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Numerical Methods
Fast Fourier Transform Part: Informal Development of Fast Fourier Transform
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Click on Keyword Click on Fast Fourier Transform
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Chapter 11.05: Informal Development of Fast Fourier
Transform
Major: All Engineering Majors
Authors: Duc Nguyen
http://numericalmethods.eng.usf.edu
Numerical Methods for STEM undergraduateshttp://numericalmethods.eng.usf.edu04/21/23 5
Lecture # 11
Informal Development of Fast Fourier Transform
Recall the DFT pairs of Equations (20) and (21) of Chapter 11.04 and swapping the indexes , one obtains
n k
1
0
20
)(~ N
k
kN
win
n ekfC
1
0
2~ 01
)(N
n
kN
win
n eCN
kf
(1)
(2) where
1,...,3,2,1,0, Nkn (3)
Ni
eE2
12 iN eEhenceLet (4)6
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Informal Development cont.Then Eq. (1) and Eq. (2) become
1
0)()(
~~ N
k
nkn EkfnCC
1
0
~1)(
N
n
nknECN
kf
(5)
Assuming , then)2(24 rN
)3(
)2(
)1(
)0(
)3(~)2(
~)1(
~)0(
~
1
1
1
1111
1
963
642
321
f
f
f
f
C
C
C
C
EEE
EEE
EEE
N
(5A)
7
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Informal Development cont.
To obtain the above unknown vector for a given vector , the coefficient matrix can be easily converted as
f C~
963
642
321
1
963
642
321
1
1
1
1111
1
1
1
1111
1
EEE
EEE
EEE
EEE
EEE
EEE
N
8
http://numericalmethods.eng.usf.edu
http://numericalmethods.eng.usf.edu9
Hence, the unknown vector can be computed as with matrix vector operations, as following
C~
)3(
)2(
)1(
)0(
)3(~)2(
~)1(
~)0(
~
)3)(3()2)(3()1)(3()0)(3(
)3)(2()2)(2()1)(2()0)(2(
)3)(1()2)(1()1)(1()0)(1(
)3)(0()2)(0()1)(0()0)(0(
f
f
f
f
EEEE
EEEE
EEEE
EEEE
C
C
C
C
(6)
Informal Development cont.
Informal Development cont.
)3(
)2(
)1(
)0(
)3(~)2(
~)1(
~)0(
~
9630
6420
3210
0000
f
f
f
f
EEEE
EEEE
EEEE
EEEE
C
C
C
C
(7)
For and , then:3,4 kN 2n
22222
2)4(6 EEeEeEEEE i
N
Ni
Nnk
10
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http://numericalmethods.eng.usf.edu11
Thus, in general (for ) Nnk
where Unk EE
N
nkremainderU
),mod( NnkU (8)
Informal Development cont.
Informal Development cont.
Remarks:a) Matrix times vector, shown in Eq. (7), will require 16 (or ) complex multiplications and 12 (or ) complex additions.
b) Use of Eq. (8) will help to reduce the number of operation counts, as explained in the next section.
2N)1( NN
12
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The End - Really
Numerical Methods
Fast Fourier Transform Part: Factorized Matrix and Further Operation Count
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Click on Keyword Click on Fast Fourier Transform
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Chapter 11.05: Factorized Matrix and Further Operation Count
(Contd.)Equation (7) can be factorized as
)3(
)2(
)1(
)0(
010
001
010
001
100
100
001
001
)3(~)1(
~)2(
~)0(
~
2
2
0
0
3
1
2
0
f
f
f
f
E
E
E
E
E
E
E
E
C
C
C
C
Let’s define the following “inner – product”
)3(
)2(
)1(
)0(
010
001
010
001
)3(
)2(
)1(
)0(
2
2
0
0
1
1
1
1
f
f
f
f
E
E
E
E
f
f
f
f
(9)
(10)
21
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Lecture # 12
Factorized Matrix cont.
)2()0()0( 01 fEff
)3()1()1( 01 fEff
)2()0()2( 21 fEff
)2()0( 0 fEf
From Eq. (9) and (10) we obtain (11A)
(11B)
(11C)
22
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02*42
2 1 EeeE ii
with
)3()1()3( 21 fEff
)3()1( 0 fEf (11D)
http://numericalmethods.eng.usf.edu23
Equations(11A through 11D) for the “inner” matrix times vector requires 2 complex multiplications and 4 complex additions.
Factorized Matrix cont.
Factorized Matrix cont.
Finally, performing the “outer” product (matrix times vector) on the RHS of Equation(9), one obtains
)3(
)2(
)1(
)0(
100
100
001
001
)3(
)2(
)1(
)0(
)3(~)1(
~)2(
~)0(
~
1
1
1
1
3
1
2
0
2
2
2
2
f
f
f
f
E
E
E
E
f
f
f
f
C
C
C
C
(12)
24
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http://numericalmethods.eng.usf.edu25
)1()0()0( 10
12 fEff (13A)
)1()0()1()0()1( 10
112
12 fEffEff )3()2()2( 1
112 fEff
)3()2()3()2()3( 112
113
12 fEEffEff
)3()2( 11
1 fEf
(13B)
(13C)
(13D)
Factorized Matrix cont.
Factorized Matrix cont.Again, Eqs (13A-13D) requires 2 complex multiplicationsAnd 4 complex additions. Thus, the complete RHS of Eq.(9) Can be computed by only 4 complex multiplications (or ) and 8 complex additions (or ),
where Since computational time is mainly controlled by the number of multiplications, implementing Eq. (9) will significantly reduce the number of multiplication operations, as compared to a direct matrix times vector operations. (as shown in Eq. (7)).
2
24
2r
N 2*4Nr
26
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.2rN
Factorized Matrix cont.For a large number of data points,
r
NNrN
Ratio2
2
2
(14)
,22048 )11( rN Equation (14) gives:For
36.37211
)2048(2Ratio
This implies that the number of complex multiplications involved in Eq. (9) is about 372 times less than the one involved in Eq. (7).27
http://numericalmethods.eng.usf.edu
Graphical Flow of Eq. 9
Consider the case
422 2 rN
Figure 1. Graphical form of FFT (Eq. 9) for the case
28
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422 2 rN
)0(f
)1(f
)2(f
)3(f
)0(1f
)1(1f
)2(1f
)3(1f
)0(2f
)1(2f
)2(2f
)3(2f
0E
Initial data
Vector )(kf
Vector 1 (l=1)
)(1 kf
Vector 2 (l=2=r)
)(2 kf
Computational Vectors
0E
2E
2E
2E
1E
3E
0E
)15(
)14(
)13(
)12(
)11(
)10(
)9(
)8(
)7(
)6(
)5(
)4(
)3(
)2(
)1(
)0(
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
)15(
)14(
)13(
)12(
)11(
)10(
)9(
)8(
)7(
)6(
)5(
)4(
)3(
)2(
)1(
)0(
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
)15(
)14(
)13(
)12(
)11(
)10(
)9(
)8(
)7(
)6(
)5(
)4(
)3(
)2(
)1(
)0(
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
)15(
)14(
)13(
)12(
)11(
)10(
)9(
)8(
)7(
)6(
)5(
)4(
)3(
)2(
)1(
)0(
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
)15(
)14(
)13(
)12(
)11(
)10(
)9(
)8(
)7(
)6(
)5(
)4(
)3(
)2(
)1(
)0(
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
Data Array )(0 kf )(1 kf )(2 kf )(3 kf )(4 kf
L=1 L=2 L=3 L=4
Computation Arrays
8
8
8
8
8
8
8
8
0
0
0
0
0
0
0
0
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
12
12
12
12
4
4
4
4
8
8
8
8
0
0
0
0
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
14
14
6
6
10
10
2
2
12
12
4
4
8
8
0
0
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
15
7
10
3
13
5
9
1
14
6
11
2
12
4
8
0
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
Figure 2. Graphical Form of FFT (Eq. 9) for the case 29
http://numericalmethods.eng.usf.edu1622 4 rN
THE ENDhttp://numericalmethods.eng.usf.edu
This instructional power point brought to you byNumerical Methods for STEM undergraduatehttp://numericalmethods.eng.usf.eduCommitted to bringing numerical methods to the undergraduate
Acknowledgement
For instructional videos on other topics, go to
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This material is based upon work supported by the National Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
The End - Really
Numerical Methods
Fast Fourier Transform Part: Companion Node Observation
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Click on Keyword Click on Fast Fourier Transform
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Noncommercial — You may not use this work for commercial purposes.
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Chapter 11.05 : Companion Node Observation (Contd.)
Careful observation of Figure 2 has revealed that each computed vector (where and ) we can always find two (companion) nodes which came from the same pair of nodes in the previous vector.
thl rl ,...,2,11622 4 rN
For example, and are computed in terms of and .
)0(1f)0(f
)8(1f)8(f
Similarly, the companion nodes and are computed from the same pair of nodes as and .
)8(2f)8(1f
)12(2f
)12(1f38
http://numericalmethods.eng.usf.edu
Lecture # 13
http://numericalmethods.eng.usf.edu39
)15(
)14(
)13(
)12(
)11(
)10(
)9(
)8(
)7(
)6(
)5(
)4(
)3(
)2(
)1(
)0(
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
)15(
)14(
)13(
)12(
)11(
)10(
)9(
)8(
)7(
)6(
)5(
)4(
)3(
)2(
)1(
)0(
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
)15(
)14(
)13(
)12(
)11(
)10(
)9(
)8(
)7(
)6(
)5(
)4(
)3(
)2(
)1(
)0(
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
)15(
)14(
)13(
)12(
)11(
)10(
)9(
)8(
)7(
)6(
)5(
)4(
)3(
)2(
)1(
)0(
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
)15(
)14(
)13(
)12(
)11(
)10(
)9(
)8(
)7(
)6(
)5(
)4(
)3(
)2(
)1(
)0(
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
Data Array )(0 kf )(1 kf )(2 kf )(3 kf )(4 kf
L=1 L=2 L=3 L=4
Computation Arrays
8
8
8
8
8
8
8
8
0
0
0
0
0
0
0
0
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
12
12
12
12
4
4
4
4
8
8
8
8
0
0
0
0
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
14
14
6
6
10
10
2
2
12
12
4
4
8
8
0
0
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
15
7
10
3
13
5
9
1
14
6
11
2
12
4
8
0
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
E
Figure 2. Graphical Form of FFT (Eq. 9) for the case 1622 4 rN
Companion Node Observation cont.
Furthermore, the computation of companion nodes are independent of other nodes (within the –vector). Therefore, the computed and will override the original space of and .
)0(1f )8(1f)0(f )8(f
thl
Similarly, the computed and will override the space occupied by and which in turn, will occupy the original space of and .
)8(1f)8(2f
)8(f
)12(2f)12(1f
)12(f
Hence, only one complex vector (or 2 real vectors) of length are needed for the entire FFT process.N
40
http://numericalmethods.eng.usf.edu
Companion Node Spacing
Observing Figure 2, the following statements can be made:
a) in the first vector ( ), the companion nodes and are separated by
1l )0(1f)8(1f
12
16
28
l
Nork
b) in the second vector ( ), the companion nodes and are separated by .
2l)8(2f )12(2f 4k
.),4
16
2
16
2(
2etc
Nor
l
41
http://numericalmethods.eng.usf.edu
Companion Node ComputationThe operation counts in any companion nodes (of the vector), such as and can be explained as (see Figure 2).
ndthl 2 )8(2f )12(2f
4112 )12()8()8( Efff
12112 )12()8()12( Efff
4811 )12()8( EEff
4
8
)16(2
11 )12()8( Eeff Ni
411 )12()8( Eeff i4
112 )12()8()12( Efff
(15)
(16)42
http://numericalmethods.eng.usf.edu
Companion Node Computation cont.
Thus, the companion nodes and computation will require 1 complex multiplication and 2 complex additions (see Eq. (15) and (16)). The weighting factors for the companion nodes ( and ) are (or ) and (or ), respectively.
)8(2f )12(2f
)8(2f)12(2f 4E UE 12E 2NUE
)2
()()( 11 llU
llN
kfEkfkf
)2
()()2
( 11 llU
lllN
kfEkfN
kf
48
(17)
(18)43
http://numericalmethods.eng.usf.edu
Skipping Computation of Certain Nodes
Because the pair of companion nodes and are separated by the “distance” , at the
level, after every node computation, then
the next nodes will be skipped. (see Figure 2)
L
Nk2k
L
N
2thLL
N
2L
N
2
44
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Acknowledgement
For instructional videos on other topics, go to
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This material is based upon work supported by the National Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
The End - Really
Numerical Methods
Fast Fourier Transform Part: Determination of
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UE
For more details on this topic
Go to http://numericalmethods.eng.usf.edu
Click on Keyword Click on Fast Fourier Transform
You are free
to Share – to copy, distribute, display and perform the work
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Under the following conditions Attribution — You must attribute the
work in the manner specified by the author or licensor (but not in any way that suggests that they endorse you or your use of the work).
Noncommercial — You may not use this work for commercial purposes.
Share Alike — If you alter, transform, or build upon this work, you may distribute the resulting work only under the same or similar license to this one.
http://numericalmethods.eng.usf.edu
Chapter 11.05: Determination of
The values of ""U
Step 1: Express the index in binary form, using bits. For and 01231 2)0(2)0(2)0(2)1(0,0,0,18 rk
)1,...,2,1,0( Nkr ,4,2,8 rLk
,1622 4 rN
53
Lecture # 14
;2
)()( 11
ll
Ull
NkfEkfkf
ll
Ulll
NkfEkf
Nkf
2)(
211
can be determined by the following steps:
one obtains:
UE
http://numericalmethods.eng.usf.edu54
0,1,0,00,1,,0,0,0,1 XX
Step 2: Sliding this binary number positions to the right, and fill in zeros, the results are
224 Lr
It is important to realize that the results of Step 2 (0,0,1,0) are equivalent to expressing an integer
22
8
2 24
Lr
kM in binary format. In other
words
)0,1,0,0(2M
Determination of cont.UE
Determination of cont.
Step 3: Reverse the order of the bits, then (0,0,1,0) becomes (0,1,0,0) = . Thus,
U
42)0(2)0(2)1(2)0( 0123 U
It is “NOT” really necessary to perform Step 3, since the results of Step 2 can be used to compute “ “ as followingU
42)0(2)1(2)0(2)0( 3210 U55
http://numericalmethods.eng.usf.edu
UE
Computer Implementation to find
Based on the previous discussions (with the 3-step procedures), to find the value of “ ”, one only needs a procedure to express an integer
Lr
kM
2
r
U
Assuming (a base 10 number) can be expressed as (assuming bits)
M4r
11234 JaaaaM (19)
56
http://numericalmethods.eng.usf.edu
in binary format, with bits.
UE
http://numericalmethods.eng.usf.edu57
Divide by 2, then multiply the truncated result by 2 ( ), and compute the difference between the original number and the new number.
212 JJ 222 JJJ
M
Computer Implementation cont.
Computer Implementation cont.
Compute the difference between the original number and the new number :&&)( 21 JJJM
2221
Truncated
MMJJJIDIFF (20)
If IDIFF = 0, then the bit a1 = 0If IDIFF ≠ 0, then the bit a1 = 1
58
http://numericalmethods.eng.usf.edu
http://numericalmethods.eng.usf.edu59
Once the bit has been determined, the value of is set to (or value of is reduced by a factor of 2; since the previous .
A similar process can be used to determine the value of process can be used to determine the next bit
1a 1J2J
12341 aaaaMJ
1J
)2()2()2()2( 0
1
1
2
2
3
3
41 aaaaMJ
2aetc.
Computer Implementation cont.
Example 1For , , bits and . Find the value of .
8k rN 216 4r 2LU
1242
2
8
2J
kM
Lr
Determine the bit (Index )1a 1IInitialize 0U
12
2
21
2 JJ
0)2)(1(2)2( 221 JJJJIDIFF
01 a
00202 IDIFFUU
Thus
60
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Example 1 cont.
Determine the bit (Index )2a 2I
02
1
21
2 JJ
1)20(1)2( 221 JJJJIDIFF
12 a
11202 IDIFFUU
Thus
121 JJ
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http://numericalmethods.eng.usf.edu62
Determine the bit (Index )3a 3I
02
0
21
2 J
J
0)2)(0(0)2( 221 JJJJIDIFF
03 a20212 IDIFFUU
Thus
021 JJ
Example 1 cont.
Example 1 cont.
02
1
21
2 J
J
0)2)(0(0)2( 221 JJJJIDIFF
04 a
40222 IDIFFUU
Thus
021 JJ
Determine the bit (Index )4a 4I
63
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Numerical Methods
Fast Fourier Transform Part: Unscrambling the FFT
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Chapter 11.05: Unscrambling the FFT (Contd.)
)0000(4f
)0001(4f
)0010(4f
)0100(4f
)0101(4f
)0110(4f
)0111(4f
)1000(4f
)1001(4f
)1010(4f
)1011(4f
)1100(4f
)1101(4f
)1110(4f
)0000(~
C
)0001(~
C
)0010(~
C
)0100(~
C
)0101(~
C
)0110(~
C
)0111(~
C
)1000(~
C
)1001(~
C
)1010(~
C
)1011(~
C
)1100(~
C
)1101(~
C
)1110(~
C
)1111(~
C
)0011(~
C
0
1
2
4
5
6
7
8
9
10
11
12
13
14
15
3
= skip the operation
)(4 kf )(~
nC
For the case
, (see Figure 2), the final “bit-reversing” operation for FFT is shown in Figure 3.
4216 rN
Figure 3. Final “bit-reversing” for FFT (with
)1622 4 rN72
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Lecture # 15
For do-loop index k = 0 = (0, 0, 0, 0) i = (0, 0, 0, 0) = bit-reversion = 0If (i.GT.k) ThenT = f4(k) f4(k) = f4(i) f4(i) = TEndifHence, f4(0) = f4(0) no swapping.
55
73
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http://numericalmethods.eng.usf.edu74
For k = 1 = (0,0,0,1) i = (1,0,0,0) = bit-reversion = 8If (i.GT.k) ThenT = f4(k=1) f4(k=1) = f4(i=8) f4(i=8) = TEndifHence, f4(1) = f4(8) are swapped.
http://numericalmethods.eng.usf.edu75
. For k=4=(0,1,0,0) i=(0,0,1,0)=2
In this case, since “i” is not greater than “k”.Hence, no swapping, since f4 (k = 2) and f4 (i = 4); had already been swapped earlier! .etc.
.For k=3=(0,0,1,1) i = (1,1,0,0) = 12Hence, f4(3) = f4(12); are swapped.
.For k=2=(0,0,1,0) i = (0,1,0,0) = 4Hence, f4(2) = f4(4); are swapped.
Computer Implementation of FFT case for N=2r
The pair of companion nodes computation are givenby Eqs.(17) and (18). To avoid “complex number” operations,Eq.(17) can be computed based on “real number” operations, as following
)()()()( 11 kifkfkifkf I
L
R
L
I
L
R
L
)
2()
2( 11
,,L
ILL
RL
IURU Nkif
NkfiEE
(21)In Eq. (21), the superscripts and denote real and imaginary components, respectively.
R I76
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Computer Implementation cont.
Multiplying the last 2 complex numbers, one obtains
)()()()( 11 kifkfkifkf I
L
R
L
I
L
R
L
)
2()
2( 1
,1
,L
IL
IUL
RL
RU NkfE
NkfE
)
2()
2( 1
,1
,L
RL
IUL
IL
RU NkfE
NkfEi (22)
Equating the real (and then, imaginary) components on the Left-Hand-Side (LHS), and the Right-Hand-Side (RHS) of Eq. (22), one obtains
77
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http://numericalmethods.eng.usf.edu78
)
2()
2()()( 1
,1
,1 L
IL
IUL
RL
RURL
RL
NkfE
NkfEkfkf
)
2()
2()()( 1
,1
,1 L
RL
IUL
IL
RUIL
IL
NkfE
NkfEkfkf
(23A)
(23B)
Computer implementation cont.
http://numericalmethods.eng.usf.edu
Computer implementation cont.
Recall Eq. (4)Ni
eE2
Hence
)sin()cos(22
ieeeE iNU
iU
NiU
(24)
where
N
U
N
U 28.62
(25)
Thus:)cos(, RUE
)sin(, IUE
(26A)
(26B)79
Computer Implementation cont.Substituting Eqs. (26A) and (26B) into Eqs. (23A) and (23B), one gets
)
2()sin()
2()cos()()( 111 L
I
LL
R
L
R
L
R
L
Nkf
Nkfkfkf
)
2()sin()
2()cos()()( 111 L
R
LL
I
L
I
L
I
L
Nkf
Nkfkfkf
(27B)
(27A)
Similarly, the single (complex number) Eq. (18) can be expressed as 2 equivalent (real number) Eqs. Like Eqs. (27A) and (27B).
80
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This instructional power point brought to you byNumerical Methods for STEM undergraduatehttp://numericalmethods.eng.usf.eduCommitted to bringing numerical methods to the undergraduate
Acknowledgement
For instructional videos on other topics, go to
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This material is based upon work supported by the National Science Foundation under Grant # 0717624. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
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