Page 206 - FUNDAMENTALS OF COMPUTER

NPP

206 Fundamentals of Computers NPP

Here, the first square corresponds to (00), Bg_| nhbm dJ© (00) Ho$ g§JV h¡Ÿ& Xygam grYo hmW
top right square corresponds to (01) and bottom na (01) Ho$ g§JV h¡ VWm ZrMo-grYo hmW dmbm (11)
right corresponds to (11). The values of output
variables are put in accordance with these Ho$ g§JV h¡Ÿ& Imbr dJ© _| h_ AmCQ>nwQ> Ho$ _mZ {bIVo
inputs. h¢Ÿ&

Problem 3.57 àíZ 3.57
Draw the K-map for the following {ZåZ \$bZ hoVw K-_on ~ZmAmo:
function:

F = B . A + B . A
Solution: hc:
The expression is in SOP form and there {X`m J`m ì`O§H$ SOP ê$n _| h¡Ÿ& Bg_| Xmo {_ÝQ>_©
are two minterms A.B and B . A . These terms h¡ A. B VWm B . A Omo {H$ {ZåZ g§»`mAm| Ho$ g§JV h¡:
corresponds to:
10
B . A  ( ) and .A B → ( )
→
11
Therefore two 1s will be there in K-map BZHo$ gmnoj Xmo ñWmZm| na 1 AmEJm:
B
A 0 1

0 0 0

1 1 1

Before drawing a K-map one thing must K _on ~ZmZo go nhbo Bg ~mV H$m Ü`mZ aIZm
be kept in mind that we must have the Boolean Mm{hE {H$ ì`§OH$ Ho$Zmo{ZH$b ñdê$n _| hmoZm Mm{hEŸ& Bg
function in canonical form. All terms must
contain equal number of variables. ñdê$n _| ha nX _| g_mZ g§»`m _| Ma hmoZm Mm{hE &
Three Variable K-map VrZ Ma K-_on
Suppose we have three variables A, B and _mZm {H$ h_mao nmg VrZ BZnwQ> Ma A, B, C h¢Ÿ&
C. The K-map for these three variables will BZ VrZ Mam| Ho$ {bE K- _on {ZåZ àH$ma hmoJm…
look as below:

BC
A 00 01 11 10
0

1