BASIC REVISON OF LOGIC GATES
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Published by :
Syaqira Liyana Binti Ahmad Ghazali
( B031310568 )
Operand
|
Overflow
|
|
Addition
|
Subtraction
|
|
+ve
, -ve
|
no
|
no
|
+ve
, +ve
|
If
result sign is 1
|
If
result sign is 0
|
-ve
, -ve
|
If
result sign is 0
|
If
result sign is 1
|
Actual
|
Floating Point
|
0.0000000478
|
4.78 × 10-8
|
0.00000001
|
0.1 × 10-7
|
-1000000000
|
-1.0 × 109
|
-0.00111
|
-1.11× 2-3
|
0.3125
|
× 2 =
|
0.625
|
0
|
Generate
0 and continue.
|
0.625
|
× 2 =
|
1.25
|
1
|
Generate
1 and continue with the rest.
|
0.25
|
× 2 =
|
0.5
|
0
|
Generate
0 and continue.
|
0.5
|
× 2 =
|
1.0
|
1
|
Generate
1 and nothing remains.
|
Sign [1bit]
|
Exponent [8bits]
|
Fraction [23bits]
|
1 (-)
|
10001001
|
01001000010101
000000000 |
Sign [1bit]
|
Exponent [11bits]
|
Fraction [52bits]
|
1 (-)
|
10000001001
|
01001000010101
0000000000000000
0000000000000000
000000 |
Sign [1bit]
|
Exponent [8bits]
|
Fraction [23bits]
|
0(+)
|
10001000
|
011011000010
00000000000 |
If we
add biased exponents, bias will be added twice. Therefore we
need to subtract it once to compensate:
(10 + 127) + (-5 + 127) = 259
259 - 127 = 132 which is (5 +
127) = biased
new exponent
|
Copyright © 2013 | by BITS STUDENT SIG2