## Abstract

Wave-front reconstruction with the use of the fast Fourier transform (FFT) and spatial filtering is shown to be computationally tractable and sufficiently accurate for use in large Shack–Hartmann-based adaptive optics systems (up to at least 10,000 actuators). This method is significantly faster than, and can have noise propagation comparable with that of, traditional vector–matrix-multiply reconstructors. The boundary problem that prevented the accurate reconstruction of phase in circular apertures by means of square-grid Fourier transforms (FTs) is identified and solved. The methods are adapted for use on the Fried geometry. Detailed performance analysis of mean squared error and noise propagation for FT methods is presented with the use of both theory and simulation.

© 2002 Optical Society of America

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### Equations (44)

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(1)
$${s}_{x}[m,n]=\varphi [m+1,n]-\varphi [m,n],$$
(2)
$${s}_{y}[m,n]=\varphi [m,n+1]-\varphi [m,n].$$
(3)
$$X[k,l]=\mathcal{F}\{x[m,n]\}=\frac{1}{{N}^{2}}\sum _{p=0}^{N-1}\sum _{q=0}^{N-1}x[p,q]exp\left[-\frac{j2\pi}{N}(\mathit{kp}+\mathit{lq})\right].$$
(4)
$${S}_{x}[k,l]=\mathrm{\Phi}[k,l]\left[exp\left(\frac{j2\pi k}{N}\right)-1\right],$$
(5)
$${S}_{y}[k,l]=\mathrm{\Phi}[k,l]\left[exp\left(\frac{j2\pi l}{N}\right)-1\right].$$
(6)
$$\stackrel{\u02c6}{\mathrm{\Phi}}[k,l]=\left\{\begin{array}{ll}0,& k,l=0\\ \left\{\left[exp\left(-\frac{j2\pi k}{N}\right)-1\right]{S}_{x}[k,l]\right.\\ +\left.\left[exp\left(-\frac{j2\pi l}{N}\right)-1\right]{S}_{y}[k,l]\right\}& \\ \times {\left[4\left({sin}^{2}\frac{\pi k}{N}+{sin}^{2}\frac{\pi l}{N}\right)\right]}^{-1},& \mathrm{else}\end{array}\right..$$
(7)
$${s}_{x}[m,n]={i}_{x}[m,n]+{b}_{x}[m,n],$$
(8)
$${s}_{y}[m,n]={i}_{y}[m,n]+{b}_{y}[m,n].$$
(9)
$$-u1+u2=a,\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}u2+u3=0,$$
(10)
$$u3+u4=c+b\hspace{1em}\hspace{1em}u4+u5=0\hspace{1em}\hspace{1em}u5-u6=d.$$
(11)
$$\mathbf{Mu}=\mathbf{c}.$$
(12)
$${s}_{x}[m,n]=\frac{1}{2}(\varphi [m+1,n]-\varphi [m,n]+\varphi [m+1,n+1]-\varphi [m,n+1]),$$
(13)
$${s}_{y}[m,n]=\frac{1}{2}(\varphi [m,n+1]-\varphi [m,n]+\varphi [m+1,n+1]-\varphi [m+1,n]).$$
(14)
$$\stackrel{\u02c6}{\mathrm{\Phi}}[k,l]=\left\{\begin{array}{ll}0,& \hspace{1em}k,l=0,k,l=N/2\\ \left\{\left[exp\left(-\frac{j2\pi k}{N}\right)-1\right]\left[exp\left(-\frac{j2\pi l}{N}\right)+1\right]{S}_{x}[k,l]\right.& \\ \left.+\left[exp\left(-\frac{j2\pi l}{N}\right)-1\right]\left[exp\left(-\frac{j2\pi k}{N}\right)+1\right]{S}_{y}[k,l]\right\}& \\ \hspace{1em}\times {\left[8\left({sin}^{2}\frac{\pi k}{N}{cos}^{2}\frac{\pi l}{N}+{sin}^{2}\frac{\pi l}{N}{cos}^{2}\frac{\pi k}{N}\right)\right]}^{-1},& \hspace{1em}\mathrm{else}\end{array}\right..$$
(15)
$${S}_{\varphi}(k)=0.023{k}^{-11/3}{r}_{0}^{-5/3},$$
(16)
$${\sigma}_{{\varphi}_{w}}^{2}\approx 4{S}_{\varphi}(1/2d)\mathrm{\Delta}{k}^{2}.$$
(17)
$${\sigma}_{\varphi}^{2}=1.03(D/{r}_{0}{)}^{5/3}$$
(18)
$${\sigma}_{{\varphi}_{w}}^{2}\approx 1.13(d/D{)}^{11/3}{\sigma}_{\varphi}^{2}.$$
(19)
$${s}_{a}[m,n]={s}_{x}[m,n]+{s}_{y}[m,n],$$
(20)
$${s}_{b}[m,n+1]={s}_{x}[m,n]-{s}_{y}[m,n].$$
(21)
$${c}_{v}=\frac{{\displaystyle \sum _{m=0}^{N-1}}{\displaystyle \sum _{n=0}^{N-1}}\stackrel{\u02c6}{\varphi}[m,n]v[m,n]}{{\displaystyle \sum _{m=0}^{N-1}}{\displaystyle \sum _{n=0}^{N-1}}v[m,n]v[m,n]}.$$
(22)
$${\stackrel{\u02c6}{\varphi}}_{-v}[m,n]=\stackrel{\u02c6}{\varphi}[m,n]-{c}_{v}v[m,n].$$
(23)
$$\mathbf{g}=\mathbf{H}\varphi +\mathbf{n}.$$
(24)
$$\stackrel{\u02c6}{\varphi}=\mathbf{Mg}.$$
(25)
$$\u220a=\mathbf{Mg}-\varphi $$
(26)
$$\mathbf{b}=E(\u220a).$$
(27)
$${\mathbf{\Lambda}}_{\u220a}=E[(\u220a-\mathbf{b})(\u220a-\mathbf{b}{)}^{\mathrm{T}}].$$
(28)
$$\mathrm{mse}=\frac{E({\u220a}^{\mathrm{T}}\u220a)}{a}.$$
(29)
$$\mathbf{b}=(\mathbf{MH}-\mathbf{I})\varphi ,$$
(30)
$${\mathbf{\Lambda}}_{\u220a}=\mathbf{M}{\mathbf{\Lambda}}_{\mathbf{n}}{\mathbf{M}}^{\mathrm{T}}.$$
(31)
$${\mathrm{mse}}_{\mathit{np}}=\frac{\mathrm{mse}}{{\sigma}_{n}^{2}}=\frac{\mathrm{Trace}({\mathbf{MM}}^{\mathrm{T}})}{a}$$
(32)
$$\mathbf{b}=(\mathbf{MH}-\mathbf{I}){\mathbf{m}}_{\varphi},$$
(33)
$${\mathbf{\Lambda}}_{\u220a}=(\mathbf{MH}-\mathbf{I}){\mathbf{\Lambda}}_{\varphi}({\mathbf{H}}^{\mathrm{T}}{\mathbf{M}}^{\mathrm{T}}-\mathbf{I})+\mathbf{M}{\mathbf{\Lambda}}_{\mathbf{n}}{\mathbf{M}}^{\mathrm{T}}.$$
(34)
$$\mathrm{mse}=\frac{\mathrm{Trace}({\mathbf{\Lambda}}_{\u220a})}{a}.$$
(36)
$$=\frac{\mathrm{Trace}[(\mathbf{MH}-\mathbf{I}){\mathbf{\Lambda}}_{\varphi}({\mathbf{H}}^{\mathrm{T}}{\mathbf{M}}^{\mathrm{T}}-\mathbf{I})]+\mathrm{Trace}(\mathbf{M}{\mathbf{\Lambda}}_{\mathbf{n}}{\mathbf{M}}^{\mathrm{T}})}{a}.$$
(37)
$$\mathrm{mse}={\mathrm{mse}}_{\varphi}+{\mathrm{mse}}_{n}({\mathbf{\Lambda}}_{\mathbf{n}}).$$
(38)
$$\mathrm{mse}={\mathrm{mse}}_{\varphi}+{\sigma}_{n}^{2}{\mathrm{mse}}_{\mathit{np}}.$$
(39)
$${\mathrm{mse}}_{\mathit{np}}=0.46+0.087lna.$$
(40)
$${\mathrm{mse}}_{\mathit{np}}=0.09753+\frac{1}{\pi}lnN.$$
(41)
$${\mathrm{mse}}_{\mathit{np}}=0.17+0.13lna$$
(42)
$${\mathrm{mse}}_{\mathit{np}}=0.6558+0.1603lna.$$
(43)
$${\mathrm{mse}}_{\mathit{np}}=c+\frac{3}{\pi}ln(N-1),$$
(44)
$${\mathrm{mse}}_{\mathit{np}}=0.1456{ln}^{2}a-1.7922lna+7.6175$$