Page 232 - FUNDAMENTALS OF COMPUTER

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232 Fundamentals of Computers NPP

(c) The given expression is (c) {XE JE ì`§OH$:
F = (A + B + C ) (A. + B + C ) (A. + B + C )
The expression has three maxterms and the g_rH$aU _| VrZ _¡ŠgQ>_© h¢ VWm H$aZm°\$ _¡n Ed§
karnaugh map and the groups can be drawn g_yhm| H$mo {ZåZ àH$ma go ~Zm`m Om gH$Vm h¡:
as:

A B C 00 01 11 10

0 0 1 1 0

1 1 1 1 0

There are two pairs as shown above. Con- Bg_| Xmo no`a h¢ O¡gm D$na Xem©`m J`m h¡Ÿ& d{Q>©H$b
sider the vertical pair. A changes. Therefore no`a na {dMma H$a| A n[ad{V©V hmoVm h¡Ÿ& AV: A H$mo hQ>m
eliminate A. The values of B.C is 10, therefore
write it as:(B + C ) X|Ÿ& B.C H$m 10 h¡ AV: Bgo (B + C ) {bI|Ÿ&
Consider the second pair. A is equal to ‘0’ Xygao no`a na {dMma H$a|Ÿ& A H$m _mZ 0 h¡ Bg{bE
therefore put A and move from left to right (00 A aI| VWm ~m§`o go Xm`| (00 go grYo 10) Om`|Ÿ& C H$m
to directly 10) C is constant at ‘0’. Therefore
write C. The term is: (A + C) _mZ '0' na pñWa h¡Ÿ& AV: C {bI|Ÿ& nX (A + C) h¡Ÿ&
The simplified expression in POS form is POS \$m°_© _| gabrH¥$V g_rH$aU Xmo nXm| H$m JwUm
product of the two terms- F = (B + C ) (A. + C ) h¡: F = (B + C ) (A. + C )

(d) The given expression is Y = Σm (4, 5, 7) (d) Cnamoº$ ì`§OH$ Y = Σm (4, 5, 7) go K-_on
The K-map and the groups can be drawn as: VWm g_yh {ZåZmZwgma ~Zm gH$Vo h¢:

A B C 00 01 11 10

0 0 0 0 0

1 1 1 1 0

The solution gives one quad overlapping EH$ ŠdmS> VWm EH$ no`a AmodaboqnJ h¡ & ŠdmS> _|
with one pair. The term for quad is A because ~mE§ go XmE§ OmZo na B d C XmoZm| n[ad{V©V hmo aho h¢ d A
B and C change and A is constant at ‘0’. Con-
sider the pair, A is eliminated when we move = 0 na pñWa h¡ & AV… A àmßV hmoJm Ÿ& no`a _| D$na go
downward. BC is constant at 10. Thus, write ZrMo AmZo na A n[ad{V©V hmo ahm h¡ VWm BC = 10 na
(B+ C ). pñWa h¡ Ÿ& AV… (B+ C ) àmßV hmoJmŸ&