Page 159 - FUNDAMENTALS OF COMPUTER

NPP Number System, Boolean Algebra and Logic Circuits 159

Now put the equivalent octal digit for each A~ àË`oH$ g_yh Ho$ ñWmZ na CgH$m Am°ŠQ>b A§H$
group : {bIZo na Omo g§»`m àmá hmoVr h¡,

Thus the result is (32) 8 `hr n[aUm_ h¡ …(32) 8
Note : Memorise table 3.2 ZmoQ >… Vm{cH$m 3.2 H$mo AÀN>r Vah go `mX H$a b|Ÿ&
But in case of fractional Binary numbers bo{H$Z nyU©V… Am§{eH$ g§»`mAm| H$s pñW{V _| h_
start from left and put trailing zeros. For ~m§E hmW H$s Va\$ go ewê$ H$aVo h¢ Ÿ& A§V _| `{X VrZ {~Q>
example, the Binary number 0.11010 can be Zht ~ZVo h¢ Vmo nrN>o Amoa eyÝ` bJm gH$ Vo h¢ Ÿ& CXmhaUV…
converted to octal as:
~mBZar g§»`m 0.11010 H$mo Am°ŠQ>b _| Bg Vah ^r
n[ad{V©V H$a gH$Vo h¢ …
0. 110 100 = (0.64 ) 8

Consider few more problems for Binary to A~ {d{Y II H$s ghm`Vm go Hw$N> ~mBZar g§»`mAm|
octal conversion using Direct method:
H$mo Am°ŠQ>b _| ~XbVo h¢ …
Problem 3.17 NPP àíZ 3.17
Convert following Binary Numbers into ZrMo Xr JB© ~mBZar g§»`mAm| H$mo Am°ŠQ>b _|
their octal equivalent : n[ad{V©V H$amo…

(a) 101011 (b) 11001 (c) 0.111011 (d) 1101.0101
Solution : hc :
(a) Start from right and make groups by (a) grYo hmW H$s Va\$ go VrZ-VrZ Ho$ g_yh
three bits :
~ZmAmo Ÿ&
101 011
put octal equivalent for each group. àË`oH$ g_yh Ho$ ñWmZ na Am°ŠQ>b A§H$ aIZo na
Thus, (101011) = (53) 8
2
(b) The given binary number is 11001, start (b)grYo hmW go Mbmo Am¡a VrZ-VrZ {~Q²> Ho$ g_yh
from right and put one leading zero as shown ~ZmAmo& Bg hoVw EH$ eyÝ` AmJo Omo‹S>Zm hmoJmŸ&
below
011 001:
Putting octal equivalents: àË`oH$ g_yh Ho$ ñWmZ na Am°ŠQ>b A§H$ {bImo …
(11001) = (31) 8
2
(c) The given binary number 0.111011 is a (c)Bg g§»`m _| ~mE§ hmW H$s Amoa go ewê$ hmoH$a
purely fractional number. Therefore start from VrZ-VrZ {~Q> Ho$ g_yh ~ZmAmo VWm CZHo$ Am°ŠQ>b A§H$
left and make groups of three bits and put their
octal equivalents : {bImo …
0.1110 11 = (0.73 ) 8