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How CD/DVD Encryption works - need details

  • Thread starter JPC
  • Start date
  • #1
JPC
195
1
hello
We are currently doing a TIPE Project on our first year of Prepa ( http://en.wikipedia.org/wiki/Classe_Préparatoire_aux_Grandes_Écoles )
We are thinking of doing our project on DVD/CD's (The theme is Surfaces), we would need to explain with precision how the encryption is made and how the CD Drive reads it.
But this project requires very precise details, and our mathematical/Physical reflexion.

We have tried looking for information on Howstuffworks.com, but it doesn't go very much into the details.
Do you know any websites that would satisfy our needs ?
thank you
 

Answers and Replies

  • #2
MATLABdude
Science Advisor
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Are you asking about actual encryption (for copy protection / security), or how information is encoded on a CD / DVD?

Encryption / Cryptography:
http://en.wikipedia.org/wiki/Encryption

Encoding:
http://en.wikipedia.org/wiki/Encoding

If you're working with surfaces, you probably mean the latter. In which case, you might want to start with how a CD works:
http://en.wikipedia.org/wiki/Compact_Disc#Physical_details

A DVD works in nearly the same fashion, except the pits are smaller, and you need a higher-powered laser to read these smaller pits. Note that recordable / rewritable media work on a completely different principle, but can be read in a similar fashion:
http://en.wikipedia.org/wiki/Cd-r
http://en.wikipedia.org/wiki/CD-RW

These are good places to *start*--if this is for a university-level project, you can probably find some of the original papers in some IEEE journal (talk to librarians at your technical library).
 
  • #3
minger
Science Advisor
1,495
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The actual CSS decryption is well documented. It would probably violate forum rules to link to any pages or anything like that with information, but the information is quite readily available.

I recall a website that had the procedure for decrypting the CSS like 80 different ways (there was a quite long poem which is actually pretty cool).
 
  • #4
JPC
195
1
Thank you for the links and information

Actually right now, in the end of our first year, we must search information and see what project we present next year on that theme
The thing is, we have to make our projects very precise, they don't want the usual "talking" that everyone understands, and they expect to see how personal reflexion on the subject (using Maths/Physics/Engineering theories, going really into the details, asking companies for documents, etc).

Do you know to what theories, or properties we could link this to ?
 
  • #5
minger
Science Advisor
1,495
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Google it brotha. The actual algorithms are readily available.
 
  • #6
JPC
195
1
I saw on this page http://en.wikipedia.org/wiki/Compact_Disc : "The change in height between pits and lands results in a difference in intensity in the light reflected."
: i do not really understand why, and i cannot find any details concerning this phenomena.
The source also stated that the front side of the disc is composed of a plastic layer with the "holes" and "bumps", and of a reflecting sheet behind.
Why is there a big difference in intensity between a beam emitted on a bump and a beam emitted on a hole ?

I thought that the answer would be (but i am not sure) :
the plastic layer's thickness changes with the presence of a bump or hole, and when there is a bigger thickness, the beam travels a bigger distance in the material, so more of the laser beam's intensity is absorbed (due of the absorbance of the material) ?
is this correct ?
 
  • #7
MATLABdude
Science Advisor
1,655
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It might be, but the depth of the pits--125 nm of additional (optically transparent polycarbonate) would absorb a negligible amount of the light. It might be interference effects (a round trip into/out of the pit is 250 nm, or about half the wavelength of the 500-ish nm red laser. However, it might just be what angle the light reflects off the pits vs. the rest of the surface:
http://electronics.howstuffworks.com/cd5.htm

Since a pit would cause the reflected light to shift slightly off the detector element, you'd see lower intensity. When you have CD-Rs or CD-RWs, which actually have regions that have higher or lower absorbance, you'd have actually lowered intensity, and not just changing where the light lands.
 

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