DSP Chapter4

7/30/2019 DSP Chapter4

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Chapter 4. Sampling of

Continuous-Time Signals

. n ro uc on

4.1 Periodic Sampling4.2 Frequency Representation of Sampling

. -

4.4 Changing the Sampling Rate

4.5 Digital Processing of Analog Signals

DSP 4-1


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4.0 Introductionues on o answer: ow o approx ma e a con nuous

(analog) linear system by a digital system?

Signals: continuous-time discrete-time

o a ons:

time-domain

frequency-domain

)(txc ][nxj

eX

Systems: continuous-time discrete-time

-

c

frequency-domain

tc

)( jH

n

jeH

DSP 4-2


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4.1 Periodic Sampling-

continuous-time signal is through periodic sampling.

T is the sampling period

.-, = nnxnx c

is the sampling frequency (radians per second)

An ideal continuous-to-discrete-time C/D converte

s=

Ts /2=

DSP 4-3


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Mathematical Representation of Sampling

Conversion

from impulse

)(ts

train to discrete-

time sequence

c c)(txs

train)impulseperiodic(the)()(

=

=n

nTtts

n)(modulatio)()()()()(

==ccs nTttxtstxtx

property)(sifting)()()(

= cs nTtnTxtx

DSP 4-4

=n


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Periodic Sampling Examples

DSP 4-5


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4.2 Frequency-Domain Representation

of Sampling: Time-Domain

continuous-time signals, obtaining

= cs tstxtx )()()(

= c nTttx )()(

==n

c

DSP 4-6


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Frequency-Domain Representation

)2(where)(2)()()( kjSnTtts ss

===

Since

kn ==

)(*)(21)()()()( == jSjXjXtstxtx cscs

= kXX 2*1

=

=

kscs

kX

T

1

2

DSP 4-7

=kT


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Observations of Frequency-Domain

Representation of Sampling

s equa on prov es e rea ons p e ween e ourer

transform of continuous-time signal and discrete-time signal

= scs kjXT

jX ))((1

)(

consists of eriodicall re eated and scaled co ies of the

=

XFourier transform of .

The copies of are shifted by integer multiples of the

s

)(i.e.,),( jXtx cc)( jXc

sampling frequency .

All copies of replicated spectrums are superimposed to produce

s

DSP 4-8

.


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DSP 4-9


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Sampling Rate and Bandwidth

-

NjX >= ,0)(There is no overlap between replicated spectrums, whenwe have the sam lin rate as followin

That means we CAN reconstruct the continuous-time signal withNs > 2

an ideal low-pass filter.

There will be aliasing distortion, or aliasing when

That means we CANNOT reconstruct the continuous-time signalfrom its samples.

Ns

DSP 4-10


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How to Reconstruct a Signal?

Ideal low-pass Filtering

DSP 4-11


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How to Reconstruct a Signal? (Cont'd)

sampling

-

Ideal low-passfiltering

Ideal low- ass Filterin Reconstructed Si nal

DSP 4-12


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Sampling and Reconstruction Example

ven a s gna

What is the Fourier transform of the given signal?

ttxc 0cos=

se e u er equa on, we now a

tjtjeettx 00

1cos +==

According to continuous Fourier transform, we know2

Therefore, the Fourier transform of the given signal is

)(2)()( 00 == jXetx tj

++= X

DSP 4-13


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(No Aliasing)

Original Signal

Reconstructed Signal

teetxtjtj

cos1

00 =+=

DSP 4-14

2


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(With Aliasing)

Original Signal

Reconstructed Signal

teetxtjtj ss cos

1 )()( 00 =+=

DSP 4-15

2


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Nyquist Sampling Theorem

uppose a s an - m e o a requency

interval , i.e.,

)()( aa Xtx

[ ]NN ,

Then x t can be exactl reconstructed from e uidistant

NX = for0)(

samples , if)/2()(][ sasad nxnTxnx == ,22 Ns

sT

=

,

sampling frequency (radians per second), is referred

ssT = /2 s

N

to as the Nyquist frequency, and is called the Nyquist

rate.

N2

DSP 4-16


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How to obtain discrete-time Fourier

transform (DTFT)?

ven e sampe s gna as

=

Since we have the following continuous-time Fourier

=ncs

transform (CTFT) pair ( ) nTjenTt

Thus we have the continuous-time Fourier transform of the

sampled signal as

=

=n

Tnj

cs enTxjX )()(

DSP 4-17


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transform (DTFT)? (Cont'd)

nce we now e rea ons p e ween e sampe

signal and the discrete-time sequence)(nTxc ][nx

We also have the DTFT of is defined as

nxnx c=

][nx

= njj enxeX ][)(

By comparing with

=

= Tnjcs enTxjX )()(

DSP 4-18

How to obtain discrete time Fourier


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transform (DTFT)? (Cont'd)

As we compare the following two equations

=

=n

enxe ][=

=n

cs enx

).()()(T

s eXeXjX ===

21 k 1

=

= TTTe

k

c s

k

csT

==

)( T=

DTFT representation of sampling !

is simply a frequency-scaled version of with the frequencyscaling specified by . This scaling is a normalization of the

)( jXsT=

)( j

eX

DSP 4-19

for .s

s

2= )( jeX


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Example 4.1 (Without Aliasing)-

with sampling period T=1/6000.

Continuous-time Fourier transform

cosxc =

)( jXs Discrete-time Fourier transform

Problem Analysis)( jeX

Fourier transform of the original signal

)4000()4000()( ++=jXc.40000

=

Sampling frequency

Fourier transforms of the sampled signal

.12000/2 == Ts

= scs kjXjX ))((1

)(

= cj

kjXeX ))2

((1

)(

DSP 4-20

= =


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Example 4.1 (Cont'd)

)()/( TT =)( T=

DSP 4-21


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Example 4.2 (With Aliasing)-



cosxc =

)( jXs Discrete-time Fourier transform


Fourier transform of the original signal)16000()16000()( ++=jXc.160000

=

Sampling frequency

Fourier transforms of the sampled signal are exactly same as the

.12000/2 == Ts

prevous one, w y

===

DSP 4-22


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k=1 k=-1 k=0k=2k=-2

=

)()/( TT =)( T=

160000 =

DSP 4-23


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Example 4.3 (with Aliasing)-



cosxc =

X Discrete-time Fourier transform


Fourier transform of the original signal

)4000()4000()( ++=jXc

Sampling frequency

The discrete-time Fourier transform is the same as previous one.

.3000/2 == Ts

y

)3/2cos()1500/10002cos()1500/4000cos( nnnn =+=

DSP 4-24


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k=1 k=-1 k=0

=

k=2k=-2

)( T= )()/( TT =

DSP 4-25


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4.3 Reconstruction of a Band-limited

Signal from Its Samples,

modulated impulse train is filtered by an appropriate low-pass

filter, then the Fourier transform of the filter output will beidentical to the Fourier transform of the original signal.

Given a sequence of samples x[n], we form the impulse train

==

n

s nTtnxtx )(][)(

If the impulse train is the input to an ideal low-pass continuous-

time filter with impulse response)(

thr

=

=

=

==n

rr

n

rsr nTthnxthnTtnxthtxtx )(][)(*)(][)(*)()(

DSP 4-26


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4.3.1 Ideal Reconstruction System

DSP 4-27


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Ideal Reconstruction System (Cont'd)

==TTt

thr

sinc/

)(

DSP 4-28


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Ideal Reconstruction System (Cont'd)

)(][)(nTthnxtx

rr=

n =

==tTt

thr

sinc)/sin(

)(

=

TnTtnxtx

]/)(sin[

If and then

= n TnTt /)(

nTxnx = =we have

c

txtx =

c

DSP 4-29


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Ideal Band-limited Interpolatione ea ow-pass er n erpo a es e ween e mpuses

of to construct a continuous-time signal][nx

=r

TnTtnxtx

]/)(sin[][)(

If there is no aliasing, the ideal low-pass filter interpolates

=

correct reconstruction between the samples.

However, the ideal low-pass filter has infinite length whichis not realizable in practice. Finite length low-pass

filtering wil l result in some reconstruction error.

DSP 4-30


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Ideal Band-limited Interpolation (Cont'd)

-

Sampled signalReconstructed signal

DSP 4-31


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Ideal D/C Converter

DSP 4-32


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Ideal D/C Converter (Cont'd)

seen in the frequency-domain.

TnTt ]/)(sin[ntnxtx rn

r == =

=n

rTnTt

nxx/)(

= Tnj

Linearity of continuous-timeFourier transform

=nrr

Time shifting leads to anexponential factor in theFourier transform

).()()( Tjrr eXjHjX=

Discrete-time Fourier

DSP 4-33


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Can you get the original signal back?e ea ow-pass er se ec s e ase pero o e resu ng

periodic Fourier transform and compensates for the)(Tj

eX

.

If the sequence x[n] has been obtained by sampling a band-

limited si nal at the N uist rate or hi her the reconstructed

signal will be equal to the original band-limited signal.

If there is aliasing during the sampling, the reconstructed signal

will be distorted, see Examples 4.2 and 4.3.

In any case, the output of the ideal D/C converter is alwaysband-limited to at most the cut-off frequency of the low-pass

filter, which is taken to one-half the sampling frequency.

DSP 4-34


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4.3.2 Zero-order Hold D/A Conversion-

digital output is carefully transformed into analog form.

t tje d

aa

Comments:

ompensa on w e er or - no o .

The D/A block is not filtering - it is weighting.

d e )(a)(tga

DSP 4-35

.a


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Impulse Response of Zero-order Hold,

)(tg a

1

==

n

sdda ngnxy

sT0

It holds at a constant level over each sampling period.][nyd

d

1

a

1

. .

0 1 2 3-2 -1 n-3 0 1 2 3-2 -1 n-3

DSP 4-36

Note: Z.O.H. introduces high frequencies, see sharp edges.


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Frequency Response of ZOH.

2/)2/sin()( s

Tjs

a

eT

jG=

The high frequencies in the reconstructed signal (sharp steps)

are introduced from side-lobes as follows.

Ideal)(aG

Z.O.H

2

2

DSP 4-37

ss s s

Frequency response of Z.O.H


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Compensation of ZOH

shit of T/2 which cannot be compensated and usually neglected.

The magnitude response can be compensated as follows.


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Physical Configuration for

ZOH D/A Conversion

reconstruction filter is shown as follows.

DSP 4-39


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4.4 Changing the Sampling Rate

s o en necessary o c ange e samp ng ra e o a scre e-

time signal, i.e., to obtain a new discrete-time representation of

- .

''' cc

rate that involve discrete-time operations.

]['][ nxnx

DSP 4-40

S li R t Ch E l


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Sampling Rate Change Examples

(Down-sampling)

What happened during

DSP 4-41

-

4 4 1 Sampling Rate Reduction by an


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4.4.1 Sampling Rate Reduction by an

Integer Factor (Down-sampling)

"sampling it" by defining a new sequence

).(][][ nMTxnMxnx cd ==

DSP 4-42


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Frequency Representation of Down-Sampling [W-D]

nxnx c=

=

j k 21

= kc

TTT

==, c

=jr

XeX 21

uestions: what is the relationshi between them?

= rc

TTT

)()( jdj

eXeX

DSP 4-43

Frequency Representation of


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Down-Sampling (Cont'd)

=c

j

d

rjXeX

21)(

ris still an interger ranging

from -inf and inf

=r

)( kMir +=,


45/66



=

=

=

1

0

2211)(

M

i k

c

j

dT

k

MT

ijX

TMeX

We know that

CTFT)from(DTFT21

)(

= k

jXeX cj

21)(

= k

jXeX cj

=k =k

( )

=

=

k

c

Mij

T

k

MT

ijX

TeX

221/)2(

Therefore, we have

=1

/)2(1M

Mrjj eXeX

DSP 4-45=0r



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=

1/)2(1

MMrjj

=0r

jj

can be composed ofM copies of the periodic Fouriertransform , frequency scaled by M and shifted by integer

.d

)(

jd eX)( jeX

mu pes o .

is periodic with period 2.)( jd eX

= limited)-(band,0)( N

jeX

DSP 4-46

an e-narrowN


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NT = 4/2

Ns = 4

2/ == T

DSP 4-47

4-4



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Down-Sampling: Example

DSP 4-48


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Down-sampling after Pre-filtering

- ,the band-width of signal x[n] prior to down-sampling.

- -cut-off frequency /M.

DSP 4-49


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Down-sampling after Pre-filtering (Example)

Mnn /sinsinnn

n clp

==

DSP 4-50

4.4.2 Increasing the Sampling Rate


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4.4.2 Increasing the Sampling Rate

by an Integer Factor

increase by a factor ofL (up-sampling).

==

otherwise,0

,][nxe

For example, ( )4321054321][

==

nnx

( )

2and9876543210

0504030201][

==

=

Ln

nxe

Or equivalently,

=

DSP 4-51

=k

e

Sampling Rate Change Examples


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p g g p

(Up-sampling) Demo

-

DSP 4-52


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Frequency Representation of Up-Sampling

e ourer rans orm o e up-sampe s gna s

][][)( n

n k

e ekLnkxeX

=

= =

)(][Lj

k

kLj

eXekx

== =

We can see the Fourier transform of the output of the

-transform of the input, i.e, is replaced by L.

DSP 4-53



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(Example)

DSP 4-54


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General System for Up-sampling

, -

therefore considered to be synonymous with interpolation.

LnnLnh clp

/][

=

=

DSP 4-55

a n


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Ideal Low-pass Filtering after Up-sampling

,interpolation formula with an ideal low-pass filter as

==

==kk

lpiLkLn

kxkLnhkxnx/)(

][][][][

The impulse response of the low pass filter has properties

Lnnhi

/][

=

,....3,2,,0][ LLLnnhi

i

==

Thus for the ideal low-pass interpolation filter, we have

==

DSP 4-56

,...,,,i



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and Ideal Low-pass Filtering (Example)

DSP 4-57


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Linear Interpolation after Up-sampling

,/1 LnLn

= otherwise,0

lin

DSP 4-58



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(Example)

2

lin

)2/sin(

)2/sin(1)(

=

L

L

eH j

DSP 4-59



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(Example, Cont'd)

1]0[hi =

=

,/1 LnLn

,....3,2,,0][ LLLnnhi == otherwise,0n

= kLnhkxnx

,...3,2,,/ LLLnLnxnx ==

=k

The amount of distortion in the intervening samples can be

gauged by comparing the frequency response of the linearinterpolator with that of the ideal low-pass interpolator, as

2

)2/sin(1 =

Lj

DSP 4-60

n

)2/sin(

L

4 5 Di it l P i f A l Si l


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4.5 Digital Processing of Analog Signals

Sample at a faster rate - perhaps not possible (why?).

Use an anti-aliasin filter.

DSP 4-61

Ho to red ce aliasing?


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How to reduce aliasing?

- - is applied to the continuous signal prior to sampling.

The idea is sam le: remove the hi h fre uencies . The ideal frequency response of the anti-aliasing filter is an ideal

low-pass filter as ,2/s>

where the cut off >=

cF

0;1 sc

T=


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Anti-aliasing: Formulation

- ,

mm

221sasad nnxnx = = = s

a

m s

a

s

dTTT

e

The repeated spectra will not fold or overlap.

If is an ideal LPF with cutoff , then

)()( aa FX

)(aF c

=

=

ca

aa

j

d

XX

mXeX

);(where

21)(

Usually, an ideal LPF cannot be realized and must be

= csms

DSP 4-63approximated.

Anti aliasing E ample 1


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Anti-aliasing: Example-1

)(aX

Analog Signal Spectrum

0

)()( aa FX

Anti-aliased Spectrum

)(

j

d eX

0

c c

Sampled Signal Spectrum

2/


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Anti-aliasing: Example-2

- -with aliasing

with anti-aliasing

DSP 4-65

Anti aliasing: Digital Filter Output


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Anti-aliasing: Digital Filter Output

eca e overa sys em o n eres :

j

][nxd ][nyd

da a

The res onse of the di ital filte)(

aH

jeY

jeH

)(21

)( j

da

j

d eH

T

mX

T

eY

= without anti-aliasing

)(221

)( j

daa

j

d eHm

Fm

XeY

= with anti-aliasing

DSP 4-66

m sss =

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