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NPP Number System, Boolean Algebra and Logic Circuits 227
Four quads are made with the help of over- Mma ŠdmS> AmodaboqnJ go ~Zo h¢ VWm Xmo noAa amoqbJ
lapping and two pairs are made with the help go ~Zr h¡Ÿ& gab ì`§OH$ Bg àH$ma Hw$b N>… nXm| H$m `moJ
of rolling. Rolling can be done when the 1’s are
in the same row or same column and at the cor- hmoJm:
ners. The simplified expression is:
F = B . A + D . A + C . B + D . C + B . C . D + D . B . A
Problem 3.67 àíZ 3.67
Simplify the following Boolean expres- {ZåZ ~y{b`Z ì`§OH$m| H$mo K- _on go hb H$amo:
sion using K-map method:
(a) f = Σm (0, 1, 6, 7)
(b) f = Σm (2, 3, 5, 9, 10, 11, 14, 15)
Solution: hc:
(a) The given expression is Σm (0, 1, 6, 7) (a) Cnamoº$ ì`§OH$ f = Σm (0, 1, 6, 7) hoVw K-_on
The K-map can be drawn and grouping can be VWm {d{^ÝZ J«wn Bg àH$ma ~Z|Jo:
done- NPP
A B C 00 01 11 10
0 1 1 0 0
1 0 0 1 1
There are two pairs in the K-map. The sim- Xmo noAam| Ho$ H$maU gab ì`§OH$ Bg àH$ma hmoJm:
plified expression is:
f = B . A + B . A
(b) f = Σm (2, 3, 5, 9, 10, 11, 14, 15) (b) f = Σm (2, 3, 5, 9, 10, 11, 14, 15)
This is a four variable problem because the `h EH$ Mma am{e`m| dmbr g_ñ`m h¡ Š`m|{H$ g~go
largest number is 15 (1111). The K-map can be ~‹S>r g§»`m 15(1111) h¡ Ÿ& AmR> {_ÝQ>_m] hoVw AmR> 1
drawn by filling eight 1’s corresponding to eight aIH$a VWm J«wn ~ZmZo na K-_on {ZåZmZwgma hmoJm:
minterms-