## Abstract

In an optical disk drive, it is well known that a tilt of the disk
causes an offset in the tracking-error signal (TES). One effect of
disk tilt is the introduction of a dc component to the TES, which can
be largely corrected by operation of the tracking system at the
midpoint between the maximum and the minimum values of the open-loop
TES. However, this method of correcting for the dc shift in the TES
does not correct for the effect of coma in the focused spot, which
leads to track offset. The track offset of a system is defined as
the distance between the peak irradiance in the focused spot and the
center of the groove when the tracking system is operating at the
midpoint between the maximum and the minimum values of the open-loop
TES in the presence of disk tilt. Calculations are performed that
show the dependence of track offset on various system parameters,
including track pitch, wavelength, and numerical aperture and rim
intensity of the objective lens, and on the regions of the beam used to
generate the TES. The track offsets for several beam-segmentation
schemes are calculated for a digital versatile disk that uses
push–pull and differential phase tracking. It is shown that for
differential phase tracking the value of track offset depends on the
mark length.

© 1998 Optical Society of America

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### Equations (9)

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(1)
$${W}_{31}=\left(\frac{1}{2n}-\frac{1}{2{n}^{3}}\right){\mathrm{NA}}^{3}\frac{t}{\mathrm{\lambda}}\mathrm{\delta}\mathrm{\theta},$$
(2)
$${W}_{31}\left(\mathrm{\text{in waves}},\mathrm{\text{for CD-R}}\right)=0.715\times \mathrm{\delta}\mathrm{\theta}\left(\mathrm{in}\mathrm{degrees}\right).$$
(3)
$${W}_{31}\left(\mathrm{in}\mathrm{waves},\mathrm{for}\mathrm{DVD-R}\right)=0.676\times \mathrm{\delta}\mathrm{\theta}\left(\mathrm{in}\mathrm{degrees}\right).$$
(4)
$$\frac{D}{R}=\frac{2}{\mathrm{NA}}\mathrm{\delta}\mathrm{\theta},$$
(5)
$$\mathrm{\delta}x=-\frac{\mathrm{\lambda}}{\mathrm{NA}}{W}_{11},$$
(6)
$$\mathrm{\delta}x=\left(-0.653\right)\frac{\mathrm{\lambda}}{\mathrm{NA}}{W}_{31}.$$
(7)
$$\frac{{W}_{11}}{{W}_{31}}=\left\{\begin{array}{ll}-0.616& \mathrm{RI}\mathrm{of}30\%\\ -0.644& \mathrm{RI}\mathrm{of}50\%\\ -0.653& \mathrm{RI}\mathrm{of}70\%\u2013100\%\end{array}\right.,$$
(8)
$${\mathrm{TO}}_{\mathrm{CD}-\mathrm{R}}\left(\mathrm{nm}/\mathrm{deg}\mathrm{of}\mathrm{disk}\mathrm{tilt}\right)=\left({W}_{11}/{W}_{31}+0.653\right)\left(1070\mathrm{nm}/\mathrm{deg}\right).$$
(9)
$${\mathrm{TO}}_{\mathrm{DVD}-\mathrm{R}}\left(\mathrm{nm}/\mathrm{deg}\mathrm{of}\mathrm{disk}\mathrm{tilt}\right)=\left({W}_{11}/{W}_{31}+0.653\right)\left(730\mathrm{nm}/\mathrm{deg}\right).$$