You made one error, which turns out to be critical: 00 84 represents a length of 132 bytes, not 131.

As shown below, the length would be 131 for a DQT segment containing a 16-bit table. However, the length would be 132 for a DQT segment containing two 8-bit tables. This appears to be what you are seeing.

A DQT segment is not limited to containing a single quantization table. At the bottom of Figure B.6 is a curved line showing the table data representation; the notation below that says, "Multiple (t = 1, ..., n)." Likewise, below Table B.4 is the notation, "The value n in Table B.4 is the number of quantization tables specified in the DQT marker segment."

If Pq (precision) is 0, that means 8 bit precision, consistent with the baseline standard. The formula for Lq (length of the table data) evaluates to 2 + (65 + 64 * 0) + (65 + 64 * 0) = 132. If there were a single 16-bit table in the segment (non-baseline), Pq would have a value of 1, and the formula for Lq would evaluate to 2 + (65 + 64 * 1) = 131. In each case, this represents the total number of bytes needed to hold 64 entries (whether 8 or 16 bytes) plus 1 byte to contain both Pq and Tq, plus 2 bytes to hold the length Lq.

As a result, it would seem that your JPEG baseline file actually contains five quantization tables. Since there are only four destinations for quantization tables, at least one of these is a replacement of a prior table for a certain segment of the scan. Tq (the second 4 bits of the 00 byte preceding the table data) will tell you which destination: 0, 1, 2, or 3. Since two of your DQT segments contain two tables, you will have to look in the middle to find this for the second table in each.

I used the 09/92 edition of the T.81 standard, available free here <http://eu.sabotage.org/www/ITU/T/T0081e.pdf>. I doubt the figure, table, and section numbers have changed much in other editions.

I hope this is helpful.

Tim_Bate

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Photoshop JPEG files (Quantization Tables)