0 0 50 50 100 100 150 150 200 200 250 0 250 50 100 150 200 250 300 350 0 50 100 150 200 250 300 350 Wavelet Based Methods in Image Processing Lecture 2.a - Fourier Transform Applied Mathematics Seminar S. Allen Broughton 1 Lecture 2.a • • • • • • • signal samples and time domain model wave forms and frequency domain signal analysis & synthesis Discrete Fourier transform - DFT extension to 2D convolution theorem restoration 2 Signal Samples and Time Domain • • • • X = [X(0),...,X(N-1)] vector of length N X(i) = f(i/N): sample analog signal f X(i+N) = X(i): periodic extension (mod N) time domain {0,1/N,...,(N-1)/N} or {0,1,...,N-1} 3 Model Wave Forms • model wave forms E • k n (n ) = exp(2π i(k ⋅ )) N frequency domain k = 0,..., N − 1 • Maple script - waveform.mws 4 Signal Analysis & Synthesis - 1 • energy of a signal (* = conjugate transpose) μ(X , X ) = X ⋅ X = X *X = N −1 ∑ | X ( n )|2 n=0 • analysis of a signal -“energy content at a frequency” ~ X ( r ) = E k ⋅ X = E k* X = N −1 ∑ n=0 n X ( n ) exp( − 2 π i ( k ⋅ )) N 5 Signal Analysis & Synthesis - 2 • synthesis or reconstruction for a signal X = N −1 ∑ k =0 ak Ek ~ X (k ) ak = N 6 Signal Analysis & Synthesis - 3 • additivity of energy 1 μ(X , X ) = N N −1 ∑ k =0 ~ | X ( k ) |2 = N N −1 ∑ k =0 | a k |2 7 Discrete Fourier Transform (DFT) - Matrix definition • set F = [ E 0 , E 1 , ... , E N −1 ] * kl F (k ,l) = exp(− 2π i ) N • • • then and ~ X = FX 1 F F = I N * hence (inversion formula) 1 ~ X = FX N 8 Discrete Fourier Transform Pictures • Matlab scripts – dft1demo1.m, dft1demo2 – dft2demo1.m, dft2demo2.m ,dftdemo3.m 9 Convolution Theorem • Fourier transform takes convolution to pointwise multiplication ~ ~ X *Y = X •Y • Eigenvector interpretation ~ X * E k = H X E k = X (k ) E k = λE k g a in = | λ | , p h a s e s h ift = ∠ λ 10 2D Discrete Fourier Transform • Let X be an mxn image then ~ t X = Fm XFn = Fm XFn • 2D convolution theorem holds ~ ~ X *Y = X •Y 11 Simple Restoration -1 • Let X be an true image and Y the blurred and noisy image then Y = M*X + E • • where M is a blurring mask and E is the noise ideally, find another mask R such that R * M = id • R* E = 0 can’t be done in practice but you try your best 12 Simple Restoration -2 ~ process ~ F −1 Y⎯ ⎯→ Y ⎯⎯⎯→ Z ⎯⎯→ Z F • • • ~ ~ ~ ~ ~ ~ Z = R• M• X + R•E where work in Fourier domain ~ ~ R * M = id → R • M = id or ~ ~ −1 pointwise R=M 13 Simple Restoration - 3 • Matlab Example - ansmid3.m 14 Issues in Restoration - 1 • Model issues – Good information on the blurring model M needs to be available, it can sometimes be found by experimentation – The distribution of E needs to be known – If M and E are unknown then you need to estimate them from the degraded image 15 Issues in Restoration - 2 • Computational issues – ~ M – thus has zeros or small entries ~ ~ −1 R=M – and so image ~ ~ R•E doesn’t exist or is very large ‘‘noises up’’ the restored 16