1 DSP Fundamentals

8/20/2019 1 DSP Fundamentals

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Fundamentals of Digital Signal Processing


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Fourier Transform of continuous time signals

{ } ∫

+∞

∞−

−

== dt et xt x FT F X Ft j π 2

)()()(

{ } ∫ +∞

∞−

== dF e F X F X IFT t x Ft j π 2)()()(

with t in sec and F in Hz (1/sec).

Examples:

( ){ } ( )0 0 0

sinct T

FT rect T FT =

{ } ( )0

2 0

F F e FT

t F j

−=δ

π

( ){ } ( ) ( )021

021

02cos F F e F F et F FT j j ++−=+ − δ δ α π α α


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Discrete Time Fourier Transform of sampled signals

{ } ∑+∞

−∞=

−==n

fn jen xn x DTFT f X π 2][][)(

{ } ∫ −== 2

1

21

2)()(][ df e f X f X IDTFT n x fn j π

with f the digital fe!"enc# (no dimensions).

Example:

{ }02 0( ) j f n

k

DTFT e f f k π

δ

+∞

=−∞

= − −∑

since$ "sing the %o"ie &eies$

2( ) j nt

k n

t k e π δ +∞ +∞

=−∞ =−∞

− =∑ ∑


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Property of DTFT

' f is the digital fe!"enc# and has no dimensions

' is periodic with peiod f 1.)1()( += f X f X

21

21


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t F j

et x 02

)( π

=

n j

s

s F

F

enT xn x02

)(][ π

==

s s T F /1=

Sampled Complex Exponential: no aliasing

0 F F

)( F X

f

( ) X f

2

1−

1

2

1. o *liasing2

0 s F F <

2

s F −2

s F

0 f

s F

F f 00 =digital fe!"enc#


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t F j

et x 02

)( π

=

n j

s

s F

F

enT xn x02

)(][ π

==

s s T F /1=

Sampled Complex Exponential: aliasing

0 F F

)( F X

f

( ) X f

2

1−

1

2

2. *liasing 02

s F F >

2

s F −2

s F

0 f

0 00

s s

F F f round

F F

= − ÷

digital fe!"enc#


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t F jet x 02)( π = 02[ ] ( ) j f n s x n x nT e π = =

s s T F /1=

Mapping between Analog and Digital Freuency

−=

s s F

F round

F

F f 000


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Example

t jet x 10002)( π

= kHz F s +=

,hen:

' analog fe!"enc#

' %,:

' digital fe!"enc#

'-,%,: fo

)1000()( −= F F X FT δ

Hz F 10000 =

( ) ( )0 0 1 1 10 + + + s s F F

F F f round round = − = − =

( )+1)( −= f f X DTFT δ 2

1


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Example

t jet x 20002)( π =kHz F s +=

,hen:

' analog fe!"enc#

' %,:

' digital fe!"enc#

'-,%,: fo

)2000()( −= F F X FT δ

Hz F 20000 =

( )+1)( += f f X DTFT δ 2

1


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Example

t j jt j j eeeet t x π π π π π π 0001.02

10001.0

2

1)1.0000cos()( −−

+=+=kHz F s +=

,hen:

' analog fe!"encies

' %,:

' digital fe!"encies

'-,%,

)000()000()( 1.0211.0

21 ++−= − F e F e F X j j FT δ δ

π π

0 1000 $ 000 F Hz F Hz = = −

21


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!inear Time "n#ariant $!T"% Systems and &'Transform

][n x ][n y][nh

f the s#stem is , we comp"te the o"tp"t with the con3ol"tion:

∑

+∞

−∞=

−==m

mn xmhn xnhn y ][][][4][][

f the imp"lse esponse has a finite d"ation$ the s#stem is called %5

(%inite mp"lse 5esponse):

][][...]1[]1[][]0[][ N n x N hn xhn xhn y −++−+=


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{ } ∑+∞

−∞=

−==n

n z n xn x Z z X ][][)(

('Transform

%acts:

)()()( z X z H z Y =

][n x ][n y)( z H

%e!"enc# 5esponse of a filte:

f j

e z

z H f H π 2)()(=

=


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Digital Filters

)( z H

][n x ][n y

deal ow 6ass %ilte

f 21

21−

)( f H

f 21

21−

)( f H ∠

P f

P f

pass7and

constant magnit"dein pass7and8

8 and linea phase

A


14/26

"mpulse )esponse of "deal !PF

∫ ∫ −− == P

P

f

f

fn j fn j

idea df Aedf e f H nh

π π 2221

21 )(][

*ss"me zeo phase shift$

⇒ ( )n f sinc Af nh P P idea 22][ =

,his has nfinite mp"lse 5esponse$ non ec"si3e and it is non9

ca"sal. ,heefoe it cannot 7e ealized.

-50 -40 -30 -20 -10 0 10 20 30 40-0.05

0

0.05

0.1

0.15

0.2

n

fp=0.1

[ ]h n

n

0.1

1

P f

A

==


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*on "deal "deal !PF

,he good news is that fo the deal 6%

0][lim =±∞→ nhidea n

n

][nh

!− !

n !2 !

][nh


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Freuency )esponse of t+e *on "deal !PF

P f "T#P f

passstop stop

f

tansition egion

attenuation

ri$$e

)( f H 11 δ +

11 δ −2δ

6% specified 7#:

' pass7and fe!"enc#

' pass7and ipple o

' stop7and fe!"enc#

' stop7and atten"ation o

P f

1δ d% & P 11

1

1

10log20 δ δ

−+=

"T#P f

2δ

d% & '" δ

10log20=


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est -esign tool fo %5 %iltes: the E!"iipple algoithm (o 5emez). t

minimizes the maxim"m eo 7etween the fe!"enc# esponses of the

ideal and act"al filte.

1 f 2 f 21

attenuation

ri$$e

)( f H 11 δ +

11 δ −

2δ

[ ] [ ] [ ]( )1 2 + + 1 2$ 0$ $ $ / $ 1$1$0$0 $ $h fir$m N f f f f ( (=

imp"lse esponse

[ ]][]$...$0[ N hhh =

1 f 2 f 21

+ = f 0

inea ntepolation

1

1/ (δ

2/ (δ


18/26

,he total imp"lse esponse length N ;1 depends on:

' tansition egion

' atten"ation in the stop7and

1 f 2 f

)( f H

2δ

12 f f f −=∆

( ) N f 1

< 22)(log20 210 δ

∆

Example:

we want

6ass7and: +=Hz

&top7and: +.>=Hz

*tten"ation: ?0d&ampling %e!: 1> =Hz

,hen: fom the specs

@e detemine the ode the filte

+01

0.1>0.+>.+ ==∆ − f

2+0<22

?0 =× N


19/26

%e!"enc# esponse

0 0.1 0.2 0.3 0.4 0.5-100

-80

-60

-40

-20

0

20magnitude

digital frequency

d B

0 0.1 0.2 0.3 0.4 0.5-120

-100

-80

-60

-40

-20

0

20magnitude

digital frequency

d B

2

A


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Example: ow 6ass %ilte

0 0.1 0.2 0.3 0.4 0.5-120

-100

-80

-60

-40

-20

0

20 magnitude

digital frequency

d B

6ass7and f 0.2

&top7and f ) 0.2> with atten"ation 0d

Bhoose ode 0/(224(0.2>90.20))+C

( ) H f

f

Amost *+d%,,,


21/26

Example: ow 6ass %ilte

6ass7and f 0.2

&top7and f ) 0.2> with atten"ation 0d

Bhoose ode 0 D +C

( ) H f

f

0 0.1 0.2 0.3 0.4 0.5-80

-70

-60

-50

-40

-30

-20

-10

0

10magnitude

digital frequency

d B

FGGG

l %5 %il f 7i % 5


22/26

eneal %5 %ilte of a7ita# %e!"enc# 5esponse

]$...$$[

]$...$$$0[

10

21

-

-

H H H H

f f f f

=

=

01 f 2 f + f 1− - f 2

1= - f

0 H 1 H 2 H

+ H 1− - H - H

@eights fo Eo:

1(

2( 2/)1( + - (

]$...$$[ 2/)1(21 += - ((((

,hen appl#:

( )$ / $ $ - h fir$m N f f H (=8 and alwa#s chec= fe!"enc# esponse if it is what #o" expectG

E l


23/26

Example:( ) 1/ sinc( ) H f f = fo 0 0.2 f ≤ <

( ) 0 H f = 0.2> 0.> f ≤ <

0 0.2 0.2> 0.> f

f$)+.+/+0.+/'1 2 3ector of $ass4and fre5uencies

fs)[0.2>$0.>]I 2 sto$4and fre5uencies

-)[1./sinc(fp)$ 0$ 0]I 2 desired ma6nitudes Df 0.2>90.2I 2 transition re6ion

N ceil(*/(224 Df ))I 2 first 6uess of order

hfipm( N $ [ f$$ fs]/0.>$ - )I 2 im$use res$onse

0 A d%=


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0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1

1.2

1.4magnitude

digital frequency

0 0.1 0.2 0.3 0.4 0.5-90

-80

-70

-60

-50

-40

-30

-20

-10

0

10

not 3e# good heeG

d%

+C N =


25/26

,o impo3e it:

1. ncease ode

2. *dd weights

0 0.2 0.2> 0.> f

0 A d%=

1( = 0.2( =

w[14ones(1$length(fp)/2)$ 0.24ones(1$ length(fs)/2)]I

hfipm($ [fp$ fs]/0.>$J$w)I

magnitude


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0 0.1 0.2 0.3 0.4 0.50

0.2

0.4

0.6

0.8

1

1.2

1.4magnitude

digital frequency

0 0.1 0.2 0.3 0.4 0.5-160

-140

-120

-100

-80

-60

-40

-20

0

20

100 N =d%

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1 DSP Fundamentals