Page 207 - FUNDAMENTALS OF COMPUTER

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NPP Number System, Boolean Algebra and Logic Circuits 207

Note, the values of BC are written as Ü`mZ Xo {H$ BC H$m _mZ Bg àH$ma {bIm J`m h¡…
00 → 01 → 11 → 10

This is essential because when we move `h Bg{bE Oê$ar h¡ {H$ O~ h_ EH$ g§»`m go Xygar
from one number to another number there must g§»`m na OmE Vmo EH$ hr {~Q> Ho$ _mZ _| n[adV©Z AmZm
be a change in single bit position only. Then Mm{hEŸ& V^r h_ ì`§OH$ H$mo gab H$a gH$Vo h¡Ÿ& Omo Ma
only we can simplify the Boolean expression.
Eliminate those variables which change their n[ad{V©V hmo CÝh| N>mo‹S> XoVo h¢Ÿ& g~go D$na ~m`m± dJ©
values. The top-left square corresponds to (000) (000) Ho$ g§JV VWm g~go ZrMo, Xm`m± dJ© (110) Ho$
and the bottom right square corresponds to g§JV h¡& dJm] _| AmCQ>nwQ> H$m _mZ {bIm OmVm h¡ &
(110). The value of output variables is put in
the square.
Problem 3.58 àíZ 3.58
Draw K-map for the following Boolean {ZåZ \$bZ hoVw K-map ~ZmAmo:
function:
F ( y,n ) z , = Σ m ( 3,0 ) 7 , 4 ,
Solution: hc:

Here F is a function of n, y, z. Therefore we My±{H$ F \$bZ h¡ n, y, z H$m, AV… BZH$mo hr BZnwQ>
will have to take n, y, z as input variables. Ma _mZZm hmoJm & gmW hr Mma {_ÝQ>Z© {ZåZ g§»`mAm|
There are four minterms in the expression and
corresponds to: Ho$ g§JV h¢…

,
0 → 000 3 → 011 ,
,
4 → 100 7 → 111
Therefore the K-map will be drawn to have AV… K- _on _| Mma 1 VWm ~mH$s 0 hm|JoŸ&
four ones and four zeros.
yz
n 00 01 11 10

0 1 0 1 0

1 1 0 1 0

Similarly, a four variable K-map can be Bgr Vah go Mma Mam| H$m K- _on ~Zm`m Om gH$Vm
drawn. h¡Ÿ&
Simplification of Boolean Expression us- K-_on H$s ghm`Vm go ~y{b¶Z 춧OH$ gab H$aZm
ing K-map
K-map provides us a technique to simplify K-_on H$s ghm`Vm go h_ {H$gr ^r ~‹S>o ì`§OH$ H$mo
a Boolean function. The following steps are gab H$a gH$Vo h¢Ÿ& BgH$s {d{Y Bg àH$ma h¡…
involved while simplifying a Boolean function: