CU Boulder 03-17-2003
ECE Department
Data Storage Technology
ECEN 4018/5018
Homework #5 (Solution)
Problem 1
The depth of pits on optical media should be equal to

/4, where

is the optical laser wavelength. The

/4 pit depth is used to have the reflected beam from the pit and the reflected beam from the surrounding land out of phase, so that it will cause a destructive interference. The total beam reflected when reading a pit will be cancelled or nearly cancelled and it appears on the photo detector as a dark or nearly dark spot. By using such a technique we can distinguish between a pit and a land. A land, on the other hand, will appear as a bright spot on the photo detector.
For a CD-ROM,

=780 nm in the air. However, this wavelength gets shorter inside a material with larger refraction index

. Since the pits are inside the disc substrate made of polycarbonate with

=1.55, the wavelength becomes
 m
=

/1.55.
Therefore, for a CD-ROM, the depth of pits is:

And for a DVD-ROM, the depth of pits is:
 m
/4 =

/(4

1.55) = 780/(4

1.55)

126 nm. m
/4 =

/(4

1.55) = 650/(4

1.55)

105 nm.
Problem 2
In phase change technology, the write process is based on the change of state of the media. In this technology the rewritable media state can be changed from crystalline state to amorphous state and vice versa.
Before writing any information on the disc, the medium is in the crystalline state. During the write process a high power laser beam is focused on the track to change its state into amorphous state. The latter state is less reflective than the crystalline state. So, at the place of a pit the state is amorphous and at a land the state is crystalline.
Thus, as in the case of non-rewritable media, the pits are less reflective than the lands and will appear darker on a photo detector.
To erase an already-recorded information, a laser beam is heated to an intermediate level of power (less than write level) which re-crystallizes the amorphous domains and therefore transforms the medium into its original state (crystalline state).
Problem 3
The substrate aperture radius is given by: R = d tan(sin -1 (NA/

)) a) For CDs: d =1.2 mm, NA = 0.45, and

=1.55 (for polycarbonate substrate)
R = 1.2
 tan(sin -1 (0.45/1.55))

0.364 mm
Therefore the diameter of the substrate aperture for CDs is:
D = 2

R = 0.73 mm.
For DVDs: d =0.6 mm, NA = 0.6, and

=1.55 (for polycarbonate substrate)
R = 0.6
 tan(sin -1 (0.6/1.55))

0.252 mm
Therefore the diameter of the substrate aperture for DVDs is:
D = 2

R = 0.5 mm. b) The substrate aperture is the area through which the laser beam enters the substrate layer of the disc. The advantage of having a substrate aperture much larger than the laser spot size resides in the fact that any small scratches or dust particles won't be an obstacle for the read process. In other words, even if we have scratches or dust particles much larger than the laser spot size (but still less than the substrate aperture), we can still perform the read process without any problem.

Problem 4
The intensity profile of a pickup head (optical head) is a spatial function, given by g ( d )
 sin
2
(

F c
(

F c d ) 2 d ) where F c
= 2NA/

is the spatial cutoff frequency.
When a disc is rotating and the information is read, the spatial information is transformed into a temporal information.
The corresponding temporal intensity profile is called "optical impulse response" and is given by h ( t )
 g ( d
 vt )
 sin
2
(

F c
(

F c vt ) vt )
2
 sin
2
(

F oc t )
(

F oc t )
2 where v is the linear velocity of the disc given in m/s and t is the time variable. F oc
= F c v is the temporal cutoff frequency given in Hz (number of cycles/s).
Usually h ( t ) is approximated by a Gaussian function of the form h ( t )
 e

( t /

)
2
In the spatial domain, the laser spot diameter is D =1.2
 m.
In the time domain, the laser spot diameter is represented as a time interval.
Thus the time interval (or time duration) of the laser spot is given by
T = D / v = 1.2

10 -6 /1.2 = 10 -6 s. t and the time
1 t
1
corresponding to h ( t ) = 0.5 is
= T /2 = 0.5

10 -6 = 0.5
 s
This gives
0 .
5
 e

( t
1
/

)
2
 
Ln ( 2 )
 
( t
1
/

) 2
   t
1
2
Ln ( 2 )
 t
1
Ln ( 2 )

0 .
60
 s