CSI defines the largest allowable range of a
high resolution number to be 99999.
Interpretation of the decimal locator for a 4 byte
data value is given below. The decimal
equivalent of bits GH is the negative exponent
to the base 10.
BITS
G H A
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
C.3 GENERATION OF SIGNATURE
At the end of a binary transmission, a signature
is sent. The signature is a 2 byte integer value
which is a function of the data and the
sequence of data in the Output Array. It is
derived with an algorithm that assures a
99.998% probability of detecting a change in
the data or its sequence. The CR10 calculates
the signature using each transmitted byte
beginning with the Final Storage format data
(for K command, echo and carriage return line
feed are not included) until the 2 byte signature
itself. By calculating the signature of the
received data and comparing it to the
transmitted signature, it can be determined
whether the data was received correctly.
SIGNATURE ALGORITHM
•
S1,S0 - represent the high and low bytes of
the signature, respectively
•
M - represents a transmitted data byte
•
n - represents the existing byte
•
n+1 - represents the new byte
•
T - represents a temporary location
•
C - represents the carry bit from a shift
operation
1. The signature is initialized with both bytes
set to hexadecimal AA.
S
(n) = S
1
DECIMAL FORMAT
5 digits
XXXXX.
XXXX.X
XXX.XX
XX.XXX
X.XXXX
.XXXXX
(n) = AA
0
APPENDIX C. BINARY TELECOMMUNICATIONS
2. When a transmitted byte, M(n+1), is
received, form a new highsignature byte by
setting it equal to the existing low byte.
Save the old high byte for later use.
T
S
(n+1) = S
1
3. Form a temporary byte by shifting the old
low signature byte one bit to the left and
adding any carry bit which results from the
shift operation. A "shift left" is identical to a
multiply by 2. Ignore any carry bit resulting
from the add.
T
= shift left (S
2
4. Form the new low signature byte by adding
the results of operation 3 to the old high
signature byte and the transmitted byte.
Ignore any carry bits resulting from these
add operations.
S
(n+1) = T
0
As each new transmitted byte is received, the
procedure is repeated.
= S
(n)
1
1
(n)
0
(n)) + carry
0
+ S
(n) + M(n+1)
2
1
C-5