Page 223 - FUNDAMENTALS OF COMPUTER

NPP

NPP Number System, Boolean Algebra and Logic Circuits 223

A B 0 1

0 0 0

1 1 1

It is clear that there are two adjacent 1’s and ñnîQ> h¡ {H$ Xmo 1 H$s ghm`Vm go EH$ no`a ~ZoJm :
therefore a pair will be formed as follows:
A B 0 1

0 0 0

1 1 1

Now we can write the simplified expres- A~ h_ Bg noAa hoVw gab ì`§OH$ {bI|JoŸ& D$na go
sion for pair. Go from top to bottom and left to ZrMo VWm ~mE§ go XmE§ OmZo na h_ XoIVo h¢ {H$ A = 1 na
right and see which variables change. Thus, we
have seen that B changes form 0 to 1. Therefore pñWa h¡ naÝVw B H$m _mZ n[ad{V©V hmo ahm h¡Ÿ& AV… B H$mo
remove that variable A is constant at 1, there- N>mo‹S> XoVo h¢Ÿ& My±{H$ A = 1 na pñWa h¡ AV… ì`§OH$ A hr h¡Ÿ&
fore the term given is: A which is the only sim- `hr y hoVw gab ì`§OH$ h¡- Y = A
plified solution Y = A.
(b) The given expression is (b) {X`m J`m ì`§OH$ h¡:

F = C . B . A + C . B . A + C . B . A
The Karnaugh map will contain three 1’s Bg_| VrZ {_ÝQ>_© hoVw VrZ 1 AmE§Jo & K-_on
and it will be drawn as follows: {ZåZmZwgma hmoJm:

A B C 00 01 11 10

0 0 0 0 0

1 0 1 1 1

First draw one pair taking middle 1 with gd©àW_ H$moB© ^r nmg-nmg Ho$ Xmo 1 H$mo boH$a EH$
any of the 1s. noAa ~Zm br{OE Ÿ&