MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.161 Signal Processing – Continuous and Discrete Introduction to Two-Dimensional/Image Processing 1 1 Introduction In this set of notes we extend the concepts of one-dimensional signal processing to two dimensions. Typical fields that involve the topics introduced here are imaging, including optics, and image processing. The most important concept to grasp is that we are no longer dealing with functions of time, and the primary domain is the spatial domain. Therefore, instead of dealing with f(t) we deal with f(x), where x is a spatial distance, or in the two-dimensional situation f(x, y). The methods developed here use Fourier techniques to transform the spatial representa-tion f(x, y) to a frequency domain space F (j u, j v), where u, and v are spatial frequencies with units of radians/unit distance. As in temporal processing the difficult operation of convolution is replaced by simple multiplication in the frequency domain. In addition, an understanding of Fourier methods gives qualitative insights to image processing techniques such as “sharpening” and “blurring”. In these notes we begin by thoroughly examining one-dimensional Fourier methods and properties to gain insights on the Fourier transform and frequency domain processing, and then generalize to two-dimensional imaging. 2 Two-dimensional Imaging We now address the more usual situation of processing a two-dimensional object function f(x, y) through a linear image processing system, producing a two-dimensional image g(x, y). The sifting property of the Dirac delta function may be extended to two dimensions: f ( x , y ) g ( x , y ) = f ( x , y ) Ä h ( x , y ) L i n e a r I m a g e P r o c e s s o r h ( x , y ) I n p u t I m a g e O u t p u t I m a g e Figure 1: A two-dimensional linear image processing system. ��∞ f(x, y)δ(x − a)δ(y − b)dxdy = f(a, b) (1) −∞ 1D. Rowell, November 20, 2008 1� � so that we can write the input f (x, y) as a superposition of two-dimensional impulses δ2(x, y)= δ(x)δ(y): ��∞ f(x, y)= O(σ, ρ)δ(x − σ)δ(y − ρ)dσdρ (2) −∞ Assume that the action of the image processor can be expressed as a two-dimensional linear system operator �L2 {} that maps the input image f (x, y) to the output image g(x, y), that is g(x, y)= �L2 {f(x, y)} (3) so that we can write the image as ��∞ g(x, y)= �L2 f(σ, ρ)δ(x − σ)δ(y − ρ)dσdρ (4) −∞ As in one-dimensional signal processing, we can reverse the order of the operator and inte-gration ��∞ g(x, y)= f(σ, ρ)�L2 {δ(x − σ)δ(y − ρ)} dσdρ (5) −∞ For a spatially invariant system, we define the two-dimensional point-spread function h(x, y)asthe response to a two-dimensional delta function δ2(x, y)= δ(x)δ(y)at the origin of the object space: h(x, y)= �L2 {δ(x)δ(y)} (6) Equation (5) can be written ��∞ g(x, y)= f(σ, ρ)h(x − σ, y − ρ)dσdρ (7) −∞ which is a two-dimensional convolution between the point-spread function h(x, y) and the object space f(x, y). As in the one-dimensional case, we conclude that the point-spread function completely characterizes the imaging system, in the sense that the imaging fidelity is completely deter-mined by h(x, y). Example 1 Determine the image formed by convolving a square uniform white image of size 20×20 by a linear image processing filter with point-spread function that is cir-cularly symmetric with a radius of 5 units and height of unity. Solution: The resulting images shown in Fig. 2 were computed using the fol-lowing Matlab script: 2% Matlab script to demonstrate 2-D convolution % First create an array containing the point-spread function psf = zeros(40,40); for i=1:40 for j=1:40 r = sqrt((i-20)^2 + (j-20)^2); if (r < 5) psf(i,j)= 1; end; end; end; % % Now create the object object=zeros(40,40); for i=10:29 for j=10:29 object(i,j)=1; end; end; % % Set up a simple colormap so that all plots are in black and white bw=[0 0 0]; colormap(bw); % % Make 3-D plots of the two input functions mesh(psf); title(’Point-spread function’); pause; mesh(object); title(’Rectangular object’); pause; % % Use the conv2 function to do the 2-D convolution image=conv2(psf,object,’same’); mesh(image); title(’Resulting image’); pause; The Two-Dimensional Fourier Transform In imaging processing work, representing two-dimensional functions, such as images, in the Fourier domain is of great importance. Let f(x, y) be a function of two independent variables x and y; the Fourier transform of f (x, y)is � ∞ � ∞ F (u, v)= f(x, y)e −j(ux+vy)dxdy (8) −∞ −∞ 3 31 0.8 0.6 0.4 0.2 0 40 30 40 30 20 20 10 10 00 1 0.8 0.6 0.4 0.2 0 40 30 40 30 20 20 10 10 00 70 60 50 40 30 20 10 0 40 30 40 30 20 20 10 10 00 Figure 2: Two-dimensional convolution: (a) A circular point-spread function, (r = 5), (b) a uniform rectangular 20 ×20 object, and (c) the image formed by convolution. 4� � � � � � � � where u and v are spatial frequencies with units of radians per unit length. By splitting the exponential into two halves it can be seen that the transform can be expressed as two one-dimensional transforms; first with respect to x and then with respect to y, � ∞ � ∞ F (u, v)= f(x, y)e −juxdx e −jvydy. (9) −∞ −∞ In general F (u, v) is a complex-valued function of u and v. Example 2 Find the two-dimensional Fourier transform of the two-dimensional impulse f(x, y)= δ2(x, y)= δ(x)δ(y). Solution: Separating the variables as in Eq. 9, � ∞ � ∞ Δ(u, v)= δ(x)e −juxdx δ(y)e −jvydy −∞ −∞ and by the sifting property of the delta function, Δ(u, v)=1 Example 3 Find the Fourier transform of the function f(x, y) = rect(x, y). Solution: The function has a value of unity within the square defined by −0.5 > x< 0.5,y > −0.5 < 0.5. The transform is therefore 1/2 1/2 F (u, v)= e −j(ux+vy)dxdy −1/2 −1/2 sin u sin v = = sinc(u, v) u v as shown in Fig. 3. The two-dimensional inverse Fourier