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230 Fundamentals of Computers NPP

F = + A.B.C + A.B.C + A.B.C + A.B.C + A.B.C + A.B.C A.B.C
The expression contains seven minterms Bg ì`§OH$ hoVw {ZåZ K-_on VWm J«wßg ~Zm`m Om
and the K-map and the groups can be drawn gH$Vm h¡:
as follows-
A B C 00 01 11 10

0 1 1 1 1

1 1 1 1 0

The simplified expression for three quads AV… gab ì`§OH$ {ZåZmZwgma hmoJm:
can be written as

F = A + B + C
3.25 To Obtain the Solution in POS Form 3.25 POS ñdê$n _| gab H$aZm
The solutions which we have obtained in A^r VH$ {OVZo ^r hb àmßV {H$E JE h¢ g^r SOP
previous problems, are in SOP Form. We can (Sum of product) Ho$ ê$n _| àmßV hþE h¢Ÿ& h_Zo Bg hoVw
also write the simplified expression in POS "1" Ho$ g_yh ~ZmE WoŸ& K-_on go h_ grYo-grYo POS
(Product of Sum) form. For getting this, pairs, (Product of sum) Ho$ ê$n _| ^r hb àmßV H$a gH$Vo
quads and octets are formed using 0’s . The fol- h¢Ÿ& Bg hoVw 0 Ho$ Cn`moJ go gmao g_yh ~ZmZo hm|JoŸ& {ZåZ
lowing steps are taken: {d{Y AnZmB© Om gH$Vr h¡:
(1) Cover all the 0’s in pair, quads and octets. (1) gmao 0 H$mo noAa, ŠdmS> VWm Am°ŠQ>oQ> Am{X _| H$da
H$amo&

(2) Go from top to bottom and left to right for (2) àË`oH$ J«wn _| D$na go ZrMo VWm ~mE§ go XmE§ MbmoŸ&
each group.
(3) Eliminate the variable which changes its (3) Omo am{e AnZm _mZ ~XbVr h¡, Cgo hQ>m Xmo &
value.
(4) If a variable is constant at ‘1’ put a bar and (4) `{X H$moB© Ma ‘1’ na pñWa h¡ Vmo CgHo$ D$na ~ma
if this is constant at ‘0’ do not put a bar. Put bJmAmo (O¡go ) AÝ`Wm ZhtŸ& gmao Mam| Ho$ _Ü`
‘+’ between all the constant variables. You A
will get a term for one group. (+) H$m {M• bJmAmoŸ& Bg Vah `moJ (sum) Ho$ ê$n
_| EH$ nX àmßV hmoJmŸ&
(5) Put a ‘ . ’ (dot) between all the terms ob- (5) gmao nXm| Ho$ _Ü` (.) S>m°Q> bJmH$a POS Ho$ ê$n _|
tained. ì`§OH$ àmßV {H$`m OmVm h¡Ÿ&

Problem 3.69 àíZ 3.69
Simplify the following Boolean expres- K- _on {d{Y H$m Cn`moJ H$a {ZåZ hoVw POS Ho$
sion in POS Form using K-map method: ê$n _| gab ì`§OH$ kmV H$amo: