Abstract

A two-frequency lidar inversion on the assumptions of a range-independent relationship between the extinction coefficients of the two considered lidar wavelengths and of constant extinction-to-backscatter ratios was originally developed by Potter [Appl. Opt. 26, 1250 (1987)]. It is an iterative procedure to retrieve the boundary value for solution of the single-scatter lidar equation. This boundary value is expressed by the aerosol transmission along the evaluated lidar path. Recently, Kunz [Appl. Opt. 38, 1015 (1999)] stated that there is not enough information in the lidar signals of two wavelengths to obtain a unique solution for the boundary value and hence for the aerosol extinction profile. It is shown that a unique solution of the two-frequency lidar inversion exists, for which an analytical expression of the boundary value and, hence, the aerosol extinction profile, is given.

© 1999 Optical Society of America

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References

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  1. J. F. Potter, “Two-frequency lidar inversion technique,” Appl. Opt. 26, 1250–1256 (1987).
    [CrossRef] [PubMed]
  2. J. Ackermann, “Two-wavelength lidar inversion algorithm for a two-component atmosphere,” Appl. Opt. 36, 5134–5143 (1997).
    [CrossRef] [PubMed]
  3. J. Ackermann, “Two-wavelength lidar inversion algorithm for a two-component atmosphere with variable extinction-to-backscatter ratios,” Appl. Opt. 37, 3164–3171 (1998).
    [CrossRef]
  4. G. Kunz, “Two-wavelength lidar inversion algorithm,” Appl. Opt. 38, 1015–1020 (1999).
    [CrossRef]
  5. G. Kunz, “Transmission as an input boundary value for an analytical solution of a single-scatter lidar equation,” Appl. Opt. 35, 3255–3260 (1996).
    [CrossRef] [PubMed]

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Figures (1)

Fig. 1
Fig. 1

(a) Contour plot of the variance criterion ∊ [Eq. (22)]. The lidar signals L L *(R) were computed from the extinction profiles α L *(R), which were generated with Eq. (12) for distinct combinations of α L *(R 0) and k*. The asterisk marks the true values of α L (R 0) and k. (b) Same as (a) but for Eq. (21) with k* and T L *(R 0, R F ). The asterisk marks the true values of T L (R 0, R F ) and k. It coincides with the solution of the analytical expressions [Eqs. (34) and (28)].

Equations (35)

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k=αSRαLR.
LLR=KLαLRSL-1exp-2 0R αLrdr,
LSR=KSkαLRSS-1exp-2k 0R αLrdr.
TLR0, RF=exp-R0RF αLrdr,
αLR=LLRBLTLR0, RF-2-1+DLR,
BL=2 R0RF LLrdr,
DLR=2 RRF LLrdr.
αSR=LSRBSTLR0, RF-2k-1+DSR,
BS=2 R0RF LSrdr,
DSR=2 RRF LSrdr.
αLRαLR0=LLRLLR0kk-1LSR0LSR1k-1.
αL*RαL*R0=LLRLLR0k*k*-1LSR0LSR1k*-1.
αLRαL*R0αLR0αL*R=LLRLSR0LLR0LSRk*-kk*-1k-1.
LL*RLLR=const.
αLRTLR0, R2αL*RTL*R0, R2=αLR0αL*R0.
TL*R0, R2TLR0, R2=LLRLSR0LLR0LSRk*-kk*-1k-1.
TLR0, Rk-1=TL*R0, Rk*-1.
lnTLR0, RlnTL*R0, R=k*-1k-1.
R0R αLrdrR0R αL*rdr=const.
αLRαL*R=const.
αLR=LLRR0RF LLrdrkk-1×k1-TLR0, RF2kR0RF LSrdr1-TLR0, RF2kLSR1k-1.
=i=1NLL*RiLLRi-LL*RiLLRi¯2.
αLR0=LLR01-TLR0, RF2BL,
αSR0=LSR01-TLR0, RF2kBS,
αLRF=LLRF1-TLR0, RF2BLTLR0, RF2,
αSRF=LSRF1-TLR0, RF2kBSTLR0, RF2k.
CR0, RF=LLR0LSRFLSR0LLRF.
CR0, RF=TLR0, RF2k-1.
αLRX=LLRXBLTLR0, RF-2-1+DLRX,
αSRX=LSRXBSTLR0, RF-2k-1+DSRX.
CR0, RX=LLR0LSRXLSR0LLRX.
CR0, RX=BLBSTLR0, RF2k+DSRXBL1-TLR0, RF2kBLBSTLR0, RF2+DLRXBS1-TLR0, RF2.
TLR0, RF2=DSRXBL-DLRXBSCR0, RXBLBSCR0, RX-DLRXBSCR0, RX-BLBSCR0, RF+DSRXBLCR0, RF.
TLR0, RF2=DLRXBSLSRXLL-1RX-DSRXBLLSR0LL-1R0BSLSRXLL-1RXDLRX-BL-BLLSRFLL-1RFDSRX-BS.
αLR=DSRLSR0LL-1R0LLR-BSLSR-LSRFLL-1RFLLRDSR-BSDSRLSR0LL-1R0DLR-BL-DLRLSRFLL-1RFDSR-BS.

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