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Two-Dimensional Filters Digital Image Processing Instructor: Dr. Cheng-Chien LiuCheng-Chien Liu Department of Earth Sciences National Cheng Kung University Last updated: 24 September 2003 Chapter 5
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Introduction Filter an image Kill or modify certain frequency components Multiply the Fourier transform of the image with a filter function Use convolution to avoid using the Fourier transform itself Omit or enhance some details
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Definition 2D filter (the system function) H( , ) = [1/(2 ] - - h(n,m)e -j( n+ m) The unit sample response h(k,l) = [1/(2 ] - - H( , )e j( k+ l) d d The filter function in the real domain h(k,l) = [1/(2 ] - - H( , )cos( k+ l)d d H( , ) = [1/(2 ] - - h(n,m)cos( n+ m) Example 5.1 Convolution instead of using the Fourier transform 1D case, analogous to 2D case (Fig 5.1)
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Convolution filter For h(k,l) to be a convolution filter h(k,l) = 0 for k > K and l > L The filter must be a finite array of numbers Idea filters don’t fulfill this condition The unit sample response of the ideal lowpass filter B5.1: h(k,l) = R(k 2 +l 2 ) -1/2 J 1 (R(k 2 +l 2 ) 1/2 ) 2D h(k) = sink / k 1D (Fig 5.2: comparison of 1D and 2D) The unit sample response of the ideal bandpass filter Example 5.1: h(k,l) = (k 2 +l 2 ) -1/2 [R 2 J 1 (R 2 (k 2 +l 2 ) 1/2 ) – R 1 J 1 (R 1 (k 2 +l 2 ) 1/2 )] The unit sample response of the ideal highpass filter Example 5.2: there is no real domain function that has as Fourier transform the ideal highpass filter
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z-transform Definition X(z) = k=1 m x k z -k l and m are defined according to which term of the string of number {x k } is assumed to be the k = 0 term If the filter is of infinite extent X(z) = - x k z -k Usually express X(z) in closed form as H(z) = i=0 Ma a i z i / i=0 Mb b i z i Advantage of the z-transform Can be easily realized in hardware Obey the convolution theorem
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2D z-transform Definition H(z 1, z 2 ) = i=0 M j=0 N c ij z 1 i z 2 j Usually express H(z 1, z 2 ) as H(z 1, z 2 ) = i=0 Ma j=0 Na a ij z 1 i z 2 j / i=0 Mb j=0 Nb b ij z 1 i z 2 j M a, N a, M b, N b are some integers Conventionally we choose b 00 = 1
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Recursive filter The extent of a filter recursive or not R(z 1, z 2 ) = i=0 M j=0 N r ij z 1 i z 2 j = H(z 1, z 2 ) D(z 1, z 2 ) D(z 1, z 2 ): z-transform of input image H(z 1, z 2 ): z-transform of filter R(z 1, z 2 ): z-transform of output image R(z 1, z 2 ) i=0 Mb j=0 Nb b ij z 1 i z 2 j = i=0 Ma j=0 Na a ij z 1 i z 2 j D(z 1, z 2 ) R(z 1, z 2 ) = i=0 Ma j=0 Na a ij z 1 i z 2 j D(z 1, z 2 ) - [ i=0 Mb j=0 Nb b ij z 1 i z 2 j ]R(z 1, z 2 ) [ i=0 Mb j=0 Nb b ij z 1 i z 2 j ]: i and j are not both zero Recursive: the value of r mn can be calculated in terms of the previously calculated values of r mn In the case of a finite filter all b kl = 0, except b 00 = 1 no recursive Example 5.3
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Approximation theory An alternative to using infinite H( , ) Decide upon the desired system function F( , ) Choose a finite filter H( , ) that approximate F awap Chebysheu norm Error = ||F( , ) - H( , )|| max ( , ) |F( , ) - H( , )| The best approximation Error = min{max ( , ) |F( , ) - H( , )|} Fig 5.5 Total square error F( , ) - H( , ) 2 d d 3 techniques of designing 2D filters Windowing, frequency sampling, linear programming
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Windowing Windowing method Truncate an infinite impulse response at some desired size Problem Sharp edge Fourier transform high frequency ripples Solution Replace the sharp-edge window by a smooth window Several such smooth windows have been developed for the case of 1D signals Extend 1D window to 2D not hazard free Not an optimal approximation See Fig 5.2 and discussion
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Linear programming Linear programming problem (LPP) An optimization method to solve the problem Minimize: z = c 1 T x 1 + c 2 T x 2 + d Subject to: A 11 x 1 + A 12 x 2 B 1 (p 1 inequality constraints) A 21 x 1 + A 22 x 2 = B 2 (p 2 equality constraints) x 1i 0 (n 1 variables x 1 ) x 2j free (n 2 variables x 2 ) Where: c 1, c 2, B 1, B 2 are vectors A 11, A 12, A 21, A 22, are matrices x 1, x 2 are vectors made up from variables x 1i and x 2j respectively
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Linear programming (cont.) Filter design problem LPP H( , ) = n=-N N m=-N N h(n,m)e -j( n+ m) Where h(n,m) is the digitized finite filter that we want to use, and h(n,m) are real numbers If h( , ) = h(-, ) H( , ) is real B5.3: H( , ) = 2 n=-N N m=1 N h(n,m) cos( n+ m) + 2 n=1 N h(n,0) cos( n) + h(0,0) Free variables: (2N+1)N + N + 1 = 2N 2 + 2N + 1 Choose h(i, j) so that max ( , ) |H( , ) - F( , )| |H( i, i ) - F( i, i )| for i = 1, 2, …, p 1 Inequality constraints: 2
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Linear programming (cont.) Filter design problem LPP (cont.) Minimize: z = x 11 Under the constraints: A 11 x 11 + A 12 x 2 B 1 (2p 1 inequality constraints) x 11 0 (one non-negative variable) x 2j free (2N 2 + 2N + 1 free variables) Drawback Lots of constraint points are required for a given number of required coefficients
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Iterative approach Philosophy of the iterative approach Breaking the LPP into a series of small ones Fig 5.6: fit a surface gradually Maximizing algorithm Simultaneously increase the lower limit of the error and decrease its upper limit Works by making use of two concepts Limiting set of equations La Valle Poussin theorem
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Iterative approach (cont.) Limiting set of equations m F( m, m ) - i=1 n c i g i ( m, m ) m = 1, 2, …, M Limiting if All m 0 | m | cannot simultaneously be reduced for any choice of c La Valle Poussin theorem min X |P - F| max X |P * - F| max X |P - F| The best approximation: P * i=1 n c i * g i ( m, m ) Any other approximation: P i=1 n c i g i ( m, m ) Error of the best Chebyshev approximation: max X |P * - F|
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Iterative approach (cont.) Steps Choose a subset X k of the set of points X Choose the best approximation in X k P k ( , ) i=1 n c i k g i ( , ) Error in X k : k max Xk |P - F| Error in the whole set X: E k max X |P - F| Include an extra point in a new subset X k+1 Error in X k+1 : k+1 max Xk+1 |P - F| La Vallee Poussin theorem: k k+1 E k Narrow the double inequality by increasing its lower limit
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Iterative approach (cont.) Working in the frequency domain B5.4 Example 5.4
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