FILTERING IN THE FREQUENCY DOMAIN

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Transcription:

1 FILTERING IN THE FREQUENCY DOMAIN Lecture 4

Spatial Vs Frequency domain 2 Spatial Domain (I) Normal image space Changes in pixel positions correspond to changes in the scene Distances in I correspond to real distances Frequency Domain (F) Changes in image position correspond to changes in the spatial frequency This is the rate at which image intensity values are changing in the spatial domain image I

The Fourier Series 3 Periodic functions can be expressed as the sum of sines and/or cosines of different frequencies each multiplied by a different coefficient

Image processing 4 Spatial Domain (I) Directly process the input image pixel array Frequency Domain (F) Transform the image to its frequency representation Perform image processing Compute inverse transform back to the spatial domain

Frequencies in an Image 5 Any spatial or temporal signal has an equivalent frequency representation What do frequencies mean in an image? High frequencies correspond to pixel values that change rapidly across the image (e.g. text, texture, leaves, etc.) Strong low frequency components correspond to large scale features in the image (e.g. a single, homogenous object that dominates the image) We will investigate Fourier transformations to obtain frequency representations of an image

Properties of a Transform 6 A transform maps image data into a different mathematical space via a transformation equation Most of the discrete transforms map the image data from the spatial domain to the frequency domain, where all the pixels in the input (spatial domain) contribute to each value in the output (frequency domain)

Spatial Frequency 7 Rate of change Faster the rate of change over distance, higher the frequency

Image Transforms 8 Image transforms are used as tools in many applications, including enhancement, restoration, correlation and SAR data processing Discrete Fourier transform is the most important transform employed in image processing applications Discrete Fourier transform is generated by sampling the basis functions of the continuous transform, i.e., the sine and cosine functions

Concept of Fourier Transform 9 The Fourier transform decomposes a complex signal into a weighted sum of sinusoids, starting from zerofrequency to a high value determined by the input function The lowest frequency is also called the fundamental frequency

Frequency Decomposition 10 The base frequency or the fundamental frequency is the lowest frequency. All multiples of the fundamental frequency are known as harmonics. A given signal can be constructed back from its frequency decomposition by a weighted addition of the fundamental frequency and all the harmonic frequencies

Different forms of Fourier Transform 11 Continuous Fourier Transform Fourier Series 1 2 j2 ux F( u) f ( x) e dx f ( x) a a cos(2 nx) b sin(2 nx) 0 n where 1 an f ( x)cos(2 nx) dx 2 1 bn f ( x)sin(2 nx) dx 2 n n

Continuous Fourier Transform 12 In the continuous domain, the basis functions of the Fourier transform are the complex exponentials e -j2pux These functions extend from - to + These are continuous functions, and exist everywhere

Real and Imaginary Parts of Fourier Transform 13 1 2 j2 ux F ( u) f ( x) e dx 1 1 F( u) f ( x) cos(2 ux) dx j f ( x)sin(2 ux) dx 2 2 Real part Imaginary Part

The Discrete Fourier Transform 14

The Discrete Fourier Transform 15

The 2-D Discrete Fourier Transform 16

The 2-D Discrete Fourier Transform 17

The 2-D Discrete Fourier Transform 18

The 2-D Discrete Fourier Transform 19

Properties of the Fourier Transform 20

21 Filtering Example Smooth an Image with a Gaussian Kernel

22 Filtering Example Smooth an Image with a Gaussian Kernel

23 Filtering Example Smooth an Image with a Gaussian Kernel

24 Filtering Example Smooth an Image with a Gaussian Kernel

25 Filtering Example Smooth an Image with a Gaussian Kernel

26 Filtering Example Smooth an Image with a Gaussian Kernel

The Fourier Transform 27

Properties of the Fourier Transform 28

Some Fundamental Transform Pairs 29

Some Fundamental Transform Pairs 30

Example 31 Given f(n) = [3,2,2,1], corresponding to the brightness values of one row of a digital image. Find F (u) in both rectangular form, and in exponential form

Example Contd. 32 1 F(0) [3 2 2 1] 2 4 1 j 2 1.1/ 4 j 2 2.1/ 4 j2 3.1/ 4 F(1) [3 2e 2e 1. e ] 4 1 1 [3 2 2 j j] [1 j] 4 4

Example Contd. 33 1 1 F(2) [3 2 2 1] 4 2 1 j2 3.1/ 4 j 2 3.2/ 4 j2 3.3/ 4 F(3) [3 2e 2e 1. e ] 4 1 1 [3 2 j 2 j] [1 j] 4 4 Therefore F(u) = [2 ¼ (1-j) ½ ¼ (1+j) ]

Magnitude-Phase Form 34 F(0)= 2 = 2 + j0 Mag=sqrt(2 2 + 0 2 )=2; Phase=tan -1 (0/2)=0 F(1) = ¼ (1-j) = ¼ - j ¼ Mag= ¼ sqrt(1 2 + (-1) 2 )=0.35; Phase = tan -1 (-(1/4) / (1/4)) = tan -1 (-1) = -p/4 F(2) = ½ = ½ + j0 Mag = sqrt(( ½ ) 2 + 0 2 ) = ½ Phase = tan -1 ( 0 / (1/2) ) = 0 F(3) = ¼ (1+j) = ¼ - j ¼ Mag= ¼ sqrt(1 2 + (-1) 2 )=0.35; Phase = tan -1 ((1/4) / (1/4)) = tan -1 (1) = p/4

Fourier Transform Calculation 35 Given f(n) = [ 3 2 2 1 ] F(u) = [2 ¼ (1-j) ½ ¼ (1+j) ] In phase magnitude form, M(u) = [2 0.35 ½ 0.35 ] F(u) = [0 p/4 0 p/4 ] Calculate the above for f(n) = [ 0 0 4 4 4 4 0 0] Plot f(n), F(u), M(u) and F(u) graphically