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The following examples illustrate how to obtain
ordinary (base-10) numbers from the compacted
binary data.
EXAMPLE 1. (Bytes are shown separated by spaces.)
Byte 1: Normal measurement. Range 2. RLC, QDR,
and bin number data are needed .
Byte 2, bit 7: Sign of C value is +
Byte 2, other bits: C exponent is 1100101 in 2s
complement notation, which is the same as a negative
exponent of 011011 (i.e., -27, base 10).
NOTE 1: bit 6 is always 1 for a negative exponent ,
0 for a positive exponent. Bytes 3, 4: C mantissa is
.879044 (directly from the 16-bit binary number).
NOTE 2: In the mantissa, the first bit has the weight
of 0.5 , the next bit 0.25, the next bit 0.125, and each
other bit half of the one before it, to the 16th bit. .
Therefore, the C value is + (2 to the -27th power)
* (.879044) = (7.45058) * (10 to the -9 power) *
(.879044) = 6.54939 * 10 e(-9) farads = 6.5494 nF.
Bytes 5, 6, and 7 (by the same method as bytes 2, 3,
and 4): D = .0003
Byte 8: Parameters are C/D. Data numbers are values
(normal). Bin 7 assignment.
Operation
EXAMPLE 2. (Bytes are shown separated by spaces.]
The example-2 interpretation is as follows:
Byte 1: Normal measurement. Range 3. RLC data
are needed.
Byte 2, bit 7: Sign of R value is +.
Byte 2, other bits: R exponent is 0001010 in 2s
complement notation, which is the same as a positive
exponent of 001010 (i.e., +10, base 10). (See NOTE
1, above.) Bytes 3,4: R mantissa is .637146 (from the
16-bit binary number; see NOTE 2, above).
Therefore, the R value is + (2 to the +10th power) *
(.637146) = (1024) * (.637146) = 652.44 ohms.
1693 RLC Digibridge
99
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