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PSF reconstruction for Keck AO - Laboratory for Adaptive Optics

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5 FOURIER DOMAIN PSD MODELING 355.2 OperatorsThe operator <strong>for</strong>malism used in the derivations are reviewed here.Operators: spatialFor a mean-gradient type wavefront sensor, with rectangular sub-apertures in a regular rectangular grid, the measurementoperator M is (see [17]):M =comb× [Π ∗∇], (156)which has the Fourier trans<strong>for</strong>mF[M] =comb∗ [˜Π × ˜∇], (157)since the comb function is its own Fourier trans<strong>for</strong>m (see below). The corresponding functions (and their abbreviatednotations) are:( x) {1, |x|, |y| ≤d/2Π =, (158)d 0, otherwise∇ = ∂ [ ∂∂x = ∂x , ∂ ], (159)∂y( x)combd= ∑ ( x)δd − m =m+∞∑+∞∑m=−∞ n=−∞( x) ( y)δd − m δd − n , (160)where boldface denotes a (2-element) vector. The spatial plane coordinate is x =(x, y), with the Fourier conjugatespatial frequency variable f =(f x ,f y ). The Fourier trans<strong>for</strong>ms of (158)-(160) are:˜Π(fd) = sinc(fd) = sin(πf xd)πf x d× sin(πf yd), (161)πf y d˜∇ = 2πif = 2πi[f x ,f y ], (162)F[comb](fd) = ∑ mδ(fd − m). (163)A wavefront reconstructor may be defined as the operator R that fulfills R[M(ϕ ‖ )] = ϕ ‖ ). The Fourier domainrecostructor that does this isf˜R −1=4πi sinc(fd) . (164)Operators: temporalThe measurement operator M in (156) only represents the spatial wavefront sampling. To account <strong>for</strong> the finiteintegration time t i of the WFS we must evaluate various integrals of the type given below, represented by the WFStemporal integration operator I:I n ( ˜ϕ) = 1 ∫ +ti/2dτ ˜ϕ(f,t− t d − nt i − τ), (165)t i−t i/2where t d is an additional temporal delay due to e.g. CCD read-out, centroiding and <strong>reconstruction</strong> computations.We expressed this directly in the spatial Fourier domain, because that is the most useful <strong>for</strong>m <strong>for</strong> evaluating thePSDs. The integer index n was included <strong>for</strong> generality and to show the relation to the closed loop function G n (seesection 5.2), but <strong>for</strong> the remainder of this section we set n = 0 <strong>for</strong> brevity. Using the result (154) givesI o ( ˜ϕ) = 1 t i∑N ll=1∫ +ti/2˜ϕ l (f,t)exp(2πif · v l t d ) dτ exp(2πif · v l τ) . (166)−t} i/2{{ }I ′

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