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KBR multicomp F144-MS-1V1C1TI6DO6RO-3 - Page 52

KBR multicomp F144-MS-1V1C1TI6DO6RO-3
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Rev. 1.00
22586_EDEBDA0253-0318-1_EN
52
KBR multicomp F144-3 6DO/6RO
Example 1: -12.5 decimal = 0xC1480000 hex
M: 24 bit-mantissa
E: Exponent with oset of 127
S: Sign for mantissa (S=1 neg.; S=0 pos.)
Address +0 +1 +2 +3
Format SEEEEEEE EMMMMMMM MMMMMMMM MMMMMMMM
Binary 1 1 0 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Hex C1 48 00 00
The byte sequence is dened as follows:
The byte with the "S sign bit" is transmitted over the bus as the rst byte.
The sequence of the oat bytes of the bus can be reversed, if necessary, using
the device parameter 0xD02C (see table 1).
The register value 0xD02C in this case means:
with 1 -> sign bit S in 1
st
byte (sequence as dened)
with 0 -> sign bit S in 4
th
byte (sequence reversed)
The following information can be derived from this:
The sign bit is 1 => negative mantissa
The value of the exponent amounts to 10000010 bin or 130 dec.
This results in an exponent value of: 130 - 127 = 3
The mantissa has the following value: 10010000000000000000000
The decimal point can be found at the left end of the mantissa, preceded by a
1. This position does not appear in the hexadecimal numeric notation. If you
add 1 and set the decimal point at the beginning of the mantissa, the following
value is obtained:
1.10010000000000000000000
Now, the mantissa needs to be adjusted to the exponent. A negative exponent
shifts the decimal point to the left, a positive exponent shifts it to the right.
Since the exponent is 3, this is represented as: 1100.10000000000000000000
The number obtained corresponds to the binary oating-point number.
Binary digits to the left of the decimal point result in values > 1. In this example,
1100 bin results in the number 12 dec. {(1x23)+ (1x22)+ (0x21)+ (0x20)}
Binary digits to the right of the decimal point result in values < 1. In this ex-
ample, .100....... bin results in the number 0.5 dec. {(1x2-1)+ (0x2-2)+ (0x2-3)+
(0x2-4)}
By adding the individual values, 12.5 is obtained. As the sign bit was set, it is a
negative value, -12.5. The hexadecimal number 0xC1480000 thus corresponds
to -12.5.

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