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Integral Transform. 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. - PowerPoint PPT Presentation
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Integral TransformIntegral Transform
Integral TransformsIntegral 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.
dttftKgb
a )(),()(
Problem in
Transform space
Original
problem
Solution in
Transform space
Solution of
original problem
Integral transform
Relatively easy solution
Difficult solution
Inverse transform
Fourier Transform, IFourier 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 =2f,
dfefHth
dtethfH
ift
ift
2
2
)()(
)()(
deHth
dtethH
ti
ti
)(2
1)(
)()(
n
n
ixtnn
ndte
x
nxxxx
2
1sin)( ),(lim)(
)()()( afxfax
Examples, IExamples, I FT of Dirac delta function:
FT of cos(0t)
t
(t)
1)( 2 dtet ift
0{cos( )}tFcos(0t)
t
Examples, IIExamples, II FT of exp(2if0t)=exp(i0t)
Sum
F {exp(i0t)}
exp(i0t)
t
t Re
Im
FT
Fast Fourier Transform (FFT), IFast Fourier Transform (FFT), I DFT appears to be an O(N2) 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 Hk0 to the transform of its
N/4 even-numbered input data and N/4 odd-numbered data. For N = 2R, we can continue applying the reduction until we
subdivide the data into the transforms of length 1. For every pattern of log2N number of 0’s and 1’s, there is one-
point transformation that is just one of the input number hn
)( , /2101/20
12/
012
)2//(2/212/
02
)2//(2
12/
012
/)12(212/
02
/)2(2
1
0
/2
Nik
kkk
Nikk
N
kk
NijkNikN
jj
Nijk
N
jj
NkjiN
jj
Nkji
N
jj
Nijkk
eWHWHHeH
heehe
hehe
heH
nhH nk somefor 001...1001
Fast Fourier Transform (FFT), IIFast Fourier Transform (FFT), II For N=8
Since WN/2 = -1, Hk0 and Hk
1 have period N/2,
Diagrammatically (butterfly),
There are N/2 butterflies for this stage of the FFT, and each butterfly requires one multiplication
12/,...,1,0for , 102
10 NkHWHHHWHH kk
kN/kkk
kk
)12/,...,1,0(
101
100
NkHWHH
HWHH
kk
kk
kk
kk
)]]([)()[()]([)()(
)]()[()()(
)()(
74
32
54
164
22
44
0
74
32
54
164
22
44
0
76
54
32
166
44
22
0
77
66
55
44
33
22
10
7
0
hWhWhWhWhWhWhWh
hWhWhWhWhWhWhWh
hWhWhWhWhWhWhWh
hWhWhWhWhWhWhWhhWH
kkkkkkk
kkkkkkk
kkkkkkk
kkkkkkk
jj
jkk
Fast Fourier Transform (FFT), Fast Fourier Transform (FFT), IIIIII So far,
The splitting of {Hk} into two half-size DFTs can be repeated on Hk
0 and Hk1 themselves,
713
303
13
612
202
12
511
101
11
4100
010
313
303
03
212
202
02
111
101
01
010
000
00
HHWHH
HHWHH
HHWHH
HHWHH
HHWHH
HHWHH
HHWHH
HHWHH
k
14/,...,1,0for ,
,112101
4112101
0120004
012000
NkHWHHHWHH
HWHHHWHH
kk
kN/kkk
kk
kk
kN/kkk
kk
Fast Fourier Transform (FFT), Fast Fourier Transform (FFT), IIIIII So far,
– {Hk00} is the N/4-point DFT of {h0, h4,…, hN-4},
– {Hk01} is the N/4-point DFT of {h2, h6,…, hN-2},
– {Hk10} is the N/4-point DFT of {h1, h5,…, hN-3},
– {Hk11} is the N/4-point DFT of {h3, h7,…, hN-1},
– Note that there is a reversal of the last two digits in the binary expansions of the indices j in {hj}.
713
303
13
111
2101
111
612
202
12
110
0100
110
511
101
11
111
2101
101
410
000
10
110
0100
100
313
303
03
011
2001
011
212
202
02
010
0000
010
111
101
01
011
2001
001
010
000
00
010
0000
000
HHWHHHWHH
HHWHHHWHH
HHWHHHWHH
HHWHHHWHH
HHWHHHWHH
HHWHHHWHH
HHWHHHWHH
HHWHHHWHH
Fast Fourier Transform (FFT), Fast Fourier Transform (FFT), IVIV
Fast Fourier Transform (FFT), VFast Fourier Transform (FFT), V If we continue with this process of halving
the order of the DFTS, then after R=log2N stages, we reach where we are performing N one-point DFTs.
– One-point DFT of the number hj is just the identity hj hj
– 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 {hj} so it is listed in bit reverse order.
For each of the log2N stages, there are N/2 multiplications, hence there are (N/2)log2N 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.
Fast Fourier Transform (FFT), Fast Fourier Transform (FFT), VIVI
