Abstract

The optical implementation of fractional Fourier transforms (FractFT’s) of different orders usually requires different geometric configurations. I obtain a scaling relation for optical implementation of the FractFt that allows one to do the 1/Q1-order FractFT on the 1/Q-order optical FractFt setup, except for the usual Fourier transform. With the scaling relation, FractFt’s of different orders can be optically implemented simultaneously. The optical implementation of a FractFT in the frequency domain is also suggested.

© 1995 Optical Society of America

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References

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  1. D. Mendlovic, H. M. Ozaktas, J. Opt. Soc. Am. A 10, 1875 (1993).
    [CrossRef]
  2. A. W. Lohmann, J. Opt. Soc. Am. A 10, 2181 (1993).
    [CrossRef]
  3. S. A. Collins, J. Opt. Soc. Am. 60, 1168 (1970).
    [CrossRef]
  4. H. M. Ozaktas, B. Barshan, D. Mendlovic, L. Onural, J. Opt. Soc. Am. A 11, 547 (1994).
    [CrossRef]
  5. See the special issue on optical wavelet transforms, Opt. Eng. 31, 1823 (1992).

1994 (1)

1993 (2)

1992 (1)

See the special issue on optical wavelet transforms, Opt. Eng. 31, 1823 (1992).

1970 (1)

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Figures (2)

Fig. 1
Fig. 1

Optical setup for the 1/Q-order two-dimensional FractFt. For the one-dimensional FractFT, the lens should be replaced by a cylindrical lens. l = f tan(π/4Q) and f1/Q = f/ sin(π/2Q).

Fig. 2
Fig. 2

Optical setup for a multichannel FractFt or a one-dimensional signal. l = f tan(π/4Q), fx = f1/Q = f / sin(π/2Q), and fy = 2l for the astigmatic lens.

Tables (1)

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Table 1 Scaling Factors of Eqs. (7) and (11)

Equations (12)

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l = f tan ( π / 4 Q ) ,             f 1 / Q = f / sin ( π / 2 Q ) .
[ A B C D ] = [ cos ( π / 2 Q ) f sin ( π / 2 Q ) - sin ( π / 2 Q ) / f cos ( π / 2 Q ) ] .
E 2 ( x 2 ) = 1 λ B - + E 1 ( x 1 ) × exp [ i k 2 B ( A x 1 2 - 2 x 1 x 2 + D x 2 2 ) ] d x 1 = 1 λ f sin  ϕ - + E 1 ( x 1 ) × exp [ i k 2 f sin  ϕ ( x 1 2 cos  ϕ - 2 x 1 x 2 + x 2 2 cos  ϕ ) ] d x 1 ,
x 1 = x 1 / s ,             x 2 = x 2 / s ,
E 2 ( x 2 / s ) = 1 sin 1 / 2 ϕ - + E 1 ( x 1 / s ) × exp [ i π ( x 1 2 cot  ϕ - 2 x 1 x 2 csc  ϕ + x 1 2 cot  ϕ ) ] d x 1 .
l = f 1 / Q sin ( π / 2 Q ) tan ( π / 4 Q ) = 2 f 1 / Q sin 2 ( π / 4 Q ) .
x 1 = x 1 s cot  ϕ 1 cot  ϕ ,             x 2 = x 2 s sin  2 ϕ sin  2 ϕ 1 ,
E 2 ( x 2 s sin  2 ϕ sin  2 ϕ 1 ) = cot  ϕ 1 cot  ϕ - + E 1 ( x 1 s cot  ϕ 1 cot  ϕ ) × exp [ i π ( x 1 2 cot  ϕ 1 - 2 x 1 x 2 csc  ϕ 1 + x 2 2 cot  ϕ sin  2 ϕ sin  2 ϕ 1 ) ] d x 1 .
E 2 ( x 2 / s ) = F 1 / Q E 1 ( x 1 / s ) .
E 2 ( x 2 s sin  2 ϕ sin  2 ϕ 1 ) = S ( ϕ , ϕ 1 , x 2 ) F 1 / Q 1 E 1 ( x 1 s cot  ϕ 1 cot  ϕ ) ,
S ( ϕ , ϕ 1 , x 2 ) = cot  ϕ 1 cot  ϕ × exp [ i π x 2 2 ( cot  ϕ sin  2 ϕ sin  2 ϕ 1 - cot  ϕ 1 ) ] .
1 s cot  ϕ 1 cot  ϕ ,             1 s sin  2 ϕ sin  2 ϕ 1 ,             cos  ϕ 1 cos  ϕ .

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