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222 Fundamentals of Computers NPP
We can simplify a Boolean expression with H$m°aZm°\$ _on H$s ghm`Vm go ~y{b`Z ì`§OH$ H$mo
the help of Karnaugh map. Following steps are gab {H$`m Om gH$Vm h¡ Ÿ& {ZåZ {d{Y AnZmB© OmVr h¡:
taken to simplify the expression:
Step 1. Draw K-map from the given equation. Step 1. {X`o JE ì`§OH$ go K-_on ~ZmAmoŸ&
Step 2. Make octets, quads, pairs or singles etc. Step 2. 1 `m 0 Ho$ {bE {d{^ÝZ J«wn ~ZmAmo &
for 1 (or ‘0’)
Step 3. Write the simplified expression in one Step 3. {ZåZ _| go {H$gr EH$ ñdê$n _| gabrH¥$V ì`§OH$
of the following forms: {bImo:
SOP Form if ‘1’ is used for groups. SOP _|, `{X 1 Ho$ g_yh ~ZmE h¢ &
POS Form if ‘0’ is used for groups. POS _|, `{X 0 Ho$ g_yh ~ZmE h¢ &
Some Important Points to be remem- Hw$N> _hËdnyU© {~ÝXw …
bered while solving Karnaugh map prob-
lems:
1. Draw the biggest group first. Then, go for 1. gd©àW_ ~‹S>m J«wn ~ZmE, CgHo$ ~mX N>moQ>o J«wn ~ZmE &
the smaller one, e.g. If a quad can be made, O¡go `{X EH$ ŠdmS> ~Z ahm hmo Vmo Xmo noAa Z ~ZmEŸ&
do not make two pairs,
2. Total number of groups must be as small as 2. Hw$b J«wnm| H$s g§»`m Ý`yZV_ hmoZm Mm{hEŸ& O¡go `{X
possible. e.g. If one person makes four EH$ ì`{º$ nm±M J«wn ~ZmVm h¡Ÿ VWm Xygam ì`{º$ Mma
groups and other only three, the second one
is correct because it will provide more sim- J«wn _| hr H$ìha H$a boVm h¡ Vmo Xygam ì`{º$ ghr h¡Ÿ&
plified solution. H$_ g§»`m _| J«wn ~ZZo go ì`§OH$ N>moQ>m àmßV hmoJmŸ&
3. A group containing all don’t Care condi- 3. Eogm J«wn {Og_| gmao S>m|Q> Ho$`a H§${S>eÝg h¡, H$^r Z
tions is illegal. It must not be drawn. ~ZmE§Ÿ&
4. A redundant group is that whose all 1’s are 4. [aS>ÝQ>|Q> J«wn dh h¡ {OgHo$ gmao 1 EH$ go A{YH$ J«wnm|
used more than once. It must be removed _| Cn`moJ _| AmE h¢Ÿ& Bg àH$ma Ho$ J«wn H$mo hQ>m XoZm
before writing the simplified expression.
Mm{hE&
Problem 3.65 àíZ 3.65
Simplify the following Boolean expres- {ZåZ ~y{b`Z ì`§OH$m| H$mo K-_on H$s ghm`Vm go
sion using Karnaugh map method: hb H$amo:
(a) Y = B . A + B . A
(b) F = C . B . A + C . B . A + C . B . A
(c) Y = D . C . B . A + D . C . B . A + D . C . B . A
Solution: hc:
(a) The given expression is (a) ì`§OH$ Y = B . A + B . A . H$m k-_on {ZåZmZwgma
Y = B . A + B . A . The K-map for the expression ~Zm`m Om gH$Vm h¡:
can be drawn as:
