Fourier Optics and Image Analysis

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1 Laboratory instruction to Fourier Optics and Imae Analysis Sven-Göran Pettersson and Anders Persson updated by Maïté Louisy and Henrik Ekerfelt This laboratory exercise demonstrates some important applications of Fourier optic. It shows how Fourier transformations can be made with optics and how diffractin objects can be modified/analyzed usin appropriate masks in the Fourier plane. Preparation: See problems in the end (all must be solved before the laboratory work). If you fail to show at least an attempt to solve all problems, you will not be allowed to participate in the laboratory exercise. Theory: See the textbook Chapter 4 on Fourier optics expecially 4.2B, 4.3A and 4.4. Also review the properties of the Fourier transform in Appendix A. Set-up for the laboratory work: The principle of the optical set-up used is shown in fiure 1: Spatial filter L1 L2 Object plane FTplane L3 L4 S Imae plane to L3 Fiure 1. The optical processor used for imae processin Screen A HeNe-laser, spatially filtered by a microscope objective and a pinhole, illuminates the object coherently. The wavelenth of the laser liht is 633 nm. A mirror can be placed between the lenses L1 and L2 so that white liht can be entered into the optical system. The lenses L1 and L2 ive toether a ood point imae of the pinhole in the Fourier plane. When an object is placed at the object plane, a two-dimensional Fourier transformation is achieved at the FT-plane. This means that the transmission function is transferred into a combination of spatial frequencies. With the lens L3, an enlared imae of the Fourier plane is projected onto the screen. The lens L4 ives an enlared imae of the object instead. This lens can easily be moved in place or removed from the optical system. The advantae of this set-up is that, the Fourier plane bein accessible, it is possible to filter (or modify) in this plane some frequency components by usin an appropriate transmission mask. We can therefore filter the imae, i.e. modify its spatial properties (the same way we can modify temporal properties of an electronic sinal with low-pass, hih-pass or band-pass filters).

2 Laboratory experiments: 1. Adjustment of the set-up a) Turn on the laser and check that the pinhole is in the riht position b) Identify the different optical components in the Fourier transform set-up. 2. The Fourier transform of some simple objects a) Put the object D2-2 in the object holder. b) Adjust L4 so an imae of the object is seen on the screen. It is now easy to chane between object and transform by enterin or removin the lens L4. c) Compare the objects D2-1 and D2-4. Examine both objects and their transforms. d) Which tranform theorems can you see illustrated? 3. The Fourier transform of many objects a) Enter the object D3-1 (many circular openins). By usin the same reasonin as in the last preparation exercise (fiure P2), try to understand the features of the Fourier transform. b) Compare the transform of object D3-2 (randomly placed circular openins) with the transform of object D3-1. c) Enter the object D3-3 (many more circular openins) and compare with the object D3-2. Why is the Fourier transform like this, what do you learn about the object? 4. The Fourier transform of a periodic object. a) Enter two ratin patterns (object D4-1) in the holder and a mask, so that both ratins can be viewed separately. b) Compare the Fourier transforms of the two patterns. c) Block the zeroth order by insertin a mask (dark spot) in the Fourier plane. Compare the two imaes of the objects when the zeroth order is blocked. (Check the imaes in detail!) 5. The transfer function of a lens a) Put object D5-1 in the holder (sector star). b) Adjust for sharp imae and study the quality of the imae in the centre. What do you learn about the lenses? c) Make a low-pass, hih-pass and band-pass filterin by usin small holes and dark spots. 6. Spatial filterin a) Place the object D6-1 (cross ratin) in the holder. b) Compare the object and the transform. c) Make vertical or horizontal spatial filterin, i.e. transform the rid into a ratin. 7. Simple pattern reconition 2

3 a) Place the object D7-1 in the object plane. Increasin frequency Increasin frequency Fiure 2. Object D7-1. b) Put a metal plate with a small openin (1 mm) in the Fourier plane and study the imae when only the zeroth order is transmitted throuh the system. c) Move the hole in the Fourier plane and study the appearance of the imae. 8. Theta modulation and incoherent illumination a) Place the object D8-1 (three bi circular areas) in the object plane. b) Study the Fourier transform of the object. c) Put a mirror between L1 and L2 and direct white liht into the system. d) Study the imae in the Fourier plane. e) What is the reason for usin monochromatic liht when makin imae processin? f) Filter the Fourier plane with filter D8-2 and study the imae. Adjust so that one area will be red, one yellow and one blue. Theory. For the one-dimensional continuous Fourier transform we have: and (1) Further more for a function f(t), it s Fourier transform F(t) and a real scalin factor a the function f(t/a) has a Fourier transform a F(aν). In this lab, the Convolusion Theorem will be used. The convulsion of two functions f 1 (t) and f 2 (t) with Fourier transforms F 1 (ν) and F 2 (ν) respectively is the same as the inverse Fourier transform of ν ν. 3

4 For the preparation problems all the theorems for the two-dimensional transform are reduced to one dimension. Transforms for special functions and ordinary properties of the Fourier transform can be found in mathematical tables. The discrete Fourier transform How does a Fourier transform look like for a sampled function, e.. a function iven only by its values at a discrete number of points? In this case the Fourier transform is a discrete Fourier transform (DFT) which is a sum of functions with different frequency. If we set the samplin distance x in the object plane to be 1 then we sample the Fourier transform with a samplin distance in frequency f = 1/N, where N is the number of samplin points. We can then label the frequency components by n and the samplin points by k and obtain the DFT as: N 1 Gn k e j 2 ( ) ( ) nk / N, n =... N -1 (2) k Similarly the inverse Fourier transform (IDFT) is iven by: N nk / N k ( ) Gne ( ) j N k, k =... N -1 (3) The normalisation constant 1/N can also be included in DFT instead if one desires. Many of the ordinary properties of the Fourier transform are valid unchaned even for the discrete Fourier transform. This means for example that the DFT is linear, scalin affects both the frequency and the amplitude, and a spatial displacement only affects the phase of the transform and so on. However, some properties are specific for the DFT and IDFT and they are described below: a) The DFT is due to the samplin of a periodic function, exactly determined by its value in N points followin each other: For an inteer p we have: N 1 Gn p N ke j 2 ( ( ) ( ) n p / N ) k / N k (4) N 1 k ( k) e j2nk / N e j2pkn = 1 G( n) (5) b) In the same way samplin in the frequency domain ives rise to periodicity in space. 4

5 1/2x Fiure 3. Aliasin. c) The sample interval is not infinitesimal. This can cause aliasin (fiure 3). Accordin to the sample theorem the oriinal sinal can only be recreated exactly from the sample values if the interval x is shorter than 1/2f max, where f max is the upper frequency limit of the analoues sinal. If the samplin is made too coarse a distorted frequency function is achieved. This is aliasin. d) The number of points, N, is not infinite. This means a truncation of the point series (k) or in a way a multiplication by a "box function". x = Fiure 4. Multiplication by a box function This multiplication in space results in the frequency plane in a convolution with the Fourier transform of the box function, which is a sinc(f) function. Usin a larer number of points decreases this truncation. This reduction is paid by an increased calculation time. Another way of reducin the distortion from truncation is to apply functions with an other appearance than a box. Such functions are called window functions or apodization functions. By a clever selection of the apodization function the amplitude of the side lobes described by the sinc 2 (f)-function can be reduced. e) Convolution can normally be calculated by makin the Fourier transform of the two sinals involved, multiply their transforms and then make the inverse Fourier transform. Due to the earlier discussed periodicity in the DFT and IDFT, the circular convolution sum is achieved: N 1 k ( ) hx ( ) lhk ( ) ( l) l (6) 5

6 The circular convolution means that the two number series (k) and h(k) will be repeated periodically outside the interval [, N-1] before the sum is made. If the circular convolution shall correspond to the ordinary convolution sum it is noteworthy to remember that two sinals with the lenth p, ive a result with the lenth 2p-1. So if the two series (k) and h(k) are expanded with zeroes so that N is reater than 2p-1, then the circular convolution will correspond exactly with the ordinary within the interval [, N-1]. The reason why one is really interested in calculatin the convolution via DFT and IDFT is that it is much faster than to make the calculations directly as lon as the number of points is of the order 64 or reater. Fast Fourier Transform (FFT) To calculate DFT of a number series by writin a proram that directly calculates the N sums in the transform (2) is not wise. N does not need to be so lare before the time to calculate will be unsuitably lon. The time to calculate is N 2, and already N = 32 means 124 operations. Even before computers where available for calculations it was necessary to make Fourier transformations and a stron need to eliminate the dull calculations by hand. As early as 193 a method was presented by C. Runde to minimise the calculations of transforms with 12 and 24 points. By notin some symmetries and periodicity in the calculations of the Fourier transform it was possible for Danielson and Lanczos to present a eneral method for number series of 2 k points that reduced the number of operations from N 2 to N( 2 lon). When the computers were available this method was not known enerally and most calculations where made by the N 2 operations in (2). Not until the beinnin of the 196s was this method rediscovered and presented to a lare number of persons in the article: Cooley, J.W. and Tukey, J.W., "An Alorithm for the Machine Calculation of Complex Fourier Series.", Mathematics of Computation, Vol. 1, April 1965, pp How DFT is calculated in a 2 k points FFT-scheme is described in the followin schematic example: (Sande-Tukey alorithm for computin the DFT - decimation in frequency.) () (1) (2) (3) (4) (5) (6) (7) 1 p= p=1 p=2 p=3 2 p= p=2 p= 3 p= p= p= p=2 p= G() G(4) G(2) G(6) G(1) G(5) G(3) G(7) Fiure 5. Calculation scheme for an 8 points FFT 6

7 The Fourier transform (2) is written by: N 1 k nk Gn ( ) ( k) W where W e j2 / N (1) The calculations are divided into M = 2 lon steps where in each step the number series m (k) is transformed by two calculations. m1 () r m() r m() s (when in upper subroup) (11) p m1( s) m() r m() s W (when in lower subroup) (12) This means that no new computer memory is needed for storin the new values in the number series; m+1 replaces directly the old m. The numbers r, s and p are inteers in the interval [, N-1]. Which values they have are shown for N = 8 in fi. 5. As seen in the fiure the calculations are divided into half as bi subroups for each step in the scheme. The upper part in such a subroup ets the index r while the lower ets the index s in the calculations of (1). For example 1 (5) is iven by equation (12), where r = 1 and s = 5. Correspondinly 2 () is found in equation (11) with r = and s = 2. The exponent p always starts on zero in the columns of each lower subroup; it increases by the step 2 m-1 in the separate columns. This means that p increases by the step 1 in 1, by the step 2 in 2, by the step 4 in 3 (not applicable in this case), and so on. The elements in the G-vector appear in a peculiar order (see fiure 5). The arranement of the elements is iven by bit reversal. If a binary number ives the address of a certain vector in m (k) then this number is reversed and the correct address is found for the G-vector. As an example the element in 3 (3) = 3 (11) is the element G(11) = G(6) and in 3 (1) = 3 (1) we have G(1) = G(4). To increase the calculation speed in the FFT calculations when several vectors of the same lenth shall be transformed, it is useful to calculate a table of the exponentials e j2 n/ N for n =.N-1. Then the time-consumin sine and cosine calculations are only made once. Example Let us calculate 1 (6) if (6) = 7 and (2) = 3. In fi. 5 we have p = 2 at this point. This means that equation (12) is applicable in this case. We have as s = 6 and r = 2: It is easier to find 1 (2), which is iven by equation (11). We have: ( 2) ( 2) ( 6) ( 2 / 8)( 2) / 2 ( 6) ( 2) ( 6) e 4e 4i 1 7

8 Problems 1. Diital part Usin a proram of your choice (preferably Matlab) do the followin tasks. In Appendix I you can find example matlab code for each exercise that you are free to use. a) 1D Fourier transform of a sine wave 1. Plot a sine function with a frequency of 1Hz usin a samplin frequency of 5Hz. 2. Plot its fast Fourier transform (FFT). 3. Move the zero-frequency component to the center of the array. It is useful for visualizin a Fourier transform with the zero-frequency component in the middle of the spectrum. 4. Multiply the sine wave by a Hammin window. What is happenin to the FFT? 5. Repeat tasks 1 to 3 usin samplin frequencies of 2Hz and 21Hz. What do you learn? b) 2D Fourier transform of a rectanle 1. Plot a rectanle. 2. Plot its 2D FFT. 3. Move the zero-frequency component to the center of the array. 4. Try different rectanle sizes. What happens? c) 2D Fourier transform of a circle 1. Plot a circle. 2. Plot its 2D FFT. 3. Move the zero-frequency component to the center of the array. 4. Try different diameter sizes. What happens? 2. Analoue part Collimated liht y Q R O r x n C D Y P X A Plane wave Object (x,y) f Focal plane Fiure P1 The decomposition of the object transmission function (x,y) into Fourier components. 8

9 a) Which spatial frequency is found at a point P in the Fourier plane of a lens with the focal lenth 4 cm if the point is on the Y-axis and is.78 cm from the X- axis? = 633 nm. The fiure P1 shows a eneral Fourier transformation by a lens. b) After the focal plane (or Fourier plane), we can observe the imae of the object by the lens (see Fiure 1). It is therefore possible to filter the object, i.e. modify its spatial properties (the same way we can modify temporal properties of an electronic sinal with low-pass, hih-pass or band-pass filters). Explain how you could perform a low-pass, a hih-pass or a band-pass filter. c) Draw the two functions: a(x) and s(x) defined below. Draw also the convolution of a(x) with s(x). 2 for 1.5 for 3 4 for all other.5 5 where δ(x) is the dirac delta function. d) Fiure P2 illustrates how the Fourier transform of a complicated object is deduced from a combination of some simple transforms, thereby illustratin the convolution theorem. The complete object is shown in Fiure P2 a). We can obtain this object usin three basic components. If we start with an infinite matrix of delta functions (Fiure P2 d)). Then we perform a convolution with a hole (Fiure P2 c)). The result is an infinite matrix of holes. We limit the number of holes by multiplication with a rectanular aperture. Thus we obtain the object in Fiure P2 a). Directly below Fiure P2 a)-d) the Fourier transform of each object is presented in Fiure P2 e)-h). e.. the Fourier transform of Fiure P2 d) is P2 ). In a similar way, we can obtain the Fourier transform of Fiure P2 a) usin the Fourier transform of each basic component shown in Fiure P2 b)-d). Accordin to the Convolusion Theorem a convolusion in space is a multiplication in Fourier space and vice versa. Therefore, instead of convolutin Fiure P2 ) and f), we multiply them. Instead of multiplyin the aforementioned product with e), we convolute it. Thus obtainin the Fourier transform of the more complex object h). Keepin this process in mind, what would be the object (Fiure P2 a)) and the Fourier transform (Fiure P2 h)) be if the window in b) instead was a thin vertical rectanular openin? 9

10 Fiure P2 Combination of different objects and their Fourier transforms. 1

11 Appendix I: Matlab code for computer exercises Part a) %% Sine function %Plot a sine function F1 = 1; %Frequency of the sine wave in Hz Fs = 1; %Samplin frequency in Hz Ts = 1/Fs; %Period of the samplin ncyl = 1; %Number of sine wave cycles to enerate t = :Ts:nCyl*1/F1; %Time base H = sin(2*pi*f1*t); fiure('units','normalized','outerposition',[ 1 1]) subplot(2,3,1); plot(t,h); title(['sine Wave f=', num2str(f1), 'Hz; Samplin fs=', num2str(fs), 'Hz']); xlabel('time(s)'); ylabel('amplitude'); %Plot its FFT F_H = fft(h); %perform the FFT N = lenth(t); n = :N-1; subplot(2,3,2); plot(n,abs(f_h)); %the FFT is complex so we want to plot only the absolute value. title('double Sided FFT - without FFTShift'); xlabel('sample points (N-point DFT)') ylabel(' DFT Values '); %Move the zero-frequency component to the center of the array. F_H_shift = fftshift(f_h); f = Fs*(-N/2+.5:N/2-.5)/N; subplot(2,3,3); plot(f,abs(f_h_shift)); title('double Sided FFT - with FFTShift'); xlabel('frequency (Hz)') ylabel(' DFT Values '); %Multiplication by a Hammin window W = hammin(n); %Creates the Hammin window WH = W'.*H; %Multiplies the sine wave by a Hammin window subplot(2,3,4); plot(t,wh); title(['sine Wave multiplied by a Hammin window f=', num2str(f1), 'Hz; Samplin fs=', num2str(fs), 'Hz']); xlabel('time(s)'); ylabel('amplitude'); F_WH = fft(wh); %perform the FFT subplot(2,3,5); plot(n,abs(f_wh)); %the FFT is complex so we want to plot only the absolute value. title('double Sided FFT - without FFTShift (Hammin window)'); xlabel('sample points (N-point DFT)') ylabel(' DFT Values ');

12 F_WH_shift = fftshift(f_wh); subplot(2,3,6); plot(f,abs(f_wh_shift)); title('double Sided FFT - with FFTShift (Hammin window)'); xlabel('frequency (Hz)') ylabel(' DFT Values '); Optics and optical desin Part b) %%Rectanle %Plot a rectanle N = 3; %number of samplin points Rx = 1/2; %horizontal size of the rectanle in ratio of N Ry = 1/1; %vertical size of the rectanle in ratio of N x = linspace(-2,2,n); y = linspace(-2,2,n); R = zeros(n); Rx1 = floor(1/2*n*(1-rx)); Rx2 = N-Rx1; Ry1 = floor(1/2*n*(1-ry)); Ry2 = N-Ry1; nx = Rx2-Rx1+1; ny = Ry2-Ry1+1; R(Ry1:Ry2,Rx1:Rx2)=ones(ny,nx); %Create the rectanle fiure; imaesc(x,y,r); title('rectanle') xlabel('x') ylabel('y'); %Plot its FFT F_R = fft2(r); fiure; imaesc(x,y,abs(f_r)); %the FFT is complex so we want to plot only the absolute value. title('fft without FFTShift'); xlabel('x') ylabel('y'); %Move the zero-frequency component to the center of the array. F_R_shift = fftshift(f_r); fiure; imaesc(x,y,abs(f_r_shift)); %the FFT is complex so we want to plot only the absolute value. title('fft with FFTShift'); xlabel('x') ylabel('y'); Part c) %%Circle %Plot a circle N = 5; %number of samplin points D = 1/2; %diameter of the circle in ratio of N x = linspace(-2,2,n); 12

13 y = linspace(-2,2,n); C = zeros(n); for k=1:n for l=1:n if (N/2-k)^2+(N/2-l)^2<(D*N/2)^2 C(k,l)=1; end end end fiure; imaesc(x,y,c); title('circle') xlabel('x') ylabel('y'); %Plot its FFT F_C = fft2(c); fiure; imaesc(x,y,abs(f_c)); %the FFT is complex so we want to plot only the absolute value. title('fft without FFTShift'); xlabel('x') ylabel('y'); %Move the zero-frequency component to the center of the array. F_C_shift = fftshift(f_c); fiure; imaesc(x,y,abs(f_c_shift)); %the FFT is complex so we want to plot only the absolute value. title('fft with FFTShift'); xlabel('x') ylabel('y'); 13

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