chapter 19 fast fourier transform (fft) (theory and implementation)

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Chapter 19 Chapter 19 Fast Fourier Transform Fast Fourier Transform (FFT) (FFT) (Theory and (Theory and Implementation) Implementation)

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Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation). Learning Objectives. DFT algorithm. Conversion of DFT to FFT algorithm. Implementation of the FFT algorithm. DFT Algorithm. The Fourier transform of an analogue signal x(t) is given by:. - PowerPoint PPT Presentation

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Page 1: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Chapter 19Chapter 19

Fast Fourier Transform (FFT)Fast Fourier Transform (FFT)

(Theory and Implementation)(Theory and Implementation)

Page 2: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 2

Learning ObjectivesLearning Objectives

DFT algorithm.DFT algorithm. Conversion of DFT to FFT algorithm.Conversion of DFT to FFT algorithm. Implementation of the FFT algorithm.Implementation of the FFT algorithm.

Page 3: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 3

DFT AlgorithmDFT Algorithm

The Fourier transform of an analogue The Fourier transform of an analogue signal x(t) is given by:signal x(t) is given by:

dtetxX tj

The Discrete Fourier Transform (DFT) of a discrete-time signal x(nT) is The Discrete Fourier Transform (DFT) of a discrete-time signal x(nT) is given by:given by:

Where:Where:

1

0

2N

n

nkNj

enxkX

nxnTx

Nk

1,1,0

Page 4: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 4

0 20 40 60 80 100 120-2

-1

0

1

2Sampled signal

Sample

Am

plitu

de

0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1Frequency Domain

Normalised Frequency

Mag

nitu

de

DFT AlgorithmDFT Algorithm

If we let:If we let: NNj

We

2

then:then:

1

0

N

n

nkNWnxkX

Page 5: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 5

DFT AlgorithmDFT Algorithm

1

0

N

n

nkNWnxkX

X(0) X(0) = x[0]W= x[0]WNN00 + x[1]W + x[1]WNN

0*10*1 +…+ x[N-1]W +…+ x[N-1]WNN0*(N-1)0*(N-1)

X(1) X(1) = x[0]W= x[0]WNN00 + x[1]W + x[1]WNN

1*11*1 +…+ x[N-1]W +…+ x[N-1]WNN1*(N-1)1*(N-1)

::

X(k) X(k) = x[0]W= x[0]WNN00 + x[1]W + x[1]WNN

k*1k*1 +…+ x[N-1]W +…+ x[N-1]WNNk*(N-1)k*(N-1)

::

X(N-1)X(N-1) = x[0]W= x[0]WNN00 + x[1]W + x[1]WNN (N-1)*1(N-1)*1 +…+ x[N-1]W +…+ x[N-1]WNN (N-1)(N-1)(N-1)(N-1)

Note: For N samples of x we have N frequencies Note: For N samples of x we have N frequencies representing the signal.representing the signal.

x[n] x[n] = input= input

X[k] X[k] = frequency bins= frequency bins

WW = twiddle factors= twiddle factors

Page 6: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 6

Performance of the DFT AlgorithmPerformance of the DFT Algorithm

The DFT requires NThe DFT requires N2 2 (NxN) complex multiplications: (NxN) complex multiplications: Each X(k) requires N complex multiplications.Each X(k) requires N complex multiplications. Therefore to evaluate all the values of the DFT ( X(0) to X(N-1) ) NTherefore to evaluate all the values of the DFT ( X(0) to X(N-1) ) N22 multiplications are required. multiplications are required.

The DFT also requires (N-1)*N complex additions:The DFT also requires (N-1)*N complex additions: Each X(k) requires N-1 additions.Each X(k) requires N-1 additions. Therefore to evaluate all the values of the DFT (N-1)*N additions are required.Therefore to evaluate all the values of the DFT (N-1)*N additions are required.

Page 7: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 7

Performance of the DFT AlgorithmPerformance of the DFT Algorithm

0

20

40

60

80

100

120

0 1 2 3 4 5 6 7 8 9 10

Number of Samples

Num

ber

of M

ulti

plic

atio

ns

0

20

40

60

80

100

120

0 1 2 3 4 5 6 7 8 9 10

Number of Samples

Num

ber

of A

ddit

ions

Can the number of computations required Can the number of computations required be reduced?be reduced?

Page 8: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 8

DFT DFT FFT FFT

A large amount of work has been devoted to reducing A large amount of work has been devoted to reducing the computation time of a DFT.the computation time of a DFT.

This has led to efficient algorithms which are known as This has led to efficient algorithms which are known as the Fast Fourier Transform (FFT) algorithms.the Fast Fourier Transform (FFT) algorithms.

Page 9: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 9

DFT DFT FFT FFT

x[n] = x[0], x[1], …, x[N-1]x[n] = x[0], x[1], …, x[N-1]

10 ;1

0

NkWnxkXN

n

nkN [1][1]

Lets divide the sequence x[n] into even and odd sequences:Lets divide the sequence x[n] into even and odd sequences: x[2n] x[2n] = x[0], x[2], …, x[N-2]= x[0], x[2], …, x[N-2] x[2n+1] x[2n+1] = x[1], x[3], …, x[N-1]= x[1], x[3], …, x[N-1]

Page 10: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 10

nkN

nkNjnk

Nj

nkN

W

eeW

2

2

22

22

1

2

0

12

12

0

2 122

N

n

knN

N

n

nkN WnxWnxkX

DFT DFT FFT FFT

[2][2]

Equation 1 can be rewritten as:Equation 1 can be rewritten as:

Since:Since:

nkN

kN

knN WWW

2

12

kZWkY

WnxWWnxkX

kN

N

n

nkN

kN

N

n

nkN

12

0 2

12

0 2

122

Then:Then:

Page 11: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 11

kZWkY

WnxWWnxkX

kN

N

n

nkN

kN

N

n

nkN

12

0 2

2

12

0 2

1

DFT DFT FFT FFT

The result is that an N-point DFT can be The result is that an N-point DFT can be divided into two N/2 point DFT’s:divided into two N/2 point DFT’s:

10 ;1

0

NkWnxkXN

n

nkN N-point DFTN-point DFT

Where Y(k) and Z(k) are the two N/2 point DFTs Where Y(k) and Z(k) are the two N/2 point DFTs operating on even and odd samples respectively:operating on even and odd samples respectively:

Two N/2-Two N/2-point DFTspoint DFTs

Page 12: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 12

12

0

2

2

22

12

0

2

2

1

12

0 2

2

12

0 2

1

2

N

n

Nkn

N

Nk

N

N

n

Nkn

N

N

n

nkN

kN

N

n

nkN

WnxWWnxN

kX

WnxWWnxkX

DFT DFT FFT FFT

PeriodicityPeriodicity and and symmetrysymmetry of W can be exploited of W can be exploited to simplify the DFT further:to simplify the DFT further:

Or:Or:

And:And:

kN

kNj

jk

Nj

N

Njk

Nj

Nk

N WeeeeeW

22

2

22

2

kN

kNj

N

Njk

NjN

k

N WeeeW2

2

2

22

2

2

2

2

2

[3][3]

: Symmetry: Symmetry

: Periodicity: Periodicity

Page 13: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 13

DFT DFT FFT FFT

Symmetry and periodicity:Symmetry and periodicity:

WW880 0 == WW88

88

WW881 1 = W= W88

99

WW8822

WW8844

WW8833

WW8866

WW8877WW88

55

WWNNk+N/2 k+N/2 == - -WWNN

kk

WWN/2N/2k+N/2 k+N/2 == WWN/2N/2

kk

WW88k+4 k+4 == - -WW88

kk

WW88k+8 k+8 == WW88

kk

Page 14: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 14

kZWkY

WnxWWnxN

kX

kN

N

n

nkN

kN

N

n

nkN

12

0 2

2

12

0 2

12

DFT DFT FFT FFT

Finally by exploiting the symmetry and Finally by exploiting the symmetry and periodicity, Equation 3 can be written as:periodicity, Equation 3 can be written as:

[4][4]

Page 15: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 15

12

,0 ;2

12

,0 ;

NkkZWkY

NkX

NkkZWkYkX

kN

kN

DFT DFT FFT FFT

Y(k) and WY(k) and WNNk k Z(k) only need to be calculated once and used for both equations.Z(k) only need to be calculated once and used for both equations.

Note: the calculation is reduced from 0 to N-1 to 0 to (N/2 - 1).Note: the calculation is reduced from 0 to N-1 to 0 to (N/2 - 1).

Page 16: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 16

DFT DFT FFT FFT

Y(k) and Z(k) can also be divided into N/4 point Y(k) and Z(k) can also be divided into N/4 point DFTs using the same process shown above:DFTs using the same process shown above:

kVWkUN

kY

kVWkUkY

kN

kN

2

2

4

kQWkPN

kZ

kQWkPkZ

kN

kN

2

2

4

The process continues until we reach 2 point The process continues until we reach 2 point DFTs.DFTs.

12

,0 ;2

12

,0 ;

NkkZWkY

NkX

NkkZWkYkX

kN

kN

Page 17: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 17

DFT DFT FFT FFT

Illustration of the first decimation in time FFT.Illustration of the first decimation in time FFT.

y[0]y[0]y[1]y[1]y[2]y[2]

y[N-2]y[N-2]

N/2 point N/2 point DFTDFT

x[0]x[0]x[2]x[2]x[4]x[4]

x[N-2]x[N-2]

N/2 point N/2 point DFTDFT

x[1]x[1]x[3]x[3]x[5]x[5]

x[N-1]x[N-1]

z[0]z[0]z[1]z[1]z[2]z[2]

z[N/2-1]z[N/2-1]

X[N/2] = y[0]-WX[N/2] = y[0]-W00z[0]z[0]-1-1

X[0] = y[0]+WX[0] = y[0]+W00z[0]z[0]

WW00 X[N/2+1] = y[1]-WX[N/2+1] = y[1]-W11z[1]z[1]-1-1

X[1] = y[1]+WX[1] = y[1]+W11z[1]z[1]

WW11

Page 18: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 18

FFT ImplementationFFT Implementation

Calculation of the output of a ‘butterfly’:Calculation of the output of a ‘butterfly’:

U’=UU’=Urr’+jU’+jUii’ = X(k)’ = X(k)

L’=LL’=Lrr’+jL’+jLii’ = X(k+N/2)’ = X(k+N/2)

Y(k) = UY(k) = Urr+jU+jUii

WWNNkkZ(k) =Z(k) = (L (Lrr+jL+jLii)(W)(Wrr+jW+jWii))

-1-1

Different methods are available for calculating the outputs U’ and L’.Different methods are available for calculating the outputs U’ and L’. The best method is the one with the least number of multiplications and additions.The best method is the one with the least number of multiplications and additions.

Key:Key: U = UpperU = Upper r = realr = real

L = LowerL = Lower i = imaginaryi = imaginary

Page 19: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 19

FFT ImplementationFFT Implementation

Calculation of the output of a ‘butterfly’:Calculation of the output of a ‘butterfly’:

iiriirrririr WLWjLWjLWLjWWjLL

U’=UU’=Urr’+jU’+jUii’’

L’=LL’=Lrr’+jL’+jLii’’

UUrr+jU+jUii

(L(Lrr+jL+jLii)(W)(Wrr+jW+jWii))-1-1

iriirriirr

irriiriirr

UWLWLjUWLWL

jUUWLWLjWLWLU

riiriiirrr

riiriirrir

WLWLUjWLWLU

WLWLjWLWLjUUL

Page 20: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 20

FFT ImplementationFFT Implementation

Calculation of the output of a ‘butterfly’:Calculation of the output of a ‘butterfly’:

To further minimise the number of operations (* To further minimise the number of operations (* and +), the following are calculated only once:and +), the following are calculated only once:

temp1 = Ltemp1 = LrrWWrr temp2 = Ltemp2 = LiiWWii temp3 = Ltemp3 = LrrWWii temp4 = Ltemp4 = LiiWWrr

temp1_2 = temp1 - temp2temp1_2 = temp1 - temp2 temp3_4 = temp3 + temp4temp3_4 = temp3 + temp4

UUrr’ ’ = temp1 - temp2 + U= temp1 - temp2 + Urr = temp1_2 + U= temp1_2 + Urr

UUii’ ’ = temp3 + temp4 + U= temp3 + temp4 + Uii = temp3_4 + U= temp3_4 + Uii

LLrr’ ’ = U= Urr - temp1 + temp2 - temp1 + temp2 = U= Urr - temp1_2 - temp1_2

LLii’ ’ = U= Uii - temp3 - temp4 - temp3 - temp4 = U= Uii - temp3_4 - temp3_4

U’=UU’=Urr’+jU’+jUii’ = (L’ = (Lr r WWr r - L- Li i WWi i + U+ Urr)+j(L)+j(Lr r WWi i + L+ Li i WWr r + U+ Uii))

L’=LL’=Lrr’+jL’+jLii’ = (U’ = (Ur r - L- Lr r WWr r + L+ Li i WWii)+j(U)+j(Ui i - L- Lr r WWi i - L- Li i WWrr))

UUrr+jU+jUii

(L(Lrr+jL+jLii)(W)(Wrr+jW+jWii))-1-1

Page 21: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 21

FFT Implementation (Butterfly Calculation)FFT Implementation (Butterfly Calculation)

Converting the butterfly calculation into ‘C’ Converting the butterfly calculation into ‘C’ code:code:

temp1 = (y[lower].real * WR);temp1 = (y[lower].real * WR);

temp2 = (y[lower].imag * WI);temp2 = (y[lower].imag * WI);

temp3 = (y[lower].real * WI);temp3 = (y[lower].real * WI);

temp4 = (y[lower].imag * WR);temp4 = (y[lower].imag * WR);

temp1_2 = temp1 - temp2;temp1_2 = temp1 - temp2;

temp3_4 = temp 3 + temp4;temp3_4 = temp 3 + temp4;

y[upper].real = temp1_2 + y[upper].real;y[upper].real = temp1_2 + y[upper].real;

y[upper].imag = temp3_4 + y[upper].imag;y[upper].imag = temp3_4 + y[upper].imag;

y[lower].imag = y[upper].imag - temp3_4;y[lower].imag = y[upper].imag - temp3_4;

y[lower].real = y[upper].real - temp1_2;y[lower].real = y[upper].real - temp1_2;

Page 22: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 22

FFT ImplementationFFT Implementation

To efficiently implement the FFT algorithm a few observations are made:To efficiently implement the FFT algorithm a few observations are made: Each stage has the same number of butterflies (number of butterflies = N/2, N is number of points).Each stage has the same number of butterflies (number of butterflies = N/2, N is number of points). The number of DFT groups per stage is equal to (N/2The number of DFT groups per stage is equal to (N/2 stagestage).). The difference between the upper and lower leg is equal to 2The difference between the upper and lower leg is equal to 2 stage-1stage-1.. The number of butterflies in the group is equal to 2The number of butterflies in the group is equal to 2 stage-1stage-1..

Page 23: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 23

FFT ImplementationFFT Implementation

WW00-1-1

WW00-1-1

WW00-1-1

WW00-1-1

WW22-1-1

WW00

-1-1WW00

WW22-1-1

-1-1WW00

WW11-1-1

WW00

WW33-1-1

-1-1WW22

-1-1

Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Example: 8 point FFTExample: 8 point FFT

Page 24: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 24

FFT ImplementationFFT Implementation

WW00-1-1

WW00-1-1

WW00-1-1

WW00-1-1

WW22-1-1

WW00

-1-1WW00

WW22-1-1

-1-1WW00

WW11-1-1

WW00

WW33-1-1

-1-1WW22

-1-1

Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages:

Page 25: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 25

FFT ImplementationFFT Implementation

WW00-1-1

WW00-1-1

WW00-1-1

WW00-1-1

WW22-1-1

WW00

-1-1WW00

WW22-1-1

-1-1WW00

WW11-1-1

WW00

WW33-1-1

-1-1WW22

-1-1

Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 1 = 1

Stage 1Stage 1

Page 26: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 26

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 2 = 2

FFT ImplementationFFT Implementation

WW00-1-1

WW00-1-1

WW00-1-1

WW00-1-1

WW22-1-1

WW00

-1-1WW00

WW22-1-1

-1-1WW00

WW11-1-1

WW00

WW33-1-1

-1-1WW22

-1-1

Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Stage 2Stage 2Stage 1Stage 1

Page 27: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 27

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

FFT ImplementationFFT Implementation

WW00-1-1

WW00-1-1

WW00-1-1

WW00-1-1

WW22-1-1

WW00

-1-1WW00

WW22-1-1

-1-1WW00

WW11-1-1

WW00

WW33-1-1

-1-1WW22

-1-1

Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Stage 2Stage 2 Stage 3Stage 3Stage 1Stage 1

Page 28: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 28

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1:Stage 1:

FFT ImplementationFFT Implementation

WW00-1-1

WW00-1-1

WW00-1-1

WW00-1-1

WW22-1-1

WW00

-1-1WW00

WW22-1-1

-1-1WW00

WW11-1-1

WW00

WW33-1-1

-1-1WW22

-1-1

Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Stage 2Stage 2 Stage 3Stage 3Stage 1Stage 1

Page 29: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 29

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 1 = 1

FFT ImplementationFFT Implementation

WW00-1-1

WW00-1-1

WW00-1-1

WW00-1-1

WW22-1-1

WW00

-1-1WW00

WW22-1-1

-1-1WW00

WW11-1-1

WW00

WW33-1-1

-1-1WW22

-1-1

Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Stage 2Stage 2 Stage 3Stage 3Stage 1Stage 1

Block 1Block 1

Page 30: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 30

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 2 = 2

FFT ImplementationFFT Implementation

WW00-1-1

WW00-1-1

WW00-1-1

WW00-1-1

WW22-1-1

WW00

-1-1WW00

WW22-1-1

-1-1WW00

WW11-1-1

WW00

WW33-1-1

-1-1WW22

-1-1

Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Stage 2Stage 2 Stage 3Stage 3Stage 1Stage 1

Block 1Block 1

Block 2Block 2

Page 31: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 31

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 3 = 3

FFT ImplementationFFT Implementation

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Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Stage 2Stage 2 Stage 3Stage 3Stage 1Stage 1

Block 1Block 1

Block 2Block 2

Block 3Block 3

Page 32: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 32

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 4 = 4

FFT ImplementationFFT Implementation

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Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Stage 2Stage 2 Stage 3Stage 3Stage 1Stage 1

Block 1Block 1

Block 2Block 2

Block 3Block 3

Block 4Block 4

Page 33: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 33

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 4 = 4 Stage 2: NStage 2: Nblocksblocks = 1 = 1

FFT ImplementationFFT Implementation

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Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Stage 2Stage 2 Stage 3Stage 3Stage 1Stage 1

Block 1Block 1

Page 34: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 34

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 4 = 4 Stage 2: NStage 2: Nblocksblocks = 2 = 2

FFT ImplementationFFT Implementation

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Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Stage 2Stage 2 Stage 3Stage 3Stage 1Stage 1

Block 1Block 1

Block 2Block 2

Page 35: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 35

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 4 = 4 Stage 2: NStage 2: Nblocksblocks = 2 = 2 Stage 3: NStage 3: Nblocksblocks = 1 = 1

FFT ImplementationFFT Implementation

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Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Stage 2Stage 2 Stage 3Stage 3Stage 1Stage 1

Block 1Block 1

Page 36: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 36

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 4 = 4 Stage 2: NStage 2: Nblocksblocks = 2 = 2 Stage 3: NStage 3: Nblocksblocks = 1 = 1

(3)(3) B’flies/block:B’flies/block: Stage 1:Stage 1:

FFT ImplementationFFT Implementation

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Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Stage 2Stage 2 Stage 3Stage 3Stage 1Stage 1

Page 37: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 37

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 4 = 4 Stage 2: NStage 2: Nblocksblocks = 2 = 2 Stage 3: NStage 3: Nblocksblocks = 1 = 1

(3)(3) B’flies/block:B’flies/block: Stage 1: NStage 1: Nbtfbtf = 1 = 1

FFT ImplementationFFT Implementation

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Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Page 38: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 38

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 4 = 4 Stage 2: NStage 2: Nblocksblocks = 2 = 2 Stage 3: NStage 3: Nblocksblocks = 1 = 1

(3)(3) B’flies/block:B’flies/block: Stage 1: NStage 1: Nbtfbtf = 1 = 1 Stage 2: NStage 2: Nbtfbtf = 1 = 1

FFT ImplementationFFT Implementation

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Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Page 39: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 39

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 4 = 4 Stage 2: NStage 2: Nblocksblocks = 2 = 2 Stage 3: NStage 3: Nblocksblocks = 1 = 1

(3)(3) B’flies/block:B’flies/block: Stage 1: NStage 1: Nbtfbtf = 1 = 1 Stage 2: NStage 2: Nbtfbtf = 2 = 2

FFT ImplementationFFT Implementation

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Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Page 40: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 40

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 4 = 4 Stage 2: NStage 2: Nblocksblocks = 2 = 2 Stage 3: NStage 3: Nblocksblocks = 1 = 1

(3)(3) B’flies/block:B’flies/block: Stage 1: NStage 1: Nbtfbtf = 1 = 1 Stage 2: NStage 2: Nbtfbtf = 2 = 2 Stage 3: NStage 3: Nbtfbtf = 1 = 1

FFT ImplementationFFT Implementation

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Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Page 41: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 41

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 4 = 4 Stage 2: NStage 2: Nblocksblocks = 2 = 2 Stage 3: NStage 3: Nblocksblocks = 1 = 1

(3)(3) B’flies/block:B’flies/block: Stage 1: NStage 1: Nbtfbtf = 1 = 1 Stage 2: NStage 2: Nbtfbtf = 2 = 2 Stage 3: NStage 3: Nbtfbtf = 2 = 2

FFT ImplementationFFT Implementation

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Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Page 42: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 42

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 4 = 4 Stage 2: NStage 2: Nblocksblocks = 2 = 2 Stage 3: NStage 3: Nblocksblocks = 1 = 1

(3)(3) B’flies/block:B’flies/block: Stage 1: NStage 1: Nbtfbtf = 1 = 1 Stage 2: NStage 2: Nbtfbtf = 2 = 2 Stage 3: NStage 3: Nbtfbtf = 3 = 3

FFT ImplementationFFT Implementation

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Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Page 43: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 43

Example: 8 point FFTExample: 8 point FFT

(1)(1) Number of stages:Number of stages: NNstagesstages = 3 = 3

(2)(2) Blocks/stage:Blocks/stage: Stage 1: NStage 1: Nblocksblocks = 4 = 4 Stage 2: NStage 2: Nblocksblocks = 2 = 2 Stage 3: NStage 3: Nblocksblocks = 1 = 1

(3)(3) B’flies/block:B’flies/block: Stage 1: NStage 1: Nbtfbtf = 1 = 1 Stage 2: NStage 2: Nbtfbtf = 2 = 2 Stage 3: NStage 3: Nbtfbtf = 4 = 4

FFT ImplementationFFT Implementation

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Decimation in time FFT:Decimation in time FFT: Number of stages = logNumber of stages = log22NN Number of blocks/stage = N/2Number of blocks/stage = N/2stagestage

Number of butterflies/block = 2Number of butterflies/block = 2stage-1stage-1

Page 44: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 44

FFT ImplementationFFT Implementation

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Twiddle Factor IndexTwiddle Factor Index N/2 = 4N/2 = 4

Start IndexStart Index 00 00 00

Input IndexInput Index 11 22 44

Page 45: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 45

FFT ImplementationFFT Implementation

Twiddle Factor IndexTwiddle Factor Index N/2 = N/2 = 44 44 /2 = 2/2 = 2

Start IndexStart Index 00 00 00

Input IndexInput Index 11 22 44

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Page 46: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 46

FFT ImplementationFFT Implementation

Twiddle Factor IndexTwiddle Factor Index N/2 = 4N/2 = 4 44 /2 =/2 = 2 2 22 /2 = 1/2 = 1

Start IndexStart Index 00 00 00

Input IndexInput Index 11 22 44

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Page 47: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 47

FFT ImplementationFFT Implementation

Start IndexStart Index 00

Input IndexInput Index 11

Twiddle Factor IndexTwiddle Factor Index N/2 = 4N/2 = 4 44 /2 = 2/2 = 2 22 /2 = 1/2 = 1

Indicies UsedIndicies Used WW00 WW00

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Page 48: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 48

FFT ImplementationFFT Implementation

The most important aspect of converting the FFT diagram to ‘C’ The most important aspect of converting the FFT diagram to ‘C’ code is to calculate the upper and lower indices of each butterfly:code is to calculate the upper and lower indices of each butterfly:

GS = N/4; GS = N/4; /* Group step initial value *//* Group step initial value */step = 1;step = 1; /* Initial value *//* Initial value */NB = N/2;NB = N/2; /* NB is a constant *//* NB is a constant */

for (k=N; k>1; k>>1)for (k=N; k>1; k>>1) /* Repeat this loop for each stage *//* Repeat this loop for each stage */{{

for (j=0; j<N; j+=GS)for (j=0; j<N; j+=GS) /* Repeat this loop for each block *//* Repeat this loop for each block */{{

for (n=j; n<(j+GS-1); n+=step)for (n=j; n<(j+GS-1); n+=step) /* Repeat this loop for each butterfly *//* Repeat this loop for each butterfly */{{

upperindex = n;upperindex = n;lowerindex = n+step;lowerindex = n+step;

}}}}/* Change the GS and step for the next stage *//* Change the GS and step for the next stage */

GS = GS << 1;GS = GS << 1;step = step << 1;step = step << 1;

}}

Page 49: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 49

FFT ImplementationFFT Implementation

How to declare and access twiddle factors:How to declare and access twiddle factors:(1)(1) How to declare the twiddle factors:How to declare the twiddle factors:

struct {struct {short real; // 32767 * cos (2*pi*n) -> Q15 formatshort real; // 32767 * cos (2*pi*n) -> Q15 formatshort imag; // 32767 * sin (2*pi*n) -> Q15 formatshort imag; // 32767 * sin (2*pi*n) -> Q15 format

} w[] = { 32767,0,} w[] = { 32767,0, 32767,-201, 32767,-201,

......};};

short temp_real, temp_imag;short temp_real, temp_imag;

temp_real = w[i].real;temp_real = w[i].real;temp_imag = w[i].imag;temp_imag = w[i].imag;

(2)(2) How to access the twiddle factors:How to access the twiddle factors:

Page 50: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 50

FFT ImplementationFFT Implementation

Links:Links: Project location: Project location:

\CD ROM 2004\DSP Code 6711 Update\Chapter 19 - Fast Fourier Transform\FFT DIF _in C_Fixed point\fft_bios.pjt\CD ROM 2004\DSP Code 6711 Update\Chapter 19 - Fast Fourier Transform\FFT DIF _in C_Fixed point\fft_bios.pjt \CD ROM 2004\DSP Code for DSK6416\Chapter 19 - Fast Fourier Transform\FFT DIF _in C_Fixed point\fft_project.pjt\CD ROM 2004\DSP Code for DSK6416\Chapter 19 - Fast Fourier Transform\FFT DIF _in C_Fixed point\fft_project.pjt \CD ROM 2004\DSP Code for DSK6713\Chapter 19 - Fast Fourier Transform\FFT DIF _in C_Fixed point\fft_project.pjt\CD ROM 2004\DSP Code for DSK6713\Chapter 19 - Fast Fourier Transform\FFT DIF _in C_Fixed point\fft_project.pjt

Laboratory sheet: Laboratory sheet: FFTLabSheet.pdfFFTLabSheet.pdf

Page 51: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 51

FFT test on the DSK6713FFT test on the DSK6713

Page 52: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Dr. Naim Dahnoun, Bristol University, (c) Texas Instruments 2004

Chapter 19, Slide 52

FFT test on the DSK6416FFT test on the DSK6416

Page 53: Chapter 19 Fast Fourier Transform (FFT) (Theory and Implementation)

Chapter 19Chapter 19

Fast Fourier Transform (FFT)Fast Fourier Transform (FFT)

(Theory and Implementation)(Theory and Implementation)

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