Sampling of Continuous-Time Signalssip.cua.edu/res/docs/courses/ee515/chapter05/ch5-1.pdf · Sampling of Continuous-Time Signals • As indicated earlier, discrete-time signals in

1Copyright © 2001, S. K. Mitra

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Digital Processing ofDigital Processing ofContinuous-Time SignalsContinuous-Time Signals

• Digital processing of a continuous-timesignal involves the following basic steps:(1) Conversion of the continuous-timesignal into a discrete-time signal,(2) Processing of the discrete-time signal,(3) Conversion of the processed discrete-time signal back into a continuous-timesignal


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• Conversion of a continuous-time signal intodigital form is carried out by an analog-to-digital (A/D) converter

• The reverse operation of converting adigital signal into a continuous-time signalis performed by a digital-to-analog (D/A)converter


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• Since the A/D conversion takes a finiteamount of time, a sample-and-hold (S/H)circuit is used to ensure that the analogsignal at the input of the A/D converterremains constant in amplitude until theconversion is complete to minimize theerror in its representation


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Digital Processing ofDigital Processing ofContinuos-Time SignalsContinuos-Time Signals

• To prevent aliasing, an analog anti-aliasingfilter is employed before the S/H circuit

• To smooth the output signal of the D/Aconverter, which has a staircase-likewaveform, an analog reconstruction filteris used


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Complete block-diagram

• Since both the anti-aliasing filter and thereconstruction filter are analog lowpassfilters, we review first the theory behind thedesign of such filters

• Also, the most widely used IIR digital filterdesign method is based on the conversion ofan analog lowpass prototype

Anti-aliasing

filterS/H A/D D/ADigital

processorReconstruction

filter


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Sampling of Continuous-TimeSampling of Continuous-TimeSignalsSignals

• As indicated earlier, discrete-time signals inmany applications are generated bysampling continuous-time signals

• We have seen earlier that identical discrete-time signals may result from the samplingof more than one distinct continuous-timefunction


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• In fact, there exists an infinite number ofcontinuous-time signals, which whensampled lead to the same discrete-timesignal

• However, under certain conditions, it ispossible to relate a unique continuous-timesignal to a given discrete-time signals


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• If these conditions hold, then it is possibleto recover the original continuous-timesignal from its sampled values

• We next develop this correspondence andthe associated conditions


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Effect of Sampling in theEffect of Sampling in theFrequency DomainFrequency Domain

• Let be a continuous-time signal that issampled uniformly at t = nT, generating thesequence g[n] where

with T being the sampling period• The reciprocal of T is called the sampling

frequency , i.e.,

∞<<∞−= nnTgng a ),(][

)(tga

TTF 1=TF


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• Now, the frequency-domain representation of is given by its continuos-time Fourier

transform (CTFT):

• The frequency-domain representation of g[n]is given by its discrete-time Fourier transform(DTFT):

)(tga

dtetgjG tjaa ∫

∞∞−

Ω−=Ω )()(

∑∞−∞=

ω−ω = nnjj engeG ][)(


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• To establish the relation betweenand , we treat the sampling operationmathematically as a multiplication ofby a periodic impulse train p(t):

∑ −δ=∞

−∞=nnTttp )()(

)( ωjeG)( ΩjGa

×)(tga )(tgp

)(tp

)(tga


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• p(t) consists of a train of ideal impulseswith a period T as shown below

• The multiplication operation yields animpulse train:

∑ −δ==∞

−∞=naap nTtnTgtptgtg )()()()()(


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• is a continuous-time signal consistingof a train of uniformly spaced impulses withthe impulse at t = nT weighted by thesampled value of at that instant

)(tg p

)(nTga )(tga


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• There are two different forms of :• One form is given by the weighted sum of

the CTFTs of :

• To derive the second form, we note that p(t)can be expressed as a Fourier series:

where

)( ΩjGp

)( nTt −δ

∑∞−∞=

Ω−=Ω nnTj

ap enTgjG )()(

∑∞ −∞==∑∞ −∞== Ωk ek etp ktj

TkTTj

TT121 )/()( π

TT /π2=Ω


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• The impulse train therefore can beexpressed as

• From the frequency-shifting property of theCTFT, the CTFT of is given by

)(tg p

)()( 1 tgetg ak

ktjTp T ⋅

= ∑

∞

−∞=

Ω

)(tge aktj TΩ

( ))( Ta kjG Ω−Ω


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• Hence, an alternative form of the CTFT of is given by

• Therefore, is a periodic function ofΩ consisting of a sum of shifted and scaledreplicas of , shifted by integermultiples of and scaled by

( )∑∞

−∞=Ω−Ω=Ω

kTaTp kjGjG )()( 1

)( ΩjGp

)( ΩjGaTΩ T

1

)(tg p


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• The term on the RHS of the previousequation for k = 0 is the baseband portionof , and each of the remaining termsare the frequency translated portions of

• The frequency range

• is called the baseband or Nyquist band

)( ΩjGp

)( ΩjGp

22TT Ω

≤Ω ≤Ω−


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• Assume is a band-limited signal with aCTFT as shown below

• The spectrum of p(t) having asampling period is indicated below

)(tga)( ΩjGa

TT Ω

π= 2)( ΩjP


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• Two possible spectra of are shownbelow

)( ΩjGp


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• It is evident from the top figure on theprevious slide that if , there is nooverlap between the shifted replicas ofgenerating

• On the other hand, as indicated by the figureon the bottom, if , there is anoverlap of the spectra of the shifted replicasof generating

)( ΩjGa)( ΩjGp

)( ΩjGp)( ΩjGa

mT Ω>Ω 2

mT Ω<Ω 2


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If , can berecovered exactly from by passing itthrough an ideal lowpass filter witha gain T and a cutoff frequency greaterthan and less than as shownbelow

mT Ω>Ω 2 )(tga)(tga

)( ΩjHrcΩ

mΩ mT Ω−Ω


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• The spectra of the filter and pertinentsignals are shown below


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• On the other hand, if , due to theoverlap of the shifted replicas of ,the spectrum cannot be separated byfiltering to recover because of thedistortion caused by a part of the replicasimmediately outside the baseband foldedback or aliased into the baseband

mT Ω<Ω 2)( ΩjGa

)( ΩjGa)( ΩjGa


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Sampling theorem - Let be a band-limited signal with CTFT for

• Then is uniquely determined by itssamples , if

where

)(tga0)( =ΩjGa

mΩ>Ω

)(tga)(nTga ∞≤≤∞− n

mT Ω≥Ω 2TT /2π=Ω


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• The condition is often referred toas the Nyquist condition

• The frequency is usually referred to asthe folding frequency2

TΩ

mT Ω≥Ω 2


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• Given , we can recover exactlyby generating an impulse train

and then passing it through an ideal lowpassfilter with a gain T and a cutofffrequency satisfying

∑∞−∞= −δ= n ap nTtnTgtg )()()(

)( nTga )(tga

)( ΩjHrcΩ

( )mTcm Ω−Ω<Ω<Ω


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• The highest frequency contained inis usually called the Nyquist frequencysince it determines the minimum samplingfrequency that must be used tofully recover from its sampled version

• The frequency is called the Nyquistrate

mΩ

mT Ω=Ω 2)(tga

mΩ2

)(tga


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• Oversampling - The sampling frequency ishigher than the Nyquist rate

• Undersampling - The sampling frequency islower than the Nyquist rate

• Critical sampling - The sampling frequencyis equal to the Nyquist rate

• Note: A pure sinusoid may not berecoverable from its critically sampledversion


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• In digital telephony, a 3.4 kHz signalbandwidth is acceptable for telephoneconversation

• Here, a sampling rate of 8 kHz, which isgreater than twice the signal bandwidth, isused


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• In high-quality analog music signalprocessing, a bandwidth of 20 kHz has beendetermined to preserve the fidelity

• Hence, in compact disc (CD) musicsystems, a sampling rate of 44.1 kHz, whichis slightly higher than twice the signalbandwidth, is used


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• Example - Consider the three continuous-time sinusoidal signals:

• Their corresponding CTFTs are:

)6cos()(1 ttg π=)14cos()(2 ttg π=)26cos()(3 ttg π=

)]26()26([)(3 π+Ωδ+π−Ωδπ=ΩjG)]14()14([)(2 π+Ωδ+π−Ωδπ=ΩjG

)]6()6([)(1 π+Ωδ+π−Ωδπ=ΩjG


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• These three transforms are plotted below


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• These continuous-time signals sampled at arate of T = 0.1 sec, i.e., with a samplingfrequency rad/sec

• The sampling process generates thecontinuous-time impulse trains, ,

, and• Their corresponding CTFTs are given by

π=Ω 20T

)(1 tg p)(2 tg p )(3 tg p

( ) 31,)(10)( ≤≤Ω−Ω=Ω ∑∞−∞= lll k Tp kjGjG


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• Plots of the 3 CTFTs are shown below


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• These figures also indicate by dotted linesthe frequency response of an ideal lowpassfilter with a cutoff at anda gain T = 0.1

• The CTFTs of the lowpass filter output arealso shown in these three figures

• In the case of , the sampling ratesatisfies the Nyquist condition, hence noaliasing

π=Ω=Ω 102/Tc

)(1 tg


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• Moreover, the reconstructed output isprecisely the original continuous-timesignal

• In the other two cases, the sampling ratedoes not satisfy the Nyquist condition,resulting in aliasing and the filter outputsare all equal to cos(6πt)


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• Note: In the figure below, the impulseappearing at Ω = 6π in the positivefrequency passband of the filter results fromthe aliasing of the impulse in at

• Likewise, the impulse appearing at Ω = 6πin the positive frequency passband of thefilter results from the aliasing of the impulsein at

)(2 ΩjGπ−=Ω 14

π=Ω 26)(3 ΩjG


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• We now derive the relation between theDTFT of g[n] and the CTFT of

• To this end we compare

with

and make use of

)(tg p

∑∞−∞=

ω−ω = nnjj engeG ][)(

∑∞−∞=

Ω−=Ω nnTj

ap enTgjG )()(

∞<<∞−= nnTgng a ),(][


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• Observation: We have

or, equivalently,

• From the above observation and

Tpj jGeG

/)()(

ω=Ωω Ω=

Tj

p eGjGΩ=ω

ω=Ω )()(

( )∑∞

−∞=Ω−Ω=Ω

kTaTp kjGjG )()( 1


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we arrive at the desired result given by

TkTaT

j jkjGeG/

1 )()(ω=Ω

∞

−∞=

ω ∑ Ω−Ω=

)(1 ∑ Ω−=∞

−∞=

ω

kTTaT

jkjG

)( 21 ∑ −=∞

−∞=

πω

k Tk

TaTjjG


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• The relation derived on the previous slidecan be alternately expressed as

• From

or from

it follows that is obtained fromby applying the mapping

∑ Ω−Ω= ∞−∞=

Ωk TaT

Tj jkjGeG )()( 1

Tpj jGeG

/)()(

ω=Ωω Ω=

Tj

p eGjGΩ=ω

ω=Ω )()(

Tω=Ω

)( ΩjGp)( ωjeG


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• Now, the CTFT is a periodicfunction of Ω with a period

• Because of the mapping, the DTFTis a periodic function of ω with a period 2π

)( ΩjGp

)( ωjeGTT /2π=Ω


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Recovery of the Analog SignalRecovery of the Analog Signal• We now derive the expression for the output

of the ideal lowpass reconstructionfilter as a function of the samplesg[n]

• The impulse response of the lowpassreconstruction filter is obtained by takingthe inverse DTFT of :

)( ΩjHr

)(tga^

)( ΩjHr

=Ω)( jHr

ccT

Ω>ΩΩ≤Ω

,0,

)(thr


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Recovery of the Analog SignalRecovery of the Analog Signal

• Thus, the impulse response is given by

• The input to the lowpass filter is theimpulse train :

∫∫ΩΩ−

Ωπ

Ω∞∞−π

Ω=ΩΩ= c

cdedejHth tjTtj

rr 221 )()(

∞≤≤∞−ΩΩ

= tt

tT

c ,2/)sin(

∑∞−∞= −δ= np nTtngtg )(][)()(tgp


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• Therefore, the output of the ideallowpass filter is given by:

• Substituting in theabove and assuming for simplicity

, we get

)(tga^

∑∞

−∞=−==

nrpra nTthngtgthtg )(][)()()(^

*

)2//()sin()( ttth Tcr ΩΩ=

TTc /2/ π=Ω=Ω

)(tga^ ∑

∞

−∞= −π−π=

n TnTtTnTtng

/)(]/)(sin[][


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• The ideal bandlimited interpolation processis illustrated below


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Recovery of the Analog SignalRecovery of the Analog Signal• It can be shown that when in

and for• As a result, from

we observe

for all integer values of r in the range

2/)sin()(

tt

rT

cthΩΩ=

2/Tc Ω=Ω

1)0( =rh 0)( =nThr 0≠n

)(tga^ ∑∞

−∞= −π−π= n TnTt

TnTtng/)(

]/)(sin[][

)(][)( rTgrgrTg aa ==

∞<<∞− r


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• The relation

holds whether or not the condition of thesampling theorem is satisfied

• However, for all values oft only if the sampling frequency satisfiesthe condition of the sampling theorem

)(][)( rTgrgrTg aa ==^

)()( rTgrTg aa =^

TΩ


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Implication of the SamplingImplication of the SamplingProcessProcess

• Consider again the three continuous-timesignals: , ,and

• The plot of the CTFT of thesampled version of is shownbelow

)6cos()(1 ttg π= )14cos()(2 ttg π=)26cos()(3 ttg π=

)(1 tg)(1 tg p

)(1 ΩjG p


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• From the plot, it is apparent that we canrecover any of its frequency-translatedversions outside thebaseband by passing through an idealanalog bandpass filter with a passbandcentered at

])620cos[( tk π±)(1 tg p

π±=Ω )620( k


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• For example, to recover the signal cos(34πt),it will be necessary to employ a bandpassfilter with a frequency response

where ∆ is a small number=Ω)( jHr otherwise,0

)34()34(,1.0 π∆+≤Ω≤π∆−


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• Likewise, we can recover the aliasedbaseband component cos(6πt) from thesampled version of either orby passing it through an ideal lowpass filterwith a frequency response:

=Ω)( jHr otherwise,0

)6()6(,1.0 π∆+≤Ω≤π∆−

)(2 tg p )(3 tg p


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• There is no aliasing distortion unless theoriginal continuous-time signal alsocontains the component cos(6πt)

• Similarly, from either or wecan recover any one of the frequency-translated versions, including the parentcontinuous-time signal or as thecase may be, by employing suitable filters

)(2 tg p )(3 tg p

)(3 tg)(2 tg


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Sampling of Sampling of Bandpass Bandpass SignalsSignals• The conditions developed earlier for the

unique representation of a continuous-timesignal by the discrete-time signal obtainedby uniform sampling assumed that thecontinuous-time signal is bandlimited in thefrequency range from dc to some frequency

• Such a continuous-time signal is commonlyreferred to as a lowpass signal

mΩ


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Sampling ofSampling of Bandpass Bandpass Signals Signals• There are applications where the continuous-

time signal is bandlimited to a higherfrequency range with

• Such a signal is usually referred to as thebandpass signal

• To prevent aliasing a bandpass signal can ofcourse be sampled at a rate greater thantwice the highest frequency, i.e. by ensuring

HT Ω≥Ω 2

HL Ω≤Ω≤Ω 0>ΩL


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Sampling ofSampling of Bandpass Bandpass Signals Signals

• However, due to the bandpass spectrum ofthe continuous-time signal, the spectrum ofthe discrete-time signal obtained by samplingwill have spectral gaps with no signalcomponents present in these gaps

• Moreover, if is very large, the samplingrate also has to be very large which may notbe practical in some situations

HΩ


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• A more practical approach is to use under-sampling

• Let define the bandwidth ofthe bandpass signal

• Assume first that the highest frequencycontained in the signal is an integer multipleof the bandwidth, i.e.,

HΩ

LH Ω−Ω=∆Ω

)(∆Ω=Ω MH


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• We choose the sampling frequency tosatisfy the condition

which is smaller than , the Nyquistrate

• Substitute the above expression for in

HΩ2

TΩ

MTHΩ=∆Ω=Ω 22 )(

TΩ

( )∑∞

−∞=Ω−Ω=Ω

kTaTp kjGjG )()( 1


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Sampling ofSampling of Bandpass Bandpass Signals Signals• This leads to

• As before, consists of a sum ofand replicas of shifted by integermultiples of twice the bandwidth ∆Ω andscaled by 1/T

• The amount of shift for each value of kensures that there will be no overlapbetween all shifted replicas no aliasing

( )∑∞ −∞= ∆Ω−Ω=Ω k kjjGjG aTp )()( 21

)( ΩjGp)( ΩjGa

)( ΩjGa


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Sampling ofSampling of Bandpass Bandpass Signals Signals• Figure below illustrate the idea behind

)( ΩjGa

Ω0LΩ−HΩ− HΩLΩ

)( ΩjGp

Ω0LΩ−HΩ− HΩLΩ


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Sampling ofSampling of Bandpass Bandpass Signals Signals• As can be seen, can be recovered from

by passing it through an idealbandpass filter with a passband given by

and a gain of T• Note: Any of the replicas in the lower

frequency bands can be retained by passing through bandpass filters with

passbands , providing a translation to

lower frequency ranges

)(tga)(tgp

HL Ω≤Ω≤Ω

)(tgp)()( ∆Ω−Ω≤Ω≤∆Ω−Ω kk HL

11 −≤≤ Mk

Documents

Sampling of Continuous-Time Signalssip.cua.edu/res/docs/courses/ee515/chapter05/ch5-1.pdf · Sampling of Continuous-Time Signals • As indicated earlier, discrete-time signals in