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Radix 4 FFT algorithm and it time complexity computation
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.
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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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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 wN3 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
rN / 4
rN
X rN / 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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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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