unit - 3: fast-fourier-transform (fft) algorithms[ · fast fourier transform algorithms...

UNIT - 3: Fast-Fourier-Transform (FFT)algorithms[?, ?, ?, ?]

Dr. Manjunatha. [email protected]

ProfessorDept. of ECE

J.N.N. College of Engineering, Shimoga

October 15, 2014

Unit 3 Syllabus Introduction

Digital Signal Processing: [?, ?, ?, ?]

Slides are prepared to use in class room purpose, may be used as areference material

All the slides are prepared based on the reference material

Most of the figures/content used in this material are redrawn, someof the figures/pictures are downloaded from the Internet.

This material is not for commercial purpose.

This material is prepared based on Digital Signal Processing forECE/TCE course as per Visvesvaraya Technological University (VTU)syllabus (Karnataka State, India).

Dr. Manjunatha. P (JNNCE) UNIT - 3: Fast-Fourier-Transform (FFT) algorithms[?, ?, ?, ?]October 15, 2014 2 / 100

Unit 3 Syllabus

Unit 3 Syllabus

PART - A

UNIT - 3: Fast-Fourier-Transform (FFT) algorithms

Fast-Fourier-Transform (FFT) algorithms.

Direct computation of DFT.

Need for efficient computation of the DFT (FFT algorithms)


Fast Fourier Transform Algorithms Introduction

Fast Fourier Transform Algorithms

This unit provides computationally efficient algorithms for evaluating the DFT.

Direct computation of DFT has large number addition and multiplication operations.

The DFT has the various applications such as linear filtering, correlation analysis, andspectrum analysis.

Hence an efficient computation of DFT is an important issue in DSP.

There are two different approaches in computing efficient DFT, those are:

1 Divide and Conquer approach2 DFT as Linear filtering.



Computation ofDFT

DIT DFTAlgorithm

Radix-2FFT

Algorithm

DFT as LinearFiltering

Divide andConquerApproach

Radix-4FFT

Algorithm

Split-RadixFFT

Algorithm

Chirp-ZTransformAlgorithm

DIF DFTAlgorithm

GoertzelAlgorithm

Figure 1: FFT Algorithms



Computational complexity of DFT

Direct Computation of DFT

DFT expression is

X (k) =

N−1∑n=0

x(n)W knN k = 0, 1, . . . ,N − 1

IDFT expression is

x(n) =1

N

N−1∑k=0

X (k)W −knN n = 0, 1, . . . ,N − 1

X (k) = x(0)W 0N + x(1)W k

N + x(2)W 2kN , . . . ,+x(N − 1)W

(N−1)kN

For each value of k there are N complex multiplications and N − 1 additions.

Number of complex multiplications required to calculate X (k) is N × N = N2.

Number of complex additions required to calculate X (k) is (N − 1)× N = N2 − N.



Direct Computation of DFT for complex valued sequence x(n) (To determine number of realmultiplications, real additions and trigonometric functions)Let the sequence x(n) be of complex valued and is expressed as

x(n) = xR (n) + jxI (n)

Its DFT isX (k) = XR (k) + jXI (k)

X (k) =

N−1∑n=0

x(n)e−j 2πN

kn

=

N−1∑n=0

[xR (n) + jxI (n)]e−j 2πN

kn

=

N−1∑n=0

[xR (n) + jxI (n)]

[cos

(2π

Nkn

)− jsin

(2π

Nkn

)]

=

N−1∑n=0

[xR (n)cos

(2π

Nkn

)+ xI (n)sin

(2π

Nkn

)]

−j

N−1∑n=0

[xR (n)sin

(2π

Nkn

)− xI (n)cos

(2π

Nkn

)]



X (k) =

N−1∑n=0

[xR (n)cos

(2π

Nkn

)+ xI (n)sin

(2π

Nkn

)]

−j

N−1∑n=0

[xR (n)sin

(2π

Nkn

)− xI (n)cos

(2π

Nkn

)]

There are 4 Real multiplications and 2 real additions.

For each value of k there are 4N real multiplications and 2(N-1) real additions.

For complete DFT evaluation there are 4N.N = 4N2 real multiplications

For complete DFT evaluation there are X (k) 2(N − 1)N = 2N(N − 1) real additions.

There are 2 trigonometric functions are there in the above equation.

For each value of k there are 2N trigonometric functions are executed.

For complete DFT evaluation there are 2N.N = 2N2 trigonometric functions are executed.

Consider a two complex numbers (a+jb) and (c+jd), ”+” symbol is used represent thecomplex number, and addition of these complex number is:

(a + c) + j(b + d)



Periodicity PropertyW k+N

N = W kN

Proof:

WN = e−j 2πN

W k+NN = e−j 2π

N(k+N)

W k+NN = e−j 2π

N(k+N)

= e−j 2πN

k .e−j2π

= e−j 2πN

k .1

Since e−j2π = cos2π − jsin2π = 1

W k+NN = e−j 2π

Nk

= [e−j 2πN ]k

= W kN



Symmetry Property

Wk+N/2N = −W k

N

Proof:

WN = e−j 2πN

Wk+N/2N = e−j 2π

N(k+N/2)

= e−j 2πN

k .e−jπ

= e−j 2πN

k .1

Since e−jπ = cosπ − jsinπ = −1

Wk+N/2N = e−j 2π

Nk .(−1)

= [e−j 2πN ]k

= −W kN



W 2N = WN/2

Proof:

WN = e−j 2πN

WN/2 = e−j 2π

N/2

= [e−j 2πN ]2

= W 2N

ThereforeW 2

N = WN/2

Direct computation of the DFT does not exploit these symmetry and periodicity property.


FFT algorithms DIT-FFT algorithm

FFT algorithms



There are two methods in divide and conquer approach

1 Decimation in Time (DIT)-FFT algorithm and

2 Decimation in Frequency (DIF)-FFT algorithm



Decimation in Time (DIT)-FFT algorithm


FFT algorithms Radix-2 DIT-FFT algorithm

Radix-2 algorithm

The sequence x(n) of length N is factored in such a way that

N = r1r2r3 . . . rv

In this case r1 = r2 = r3 = . . . rv = r so that N = rv , where r is called the radix of theFFT algorithm.

r = 2 is called radix-2 algorithm, which is most widely used FFT algorithm.



The N point data sequence x(n) is splitted into two N/2 point data sequences f1(n), f2(n)These f1(n) and f2(n) data sequences contain even and odd numbered samples of x(n).

f1(n) = x(2n) n = 0, 1, . . . ,N/2− 1

f2(n) = x(2n + 1) n = 0, 1, . . . ,N/2− 1

The N point DFT is expressed as

X (k) =

N−1∑n=0

x(n)W knN k = 0, 1, . . . ,N − 1

The sequence x(n) is splitted as even and odd numbered samples and its DFT is:

X (k) =∑

n=even

x(n)W knN +

∑n=odd

x(n)W knN

=

N/2−1∑m=0

x(2m)W 2mkN +

N/2−1∑m=0

x(2m + 1)Wk(2m+1)N

In the above equation f1(m) = x(2m) and f2(m) = x(2m + 1), and its DFT is

X (k) =

N/2−1∑m=0

f1(m)(W 2N )mk +

N/2−1∑m=0

f2(m)(W 2N )mk W k

N



X (k) =

N/2−1∑m=0

f1(m)(WN/2)mk + W kN

N/2−1∑m=0

f2(m)(WN/2)mk

Comparing the above equation with DFT equation and can be expressed as:

X (k) = F1(k) + W kN F2(k) k = 0, 1, . . . ,N − 1

where F1(k) and F2(k) are N/2 point DFT of f1(m) and f2(m) respectively

F1(k) and F2(k) are periodic with period N/2 hence F1(k + N/2) = F1(k) and

F2(k + N/2) = F2(k), and also W(k+N/2)N = −W k

N . Replace k by k+N/2

X (k + N/2) = F1(k + N/2) + W(k+N/2)N F2(k + N/2)

X (k + N/2) = F1(k)−W kN F2(k)

X (k) = F1(k) + W kN F2(k) k = 0, 1, . . . ,N/2− 1

X (k + N/2) = F1(k)−W kN F2(k) k = 0, 1, . . . ,N/2− 1



8 point DFT

To Demonstrate the FFT algorithm 8 point DFT is considered as an example.

The block diagram of an 8 point DFT is as shown in Figure.

8-point DFT

(0)x (0)X

(2)X(1)X

(7)X(7)x

(2)x(1)x

DiscreteTime Signal( )x n is

DFT of( )X k

( )x n

Figure 2: Block diagram of 8 point DFT

For this Figure X (k) can be obtained from F1(k) and F2(k).

where F1(k) and F2(k) are two 4-point DFTs



The 8 point DFT can be found by combining two 4 point DFT F1(k) and F2(k).The sequences f1(m) and f2(m) are

f1(m) = x(2n) = [x(0), x(2), x(4), x(6)]

f2(m) = x(2n + 1) = [x(1), x(3), x(5), x(7)]

X (k) = F1(k) + W kN F2(k) k = 0, 1, . . . ,N/2− 1

X (k + N/2) = F1(k)−W kN F2(k) k = 0, 1, . . . ,N/2− 1

(0)X

-1-1

1(2 ) ( )x n f m=

1(0) (0)x f=

1(2) (1)x f=

1(6) (3)x f=

1(4) (2)x f=

1(0)F

1(3)F

1(2)F

1(1)F

2(2 1) ( )x n f m+ =

2(1) (0)x f=

2(2) (1)x f=

2(7) (3)x f=

2(5) (2)x f=

2 (0)F

2 (3)F

2 (2)F

2 (1)F

08W

38W

28W

18W -1

-1

(3)X

(2)X

(1)X

(0 4) (4)X X+ =

(3 4) (7)X X+ =

(2 4) (6)X X+ =

(1 4) (5)X X+ =N/2-point DFTi.e.4-point DFT

N/2-point DFTi.e.4-point DFT

Figure 3: 8 point DFT using two 4-point DFTDr. Manjunatha. P (JNNCE) UNIT - 3: Fast-Fourier-Transform (FFT) algorithms[?, ?, ?, ?]October 15, 2014 19 / 100


The process of decimation in time is repeated for the sequences f1(n) and f2(n).

f1(n) and f2(n) are two N/2 point sequences.

This is done by splitting the sequences into odd and even numbered sequences for f1(n) as

v11(n) = f1(2n) n = 0, 1, . . . ,N

4− 1

v12(n) = f1(2n + 1) n = 0, 1, . . . ,N

4− 1

Similarly splitting is done into odd and even numbered sequences for f2(n) as

v21(n) = f2(2n) n = 0, 1, . . . ,N

4− 1

v22(n) = f2(2n + 1) n = 0, 1, . . . ,N

4− 1

v11(n), v12(n), v21(n) and v22(n) are four N/4 point samples.



Apply the N/4 point DFT to obtain F1(k), based on the N/2 point DFT.

F1(k) = V11(k) + W kN/2V12(k) k = 0, 1, . . . ,

N

4− 1

F1(k + N/4) = V11(k)−W kN/2V12(k) k = 0, 1, . . . ,

N

4− 1

Similarly apply the N/4 point DFT to obtain F2(k).

F2(k) = V21(k) + W kN/2V22(k) k = 0, 1, . . . ,

N

4− 1

F2(k + N/4) = V21(k)−W kN/2V22(k) k = 0, 1, . . . ,

N

4− 1



The symbolic representation of F1(k) and F2(k) are as shown in Figure.

N/2-pointDFT i.e.4-point DFT

1(0)f1(0)F

1(2)F

1(1)F

1(3)F1(3)f

1(2)f

1(1)f

Figure 4: Symbolic representation of F1(k)


2 (0)f2 (0)F

2 (2)F

2 (1)F

2 (3)F2 (3)f

2 (2)f

2 (1)f




The 4-point sequences v11(n) and v12(n) are:

v11(n) = f1(2n) = x(4n) = [x(0), x(4)] n = 0, 1

v12(n) = f1(2n + 1) = x(4n + 2) = [x(2), x(6)] n = 0, 1

F1(k) = V11(k) + W kN/2V12(k) k = 0, 1, . . . ,

N

4− 1

F1(k + N/4) = V11(k)−W kN/2V12(k) k = 0, 1, . . . ,

N

4− 1

The detailed two N/4 point DFT for F1(k) are shown in Figure.


11 1(0) (0) (0)v f x= =

11 1(1) (2) (4)v f x= =

11(0)V

11(1)V

12 1(0) (1) (2)v f x= =

12 1(1) (3) (6)v f x= =

12 (0)V

12 (1)V

12 1( ) (2 1) (4 2)V n f n x n= + = +


11 1( ) (2 ) (4 )V n f n x n= =1(0)F

1(2)F

1(1)F

1(3)F14W

04W

-1

-1

Figure 6: 4-point DFT F1(k)



The 4-point sequences v21(n) and v22(n) are:

v21(n) = f2(2n) = x(4n) = [x(1), x(5)] n = 0, 1

v22(n) = f2(2n + 1) = x(4n + 2) = [x(3), x(7)] n = 0, 1

F2(k) = V21(k) + W kN/2V22(k) k = 0, 1, . . . ,

N

4− 1

F2(k + N/4) = V21(k)−W kN/2V22(k) k = 0, 1, . . . ,

N

4− 1



21 2(0) (0) (1)v f x= =

21 2(1) (2) (5)v f x= =

21(0)V

21(1)V

22 2(0) (1) (3)v f x= =

22 2(1) (3) (7)v f x= =

22 (0)V

22 (1)V

22 2( ) (2 1) (4 3)V n f n x n= + = +


21 2( ) (2 ) (4 1)V n f n x n= = +2 (0)F

2 (2)F

2(1)F

2 (3)F14W

04W

-1

-1




The decimation of the data sequence repeated again and again until the sequences reduceto the one point sequence.

For N = 2v , the decimation can be performed till v = log2N.

For N=8 v = log28 = 3.

The N/4 sequences are further splitted as odd and even sequences.

This will result in sequence of length N/8. The 2-point DFTs are expressed as

2-point DFT11 1(0) (0) (0)v f x= =

11 1(1) (2) (4)v f x= =

11(0)V

11(1)V

11( ), 0,1V k where k =11( ), 0,1V n where n =

Figure 8: 2-point DFT V11(k)



The n DFT is expressed as

X (k) =

N−1∑n=0

x(n)W knN k = 0, 1, . . . ,N/2− 1

The two point DFT for V11(k) is expressed as

V11(k) =

N−1∑n=0

v11(n)W kn2 k = 0, 1

For k=0

V11(0) =1∑

n=0

v11(n)W 02

=1∑

n=0

v11(n)

= v11(0) + v11(1)



For k=1

V11(1) =1∑

n=0

v11(n)W n2

= v11(0)W 02 + v11(1)W 1

2

W 02 = 1 and W 1

2 = e−j 2π2 ⇒ W 1

2 = e−jπ = cosπ − jsinπ = −1

V11(1) = v11(0)− v11(1)

V11(0) = v11(0) + W 02 v11(1)

V11(1) = v11(0)−W 02 v11(1)

2-point DFT

11(0)V

11(1)V02W

-1

11(0)v

11(1)v

Figure 9: 2-point DFT V11(k)

Similarly V12(k),V21(k), and V22(k) are estimated, and these details are as shown inFigure 11




The block diagram 8 point DFT computation is as shown in Figure

(7)X

(1)X

(2)X

(3)X

(4)X

(0)X

(6)X

(5)X

Third Stage Decimation

First Stage Decimation

Second Stage Decimation

Shuffled array inbit reversal order

Natural order ofDFT Sequence

2-point DFT(0)x

(4)x

2-point DFT(2)x

(6)x

Combine2-point DFTs

2-point DFT(1)x

(5)x

2-point DFT(3)x

(7)x

Combine2-point DFTs

Combine4 pointDFTs

Figure 10: Three stages in computation of an N=8 point DFT



11(0)V

0 02 8W W=

2-point DFT

(0)x

(4)x

-10 0

4 8W W=

Combine 2-point DFTs

1 24 8W W=0 0

2 8W W=

(2)x

(6)x

-1 -1

-1

12 (1)V

12 (0)V

11(1)V

21(0)V

0 02 8W W=

2-point DFT

(1)x

(5)x

-10 0

4 8W W=


1 24 8W W=0 0

2 8W W=

(3)x

(7)x

-1 -1

-1

22 (1)V

22 (0)V

21(1)V

-1

-1

-1

-1

28W

18W

08W

38W

1(0)F

1(2)F

1(3)F

1(1)F

2 (0)F

2 (2)F

2 (3)F

2 (1)F

(7)X

(1)X

(2)X

(3)X

(4)X

(0)X

(6)X

(5)X

Combine Two-4 point DFTs

Third Stage Decimation First Stage DecimationSecond Stage Decimation

Shuffled array inbit reversal order

Natural order ofDFT Sequence

Figure 11: Signal flow graph depicting Radix-2 DIT-FFT algorithm for N=8



Computational complexity in FFT algorithm

The basic computation of every stage of signal flow graph is represented by generalstructure is known as butterfly computation.

In this computation two complex numbers say (a,b), multiply b by W rN , and then add and

subtract the product from a to form new complex numbers (A,B).

This basic computation is as shown in Figure and is called as butterfly because the flowgraph resembles a butterfly

rNA a W b= +

rNW

-1

a

br

NB a W b= −

Figure 12: Butterfly computation in DIT-FFT algorithm

Butterfly operation requires two complex additions and one complex multiplication.

In complete signal flow graph Figure 11. there are N/2=4 butterflies in every stage ofdecimation and there are log2N = 3 = v stages of decimation (for N=8).

Total number of butterfly operations =N/2log2N = 4× 3 = 12

Total number of complex additions =2× N/2× log2N = N × log2N

Total number of complex multiplications =1× N/2× log2N = N/2× log2N



Table 1: Comparison of computational complexity for the direct computation of DFT versusthe FFT algorithm

No of Direct Computation DIT-FFT algorithm SpeedPoints Complex Complex Complex Complex improvement

multiplications additions multiplications additions for multiplications

N2 N2 − N N/2log2N Nlog2N N2

N/2log2N

4 16 12 4 8 4 times8 64 56 12 24 5.3 times

16 256 240 32 64 8 times32 1024 992 80 160 12.8 times64 4096 4032 192 384 21.3 times

256 65536 65280 1024 2048 64.0 times1024 1048576 1047552 5120 10240 204.8 times



Memory requirement in FFT algorithm

In butterfly computation, two complex numbers say (a,b), are input and (A,B) form newcomplex numbers.

Once ’A’ and ’B’ are estimated there is no need to store the old values a and b, hencesame memory locations can be used for ’A ’and ’B’.

’A’ is stored in location ’a’ and ’B ’is stored in location in ’b’, this is called in placecomputation.

For one butterfly computation four memory locations are required, two for ’a’ or ’A’ andtwo for ’b’ or ’B’ (’a’ or ’b’are two complex numbers ). Memory locations for onebutterfly = 2× 2 = 4

rNA a W b= +

rNW

-1

a

br

NB a W b= −

Figure 13: Butterfly computation in DIT-FFT algorithm

Computations are usually performed in stage wise. In every stage there are N/2 butterflies.Memory locations for one stage = 4× N/2 = 2N

2N locations are required for each stage. Usually stage by stage calculations areperformed, and the same locations are shared for next stage.

In every stage N/2 twiddle factors are required. Hence a maximum of (2N + N/2)memory locations are required.



Table 2: Shuffling of the data and bit reversal

Sl. Memory address Memory addressNo in bit reversed

Decimal Binary Binary Decimal1 0 000 000 02 1 001 100 43 2 010 010 24 3 011 110 65 4 100 001 16 5 101 101 57 6 110 011 38 7 111 111 7



Computation of twiddle factor

WN = e−j 2πN

W8 = e−j 2π8 = e−j π

4

W 08 = e0 = 1

W 18 = e−j π

41 = e−jπ/4 = cos(π/4)− jsin(π/4) = 0.7071− j0.7071

W 28 = e−j π

42 = e−jπ/2 = cos(π/2)− jsin(π/2) = −j

W 38 = e−j π

43 = e−j 3π

4 = cos

(3π

4

)− jsin

(3π

4

)= −0.7071− j0.7071


DIT-FFT algorithm Problems on DIT-FFT algorithm

Use the 8 point redix-2 DIT-FFT algorithm to find the DFT of the sequence x(n)={0.707,1,0.707,0,-0.707, -1,-0.707,0}

Solution:Based on the signal flow graph it is first we have to determine the two point DFT

V11(0) = x(0) + W 08 x4

= 0.707 + 1(−0.707) = 0

V11(1) = x(0)−W 08 x4

= 0.707− 1(−0.707) = 1.414

V11(1) = x(0)−W 08 + x4

= 0.707− 1(−0.707) = 1.414

V12(0) = x(2) + W 08 x6

= 0.707 + 1(−0.707) = 0

V12(1) = x(2)−W 08 x6

= 0.707− 1(−0.707) = 1.414



V21(0) = x(1) + W 08 x5

= 1 + 1(−1) = 0

V21(1) = x(1)−W 08 x5

= 1− 1(−1) = 2

V22(0) = x(3) + W 08 x7

= 0 + 1(0) = 0

V22(1) = x(1)−W 08 x5

= 1− 1(0) = 0



F1(0) = V11(0) + W 08 V12(0)

= 0 + 1(0) = 0

F1(1) = V11(1) + W 08 V12(1)

= 1.414 + (−j)1.414 = 1.414− j1.414

F1(2) = V11(0)−W 08 V12(0)

= 0− 1(0) = 0

F1(3) = V11(1)−W 08 V12(1)

= 1.414− (−j)1.414 = 1.414 + j1.414

F2(0) = V21(0) + W 08 V22(0)

= 0 + 1(0) = 0

F2(1) = V21(1) + W 08 V22(1)

= 2 + (−j)0 = 2

F2(2) = V21(0)−W 08 V22(0)

= 0− 1(0) = 0

F2(3) = V21(1)−W 08 V22(1)

= 2− (−j)0 = 2Dr. Manjunatha. P (JNNCE) UNIT - 3: Fast-Fourier-Transform (FFT) algorithms[?, ?, ?, ?]October 15, 2014 37 / 100


X (0) = F1(0) + W 08 F2(0)

= 0 + 1(0) = 0

X (1) = F1(1) + W 18 F2(1)

= (1.414− j1.414) + (0.7071− j0.7071)2

= 2.8284− j2.8284

X (2) = F1(2) + W 28 F2(2)

= 0 + (−j)(0) = 0

X (3) = F1(3) + W 38 F2(3)

= (1.414 + j1.414) + (−0.7071− j0.7071)2 = 0

X (4) = F1(0)−W 08 F2(0)

= 0 + 1(0) = 0

X (5) = F1(1)−W 18 F2(1)

= (1.414− j1.414)− (0.7071− j0.7071)2 = 0

X (6) = F1(2)−W 28 F2(2)

= 0− (−j)(0) = 0

X (7) = F1(3)−W 38 F2(3)

= (1.414 + j1.414)− (−0.7071− j0.7071)2

= 2.8284 + j2.8284



11(0) 0V =

0 02 8W W=

2-point DFT

(0) 0.707x =

(4) 0.707x = −

-10 0

4 8W W=


1 24 8W W=0 0

2 8W W=

(2) 0.707x =

(6) 0.707x = −

-1 -1

-1

12 (1) 2V =

12 (0) 0V =

11(1) 1.414V =

21(0)0V

0 02 8W W=

2-point DFT

(1) 1x =

(5) 1x = −

-10 0

4 8W W=


1 24 8W W=0 0

2 8W W=

(3) 0x =

(7) 0x =

-1 -1

-1

22 (1) 0V =

22 (0) 0V =

21(1) 2V =

-1

-1

-1

-1

28W

18W

08W

38W

1(0) 0F =

1(2) 0F =

1(3) 1.414 1.414F j= +

1(1) 1.414 1.414F j= −

2 (0) 0F =

2 (2) 0F =

2 (3) 2F =

2 (1) 2F =

(7) 2.8284 2.8284X j= +

Combine Two-4 point DFTs

(1) 2.8284 2.8284X j= −

(2) 0X =

(3) 0X =

(4) 0X =

(0) 0X =

(6) 0X =

(5) 0X =




X (0)X (1)X (2)X (3)X (4)X (5)X (6)X (7)

=

1 1 1 1 1 1 1 1

1 1√2− j 1√

2−j − 1√

2− j 1√

2−1 − 1√

2+ j 1√

2j 1√

2+ j 1√

21 −j −1 j 1 −j −1 j

1 − 1√2− j 1√

2j 1√

2− j 1√

21 1√

2+ j 1√

2−j − 1√

2+ j 1√

21 −1 1 −1 1 −1 1 −1

1 − 1√2

+ j 1√2

−j 1√2

+ j 1√2

−1 1√2− j 1√

2j − 1√

2− j 1√

21 j −1 −j 1 j −1 −j

1 1√2

+ j 1√2

j − 1√2

+ j 1√2

−1 − 1√2− j 1√

2−j 1√

2− j 1√

2

=

.7071.7070−.707−1−.7070

02.8284 − j2.8284000002.8284 + j2.8284



Use the 8 point redix-2 DIT-FFT algorithm to find the DFT of the sequencex(n)={1,1,1,1,0,0,0,0}Solution: Based on the signal flow graph it is first we have to determine the two point DFT

V11(0) = x(0) + W 08 x4

= 1 + 1(0) = 1

V11(1) = x(0)−W 08 x4

= 1− 1(0) = 1

V12(0) = x(2) + W 08 x6

= 1 + 1(0) = 1

V12(1) = x(2)−W 08 x6

= 0.707− 1(0) = 1



V21(0) = x(1) + W 08 x5

= 1 + 1(0) = 1

V21(1) = x(1)−W 08 x5

= 1− 1(0) = 1

V22(0) = x(3) + W 08 x7

= 1 + 1(0) = 1

V22(1) = x(1)−W 08 x5

= 1− 1(0) = 1



F1(0) = V11(0) + W 08 V12(0)

= 1 + 1(1) = 2

F1(1) = V11(1) + W 08 V12(1)

= 1 + (−j)1 = 1− j

F1(2) = V11(0)−W 08 V12(0)

= 1− (−j)(1) = 1− j

F1(3) = V11(1)−W 08 V12(1)

= 1− (−j)1.414 = 1 + j

F2(0) = V21(0) + W 08 V22(0)

= 1 + 1(1) = 2

F2(1) = V21(1) + W 08 V22(1)

= 1 + (−j)1 = 1− j



F2(2) = V21(0)−W 08 V22(0)

= 1− 1(1) = 0

F2(3) = V21(1)−W 08 V22(1)

= 1− (−j)1 = 1 + j



X (0) = F1(0) + W 08 F2(0)

= 2 + 1(2) = 4

X (1) = F1(1) + W 18 F2(1)

= (1− j1) + (0.707− j0.7071)(1− j)

= 1− j2.414

X (2) = F1(2) + W 28 F2(2)

= 0 + (−j)(0) = 0

X (3) = F1(3) + W 38 F2(3)

= (1 + j) + (−0.7071− j0.7071)(1 + j)

= 1− j0.414

X (4) = F1(0)−W 08 F2(0)

= 2− 1(2) = 0

X (5) = F1(1)−W 18 F2(1)

= (1− j1)− (0.707− j0.7071)(1− j)

= 1− j0.414

X (6) = F1(2)−W 28 F2(2)

= 0− (−j)(0) = 0

X (7) = F1(3)−W 38 F2(3)

= (1 + j)− (−0.7071− j0.7071)(1 + j)

= 1 + j2.414Dr. Manjunatha. P (JNNCE) UNIT - 3: Fast-Fourier-Transform (FFT) algorithms[?, ?, ?, ?]October 15, 2014 45 / 100

FFT algorithms DIF-FFT algorithm

Decimation in Frequency (DIF)-FFT algorithm



N point DFT is expressed as:

X (k) =

N−1∑n=0

x(n)W knN k = 0, 1, . . . ,N − 1

Split this summations into two summations with N/2 data points for each summation.

X (k) =

N/2−1∑n=0

x(n)W knN +

N−1∑n=N/2

x(n)W knN

=

N/2−1∑n=0

x(n)W knN +

N/2−1∑n=0

x(n + N/2)Wk(n+N/2)N

=

N/2−1∑n=0

x(n)W knN + W

kN/2N

N/2−1∑n=0

x(n + N/2)W knN



WkN/2N can be expressed as

WkN/2N = e−j 2π

NkN2 = e−jπk = e−jπ = (−1)k

X (k) =

N/2−1∑n=0

x(n)W knN + (−1)k

N/2−1∑n=0

x(n + N/2)W knN

=

N/2−1∑n=0

[x(n) + (−1)k x(n + N/2)

]W kn

N

Split the X (k) into even and odd numbered samples. This splitting in frequency is knownas decimation in frequency (DIF).

X (2k) =

N/2−1∑n=0

[x(n) + (−1)2k x(n + N/2)

]W 2kn

N k = 0, 1, . . .N/2− 1

X (2k + 1) =

N/2−1∑n=0

[x(n) + (−1)2k+1x(n + N/2)

]W

(2k+1)nN k = 0, 1, . . .N/2− 1



(−1)2k = 1 and W 2knN = (W 2

N )kn = W knN/2

Since W 2N = WN/2

Similarly (−1)2k+1 = −1 and W(2k+1)nN = W 2kn+n

N = W 2knN .W n

N Since W 2N = WN/2

Substituting these values in the previous equations:

X (2k) =

N/2−1∑n=0

[x(n) + x(n + N/2)] W knN/2 k = 0, 1, . . .N/2− 1

X (2k + 1) =

N/2−1∑n=0

[x(n)− x(n + N/2)] W 2knN .W n

N

=

N/2−1∑n=0

{[x(n)− x(n + N/2)] W nN}W

knN/2



Let us define N/2 point sequences g1(n) and g2(n) as

g1(n) = [x(n) + x(n + N/2)] n = 0, 1, . . .N/2− 1

g2(n) = [x(n)− x(n + N/2)] W nN n = 0, 1, . . .N/2− 1

X (2k) =

N/2−1∑n=0

g1(n)W knN/2 k = 0, 1, . . .N/2− 1

X (2k + 1) =

N/2−1∑n=0

g2(n)W knN/2 k = 0, 1, . . .N/2− 1



To Demonstrate the DIF-FFT algorithm 8 point DFT is considered as an example. Thevalues of g1(n) and g2(n) is defined as

g1(n) = x(n) + x(n + N/2) n = 0, 1, . . .N/2− 1

g1(0) = x(0) + x(4)

g1(1) = x(1) + x(5)

g1(2) = x(2) + x(6)

g1(3) = x(3) + x(7)

g2(n) = [x(n)− x(n + 4)] W n8 n = 0, 1, . . .N/2− 1

g2(0) = [x(0)− x(4)]W 08

g2(1) = [x(1)− x(5)]W 18

g2(2) = [x(2)− x(6)]W 28

g2(3) = [x(3)− x(7)]W 38



38W (7)X

(0)x

(3)x

(2)x

(1)x

1(0)g

1(1)g

1(3)g

1(2)g

(4)x

(7)x

(6)x

(5)x

2 (0)g

2 (1)g

2 (3)g

2 (2)g

4 point DFTs

4 point DFTs (5)X

(4)X

(3)X

(2)X

(1)X

(0)X

(6)X

28W

18W

08W

-1

-1

-1

-1

(2 )X K

(2 1)X K +

Figure 15: Decimation of DFT-FFT algorithm for N=8



Let us define N/4 point sequences p11(n) and p12(n) as

p11(n) = [g1(n) + g1(n + N/4)] n = 0, 1, . . .N/4− 1

p12(n) = [g1(n)− g1(n + N/4)] W nN/2 n = 0, 1, . . .N/4− 1

p21(n) = [g2(n) + g2(n + N/4)] n = 0, 1, . . .N/4− 1

p22(n) = [g2(n)− g2(n + N/4)] W nN/2 n = 0, 1, . . .N/4− 1

X (4k) =

N/4−1∑n=0

p11(n)W knN/4 k = 0, 1, . . .N/4− 1

X (4k + 2) =

N/4−1∑n=0

p12(n)W knN/4 k = 0, 1, . . .N/4− 1

X (4k + 1) =

N/4−1∑n=0

p21(n)W knN/4 k = 0, 1, . . .N/4− 1

X (4k + 3) =

N/4−1∑n=0

p22(n)W knN/4 k = 0, 1, . . .N/4− 1



The 4-point sequence g1(n) is splitted into two 2-point sequences p11(n) and p12(n) andrepresented as

p11(0) = g1(0) + g1(2)

p11(1) = g1(1) + g1(3)

p12(0) = [g1(0)− g1(2)]W 08 ∵ (W 0

4 = W 08 )

p12(1) = [g1(1)− g1(3)]W 28 ∵ (W 1

4 = W 28 )

Similarly the 4-point sequence g2(n) is splitted into two 2-point sequences p21(n) andp22(n) and represented as

p21(0) = g2(0) + g2(2)

p21(1) = g2(1) + g2(3)

p22(0) = [g2(0)− g2(2)]W 08

p22(1) = [g2(1)− g2(3)]W 28



1(0)g

1(1)g

1(3)g

1(2)g

2 (0)g

2 (1)g

2 (3)g

2 (2)g

-1

-1

28W

08W

11(0)p

12 (1)p

12 (0)p

11(1)p

-1

-1

28W

08W

21(0)p

22(1)p

22 (0)p

21(1)p

Figure 16: N=8 point decimation in frequency FFT algorithm



The 2-point DFTs are computed directly and represented as

X (4k) =

N/4−1∑n=0

p11(n)W knN/4 k = 0, 1, . . .N/4− 1

X (4k + 2) =

N/4−1∑n=0

p12(n)W knN/4 k = 0, 1, . . .N/4− 1

X (4k + 1) =

N/4−1∑n=0

p21(n)W knN/4 k = 0, 1, . . .N/4− 1

X (4k + 3) =

N/4−1∑n=0

p22(n)W knN/4 k = 0, 1, . . .N/4− 1

X (0) = P11(0) + P11(1)

X (4) = [P11(0)− P11(1)]W 08

X (2) = P12(0) + P12(1)

X (6) = [P12(0)− P12(1)]W 08

X (1) = P21(0) + P21(1)

X (5) = [P21(0)− P21(1)]W 08

X (3) = P22(0) + P22(1)

X (7) = [P22(0)− P22(1)]W 08Dr. Manjunatha. P (JNNCE) UNIT - 3: Fast-Fourier-Transform (FFT) algorithms[?, ?, ?, ?]October 15, 2014 56 / 100


(4)X

(2)X

(0)X

(6)X

-1

-1 08W

08W

11(0)p

12 (1)p

12 (0)p

11(1)p

(5)X

(3)X

(1)X

(7)X

-1

-1 08W

08W

21(0)p

22(1)p

22 (0)p

21(1)p

2 point DFT




38W

(0)x

(3)x

(2)x

(1)x

1(0)g

1(1)g

1(3)g

1(2)g

(4)x

(7)x

(6)x

(5)x

2 (0)g

2 (1)g

2 (3)g

2 (2)g28W

18W

08W

-1

-1

-1

-1

(4)X

(2)X

(0)X

(6)X

-1

-1

-1

28W

08W

-1 08W

08W

11(0)p

12 (1)p

12 (0)p

11(1)p

(5)X

(3)X

(1)X

(7)X

-1

-1

-1

28W

08W

-1 08W

08W

21(0)p

22(1)p

22 (0)p

21(1)p

DFT sequenceshuffled in bitreversal order

Input Sequence

2 point DFT




Computational Complexity Decimation in Frequency (DIF)-FFTalgorithm

A a b= +

rNW

-1

a

b ( ) rNB a b W= −

Figure 19: Butterfly structure decimation in frequency FFT algorithm







DIF-FFT algorithm Problems on DIF-FFT algorithm

Obtain the 8 point DFT of the following sequence using redix-2 DIF-FFT algorithm. Show allthe results along signal flow graph x(n)={2,1,2,1}

Solution:

W 08 = 1

W 18 = 0.707− j0.707

W 28 = −j

W 38 = −0.707− j0.707

g1(0) = x(0) + x(4)

= 2 + 0 = 2

g1(1) = x(1) + x(5)

= 1 + 0 = 1

g1(2) = x(2) + x(6)

= 2 + 0 = 2

g1(3) = x(3) + x(7)

= 1 + 0 = 1

g2(0) = [x(0)− x(4)]W 08

= [2− 0]× 1 = 2

g2(1) = [x(1)− x(5)]W 18

= [1− 0]× [0.707− j0.707]

= 0.707− j0.707

g2(2) = [x(2)− x(6)]W 28

= [2− 0]× [−j] = −j2

g2(3) = [x(3)− x(7)]W 38

= [1− 0]× [−0.707− j0.707]

= −0.707− j0.707



38W

(0)x

(3)x

(2)x

(1)x

(4)x

(7)x

(6)x

(5)x

28W

18W

08W

-1

-1

-1

-1

1(0) (0) (4) =2+0=2g x x= +

1(1) (1) (5) =1+0=1g x x= +

1(3) (3) (7) =1+0=1g x x= +

1(2) (2) (6) =2+0=2g x x= +

02 8(0) [ (0) (4)]

=[2-0] 1=2g x x W= −

×1

2 8(1) [ (1) (5)] =[1-0] (.707 .707)g x x W

j= −

× −2

2 8(2) [ (2) (6)] =[2-0] ( ) 2g x x W

j j= −

× − = −3

2 8(3) [ (3) (7)] =[1-0] ( .707 .707)g x x W

j= −

× − −



p11(0) = g1(0) + g1(2)

= 2 + 2 = 4

p11(1) = g1(1) + g1(3)

= 1 + 1 = 2

p12(0) = [g1(0)− g1(2)]W 08

= (2− 2)× 1 = 0

p12(1) = [g1(1)− g1(3)]W 08

= (1− 1)× (−j) = 0

p21(0) = g2(0) + g2(2)

= 2− j2

p21(1) = g2(1) + g2(3)

= 0.707− j0.707− 0.707− j0.707

= −j1.414

p22(0) = [g2(0)− g2(2)]W 08

= [2− (−j2)]× 1

= 2 + j2

p22(1) = [g2(1)− g2(3)]W 08

= [0.707− j0.707 + 0.707 + j0.707]× (−j)

= −j1.414



-1

-1

28W

08W

-1

-1

28W

08W

1(0) 2g =

1(1) 1g =

1(3) 1g =

1(2) 2g =

2 (0) 2g =

2 (1) (.707 .707)g j= −

2 (2) 2g j= −

2 (3) ( .707 .707)g j= − −

11 1 1(0) (0) (2) 2 2 4p g g= +

= + =

012 1 1 8(1) [ (1) (3)]

(1 1) ( ) 0p g g W

j= −

= − × − =

012 1 1 8(0) [ (0) (2)]

(2 2) 1 0p g g W= −

= − × =

11 1 1(1) (1) (3) 1 1 2p g g= +

= + =

21 2 2(0) (0) (2) 2 2p g g

j= +

= −

122 2 2 8(1) [ (1) (3)]

(.707 .707 .707 .707) 1.414

p g g Wj j j

j

= −

= − + + ×−= −

022 2 2 8(0) [ (0) (2)]

[2 ( 2)] 2 2p g g W

j j j= −

= − − ×− = +

21 2 2(1) [ (1) (3)] (.707 .707 .707 .707)

1.414

p g gj j

j

= += − − −= −



X (0) = P11(0) + P11(1)

= 4 + 2 = 6

X (4) = [P11(0)− P11(1)]W 08

= (4− 2)× 1 = 2

X (2) = P12(0) + P12(1)

= 0 + 0 = 0

X (6) = [P12(0)− P12(1)]W 08

= (0− 0)× 1 = 0

X (1) = P21(0) + P21(1)

= (2− j2− j1.414)

= 2− j3.414

X (5) = [P21(0)− P21(1)]W 08

= [2− j2− (−j1.414)]× 1

= 2− j0.586

X (3) = P22(0) + P22(1) = (2 + j2− j1.414)

= 2 + j0.586

X (7) = [P22(0)− P22(1)]W 08

= [2 + j2− (−j1.414)]× 1 = 2 + j3.414



011 11 8(4) [ (0) (1)]

(4-2)1=2X p p W= −

12 12(2) (0) (1) 0+0=0X p p= −

11 11(0) (0) (1) 4+2=6X p p= +

011 11 8(6) [ (0) (1)]

(0-0)1=0X p p W= +

-1

-1 08W

08W

021 21 8(5) [ (0) (1)]

[2-j2-(-j1.414)]1=2-j0.566X p p W= +

22 22(3) (0) (1) 2+j2-j1.414=2+j0.586X p p= +

21 21(1) (0) (1) 2-j2-j1.414=2-j3.414X p p= +

022 22 8(7) [ (0) (1)]

[2+j2-(-j1.414)]1=2+j3.414X p p W= −

-1

-1 08W

08W

2 point DFT

11(0) 4p =

12 (1) 0p =

12 (0) 0p =

11(1) 2p =

21(0) 2 2p j= −

22 (1) 1.414p j= −

22 (0) 2 2p j= +

21(1) 1.414p j= −



38W

(0) 2x =

(3) 1x =

(2) 2x =

(1) 1x =

1(0)=2g

1(1)=1g

1(3)=1g

1(2)=2g

(4) 0x =

(7) 0x =

(6) 0x =

(5) 0x =

2 (0)=2g

2 (1) (.707 .707)g j= −

2 (2) 2g j= −2

8W

18W

08W

-1

-1

-1

-1

(4) 2X =

(2) 0X =

(0) 6X =

(6) 0X =

-1

-1

-1

28W

08W

-1 08W

08W

11(0) 4p =

12 (1) 0p =

12 (0) 0p =

11(1) 2p =

(5) 2 0.586X j= −

(3) 2 0.586X j= +

(1) 2 3.414X j= −

(7) 2 3.414X j= +

-1

-1

-1

28W

08W

-1 08W

08W

21(0) 2 2p j= −

22 (1) 1.414p j= −

2 point DFT

2 (3) ( .707 .707)g j= − −

22 (0) 2 2p j= +

21(1) 1.414p j= −


FFT algorithms Inverse Radix-2 DIF-FFT algorithm

Inverse Radix-2-FFT algorithms



Inverse Radix-2-Decimation in frequency (DIF)

FFT algorithms



N point IDFT is defined as

x(n) =1

N

N−1∑k=0

X (k)W −knN n = 0, 1, . . . ,N − 1

WN = e−j2π/N and W −knN = e−j2πkn/N =

DFT X (k) is splitted into even and odd numbered values

F1(k) = X (2k) k = 0, 1, . . . ,N/2− 1

F2(k) = X (2k + 1) k = 0, 1, . . . ,N/2− 1

x(n) =1

N

[ ∑k=even

X (k)W −knN +

∑k=odd

X (k)W −knN

]

=1

N

N/2−1∑m=0

X (2m)W −2mnN +

N/2−1∑m=0

X (2m + 1)W−2(2m+1)N



F1(m) = X (2m) and F2(m) = X (2m + 1)

x(n) =1

N

N/2−1∑m=0

F1(m)(W 2N )−mn +

N/2−1∑m=0

F2(m)(W 2N )−mn(W −n

N )

W 2N = WN/2

x(n) =1

N

N/2−1∑m=0

F1(m)(WN/2)−mn + W −nN

N/2−1∑m=0

F2(m)(WN/2)−mn

x(n) =1

2

1

N/2

N/2−1∑m=0

F1(m)(WN/2)−mn + W −nN

1

N/2

N/2−1∑m=0

F2(m)(WN/2)−mn

Summations represents the N/2 IDFT of f1(n) and f2(n) for F1(m) and F2(m)respectively

x(n) =1

2

[f1(n) + W −n

N f2(n)]

n = 0, 1, . . . ,N − 1



f1(n) and f2(n) are periodic

f1

(n +

N

2

)= f1(n)

f2

(n +

N

2

)= f2(n)

Replace n by n+N/2 forx(n)

x

(n +

N

2

)=

1

2

[f1

(n +

N

2

)+ W

−(n+ N2 )

N f2

(n +

N

2

)]

W−(n+ N

2 )N = W −n

N W−N/2N

= W −nN e−j 2π

NN2

= W −nN e jπ

= = −W −nN

Hence x(

n + N2

)can be written as

x

(n +

N

2

)=

1

2

[f1(n)−W −n

N f2(n)]



x(n) =1

2

[f1(n) + W −n

N f2(n)]

n = 0, 1, . . .N

2− 1

x

(n +

N

2

)=

1

2

[f1(n)−W −n

N f2(n)]

n = 0, 1, . . .N

2− 1



Consider an 8 point Inverse DIF FFT. The sequence x(n) can be obtained from f1(n) andf1(n), where f1(n) and f2(n) are 4 point sequences.

38W −

28W −

18W −

08W −

-1

-1

-1

-1

N/2-point IDFT4-point IDFT

(0)x

(3)x

(2)x

(1)x

(4)x

(7)x

(6)x

(5)x

1(0)f

1(1)f

1(3)f

1(2)f

2 (0)f

2 (1)f

2 (3)f

2 (2)fN/2-point IDFT4-point IDFT

1(0) (0)F X=

1(3) (6)F X=

1(2) (4)F X=

1(1) (2)F X=

2 (0) (1)F X=

2 (3) (7)F X=

2 (2) (5)F X=

2 (1) (3)F X=

1( ) (2 )F k X k=

1( ) (2 )F k X k=

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

Figure 20: 8-point IDFT by combining 4-point IDFT of f1(n) and f2(n)



The decimation of X (k) is splitted into F1(k) and F2(k) are

F1(k) = X (2k) = [X (0),X (2),X (4),X (6)]

F2(k) = X (2k + 1) = [X (1),X (3),X (5),X (7)]

Next F1(k) and F2(k) are splitted into even and odd numbered samples

V11(k) = F1(2k) k = 0, 1, . . .N

4− 1

V12(k) = F1(2k + 1) k = 0, 1, . . .N

4− 1

V21(k) = F2(2k) k = 0, 1, . . .N

4− 1

V22(k) = F2(2k + 1) k = 0, 1, . . .N

4− 1



The sequences f1(n) and f1(n + N/4) are obtained from v11(n) and v12(n), where v11(n)and v12(n) are N/4 point IDFT of V11(k) and V12(k) respectively

f1(n) =1

2

[v11(n) + W −n

N/2v12(n)

]k = 0, 1, . . .

N

4− 1

f1

(n +

N

4

)=

1

2

[v11(n)−W −n

N/2v12(n)

]k = 0, 1, . . .

N

4− 1

Similarly f2(n) can be written as

f2(n) =1

2

[v21(n) + W −n

N/2v22(n)

]k = 0, 1, . . .

N

4− 1

f2

(n +

N

4

)=

1

2

[v21(n)−W −n

N/2v22(n)

]k = 0, 1, . . .

N

4− 1

V11(k) = F1(2k) = X (4k) = [X (0),X (4)] k = 0, 1

V12(k) = F1(2k + 1) = X (4k + 2) = [X (2),X (6)] k = 0, 1

V21(k) = F2(2k) = X (4k + 1) = [X (1),X (5)] k = 0, 1

V22(k) = F2(2k + 1) = X (4k + 3) = [X (3),X (7)] k = 0, 1



N/2-pointIDFT i.e.4-pointIDFT

1(0)F

1(2)F

1(1)F

1(3)F

1(0)f

1(3)f

1(2)f

1(1)f

Figure 21: 4-point IDFT

N/2-pointIDFT i.e.4-pointIDFT

2 (0)F

2 (2)F

2 (1)F

2 (3)F

2 (0)f

2 (3)f

2 (2)f

2 (1)f

Figure 22: 4-point IDFT




11(0)v

11(1)v

12 (0)v

12 (1)v

N/4-pointDFT i.e.2-point DFT 1

4W −

04W −

-1

-1

11 1(0) (0) (0)V F X= =11 1( ) (2 ) (4 )V k F k X k= =

11 1(1) (2) (4)V F X= =

12 1(0) (1) (2)V F X= =12 1( ) (2 2) (4 2)V k F k X k= + = +

12 1(1) (3) (6)V F X= =

1(1)f

1(0)f

1(2)f

1(3)f

1/ 2

1/ 2

1/ 2

1/ 2

Figure 23: 4-point sequence by combining two 2-point sequences v11(n) and v12(n)


21(0)v

21(1)v

22(0)v

22 (1)v

N/4-pointDFT i.e.2-point DFT 1

4W −

04W −

-1

-1

21 2(0) (0) (1)V F X= =21 2( ) (2 ) (4 1)V k F k X k= = +

21 2(1) (2) (5)V F X= =

22 2(0) (1) (3)V F X= =22 2( ) (2 1) (4 3)V k F k X k= + = +

22 2(1) (3) (7)V F X= =

2 (1)f

2 (0)f

2 (2)f

2 (3)f

1/ 2

1/ 2

1/ 2

1/ 2

Figure 24: 4-point sequence by combining two 2-point sequences v21(n) and v22(n)



Computation of 2-point IDFT

By definition N point IDFT is

x(n) =1

N

N−1∑k=0

X (k)W −knN n = 0, 1, . . . ,N − 1

2-point v11(n) is defined as

v11(n) =1

2

1∑k=0

V11(k)W −kn2 n = 0, 1

For n=0

v11(0) =1

2

1∑k=0

V11(k)W −02

=1

2

1∑k=0

[V11(0)W −0

2 + V11(1)W −02

]

=1

2

1∑k=0

[V11(0) + V11(1)]



For n=1

v11(1) =1

2

1∑k=0

V11(k)W −k2

=1

2

1∑k=0

[V11(0)W −0

2 + V11(1)W −12

]

=1

2

1∑k=0

[V11(0)− V11(1)]

v11(1) =1

2

1∑k=0

[V11(0)− V11(1)W −0

2

]The 2-point IDFTs are

v11(0) =1

2

1∑k=0

[V11(0) + V11(1)W −0

2

]

v11(1) =1

2

1∑k=0

[V11(0)− V11(1)W −0

2

]



02W −

11(0) (0)V X=

11(1) (4)V X=

11(0)v

11(1)v

1/ 2

1/ 2

1−Figure 25: Butterfly structure IDFT algorithm



11(0)v

0 02 8W W− −=

2-point IDFT

-10 0

4 8W W− −=

Combine 2-point IDFTs

1 24 8W W− −=

0 02 8W W− −=

-1 -1

-1

12 (1)v

12 (0)v

11(1)v

21(0)v

0 02 8W W− −=

-10 0

4 8W W− −=


1 24 8W W− −=

0 02 8W W− −=

-1 -1

-1

22 (1)v

22 (0)v

21(1)v

-1

-1

-1

-1

28W −

18W −

08W −

38W −

Combine Two-4 point IDFTs

Third Stage Decimation First Stage DecimationSecond Stage Decimation

DFT in bit reversalorder

x(n) Naturalorder order

(0)x

(1)x

(2)x

(3)x

(4)x

(5)x

(6)x

(7)x

1(0)f

1(2)f

1(3)f

1(1)f

2 (0)f

2 (2)f

2 (3)f

2 (1)f

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2(7)X

(4)X

(2)X

(6)X

(1)X

(0)X

(3)X

(5)X

Figure 26: Signal flow graph of 8-point Inverse Radix-2DIF-FFT algorithm



Inverse Decimation in Frequency (IDFT)-FFT

algorithm



First five points of the eight point DFT of a real valued sequence is given by X(0)=[0,2+j2,-j4,2-j2,0]. Determine the remaining points. Hence find the original sequence x(n) usingDecimation in frequency FFT algorithmSolution:X (0) = 0, X (1) = 2 + j2, X (2) = −j4, X (3) = 2− j2, X (4) = 0The remaining values will be determine using symmetry propertyX (N − k) = X∗(k)X (k) = X∗(N − k)X (5) = X∗(8− 5) = X∗(3) = 2 + j2X (6) = X∗(8− 2) = X∗(2) = −j4X (7) = X∗(8− 7) = X∗(1) = 2− j2

W −08

W −18 = 0.707 + j0.707

W −28 = j

W −38 = −0.707 + j0.707



v11(0) = X (0) + W −08 X (4)

= 0 + 1× 0 = 0

v11(1) = X (0) + W −08 X (4)

= 0− 1× 0 = 0

v12(0) = X (2) + W −08 X (6)

= −j4 + 1× (j4) = 0

v12(1) = X (2) + W −08 X (6)

= −j4− 1× (j4) = −j8

v21(0) = X (1) + W −08 X (5)

= 2 + j2 + 1× (2 + j2) = 4 + j4

v21(1) = X (1)−W −08 X (5)

= 2 + j2− 1× (2 + j2) = 0

v22(0) = X (3) + W −08 X (7)

= 2− j2 + 1× (2− j2) = 4− j4

v22(1) = X (3)−W −08 X (7)

= 2− j2− 1× (2− j2) = 0



f1(0) = v11(0) + W −08 v12(0)

= 0 + 1× 0 = 0

f1(1) = v11(1) + W −28 v12(1)

= 0 + j × (−j8) = 8

f1(2) = v11(0)−W −28 v12(0)

= 0− 1× (0) = 0

f1(3) = v11(1)−W −28 v12(1)

= 0− j × (−j8) = −8

f2(0) = v21(0) + W −08 v22(0)

= 4 + j4 + 1× (4− j4) = 8

f2(1) = v21(1) + W −28 v22(1)

= 0 + j × (0) = 0

f2(2) = v21(0)−W −028 v22(0)

= 4 + j4− 1× (4− j4) = j8

f2(3) = v21(1)−W −28 v22(1)

= 0− j × (0) = 0



x(0) = 1/8[f1(0) + W −08 f2(0)]

= 1/8[0 + 1× 8] = 1

x(1) = 1/8[f1(1) + W −18 f2(1)]

= 1/8[8 + (.707 + j .707)(0) = 1

x(2) = 1/8[f2(0) + W −28 f2(2)]

= 1/8[0 + j(j8)] = −1

x(3) = 1/8[f1(3) + W −38 f2(3)]

= 1/8[−8 + (−0.707 + j .707)(0) = −1

x(4) = 1/8[f1(0)−W −08 f2(0)]

= 1/8[0− 1× 8] = −1

x(5) = 1/8[f1(1)−W −18 f2(1)]

= 1/8[8− (.707 + j .707)(0) = 1

x(6) = 1/8[f1(2)−W −28 f2(2)]

= 1/8[0− j(j8)] = 1

x(7) = 1/8[f1(3)−W −38 f2(3)]

= 1/8[−8 + (−0.707 + j .707)(0) = −1


FFT algorithms Inverse Radix-2 DIT-FFT algorithm

Inverse Radix-2 Decimation in Time (DIT)-FFT

algorithm



N point IDFT is expressed as:

x(n) =1

N

N−1∑k=0

X (k)W −knN n = 0, 1, . . . ,N − 1

Split this summations into two summations with N/2 data points for each summation.

x(n) =1

N

N/2−1∑k=0

X (k)W −knN +

N−1∑k=N/2

X (k)W −knN

=

1

N

N/2−1∑k=0

X (k)W −knN +

N/2−1∑k=0

X (k + N/2)W−n(k+N/2)N

=

1

N

N/2−1∑k=0

X (k)W −knN + W

−nN/2N

N/2−1∑k=0

X (k + N/2)W −knN



W−nN/2N can be expressed as

W−nN/2N = e−j 2π

NnN2 = e jπn = [e jπ]n = (−1)n

x(n) =1

N

N/2−1∑n=0

X (k)W −knN + (−1)n

N/2−1∑k=0

X (k + N/2)W −knN

=1

N

N/2−1∑n=0

[X (k) + (−1)k X (k + N/2)

]W −kn

N

Split the x(n) into even and odd numbered samples.

x(2n) =1

N

N/2−1∑k=0

[X (k) + (−1)2nX (k + N/2)

]W −2kn

N n = 0, 1, . . .N/2− 1

x(2n + 1) =1

N

N/2−1∑k=0

[X (k) + (−1)2n+1X (k + N/2)

]W

−(2n+1)kN n = 0, 1, . . .N/2− 1



(−1)2n = 1 and W −2knN = (W 2

N )−kn = W −knN/2

Since W 2N = WN/2

Similarly (−1)2n+1 = −1 and W−k(2n+1)N = W −2kn−k

N = W −2knN/2

.W −kN Since W 2

N = WN/2

Substituting these values in the previous equations:

X (2n) =1

N

N/2−1∑k=0

[X (k) + X (k + N/2)] W −knN/2

n = 0, 1, . . .N/2− 1

X (2n + 1) =

N/2−1∑n=0

[X (k)− X (k + N/2)] W −2knN .W −k

N

=1

N

N/2−1∑n=0

{[X (k)− X (k + N/2)] W −kN }W −kn

N/2



Let us define N/2 point DFTs G1(k) and G2(k) as

G1(k) = [X (k) + X (k + N/2)] k = 0, 1, . . .N/2− 1

G2(k) = [X (k)− X (k + N/2)] W −kN k = 0, 1, . . .N/2− 1

x(2n) =1

N

N/2−1∑n=0

G1(k)W −knN/2

k = 0, 1, . . .N/2− 1

x(2n + 1) =1

N

N/2−1∑n=0

G2(k)W −knN/2

k = 0, 1, . . .N/2− 1

x(2n) =1

2

1

N/2

N/2−1∑n=0

G1(k)W −knN/2

k = 0, 1, . . .N/2− 1

x(2n + 1) =1

2

1

N/2

N/2−1∑n=0

G2(k)W −knN/2

k = 0, 1, . . .N/2− 1



To Demonstrate the DIF-FFT algorithm 8 point DFT is considered as an example. Thevalues of G1(k) and G2(k) is defined as

G1(k) = X (n) + X (k + N/2) k = 0, 1, . . .N/2− 1

G1(0) = X (0) + X (4)

G1(1) = X (1) + X (5)

G1(2) = X (2) + X (6)

G1(3) = X (3) + X (7)

G2(k) = [x(k)− X (k + 4)] W −k8 k = 0, 1, . . .N/2− 1

G2(0) = [X (0)− X (4)]W −08

G2(1) = [X (1)− X (5)]W −18

G2(2) = [X (2)− X (6)]W −28

G2(3) = [X (3)− X (7)]W −38



38W

4 point IDFTs

4 point IDFTs28W

18W

08W

-1

-1

-1

-1

(2 )x n

(2 1)x n+

(1)X

(2)X

(0)X

(3)X

(5)X

(6)X

(4)X

(7)X

1(0)G

1(1)G

1(3)G

1(2)G

2 (0)G

2 (1)G

2 (3)G

2 (2)G

(0)x

(6)x

(2)x

(4)x

(1)x

(7)x

(3)x

(5)x

Figure 27: Decimation of DFT-FFT algorithm for N=8



Let us define N/4 point sequences P11(k), P12(k) , P21(k),and P22(k) as

P11(k) = [G1(k) + G1(k + N/4)] k = 0, 1, . . .N/4− 1

P12(k) = [G1(k)− G1(k + N/4)] W −kN/2

k = 0, 1, . . .N/4− 1

P21(k) = [G2(k) + G2(k + N/4)] k = 0, 1, . . .N/4− 1

P22(k) = [G1(k)− G2(k + N/4)] W −kN/2

k = 0, 1, . . .N/4− 1

x(4n) =1

2

1

N/4

N/4−1∑n=0

P11(k)W −knN/4

n = 0, 1, . . .N/4− 1

x(4n + 2) =1

2

1

N/4

N/4−1∑n=0

P12(k)W −knN/4

n = 0, 1, . . .N/4− 1

x(4n + 1) =1

2

1

N/4

N/4−1∑n=0

P21(k)W −knN/4

n = 0, 1, . . .N/4− 1

x(4n + 3) =1

2

1

N/4

N/4−1∑n=0

P22(k)W −knN/4

n = 0, 1, . . .N/4− 1



Two 2-point sequences P11(k) and P12(k) are obtained from 4-point sequence G1(k).

P11(0) = G1(0) + G1(2)

P11(1) = G1(1) + G1(3)

P12(0) = [G1(0) + G1(2)]W −08 ∵ (W −0

4 = W −08 )

P12(1) = [G1(1)− G1(3)]W −28 ∵ (W −1

4 = W −28 )

Similarly Two 2-point sequences P21(k) and P22(k) are obtained from 4-point sequenceG2(k).

P21(0) = G2(0) + G2(2)

P21(1) = G2(1) + G2(3)

P22(0) = [G2(0)− G2(2)]W −08

P22(1) = [G2(1)− G2(3)]W −28



-1

-1

28W −

08W −

-1

-1

28W −

08W −

1(0)G

1(1)G

1(3)G

1(2)G

2 (0)G

2 (1)G

2 (3)G

2 (2)G

11(0)P

12 (1)P

12 (0)P

11(1)P

21(0)P

22 (1)P

22 (0)P

21(1)P




The 2-point IDFTs are computed directly and represented as

x(0) =1

8[P11(0) + P11(1)]

x(4) =1

8[[P11(0)− P11(1)]W −0

8

x(2) =1

8[P12(0) + P12(1)]

x(6) =1

8[P12(0)− P12(1)]W −0

8

x(1) =1

8[P21(0) + P21(1)]

x(5) =1

8[P21(0)− P21(1)]W −0

8

x(3) =1

8[P22(0) + P22(1)]

x(7) =1

8[P22(0)− P22(1)]W −0

8



(4)X

(2)X

(0)X

(6)X

-1

-1 08W

08W

11(0)p

12 (1)p

12 (0)p

11(1)p

(5)X

(3)X

(1)X

(7)X

-1

-1 08W

08W

21(0)p

22(1)p

22 (0)p

21(1)p

2 point DFT




38W −

28W −

18W −

08W −

-1

-1

-1

-1

-1

-1

-1

28W −

08W −

-1 08W −

08W −

-1

-1

-1

28W −

08W −

-1 08W −

08W −

sequence x(n) inbit reversal order

Input DFT

2 point IDFT

(1)X

(2)X

(0)X

(3)X

(5)X

(6)X

(4)X

(7)X

(0)x

(6)x

(2)x

(4)x

(1)x

(7)x

(3)x

(5)x

11(0)P

12 (1)P

12 (0)P

11(1)P

21(0)P

22 (1)P

22 (0)P

21(1)P

1(0)G

1(1)G

1(3)G

1(2)G

2 (0)G

2 (1)G

2 (3)G

2 (2)G

1/8

1/8

1/8

1/8

1/8

1/8

1/8




Computational Complexity Decimation in Frequency (DIF)-FFTalgorithm

A a b= +

rNW

-1

a

b ( ) rNB a b W= −

Figure 31: Butterfly structure decimation in frequency FFT algorithm







unit - 3: fast-fourier-transform (fft) algorithms[ · fast fourier transform algorithms...

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