Page 163 - FUNDAMENTALS OF COMPUTER

NPP Number System, Boolean Algebra and Logic Circuits 163

Now, make groups of four bits starting A~ grYo hmW go ewê$ hmoH$a 4-{~Q>m| Ho$ g_yh
from right ~ZmAmo Ÿ&
Put extra zeros if needed : Amdí`H$ hmoZo na AmJo A{V[aŠV eyÝ` cJmAmo:

0001 0001 0101
put Hexadecimal digit for each group : àË`oH$ g_yh Ho$ ñWmZ na hoŠgmS>o{g_b A§H$ {bImo…
(115) 16
Therefore (425) = (115) 16
8
(b) First converting octal number (b) nhbo {X`o JE Am°ŠQ>b H$mo ~mBZar _| ~Xbmo …

(72.14) into Binary : (111 010. 001 100 ) 2
8
Now, make groups of 4 bits. A~, Mma-Mma {~Q>m| Ho$ g_yh ~ZmAmo …
put Hexadecimal digit for each group : àË`oH$ g_yh Ho$ ñWmZ na hoŠgmS>o{g_b A§H$ {bImo…
(3A . 30) 16
Thus (72.14) = (3A.30) 16
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Problem 3.22 NPP àíZ 3.22
Convert the following Decimal numbers {ZåZ Xe_bd g§»`mAm| H$mo nhbo ~mBZar _| ~XbVo
into their hexadecimal equivalents by first hþE CZHo$ hoŠgmS>o{g_b Vwë` {ZH$mbmo:
converting them to their binary equivalents:
(i) 125.75 (ii) 97.375
Solution : hc :
(i) The given decimal number 125.75 is a (i) 125.75 EH$ {_{lV g§»`m h¡ Ÿ& nhbo BgHo$
mixed number. First take integer part and nyUmªH$ ^mJ 125 H$mo ~mBZar _| ~XbVo h¡ …
convert it into binary number :
2 125 1
2 62 0
2 31 1
2 15 1
2 7 1
2 3 1
2 1 1
0
Thus, (125) = (1111101) 2
10
Now take fractional part and convert it A~ BgHo$ Am§{eH$ ^mJ 0.75 H$mo ~mBZar _| ~XbVo h¢…
into binary :
.75 × 2 = 1.50 1
.50 × 2 = 1.00 1
.00 × 2 = 00