Working of compact disk - computer architecture, Computer Engineering

Working of compact disk:

A CD is built from 1.2 mm thick, approximately all-pure polycarbonate plastic and its weighs is approximately 15-20 grams. From the core outward components are at the center (spindle) hole, the clamping area (stacking ring), the first-shift area (clamping ring), the information (data) area the second-transition area (mirror band), , and the rim.

A thin layer of aluminum or, rarely, gold is applied to the surface to make it reflective, and is protected by a film of lacquer that is usually spin coated directly on top of the reflective layer, on which the label print is applied. General printing methods for CDs are and offset printing and screen-printing

CD data are stored as a series of tiny indentations called -pits? which is encoded in a spiral track molded into the top of the polycarbonate layer. The areas among pits are called -lands?.  Each pit is about 100 nm deep by 500 nm wide, and varies from 850 nm to 3.5 µm in the length.

The distance among the tracks, the pitch is 1.6 µm. CD is read by focusing a 780 nm wavelength (near infrared) semiconductor laser throughout the bottom of the polycarbonate layer. The modification in height among pits (in fact ridges as seen by the laser) and lands results in a difference in intensity in the light reflected. By measuring the intensity change having photodiode, the data may be read from the disc.

The lands and pits both do not directly represent the 0 and 1 of binary data. Instead of it, Non-return- to-zero, inverted (NRZI) encoding is used for this purpose: a change from pit to land or land to pit denote a 1, whereas no change denote a series of 0. There have to be at least 2 and no more than 10 zeros between each one that is defined by the length of the pit. This is decoded in turn by reversing the Eight -to-Fourteen Modulation which is used in mastering the disc, and afterwards reversing the Cross-Interleaved Reed-Solomon Coding, at last revealing the raw data stored on the disc.

CDs are susceptible to damage from both every day use and environmental contact. Pits are much nearer to the label side of a disc, so that dirt and defects on the clear side can be out of focus at the playback time. Thus, CDs suffer from scratch and damage on the label side while scratches on the clear side may be repaired by refilling them with same refractive plastic, or by polishing carefully. First music CDs were known to suffer from or "laser rot", or "CD rot", in which the internal reflective layer degrades. When this take place the CD may become unplayable.

Posted Date: 10/13/2012 6:29:28 AM | Location : United States

Related Discussions:- Working of compact disk - computer architecture, Assignment Help, Ask Question on Working of compact disk - computer architecture, Get Answer, Expert's Help, Working of compact disk - computer architecture Discussions

Write discussion on Working of compact disk - computer architecture
Your posts are moderated
Related Questions
In a two stage network there are 512 inlets and outlets, r=s=24. If the probability that a given inlet is active is 0.8, calculate: Blocking probability Given: N =M =512,

Q. Explain Redundant Array of Independent Disks levels? One such industrial standard that exists for multiple-disk database schemes is called as RAID which implies Redundant Ar

Hello Sir/Mam, Actually i am trying to implement the concept of MAC layer protocols in NS2..... But unfortunately i am not able to do that.... and not even able to get the code fo

Q. Example on Sorting using Combinational Circuit? Example: Think about a unsorted list having element values like given below {3,9,8,5,10,12,14,20,90,95,60,40,23,35,18,0}

Problem: A company requires a software application to calculate the weekly payment for its employees. The information about the employees (employees' data) is kept in a file.

Write a ‘C’ functions to arrange the elements of an integer array in such a way that all the negative elements are before the positive elements. The array is passed to it as an arg

Find how many bits of ADC are required to get an resolution of 0.5 mV if the maximum full scale voltage is 10 V Ans. Given Resolution=.5mV Full scale output=+10v %resolut

Memory-to-Memory Architecture : The pipelines can access vector operands, intermediate and final results directly in the main memory. This needs the higher memory bandwidth. How

INTERRUPT METHOD - USING PORTB CHANGE INTERRUPT By using 4 by 4 matrix keypad connected to PORTA and PORTB. The rows are connected to PORTA-Low (RA1-RA4) and the columns are co

Universal Elimination: Here for any sentence, there is A, containing a universally quantified variable, v, just for any ground term, g, so we can substitute g for v in A. Thus