Abstract

The problem of optical beam truncation through a transmitter aperture to maximize signal returns from a coherent lidar system is considered. For common transmitter and receiver apertures of the same size, the optimum ratio of receiver diameter to beam diameter is ~1.75 when viewing a far-field diffuse target. The beam diameter is defined as twice the radius at which the beam intensity is exp(−1) of the peak intensity.

© 1988 Optical Society of America

Full Article  |  PDF Article

References

  • View by:
  • |
  • |
  • |

  1. J. Y. Wang, “Heterodyne Laser Radar SNR from a Diffuse Target Containing Multiple Glints,” Appl. Opt. 21, 464 (1982).
    [CrossRef] [PubMed]
  2. J. Y. Wang, “Detection Efficiency of Coherent Optical Radar,” Appl. Opt. 23, 3421 (1984).
    [CrossRef] [PubMed]
  3. A. E. Siegman, “The Antenna Properties of Optical Heterodyne Receiver,” Proc. IEEE 54, 1350 (1966).
    [CrossRef]
  4. D. L. Killinger, N. Menyuk, W. E. DeFeo, “Experimental Comparison of Heterodyne and Direct Detection for Pulsed Differential Absorption CO2 Lidar,” Appl. Opt. 22, 682 (1983).
    [CrossRef] [PubMed]
  5. J. D. Gaskill, Linear Systems, Fourier Transforms, and Optics (Wiley, New York, 1978), p. 441.
  6. B. J. Bartholomew, General Dynamics Convair Division; private communications (1986).

1984 (1)

1983 (1)

1982 (1)

1966 (1)

A. E. Siegman, “The Antenna Properties of Optical Heterodyne Receiver,” Proc. IEEE 54, 1350 (1966).
[CrossRef]

Appl. Opt. (3)

Proc. IEEE (1)

A. E. Siegman, “The Antenna Properties of Optical Heterodyne Receiver,” Proc. IEEE 54, 1350 (1966).
[CrossRef]

Other (2)

J. D. Gaskill, Linear Systems, Fourier Transforms, and Optics (Wiley, New York, 1978), p. 441.

B. J. Bartholomew, General Dynamics Convair Division; private communications (1986).

Cited By

OSA participates in CrossRef's Cited-By Linking service. Citing articles from OSA journals and other participating publishers are listed here.

Alert me when this article is cited.


Figures (7)

Fig. 1
Fig. 1

Optical schematics of coherent receivers.

Fig. 2
Fig. 2

Normalized received signal vs beam truncation ratio—Case 1.

Fig. 3
Fig. 3

Normalized received signal vs beam truncation ratio—Case 2.

Fig. 4
Fig. 4

Normalized received signal vs beam truncation ratio—Case 3.

Fig. 5
Fig. 5

Normalized received signal vs beam truncation ratio—Case 4.

Fig. 6
Fig. 6

Normalized received signal vs beam truncation ratio—Case 5.

Fig. 7
Fig. 7

Normalized received signal vs beam truncation ratio—Case 6.

Equations (26)

Equations on this page are rendered with MathJax. Learn more.

U ( r ) = U 0 exp ( - r 2 2 α 2 - j k r 2 2 F ) W ( r ) ,
W ( r ) = { 1 r d 1 / 2 , 0 r > d 1 / 2 ,
U 0 ( ρ ) = k exp ( j k L ) 2 π j L d 2 r U 0 ( r ) exp ( j k ρ - r 2 2 L ) = 2 π U 0 λ L exp ( j k ρ 2 2 L ) 0 d 1 / 2 d r r J 0 ( k ρ r L ) × exp [ - r 2 2 α 2 + j k r 2 2 L ( 1 - L F ) ] ,
S d = P t λ 2 T d π d 2 ρ U 0 ( ρ ) 2 U l ( ρ ) 2 ,
S d = ( d 2 d 1 ) 4 32 ζ 1 2 ζ 2 2 π 0 d ( ρ d 2 ) ( ρ d 2 ) I 1 2 I 2 2 ,
S d = S d P t β 1 4 ( T d π ) ( λ d 1 ) 2 ,
I 1 = 0 1 / 2 d r r J 0 [ β 1 ( ρ d 1 ) r ] × exp [ - 2 ζ 1 2 r 2 + j 2 ( β 1 - α 1 ) r 2 ] ,
I 2 = 0 1 / 2 d r r J 0 [ β 2 ( ρ d 2 ) r ] × exp [ - 2 ζ 2 2 r 2 + j 2 ( β 2 - α 2 ) r 2 ] .
ζ 1 = d 1 2 α ; ζ 2 = d 2 2 α ; α 1 = k d 1 2 F ; α 2 = k d 2 2 F ; β 1 = k d 1 2 L ; β 2 = k d 2 2 L .
U l ( r ) = { 2 π d 2 2 r d 2 / 2 , 0 r > d 2 / 2.
S d = ( d 2 d 1 ) 4 32 ζ 1 2 π 0 d ( ρ d 2 ) ( ρ d 2 ) I 1 2 I 3 2 .
I 3 = 0 1 / 2 d r r J 0 [ β 2 ( ρ d 2 ) r ] exp ( j 2 β 2 r 2 ) .
S g = P t π 2 T g ( Δ ρ ) 4 U 0 ( 0 ) 2 U l ( 0 ) 2 ,
S g = 16 ζ 1 4 ( d 2 d 1 ) 4 I 4 I 5 ,
S g = S g d 1 4 P t T g ( Δ ρ ) 4 β 1 4 ,
I 4 = | 0 1 / 2 d r r exp [ - 2 ζ 1 2 r 2 + j 2 ( β 1 - α 1 ) r 2 ] | 2 ,
I 5 = | 0 1 / 2 d r r exp [ - 2 ζ 2 2 r 2 + j 2 ( β 2 - α 2 ) r 2 ] | 2 .
S g = 16 ζ 1 2 ( d 2 d 1 ) 2 I 4 I 6 ,
I 6 = | 0 1 / 2 d r r exp ( j 2 β 2 r 2 ) | 2 .
S d = P t | d 2 r U r ( r ) k exp ( j k f ) 2 π j f U l ( r ) | 2 ,
U l ( r ) = 2 f π k r J 1 ( k d 3 r f ) ,
S d = P t λ 4 f 2 L 4 ( T d π ) d 2 ρ U 0 ( ρ ) 2 U l ( ρ ) 2 ,
U l ( ρ ) = d 2 r U l ( r ) exp [ j k ρ - r 2 2 L ] .
S d = 16 ζ 1 2 π ( d 2 d 1 ) 4 0 d ( ρ d 2 ) ( ρ d 2 ) I 1 2 × | 0 1 / 2 d r J 1 ( Q r ) J 0 [ β 2 ( ρ d 2 ) r ] exp ( j 2 β 2 r 2 ) | 2 ,
S g = P t π 2 T g ( Δ ρ ) 4 λ 6 L 4 f 2 U 0 ( 0 ) 2 U l ( 0 ) 2 .
S g = 4 ζ 1 2 ( d 2 d 1 ) 2 I 4 | 0 1 / 2 d r J 1 ( Q r ) exp ( j 2 β 2 r 2 ) | 2 .

Metrics