmo-Xmo ~ma AmE h¢Ÿ&
in the above expression. Therefore, these can BÝh| EH$ hr ~ma {cIZo na Ho$Zmo{ZH$c ñdê$n àmßV hmoJm:
be written only once. The canonical equation
can be written as: