the data structure of an audio CD

Q

What's the data structure of an audio CD?

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A
Based on the information provided on wikipedia.org, the data structure of an audio CD (Compact Disc Digital Audio or CD-DA) can be described as: 1. The smallest entity in an audio CD is called a frame, which consists of 588 bits of data. 2. A frame is constructed from a 24-byte music data into a 33-byte core data: 24 bytes of stereo sample data: 2 channels x 6 samples x 2 bytes (16 bits)/sample = 24 bytes. 8 bytes for Cross-Interleaved Reed-Solomon Coding error correction. 1 byte as subcode for control and display. 3. The frame core data (33 bytes = 264 bits) is then processed with a translation of: Each byte (8 bits) is translated into a 14-bit word using Eight-to-Fourteen Modulation, which alternates with 3-bit merging words, resulting a 17-bit data. The frame core data of 33 bytes is translated into 33 bytes * 17 bits per byte = 561 bits Then a 27-bit unique synchronization word is added, so that the number of bits in a frame totals 588 bits. 4. 98 frames are grouped into a sector, which has about 98 * 73.5 (588 bits) = 7,203 bytes of data, but only 98 * 24 = 2,352 bytes of music (stereo sample data). 5. The play back speed is 75 sectors per second, which results in 75 * 2,352 = 176,400 bytes of music per second. Dividing this number by channels and by 2 bytes per sample, we get 176,400 / 2 / 2 = 44,100 samples per second. 6. On a standard 74-minute audio CD, there are 74 minutes * 60 minutes per second * 75 sectors per second = 333,000 sectors on the CD. 7. Since a sector has 7,203 bytes of data, there will be 7,203 bytes per sector * 333,000 sectors per CD = 2,398,599,000 bytes, about 2,287 MB, of data an audio CD. This number seems to be too high and need further validation. 8. Looking at music data only, there will be 2,352 bytes per sector * 333,000 sectors per CD = 783,216,000 bytes, about 747 MB, of music data an audio CD. This number seems to be right.

2007-10-16, 7091🔥, 0💬