Numerical Methods Fast Fourier Transform Part: Informal Development of Fast Fourier Transform http://numericalmethods.eng.usf.edu

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Numerical MethodsFast Fourier Transform
Part Informal Development of Fast Fourier
Transform http//numericalmethods.eng.usf.edu
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• For more details on this topic
• Go to http//numericalmethods.eng.usf.edu
• Click on Keyword
• Click on Fast Fourier Transform

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Chapter 11.05 Informal Development of Fast
Fourier Transform
Lecture 11
Major All Engineering Majors Authors Duc
Nguyen http//numericalmethods.eng.usf.edu Numeri
cal Methods for STEM undergraduates
http//numericalmethods.eng.usf.edu
9/20/2020
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Informal Development of Fast Fourier Transform
(1)
(2)
(3)
(4)
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Informal Development cont.
Then Eq. (1) and Eq. (2) become
(5)
(5A)
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Informal Development cont.
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Informal Development cont.
(6)
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Informal Development cont.
(7)
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Informal Development cont.
(8)
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Informal Development cont.
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• This instructional power point brought to you by
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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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Numerical MethodsFast Fourier Transform
Part Factorized Matrix and Further Operation
Counthttp//numericalmethods.eng.usf.edu
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• Click on Keyword
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Chapter 11.05 Factorized Matrix and Further
Operation Count (Contd.)
Lecture 12
Equation (7) can be factorized as
(9)
Lets define the following inner product
(10)
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Factorized Matrix cont.
From Eq. (9) and (10) we obtain
(11A)
(11B)
(11C)
(11D)
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Factorized Matrix cont.
Equations(11A through 11D) for the inner matrix
times vector requires 2 complex multiplications
and 4 complex additions.
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Factorized Matrix cont.
Finally, performing the outer product (matrix
times vector) on the RHS of Equation(9), one
obtains
(12)
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Factorized Matrix cont.
(13A)
(13B)
(13C)
(13D)
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Factorized Matrix cont.
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Factorized Matrix cont.
For a large number of data points,
(14)
This implies that the number of complex
multiplications involved in Eq. (9) is about 372
times less than the one involved in Eq. (7).
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Graphical Flow of Eq. 9
Consider the case
Figure 1. Graphical form of FFT (Eq. 9) for the
case
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Figure 2. Graphical Form of FFT (Eq. 9) for the
case
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Acknowledgement

• This instructional power point brought to you by
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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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34
Numerical MethodsFast Fourier Transform
Part Companion Node Observationhttp//numerica
lmethods.eng.usf.edu
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  commercial purposes.
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  this one.

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Chapter 11.05 Companion Node Observation
(Contd.)
Lecture 13
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Figure 2. Graphical Form of FFT (Eq. 9) for the
case
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Companion Node Observation cont.
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Companion Node Spacing
Observing Figure 2, the following statements can
be made
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Companion Node Computation
The operation counts in any companion nodes (of
the vector), such as and
can be explained as (see Figure 2).
(15)
(16)
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Companion Node Computation cont.
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(17)
(18)
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Skipping Computation of Certain Nodes
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Acknowledgement

• This instructional power point brought to you by
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• Committed to bringing numerical methods to the
  undergraduate

47

• For instructional videos on other topics, go to
• http//numericalmethods.eng.usf.edu/videos/
• 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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Numerical MethodsFast Fourier Transform
Part Determination of http//numericalmethods.
eng.usf.edu
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Chapter 11.05 Determination of
Lecture 14
can be determined by the following steps
one obtains

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Determination of cont.
It is important to realize that the results of
Step 2 (0,0,1,0) are equivalent to expressing an
integer
in binary format. In other words
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Determination of cont.
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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

in binary format, with bits.
Assuming (a base 10 number) can be expressed
as (assuming bits)
(19)
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Computer Implementation cont.
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Computer Implementation cont.
(20)
If IDIFF 0, then the bit a1 0
If IDIFF ? 0, then the bit a1 1
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Computer Implementation cont.
etc.
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Example 1
For , , bits
and . Find the value of .
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Example 1 cont.
Thus
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Example 1 cont.
Determine the bit (Index )
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Example 1 cont.
Determine the bit (Index )
Thus
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Acknowledgement

• This instructional power point brought to you by
• Numerical Methods for STEM undergraduate
• http//numericalmethods.eng.usf.edu
• Committed to bringing numerical methods to the
  undergraduate

66

• 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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68
Numerical MethodsFast Fourier Transform
Part Unscrambling the FFThttp//numericalmetho
ds.eng.usf.edu
69

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• Click on Keyword
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  commercial purposes.
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72
Chapter 11.05 Unscrambling the FFT (Contd.)
Lecture 15
For the case , (see Figure 2),
the final bit-reversing operation for FFT is
shown in Figure 3.
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For do-loop index k 0 (0, 0, 0, 0) i
(0, 0, 0, 0) bit-reversion 0 If (i.GT.k)
Then T f4(k) f4(k) f4(i) f4(i)
T Endif Hence, f4(0) f4(0) no swapping.

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For k 1 (0,0,0,1) i (1,0,0,0)
bit-reversion 8 If (i.GT.k) Then T f4(k1)
f4(k1) f4(i8) f4(i8) T Endif Hence,
f4(1) f4(8) are swapped.
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.For k2(0,0,1,0) i (0,1,0,0)
4 Hence, f4(2) f4(4) are swapped.
.For k3(0,0,1,1) i (1,1,0,0)
12 Hence, f4(3) f4(12) are swapped.
. For k4(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.
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Computer Implementation of FFT case for N2r
The pair of companion nodes computation are
given by Eqs.(17) and (18). To avoid complex
number operations,Eq.(17) can be computed based
on real number operations, as following
(21)
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Computer Implementation cont.
Multiplying the last 2 complex numbers, one
obtains
(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
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Computer implementation cont.
(23A)
(23B)
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Computer implementation cont.
Recall Eq. (4)
Hence
(24)
where
(25)
Thus
(26A)
(26B)

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Computer Implementation cont.
Substituting Eqs. (26A) and (26B) into Eqs. (23A)
and (23B), one gets
(27A)
(27B)
Similarly, the single (complex number) Eq. (18)
can be expressed as 2 equivalent (real number)
Eqs. Like Eqs. (27A) and (27B).
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Acknowledgement

• This instructional power point brought to you by
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• Committed to bringing numerical methods to the
  undergraduate

83

• For instructional videos on other topics, go to
• http//numericalmethods.eng.usf.edu/videos/
• 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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