Teledyne LeCroy WaveAce 1000_2000 Remote Control - Manual (Page 91)

Operator's Manual
In a way not following the byte boundaries, bits are then segregated as follows:
31, 30, 29 . . . . 24, 23, 22, 21 . . . . 2, 1, 0
Sign exponent bits fractional bits
bit 0.5, 0.25, 0.125 . . .
The sign bit s is 1 for a negative number and 0 for a positive number, so it is easy to construct the sign
from this:
S = (-1)^s
The 8 exponent bits have the following values: bit 23 is worth 1, bit 24 is worth 2 . . . bit 29  64, bit 3
 128, so the resulting number can range from 0 to 28 - 1, which is 255.
127 is then subtracted from this value e creating a range from -127 to +128. This is then used as an
exponent to raise two to a power that is 2^e, to create a value E.
Then we have to create the multiplying number. The values of the 23 bits are as follows: bit 22 is worth
0.5, 21 is worth 0.25, 20 is worth 0.125, 19 is worth 0.0625 . . . .
When all the bits are added together, we obtain a positive number f that can be very close to one,
differing from it only by the value of the smallest bit, if all the bits are ones. (Generally the value will be
much less than one.) Then we add one to the result, obtaining 1 + f = F. The use of the added one
extends the dynamic range of the data. Another way of calculating f is to take the 23-bit number at
face value, and divide it by 2^24.
Finally we multiply together the sign, the value E, and the value F to create the final result:
Result = (-1)^s x 2^(e-127) x (1 + f) = S x E x F
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