Instructions
71
16. APPENDIX
16.1 CODING THE RESPONSE TO A RAPID READOUT OF THE MEASUREMENT
The two bytes transmitted in response to a rapid instruction are coded according to a special law.
A byte comprises two digits coded in Hexadecimal called A
1
, A
2
for the rst byte, and, et B
1
, B
2
for the
sencond byte transmitted.
To decode this information which arrives in the form A
1
A
2
B
1
B
2
, you must start by putting the information
in order, to get a new digit in the form B
2
A
1
A
2
B
1
.
These four digits contain the measurement information in the form of a digital value with three digits,
followed by an exponential to the power 2 which multiplies the preceding digital value:
x x x X 2
x
i.e. B
2
A
1
A
2
X 2
B1
Example: the two bytes transmitted after the normal measurement readout are: AF 6D.
After putting bytes in order, the nal value becomes DAF x 2
6
After decoding this new value in decimal the measurement becomes:
DAF = (13 x 256) + (10 x 16) + 15 = 3503
3503 x 2
6
= 3503 x 64 = 224192
This gure corresponds to 80 measurements of 250µs. To get the 250µs measurement, you
must therefore divide the gure obtained by 80.
The measurement becomes 316352/80 = 2802.4
The measurement thus calculated must be linearised according to the following formula to get the true
measurement: Measurement = Xa + b
The coefcients a and b are given in the table below, they depend on the type of probe used, which can
be found from the probe code given by reading the state of the instrument.
In the example used above, if the coefcient a is 1.893 10-3 i.e. 0.00163 and coefcient b is 7.300, the
real measurement becomes:
2802.4 x 0.001893 + 7.300 = 12.60 V/m
Each probe code contains 6 linearisation slopes whose coefcients depend on the measurement.
The table below indicates the number of the linearisation table allocated to each of the 17 probe codes.