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Integral Transform Dongsup Kim Department of Biosystems, KAIST Fall, 2004

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Integral Transforms Integral transform: – K( ,t): Kernel – Mapping a function f(t) in t-space into another function g( ) in -space – Fourier, Wavelet, Z-transform, Laplace, Hilbert, Radon, etc An original problem is difficult to solve in the original coordinate. Often, the transform of the problem can be solved more easily. Then, the inverse transform returns the solution from the transform coordinates to the original system. Problem in Transform space Original problem Solution in Transform space Solution of original problem Integral transform Relatively easy solution Difficult solution Inverse transform

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Fourier Transform, I A physical process can be described either in the time domain by a function of time t, h(t), or in the frequency domain as a function of frequency f, H(f) h(t) and H(f) are two different representations of the same process. One goes back and forth between these two representations by means of the Fourier transform, – Dirac delta function: Using angular frequency =2 f,

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Examples, I FT of Dirac delta function: FT of cos( 0 t) t (t)(t) cos( 0 t) t

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Examples, II FT of exp(2 if 0 t)=exp(i 0 t) Sum F {exp(i 0 t)} exp(i 0 t) t t Re Im FT

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Properties, II Correspondence between symmetries in the two domains: Scaling and shifting

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Scaling tt t Short pulse Medium- length pulse Long pulse

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Properties, III With two functions h(t) and g(t), and their FT H(f) and G(f), the convolution, g*h, is defined by Convolution theorem: the FT of the convolution is the product of the individual FTs. The correlation, Corr(g,h) Correlation theorem (for two real functions, g and h): Autocorrelation, Wiener-Khinchin theorem: Parseval’s theorem:

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Properties, IV FT of derivatives; in general application: partial differential equation; wave equation

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Sampling theorem, I Suppose function h(t) is sampled at evenly spaced intervals in time; – 1/ : Sampling rate For any sampling interval , there is a special frequency f c, called Nyquist critical frequency, given by – ex: critical sampling of a sine wave of Nyquist critical frequency is two sample points per cycle. A function f is “bandwidth limited” if its Fourier transform is 0 outside of a finite interval [-L, L] Sampling Theorem: If a continuous function h(t), sampled at an interval , is bandwidth limited to frequency smaller than f c, i.e., H(f)=0 for all |f|>f c, then the function h(t) is completely determined by its samples h n.

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Sampling theorem, II For bandwidth limited signals, such as music in concert hall, sampling theorem tells us that the entire information content of the signal can be recorded by sampling rate -1 equal to twice the maximum frequency pass by the amplifier. For the function that is not bandwidth limited to less then the Nyquist critical frequency, frequency component that lies outside of the frequency range, -fc < f < fc is spuriously moved into that range (aliasing). – Demo applet http://www.cs.brown.edu/exploratorie s/freeSoftware/repository/edu/brown/ cs/exploratories/applets/nyquist/nyqui st_limit_java_plugin.html http://www.cs.brown.edu/exploratorie s/freeSoftware/repository/edu/brown/ cs/exploratories/applets/nyquist/nyqui st_limit_java_plugin.html

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Discrete Fourier transform, I Suppose we have N consecutive sampled values where is the sampling interval, and assume N is even. With N numbers of input, we can produce no more than N independent number of outputs. Therefore, we seek estimates only at the discrete values; Then, discrete Fourier Transform (DFT) is ; DFT maps N complex numbers (the h k ’s) into N complex numbers (the H n ’s). ;trapezoidal approximation of the integral

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Discrete Fourier transform, II It’s periodic in n, with period N; H -n = H N-n, n=1,2,… With this conversion, one lets the n in H n vary from 0 to N-1. Then n and k vary exactly over the same range. With this convention, – zero frequency n=0 – positive frequencies, 0 < f < f c 1 n N/2-1 – negative frequencies, –f c < f < 0 N/2+1 n N-1 – the value n = N/2 both f = f c and f = -f c The DFT has symmetry properties almost exactly the same as the continuous Fourier transform.

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Discrete Fourier transform, III The discrete inverse Fourier transform is – Proof: Parseval’s theorem: – Proof?

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Fast Fourier Transform (FFT), I DFT appears to be an O(N 2 ) process. Danielson and Lanczos; DFT of length N can be rewritten as the sum of two DFT of length N/2. We can do the same reduction of H k 0 to the transform of its N/4 even- numbered input data and N/4 odd-numbered data. For N = 2 R, we can continue applying the reduction until we subdivide the data into the transforms of length 1. For every pattern of log 2 N number of 0’s and 1’s, there is one-point transformation that is just one of the input number h n

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Fast Fourier Transform (FFT), II For N=8 Since W N/2 = -1, H k 0 and H k 1 have period N/2, Diagrammatically (butterfly), There are N/2 butterflies for this stage of the FFT, and each butterfly requires one multiplication

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Fast Fourier Transform (FFT), III So far, The splitting of {H k } into two half-size DFTs can be repeated on H k 0 and H k 1 themselves,

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Fast Fourier Transform (FFT), III So far, – {H k 00 } is the N/4-point DFT of {h 0, h 4,…, h N-4 }, – {H k 01 } is the N/4-point DFT of {h 2, h 6,…, h N-2 }, – {H k 10 } is the N/4-point DFT of {h 1, h 5,…, h N-3 }, – {H k 11 } is the N/4-point DFT of {h 3, h 7,…, h N-1 }, – Note that there is a reversal of the last two digits in the binary expansions of the indices j in {h j }.

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Fast Fourier Transform (FFT), IV

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Fast Fourier Transform (FFT), V If we continue with this process of halving the order of the DFTS, then after R=log 2 N stages, we reach where we are performing N one-point DFTs. – One-point DFT of the number h j is just the identity h j h j – Since the reversal of the order of the bits will continue, all bits in the binary expansion of j will be arranged in reverse order. – Therefore, to begin the FFT, one must first rearrange {h j } so it is listed in bit reverse order. For each of the log 2 N stages, there are N/2 multiplications, hence there are (N/2)log 2 N multiplications needed for FFT. – Much less time than the (N-1)2 multiplications needed for a direct DFT calculation. – When N=1024, FFT=5120 multiplication, DFT=1,046,529 savings by a factor of almost 200.

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Fast Fourier Transform (FFT), VI

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