# fft

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```					        REPRESENTATION OF THE LAPLACIAN IN THE
FOURIER DOMAIN

 f  f    2            2
 f  2  2
2

 x y
Taking the Fourier transform of this equation, we have:
 

     f exp(2j (ux  vy))dxdy
2


 
 2 f 2 f 

  2 x   2 y  exp(2j(ux  vy))dxdy




Integrating by parts, we have,
                               
 f                        f
  x  exp(2jux) |  x (2ju) exp(2jux)dx

     
       
Any function whose Fourier transform should have the
property that the function and all its derivatives should go
to 0 at infinity, therefore the first term disappears and we
evaluate the second term (the integral) by parts.
                                            

 (2ju) f ( x, y) exp(2jux) |              f ( x, y)(2ju) exp(2jux)dx
                   2

                                            

Till now, we have not written the integration over y (for
clarity). Writing all the terms together, we have
 

      4 (u  v ) exp(2j (ux  vy))dxdy
2     2       2



= - 42 (u2+v2)F(u,v)
Laplacian in the Fourier Domain
Let us now look at the number of computations involved
in calculating the Fourier transform.

1 M 1         2jux 
F (u )     f ( x) exp  M 
M x 0               
for u  0,1,2,.... - 1
.M

Calculation of F for each value of u involves M
multiplications and M-1 additions. We have to repeat this
process for M values of u, therefore the number of
operations is of the order of :
M2
HOW TO REDUCE THIS NUMBER?
The Fast Fourier Transform:
We re-write the summation (FT) as :
1 M 1
F (u )     f ( x) WM
ux

M x 0
WM  exp(2j / M )

We assume that M = 2n
Therefore M=2K ; K is some positive integer. Re-writing
the summation once again, we have:
1 2 K 1
F (u )      
ux
f ( x) W2 K
2 K x 0
W2K  exp(2j / 2 K )
1  1 K 1                   1 K 1             u ( 2 x 1) 
 K  f (2 x)W2 K           f (2 x  1)W2 K
u (2 x)

2  x 0                                                    
K x 0                         

1  1 K 1          1 K 1                u
 K  f (2 x)WK  K  f (2 x  1)WK W2 K 
ux                  ux

2  x 0              x 0                 
K 1
1
 f (2 x)W
We define :
Feven(u ) 
ux
K
K   x 0

K 1
1
Fodd (u )       f (2 x  1)W
ux
K
K   x 0
Therefore, the original equation can be written as:

1
2

F (u )  Feven(u )  Fodd (u )W2 K
u

Also let us try to evaluate F(u+K):

1
2

F (u  K )  Feven(u )  Fodd (u )W2 K
u

Therefore if we divide the original summation into even
and odd terms, we have to evaluate only one half of F(u),
the other half follows with no extra multiplications!
In fact we need not stop here, we can go on dividing each
summation into even and odd terms and this can be
continued till we reach a 2 point transform!
This process is called the successive doubling method or
Decimation.
Let us now calculate the number of operations involved in
this new method and hence the computational advantage.
Let m(n) and a(n) represent the number of multiplications
and additions respectively. For a two point transform M=2
and n=1.
For a 2-pt transform, the summation runs from 0 to k-
1(=0) and we have to compute F(0) and F(1).
Evaluation of Feven(0) and Fodd(0) requires no
multiplications since Wux =1. One addition and one
multiplication is required to evaluate
F(0) =Feven +Fodd W2K u
F(1) follows with one more addition.
Therefore :
m(1)=1        and a(1)=2
A 4-pt transform can be split into two 2-pt transforms
Therefore
m(2) = 2m(1) + 2 and
a(2) = 2a(1) + 4
Similarly:
m(3) = 2m(2) + 4 and
a(3) = 2a(2) + 8
We can generalize these results:
m(n) = 2m(n-1) + 2n-1
a(n) = 2a(n-1) + 2n
Therefore :
m(n) = 2[2m(n-2) + 2n-2] + 2n-1
= 2.2.2….2m(n-(n-1)) + 2n-1 +………2n-1
= 2n-1m(1) + (n-1)2n-1
= 2n-1 ( 1 + (n-1))= (n/2) 2n = (M/2) log2 M
a(n) = 2[2a(n-2) + 2n-1] + 2n
= 2n-1a(1) + (n-1)2n

= 2n ( 1+ n-1) = n 2n = M log2 M
The computational advantage of the FFT over a direct
implementation of the 1-D DFT is defined as:

2                      n
M        M       2
C (M )                   
M log 2 M log 2 M   n
For example, when n=15 (32,768 pixels) ,
C(M) = 2200
Therefore we can expect for this case that the FFT is
computed 2200 times faster than the DFT
over the DFT
Implementation of the FFT Algorithm

Consider a 8-pt transform for simplicity

1 7           2jux 
F (u )   f ( x) exp      
8 x 0          8 
for u  0,1,2,....
.7

1 1 3           1 3               u
   f (2 x)W4   f (2 x  1)W4 W8 
ux                ux

2  4 x 0       4 x 0             
1 1 1               1 3               2u        
   f (4 x)W2   f (4 x  2)W2 W8  
ux                ux

1  2  2 x 0          2 x 0                      

2 1 1 1                   1 1               2u 

   f (4 x  1)W2   f (4 x  3)W2 W8 W8 
ux                ux         u

 2  2 x 0
                         2 x 0                  

1 1
1 
                 
4u 1
   2u 
 2  2 f (0)  f (4)W8  2 f (2)  f (6)W8 W8   
4u

 


2 1 1
                 
4u  1

  f (1)  f (5)W8  f (3)  f (7)W8 W8 W8 
4u  2u   u

2 2                     2                       
BUTTERFLY DIAGRAM
f(0)                            F(0)

wo

f(1)                            F(4)

f(2)                            F(2)
F(6)

f(3)
f(4)                            F(1)
w4
W2
F(5)

f(6)                            F(3)

F(7)
Determination of the Wp factor:
1. Write x in binary with n bits
2. Slide this binary number n-l bits to the right & fill the
newly opened bit positions on the left with zeros
3. Bit reversal
Comparison of filtering in the spatial and Fourier
Domains:
•Filtering in the Fourier domain is supposed to be more
efficient (why?)
•Some filters can be more easily specified in the Fourier
domain (eg: Ideal low pass filter is difficult to specify in the
spatial domain)

Averaging in Spatial and Fourier Domains:
the Fourier transform of both and multiply them in the
Fourier domain and then inverse transform them to the
spatial domain.
Number of operations in 1-D convolution = N2
Number of operations involved if we use Fourier domain
filtering:
2Nlog2 N additions + Nlog2N multiplications + N
multiplications in Fourier domain + Nlog2N additions +
(N/2) log2N multiplications for the inverse
= 3Nlog2N additions + 3N/2 log2N +N multiplications
For N=1024
N2 = 1048576
In Fourier domain number of operations = 30720 additions
+ 16384 multiplications
h(x)
f(x)
3

2

0   1      2   3      x         0     1    2   3   x

g(x) = f(x) x h(x)

M 1
1
g ( x) 
M
 f ( m) h ( x  m)
m 0
1 7
g ( x )   f ( m) h ( x  m)
8 m 0

For x < 0 h(x-m) is zero

1 7
g (0)   f (m)h(m)   f (0).h(0) 
1                6
8 m 0         8                8

g (1)   f (0).h(1)  f (1)h(0) 
1                            12
8                             8
g (2)   f (0).h(2)  f (1)h(1)  f (2)h(0) 
1                                        18
8                                         8

g (3)   f (1)h(2)  f (2)h(1)  f (3)h(0) 
1                                       18
8                                        8

g (4)   f (2).h(2)  f (3)h(1) 
1                            12
8                             8

g (5)   f (3).h(2) 
1                6
8                8
g(x)

18/8

12/8
6/8

0   1   2   3   4   5   6   7   x
Periodic nature of the DFT
For correct evaluation of the convolution in the Fourier
domain, we have to pad the image with zeros such that:
P  A + B –1

Where A and B are the number of points in the
functions f and h (which we are convolving)
And P is number of points (after padding)
Computing the Inverse Transform Using a Forward
Transform Algorithm:

1   M 1
 2jux 
F (u ) 
M
 f ( x) exp  M 
x 0               

M 1
 2jux 
f ( x)   F (u ) exp       
u 0         M 
Taking the complex conjugate of both sides of the inverse
transform and dividing by M we have:

1            1   M 1
  2jux 
M
f * ( x) 
M
 F * (u) exp M 
u 0                  

Therefore we can use the forward transform algorithm for
computing the inverse transform, if we substitute F*(u)
instead of F(u). Then we take complex conjugate of the
result and then multiply by M.

```
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