Page 69 - Engineering Digital Design

40 CHAPTER 2 / NUMBER SYSTEMS, BINARY ARITHMETIC, AND CODES

EXAMPLE 2.3 139 ]0 -» Wg
N/r Q R
139/8 = 17 3
17/8 = 2 1
2/ 8 =0 2 139io =
EXAMPLE 2.4 1 000 1 0 1 1 2
2 1 3
010001 Oil = 213 BCo
EXAMPLE 2.5 213 Bco -»• WBCH
21 3 8 B
213 BCo =010 001 Oil = 1000101 1 2 = 1000 1011 = 8B 16
EXAMPLE 2.6 213 8 -» N 5

2
1
213 8 = 2x8 + lx8 +3x8 ° = 139,0
N r Q R
139/5 = 27 4
27/5 = 5 2
5/5=1 0
1/5=0 1 213 8 = 1024 5
3 2 1
Check: 1 x 5 + 0 x 5 + 2 x 5 + 4 x 5° = 125 + 0 + 10 + 4 = 139i 0

2.5.2 Conversion of Fractions
By extracting the fraction portion from Eq. (2.2) one can write

l 2 n
•N = (a-is~ + a- 2s~ + ••• + a- ms-' ) s

l
= 5- fl_ , +fl-/5- /+ 1 (2.6)
V '=2 /,
in source radix s. This is called the nested inverse radix form and provides the basis for
computerized conversion.
If the fraction in Eq. (2.6) is represented in nested inverse radix r form, then
1 2 p
•N = (fc-ir- + b- 2r~ + •••+ b^s- } r

(2 7)
-

for any fraction represented in radix r. Now, if source N s is multiplied by r, the result is of
the form

•N sxr = I + F, (2.8)