For a compact disk, audio is converted to digital with 16-bit samples, and is tr

Question

For a compact disk, audio is converted to digital with 16-bit samples, and is treated a stream of 8-bit bytes for storage. One simple scheme for storing this data, called direct recording, would be to represent a 1 by a land and a 0 by a pit. Instead, each byte is expanded into a 14-bit binary number. It turns out that exactly 256(28) of the total of 16,134 (214) 14-bit numbers have at least two 0s between every pair of 1s, and these are the numbers selected for the expansion from 8 to 14 bits. The optical system detects the presence of 1s by detecting a transition for pit to land or land to pit. It detects 0s by measuring the distances between intensity changes. This scheme requires that there are no 1s in succession; hence the use of the 8-to-14 code.
The advantage of this scheme is as follows. For a given laser beam diameter, there is a minimum-pit size, regardless of how the bits are represented. With this scheme, this minimum-pit size stores 3 bits , because at least two 0s follow every 1. With direct recording the same pit would be able to store only one bit. Considering both the number of bits stored per pit and the 8 to 14 bit expansion, which scheme stores the most bits and by what factor?

Explanation / Answer

We have to compare the direct recording scheme of storing 1 bit per pit and the 8 to`14 bit scheme of storing 3 bits in the same minimum pit size and achieve an approximate of reliability of each.

For the direct recording scheme,

1 micron=0.001mm

Say, Min pit size P=1 bit per pit

We store * 8 bit bytes during storage stage.

Lets say the approximate size of 1 bit in area=0.8 micron

Hence, area for 1 bit=(0.8 micron)2

=0.64 sq microns

=0.32*10-6 mm2 per bit (1 micron=0.001mm)

To convert to bits per mm2 ,take reciprocal.

=1/0.32*10-6 mm2 per bit=3125000 bits per mm2 ~3200000 bits per mm2

Since we have 8 bits per byte, 3200000 /8=4,00,000 bytes per mm2

=400 kb per mm2

We have to calculate the total area of the disc using pi*r2 and subtracting outside and inner diameter values.

Say, forease of calculation our value of total area=8000m2

Then, total active area=400 kb per mm2 *8000m2

   =3200Mb

Now, lets consider our 8 to 14 recording scheme which stores 3 bits in the same minimum pit size.

Now, Min pit size P=3 bit per pit

We store 8 bit bytes during storage stage.

Lets say the approximate size of 1 bit in area=0.8 micron

Hence, area for 1 bit=1/3*(0.8 micron)2

=0.21 sq microns

=0.10*10-6 mm2 per bit (1 micron=0.001mm)

To convert to bits per mm2 ,take reciprocal.

=1/0.10*10-6 mm2 per bit=10000000 bits per mm2

Since we have 8 bits per byte,10000000 /8=1250000 bytes per mm2

=1250 kb per mm2

We have to calculate the total area of the disc using pi*r2 and subtracting outside and inner diameter values.

For ease of calculation our value of total area=8000m2

Then, total active area=1250 kb per mm2 *8000m2

   =10000Mb

=10Gb

Hence, ratio of total capacities =10,000/3200=3.125.

Hence, our second scheme of 8 to 14 bits stores more number of bits than direct recording scheme (by a factor of approximately 3.125)

     

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