Digital Signal Processing 2014 pptx.pptx

7/24/2019 Digital Signal Processing 2014 pptx.pptx

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Outcome Based Education

1Dr.P.Meena,Assoc.Prof., EEE

FocusLearning, not teachingStudents, not facultyOutcomes, not inputs or

capacity

INDIA HAS BECOME APERMANENT MEMBER OF THE

WASHINGTON ACCORD


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2

Academicabilities

CourseLectures/Demonstrations/

Videos/Animations /powerpoint presentations/hand outs

Problem solving

Teacher led Students in pairs/share

Industry Visits

pen ended e!periments

Components that contriute toAca!emic Ai"ities


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Components that contriute toTrans#era"e S$i""s

3

Trans"erable s#ills

StudentActivities

Pro$ectwor#/

pen endede!periments


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The components o# course !e"i%er& that contriute

to the !e'ne! attriutes o# the course (

Attributes

TechnicalSymposia

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


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5 5

Di)ita" Si)na" Processin)

Intro!uction

Inception*+,-./ith the !e%e"opmento# Di)ita" Har!/are such as !i)ita"har!/are(

Persona" computer re%o"ution in+,01s an! +,,1s cause! DSP e2p"osion/ith ne/ app"ications.

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


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6 6

Advantages of DSP TechnologyHigh reliabilityReproducibility

Flexibility & PrograabilityAbsence of !oponent Drift proble

!opressed storage facility "especially inthe case of speech signals #hich has a lot

of redundancy$.DSP hard#are allo#s for prograable

operations.Signal Processing functions to beperfored by hard#are can be easilyodified through soft #are"efficient

algoriths$



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7 7

Advantages of DSP TechnologyHigh reliabilityReproducibility

Flexibility & PrograabilityAbsence of !oponent Drift proble

!opressed storage facility "especially inthe case of speech signals #hich has a lot

of redundancy$.DSP hard#are allo#s for prograable

operations.Signal Processing functions to beperfored by hard#are can be easilyodified through soft #are"efficient

algoriths$



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


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Digital Signal Processing#ith overlapping borders


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1010

A T%P!A' D(TA' S()A'PR*!+SS)( S%ST+,

%&tx%&txa %&nx %&ns %&ts D3A

CON4ERTER

Ana"o)pre'"ter orAntia"iasin)'"ter

A3DCON4ERTER

Di)(Si)na"Processor

Reconstruction'"ter sameas the pre'"ter

5o/ pass'"tere!si)na"

Discretetimesi)na"

Discretetimesi)na"Samp"i

n)#re6uenc&

!B

!B



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CO+* Ai"it& to app"& the $no/"e!)e o#

mathematics7 science an! #un!amenta"s o#si)na"s an! s&stems to ascertain the eha%ioro# comp"e2 en)ineerin) s&stems(

CO8*Ai"it& to I!enti#& techni6ues7 #ormu"ate

representations an! ana"&9e responses o#!i)ita" s&stems(

CO:*Ai"it& to Desi)n !i)ita" s&stemcomponents an! test their app"ication

usin) mo!ern en)ineerin) too"s7 asso"utions to en)ineerin) pro"ems.

Course Outcomes


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Course Contents

Dierent operations on a signal in thedigital domain

Dierent forms of reali!ations of aDigital System.

Design Procedures for Digital "ilters


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1313

Outcomes of this Course:By The 'nd " The Course (

Distinguish The Digital and Analog Domains)

Analyse Signals( and reconstruct)

Develop *loc# Diagrams +or Di""erent System,epresentations()

Design Analog And Digital +ilters)

,eady to Ta#e up Speciali-ed Courses in Audio( speech( image and ,eal.time Signal Processing( +urther n



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Course Out"ine

Course Delivery:

Lecture(hand outs(videos(animations(discussions(activities

Course Assessment:

ar#s0

Tests0 12 &T3 4 T1%

5ui- 6 27

Tutorials0 32Lab0 37


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1515

Revie# ofSignals

&Systes



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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 seuence !hich is a function of a

discrete variable. "mage#Represented as a function of t!o spatialvariables

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


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2020

Relation bet#een analog

fre-uency and digitalfre-uency

( )

s

1 unit is radians per second)

y8a sin ( a signal in the continuous time domain)

t8n9

-8a sin& %

sin& %

8 is the digital "re:uency in radians/sample

There"ore( given a 1 ( get

Ts

s

f

t is

n

z a n

where

f to

T

T

=

=

=

s

1 or &1 9 %T

s

ff

f

=



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f !n "er#$(ana%o&)

' !n ra!anssa*+%eD!&!#a%

0

s2-s4-s2-s ss4 32s-32s

f0,'0

'/2,fs4

'/,fs2

' -/2,f-s4

'2/'/'0'/2'-/2

'-2/ '-/ '3/'-3/

!s# !n#era%

'-/,f-s2

Sl.)o.

Fre-uencyin Hert*f thesignal

SaplingFre-uencyFs in Hert

/ inradians0cycle

%. f&' s '

. f&s*+ s ,*

3 f&-s*+ s -,*

+. f&s* s ,

. f& -s* s -,

Diagrammatic ,epresentation o" relation between analog

"re:uency and digital "re:uency0;ve angle counter cloc# wise


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amp"in) o# continuous time si)na"s

#he "ourier transform pair for continuous$timesignals is de%ned &y

( ) ( )

( ) ( )

d$1

3t

dtt$

e

e! > 8 = < $

Assignment


R l ti f th F t f


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Resolution of the Fre-uency spectru forlonger DFT


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Properties of the DFT

9inearit% : The D!T is a linear transform

D!T 0ax,0n1-x&0n112 a D!T0x,0n11 - D!T 0x&0n11

If x,0n1 and x&0n1 ha)e different durations that is3the% are #,point and #& se$uences3respecti)el%3 then choose #2max*#,3#&+ andproceed -% ta;ing # point D!Ts.

H*,*(+)T% *F TH+ DFT


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H*,*(+)T% *F TH+ DFT

F f th i l


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Fre-uency response of the ipulse response#ith padded eros.


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74P.Meena,Ass#.Prof(EE)BMSCE 74

> point DFT plots of a se-uence x9n:;9< < <


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7575

The +ight point DFT of 9< < <


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7676

+ffect of ero padding

1he 2ero padding gives a 5igh densit0Spectrum and a better displa0ed versionfor plotting but not a high resolutionspectrum.

6ore data points needs to be obtained inorder to get a high resolution spectrum



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7777

DFT plotindicating the syetry

about ? ;pi



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7878

,AT'A@ PR*(RA,n&input(7input the values of nin the form

8'#delta9#9-%:&7);


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7979

Ge#

con$ugate

comple!areG%.


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8080

123

4

5

6

78 9

10

1112

1314


D"# )(66$n779*F G66$>77

9F

8n:&8% 3 + B E %' %% % %3 %+:

8-n:mod9&8% %+ %3 % %% %' E B + 3 :

Circular Shift


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Circular Shift

1

;13

2

4

5

6 78

9

10

1211

1314

;0

;2

;3

;4

;5 ;6;7

;9

;8

;10;11;12

;13

;1

;0

;2

;3

;4

;5 ;6

;7

;9

;8

;10;11;12

;13

141

2

3

45

6

7

8

9

1011

1213x0n1

x0n6,1 mod ,/

;1;0

;2;3

;4;5

;6;7

;9

;8

;10

;11;12

;13

23

4

6

7 8

9

10

11

12

13x0n,1 mod ,/

5

114

0right shift the se$uence1,/

0left shift the se$uence1,/

=3@(3B(31(33(32(F(D(E(A(7(@(B(1(3>=> =nx


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In general3

mod

mod

> &2%( &3%()))))))) & B%( & 1%( & 3%=

> & 3%( &2%( &3%()))))))) & B%( & 1%=

> & 1%( & 3%( &2%( &3%()))))))) & B%=

> => 3=

> 1=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 n

x n

=

=

=

[ ]

> = >3 1 B @ 7 A E =

! >E 3 1 B @ 7 A=&n.1%x n =

=

[ ]

[ ]

[ ]

[ ]1(3(3@(3B(31(33(32(F(D(E(A(7(@(B=1>

3(3@(3B(31(33(32(F(D(E(A(7(@(B(1=3>

31(33(32(F(D(E(A(7(@(B(1(3(3@(3B=1>

3B(31(33(32(F(D(E(A(7(@(B(1(3(3@=3>

@mod

@mod

3@mod

3@mod

=+

=+

=

=

nx

nx

nx

nx

P i f Di F i


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8383

Properties of Discrete FourierTransfor


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8484

&.Circular !olding => modmod KXnx NNDFT =

mod@

mod@

mod@

>3 @ B 1=

32 32

.1 $1 .1 $1G= 6 > =

.1 .1

.1.$1 .1 $1

32

.1 $1

.1

.1 $1

> =

> =

> =

DFT

x n

x n

X K

= + = = +

= +

@=B1>3!>n= =let

@=B1>3!>n= =let10

-2H2

-2

-2-2

-2H2

10

-2-2

-2



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85

a.D!T of circular shifted se$uence

mod

D+T>!>n==8G=

then D+T G=!>n.m= K

N

!f

NW =

mod@!>n=8>3 1 1 2=) +ind D+T o" (mod@!>n=(!>n.3= !>n.1=!f

mod@>2 3 1 1=

3 3 3 3 2 7

3 .$ .3 ;$ 3 1

3 .3 3 .3 1 3

3 ;$ .3 .$ 1 1

!>n.3=

j

j

=

+ =



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86

a.D!T of circular shifted se$uence

mod

3

mod@

D+T G=

D+T G=@

!>n.m=!>n.3=

K

N

K

NWW

= =

@

&2% 3 3 3 3 3 7

&3% 3 .$ .3 ;$ 1 .3.$1&1% 3 .3 3 .3 1 3

&B% 3 ;$ .3 .$ 2 .3;$1

2 79@

3 .3.$19 @

H

H> 3=

X

XX

X

DFT x n

= =

=

793 7

&.3.$1%9&.$% .1;$

1 39&.3% .339@

&.3;$1%9&$% .1.$B .3;$19@

H

H

= =

1

67

-1


!n #e D: of


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!n #e D: of

Circular Con)olution

!ind the circular con)olution of 30, 3& 3&3 41 and 0,3&33/1

/.If x0n1 is a )alued real se$uence,


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;


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8989

-. ultiplication -% Exponentials or Cicular !re$uenc% Shift

> = & %

n

N NDFT x n X K W

=

If =*>+ is circularl% shifted3 the resulting in)erse transform ?ill -e themultiplication of the in)erse of =*>+ -% a complex exponential.


f &


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90

!onsider x9n:;9 < = 8 > 4 B C :>hat is

mod@!>n.@= =



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91

!ircular !onvolution

1

3 1

3 1 3 1

>3 1 1 2=6 > = >3 3 3 3=3

7 @

3 1 2> = > =

2 2

3 1 2

12

2> = > = 6 > = > =

2

2

122

2

2

> = n

jK K

j

K K !DFT K K

!DFT

n xx

X X

X X X X

= = = = +

=

3 3 3 3 12 73 ;$ .3 .$ 2 73

3 .3 3 .3 2 7@

3 .$ .3 ;$ 2 7

= =



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92

!ircular !onvolution

3 1

3 1 1 2 3 1 1 2 3 1 1 2 3 1 1 23

1 mod@

> = > = 3 1 1 2 3 3 3 3

> =

3333 3333 3333 3333> =

....... ....... ...... ......

n nx x

nx

n kx

= ==

7 7 7 7

3 1> = > =3 1> = > = !DFT K K n n X Xx x =


+valuating the DFT


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93

+valuating the DFT

Find the DFT of 94 "3


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1

2

4

56

n0

n1n2

n3

n4 n5

12

45

6n0

n1n2

n3

n4 n5

x0n1x0n6,1mod@

3

x0 n1 x0# n1mod#

n0

n1n2

n3

n4 n5

2

3

34

5

6 1

n0

n1n2

n3

n4 n5

3

45

6

21

n0

n1n2

n3

n4 n5

4

56

1

2 3x0n,1 x0n&1 x0n1

n0

n1n2

n3

n4 n5

5

61

2

34

x0n/1

x0n6,12x0n51mod#

S**e#r +ro+er# for rea% a%e ;


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S**e#r +ro+er# for rea% a%e ;


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S%mmetr% properties for real se$uences

If x0n1 is real and a # point se$uence. then x0n12x0n1. "sing the

a-o)e propert%3 =*>+2 =*6>+ #


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;(e)


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If x0n1 is imaginar% and e)en3 then its =0>1 is purel% imaginar%.

;(e)


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;


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+

[ ] [ ] [ ][ ][ ] [ ] [ ][ ]n!n!

1

3n

n!n!13n

mododd

modev

!

!=

+=

!n #e een an o +ar#s of #e seence ;


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Real and aginary Parts of G9:

[ ] [ ]

[ ] [ ][ ] [ ]

[ ] [ ]=>

=>

9

9

9

Im

9

=>13

=>1

3

=>=>1

3

=>=>1

3

KXX

KXX

xx

xx

Ni

Nre

re

KXK

KXK

nnxjn

nnxn

=

+=

=

+=

Co*+#a#!on of +o!n# D: of a rea% seence s!n& +o!n# D:


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Co*+#a#!on of +o!n# D: of a rea% seence s!n& +o!n# D:.

&


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,ultiplication2


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105

,ultiplication2

It is the dual of the circular con)olutionpropert%.

[ ] =>=>3=>=>1333

KKN

nnDFT XXxx =



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ParsevalIs Relation


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107

Parseval s Relation1his Computes the energ0 in the freuenc0

domain3 31 1

2 2

1

1

3

is called the energy spectrum o" "inite

duration se:uences)Similarly ( "or periodic se:uences( the :uantity

called the power spectrum

> = > =

> =

& %

N N

xn kN

The "#antit$N

is

x n X KE

X K

X KN

= =

= =



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108

Soe -uic6 relations2

C.3.#n

C#82

C.3 C.32

C#82 #82

C.3

#82

3!>n=8 G=

C

3 3>2= G= G=

C C

G=8C >2=6

H

H

f

x

x

= =

3 31 1

2 2

3 31 1

2 2

3> = > =

> = > =

N N

xn k

N N

k n

N

%&

N

x n X KE

X K x n

= =

= =

= =

=

1.

2.

1C.3n

n82

!>n= be a 1C valued real se:uence(

C=8 !>n=&.3%

!f

3.



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109

Probles on Juic6 Relations

E E 1

#82 #82

> = >3 1 2 B .1 @ E 7=

with a point D+T


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110

0 1 2 3 4 5 6 70

50

100

w in radians

Magnitude

0 20 40 60 80 100 120 140 1600

50

100

k

magnitude

0 20 40 60 80 100 120 140 160-100

0

100

kangleindegrees

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

50

100

w/pi

magnitude

D!T O! A 54 H SI#E AE SAF9ED AT G>H


D!T Energ% Spectrum O! A 54 H SI#E AE ?ith SAJ


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111

0 500 1000 1500 2000 25000

5

10x 10

5

EnergySpectrum

DFT and energy spectrum of a wave form with sag

0 1 2 3 4 5 6 70

500

1000

mag

nitudeofDFT

0 500 1000 1500 2000 25000

500

1000

magofDFT

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


D!T Energ% Spectrum O! A 54 H #ormal SI#E AE and a Sine ?a)e ?iths

ag


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112Dr. P.Meena, Assoc.Prof(EE) BMSCE

DT,F T*)+ Allocation


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113

DT,F T*)+ Allocation

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

D!T of DT!


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"se of D!T in 9inear !ilterin

g


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:erefore $ero +a!n& !s se #o *aFe#e s!&na%s of ea% %en.

'inear !onvolution & !ircular


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117

'inear !onvolution & !ircular!onvolution.

(!n!#e *+%se es+onse) f!%#ers are !*+%e*en#e s!n& %!nearcono%#!on.

I!en #o seences

[ ] [ ]

3Cn)convolutiolineartoidenticalisnconvolutiocircular

then the-eros(o"numbereappropriatanpadding

byClengthsametheo"madearese:uences

(Clengtho"and

13

1313

+= NN

'oth

Nandnn

xx


Ge#


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118

Ge#

[ ] [ ]

3=)33.1.3.33>0Ans

e:ual)arethat they

showandnconvolutiocirculartheCompute

n)convolutiolineartheirDetermine

3=6333>3=6113> 13 == nxnx


Error -et?een Circular con)olution 9inear


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Error -et?een Circular con)olution 9inearcon)olution due to choice in #

hen #2max*#,3#&+ is chosen for circularcon)olution then the first 6, samples arein error ?here 2min*#,3#&+.

Hence this leads to different methods ofcon)olution in -loc; processing.


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@'*! !*)K*'1T*)S

#ecessit% of Kloc; Con)olutions: To filter an input se$uence recei)ed continuousl% such as a speech

signal from a microphone and if this filtering operation is doneusing a !IR filter3 in ?hich the linear con)olution is computed usingthe D!T then there are some practical pro-lems .

A large D!T is to -e computed. Output samples are not a)aila-le until all input samples are

processed resulting in a large amount of dela%.

In Kloc; Con)olution: The speech signal is segmented into smaller sections CO#O9"TIO#S


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P.Meena, Ass#.Prof(EE) BMSCE 121

Errors in K9OC> CO#O9"TIO#S

f ;3 1= >3 1 B @ 7 =6

length o" h>n=86

h n =


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1

length o" h>n=86

length o" !>n=8L6

To evaluate the length o" the se:uences "or convolution(3(

1 3( @ 3 B) @

verlap and save method(Input se:uence overl

1

1

M N

N N N

!n

+

+ = =

3 1

aps by &.3% samples)

> = >3 1 2 2=6 ! >n=8>2 3 1 B=(! >n=8>B @ 7 =

3 1 2 2 3 1 2 2 3 1 2 2 3 1 2 2 J3 1 2 2 3 1 2 2 3 1 2 2 3 1 2 2

2 B 1 3 3 2 B 1

h n

=

1 3 2 B B 1 3 2 J B 7 @ @ B 7 7 @ B 7 @ B

.......... ........... ........... ........... J.......... ........... ........... ...........

3 @ E J 32 3B 3

......... ........... ........... ........... J.......... ..

......... ........... ...........

The linear convolution result y>n=8>3 @ E 32 3B 3=


! % # # < = !f < =


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s!n& oer%a+ an sae *e#o co*+#e =>7=B.1.3.3122>=>

1

3

=

==

nxnx

1=(3>=> =nhO)erlap and Add method of Sectional Con)olutionN


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................................................

3133B32E@3

.................................................

3133B@

E@3

..................................................

@722@72@772@31B2231BB2311B23

221322132213221322132213221322132=7>@2=6B13>=>

=2213>=>

@array)in theconsideredsi-ebloc#theis(B

631@

631

nconvolutiocircularo"si-etheis

=(7@B13>!>n=

=(>=>

=

===

+=+=

=

nx

nh

NL

L

LM

311 +=== LMNM M

Ter!f oer%a+ an Sae Me#o


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31=33B32E@3>0Ans

223133B3237

.........................................................

222222222222B@7B@77B@@7B

J22132213221322132213221322132213

........................................................

E@3.......................................................

231BB2311B2331B2

2213221322132213

=222>=>!

=7@B>=>

=B132>=>!

2=213>=>

136B6@

31(1

B

1

3

===

====

+===

n

nx

n

nh

MLN

LMNM


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content)energy

the"indhence3)and.Cn2where

1

cos=>

se:uencetheo"D+Tbtain the

2

= NnK

nx


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D!T of a S$uare a)e


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Relationship of the DFT to *therT f


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Transfors

Relationship to F- transform#

kN

N

kj

Wez

j

N

n

n

kn

N

N

n

n

n

zXKXez

znx

(e

WnxKX

znxzX

==

=

=

=

==

=

=

=

=

1%&%&(

=>n=sin

=>%&

=>%&

3

2

3

2


2=)sing2)72>2)7o"trans"orm)theFind


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2)27)27)27)27)2%B&2)37)27)27)27)2%1&

27)27)27)27)2%3&

2)37)27)27)27)2%2&

%&%&

7)27)2)27)22)-2)72)7o"trans"orm

1)B

@

1)1

@

1)3

@

1)2@

1B13.

==+= =+=+=

==+==+=+=

=

+=+++=

=

WXWX

WX

WX

zXkX

zzz

)theFind

kNWz

The 'iitations of the direct


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calculation of the DFT

"t reuires

paireach"or"ourtions(multiplicareal

@Cre:uires#each"or


133/151


TH+ FAST F*1R+R TRA)SF*R,

This is proposed by !ooley and Tu6ey @ased on decoposing 0brea6ing thetransfor into saller transfors andcobining the to give the totaltransfor. This can be done in both Tie and

Fre-uency doains.

D+!,AT*) ) T,+ FFT


134/151


:e n*er of +o!n#s !s ass*e as a +oer of 2,#a#

!s

obtained)ares"ormspoint tran.two

int@

twointos"ormpoint tran

1

each

brea#ingthenandint1

twointotrans"orms

point.thebrea#ingo"oneisapproach

1

#ntil

transfors*o

transfors*o

This

N +=

relation(theusing

3).2(3()))))(#

)=31>=1>=> 13%1/&

2

3%1/&

2

1

=

++=

=

=

for

WWxWxKX kN

N

k

N

N

k

N


135/151


}{

}{

3).1

2(3()))))(#(=(>=>%

1=>=>

bins("re:uencyhal""irstthe

yieldsG=("ore!pressiontheintothisngsubstituti

3).1

2(3()))))(#(=(

1>=>

3).1

2(3()))))(#(=(

1>=>thatnote

sint1/%31&(=31>=>

sint1/%1&(=1>=>

as("unctionsnewde"inenowwe

3).2(3()))))(#

)=31>=1>=>

that("ollowsit(

(g

%1

&

3%1/&

2 1

3%1/&

2 1

3%1/&

2 1

3%1/&

2 1

1

1

==+

=

=+=

=+=

=+=

++=

=

=

++=

=

+

=

=

=

=

forKHWK,N

K

WW

forKHWK,KX

forN

KHKH

forN

K,K,

*oNwithxDFTWxKH

*oNwithxDFTWxK,

for

WxWWxKX

W

k

N

k

N

KN

N

k

N

N

k

N

N

k

N

N

n

k

N

k

N

N

k

N

NN W


136/151


##+sNen&!neer!n&.+re.eTSEee438e*osf%asec!*a#!on.sf

&6FOI#T D!T
https://engineering.purdue.edu/VISE/ee438/demos/flash/decimation.swfhttps://engineering.purdue.edu/VISE/ee438/demos/flash/decimation.swf


137/151


2x

3x

3

1W

2

1

W%2&X

%3&X

2x

1x 31W

2

1W

3x

Bx

2

1W

31W

2

@W3

@W

1

@W

B@W

(0)

(1)

(2)

(3)

/6FOI#T D!T


138/151


!n #e D: of ;


139/151


@T R+K+RS+D S+J1+)!+ +(HT P*)T DFT

;


140/151


2x

@x 31W

2

1W

1x

Ax

2

1W

3

1W

2

@W

3

@W

1

@W

B

@W

(1)

(2)

(3)

3

1W

2

1W

2

1W

3

1W

2

@W

3

@

W

1

@W

B

@W

(4)

(5)

(6)

(7)

2

W

3

W

1

W

B

W

@

W

7

W

A

W

E

W

3x

7x

Bx

Ex

< = < , , , , , , , =

1

-1

1

-1

1

-1

-1

2

0

0

-2

0

-2

-2

0

2

2

2

-2

-2

-2

2

-2

( )

3.69H1.96

2.82-0.78

1.53-0.39

4-0.0

1.53H0.392.82H0.78

3.69-1.96

0.0-0.0


141/151


363 @2 WW ==


142/151


1

1

1

1

1

1

1

1$

1

1

1

1

1

1

1

1

.363

E

B

1

7

3

jWjW

WjW

jWjW

WW

+====

+==

==

=!!!>!Let !>n= B132=


143/151


=!(>!=(!(>!

intodecimateor

=!!!>!Let !>n=

B312

B132

Di+ide ==!(>!=(!(>!

intodecimateor

B312

Di+ide

3!2 =

17)2!1=2

1W

2

1W7)2!3 =

317)2!B=

3

1W

3

1W

2

@W

B

@W

(0)1.875

(1)0.75-0.375

(2)0.625

(3)0.75H0.375

3

@W1

@W

2)317=2)172)73>=> =nx

17)3

E7)2

A17)2

BE7)2

!oputation of nverse DFT


144/151


( ) ( )=$13.3$1.3.>7

o"ID+Tthe

)=>

3!>n=

bygivenisD+Tinverse

3

2

+

=

=

Find

WKX

The

N

k

kn

N

T?iddle !actors are negati)e po?ers ofN

W

The output is scaled -% ,


145/151


2

1W

2

1W

3

1

W

3

1

W

2

@W

B

@

W

3@W

1

@

W1

@j

2

31 =x

133

jx =

13B jx +=

@

D

D

2

1

2

1

@

3

@

3

@

3

H

-

-1 1

32 =x

2W 2W3

3

2

1

0


146/151


12 =x

33 =x

3

1W

2

1W

31 =x

1B =x

21W

3

1W

2

@W

3

@W

1

@W

B

@W

6)B

3

B

3

1H

0

1-

33 =x

3

1

W

1W

3B =x

2

1W

3

1W

@W

3

@W1

@W

B

@W

2

1

31 =x

312H2

0

2H2

2-2

2W2

W2

2=x

2

2


147/151


2

1W

2

1W

3

1

W

3

1

W

2

@W

B

@

W

3@W

1

@

W1

@j

2

21

=x

@3

jx =

@B

jx =

@

D

D

@

3

@

3

@

3

2

1

2

1

D+!,AT*) ) FR+J1+)!% FFT "DF$ FFTor F0,1UU..-1,@ere !s #e #!%e fac#or an 2,4,8,16U can e e;+ane as,


148/151


< = ( ) ( ) ( )

A&a!n !f e s+%!# #e aoe ea#!on !n#o,


149/151


f ;


150/151


2

1W

2@W

32 =x

11 =x

BB =x

@@=xD

4

3

3

6

-2

-2 M3@W

-2

-2

10

3

3

1

-2-2

2

1W

-2H2

0

2

1

3

=0>12 0,43 6&67&3 6&3 6&7&1

2o.of s#a&es N


151/151

o.of s#a&es No.Qfco*+%e; *%#!+%!ca#!ons !n eac ##erf%2o.of B##erf%!es !n eac s#a&e 2n*er of co*+%e; *%#!+%!ca#!ons !n eac s#a&e.

no.of co*+%e; *%#!+%!ca#!ons N

Documents

Digital Signal Processing 2014 pptx.pptx