The document presents an analysis of the radix-4 FFT algorithm, which improves computational efficiency compared to the traditional radix-2 FFT. The study examines the arithmetic costs associated with the radix-4 approach, detailing the number of complex multiplications and additions required at each computation stage. Ultimately, the radix-4 FFT demonstrates a 15% cost savings compared to its radix-2 counterpart.

1. Radix-4 FFT Algorithm
&
Analysis of Time Complexity
Raj K Jaiswal
Research Scholar
National Institute of Technology, Karnataka
Surathkal, Mangalore
Email: jaiswal.raaj@gmail.com

2. Outline
• Need of Redix-4 FFT.
• Analysis of Time Complexity for Radix-4 FFT.
11/8/2013
RKJ@DIT,NITK

3. DFT
The DFT is defined as
N 1
Xr   xl w , r  0,1,........N  1
l 0
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rl
N
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4. Radix-4 FFT
•
The radix-4 FFT is derived from DFT as shown in
above equation, Which Defines the DFT of a
complex time series.
•
The above Equation can be written as follows..
Xr 
N / 4 1
 x4 kw
k 0

r (4k )
N

N / 4 1
 x 4 k  1w
k 0
r ( 4 k 1)
N

N / 4 1
 x4k  2w
k 0
r ( 4 k  2)
N

N / 4 1
r
x 4 k  3 wN( 4 k 3)

k 0
N / 4 1
N / 4 1
N / 4 1
N / 4 1
k 0
k 0
k 0
k 0
r
r
x 4 kwN( 4 k )  wN

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r
2
x 4 k  1wN( 4 k )  wNr

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r
3
x 4 k  2 wN( 4 k )  wNr

r
x 4 k  3 wN( 4 k )


5. Radix-4 FFT
•
By decimating the Time series into four sets
mentioned as follows.
{ yk | yk
{ zk | zk
{ gk | gk
{hk | hk
x 4 k ,0  k 

x4k



x4k
x4k
N
4
,0  k 
 1
 2
,0  k 
,0  k 
 3
1
}
N
4
N
4
N
4
1
}
1
}
1
}
The four sub problems with period of N/4 can be
defined after the appropriate twiddle factor is
4
w  wN
identified by
N / 4
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RKJ@DIT,NITK

6. Radix-4 FFT
Hence the above four sub problems can be written as
N / 4 1
N / 4 1
k 0
Yr 
N / 4 1
k 0
k 0
r
x 4 kwN( 4 k ) 

4
x 4 k ( wN ) rk 

rk
ykwN / 4 , r  0,1,....N / 4  1

Similarly
Zr  .......
Gr  ..........
N / 4 1
N / 4 1
k 0
Hr 
N / 4 1
k 0
k 0
r
h 4 k  3 wN( 4 k ) 

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4
x 4 k  3( wN ) rk 

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rk
hkwN / 4 , r  0,1,....N / 4  1


7. Radix-4 FFT
•Now The size of each sub problem is N/4,Which is
equal to input size.
Above sub problem can be solved recursively to
obtain the solution for main problem with size of N.
Since the series Yr has a period of N/4,so
Y r =Yr+n/4 = Yr+n/2 = Yr+3n/4 this applies to Zr, Gr H r
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RKJ@DIT,NITK

8. Radix-4 FFT
Above sub problem can be written in form of
r
2
3
X r  Yr  wN Z r  wNr Gr  wNr H r
r
2
3
X r  N / 4  Yr  wN N / 4 Z r  wN( r  N / 4 )Gr  wN( r  N / 4 ) H r
r
2
3
X r  N / 2  Yr  wN N / 2 Z r  wN( r  N / 2 )Gr  wN( r  N / 2 ) H r
r
2
3
X r 3 N / 4  Yr  wN3 N / 4 Z r  wN( r 3 N / 4 )Gr  wN( r 3 N / 4 ) H r
By noting twiddle factors
N
wN / 4  w4  e 
j 2 / 4
N
  j , wN / 2  (  j ) 2  1
and
3
wNN / 4  (  j ) 3  j
Above four equation can be written in form of
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9. Radix-4 FFT
X r  Yr  w Z r  w Gr  w H r
r
N
rN / 4
rN
X rN / 4  Y
2r
N
3r
N
r
2
3
 jwN Z r  wNr Gr  jwNr H r
r
2
3
X r  N / 2  Yr  wN Z r  wNr Gr  wNr H r
X r 3 N / 4  Yr  jw Z r  w Gr  jw H r
r
N
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2r
N
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3r
N

10. Radix-4 FFT
Analyzing the arithmetic cost
1. To analyze the arithmetic cost of radix-4 make
2
r
3
wNr Gr , wN Z r , wNr H r
sure that
Is calculated before the partial sum
2.Since the size of each sub problem is N/4.3N/4
complex multiplication and N complex addition is
required in 1st stage of butterfly computation.
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11. Radix-4 FFT
3.The 2nd Stage of butterfly computation involves no
multiplication by twiddle factor, so only N complex
additions are needed.
4.Hence 3N/4 complex multiplication & 2N complex
addition is required to perform two stages of butterfly
computation.
5.Since One complex addition incurs 2 real addition and
one complex multiplication incurs 3 real multiplication
and 3 real addition.
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RKJ@DIT,NITK

12. Radix-4 FFT
6.Accordingly 9N/4 real multiplication and 25N/4 real
additions are required per step of Radix-4.
7.Total flop count required is 17N/2
8.After considering the special cases we need 17N/2-32
flop counts.
9.Since special factors also occur in every subsequent
step, the savings can be incorporated in setting up
recurrence equation.
10.The cost of Radix-4 FFT algorithm can be
represented by the following recurrence
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13. Radix-4 FFT
T(N)={ 4T(N/4)+17/2N-32 if N=4n >4,
16
if N=4.
we get following result on solving above equation.
T(N)=4(1/4)Nlog2 N-43/6N+32/3
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14. Radix-4 FFT:Conclusion
Compare to arithmetic cost of T(N)=N log2 N of radix-2
FFT,the saving by radix-4 is 15 percent.
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RKJ@DIT,NITK