Page 154 - FUNDAMENTALS OF COMPUTER

NPP

154 Fundamentals of Computers NPP

1 1 0 1
x x x x
2 3 2 2 2 1 2 0
8 + 4 + 0 + 1 = 13
Thus, (1101) = (13) 10
2
(b) The given Binary Number is 101.11. (b) Xr JB© ~mBZar g§»`m 101.11 h¢ Ÿ& BgH$s {~Q>m|
Write the bits and the Binary point giving some d q~Xw H$mo Wmo‹S>r Xya-Xya {bI H$a ZrMo {MÌ _| Xem©E
space.
AZwgma hb H$amoŸ&
1 0 1 . 1 1
2 2 2 1 2 0 2 -1 2 -2
The decimal equivalent can be calculated Xe_bd Vwë` H$s JUZm Bg àH$ma go H$s Om gH$Vr
as below :
h¡ …
= 1 × 2 + 0 × 2 + 1 × 2 + 1 × 2 + 1 × 2 -2
1
0
-1
2
= 4 + 0 + 1 + .5 + .25
= 5.75
Thus, (101.11) = (5.75) 10
2
(c) The given Binary number is 101011 write (c) Xr JB© ~mBZar g§»`m 101011 h¡ Ÿ& BgH$s
the bits and their positional weights below it:
{~Q>m| d ^mam| H$mo ZrMo Xem©E AZwgma {bImoŸ&
1 0 1 0 1 1
2 5 2 4 2 3 2 2 2 1 2 0
32 + 0 + 8 + 0 + 2 + 1 = 43
Thus, (101011) = (43) 10
2
Octal to Decimal Conversion Am°ŠQ>b go Xe_bd _| n[adV©Z
The Base of octal number system is 8. Am°ŠQ>b H$m AmYma 8 h¡ Ÿ& AV… gmar JUZmE± 8
Therefore all the calculations are performed
with 8. The following example illustrates the H$mo boH$a H$s OmE§Jr Ÿ& ZrMo Xem©E CXmhaUm| go `o ñnï>
procedure : hmoJm…
Problem 3.13 àíZ 3.13
Convert following Octal number into Decimal : Am°ŠQ>c go Xe_bd _| ~Xbmo …
Solution : hc :
(a) The given octal number is (27) . The (a) Am°ŠQ>b g§»`m (27) H$m Xe_bd Bg àH$ma
8
decimal equivalent can be calculated as below: go àmá {H$`m Om gH$Vm h¡: 8
2 7
x x
8 1 8 0
16 + 7 = 23
Thus, (27) = (23) 10
8