Page 161 - FUNDAMENTALS OF COMPUTER

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NPP Number System, Boolean Algebra and Logic Circuits 161

Binary to Hexadecimal Conversion ~mBZar go hoŠgmS>o{g_b _| n[adV©Z
Given a Binary Number and it is needed to AmnH$mo `{X ~mBZar g§»`m Xr hmo Am¡a Bgo
find the hexadecimal equivalent. This problem hoŠgmS>o{g_b _| n[ad{V©V H$aZm h¡ Vmo Bg g_ñ`m H$mo Xmo
can be solved in two ways :
àH$ma go hb {H$`m Om gH$Vm h¡ …
Method I : First convert the binary number {d{Y I : nhbo Xr JB© ~mBZar g§»`m H$mo Xe_bd _|
into decimal and then convert the obtained VWm Bg Xe_bd H$mo hoŠgmS>o{g_b _| ~XbmoŸ& _mZm {H$
decimal into hexadecimal. Suppose the given binary Xr JB© ~mBZar g§»`m h¡ …
number is
(10101) 2

1
2
0
4
3
1 × 2 + 0 × 2 + 1 × 2 + 0 × 2 + 2 × 1
16 + 0 + 4 + 0 + 1 = (21) 10
Now convert this decimal into Bg Xe_bd H$mo A~ hoŠgmS>o{g_b _| ~XbVo h¢ …
hexadecimal:
16 21 5
16 1 1
0
Thus, (21) = (15) 16
10
and therefore (10101) = (15) 16
2
Method II : (Direct Method) In this {d{Y II : Bg {d{Y _| _| h_ Mma-Mma {~Q>m| Ho$
method, make groups of four bits. In integer g_yh ~ZmVo h¢ Ÿ& nyUmªH$ _| XmE± go VWm AnyUmªH$ _| ~mE± go
part of the number start from right and in àma§^ H$aVo h¢ Am¡a `{X eyÝ` H$s Amdí`H$Vm hmoVmo Cgo
fractional part of the binary number start from
left and put zeros if necessary. Consider the A§V _| bJm XoVo h¢ Ÿ& àË`oH$ g_yh Ho$ ñWmZ na CgH$m Mma
following examples: {~Q> _| hoŠgmS>o{g_b A§H$ {bI XoVo h¢ Ÿ& ZrMo {bI|
CXmhaUm| go `h ñnï> h¡ …
Problem 3.19 àíZ 3.19
Convert following Binary numbers into ZrMo {bIo ~mBZar H$mo hoŠgmS>o{g_b _| ~Xbmo …
Hexadecimal:
(a) 11101101 (b) 10101.110101 (c) 01011.110 (d) 0.10101101
Solution : hc :
In each case make group of four bits and Mma-Mma {~Q>m| Ho$ g_yh ~ZmH$a hoŠgmS>o{g_b Vwë`
put extra zeros if necessary:
{bImo Ÿ& (`{X Amdí`H$ hmo Vmo eyÝ` bJmAmo) …

(a) 1110 1101 = (ED ) 16 (b) 0001 0101. 1101 0100 = (15.D4 ) 16
(c) 0000 1011. 1100 = (OB.C ) 16 (d) 0. 1010 1101 = (0.AD ) 16
(Note : Memorise table 2.3) (ZmoQ>: Vm{cH$m 2.3 H$mo AÀN>r Vah `mX H$a b|Ÿ&)