Hossein Sameti
Department of Computer Engineering
Sharif University of Technology


Many real-life systems can be modeled by LTI systems  use
convolution for computing the output  use DFT to compute
convolution
Fast Fourier Transform (FFT) is a method for calculating Discrete


Fourier Transform (DFT) Only faster!

0k  N
X ( k )    x ( n )e
 0
Definition of DFT:
otherwise
How many computations?
N 1


j
2 kn
N
n 0
N pt. DFT of x(n)
Q: For each k: How many adds and how many mults?
A: (N-1) complex adds and N complex mults.
How many k values do we have?
N
Num.Comput  N ( N 1)  N 2  2N 2
O( N 2 )
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
2
2kn
 N 1
j
  x ( n )e N
X (k )   n  0
 0
0k  N
otherwise
Direct computation: O( N 2 )
Ideal case:
O(Const.)
O( N log 2N )
FFT:
Example:
N
Direct
FFT
10^3
O(10^6)
O(10^3*log10^3)=O(10^4)
10^6
O(10^12)
O(10^6*log10^6)=O(2*10^7)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
3
FFT
Decimation in
time
Decimation in
frequency
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
4
2kn
 N 1
j
  x ( n )e N
X (k )   n  0
 0
0k  N
otherwise
• The main idea: use the divide and conquer method
•It works by recursively breaking down a problem into two or
more sub-problems of the same (or related) type, until these
become simple enough to be solved directly.
•The solutions to the sub-problems are then combined to give
a solution to the original problem.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
5
2kn
 N 1
j
  x ( n )e N
X (k )   n  0
 0
N 1
X (k )   x(n)e
0k  N
N: power of 2
otherwise
j
n 0
2kn
N
N 1
  x(n)e
j
2kn
N
n 0
n: even
n: even
n=2r
n: odd
n:0N-2
r:0N/2-1
n: odd
n=2r+1
n:1N-1
r:0N/2-1
N
1
2
X ( k )   x ( 2r )e
r 0
2k ( 2 r )
j
N
N
1
2
  x(2r  1)e
j
2k ( 2 r 1)
N
r 0
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
6
N
1
2
X ( k )   x ( 2r )e
N
1
2
2k ( 2 r )
j
N
  x(2r  1)e
r 0
X ( k )   g ( r )e
2k ( 2 r )
j
N
r 0
2kr
j
X ( k )   g ( r )e N / 2
r 0
Suppose: G (k )   g (r )e
j
2kr
N /2
r 0
j
N
1
2
  h( r )e
j
2k ( 2 r 1)
N
r 0
N
1
2
X (k )  G (k )  e
2k ( 2 r 1)
N
r 0
N
1
2
N
 2 1
j
2k
N
N
2k 1
2kr
j

j
2
N
e
 h ( r )e N / 2
r 0
N
2kr
 2 1
j
H ( k )   h ( r )e N / 2
r 0
H (k )
•What are G(k) and H(k)?
7
N
 2 1
G ( k )   g ( r )e
X (k )  G (k )  e
j
2k
N
H (k )
r 0
N
 2 1
j
2kr
N /2
j
2k
N /2
H ( k )   h( r )e
r 0
•In G(k) and H(k), k varies between 0 and N/2-1.
•However, in X(k) , k varies between 0 and N-1.
Solution: use the relationship between DFS and DFT.
We thus need to replicate G(k) and H(k) “once”, to get X(k).
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
8
N
2
g(r)
x(n)
G (k )
N
2
pt. DFT
After
replication +
N
2
h(r)
N
2
X (k )

pt. DFT
H (k )
W
k
N
e
j
2k
N
(twiddle factor)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
9
X (0)  G(0)  WN0 H (0)
N
2
g(r)
G ( 0)
N
X ( )  G(0)  WNN / 2 H (0)
2
0
pt. DFT
G(
N
1
2
N
)  G(0)
2
N
 N 1
2
x(n)
H (0)
N
2
pt. DFT
h(r)
0
H(
X (k )  G (k )  e
j
2k
N
H (k )
N
1
2
N
)  H (0)
2
N
 N 1
2
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
10
X (k )  G(k )  WNk H (k )
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
11
N/2 pt. DFT block
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
12
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
13
r(0)
r(1)
N 1
R(k )   r (n)e
N 2
j
2kn
N
n 0
R(k )  r (0)e
j
2k ( 0 )
2
 r (1)e
j
2k (1)
2
R(k )  r (0)  r (1)e  jk
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
14
r(0)
r(1)
R(k )  r (0)  r (1)e  jk
k 0
R(0)  r (0)  r (1)
R(0)  x(0)  x(4)
k 1
R(1)  r (0)  r (1)
R(1)  x(0)  x(4)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
15
Flow graph of a the 2-pt. DFT
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
16
N
How many stages do we have? log 2
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
17
W
N
r
2
N
e
2
N
j
(r  )
N
2
e
j
 e
2
r
N
j
.e
2
r
N
j
2 N
( )
N 2
 W
r
N
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
18
W
N
r
2
N
 W
r
N
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
19
2 mults+ 2 adds
1 mult+ 2 adds
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
20
In-place computation (only N storage locations are needed)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
21
•How many stages do we have
log 2N
•Each stage has N inputs and N outputs.
•Each butterfly has 2 inputs and 2 outputs.
N
•Each stage has
2
butterflies.
•Each butterfly needs 1 mult and 2 adds.
Total number of
operations:
N
 2  log 2N  N log N adds
2
N
N
N
 log 2  log N mults
2
2
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
22
input
indexing
is
shuffled.
Output
indexing
is in
order.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
23
[1 0 0]
[0 0 1]
X (0)
0
[0 0 0]
[0 0 0]
0
X (1)
1
[0 0 1]
[1 0 0]
4
X (2)
2
[0 1 0]
[0 1 0]
2
X (3)
3
[0 1 1]
[1 1 0]
6
X (4)
4
[1 0 0]
[0 0 1]
1
X (5)
5
[1 0 1]
[1 0 1]
5
X (6)
6
[1 1 0]
[0 1 1]
3
X (7)
7
[1 1 1]
[1 1 1]
7
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
24
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
25
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
26
2kn
 N 1
j
  x ( n )e N
X (k )   n  0
 0
0k  N
otherwise
• The main idea: use the divide and conquer method (this
time in the frequency domain)
• Divide the computation into two parts: even indices of k
and odd indices of k.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
27
2kn
 N 1
j
  x ( n )e N
X (k )   n  0
 0
0k  N
otherwise
N 1
k  2r
X (2r )   x(n)e
j
2 ( r ) n
N /2
n 0
N
1
2
X ( 2 r )   x ( n )e
j
2 ( r ) n
N /2
n 0
n  m
N
2
N
1
2
X ( 2r )   x ( n)e
n0
j
2 ( r ) n
N /2
N 1
  x ( n) e
N
n
2
N
1
2
N
0r 
2
j
2 ( r ) n
N /2
N j
  x ( m  )e
m0
2
1
2 ( r )(m 
N
)
2
N /2
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
28
N
1
2
X ( 2 r )   x ( n )e
2 ( r ) n
j
N /2
n 0
N
1
2
X ( 2 r )   x ( n )e
N
1
2
N j
  x ( m  )e
m0
2
2 ( r ) n
j
N /2
n 0
N
1
2
2 ( r ) m
N /2
N
1
2
N
  x ( n  )e
n 0
2
N
X (2r )   ( x(n)  x(n  )) e
n 0
2
j
2 ( r ) n
N /2
j
2 ( r ) n
N /2
N
0r
2
g (n)
N/2 pt. DFT of g(n)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
29
N
1
2
N j
X (2r )   ( x(n)  x(n  )) e
n 0
2
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
2 ( r ) n
N /2
30
N 1
k  2r  1
X (2r  1)   x(n)e
j
2 ( 2 r 1) n
N
n 0
N
1
2
X (2r  1)   x(n)e
j
2 ( 2 r 1) n
N
n 0
n  m
N
1
2
X (2r  1)   x(n)e
n 0
j
N 1
  x(n)e
n
N
2
2 ( 2 r 1) n
N
N
1
2
N
0r 
2
j
2 ( 2 r 1) n
N
N
2
N j
  x ( m  )e
m0
2
2 ( 2 r 1)(m 
N
)
2
N
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
31
N
1
2
X (2r  1)   x(n)e
j
2 ( 2 r 1) n
N
n 0
N
1
2
X (2r  1)   x(n)e
2 ( 2 r 1) n
j
N
n 0
e
N
1
2
N j
  x ( m  )e
m0
2
j
2
N
( 2 r 1)
2
N
2 ( 2 r 1)(m 
N
1
2
N
)
2
N
N j
 x ( m  )e
m0
2
2 ( 2 r 1)(m )
N
-1
N
1
2
X (2r  1)   x(n)e
n 0
2 ( 2 r 1) n
j
N
N
1
2
N j
  x ( n  )e
n 0
2
2 ( 2 r 1)(n )
N
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
32
N
1
2
X (2r  1)   x(n)e
2 ( 2 r 1) n
j
N
n 0
N
1
2
N
1
2
N j
  x ( n  )e
n 0
2
N j
X (2r  1)   ( x(n)  x(n  )) e
n 0
2
N
1
2
2 ( 2 r 1)(n )
N
2 ( 2 r 1) n
N
N j
X (2r  1)   [( x(n)  x(n  )) e
n 0
2
2n
N
]e
j
2 ( r ) n
N /2
h(n)
N/2 pt. DFT of h(n) e
j
2n
N
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
33
N
1
2
N j
X (2r  1)   [( x(n)  x(n  )) e
n 0
2
2n
N
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
]e
j
2 ( r ) n
N /2
34
35
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
36
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
37
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
38
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
39




Change x with X (i.e., input nodes with output nodes)
Change X with x (i.e., output nodes with input nodes)
Reverse the order of the flow graphs.
The same system function is achieved.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
40
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
41
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
42
How can we deal with twiddle factors?
 Should we store them in a table (i.e, use a
lookup table) or should we calculate them?
 What happens if N is not a factor of 2?
 It can be shown that if N=RQ, then an N pt.
DFT can be expressed in terms of R Q-pt. DFT
or Q R pt. DFTs (Cooley-Tukey algorithm).

Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
43