GS1 EPC Tag Data Standard 1.6 - Indicod-Ecr

5517551855195520552155225523552455255526552755285529553055315532553355345535553655375538553955405541554255435544554555465547554855495550555155525553555455555556555755585559 The next data item is for OID 1, but since the table indicates that this OID’s data isalphanumeric, encoding into the Packed Object is deferred until after all of theknown-length numeric data is encoded. Next, the encoder finds that OID 3n is defined by Table L-1 as all-numeric, whoselength of 9 (in this example) was encoded as (9 – 4 = 5) into four bits within the AuxFormat subsection. Thus, a Known-Length-Numeric subsection is encoded for thisdata item, consisting of a binary value bit-pattern encoding 9 digits. Using Figure K-1 in Annex K, the encoder determines that 30 bits need to be encoded in order torepresent a 9-digit number as a binary value. In this example, the binary valueequivalent of “978123456” is the 30-bit binary sequence:“111010010011001111101011000000” At this point, encoding of the Known-Length Numeric subsection of the Data Sectionis complete.Note that, so far, the total number of encoded bits is (3 + 6 + 1 + 7 + 7 + 4 + 5 + 20 + 30)or 83 bits, representing the IDLPO Length Section (assuming that a single EBV-6 vectorremains sufficient to encode the Packed Object’s length), two 7-bit ID Values, theSecondary ID and Aux Format sections, and two Known-Length-Numeric compactedbinary fields.At this stage, only one non-numeric data string (for OID 1) remains to be encoded in theAlphanumeric subsection. The 10-character source data string is “1A23B456CD”. Thisstring contains no characters requiring a base-30 Shift out of the basic Base-30 characterset, and so Base-30 is selected for the non-numeric base (and so the first bit of theAlphanumeric subsection is set to ‘0’ accordingly). The data string has no substringswith six or more successive characters from the same base, and so the next two bits areset to ‘00’ (indicating that neither a Prefix nor a Suffix is run-length encoded). Thus, afull 10-bit Character Map needs to be encoded next. Its specific bit pattern is‘0100100011’, indicating the specific sequence of digits and non-digits in the source datastring “1A23B456CD”.Up to this point, the Alphanumeric subsection contains the 13-bit sequence ‘0 000100100011’. From Annex K, it can be determined that lengths of the two final bitsequences (encoding the Base-10 and Base-30 components of the source data string) are20 bits (for the six digits) and 20 bits (for the four uppercase letters using Base 30). Thesix digits of the source data string “1A23B456CD” are “123456”, which encodes to a 20-bit sequence of:“00011110001001000000”which is appended to the end of the 13-bit sequence cited at the start of this paragraph.The four non-digits of the source data string are “ABCD”, which are converted (usingTable K-1) to a sequence of four Base-30 values 1, 2, 3, and 4 (denoted as values v 3through v 0 in the formula below. These are then converted to a single Binary value, usingthe following formula:30 3 * v 3 + 30 2 * v 2 + 30 1 * v 1 + 30 0 * v 0In this example, the formula calculates as (27000 * 1 + 900 * 2 + 30 * 3 + 1 * 4) which isequal to 070DE (hexadecimal) encoded as the 20-bit sequenceCopyright ©2005- 2011 GS1 AISBL, All Rights Reserved. Page 209 of 218