Reverse Engineering Of A Barcode Job - Pitney Bowes FastPac DI900 Manuallines

Reverse
Engineering
a Barcode
Job
SV61314 Rev. A
So from this you know that the first character is used as page number. You
can assume that since there is only 1 byte used and since the maximum num-
ber of pages used is so far is 3 that a max value of 31 pages for page number
should be sufficient (Number of pages that can be counted to is (2^(size in
bits) – 1, or (2^5)-1 = 31).
Page Number: Char = 1, Bit = 4, size = 5, encoding = binary, code base = 32.
You don't know what the second or third bytes are used for yet. You need to
find the highest value for the Wrap Around Sequence to know if those bits are
used. You do this by finding the point where the Wrap Around Sequence actu-
ally wraps.
For now you will set the WAS to start at character 4. You will set the start bit
to 4 (the highest number that five bits can count to is 31). You will set the en-
coding type to CCD since the Wrap Around Sequence appears as a decimal
number.
WAS: Char = 4, Bit = 4, Size = 10, encoding = CCD, code base = 10, wrap at=
99, include 0 = false.
You also know that the last byte is being used as an EOC. The value that the
EOC always has in this job is 1, so the EOC is in bit 0.
EOC: Char = 7, Bit = 0, Size = 1, is controlling = true.
After running for a while, you get a wrap around sequence error. The message
given is:
Expected 1, found 0.
You remove the sheet from the accumulator and place it back in the feeder.
You run the pre-run adjustments scanner setup on the sheet and the following
code is displayed:
Sheet #
100
Decoded Barcode Data
1010001
Barcode Theory • 3
Rational /Analysis
There is a 1 in byte 3.
3-23