Page 241 - FUNDAMENTALS OF COMPUTER

NPP

NPP Number System, Boolean Algebra and Logic Circuits 241

A B .C 00 01 11 10

0 1 0 1 0

1 0 1 1 1

SOP Form : Take the same K-map, use ‘1’ SOP \$m°_©: Bg hoVw 1 Ho$ J«wn ~ZmZo na:
to form groups.

A B .C 00 01 11 10

0 1 0 1 0

1 0 1 1 1

Three overlapping pairs and one single SOP \$m°_© VrZ no`am| d EH$ qgJb go {ZåZ gab
gives rise to the SOP form solution as: ì`§OH$ àmßV hmoJm:

F = C . B . A + C . A + B . A + C . B
POS Form: Take the same K-map. Use ‘0’ POS \$m°_©: Cgr K--‘¡n H$mo br{OE& J«wn ~ZmZo Ho$
for making group
{bE ‘0’ H$m Cn¶moJ H$a|&
A B .C 00 01 11 10

0 1 0 1 0

1 0 1 1 1

There are three singles. This shows that no Bg_| VrZ qgJëg h¢Ÿ& `h Xem©Vm h¡ {H$ H$moB© ^r
variable will be eliminated and therefore the d¡[aE~b Zht hQ>m`m Om`oJm VWm Bg{bE {X`o J`o g_rH$aU
given expression cannot be simplified further. H$mo AmJo hb Zht {H$`m Om gH$Vm h¡Ÿ& Bg{bE POS \$m°_©
Therefore in POS form the simplified expres-
sion is same as the given expression: _| gabrH¥$V g_rH$aU {X`o J`o g_rH$aU Ho$ g_mZ h¢:
F = (A + B + C ) (A. + B + C ) (A. + B + C )