Hardware Reference
In-Depth Information
Handling DVD Errors
DVDs use more powerful error-correcting codes than were first devised for CDs. Unlike CDs, which
have different levels of error correction depending on whether audio/video or data is being stored,
DVDs treat all information equally and apply the full error correction to all sectors.
The main error correcting in DVDs takes place in the ECC frame. Parity Outer (column) and Parity
Inner (row) bits are added to detect and correct errors. The scheme is simple yet effective. The
information from the data frames is first broken up into 192 rows of 172 bytes each. Then a
polynomial equation is used to calculate and add 10 PI bytes to each row, making the rows 182 bytes
each. Finally, another polynomial equation is used to calculate 16 PO (Parity Outer) bytes for each
column, resulting in 16 bytes (rows) being added to each column. What started out as 192 rows of
172 bytes becomes 208 rows of 182 bytes with the PI and PO information added.
The function of the PI and PO bytes can be explained with a simple example using simple parity. In
this example, 2 bytes are stored (01001110 = N, 01001111 = O). To add the error-correcting
information, they are organized in rows, as shown here:
Data bits
1 2 3 4 5 6 7 8
-------------------------------------
Byte 1 0 1 0 0 1 1 1 0
Byte 2 0 1 0 0 1 1 1 1
-------------------------------------
Then one PI bit is added for each row, using odd parity. This means you count up the 1 bits: In the
first row there are four, so the parity bit is created as a 1, making the sum an odd number. In the
second row, the parity bit is a 0 because the sum of the 1s was already an odd number. The result is
as follows:
Data bits |
1 2 3 4 5 6 7 8 | PI
----------------------------- |---------
Byte 1 0 1 0 0 1 1 1 0 | 1
Byte 2 0 1 0 0 1 1 1 1 | 0
----------------------------- |---------
Next, the parity bits for each column are added and calculated the same as before. In other words, the
parity bit will be such that the sum of the 1s in each column is an odd number. The result is as
follows:
Data bits |
1 2 3 4 5 6 7 8 | PI
----------------------------- |---------
Byte 1 0 1 0 0 1 1 1 0 | 1
Byte 2 0 1 0 0 1 1 1 1 | 0
----------------------------- |---------
PO 1 1 1 1 1 1 1 0 | 1
Now the code is complete, and the extra bits are stored along with the data. So, instead of just the 2
bytes being stored, 11 additional bits are stored for error correction. When the data is read back, the
