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SYDE 575: Image Processing Spatial-Frequency Domain: Discrete-Time Fourier Transform (DTFT) & Discrete Fourier Transform (DFT) Textbook: Chapter 4

Signal Representation As in continuous domain, exploit linearity and shift invariance to find output as superposition of responses to input components Use the sifting property: f(m,n) =   f( k, l )  ( m - k, n - l ) Sketch:

Apply Transformation g(m,n) = T [ f(m,n) ] = T [   f( k, l )  ( m - k, n - l ) ] =   f( k, l ) T [  ( m - k, n - l ) ] =   f( k, l ) h( m - k, n - l ) = f(x,y) * h(x,y) Once again, if we want to obtain the original signal f(x,y), deconvolution is a difficult operation

Another Approach Just like the continuous domain, we can use a discrete version of the frequency domain In this case, use complex exponential e j2  un A complex exponential remains untouched by an LTI system. Why? Use e j2  un as the input: e j2  un * h(n) =  h(k) e j2  u(n-k) = [  h(k) e -j2  uk ] e j2  un = H(e j2  u ) e j2  un where H(e j2  u ) is known as the discrete-time Fourier transform (DTFT)

Notes on DTFT Forward: H(e j2  u ) =  h(k) e -j2  uk Inverse: h(n) =  H(u)e j2  un du (integrate 0 to 1) 1)‘u’ represents continuous spatial-frequency 2)H(e j2  u ) is periodic with period 1 since e j2  u is periodic with period 1 3)If H(e j2  u ) has no imaginary component, then there is zero phase Example: h(n) = [ 1/3 1/3 1/3 ]

Notes on DTFT (cont.) 4) If g(n) = f(n) * h(n), then G(e j2  u ) = F(e j2  u ) H(e j2  u ) 5) Since we only use finite extent signals h(n) where 0 <= n <= N-1 Then DTFT is also a finite sum H(e j2  u ) =  h(n) e -j2  uk (sum n=0 to N-1) N samples!

Extending DTFT to DFT We only have N samples, we only need N samples for F(e j2  u ) Instead of using range [0,1], stretch and sample to range [0,N-1] to create discrete Fourier transform (DFT) Forward: F(u) =  f(n)e -j2  un/N (sum n=0 to N-1) Inverse: f(n) = (1/N)  F(u)e j2  un/N (sum u=0 to N-1) Both F(u) and f(n) are periodic

Implicit Periodicity Implicit periodicity to f(n) = DFT -1 [F(u)] Example: 3pt average (periodic on N) So, while we define discrete arrays over a finite interval, the DFT creates an implicit periodicity outside the interval which complicates the use of the DFT to implement LSI systems Example: f(n) = h(n) = 1 for n = 0,1,…,4 What is the impact for digital image processing?

DFT in 2D For an MxN image – Forward DFT F(u,v) =  f(x,y) e -j2  ux/M+vy/N) (sum x = 0 to M-1 and y = 0 to N-1) – Inverse DFT f(x,y) =  F(u,v) e j2  ux/M+vy/N) (sum u = 0 to M-1 and v = 0 to N-1)

How can we visualize the spatial-frequency domain?

Fourier Spectra Characteristics of Images Most energy resides in low frequency components  Implies that the original image can be reconstructed with good approximation with the low frequency components Low frequency components correspond to coarse details  e.g., DC component represents average intensity of an image High frequency components correspond to fine details  e.g., edges, noise

Fourier Analysis: Spectra In image coordinates, origin (0,0) refers to top- left corner Results in spectra being centered at corner Common to multiply input image by (-1) x+y. Why? Brings origin of spectra to center of image

Fourier Spectra Example Source: Gonzalez and Woods

Fourier Spectra Example Source: Gonzalez and Woods

Fourier Analysis: Phase Characterizes structural information within an image Source: Gonzalez and Woods

Recall: Image Characteristics in Frequency Domain Low frequencies responsible for general appearance of image over smooth areas High frequencies responsible for detail (e.g., edges and noise) Intuitively, modifying different frequency coefficients affects different characteristics of an image

Example: DC component removal Source: Gonzalez and Woods Suppose we remove the DC component from the Fourier transform of an image

Why does it look like that? DC component characterizes the mean of the image intensities

Basic Steps of Filtering in Spatial-Frequency Domain 1)Multiply input f(x,y) by (-1) x+y to center transform 2)Compute DFT of image, F(u,v) 3)Multiply F(u,v) by filter function H(u,v) to get G(u,v) 4)Compute inverse DFT of G(u,v) to get g(x,y) 5)Multiply g(x,y) by (-1) x+y to get filtered image

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