Fourier Transformer

1

Fourier Transform, FT

Adawy

2

Transform domain representation of discrete time signals

Adawy

3

Representation of continuous systemsMost of continuous systems can be represented by:

Linear constant coefficients differential equation

Transfer function or system function

Impulse response

Frequency response

1- Linear constant coefficient differential equation

)(...)()(

01

1

1 tycdt

tydc

dt

tydc

N

N

NN

N

N

)(...)()(

01

1

1 txddt

txdM

dt

txdd

M

M

MM

M

M

y(t) is the system output, x(t) is the system input

Adawy

4

Take Laplace transform with all initial conditions zeros, then the last equation can be written as:

N

i

M

i

ii

ii sXSdsYSC

0 0

)()(

M

i

ii

N

i

ii

Sd

Sc

sX

sYsH

0

0

)(

)()(

System function or Transfer

function

Y(s)=H(s)X(s) If x(t)=δ(t), impulse function, X(s)=1

Y(s)=H(s), y(t)=L-1Y(s)=L-1H(s)=h(t)

Impulse response

Adawy

5

In general y(t)=L-1H(s)X(s)= h(t) * x(t)

Convolution operation

dthxdtxhty )()()()()(

A system is stable if all the poles of H(s) are in the left hand side of the S plane

Adawy

6

Difference equations

General form of a system difference equation is:

N

k

M

rrk rnxbknya

0 0

][][

][][][0 01 0

rnxa

bkny

a

any

M

r

rN

k

k

Present output Present and previous M inputs

Previous N output

A sequential method can be used to get a closed form solution for y[n]

Adawy

7

Example

y[n]=ay[n-1]+x[n], y[n]=0 for all n<0, x[n]=δ[n]

y[0]=ay[-1]+x[0]=1

y[1]=ay[0]+x[1]=a

y[2]=ay[1]+x[2]=a2

…………………

y[n]=an

The Z transform can be used to solve difference equation

Adawy

8

What Are Sinusoids?

Cosine wave Sine wave

Low frequency Middle frequency High frequency

Adawy

9

Fourier Transform

عاما عندما كان أحد أعضاء الحملة 30لقد كان العالم فورير عمره وعين حاكما على مصر العليا ، 1798الفرنسية على مصر عام

وتقدم بعدة أبحاث علمية للمعهد العلمى المصرى الذى أنشأه نابليون فى مصر . ثم عاد بعد ذلك إلى فرنسا حيث قدم أفكاره

. لقد 1807لحل مشاكل االنتقال الحرارى للمعهد الفرنسى عام تركزت أفكاره على أن اإلشارة الدورية يمكن تمثيلها بمجموعة محدودة من اإلشارات الجيبية المتوافقة . كما أوضح أيضا أن

اإلشارة الغير دورية يمكن تمثيلها بالمجموع التكاملى من اإلشارات الجيبية الغير متوافقة . هاتين الفكرتين هما األساس لما عرف فيما

Fourier ومحول فورير Fourier seriesبعد ويعرف اآلن بتتابع فورير Transform اللذين كان لهما أكبر األثر على كل فروع الهندسة

وبالذات هندسة االتصاالت واإللكترونيات ومعالجة اإلشارات .

لقد تم استخدام كل من تتابع فورير ومحول فورير لما يقرب من عام لتمثيل الظواهر االنسيابية فى الزمن االنسيابى . مع ظهور 200

األنظمة الرقمية وأزمنة الزمن المقطع ظهرت الحاجة إلى محول .Discrete Fourier Transform, DFTفورير المقطع

Adawy

10

ما هى الفائدة من تحويل أو وضع إشارة مركبة فى صورة مجموعة من اإلشارات الجيبية ؟

ولماذا اإلشارة الجيبية بالذات ولم تكن المربعة مثال ؟

فى الكثير من األحيان يصعب التعامل مع هذه اإلشارات المركبة ، فال يمكن مثال حساب استجابة )خرج( نظام لمثل هذه اإلشارات .

عند وضع اإلشارة فى صورة مجموعة من اإلشارات البسيطة يسهل حساب استجابة النظام لكل واحدة من هذه اإلشارات على

حده ثم تجميع هذه االستجابات فنحصل على استجابة النظام .Superpositionلإلشارة المركبة . هذا ما يسمى بنظرية التجميع

أما لماذا اإلشارة الجيبية بالذات فألن اإلشارة الجيبية لها ميزة استجابة أى نظام ليست موجودة فى أى إشارة أخرى وهى أن

خطى لإلشارة الجيبية هى أشارة جيبية أيضا تختلف فقط فى المقدار والزاوية ولها نفس التردد .

Adawy

11

This is our original One sine Two sines

Four sines Seven sines Fourteen sines

Adawy

Getting a square wave from a series of sinusoids

12

F(t)=…+(2/pi*45)*cos(45*pi*t)

Getting a series of pulses from a series of sinusoids

Adawy

13

Classification of signals

Signal

ContinuousDiscrete

Periodic

Aperiodic

Periodic

Aperiodic

Adawy

14

Fourier Transform for signals

Continuous periodic

Fourier series

Continuous aperiodic

Discrete aperiodic

Fourier Transform

Discrete Time Fourier Transform

Adawy

15

Continuous Fourier Series

k

otjkkeFtf )(

o

To /2

Any periodic continuous function f(t), with a periodic time T, can be represented as a sum of exponential (sinusoids)

functions of T as follows:

For all values of t

Fk is called the series coefficients, is the fundamental frequency, and it is given by:

Adawy

16

T otjk

k dtetfT

F0

)(1

1

)()(()(k

kk otkSinbotkCosaaotf

T

dttfT

ao0

)(1

T

k dtotkCostfT

a0

)()(2

T

k dtotkSintfT

b0

)()(2

Fk are given by:

These are the coefficients in the exponential form. They can be represented in a sinusoidal form as follows:

Adawy

17

Fourier Transform for continuous aperiodic functions

Fourier Transform, FT

Inverse Fourier Transform, IFT

Adawy

18

Any signal can be represented in one of these two domains:

Time domain

Frequency domain

Oscilloscope

Time

fAmplitude

Frequency

Spectrum analyzer

FT

IFT

fAmplitude

Adawy

19

The Discrete Time Fourier Transform, DTFT

It is the Fourier transform of aperiodic discrete time sequences.

n

jwnjw enxeX ][)(

)()()( jwim

jwre

jw ejXeXeX )()()( wjjwjw eeXeX

MagnitudeAngle or phase

This is called frequency spectrum, amplitude spectrum, and phase spectrum Adawy

20

We can see that:

X(ejw) is a continuous function of w

X(ejw) is periodic of w

e jӨ(w) =e j(Ө(w)+2pi k) =e jӨ(w)

Example X[n]=(0.5)n u[n]

0

)5.0(][)5.0()(n

jwnn

n

jwnnjw eenueX

jwn

njw

ee

5.01

1)5.0(

0

Adawy

21

jwjw

eeX

5.01

1)(

22

The inverse Discrete Time Fourier Transform, IDTFT

dweeXnx jwnjw)(

2

1][

Convergence of the DTFT

n

jwnjw enxeX ][)(

If the sequence x[n] is absolutely summable, then the DTFT converges to a continuous function of w.

n

nx ][ )( jweX

Adawy

23

DTFT properties

1- Linearity

A g[n] + B h[n] A G(ejw) + B H(ejw)DTFT

2- Time shift

g[n-n0] e-jwn0 G(ejw)DTFT

3- Frequency shifting

ejwn0 g[n-n0] G(ej(w-w0))DTFT

4- Differentiation in frequency

n g[n] jdG(ejw)/dwDTFT

Adawy

24

5- Convolution

g[n] * h[n] G(ejw) H(ejw)DTFT

6- Modulation

g[n] h[n]

deHeG wjj )()(2

1 )(DTFT

The DTFT of the convolution of two sequences equal to the product of the DTFT of both sequences. In some applications (when it is difficult to calculate or even draw

the convolution of two sequences) its convenient to calculate the convolution of the two sequences by

calculating the multiplication of the DTFT of the two sequences and find the IDTFT.

Adawy

25

Discrete Fourier Transform, DFT

It is applied to the aperiodic discrete time signals with definite lengths. The produced DFT is itself discrete.

Because it is discrete, it is the one that is used to represent discrete signals in the frequency domain. There are many special algorithms to speed up the calculation of

the DFT.It can be viewed as a special case of the DTFT. For a finite

length sequence x[n], 0<=n<=N-1, it is found that N samples of the DTFT X(ejw) are sufficient to determine x[n]. So it is obtained by uniformly sampling X(ejw) on the w axis between 0<=w<=2pi at wk=2pi k/N, k=0, 1, 2, …, N

1

0

/2

/2][)(][

N

n

Nknj

Nkw

jw enxeXkX

K=0,1,…N-1Adawy

26

X[k] is called the DFT, it is a finite length sequence of length N.

NjN eW /2

1

0

][][N

n

knNWnxkX

Inverse DFT

1

0

][1

][N

k

knNWkXN

nx n=0,1,2,…..N-1

As it will be seen computation of DFT or IDFT requires N2 complex multiplications, and N(N-1)

complex additions. Any algorithm used to calculate X[k] can be used to calculate x[n].

Adawy

27

The DFT can be put in a matrix form as follows:

X = D xX[0]

X[1]

.

.

.

X[N-1]

1 1 1 . . . . 1

1NW

2NW

)1(2 NNW2

NW4NW

1NNW

1NNW

)1(2 NNW

)1)(1( NNNW

………

………

………

.

.

1

1

1

=

x[0]

x[1]

.

.

.

x[N-1]

The IDFT can be put in a matrix form as follows:

x=D-1X D-1=(1/N)*ND

Adawy

1

0

/2][][N

n

NknjenxkX

1

0

][N

n

knNWnx

28

Properties of DFT

1- Because X(ejw) is periodical then X[k], and x[n] are periodical. So, both the DFT, and IDFT are both a finite length sequence that represent one period of a periodic

sequence.

2- The only difference between the calculation of the DFT and the IDFT is the scaling factor, (1/N) and the sign change in the exponent term. So, one algorithm can be modified to

calculate any of them.

3- Linearity If x1n] X1[k], and x2[n] X2[k]

Then Ax1[n]+Bx2[n] AX1[k] + BX2[k]

4- Time shift x[n-no] knoN

Nknoj WkXekX ][][ /2

Adawy

29

Circular shift of a sequenceImagine the sequence x[n] is displayed on the

circumference of a cylinder at N equally spaced points, then the circular shift operation can be viewed as rotating

the cylinder clockwise or anticlockwise.

1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5X[n] X[n-1] X[n-2] X[n-3]

Adawy

5- Circular convolution

1

0

][2][1N

m

nmxnx X1[k]X2[k]

6- Modulation

1

0

][2][11 N

m

mkXmXNx1[n]x2[n]

30

Circular convolution

1

0

][][][N

kL knhkgny

linear convolution

n=0,1,2,3,….N+M-1

1

0

][][][N

kNC knhkgny

Circular convolution

g[n]=[1 2 0 1], h[n]=[2 2 1 1]

14

04][][][

kC knhkgny 0<=n=<3

yc[0]=g[0]h[0]+g[1]h[-1]+g[2]h[-2]+g[3]h[-3]=1x2+2x1+0x1+1x2=6

g[k]=[1 2 0 1]

h[-k]=[2 1 1 2]]

Adawy

n=0,1,2,3,….N-1

31

yc[1]=g[0]h[1]+g[1]h[0]+g[2]h[-1]+g[3]h[-2]=1x2+2x2+0x1+1x1=7g[k]=[1 2 0 1]

h[1-k]=[2 2 1 1]

yc[2]=g[0]h[2]+g[1]h[1]+g[2]h[0]+g[3]h[-1]=1x1+2x2+0x2+1x1=6

g[k]=[1 2 0 1]

h[2-k]=[1 2 2 1]

yc[2]=g[0]h[3]+g[1]h[2]+g[2]h[1]+g[3]h[0]=1x1+2x1+0x2+1x2=5g[k]=[1 2 0 1]

h[3-k]=[1 1 2 2]

yc[n]=[6 7 6 5]

Circular convolution of two equal length sequences gives a sequence with the same length.

Adawy

32

Calculation of the DFT

We defined the N point DFT of an N point sequence x[n] as the N samples of the Fourier transform X(ejw) of that

sequence evaluated uniformly on the w axis at w=2πk/N, where 0≤k≤N-1:

1

0

1

0

/2/2 ][][|)(][

N

n

N

n

knN

NknjNkw

jw WnxenxeXkX

NjN eW /2 ,K=0,1,2,…..N-1

1

0

1

0

/2 ][1

][1

][N

k

N

k

knN

Nknj WkXN

ekXN

nx

DFT

IDFT

Adawy

33

)Im()Re(])[Im(])[Re(][1

0

knN

knN

N

n

WjWnxjnxkX

1

0

1

0

)Im(]).[Im()Re(]).[Re(N

n

N

n

knN

knN WnxWnx

1

0

1

0

)Re(]).[Im()Im(]).[Re(N

n

N

n

knN

knN WnxWnxj

Adawy

34

1

0

1

0

)Im(]).[Im()Re(]).[Re(][N

n

N

n

knN

knN WnxWnxkX

1

0

1

0

)Re(]).[Im()Im(]).[Re(N

n

N

n

knN

knN WnxWnxj

N real multiplications N real multiplications

N real multiplications N real multiplications

Each term of X[k] requires 4N real multiplications. As X[k] consists of N terms (k varies from 0 to N-1), then to evaluate

X[k] for all k requires 4NXN=4N2 real multiplications.

By the same way direct evaluation of X[k] requires NX(4N-2) real additions.

Simply, direct evaluation of X[k] requires N2 complex multiplications, and N(N-1) complex additions.

Adawy

35

1

0

/2][][N

n

NknjenxkX

The good news is that, many of these multiplications turns out to be redundant, because the exponential term is

periodic and will give no more than N distinct values. So, the same products are calculated many times as k and n varies

from 0 to N-1.Efficient FFT algorithms try to reduce these redundancy.

Another problem with direct implementation of the DFT is the calculation of the exponential term itself directly ( or as

sine and cosine terms). This is a very time consuming problem and an efficient FFT algorithm must reduce the use

of such functions.

Adawy

36

1

0

1

0

/2 ][][][N

n

knN

N

n

Nknj WnxenxkX

n

k

01234567

011111111

11(1-j)/√2-j-(1+j)/ √2-1-(1-j)/√2J(1+j)/ √2

21-j-1j1-j-1j

31-(1+j)/ √2J(1-j)/√2-1(1+j)/ √2-j-(1-j)/√2

41-11-11-11-1

51-(1-j)/√2-j(1+j)/ √2-1(1-j)/√2J-(1+j)/ √2

61j-1-j1j-1-j

71(1+j)/ √2j-(1-j)/√2-1-(1+j)/√2-j(1-j)/√2

For k varies from 0 to 7, and n varies from 0 to 7 we demonstrate the redundancy of the exponential term exp(-j2πkn/N)

Adawy

37

Fast Fourier Transform, FFT

Many methods for reducing the number of multiplications have been investigated since 50 years ago.

Multiplication is more critical than addition because it consumes more time.

The most widely used methods of FFT are based on decomposing or breaking the transform into smaller

transforms, and combining them to give the total transform. There are two methods to do this decomposition or

decimation:

Decimation In Time FFT

Decimation In Frequency FFT

Adawy

38

Decimation In Time FFT

The number of points in the input sequence are assumed to be power of 2; N=2s.

In this approach we break the N, x[n] points into two groups, each of them is N/2 points. The first group is the

even indexed points x[0], x[2], x[4] , … x[N-2]. The second group is the odd indexed points x[1], x[3], …, x[N-

1]. Find the DFT of each group.

Each one of the previous groups is further decimated into two N/4 points by the same way and find the DFT of

each decimated group

This process continues until you end up with 2 points in each group and find the DFT for those two points

Adawy

39

1

0

][][N

n

knNWnxkX

Break the sequence x[n] into even and odd indexed points

1

,1

1

,0

][][][N

oddn

knN

N

evenn

knN WnxWnxkX

Put n=2r in the first sum, n=2r+1 in the second sum, to get even and odd indexed points

12/

0

)12(12/

0

2 ]12[]2[][N

r

krN

N

r

rkN WrxWrxkX

rkN

rkNj

rkNjrk

Nrk

N WeeWW 2/2/

22)/2(22

Adawy

40

12/

02/

12/

02/ ]12[]2[][

N

r

rkN

kN

N

r

rkN WrxWWrxkX

N/2 point DFT of the even

indexed pints.

N/2 point DFT of the odd

indexed pints.

][][][ kHWkGkX kN k=0, 1, 2, .., N-1

This represents the first step in this algorithm. Breaking the sequence into two N/2 sequences and find the DFT

of each of them. The total number of complex multiplication P=(N/2)2+(N/2)2+N=N2/2 +N

In the last equation for k=N/2 to N, both G[k], and H[k] are periodic. So, G[N/2]=G[0], G[N/2+1]=G[1], and so on.

Adawy

41

][][][ kHWkGkX kN k=0, 1, 2, .., N-1

N/2 point DFT

N/2 point DFT

x[0]

x[2]

x[4]

x[N-2]

x[1]

x[3]

x[5]

x[N-1]

G[0]

G[N/2-1]

G[2]

G[1]

H[0]

H[N/2-1]

H[2]

H[1]

0NW1NW2NW

12/ NNW

2/NNW

12/ NNW

22/ NNW

1NNW

X[0]

X[1]

X[2]

X[N/2-1]

X[N/2]

X[N/2+1]

X[N/2+2]

X[N-1]

N point combining algebra

Adawy

42

Each one of the N/2 point sequences can be decimated further into two sequences of lengths N/4

N/4 point DFT

x[0]

x[4]

x[N-4]

x[2]

x[1]

x[5]

x[N-1]

0NW1NW2NW

12/ NNW

2/NNW

12/ NNW

22/ NNW

1NNW

X[0]

X[1]

X[2]

X[N/2-1]

X[N/2]

X[N/2+1]

X[N/2+2]

X[N-1]

N/4 point DFT

N/4 point DFT

N/4 point DFT

x[6]

x[N-3]

x[N-2]

x[3]

x[7]

][][][][][ 2/2/ kHoWkHekGoWkGekX kN

kN

P=4(N/4)2+2(N/2)+N=N2/4+2N

N/2 point combine algebra

Adawy

43

X[0]

X[1]

X[2]

X[N/2-1]

X[N/2]

X[N/2+1]

X[N-1]

N point combining

algebra

N/2 point combining

algebra

N/2 point combining

algebra

N/4 point combining

algebra

N/4 point combining

algebra

N/4 point combining

algebra

N/4 point combining

algebra

2 point

DFT

2 point

DFT

2 point

DFT

2 point

DFT

Final concept for decomposition in the decimation in time FFT

Stage 1 Stage log2(N)=v, N=2v

The combining algebra at each stage takes N complex multiplication, so total multiplications P= N.log2(N). Adawy

44

We reduced the number of complex multiplication from N2 to N.log2(N) which is a great time saving.

Example Flow graph for an 8 point FFT

x[0]

x[4]

x[2]

x[6]

x[1]

x[5]

x[3]

x[7]

X[0]

X[1]

X[2]

X[3]

X[4]

X[5]

X[6]

X[7]

02W

12W

12W

12W

12W

02W

02W

02W

04W

14W

24W

34W

04W

14W

24W

34W

08W

18W

28W

38W

48W

58W

68W

78W

Adawy

45

An interesting feature in all previous butterflies

The horizontal scale factor can be rewritten as the negative of the diagonal scale factor

02W

12W

102 W

12/1212 jj eeW

38W

78W

18

18

18

48

148

38 .1. WWWWWW

18

18

18

88

188

78 .1. WWWWWW

02W -1

38W -1

Redraw previous

figure

Adawy

46

x[0]

x[4]

x[2]

x[6]

x[1]

x[5]

x[3]

x[7]

X[0]

X[1]

X[2]

X[3]

X[4]

X[5]

X[6]

X[7]

02W

02W

02W

02W

04W

14W

04W

14W

08W

18W

28W

38W

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

-1

This gives a further reduction in the number of complex multiplication to (N/2).log2(N)

Adawy

47

Number of stages

v

Number of points

N

N2(N/2)log2(N)Ratio of N2 to (N/2)log2(N)

241644

3864125.333

416256328

53210248012.8

664409619221.33

71281638444836.57

825665536102464

95122621442304113.77

10102410485765120204.8

Comparison for number of complex multiplications obtained by direct method and FFT method

Adawy

48

Shuffling the input data

Natural order of

input data

3 bit codeBit reversal

code

Shuffled data

x[0]000000X[0]

x[1]001100X[4]

x[2]010010X[2]

x[3]011110X[6]

x[4]100001X[1]

x[5]101101X[5]

x[6]110011X[3]

x[7]111111X[7]

Adawy

49

FFT using MATLAB

fft(x) fft(x,N)

ifft(x) ifft(x,N)

The function fft(x) computes the DFT of a vector x with the same length as x. The function fft(x,N) computes an N points DFT of the vector x. If N<x then x is padded with zeros, if N<x

then x is truncated to the first N points. The functions ifft() works the same way.

MATLAB uses the decimation in time radix 2 algorithm in calculating the DFT if the input sequence is a power of 2. If x

is not power of 2, it uses mixed algorithms with the radix 2 algorithm, but it takes more time in this case.

Adawy

50

Example

Calculation of the DFT of a sinusoidal sequence with certain frequency.

% calculation of the fftfreq=10; % frequency of the input sinusoidfs=64; %sampling frequencyN=32; %length of the DFTk=0:N-1;f=sin(2*pi*freq*k/64);subplot(2,1,1);stem(k,f);grid;ylabel('X[n]');F=fft(f);subplot(2,1,2);stem(k,abs(F));grid;xlabel('frequency index k');ylabel('|x[k]|');

Sampling frequency much larger than 20Hz the Nyquest rate for the 10Hz freq. There will be

no aliasing

Adawy

51

Input freq. F=10 Hz

Input Ω=2pi10 rad/sec, Analog frequency

Digital frequency w=ΩTs=2pi10/64=2pik/32

From which k=5

So, the impulse at f=10Hz in the analog frequency will appear at k=5 in the digital freq. domain

Adawy

52

As the input sequence x[n] is pure sinusoid, its Fourier transform should be two impulses at +10Hz, and -10Hz,

and zero elsewhere. As we saw in the last figure the 10Hz appears as X[5] in the DFT frequency domain at location

k=5. The impulse at -10Hz appears at X[27] at location 32-5=27. So, the DFT is correctly calculated.

Remember that the digital spectrum is a repeated versions of the analog spectrum around half the sampling frequency fs. So the first half of the DFT samples for k=0 to k=N/2-1 corresponds to the positive analog frequency axis f=0 to

f=fs/2. While the second half for k= N/2 to k=N-1 corresponds to f=-fs/2 t0 f=0.

f10-10

Analog spectrum

fs/2fs/2 k0 N-1

Adawy

53

Previous example with N=64Input freq. F=10 Hz



From which k=10

So, the impulse at f=10Hz in the analog frequency will appear at k=10 in the digital freq. domain

Adawy

54

Same previous example with

f=11Hz, N=30, fs=64Hz

Input freq. F=11 Hz



From which k=11/2

So, the impulse at f=11Hz in the analog frequency will appear at k=5 and 6 in the digital freq. domain Adawy

55

%demonistration of the IFFTg=[1 2 0 1];ga=[g zeros(1,length(h)-1)]subplot(3,1,1);stem(ga);ylabel('First sequence');h=[2 2 1 1];ha=[h zeros(1,length(g)-1)]subplot(3,1,2);stem(ha);ylabel('Second sequence');G=fft(ga);H=fft(ha);Y=G.*H;y=ifft(Y);subplot(3,1,3);stem(real(y));ylabel('Result sequence');disp('New sequence');disp(real(y));

Using fft to calculate convolution of two sequences

New sequence 2.0000 6.0000 5.0000 5.0000 4.0000 1.0000 1.0000

Adawy

Documents

Fourier Transformer