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DSP-3 (FFT) 1 of 17 Dr. Ravi Billa

Digital Signal Processing 3 December 26, 2009

III. Fast Fourier transform

2007 Syllabus: Fast Fourier Transform (FFT) Radix-2 decimation in time and decimation in

frequency FFT Algorithms, Inverse FFT, and FFT for composite N.

Contents:3.1 Introduction

3.2 Radix-2 decimation-in-time FFT (Cooley-Tukey)3.3 Radix-2 decimation-in-frequency FFT (Sande-Tukey)

3.4 Inverse DFT using the FFT algorithm

3.5 *Decimation-in-time algorithm forN= 4 (Cooley-Tukey formulation)

3.6 *Decimation-in-frequency algorithm forN= 4 (Sande-Tukey formulation)3.7 FFT with general radix

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3.1 Introduction

For a finite-duration sequencex(n) of lengthN, the DFT sum may be written as

X(k) =

1

0

)(N

n

nk

NWnx , k= 0, 1, ,N1

where NW =Nje /2 . There are a total ofNvalues ofX(.) ranging fromX(0) toX(N1). The

calculation ofX(0) involves no multiplications at all since every product term involves 0NW =

0je = 1. Further, the first term in the sum always involves 0NW or0je = 1 and therefore does

not require a multiplication. EachX(.) calculation other thanX(0) thus involves (N1) complex

multiplications. And eachX(.) involves (N1) complex additions. Since there areNvalues ofX(.)

the overall DFT requires 2)1( N complex multiplications andN(N1) complex additions. For

largeNwe may round these off to 2N complex multiplications and the same number of complexadditions.

Each complex multiplication is of the form

(A + jB) (C + jD) = (ACBD) + j(BC + AD)

and therefore requires four real multiplications and two real additions. Each complex addition isof the form

(A + jB) + (C + jD) = (A + C) + j(B + D)

and requires two real additions. Thus the computation of all Nvalues of the DFT requires 4N2

real multiplications and 4 2N (= 2 2N + 2 2N ) real additions.

Efficient algorithms which reduce the number of multiply-and-add operations are known

by the name offast Fourier transform (FFT). The Cooley-Tukey and Sande-Tukey FFT

algorithms exploit the following properties of the twiddlefactor (phase factor), WN=Nje /2

(the factor

Nj

e

/2

is called theN

th

principal root of 1):

1. Symmetry property 2N

k

NW

= kNW

2. Periodicity property NkNW = kNW

To illustrate, for the case ofN= 8, these properties result in the following relations:

0

8W =4

8W = 11

8W =5

8W =2

1 j

2

8W =6

8W =j3

8W =7

8W =

2

1 j

The use of these properties reduces the number of complex multiplications fromN2

to

NN

2log2

(actually the number of multiplications is less than this because several of the

multiplications by rNW are really multiplications by 1 or jand dont count); and the number of

complex additions are reduced from 2N to NN 2log . Thus, with each complex multiplication

requiring four real multiplications and two real additions and each complex addition requiringtwo real additions, the computation of allNvalues of the DFT requires

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Number of real multiplications = 4

N

N2log

2= NN 2log2

Number of real additions = NN 2log2 + 2

N

N2log

2= NN 2log3

We can get a rough comparison of the speed advantage of an FFT over a DFT by

computing the number of multiplications for each since these are usually more time consuming

than additions. For instance, forN= 8 the DFT, using the above formula, would need 82

= 64

complex multiplications, but the radix-2 FFT requires only 12 (= 8log28

2 = 4 x 3).

Number of multiplications: DFT vs. FFT

No. of points

N

No. of complex multiplications No. of real multiplications

DFT FFT DFT FFT

32 1024 80 4096 320

128 16384 448 65536 1792

1024 1048576 5120 4194304 20480

We consider first the case where the lengthNof the sequence is an integral power of 2,

that is,N = 2

where is an integer. These are called radix-2 algorithms of which the

decimation-in-time (DIT) version is also known as the Cooley-Tukey algorithm and the

decimation-in-frequency (DIF) version is also known as the Sande-Tukey algorithm. We

show first how the algorithms work; their derivation is given later.

For a radix of (r= 2), the elementary computation (EC) known as the butterfly consists

of a single complex multiplication and two complex additions.

If the number of points,N, can be expressed asN= mr , and if the computation algorithm

is carried out by means of a succession ofr-point transforms, the resultant FFT is called a radix-ralgorithm. In a radix-rFFT, an elementary computation consists of an r-point DFT followed

by the multiplication of the rresults by the appropriate twiddle factor. The number ofECsrequired is

Cr= Nr

Nrlog

which decreases as rincreases. Of course, the complexity of anECincreases with increasing r.

Forr= 4, theECrequires three complex multiplications and several complex additions.

Suppose that we desire anN-point DFT whereNis a composite number that can befactored into the product of integers

N=N1 N2 Nm

If, for instance,N = 64 and m = 3, we might factorNinto the product 64 = 4 x 4 x 4, and the 64-point transform can be viewed as a three-dimensional 4 x 4 x 4 transform.

IfNis a prime number so that factorization ofNis not possible, the original signal can bezero-paddedand the resulting new composite number of points can be factored.

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3.2 Radix-2 decimation-in-time FFT (Cooley-Tukey)

Procedure and important points

1. The number of input samples isN = 2

where is an integer.2. The input sequence is shuffled through bit-reversal. The index n of the sequencex(n) is

expressed in binary and then reversed.

3. The number of stages in the flow graph is given by = N2log .

4. Each stage consists ofN/2 butterflies.5. Inputs/outputs for each butterfly are separated as follows:

Separation = 2m1

samples where m = stage index, stages being numbered from

left to right (that is, m = 1 for stage 1, m = 2 for stage 2 etc.).

This amounts to separation increasing from left to right in the order 1, 2, 4, ,N/2.

6. The number of complex additions = NN 2log and the number of complex multiplications

is NN

2log2

.

7. The elementary computation block in the flow graph, called the butterfly, is shown here.

This is an in-place calculation in that the outputs (A +B kNW ) and (ABk

NW ) can be

computed and stored in the same locations asA andB.

A +Bk

NW

AB kNW k

NW

A

B

Separation of 1

Separation of 2

Separation of 4

Stage 1 Stage 2 Stage 3

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Example 3.2.1 [2002] Radix-2, 8-point, decimation-in-time FFT for the sequence

n 0 1 2 3 4 5 6 7

x(n) = {1, 2 3 4 4 3 2 1}

Solution The twiddle factors are

0

8W = 11

8W =8/2je = 4/je =

2

1j

2

1

2

8W = 28/2je = 2/

je =j 38W = 38/2je = 4/3

je =2

1j

2

1

One of the elementary computations is shown below:

The signal flow graph follows:

A +B kNW

= 14 =3

ABk

NW = 1 + 4 = 5

0

8W= 1

A

= 1

B

=4

0

8W

2

8W

1

8W

3

8W

2

8W

0

8W

2

8W

0

8W

0

8W

0

8W

0

8W

08W

000 000 x(0) = 1 X(0) = 0

001 100 x(4) =4

111 111 x(7) =1

110 011 x(3) = 4

101 101 x(5) =3

100 001 x(1) = 2

010 010 x(2) = 3

011 110 x(6) =2

X(7) =

X(6) =

X(5) =

X(4) =

X(1) =

X(2) =

X(3) =

Stage 3 Bit-reversed

order

Natural

orderStage 2 Stage 1

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8-point FFT using DIT

Results of the first stage

Input Stage 1 Stage 2 Stage 3 (Output)

1 1 + (4) . 1 =3

4 1(4) . 1 = 5

3 3 + (2) . 1 = 1

2 3(2) . 1 = 5

2 2 + (3) . 1 =1

3 2(3) . 1 = 5

4 4 + (1) . 1 = 3

1 4(1) . 1 = 5

Results of the second stage


1 3 3 + 1 . 1 =2

4 5 5 + 5 . (j) = 5 4/2je

3 1 31 . 1 =4

2 5 55 . (j) = 5 4/2

je 2 1 1 + 3 . 1 = 2

3 5 5 + 5 . (j) = 5 4/2 je

4 3 13 . 1 =4

1 5 55 . (j) = 5 4/2je

Results of the third stage


1 3 2 2 + 2 . 1 = 0

4 5 5 4/2je 5 4/2

je + 5 4/2je . 4/je = 5j12.07

3 1 4 4 + (4) . (j) =4 + j4 = 4 4/32 je

2 5 5 4/2 je 5 4/2 je + 5 4/2 je . 4/3je = 5j 2.07

2 1 2 22 . 1 =4

3 5 5 4/2je 5 4/2

je 5 4/2je . 4/je = 5 + j2.07

4 3 4 4(4) . (j) =4j4 = 4 4/32je

1 5 5 4/2je 5 4/2

je 5 4/2je . 4/3je = 5 + j12.07

The DFT isX(k) = {0, (5j12.07), (4 + j4), (5j2.07),4, (5 + j2.07), (4j4), (5 + j12.07)}

The MATLAB program:

x = [1, 2, 3, 4, -4, -3, -2, -1], X = fft(x)

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3.3 Radix-2 decimation-in-frequency FFT (Sande-Tukey)

Procedure and important points

1. The number of input samples isN = 2

where is an integer.2. The input sequence is in natural order; the output is in bit-reversed order.

3. The number of stages in the flow graph is given by = N2log .

4. Each stage consists ofN/2 butterflies.5. Inputs/outputs for each butterfly are separated in the reverse order from that of the DIT.

The separation decreasesfrom left to rightin the orderN/2, , 4, 2, 1.

6. The number of complex additions = NN 2log and the number of complex multiplications

is NN

2log2

.

7. The basic computation block in the flow graph of the DIF FFT is the butterfly shown

here. This is an in-place calculation in that the two outputs (A + B) and (A B) kNW can

be computed and stored in the same locations asA andB.

A +B

(AB) kNW

k

NW A

B

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Example 3.3.1 [Cf. Example 2.1] Radix-2, 8-point, decimation-in-frequency FFT for the

sequence

n 0 1 2 3 4 5 6 7

x(n) = {1, 2 3 4 4 3 2 1}

Solution The twiddle factors are the same as in the DIT FFT done earlier (both being 8-point

DFTs):

0

8W = 11

8W =8/2je =

2

1 j2

1

2

8W = 28/2je = 2/

je =j3

8W = 38/2je = 4/3

je =2

1j

2

1

One of the elementary computations is shown below:


X(7) =

A +B

= 14 =3

(AB) kNW

= (1 + 4) 1 = 5

0

8W = 1A

= 1

B

=4

2

8W

1

8W

0

8W

0

8W 0

8W

2

8W

0

8W

0

8W 0

8W

2

8W

0

8W

3

8W

x(0) = 1

Natural

order Stage 1 Stage 2 Stage 3

x(7) =1

x(6) =2

x(5) =3

x(4) =4

x(3) = 4

x(2) = 3

x(1) = 2

X(0) =

Bit-reversed

order

X(4) =

X(2) =

X(6) =

X(1) =

X(5) =

X(3) =

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8-point FFT using DIF



1 1 + (4) =3

2 2 + (3) =1

3 3 + (2) = 1

4 4 + (1) = 3

4 (1(4)) 1 = 5

3 (2(3)) 4/je = 5 4/je

2 (3(2)) (j) =j5

1 (4(1)) 4/3je = 5 4/3je



1 3 3 + 1 =2

2 1 1 + 3 = 2

3 1 (31) 1 =4

4 3 (13) (j) = j44 5 5 + (j5) = 5 4/2 je

3 5 4/je 5 4/je + 5 4/3je =j5 2

2 j5 (5(j5)) 1 = 5 4/2je

1 5 4/3je (5 4/je 5 4/3je ) (j) =j5 2



1 3 2 2 + 2 = 0

2 1 2 (22) 1 =4

3 1 4 4 + j4 =4 + j4 = 4 4/32je

4 3 j4 (4j4) 1 =4j4 = 4 4/32 je

4 5 5 4/2 je 5 4/2 je + (j5 2 ) = 5j12.07

3 5 4/je j5 2 (5 4/2 je (j5 2 )) 1 = 5 + j2.07

2 j5 5 4/2 je 5 4/2 je + (j5 2 ) = 5j2.07

1 5 4/3je j5 2 (5 4/2je (j5 2 )) 1 = 5 + j12.07

The DFT isX(k) = {0, (5j12.07), (4 + j4), (5j2.07),4, (5 + j2.07), (4j4), (5 + j12.07)}

The MATLAB progarm is the same as shown in Example 1.

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(DIT Template)

The elementary computation (Butterfly):

The signal flow graph:

A +B kNW

AB kNW kNW

A

B

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(DIF Template)

The elementary computation (Butterfly):

The signal flow graph:

A +B

(AB) kNW

k

NW A

B

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16-point DIF FFT

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3.4 Inverse DFT using the FFT algorithm

The inverse DFT of anN-point sequence {X(k), k= 1, 2, , (N1)} is defined as

x(n) =N

1

1

0

)(N

k

nk

NWkX , n= 0, 1, ,N1

where WN=Nje /2 . Take the complex conjugate ofx(n) and multiply byNto get

Nx

*

(n) =

1

0

*

)(

N

k

nk

NWkX

The right hand side of the above equation is simply the DFT of the sequenceX*(k) and can be

computed by using any FFT algorithm. The desired output sequence is then found by taking the

conjugate of the result and dividing byN

x(n) =N

1

*1

0

* )(

N

k

nk

NWkX

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Example 3.4.1 Given the DFT sequenceX(k) = {0, (1j), j, (2+j), 0, (2j),j, (1+j)} obtain the

IDFTx(n) using the DIF FFT algorithm.

Solution This is an 8-point IDFT. The 8-point twiddle factors are, as calculated earlier,

0

8W = 11

8W =8/2je =

2

1j

2

1

2

8W = 28/2je = 2/

je =j 38W = 38/2je =

4/3je =2

1j

2

1

The elementary computation (Butterfly) is shown below:


A +B

(AB) kNW

k

NW A

B

X*(0) = 0

Natural

order Stage 1 Stage 2 Stage 3

X*(7) =

X

*

(6) = j

X*(5) =

X*(4) = 0

X*(3) =

X*(2) =j

X*(1) =

8 x*(0) =

Bit-reversed

order

8 x*(4) =

8 x*(2) =

8 x*(6) =

8 x*(1) =

8 x*(5) =

8 x*(3) =

8 x*(7) =

0

8W

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8-point IDFT using DIF FFT


Input

X*(k)

Stage 1 Stage 2 Stage 3 (Output)

0 0 + 0 = 0

1+j 1+j + 2+j = 1+j2

j j + j = 0

2j 2j + (1j) = 1j2

0 (00) 1 = 0

2+j (1+j(2+j)) 4/je =3 4/je

j (jj) (j) =2

1j (2j(1j)) 4/3je = 3 4/3je



0 0 0 + 0 = 0

1+j 1+j2 1+j2 + 1j2 = 2

j 0 (00) 1 = 02j 1j2 (1+j2(1j2)) (j) = 4

0 0 0 + (2) =2

2+j 3 4/je 3 4/je + 3 4/3je =3 2

j 2 (0(2)) 1 = 2

1j 3 4/3je (3 4/je 3 4/3je ) (j) = 3 2



0 0 0 0 + 2 = 2 8x*(0)

1+j 1+j2 2 (02) 1 =2 8x

*

(4)j 0 0 0 + 4 = 4 8x*(2)

2j 1j2 4 (04) 1 =4 8x*(6)

0 0 2 2 + (3 2 ) =6.24 8x*(1)

2+j 3 4/je 3 2 (2(3 2 )) 1 = 2.24 8x*(5)

j 2 2 2 + 3 2 = 6.24 8x (3)

1j 3 4/3je 3 2 (23 2 ) 1 =2.24 8x

*(7)

The output at stage 3 gives us the values )(8 * nx in bit-reversed order:

orderrevbitnx )(8 * = {2,2, 4,4,6.24, 2.24, 6.24,2.24}

The IDFT is given by arranging the data in normal order, taking the complex conjugate of the

sequence and dividing by 8:

ordernormal

nx )(8 * = {2,6.24, 4, 6.24,2, 2.24,4,2.24}

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x(n) =

4

1,

8

24.6,2

1,

8

24.6,

4

1,

8

24.2,

2

1,

8

24.2

x(n) = {0.25,0.78, 0.5, 0.78,0.25, 0.28,0.5,0.28}

Note Because of the conjugate symmetry of {X(k)}, we should expect the sequence {x(n)} to bereal-valued.

The MATLAB program:

X = [0, (-1-j), j, (2+j), 0, (2-j), -j, (-1+j)], x = ifft(X)

Example 3.4.2 [2002] Given the DFT sequenceX(k) = {0, (1j), j, (2+j), 0, (2j), (1+j),j}

obtain the IDFTx(n) using the DIF FFT algorithm.

Solution There is no conjugate symmetry in {X(k)}. Using MATLAB

X = [0, 1-1j, 1j, 2+1j, 0, 2-1j, -1+1j, -1j]

x = ifft(X)The IDFT is

x(n) = {0.5, (-0.44 + 0.037i), (0.375 - 0.125i), (0.088 + 0.14i), (-0.75 + 0.5i),(0.44 + 0.21i), (-0.125 - 0.375i), (-0.088 - 0.39i)}

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DSP-3 (FFT) 17 of 17 Dr Ravi Billa

3.7 FFT with general radix

As mentioned in the introduction, if the number of points, N, can be expressed asN= mr , and if

the computation algorithm is carried out by means of a succession ofr-point transforms, the

resultant FFT is called a radix-ralgorithm. In a radix-rFFT, an elementary computation (EC)

consists of an r-point DFT followed by the multiplication of the rresults by the appropriatetwiddle factor. The number ofECs required is

Cr= Nr

Nrlog

which decreases as rincreases.

Of course, the complexity of anECincreases with increasing r. Forr= 2, theEC(the

butterfly) consists of a single complex multiplication and two complex additions; forr= 4, theECrequires three complex multiplications and several complex additions.

Suppose that we desire anN-point DFT whereNis a composite number that can be

factored into the product of integers

N=N1 N2 Nm

If, for instance,N = 64 and m = 3, we might factorNinto the product 64 = 4 x 4 x 4, and the 64-point transform can be viewed as a three-dimensional 4 x 4 x 4 transform.

IfNis a prime number so that factorization ofNis not possible, the original signal can be

zero-paddedand the resulting new composite number of points can be factored.We illustrate in the table below the situation forN= 64. Since 64 = 2

6, we can have a

radix-2 FFT; alternatively, since 64 = 43, we can also have a radix-4 FFT.

N= 64 = 26

= 43

= 82

Radix-2 Radix-4 Radix-8

No. of stages 64log2 = 6 64log4 = 3 64log8 = 2

No. ofECs per stage 64/2 = 32 64/4 = 16 64/8 = 8

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