From Diffraction Patterns to Lecture 6 From Diffraction Patterns to Image Analysis

Recap…. Outline of Lecture 6 Diffraction and convolution Spatial frequencies and image analysis Outline of Lecture 6 The diffraction grating 2D Fourier transforms Image processing via (spatial) frequency analysis

No lecture next Monday (March 19)! Problems classes will run at usual times. Matlab – B11 (Monday and Friday), B15 (Thursday)

The diffraction grating (Problems Class 5) An (infinite) diffraction grating has a transmission function which looks like: We saw earlier how the double slit transmission function could be represented as a convolution of two functions. The grating transmission function can be treated similarly. ? The transmission function above can be represented as the convolution of two functions. Sketch them.

The diffraction grating The ‘train’ of delta functions is known as a Dirac ‘comb’ (or a Shah function). …whose Fourier transform is another Dirac comb: where:

The diffraction grating has Fourier transform: ? At what value of k is the first zero in G(k) located? ? Sketch the Fourier transform (i.e. the diffraction pattern) of the transmission function for the infinite diffraction grating. See Problems Class 5

The diffraction grating Now, what happens if we want to consider a real diffraction grating (i.e. one that is not infinite in extent)? ? The slits in the infinite grating above are spaced by an amount L. Imagine that we want to determine the Fourier transform of a grating which is 50L wide. How do we convert the transmission function for the infinite grating into that for a real grating which is 50L wide?

The diffraction grating

Reciprocal space and spatial frequencies Just as we can build up a complex waveform from a variety of sinusoids of different amplitudes and phases, so too can we generate an image from a Fourier integral.

Image Processing via Fourier Analysis Extracting information from an image Enhancing an image James Bouwer, UCSD https://ncmir.ucsd.edu/Training/UCSDCourses/Microscopy_NEU259

Patterns: Here, There, and Everywhere Is there a characteristic spatial frequency…?

Fourier and Fingerprints 1cm 1cm 0=black 255=white shades of grey 2 D a r y ( 5 6 £ ) Find average spacing between ridges 2. Remove noise

Fourier and Fingerprints Average spacing between ridges (a) Manually: 9 r i d g e s ¼ 4 : 8 m R p a c n 5 3 1cm

Fourier and Fingerprints (b) Automatically: Simplifying observations: - Ridges are “locally parallel”

Fourier and Fingerprints: 2D cosine waves 2. Locally parallel regions are similar to 2D cosine waves Plot 3D surface 2D cos wave as image

A recap of FT of 1D cosine wave ( x ) = A c o s k + Á f ( x ) = A c o s k + Á 1 2 £ e i ¡ ¤ B u t F T f e i k x g = p 2 ¼ ± ( ¡ ) + e ¡ i Á ± ( k ) ¤ F = p ¼ 2 A £ F ( k ) ¡ A p ¼ = 2 e i Á

2D cosine wave ¸ y y µ = d i r e c t o n o f t r a v e l x x µ y x ( ; ( ; ) [ k = 2 ¼ ¸ ] f A c o s + Á x = c o s µ + y i n f ( ; ) A [ k Á ]

2D Fourier Transform F T : ( k ; ) = 1 2 ¼ R f e i d F ( k ; ) = 1 p 2 x ; y ) = 1 2 ¼ R ¡ f e i d F ( k x ; y ) = 1 p 2 ¼ R ¡ · f e i d ¸ = F T y f x ( ; ) g I f ( x ; y ) = X Y F k T g [ s e p a r b l ] = F T y f Y ( ) x X g = F T x f X ( ) g y Y

Fourier Transform of 2D cosine wave ( x ; y ) = A c o s [ k µ + i n Á ] = A c o s [ k ( x + y ) Á ] = A 2 h e i [ k ( x c + y s ) Á ] ¡ = A 2 £ e i Á k c x s y + m l a r ¤ X ( ) Y F ( k x ; y ) = A 2 £ e i Á p ¼ ± ¡ c s + m l a r ¤ F ( k x ; y ) = ¼ A £ e i Á ± ¡ c o s µ n + ¤ k y x ¼ A e i Á ¡ k k s i n µ c o µ Fourier Transform of 2D cosine wave

k y F ( k ; ) k j F ( x ; y ) [ c o m p u t e r ] k x f ( x ; y ) ¹ ¸ B 2 ¼ ¸ A A 2D FT k x A j F ( x ; y ) [ c o m p u t e r ] A B C k 2 ¼ ¸ x f ( x ; y ) ¹ ¸ ¼ : 5 4 m [ c f . p r e v i o u s a l 3 ]

Nanoparticle networks on silicon 2D FFT and radially averaged FFT for 20 x 20 mm2 image Fourier analysis Intensity (a.u.) k (nm-1) 2 mm P. Moriarty, MDR Taylor, and M. Brust Phys. Rev. Lett. 89 248303 (2002) Peak at k ~ 7 x 10-3 nm-1 (i.e. l ~ 880 nm) but otherwise structure factor is featureless - no orientational order

) Key points F e a t u r o f n i m g v s l h - c ¸ d µ P k p ¼ 2 = , . x . ² P e a k c o r s p n d t m h i g : f b u . w [ ] ¯ l

High frequency “noise” 2. Removing noise Top right corner of filtered fingerprint Top right corner of fingerprint FT High frequency “noise” Low frequency “noise” 2D FT of whole fingerprint Band pass filter

Original Band-pass filtered

Things to think about: W h a t d o e s p k ( ; ) = r n i c l ? ® w u m x ; y ) = r n i c l ? ® w u m v g f I s t h e r i n g o f ¯ w d l y b c a u p m ? I f t h e r i d g s w m o a n , ® c u l v 2 D F T ?

2D Images and 2D Fourier Transforms Consider an aperture: ? f(x,y) in this case can be broken down into two functions f(x) and f(y). Sketch those functions.

2D Images and 2D Fourier Transforms So, for a square aperture we have two sinc functions, one along kx and one along ky Figures taken from Optics, Hecht (Addison-Wesley, 2nd Ed. 1987)

2D Images and 2D Fourier Transforms ? Which area of the diffraction pattern is associated with low spatial frequencies? With high spatial frequencies?

2D Images and 2D Fourier Transforms Aperture function (2 slits) 2 slit pattern ? What is the effect on the image if we only pass the spatial frequencies within the circle shown?

2D Images and 2D Fourier Transforms ? What is the effect on the image if we block the spatial frequencies within the circle shown?

Niamh’s Fourier transform (modulus2) Complex images: Fourier transforming and spatial filtering Niamh Niamh’s Fourier transform (modulus2)

Complex images: Fourier transforming and spatial filtering

Complex images: Fourier transforming and spatial filtering Optical computer