incremental recording for photorefractive hologram multiplexing: comment

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July 1, 1992 / Vol. 17, No. 13 / OPTICS LETTERS 961 Incremental recording for photorefractive hologram multiplexing: comment Eric S. Maniloff* and Kristina M. Johnson Optoelectronic Computing Systems Center, University of Colorado at Boulder, Boulder, Colorado 80309-0425 Received February 10, 1992 Recently it was claimed in Ref. 1 that incremental recording in photorefractive crystals requires less total recording time than sequential recording does. This result is shown to be in error. It is assumed that during recording the index modulation An increases as [1 - exp(-t/r)] to its saturation value and that during erasure it decays as [exp(-t/T)]. The solution for the nth recording time, tn, is 2 [(n - 1)/ + Ii tlT lfl[(n- 2 ) 3 J + ' (1) where 3 = An,/An,, the ratio of the index modula- tion recorded during the first exposure to the satu- ration value. An depends on the number of exposures N as 2 =A[ 1 + (N - 1)131(2 and the total recording time T is N T = ti + X tm= ti + T ln[1 + (N 1)P). (3) M=2 After one cycle with the use of incremental recording, in which each of the holograms is recorded for a time 8t, the index modulation for the first hologram in the superposition is [ P~ ~t (] - Wt] An 1 , 1 = An,{1 - exp( - )Jexp[( -)B] (4) The subscripts i andj on Anij indicate the number of the hologram and the number of completed recording cycles, respectively. After a second recording cycle, the index modulation is An 1 , 2 = AIns{1 - exp[ (at + ]} X exp[- (N -) ] (5) where t4 is the time to reach An 1 , 1 in a single record- ing. By including t4 in Eq. (5) it is found that Anl, 2 = Ans[1 - exp( - ] [ + x - (N - 1)5t X exp_. Proceeding in this manner, it is show that (6) straightforward to Ankt = An[,1 - exp( T)] [0 exp( - T) (7) The last exponential in Eq. (7) is approximately equal to one for small At. Evaluation of the summa- tion in Eq. (7) and assuming that N8t << leads to the result Ank,l An [1 -Nexp - ) N [l e x k r / (8) Since INbt is the total time used, it is replaced by T, as given by Eq. (3), so that the two recording meth- ods can be compared. This leads to the result An [ exptr N 1 + (N A-1 Equation (9) can be simplified to obtain Eq. (2), which shows that given equal recording times the two methods produce equal index modulations. For the case of unequal recording and erasure time constants, the two methods were compared nu- merically and were again found to be equivalent. *Present address, Physical Chemistry Laboratory, Swiss Federal Institute of Technology, ETH- Zentrum, CH-8092 Zurich, Switzerland. References 1. Y. Taketomi, J. E. Ford, H. Sasaki, J. Ma, Y. Fainman, and S. H. Lee, Opt. Lett 16, 1774 (1991). 2. E. S. Maniloff and K. M. Johnson, J. Appl. Phys. 70, 4702 (1991). 0146-9592/92/130961-01$5.00/0 © 1992 Optical Society of America (N - Wt X exp T

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Page 1: Incremental recording for photorefractive hologram multiplexing: comment

July 1, 1992 / Vol. 17, No. 13 / OPTICS LETTERS 961

Incremental recording for photorefractive hologrammultiplexing: comment

Eric S. Maniloff* and Kristina M. Johnson

Optoelectronic Computing Systems Center, University of Colorado at Boulder, Boulder, Colorado 80309-0425

Received February 10, 1992

Recently it was claimed in Ref. 1 that incrementalrecording in photorefractive crystals requires lesstotal recording time than sequential recording does.This result is shown to be in error.

It is assumed that during recording the indexmodulation An increases as [1 - exp(-t/r)] to itssaturation value and that during erasure it decaysas [exp(-t/T)]. The solution for the nth recordingtime, tn, is2

[(n - 1)/ + IitlT lfl[(n-2)3 J + ' (1)where 3 = An,/An,, the ratio of the index modula-tion recorded during the first exposure to the satu-ration value. An depends on the number ofexposures N as2

=A[ 1 + (N - 1)131(2and the total recording time T is

NT = ti + X tm = ti + T ln[1 + (N 1)P). (3)

M=2

After one cycle with the use of incrementalrecording, in which each of the holograms isrecorded for a time 8t, the index modulation for thefirst hologram in the superposition is

[ P~ ~t (] - Wt]An1,1 = An,{1 - exp( - )Jexp[( -)B]

(4)

The subscripts i andj on Anij indicate the numberof the hologram and the number of completedrecording cycles, respectively. After a secondrecording cycle, the index modulation is

An1,2 = AIns{1 - exp[ (at + ]}

X exp[- (N -) ] (5)

where t4 is the time to reach An1,1 in a single record-ing. By including t4 in Eq. (5) it is found that

Anl,2 = Ans[1 - exp( - ] [ + x -

(N - 1)5tX exp_.

Proceeding in this manner, it isshow that

(6)

straightforward to

Ankt = An[,1 - exp( T)] [0 exp( - T)

(7)

The last exponential in Eq. (7) is approximatelyequal to one for small At. Evaluation of the summa-tion in Eq. (7) and assuming that N8t << leads tothe result

Ank,l An [1 -Nexp - )N [l e x k r / (8)

Since INbt is the total time used, it is replaced by T,as given by Eq. (3), so that the two recording meth-ods can be compared. This leads to the result

An [ exptr N 1 + (N A-1/Equation (9) can be simplified to obtain Eq. (2),which shows that given equal recording times thetwo methods produce equal index modulations.

For the case of unequal recording and erasuretime constants, the two methods were compared nu-merically and were again found to be equivalent.

*Present address, Physical Chemistry Laboratory,Swiss Federal Institute of Technology, ETH-Zentrum, CH-8092 Zurich, Switzerland.

References

1. Y. Taketomi, J. E. Ford, H. Sasaki, J. Ma, Y. Fainman,and S. H. Lee, Opt. Lett 16, 1774 (1991).

2. E. S. Maniloff and K. M. Johnson, J. Appl. Phys. 70,4702 (1991).

0146-9592/92/130961-01$5.00/0 © 1992 Optical Society of America

(N - WtX exp

T