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NPP Number System, Boolean Algebra and Logic Circuits 237

Problem 3.71 àíZ 3.71
Simplify the following Boolean expres- SOP VWm POS _| gab H$a `y{Zdg©b JoQ>m| go
sion in SOP and POS Forms and Implement ~ZmAmo:
using Universal Gates:
(a) F = C . B . A + C . B . A + C . B . A + C . B . A (b) F = (A+ B+ C ) (A. + B+ C ) (A. + B+ C )

(c) F = Σ m ( 5,4 ) 7 , (d) F = π M (0,1, 7,9,12,15 )
)
)
(
(e) F = Σ m ( 1,0 ) 4 , 2 , + d ( 5,3 ) 7 , (f) F = π M (4, 7 . d 0,1,5 .
Solution: hc:
(a) The given expression is: (a) {X`m J`m ì`§OH$:

F = C . B . A + C . B . A + C . B . A + C . B . A

This is a three variable problem. The Cnamoº$ ì`§OH$ _| Mma {_ÝQ>_© h¡ & AV… K-_on _|
K-map contains four 1’s corresponding to four Mma 1 aIZo na…
minterms in the expression. first draw the K-
map:
A B C 00 01 11 10

0 0 0 0 1

1 1 0 1 1

SOP Form Solution: To get the simplified SOP \$m_©: Cnamoº$ K-_on _| 1 Ho$ g_yh ~ZmZo
expression SOP form, take the same Karnaugh na …
map and make pairs, quad etc. taking ‘1’.
A B C 00 01 11 10

0 0 0 0 1

1 1 0 1 1

The simplified expression in SOP Form for VrZ no`am| go àmßV gab ì`§OH$ {ZåZmZwgma hmoJm:
the three pairs will be as follows:
F = C . B + B . A + C . A