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220 Fundamentals of Computers NPP

)
)
(d) F = π M (3,6,9,12,15 . The above ex- (d) F = π M (3,6,9,12,15 . ì`§OH$ H$s 5 _oŠñQ>_©
pression contains five maxterms which corre- {ZåZmZwgma nm±M g§»`mAm| go g§~§{YV h¢:
sponds to the following combinations:
3  → 0011
6  → 0110
9  → 1001
12  → 1100
15  → 1111
Filling five zeros for the above numbers, AV… K-_on _| BZ ñWmZm| na nm±M 0 VWm AÝ`
the four variable K-map would look like: ñWmZm| na 1 aIZo na:

PQ R Z 00 01 11 10

00 1 1 0 1

01 1 1 1 0

11 0 1 0 1

10 1 0 1 1

Problem 3.64 àíZ 3.64
Draw the Karnaugh map for the follow- {ZåZ ì`§OH$m| hoVw K-_on ~ZmAmo:
ing Boolean expressions:
(a) F = ∑ m (0, 1, 2, 7) + d (4, 6)
(b) Y = πM (3, 5) . d (0, 1, 7)

(c) Q = C . B . A + C . B . A + C . B . A + B . A
Solution: hc:
(a) The given expression is F = ∑ m (0, 1, 2, 7) (a) Cnamoº$ g_rH$aU F = ∑ m (0, 1, 2, 7) + d(4, 6)_|
+ d (4, 6). It contains four minterms which Mma {_ÝQ>_© h¢ {OÝh| m go Xem©`m J`m h¡ VWm Xmo S>m|Q>
are represented by m and two don’t care
conditions which are represented by d. Ho$`a H§$S>reZ h¢ {OÝh| d go Xem©`m J`m h¡Ÿ& {_ÝQ>_©
Put a ‘1’ corresponding to minterms and Ho$ ñWmZ na 1 VWm S>m|Q> Ho$`a H§$S>reZ Ho$ ñWmZ na
put a ‘X’ corresponding to don’t care con- 'X' aIZo na:
ditions. The Karnaugh map is as follows: