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                  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:
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