Page 231 - FUNDAMENTALS OF COMPUTER

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NPP Number System, Boolean Algebra and Logic Circuits 231

(a) F = B . A + A B . (b) F = C . B . A + C . B . A + C . B . A + C . B . A

(c) F = (A+ B+ C ) (A. + B + C ) (A. + B+ C ) (d) Y = Σ m ( 5,4 ) 7 ,

(e) Y = Σ m ( 1,0 , 9 , 8 , 7 , 3 , 13 , 14 , 12 ) (f) F = π M (1, 3,6,7 )
)
(g) F = π M (2, 4, 7, 9, 12, 13, 15 .
Solution: hc:

(a) The given expression is (a) Cnamoº$ g_rH$aU F = B . A + A B . hoVw K-
F = B . A + A B . . The Karnaugh map and group _on VWm eyÝ` Ho$ J«wn Bg àH$ma ~Z|Jo:
of 0’s are as follows:
A B 0 1

0 1 0

1 1 0

There is only one pair. When we move form Cnamoº$ K- _on _| EH$ hr no`a h¡ Ÿ& D$na go ZrMo
top to bottom, A changes, therefore eliminate OmZo na A n[ad{V©V hmoVm h¡ AV… Bgo hQ>m XmoŸ& My±{H$ B =
A. B is constant at ‘1’. Therefore write B . 1 na pñWa h¡, Bg{bE B {bImoŸ& `hr hb h¡:

F = B
(b) The given expression is: (b) {X`m J`m ì`§OH$:

F = C . B . A + C . B . A + C . B . A + C . B . A
The K-map and group of 0’s can be drawn _| Mma {_ÝQ>_© h¢ AV… Mma 1 VWm Mma 0 H$s
as:
ghm`Vm go K- _on ~ZoJm d {ZåZmZwgma EH$ ŠdmS> ~ZoJm:
A B C 00 01 11 10

0 1 1 0 0

1 1 1 0 0

There is only one quad for 0’s. For this quad My±{H$ ŠdmS> _| D$na go ZrMo OmZo na A d ~mE§ go XmE§
A and C change. Therefore these variables are OmZo na C n[ad{V©V hmo ahm h¡, AV… A d C XmoZm| H$mo
eliminated. B is constant at ‘1’. Therefore write
B This is the solution. F = B hQ>m Xmo & B=1 na pñWa h¡ AV… B {bImo & `hr gab
ì`§OH$ h¡… F = B