Outcome Based Education 1 Dr.P.Meena,Assoc.Prof., EEE Focus Learning, not teaching Students, not faculty Outcomes, not inputs or capacity INDIA HAS BECOME.

Outcome Based Education

1

Dr.P.Meena,Assoc.Prof., EEE

FocusLearning, not teachingStudents, not facultyOutcomes, not inputs or capacity

INDIA HAS BECOME A PERMANENT MEMBER OF

THE WASHINGTON ACCORD

2

Academic abilities

Course Lectures/Demonstrations/

Videos/Animations /power point presentations/hand outs

Problem solving • Teacher led• Students in pairs/share

Industry Visits

Open ended experiments

Components that contribute to Academic Abilities

Components that contribute to Transferable Skills

3

Transferable skills

Student Activities

Project work/

Open ended experiments

The components of course delivery that contribute to the defined attributes of the course .

Attributes

Technical Symposia

Community

connect

4Dr.P.Meena,Assoc.Prof.,EEE

55

Digital Signal ProcessingIntroduction

Inception:1975 with the development of Digital Hardware such as digital hardware.

Personal computer revolution in 1980s and 1990s caused DSP explosion with new applications.

Dr. P.Meena, Assoc.Prof(EE) BMSCE

66

Advantages of DSP TechnologyHigh reliabilityReproducibility

Flexibility & ProgrammabilityAbsence of Component Drift problem

Compressed storage facility (especially in the case of speech signals which has a lot

of redundancy).DSP hardware allows for programmable

operations.Signal Processing functions to be performed by hardware can be easily modified through

soft ware(efficient algorithms)Dr. P.Meena, Assoc.Prof(EE) BMSCE

77

Advantages of DSP TechnologyHigh reliabilityReproducibility

Flexibility & ProgrammabilityAbsence of Component Drift problem

Compressed storage facility (especially in the case of speech signals which has a lot

of redundancy).DSP hardware allows for programmable

operations.Signal Processing functions to be performed by hardware can be easily modified through

soft ware(efficient algorithms)Dr. P.Meena, Assoc.Prof(EE) BMSCE

Dr. P.Meena, Assoc.Prof(EE) BMSCE 8


Digital Signal Processing with overlapping borders

1010

A TYPICAL DIGITAL SIGNAL PROCESSING SYSTEM

)(tx)(txa )(nx )(ns )(ts D/A

CONVERTER

Analog prefilter orAntialiasingfilter

A/DCONVERTER

Dig.SignalProcessor

Reconstructionfilter same as the pre filter

Low pass filtered signal

Discrete time signal

Discrete time signal

Samplingfrequency

dB dB



CO1: Ability to apply the knowledge of mathematics, science and fundamentals of signals and systems to ascertain the behavior of complex engineering systems.

CO2:Ability to Identify techniques, formulate representations and analyze responses of digital systems.

CO3:Ability to Design digital system components and test their application using modern engineering tools, as solutions to engineering problems.

Course Outcomes


Course Contents

• Different operations on a signal in the digital domain

• Different forms of realizations of a Digital System.

• Design Procedures for Digital Filters

1313

Outcomes of this Course: By The End Of The Course ,

• Distinguish The Digital and Analog Domains.

• Analyse Signals, and reconstruct.

• Develop Block Diagrams For Different System Representations,.

• Design Analog And Digital Filters.

• Ready to Take up Specialized Courses in Audio, speech, image and Real-time Signal Processing, Further On



Course Outline

Course Delivery:Lecture,hand outs,videos,animations,discussions,activities

Course Assessment:

Marks: Tests: 20 (T1 & T2)Quiz ; 05Tutorials: 10Lab: 15

1515

Review of Signals

& Systems


1616

Signals

Audio Video (Represented as a function of 3 variables.) Speech-

Continuous-represented as a function of a single (time) variable). Discrete-as a one dimensional sequence which is a function of a discrete variable.

Image:Represented as a function of two spatial variables

Electrical - Dr. P.Meena, Assoc.Prof(EE) BMSCE




2020

Relation between analog frequency and digital frequency

s

2 unit is radians per second.

y=a sin , a signal in the continuous time domain.

t=n*

z=a sin( )

sin( )

= is the digital frequency in radians/sample

Therefore, given a 2 , get

Ts

s

f

t is

n

z a n

where

f to

T

T

s

2 or (2 * )T

s

ff

f



f in Hertz(analog)

Ω in radians/sampleDigital

0

Fs/2-Fs/4-Fs/2-Fs FsFs/4 3/2Fs-3/2Fs

f=0,Ω=0

Ω=∏/2,f=Fs/4

Ω=∏,f=Fs/2

Ω= -∏/2,f=-Fs/4

Ω=2∏Ω=∏Ω=0Ω=∏/2Ω=-∏/2

Ω=-2∏ Ω=-∏ Ω=3∏Ω=-3∏

Nyquist interval

Ω=-∏,f=-Fs/2

Sl.No.

Frequencyin HertzOf the signal

Sampling FrequencyFs in Hertz

Ω in radians/cycle

1. f=0 Fs 0

2. f=Fs/4 Fs ∏/2

3 f=-Fs/4 Fs -∏/2

4. f=Fs/2 Fs ∏

5. f= -Fs/2 Fs -∏

Diagrammatic Representation of relation between analog frequency and digital frequency:+ve angle counter clock wise


Sampling of continuous time signals

The Fourier transform pair for continuous-time signals is defined by

dj2

1t

dttj

eXx

exX

tj

aa

tj

aa

dj2

1t

dttj

eXx

exX

tj

aa

tj

aa

If txa is sampled uniformly at times T seconds apart

from to a discrete- time signal x[n] is obtained. )t(xa]n[x

nTt


The Fourier transform of the resulting sequence x[n] can be shown to be

r 2T

1j

T

1X

ra

j

Xe

Thus ej

X

is the sum of an infinite number of amplitude-scaled, frequency-scaled, and translated versions of jXa

The Fourier transform of the continuous-time signal

txa

is illustrated in the following figure..

Tlogana

T



From the figure it is easy to see that the triangles of ej

X

Will not overlap if T0 <π

This inequality can be re arranged to give,

0

T

1

If we let,Ω0 equal 2πf0 where f0 is in Hertz, the above inequality becomes,

f 0

2T

1

Therefore, if the sampling rate I/T is greater than 2 f0 no overlap occurs. If there is no overlap, the spectrum

Can be found and by the inverse transform can be reconstructed. If however,

jXa

txa

T0 >π

The triangles will overlap and the spectrum of the continuous signal cannot be reconstructed


The Sampling Theorem A signal xa(t) can be

reconstructed from its sample values xa(nT) if the sampling rate 1/T is greater than twice the highest frequency (f0 in Hertz ) present in xa(t).• The sampling rate 2f0 for

an analog band limited signal is referred to as the Nyquist Rate.


Effect of Variation of Sampling frequency on the sampling of a sine wave of frequency 50Hz.


2929

Periodic Signals


,2 N

M

What is the equation to this sequence ???

3030

APeriodic Signals

Extract from cnx.orgDr. P.Meena, Assoc.Prof(EE) BMSCE


n

nn

nn

nn

Consider

xx

4.0cos

24.0cos

4.2cos

4.0cos

,

2

1

The two signals shown though of different digital frequencies have the same sampled sequence .

n

nn

nn

nn

Consider

xx

4.0cos

24.0cos

6.1cos

4.0cos

,

2

1


Therefore we find that a sampled sine wave of frequency f is indistinguishable from a sine wave of frequency fs-f, fs+f,2fs-f,2fs+f….Or in general kfs ±f for any integer k.•These set of frequencies that are indistinguishable from one another are called aliases and the phenomenon is called “Aliasing”


enj ee

njnjH

Linear Shift Invariant System

enj

H

Is the frequency response of the system is recognized as the Fourier transform of the system’s impulse response

n

nj

n

n2jnj

n

n2j)2(j

n

njj

ee

ee

ee

)n(h)n(h

)n(hH

)n(hH


Frequency spectrum of a 100Hz sine wave sampled at 500Hz

Frequency spectrum of a 100Hz sine wave sampled at 80Hz

Under Sampling


Alias during under sampling

3636

0 5 10 15-1

0

1

n

x1[n

]

Plot of Cos(0.4*pi*n)

0 5 10 15-1

0

1

n

x1(t

)

Cosine wave of frequency 100Hz sampled at 500Hz

0 5 10 15-1

0

1

n

x2[n

]

Cos(2.4*pi*n)

0 5 10 15-1

0

1

nx2(t

)

Cosine wave of frequency 600Hz sampled at 500Hz

Effect of aliasing

Effect of Aliasing


3737

Real &Imag parts of j te


3838

Representation of a Complex Exponential Function

Extract from cnx.orgDr. P.Meena, Assoc.Prof(EE) BMSCE

3939

Two dimensional plot of j te


4040

Frequency content of a signal

Continuous time :Periodic - Fourier seriesNon-periodic- Fourier transform

Discrete time:Discrete Time Fourier Transform-DTFTDiscrete Fourier Transform -DFT



freq.dig,)n(xXn

njj

ee

THE DISCRETE TIME FOURIER TRANSFORM

The discrete time Fourier transform of a sequence:

The discrete time Fourier transform of a system:

n

nj

n

n2jnj

n

n2j)2(j

n

njj

ee

ee

ee

)n(h)n(h

)n(hH

)n(hH

It seen that the discrete time Fourier transform of a system is 2π periodic and continuous in Ω


Inverse Discrete Time Fourier Transform (iDTFT)

deeX2

1nx njj

43

DTFT of a Sinusoidal signal of analog frequency 50Hz


e5e4e32e.1enxeX

5,4.3,2,1)n(x If.2

.1e]n[e:DTFT Its],n[]n[x If.1

3j-2j-j-jnjj

n

nj-j

DTFT of a Discrete Impulse


Discrete Unit Step Function

Magnitude of the DTFT


Disadvantage of DTFT


1.The Transfer function is a function of the continuous variable Ω.2. This needs computation of infinite sum at uncountable infinite frequencies.3.Hence the above transform is not numerically computable. 4.The transform is defined for aperiodic sequences.

A repeated sequence of finite length N


nx

x

N

r

N) ulomod n(x

rNnx)n(

x[n]=0 for n<0 and >N;≠0 otherwise


Discrete Fourier Transform, DFT

10,1

is, IDFT

.2

,

10,

frequency digital is

0

0

1

0

0

1

0

NkkXN

nx

THEN

where

NknxkX

e

e

njkN

k

njkN

n


210.1

2x1x0x0X

3

2.

N

2

1] 0 1[)n(x.;N

2, where

1Nk0,nxkX

eee

e

2.3

20j1.

3

20j0.

3

2jk

0

0

njk1N

0n

0

51P.Meena,Asst.Prof(EE)BMSCE 51

Discrete Fourier Transform

1

0

2

2( ) ( ) .... 0,1,2,.... 1

.

nk

N

n

jN

N

jX k x n k NNe

W e

1

0

2 /

( )N kn

Nn

j N

N

X k x n

whereW

W e


5252

1

0

1

0

2( ) ( ) .... 0,1,2,.... 1.

( ) ( ) .... 0,1,2,.... 1.

nk

N

n

nkN

n

jX k x n k NN

X k x n k NN

e

W

2

.jN

NW e

is known as twiddle factor.

21

1

N

N

NN

jNN eW

W


N roots of Unity.

5353

The Locus of the Twiddle Factor for N roots traces N points on a unit circle.


5454

2.Find the four point DFT of the sequence :

x[n]=(1,0.5,0.25,0.125)

2

244

= 12 2

j

j

Cos jSin j

eW

e


5555

Location of Roots

1=W4 0

j=√-1=W4 3

-1=W4 2

-j=-√-1=W4 1


jjW

jW

jjW

jW

334

224

114

004

1

1


1 1 1 1 (0) 1 1.875

1 1 (1) 0.5 0.75-j0.37

1 1 1 1(2) 0.25

1 1 (3) 0.125

4

Xj j

Xk

Xj j

X

n

knW

5

0.625

0.75+j0.375

1

5757

Solution:

1 1 1 1 (0) 1 1.875

1 1 (1) 0.5 0.75-j0.37

1 1 1 1(2) 0.25

1 1 (3) 0.125

4

Xj j

Xk

Xj j

X

n

knW

5

0.625

0.75+j0.375



123

4

5

6

78 9

10

1112

1314

mod

mod

mod

[ ] =[1 2 3 4 5 6 7 8 9 10 11 12 13 14]

[1 14 13 12 11 10 9 8 7 6 5 4 3 2]

[2 1 14 13 12 11 10 9 8 7 6 5 4 3 ]

[3 2 1 14 13 12 11 10 9 8

[ ][1 ][2 ]

N

N

N

x n

x kx kx k

7 6 5 4 ]

........... so on

N

Circular Folding

5959

Discrete Fourier TransformDFT

It’s a numerically computable transform.Its obtained by sampling the discrete

time Fourier Transform in the frequency domain.

This is developed by analyzing periodic sequences such as the Discrete Fourier Series.

DFS is then extended to finite duration sequences, leading to DFT


Discrete Fourier Transform, DFT






6565

The importance of DFT

( )j

X e

0 1n N

From the frequency sampling theorem it can be concluded that N equispaced samples of the Discrete Time Fourier Transform of the N point sequence x(n) can uniquely reconstruct

These N samples around the unit circle are called the discrete Fourier transform coefficients.

It is clear that the DFS is practically equivalent to the DFT when

If x(n) is a sequence defined only over the interval from0 to N-1, the DFT,X(k) of x(n) is defined only over the same interval from 0 to N-1.


6666

2

1

0

1, 0, 1,.......

N j knN

kn k n

Nx eX

The IDFT-Inverse DFT


1,.......0,1

)(1

0

NnWKX

Nnx Kn

N

N

k

6767

3.Find the eight point DFT of the sequence :

x[n]=(1 2 3 4 4 3 2 1 )

Assignment



Resolution of the Frequency spectrum for longer DFT


Properties of the DFT

Linearity : The DFT is a linear transform

DFT [ax1[n]+bx2[n]]= a DFT[x1[n]]+ b DFT [x2[n]]

If x1[n] and x2[n] have different durations that is, they are N1 point and N2 sequences, respectively, then choose N3 =max(N1,N2) and proceed by taking N3 point DFTs.


HOMOGENITY OF THE DFT


Frequency response of the impulse response with padded zeros.



74P.Meena,Asst.Prof(EE)BMSCE 74

4 point DFT plots of a sequence x[n]=[1 1 1 1] in Matlab


7575

The Eight point DFT of [1 1 1 1]obtained by appending with zeros


7676

Effect of zero padding

The zero padding gives a High density Spectrum and a better displayed version for plotting but not a high resolution spectrum.

More data points needs to be obtained in order to get a high resolution spectrum


7777

DFT plot indicating the symmetry about ω =pi


7878

MATLAB PROGRAM

n=input('input the values of nin the form [0:deltaN:N-1]=');

k=input('input the values of kin the form [0:deltak:k-1]=');

xn=input('input the sequence of signal values=');w=(2*pi/(length(n))*k);WN=exp(-j*2*pi/(length(n)));nk=n'*k;WNnk=WN.^nk;Xk=xn*WNnk;stem(w/pi,abs(Xk))


7979

Let

conjugate

complex are K)-X(N and K)X(N eff-co DFT

sequence real valued2N a be ][

nx


8080

Circular Folding

123

4

5

6

78 9

10

1112

1314


DFT [x((-n))N]= X((-k)) N =

x[n]=[1 2 3 4 5 6 7 8 9 10 11 12 13 14]

x[-n]modN=[1 14 13 12 11 10 9 8 7 6 5 4 3 2]


Circular Shift

1

x13

2

4

5

6 78

9

10

12

11

1314

x0

x2

x3

x4x5

x6x7

x9

x8

x10x11

x12x13

x1x0

x2

x3

x4x5

x6x7

x9

x8

x10x11

x12x13

141

2

3

4

56

7

8

9

10

1112

13x[n]x[n-1] mod 14

x1x0

x2

x3

x4x5

x6x7

x9

x8

x10x11

x12x13

23

4

6

7 89

10

11

12

13x[n+1] mod 14

5

114

[right shift the sequence]14

[left shift the sequence]14


In general,

mod

mod

[ (0), (1),........ ( 3), ( 2), ( 1)]

[ ( 1), (0), (1),........ ( 3), ( 2)]

[ ( 2), ( 1), (0), (1),........ ( 3)]

[ ][ 1][ 2]

N

N

x x x N x N x N

x N x x x N x N

x N x N x x x N

x nx nx n

8

[ ] [1 2 3 4 5 6 7 8]

x [7 8 1 2 3 4 5 6](n-2)x n

2,1,14,13,12,11,10,9,8,7,6,5,4,3]2[

1,14,13,12,11,10,9,8,7,6,5,4,3,2]1[

12,11,10,9,8,7,6,5,4,3,2,1,14,13]2[

13,12,11,10,9,8,7,6,5,4,3,2,1,14]1[

]14,13,12,11,10,9,8,7,6,5,4,3,2,1[ ][

4mod

4mod

14mod

14mod

nx

nx

nx

nx

nx

8383

Properties of Discrete Fourier Transform

1.Linearity: The DFT is a linear transform

1

29)()(a

4

20)(;

3

9)(a

4b3;a

1

5)(;

1

3)(

1

2915] 14DFT[)]()](DFT[a

12] 8[)(b 4,b

,3]; 6[)(a 3,a If 3]; 2[)( ;1 2)(

)]([)]([a

)]([)]([ )]()([

21

21

21

21

2

121

21

2121

kbk

kbk

Let

kk

nbn

n

nnn

kbk

nbDFTnDFTanbnaDFT

XX

XX

XX

xx

xxxx

XXxxxx


8484

2.Circular Folding /Shift :

][][modmod

KXnxNN

DFT

mod 4

mod 4

mod 4

[1 4 3 2]

10 10

-2 j2 -2 j2X[K] ; [ ]

-2 -2

-2-j2 -2 j2

10

-2 j2

-2

-2 j2

[ ]

[ ]

[ ]

DFT

x n

x n

X K

4] 3 2 [1 x[n] let

4] 3 2 [1 x[n] let10

-2+j2

-2

-2-j2

-2+j2

10

-2-j2

-2


85

3a.DFT of circular shifted sequence

mod

DFT[x[n]]=X[K]

then DFT X[K]x[n-m]Km

N

If

NW

mod 4 x[n]=[1 2 2 0]. Find DFT of ,mod 4x[n],x[n-1] x[n-2]If

mod4[0 1 2 2]

1 1 1 1 0 5

1 -j -1 +j 1 2

1 -1 1 -1 2 1

1 +j -1 -j 2 2

x[n-1]

j

j


86

3a.DFT of circular shifted sequence

mod

1

mod 4

DFT X[K]

DFT X[K]4

x[n-m]

x[n-1]

Km

N

K

NW

W

4

(0) 1 1 1 1 1 5

(1) 1 -j -1 +j 2 -1-j2

(2) 1 -1 1 -1 2 1

(3) 1 +j -1 -j 0 -1+j2

0 5*4

1 -1-j2*4

W

W[ 1]

X

X

X

X

DFT x n

5*1 5

(-1-j2)*(-j) -2+j

2 1*(-1) -11*4

(-1+j2)*(j) -2-j3 -1+j2*4

W

W

1

j

-j

-1



Find the DFT of

Circular Convolution

Find the circular convolution of ,[1 ,2 ,2, 0] and [1,2,3,4]

4.If x[n] is a 2N valued real sequence,

X[N+k] and X[N-k] are complex conjugate. X[0] X[1] X[2] X[3] X[4] X[5] X[6] X[7]


x[n]=x[n]=

Inverse DFT of a sequence

Find the inverse DFT of [ 5 (-2+j1) -1 (-2-j1) ]

8989

3b. Multiplication by Exponentials or Cicular Frequency Shift

[ ] ( )mn

N NDFT x n X K mW

If X(K) is circularly shifted, the resulting inverse transform will be the multiplication of the inverse of X(K) by a complex exponential.


If g[n] and h[n] are two sequences of length 6. They have six point DFTs G[K] and H[K] . The sequence g[n] is given by

g[n]= 4.1, 3.5, 1.2, 5, 2, 3.3

If DFTs G[k] and H[k are related by circular frequency shift as,

H[k]=G[k-3] 6 Find h[n].

90

Consider x[n]=[ 1 2 3 4 5 6 7 8 9]What is

mod4?????????x[n-4]


91


2

1 2

1 2 1 2

[1 2 2 0]; [ ] [1 1 1 1]1

5 4

1 2 0[ ] [ ]

0 0

1 2 0

20

0[ ] [ ] ; [ ] [ ]

0

0

20

0

0

0

[ ] n

jK K

j

K K IDFT K K

IDFT

n xx

X X

X X X X

1 1 1 1 20 5

1 +j -1 -j 0 51

1 -1 1 -1 0 54

1 -j -1 +j 0 5


92


1 2

1 2 2 0 1 2 2 0 1 2 2 0 1 2 2 01

2 mod 4

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

[ ] 1111 1111 1111 1111[ ]

------- ------- ------ ------

n nx xnxn kx

5 5 5 5

1 2[ ] [ ]1 2[ ] [ ] IDFT K Kn n X Xx x


93

Evaluating the IDFTFind the IDFT of [5 (-1-j2) 1 (-1+j2)]

1 2

1 2 2 0 1 2 2 0 1 2 2 0 1 2 2 01

2 mod 4

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

[ ] 1111 1111 1111 1111[ ]

------- ------- ------ ------

n nx xnxn kx

5 5 5 5

1

+j

-j

-1

kn

NW



1

2

4

56

n=0

n=1n=2

n=3

n=4 n=5

12

45

6n=0

n=1n=2

n=3

n=4 n=5

x[n] x[n-1]mod6

3

x[-n]=x[N-n]modNCircular Shift

n=0

n=1n=2

n=3

n=4 n=5

2

3

34

5

6 1

n=0

n=1n=2

n=3

n=4 n=5

3

45

6

21

n=0

n=1n=2

n=3

n=4 n=5

4

56

1

2 3x[n+1] x[n+2] x[n+3]

n=0

n=1n=2

n=3

n=4 n=5

5

61

2

34

x[n+4]

x[n-1]=x[n+5]modN


Symmetry property for real valued x[n]

If the sequence is real, X[N-k] =X*[k]=X[-k]

If the first 5 points of the 8 point DFT of a real valued sequence areX[k]=[ 0.5 (1-j) 0 (1-j1.72) 0 ],find the remaining 3 points of the DFT.

From the symmetry property,

X[N-k]=X*[k]

If x[n]=Find the four point DFT of x[n].[ 4 ( -1+j) 2 (-1-j ) ]


If x[n] is real ,We know that DFT of x[-n] = X[-k] = X[N-k= X*[K]Example DFT of[1 ,2, 3 ,4]

Which implies that, for real valued time signals,

X*[K]= X[-K]

And for imaginary valued time signals,

X*[K]= -X[-K]

Conjugation: DFT [x*(n)]= X* ((-K))N


Symmetry properties for real sequences

If x[n] is real and a N point sequence. then x[n]=x*[n]. Using the above property, X(K)= X*(-K) N


x(ev)[n]= 1/2[ x[n]+x[-n]]

X (ev) (K)=1/2[ X[K]+X[-K]]If signal is real, x[-n]=x[n]

Therefore X[-K]= X*[K]

Hence X[K] is real and even=

½[ X[K]+X*[K]]=X(real)[K].Case 2:If x[n] is real and odd, then its X[K] is purely imaginary X(odd)[n]= 1/2[ x[n]-[-n]]Therefore, X(odd[K])=1/2[X[K]-X[-K]]=jX(imag)[K].


Case 3 :If x[n] is imaginary and even, then its X[K] is purely imaginary.

x(ev)[n]= 1/2[ x[n]+x[-n]]

X (ev) (K)=1/2[ X[K]+X[-K]]

If the signal is imaginary,Then we know that X*[K]=-X[-K].Hence X[K] imaginary and even,=1/2[X[K]-X*[K]]= jX(imag)[K].Case 4 :If x[n] is imaginary and odd , then its X[K] is purely real.X(odd)[n]= 1/2[ x[n]-[-n]]X(odd[K])=1/2[X[K]-X[-K]]If signal is imaginary

X*[K]=-X[-K],therefore½[X[k]+x*[-K]]=1/2[ X(real)[K + j X(imag)[K] + [ X(real)[K - j X(imag)[K]= X(real[K]).


Complex Conjugate Properties

x*[n] has a DFT X*[N-k]= X*[-K] N


Even and odd parts of a sequence

nxnx

2

1n

nxnx2

1n

Nmododd

Nmodev

x

x

Find the even and odd parts of the sequence x[n]=[1 2 2 0]


Real and Imaginary Parts of X[K]

][

][*

*

*

Im

*

][2

1

][2

1

][][2

1

][][2

1

KXX

KXX

xx

xx

Nim

Nre

re

KXK

KXK

nnxj

n

nnxn


Computation of N point DFT of a real sequence using N point DFT.

g[n]=[1 2 0 1]; h[n]=[2 2 1 1]x[n]=


105

Multiplication:

It is the dual of the circular convolution property.

][][1

][][2111KK

NnnDFT XXxx



107

Parseval’s RelationThis Computes the energy in the frequency

domain1 12 2

0 0

2

2

1

is called the energy spectrum of finite

duration sequences.Similarly , for periodic sequences, the quantity

called the power spectrum

[ ] [ ]

[ ]

( )

N N

xn kN

The quantityN

is

x n X KE

X K

X KN


108

Some quick relations:

N-1-kn

Nk=0

N-1 N-10

Nk=0 k=0

N-1

k=0

1 x[n]= X[K]

N

1 1[0] X[K] X[K]

N N

X[K]=N [0];

W

W

f

x

x

1 12 2

0 0

1 12 2

0 0

1[ ] [ ]

[ ] [ ]

N N

xn k

N N

k n

N

OR

N

x n X KE

X K x n

1.

2.

2N-1 n

n=0

x[n] be a 2N valued real sequence,

X[N]= x[n](-1)

If

3.


109

Problems on Quick Relations

7 7 2

k=0 k=0

[ ] [1 2 0 3 -2 4 7 5]

with a 8 point DFT X(K). Evaluate the following without

explicitly computing

X(K).

1)x[0] 2)x(4) 3) ( ) 4) ( )

x n

X K X K

Ans: 1)20 2)-8 3)8 4) 864


110

0 1 2 3 4 5 6 70

50

100

w in radians

Mag

nitud

e

0 20 40 60 80 100 120 140 1600

50

100

k

mag

nitud

e

0 20 40 60 80 100 120 140 160-100

0

100

kangle

in d

egre

es

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

50

100

w/pi

mag

nitud

e

DFT OF A 50 Hz SINE WAVE SAMPLED AT 8KHZ


111

0 500 1000 1500 2000 25000

5

10x 10

5

Ene

rgy

Spe

ctru

m

DFT and energy spectrum of a wave form with sag

0 1 2 3 4 5 6 70

500

1000

mag

nitu

de o

f DFT

0 500 1000 1500 2000 25000

500

1000

mag

of D

FT

0 500 1000 1500 2000 2500-100

0

100

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

500

1000

0 500 1000 1500 2000 2500-1

0

1

DFT & Energy Spectrum OF A 50 Hz SINE WAVE with SAG


112

DFT & Energy Spectrum OF A 50 Hz Normal SINE WAVE and a Sine wave with sag


113

DTMF TONE Allocation

P.Meena, Asst.Prof(EE) BMSCE 113Dr. P.Meena, Assoc.Prof(EE) BMSCE

114

DFT of DTMF



Use of DFT in Linear Filtering


Therefore zero padding is used to make the signals of equal length.

117

Linear Convolution & Circular Convolution.

FIR(Finite Impulse Response) filters are implemented using linear convolution.

Given two sequences

1N

n.convolutiolinear toidentical isn convolutiocircular

then thezeros, ofnumber eappropriatan padding

by Nlength same theof made are sequences

, Nlength of and

21

2121

NN

Both

Nandnn xx


118

Let

1]. 1 1- 2- 1- 1 1 [:Ans

equal. arethat they

show andn convolutiocircular theCompute

n.convolutiolinear their Determine

1]; 1 1 1[1]; 2 2 1[ 21 nxnx



Error between Circular convolution & Linear convolution due to choice in N

When N=max(N1,N2) is chosen for circular convolution then the first M-1 samples are in error where M=min(N1,N2).

Hence this leads to different methods of convolution in block processing.


BLOCK CONVOLUTIONS

Necessity of Block Convolutions:• To filter an input sequence received continuously such as a speech

signal from a microphone and if this filtering operation is done using a FIR filter, in which the linear convolution is computed using the DFT then there are some practical problems .

• A large DFT is to be computed. • Output samples are not available until all input samples are

processed resulting in a large amount of delay.

In Block Convolution:• The speech signal is segmented into smaller sections /blocks .• Each section is processed using the DFT • Finally the output sequence is assembled by assembling the

outputs of each section.

P.Meena, Asst.Prof(EE) BMSCE 121

Errors in BLOCK CONVOLUTIONS

• If x[n] is sectioned into N point sequences and• The impulse response of the filter is an M point sequence ,

where M <N. • Then the N point circular convolution between the input block

and the impulse response will yield a block output sequence in which the first M-1 samples are not the correct output values.

• Therefore x[n] is partitioned into sections by adding M-1 zeros into first section each overlapping with the previous one by exactly (M-1) samples, save the last (N-M+1) output samples and finally concatenate these outputs into a sequence.

• To correct for the first M-1 samples in the first output block, set the first (M-1) samples in the first input block to zero. This procedure is called the Overlap –save method

121Dr. P.Meena, Assoc.Prof(EE) BMSCE

M

2

[ ] [1 2] [1 2 3 4 5 6];

length of h[n]=M;

length of x[n]=L;

To evaluate the length of the sequences N for convolution,

1,

2 1, 4 1 3. 4

Overlap and save method,Input sequence overl

22

h n

M N

N N N

In

1 2

aps by (M-1) samples.

[ ] [1 2 0 0]; x [n]=[0 1 2 3],x [n]=[3 4 5 6]

1 2 0 0 1 2 0 0 1 2 0 0 1 2 0 0 |1 2 0 0 1 2 0 0 1 2 0 0 1 2 0 0

0 3 2 1 1 0 3 2

h n

2 1 0 3 3 2 1 0 | 3 6 5 4 4 3 6 5 5 4 3 6 6 5 4 3

---------- ----------- ----------- ----------- |---------- ----------- ----------- -----------

1 4 7 | 10 13 16

--------- ----------- ----------- ----------- |---------- --

--------- ----------- -----------

The linear convolution result y[n]=[1 4 7 10 13 16]



Using overlap and save method compute y[n] if h[n]=[3 2 1]And input x[n]=[ 2 1 -1 -2 -3 5 6 -1 2 0 2 1 ].

Use only 8 point circular convolution.M=3;

1] 2 [3 h[n]

1] 2 0 2 1- 6 5 3[][

5] 3- 2- 1- 1 2 0 0[][

2

1

nx

nx


------------------------------------------------

12 16 13 10 7 4 1

-------------------------------------------------

12 16 13 4

6 7 4 1

----- ------ ------ ------- ----- - ------ -- ------ ------

4 5 6 0 0 4 5 6 6 0 4 5 5 6 0 4 1 2 3 0 0 1 2 3 3 0 1 2 2 3 0 1

0 0 2 1 0 0 2 1 0 0 2 1 0 0 2 1 0 0 2 1 0 0 2 1 0 0 2 1 0 0 2 1

0] 6 5 [40]; 3 2 1[][

]0 0 2 1[][

4array. in the considered sizeblock theis ,3

;124

;12

nconvolutiocircular of size theis N

6], 5 4 3 2 1 [x[n]

2] ,1[][

M

nx

nh

NL

L

LM

nhOverlap and Add method of Sectional Convolution:


12] 6 16 13 10 7 4 1 [:Ans

0 0 12 6 16 13 10 15

---------------------------------------------------------

6 0 0 0 60 0 0 0 0 6 0 0 0 0 6 3 4 5 6 6 3 4 5 5 6 3 4 4 5 6 3

| 00 2 1 0 0 2 1 0 0 2 1 0 0 2 1 00 2 1 0 0 2 1 0 0 2 1 0 0 2 1

------------------------------ ------------- ------- ------

7 4 1 6

----------------------------- ------------ -------- ------

0 1 2 3 3 0 1 2 2 3 0 1 1 2 3 0

00 2 1 0 0 2 1 0 0 2 1 0 0 2 1

] 0 0 0 6[][x

6] 5 4 3[][

]3 2 1 0[][x

0] 0 2 1[][

21;3;4

12 ,2

3

2

1

n

nx

n

nh

MLN

LMNM M

Verify by overlap and Save Method


content.energy

thefind hence 1.and-Nn0 where2

cos][

sequence theof DFT Obtain the

0

N

nKnx




DFT of a Square Wave

Relationship of the DFT to Other TransformsRelationship to Z- transform:

kN

N

kj

Wez

j

N

n

n

knN

N

n

n

n

zXKXez

znx

ce

WnxKX

znxzX

2)()(,

][X(z)

1,-Nn0 range theoutside 0 x[n]sin

][)(

][)(

1

0

1

0



0.05.05.05.05.0)3(

0.15.05.05.05.0)2(

05.05.05.05.0)1(

0.15.05.05.05.0)0(

)()(

5.05.0.05.00.z0.5X(z)

DFT. its find transform Zthe

0].Using 0.5 0 [0.5 of transform

2.34

2.24

2.14

2.04

2321-

WX

WX

WX

WX

zXkX

zzz

ZtheFind

kNWz

The Limitations of the directcalculation of the DFT

It requires

pair each for fourtions,multiplica real

4N requiresk each for X(k) of evaluationdirect For

additions.

complex 1)-N(N and tionsmultiplicacomplex 2N



THE FAST FOURIER TRANSFORM

• This is proposed by Cooley and Tukey• Based on decomposing /breaking the transform into smaller transforms and combining them to give the total transform.• This can be done in both Time and Frequency domains.


DECIMATION IN TIME FFT

The number of points is assumed as a power of 2,that is

obtained. are sformspoint tran- two

int4

N twointo sformpoint tran

2

Neach

breaking then and int 2

N twointo transforms

point - N thebreaking of one isapproach

2

until

transformspo

transformspo

This

N v


1.-2

N0,1,.....,k ,],[][)

2X(

follows, as computed be can binsfrequency of half second thethen,

equations, above theusing andfact following thegConsiderin

1.-2

N0,1,.....,k ,],[][][

bins,frequency halffirst the

yieldsX[K],for expression theinto thisngsubstituti

1.-2

N0,1,.....,k ,],

2[][

1.-2

N0,1,.....,k ,],

2[][ that note

sint2/ )12(,]12[][

sint2/ )2(,]2[][

as, functionsnew definenow we

1.-N0,1,.....,k

.]12[ ]2[][

that,followsit ,

relation, theusing

1.-N0,1,.....,k

.]12[]2[][

)2

(

1)2/(

0 2

1)2/(

0 2

1)2/(

0 2

1)2/(

0 2

2

2

21)2/(

0

1)2/(

0

2

forKHWKGN

K

WW

forKHWKGKX

forN

KHKH

forN

KGKG

poNwithmxDFTWmxKH

poNwithmxDFTWmxKG

for

WmxWWmxKX

W

for

WWmxWmxKX

kN

kN

KN

N

kN

N

m

mkN

N

m

mkN

N

n

mkN

kN

N

m

mkN

NN

mkN

N

m

kN

N

m

mkN

W


https://engineering.purdue.edu/VISE/ee438/demos/flash/decimation.swf

https://engineering.purdue.edu/VISE/ee438/demos/flash/decimation.swf


0x

1x

12W

02W

)0(X

)1(X

0x

2x 12W

02W

1x

3x

02W

12W

04W

14W

24W

34W

X(0)

X(1)

X(2)

X(3)

2-POINT DFT

4-POINT DFT


Find the DFT of x[n]=[1 2 3 4] using DIT algorithm

Ans:[10 -2+j2 -2 -2-j2].

Find the inverse of :[10 -2+j2 -2 -2-j2]. using DITFFT.



0x

4x 12W

02W

2x

6x

02W

12W

04W

14W

24W

34W

X(1)

X(2)

X(3)

12W

02W

02W

12W

04W

14W

24W

34W

X(4)

X(5)

X(6)

X(7)

08W

18W

28W

38W

48W

58W

68W

78W

1x

5x

3x

7x

BIT REVERSED SEQUENCE EIGHT POINT DFT

x[n]=[1, -1, -1, -1, 1, 1, 1, -1]1

1

-1

1

-1

1

-1

-1

2

0

0

-2

0

-2

-2

0

2

2j

2

-2j

-2

-2

2

-2

X(0)

3.69+j1.96

2.82-j0.78

1.53-j0.39

4-j0.0

1.53+j0.39

2.82+j0.78

3.69-j1.96

0.0-j0.0



2

2

2

2

2

2

2

2

j

2

2

2

2

2

2

2

2

-1 ;1

78

38

68

28

58

18

48

08

jWjW

WjW

jWjW

WW


]x,[x ],x,[x

into decimateor

] x x x[x Let x[n]

3120

3210

Divide

]x,[x ],x,[x

into decimateor

] x x x[x Let x[n]

3120

3210

Divide

1x0 25.0x2

02W

02W

5.0x1

125.0x3

12W

12W

04W

34W

X(0)=1.875

X(1)=0.75-j0.375

X(2)=0.625

X(3)=0.75+j0.375

14W

24W

0.125] 0.25 0.5 1[][ nx

25.1

75.0

625.0

375.0


Computation of Inverse DFT

]j21- 1 j2-1- [5

of IDFT the

.][N

1x[n]

bygiven is DFT inverse 1

0

Find

WKX

TheN

k

knN

Twiddle Factors are negative powers of NW

The output is scaled by 1/N


02W

02W

12W

12W

04W

34W

14W

24W2

4j

50 x

12 x

211 jx

213 jx

6

4

4

8

8

0

1

2

0

2

4

1

4

1

4

1

4

1

+j

-j

-1 1


20x

11x

12W

02W

12

x

23x

02W

12W

04W

14W

24W

34W

6)3

1

3

1

1+j

0

1-j

10x

11

x12W

02W

13x

02W

12W

04W

14W

24W

34W

1

0

2

12x

1

0

20

2+j2

0

2+j2

2-j2


02W

02W

12W

12W

04W

34W

14W

24W2

4j

00x

02x

41jx

43

jx

4

8

8

4

1

4

1

4

1

0

0

0

2

0

2


DECIMATION IN FREQUENCY FFT (DIF) FFT

or k=0,1……..N-1,Where = is the twiddle factor and N=2,4,8,16… can be expanded as,X[K]=x(0)+x(1)k +………+x(N-1)k(N-1)

Again if we split the above equation into,X[K]=x(0)+x(1)k+…x(-1)k(-1)+x()WN

k +… …+x(N-1) k(N-1)

Then we can rewrite as a sum of the following two parts,+Which can be written as ,+=-1 , then we have ,+Now letting k=2m as an even number achieves,X[2m]=+ ,X[2m+1]=-,a[n]=x[n]+x[n+N/2], for n=0,1,………N/2 -1,b[n]=x[n]-x[n+N/2], for n=0,1….N/2 -1We get DFTx[n]with N points=



If x[n]=[1 2 3 4]. Find X[K] using DIF FFT.

02W

04W

10 x

22 x

33x

44 x8

4

1

1

6

-2

-2 `14W

-2

-2j

10

1

1

2

-2-j2

02W

-2+j2

X0

X2

X1

X3

X[K]= [10, -2-j2, -2, -2+j2]


N=2No.of stages =NNo.Ofcomplex multiplications in each butterfly=2 No.of Butterflies in each stage = N/2 number of complex multiplications in each stage=N.no.of complex multiplications =NN

Outcome Based Education 1 Dr.P.Meena,Assoc.Prof., EEE Focus Learning, not teaching Students, not faculty Outcomes, not inputs or capacity INDIA HAS BECOME.

Documents

eee slide

profee bmsce slide

digital filters

digital system components

dsp hardware

signal processing functions

responses of digital

washington accord slide