Session 2. (1) Spatial Frequency and Fourier Transformation

2005 Autumn semester Pattern Information Processing Topic 1. Sampling and digital processing of images Session 2. (1) Spatial frequency and Fourier transformation Organization of digital images is explained in this Diffraction of light and imaging and the next sessions. Although an image is natu- We start from the optical phenomenon of imaging. rally a continuous distribution of brightness, it has to The imaging is defined as a collection of diverged be converted into a discrete set of integers for com- light spread from a point of object into one point by puter processing. The conversion into discrete pixels a lens. This phenomenon can be observed from the is called sampling, and the conversion into integers following point of view: The light has the property of is called quantization. The sampling period is quite diffraction. The diffraction of wave is a phenomenon important issue, and it is evaluated by the concept of that the wave reaches beyond an opaque object ob- spatial frequency. The concept of spatial frequency stracting its path. For example, even if the progress and Fourier transformation are explained in this ses- of wave on a water surface is obstructed by a board, sion. it reaches beyond the board. Since the light is a kind of electromagnetic wave, the light has this property. The radio wave reaches beyond obstructing objects from the broadcasting station by diffraction. wavefronts reach beyond divergence of light the obstruction collection of light Opaque object wavefronts Fig. 2: Diffraction. Fig. 1: Imaging. If the light passes through a diffraction grating, of first order, is obtained in this direction. The smaller which is an object whose transparence is changing the period of bands, the larger the angle between the periodically or where transparent and opaque bands diffracted light and the light passing directly through are aligned one after another, the light that passes the grating (called the zeroth order light). If the grat- through a transparent band interferes with the light ing contains pure opaque and pure transparent bands that passes the other transparent bands. Since the only, diffracted lights along several directions are ob- light waves passing through adjacent bands empha- tained. However, if the transparence of grating is si- size each other along the direction such that the phase nusoidally distributed, diffracted lights of zeroth and shift of the light waves is exactly the same as the first orders only are obtained. wavelength, a bright light, called the diffracted light A. Asano / Pattern Information Processing (2005 Autumn semester) Session 2 (Oct. 14, 2005) Page 1/4 sparce grating fine grating diffracted light diffracted light of first order of first order direct light direct light wavelength wavelength wavelength wavelength Fig. 3: Diffraction grating and diffracted light of first order. Suppose that a figure on a transparent film is il- If these diffracted lights pass through the imaging luminated by a plain light wave1. Suppose also the lens, each diffracted light is refracted and interferes figure is organized by a superposition of many si- with the zeroth order light at the image plane. This nusoidal wave of transparence, i.e. superposition of interference produces a stripe, or interference fringe, many diffraction gratings. Each grating diffracts the on the image plane. The larger the angle of diffracted incident light and produces the diffracted light. The light is, the smaller the period of stripe is. The distri- smaller the period of grating is, the larger the angle bution of transparence on the film is reconstructed on of diffraction is. the image plane as the superposition of these stripes. figure 1 = superpositon νx of gratings x diffraction interference 1 νy y Fig. 4: Imaging by diffraction and interference. Fig. 5: Spatial frequency. Spatial frequency tition of the sinusoidal wave per unit length is called spatial frequency. The unit of spatial frequency is cy- Understanding the process of imaging, the figure cle/m in the MKSA unit system. Note that the wave on the film is regarded as a superposition of sinu- is “a wave on the plane.” Since the wave on the plane soidal waves of transparence. The number of repe- 1The following explanation is in case of coherent illumination such as lasers. It is more complicated in the case of ordinary infoher- ent illumination. A. Asano / Pattern Information Processing (2005 Autumn semester) Session 2 (Oct. 14, 2005) Page 2/4 has its direction, the spatial frequency is described where θ1,θ2,...θn are phase shifts of each wave. Ap- as a set of νx and νy which are frequencies along plying the following operation, x-direction and y-direction, respectively. The figure ∞ f (x) exp(−i2πν1 x)dx, (4) on a film is decomposed into a set of waves of vari- −∞ ous spatial frequencies. The amplitude of wave at a to Eq. (3), we get specific spatial frequency is called component corre- ∞ sponding to the spatial frequency. f (x) exp(−i2πν1 x)dx −∞ Fourier transformation ∞ = a1 exp(i(2πν1 x + θ1)) exp(−i2πν1 x)dx −∞ It is shown in the previous section that a figure on a ∞ film can be decomposed into spacial frequency com- + a2 exp(i(2πν2 x + θ2)) exp(−i2πν1 x)dx −∞ ponents. Fourier transformation is the operation to ∞ calculate the components. The principle of Fourier + ···+ an exp(i(2πνn x + θn)) exp(−i2πν1 x)dx −∞ transformation will be explained in the following + ··· . (5) way: The transparence distribution of the figure on the film can be regarded as a mathematical function. From Eq. (1), we get that the first term of right side Here we assume onedimensional functions for sim- of Eq. (5) is plicity. The function f (x) is assumed as a superposi- ∞ a1 exp(i(2πν1 x + θ1)) exp(−i2πν1 x)dx tion of sinusoidal waves of various frequencies. The −∞ ∞ sinusoidal wave of frequency ν1 is expressed in ex- = a1 exp(iθ1)) exp(i2πν1 x) exp(−i2πν1 x)dx ponential form as exp(i2πν1 x). The multiplication of −∞ 2π expresses the frequency in radian per unit length. = a1 exp(iθ1)δ(0) (6) This value, 2πν1, is called angular frequency. and all the other terms are zero. Consequently, Eq. This exponential function has the following prop- (4) means the operation of extracting the amplitude of erty: the sinusoidal wave of frequency ν , i. e. the compo- ∞ 1 nent of frequency ν , as its real part. The imaginary exp(i2πν1x) exp(−i2πν2 x)dx = δ(ν1 − ν2). (1) 1 −∞ part indicates the phase shift of the wave. The right side of Eq. (1) is called Dirac’s delta Applying the operation of Eq. (4) for various ν,we function, defined as get various frequency components. Regarding these ∞ components as a function of ν, we denote as δ(x) = 0(x 0), δ(x) = 1. (2) ∞ −∞ F(ν) = f (x) exp(−i2πνx)dx. (7) Equation (1) states that the integral has a nonzero −∞ value only if two waves of the same frequency are This is the definition of Fourier transformation, superposed, otherwise it has zero. Such kind of func- and applying this operation to the function in Eq. (3) tion set is called the orthogonal function system. This yields peaks of heights a1, a2,...,an,... at positions is explained again in the section for image compres- ν1,ν2,...,νn,... on the real part of ν axis, respec- sion in Topic. 2. tively. In two dimensional case, it is expressed as Assuming that the function f (x) is a superposition ∞ F(νx,νy) = f (x, y)exp{−i2π(νx x + νyy)}dxdy. of sinusoidal waves, it can be expressed as follows: −∞ (8) f (x) = a exp(i(2πν x + θ )) 1 1 1 The original (x, y) plane is called real domain, and + πν + θ + ··· a2 exp(i(2 2 x 2)) (νx,νy) plane yielded by Fourier transformation is + an exp(i(2πνn x + θn)) + ··· (3) called frequency domain. A. Asano / Pattern Information Processing (2005 Autumn semester) Session 2 (Oct. 14, 2005) Page 3/4 Is it really able to express the original function f (x) How is the case where f (x) is not periodic? by a superposition of sinusoidal waves as Eq. (3)? It is considered as the case where the period L Let us assume that f (x) is a periodic function of tends to the infinity. When L tends to the infin- period L. In this case, the period of all the super- ity, the intervals between the basic periods of the / , / , / ,..., / ,... posed waves in Eq. (3) should be also L. Thus the waves, L 2 L 3 L 4 L n , become smaller right side of Eq. (3) can contain the waves of ba- and smaller, and finally the intervals disappear. It in- sic periods L/2, L/3, L/4,...,L/n,... only, where n dicates that the original function cannot be expressed is an integer. The infinite number of such waves can by a summation such as Eq. (3) since the intervals be composed, however, the basic periods are discrete. are not discrete but continuous and the waves are not ν Thus the number of such waves is the countable infin- countable. In this case, the Fourier transform F( )as ity, and it is possible to write the function by a sum of in Eq. (7) is not regarded as an arrangement of peaks the infinite terms, called series, as shown in Eq. (3). but a continuous function composed by connecting The series in Eq. (3) is called Fourier series expan- adjacent peaks. sion. Cosine wave and exponential function get From Euler’s equation, a1 cos(2πν1 x + θ) exp(iω) = cos ω + i sin ω, (9) a = 1 exp(i(2πν + θ)) 2 1 we get the following relationship between the a1 + exp(−i(2πν1 + θ)) trigonometric functions and the exponential function: 2 a1 ω + − ω = exp(i2πν1) exp(iθ) ω = exp(i ) exp( i ) 2 cos a 2 + 1 exp(i2π(−ν ) exp(−iθ) (12) exp(iω) − exp(−iω) 2 1 sin ω = .

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