Page 214 - FUNDAMENTALS OF COMPUTER

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214 Fundamentals of Computers NPP

total four values are possible (00, 01, 10, 11). Xmo n§{º$`m± d BC Ho$ gmnoj Mma ñVå^ ~ZmE JE h¢ Ÿ& naÝVw
But order of entry is 00, 01, 11,10. Because only BC Ho$ _mZm| H$mo Ü`mZ go XoImo (00,01,11,10) Ÿ& Eogm
one bit position should change, when we move
form one number to another. After filling the Bg{bE {bIm J`m h¡ {H$ EH$ g§»`m go Xygar g§»`m na
values of Y the K-map would be: OmZo na Ho$db EH$-{~Q> _| n[adV©Z hmoZm Mm{hE Ÿ& Y Ho$
{d{^ÝZ _mZ aIZo na gånyU© K-_on {ZåZmZwgma hmoJm:
A B C 00 01 11 10

0 0 1 0 0

1 1 0 0 1

Problem 3.61 àíZ 3.61
Draw the K-map for the following: {ZåZ ì`§OH$m| hoVw K-_on ~ZmBE…
(a) Y = B . A + B . A
(b) Y = C . B . A + C . B . A + C . B . A

(c) Y = D . C . B . A + D . C . B . A + D . C . B . A
Solution: hc:

(a) The given equation is Y = B . A + B . A . (a) {X`m J`m g_rH$aU Y = B . A + B . A EH$
This is a two variable equation. The blank Xmo Mam| dmbm g_rH$aU h¡ Ÿ& BgH$m [aº$ K-_on {ZåZmZwgma
Karnaugh map will be as follows:
hmoJm…
A B 0 1

0

1

The product .A B is a minterm and it cor-- JwUZ\$b A EH$ {_ZQ>_© h¡ Omo {H$ (0 1) Ho$
B .
responds to the combination (01). (put a ‘0’ gmnoj h¡ (Ohm± ~ma Am`m h¡ dhm± 0 VWm Ohm± ~ma Zht
where, there is a bar else put "1"). Therefore,
B .
we will put ‘1’ corresponding to the minterm Am`m h¡ dhm± 1 aI|)& Bgr àH$ma A Ho$ gmnoj (00)
B . A . The second product .A B corresponds to àmßV hmoJm Ÿ& BZ XmoZm| g§»`mAm| Ho$ ñWmZ na 1 aIZoo na
00. Therefore two 1’s will be there correspond- {ZåZmZwgma K-_on àmßV hmoJm:
ing to two minterms. The K-map would look
like this: