CMC TECHNICAL REFERENCE MANUAL
1X36003 Version 2.52
1996-1999 Ingersoll-Rand Company
Date of Issue: 18-Oct-1999
125
3. From the result obtained from step 2, subtract 1. Then, multiply this result by 2. If the
result is less than 1, then the value of the first mantissa bit is 0. Otherwise, the mantissa
bit is 1. If the result is greater than or equal to 1, then subtract 1 from the result and
proceed with step 3 until the result is 0 or you have gone through this process 23 times.
4. Combine all 32 bits from the steps above and convert this value to hexadecimal. These
32 bits are the 4 hexadecimal data bytes needed for the command.
As an example, we will start with the decimal value of 105.4.
1. Since this is a positive number, the first bit is
0
.
2. Determine the exponent bits by ...
It took us six iterations to get the result to a number that is less than two and greater than or
equal to one. Now, we must add 127 for an exponent of 133. Converting this to binary, the
next eight bits are represented as
10000101
.
3. Determine the mantissa bits by ...
From the table above, 10100101100110011001100 represent the next 23 bits.
Iteration
Decimal
Result
1
105.40000
/
2 =
52.700000
2
52.70000
/
2 =
26.350000
3
26.35000
/
2 =
13.175000
4
13.17500
/
2 =
6.587500
5
6.58750
/
2 =
3.293750
6
3.29375
/
2 =
1.646875
Iteration
Decimal Operation Result
Bit
1
1.646875
- 1 * 2 = 1.29375
1
2
1.29375
- 1 * 2 =
0.5875
0
3
0.5875
* 2 =
1.175
1
4
1.175
- 1 * 2 =
0.35
0
5
0.35
* 2 =
0.7
0
6
0.7
* 2 =
1.4
1
7
1.4
- 1 * 2 =
0.8
0
8
0.8
* 2 =
1.6
1
9
1.6
- 1 * 2 =
1.2
1
10
1.2
- 1 * 2 =
0.4
0
11
0.4
* 2 =
0.8
0
12
0.8
* 2 =
1.6
1
13
1.6
- 1 * 2 =
1.2
1
14
1.2
- 1 * 2 =
0.4
0
15
0.4
* 2 =
0.8
0
16
0.8
* 2 =
1.6
1
17
1.6
- 1 * 2 =
1.2
1
18
1.2
- 1 * 2 =
0.4
0
19
0.4
* 2 =
0.8
0
20
0.8
* 2 =
1.6
1
21
1.6
- 1 * 2 =
1.2
1
22
1.2
- 1 * 2 =
0.4
0
23
0.4
* 2 =
0.8
0
