Non-Redundant Aperture Masking Robert J. Vanderbei 2009 November 6 - PowerPoint PPT Presentation

Non-Redundant Aperture Masking Robert J. Vanderbei 2009 November 6 TPF Group Meeting http://www.princeton.edu/ ∼ rvdb

Forget PICTURES Today we’re doing MATH

Aperture Mask Consider a telescope with focal length f and an aperture that has been masked into a disjoint collection of small circular subapertures: A2 −8 � −6 A ( x, y ) = A k ( x, y ) , −4 k −2 0 where 2 4 6 A k ( x, y ) = 1 ( x − x k ) 2 +( y − y k ) 2 ≤ a 2 / 4 8 −8 −6 −4 −2 0 2 4 6 8 is the indicator function of a circular subaperture cen- tered at ( x k , y k ) and having diameter a .

Electric Field at Image Plane The monochromatic electric field at the image plane is then given by �� �� � e 2 πi ( xξ + yη ) /λf A ( x, y ) dxdy = e 2 πi ( xξ + yη ) /λf A k ( x, y ) dxdy E ( ξ, η ) = k �� �� � � e 2 πi (( x k + x ) ξ +( y k + y ) η ) /λf dxdy = e 2 πi ( x k ξ + y k η ) /λf e 2 πi ( xξ + yη ) /λf dxdy = k k � � � e 2 πi ( x k ξ + y k η ) /λf , = E a,f ( ξ, η ) k �� where λ is the wavelength of the light, denotes integration over the set � { ( x, y ) : x 2 + y 2 ≤ a 2 / 4 } , and �� e 2 πi ( xξ + yη ) /λf dxdy = a E a,f ( ξ, η ) = 2 ρJ 1 ( πaρ/λf ) � is the electric field associated with the usual Airy pattern for an aperture of diameter a (note that ρ = √ ξ 2 + η 2 simply denotes radius in the image plane).

The Image As usual, the image is the magnitude squared of the electric field � I ( ξ, η ) = | E ( ξ, η ) | 2 = P a,f ( ρ ) e 2 πi (( x k − x k ′ ) ξ +( y k − y k ′ ) η ) /λf k,k ′   �  ,  n + 2 = P a,f ( ρ ) cos(2 π (( x k − x k ′ ) ξ + ( y k − y k ′ ) η ) /λf ) k<k ′ where a,f ( ξ, η ) = a 2 P a,f ( ρ ) = E 2 4 ρ 2 J 2 1 ( πaρ/λf ) is just the ”wide” PSF of associated with the small apertures. Note that the image is this broad PSF modulated by tight fringe patterns associated with the longer baselines ( ( x k − x k ′ , y k − y k ′ ) ).

Full Aperture, One Subaperture, Two Subs A2 A2 A2 −8 −8 −8 −6 −6 −6 −4 −4 −4 −2 −2 −2 0 0 0 2 2 2 4 4 4 6 6 6 8 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 PSF PSF PSF −8 −8 −8 −6 −6 −6 −4 −4 −4 −2 −2 −2 0 0 0 2 2 2 4 4 4 6 6 6 8 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8

Full Aperture, Two Subapertures, Eleven Subs A2 A2 A2 −8 −8 −8 −6 −6 −6 −4 −4 −4 −2 −2 −2 0 0 0 2 2 2 4 4 4 6 6 6 8 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 PSF PSF PSF −8 −8 −8 −6 −6 −6 −4 −4 −4 −2 −2 −2 0 0 0 2 2 2 4 4 4 6 6 6 8 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8

Full Aperture, Two Subapertures, Eleven Subs A2 A2 A2 −8 −8 −8 −6 −6 −6 −4 −4 −4 −2 −2 −2 0 0 0 2 2 2 4 4 4 6 6 6 8 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 PSF PSF PSF −8 −8 −8 −6 −6 −6 −4 −4 −4 −2 −2 −2 0 0 0 2 2 2 4 4 4 6 6 6 8 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8

Phase Errors Now, suppose that atmospheric seeing has introduced some phase errors across the large aperture. For simplicity, we assume that the subapertures are small enough that we can regard the phase errors as fixed over each subaperture. Then, the electric field formula given above has to be modified to account for the phase shifts: �� e 2 πi ( xξ + yη ) /λf e 2 πiφ ( x,y ) A ( x, y ) dxdy E ( ξ, η ) = � e 2 πi ( x k ξ + y k η ) /λf e 2 πiφ k /λ , = · · · = E a,f ( ξ, η ) k where φ k = φ ( x k , y k ) ≈ φ ( x, y ) for all ( x, y ) in the k -th subaperture. And, the image is the magnitude squared of the electric field: � e 2 πi (( x k − x k ′ ) ξ +( y k − y k ′ ) η ) /λf +2 πi ( φ k − φ k ′ ) /λ I ( ξ, η ) = P a,f ( ρ ) k,k ′

Fourier Analysis Non-redundant aperture masking makes it possible to automatically remove the phase dis- tortions introduced by the aperture. To do this, one starts by taking the Fourier transform of the image: �� e 2 πi ( xξ + yη ) /λf I ( ξ, η ) dξdη F ( x, y ) = �� � e 2 πi ( xξ + yη ) /λf P a,f ( ρ ) e 2 πi (( x k − x k ′ ) ξ +( y k − y k ′ ) η ) /λf +2 πi ( φ k − φ k ′ ) /λ dξdη = k,k ′ �� � e 2 πi ( φ k − φ k ′ ) /λ e 2 πi [( x + x k − x k ′ ) ξ +( y + y k − y k ′ ) η ] /λf P a,f ( ξ, η ) dξdη = k,k ′ � e 2 πi ( φ k − φ k ′ ) /λ � = P a,f ( x + x k − x k ′ , y + y k − y k ′ ) . k,k ′

Fourier Transform of Image—Evenly Spaced No Phase Errors A2 mag of FT of PSF −8 −8 −6 −6 −4 −4 −2 −2 0 0 2 2 4 4 6 6 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8

Fourier Transform of Image—Irregularly Spaced No Phase Errors A2 mag of FT of PSF −8 −8 −6 −6 −4 −4 −2 −2 0 0 2 2 4 4 6 6 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8

Non-Redundancy and Closure Phase The function � P a,f is the subaperture pupil convolved with itself. Since the pupil has compact support (a disk of diameter a ), so does its convolution with itself (having diameter 2 a ). Therefore, F ( x, y ) is a sum of n 2 terms with each term having compact support. The pupil mask is called non-redundant if these terms have disjoint support for all k � = k ′ . In that case, we can extract n 2 − n distinct terms and register them (i.e., center them at (0 , 0) ): k, k ′ = 1 , 2 , . . . , n, F k,k ′ ( x, y ) = e 2 πi ( φ k − φ k ′ ) /λ � ( k � = k ′ ) . P a,f ( x, y ) , We can remove the phase errors by multiplying these functions pairwise (I don’t understand at the moment why the literature says we have to do a three-factor multiplication): F k,k ′ ( x, y ) F k ′ ,k ( x, y ) = � P 2 a,f ( x, y ) . Next, we take the square root and rebuild F ( x, y ) using these corrected factors and then inverse Fourier transform to get back to a real image.

Half-Wave Error Peak-To-Valley A2 Phase error Masked Phase error −8 −8 −8 −6 −6 −6 −4 −4 −4 −2 −2 −2 0 0 0 2 2 2 4 4 4 6 6 6 8 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 PSF mag of FT of PSF PSF2 −8 −8 −8 −6 −6 −6 −4 −4 −4 −2 −2 −2 0 0 0 2 2 2 4 4 4 6 6 6 8 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8

Half-Wave Error Peak-To-Valley, Full Aperture A2 Phase error −8 −8 −6 −6 −4 −4 −2 −2 0 0 2 2 4 4 6 6 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 PSF −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8

One-Wave Error Peak-To-Valley A2 Phase error Masked Phase error −8 −8 −8 −6 −6 −6 −4 −4 −4 −2 −2 −2 0 0 0 2 2 2 4 4 4 6 6 6 8 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 PSF mag of FT of PSF PSF2 −8 −8 −8 −6 −6 −6 −4 −4 −4 −2 −2 −2 0 0 0 2 2 2 4 4 4 6 6 6 8 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8

One-Wave Error Peak-To-Valley, Full Aperture A2 Phase error −8 −8 −6 −6 −4 −4 −2 −2 0 0 2 2 4 4 6 6 8 8 −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8 PSF −8 −6 −4 −2 0 2 4 6 8 −8 −6 −4 −2 0 2 4 6 8

Final Remarks/Questions • No photon noise • Monochromatic • What is the optimal distribution of subapertures? • Is NRM better than lucky imaging ? • Is NRM better than dynamic deconvolution ?

Off-Axis Source: Angle = ( α, β ) Radians Image-Plane Electric Field: � e 2 πi ( x k ξ + y k η ) /λf e 2 πiφ k /λ e 2 πi ( αx k + βy k ) /λ E ( ξ, η ) = E a,f ( ξ + αf, η + βf ) k Image: � e 2 πi ( ( x k − x k ′ ) ξ +( y k − y k ′ ) η ) /λf e 2 πi ( φ k − φ k ′ ) /λ e 2 πi ( α ( x k − x k ′ )+ β ( y k − y k ′ ) ) /λ P ( ξ, η ) = P a,f ( ξ + αf, η + βf ) k,k ′ Fourier Transform of Image: � e 2 πi ( φ k − φ k ′ ) /λ e 2 πi ( α ( x k − x k ′ )+ β ( y k − y k ′ )) /λ � F ( x, y ) = P a,f,α,β ( x + x k − x k ′ , y + y k − y k ′ ) k,k ′ � = F k,k ′ ( x + x k − x k ′ , y + y k − y k ′ ) k,k ′ where F k,k ′ ( x, y ) = e 2 πi ( φ k − φ k ′ ) /λ e 2 πi ( α ( x k − x k ′ )+ β ( y k − y k ′ )) /λ � P a,f,α,β ( x, y ) and � P a,f,α,β is the Fourier transform of ( ξ, η ) → P a,f ( ξ + αf, η + βf ) . Multiplying F k,k ′ by F k ′ ,k wipes out both phase error and tilt .