Chapter 6 Sampling and Reconstruction of Signalscwlin/courses/dsp/notes/DSP... · 2010-05-12 · 1 Sampling and Reconstruction of Signals 清大電機系林嘉文 [email protected]

1

Sampling and Reconstruction of Signals

清大電機系林嘉文[email protected]

Chapter 6

Ideal Sampling and Reconstruction of Continuous-Time Signals

2010/5/12 Introduction to Digital Signal Processing 2

� To process a continuous-time signal using digital signal processing techniques, it is necessary to convert the signal into a sequence of numbers.

� : a function of sampling the analog signal

� : period of a discrete-time signal

� If the spectrum of the analog signal can be recovered from the spectrum of the discrete-time signal, there is no loss of information.

( )ax t( )x nT

( ) ( ),ax n x nT= n−∞ < < ∞

2



� If is an periodic signal with finite energy, its (voltage) spectrum is given by the Fourier transform relation

� Whereas the signal can be recovered from its spectrum by the inverse Fourier transform

� Infinite frequency range is necessary if the signal is not bandlimited.

2( ) ( ) j Fta aX F x t e dtπ∞ −

−∞= ∫

( )ax t

2( ) ( ) j Fta ax t X F e dFπ∞

−∞= ∫

F−∞ < < ∞( )ax t



� The spectrum of a discrete-time signal , obtained by sampling , is given by the Fourier transform relation

� or, equivalently,

( )x n( )ax t

( ) ( ) j n

n

X x n e ωω∞

−

=−∞

= ∑

2( ) ( ) j fn

n

X f x n e π∞

−

=−∞

= ∑

3



� The sequence can be recovered from its spectrum or by the inverse transform

� A relationship between the independent variables and in the signals and , respectively.

� This relationship in the time domain implies a corresponding relationship between the frequency and in and , respectively.

( )x n( )X f

( )X ω

1/2 2

1/2

1( ) ( ) ( )

2sj n j fnx n X e d X f e df

π ω π

πω ω

π − −= =∫ ∫

t n( )ax t ( )x n

F f( )aX F ( )X f

s

nt nT

F= =



� This relationship between the variables and of the corresponding analog and discrete-time signals, respectively. That is,

� Simple change in variable obtain the result

2 /( ) ( ) ( ) sj nF Fa ax n x nT X F e dFπ∞

−∞≡ = ∫

F f

s

Ff

F=

1/2 2 /2

1/2( ) ( ) sj nF Fj fn

aX f e df X F e dFππ∞

−−∞

=∫ ∫

2 //2 2 /

/2

1( ) ( )

sss

s

j nF FF j nF FaF

s

X F e dF X F e dFF

ππ ∞

− −∞=∫ ∫

4



� The integration range of this integral can be divided into an infinite number of intervals of width .

� Where we have used the periodicity of the complex exponential, namely,

sF2 /( 1/2)2 /

( 1/2)( ) ( )

sss

s

j nF Fk Fj nF Fa ak F

k

X F e dF X F e dFπ

π∞∞ +

−∞ −=−∞

= ∑∫ ∫2 //2 /2 2 /

/2 /2( ) ( )

ss ss

s s

j nF FF F j nF Fa s a sF F

k k

X F kF e dF X F kF e dFπ

π∞ ∞

− −=−∞ =−∞

= − = −

∑ ∑∫ ∫

2 ( )/ 2 /s s sj n F kF F j nF Fe eπ π+ =



� Between the spectrum or of the discrete-time signal and spectrum of the analog signal, we have

� The spectrum of the discrete-time signal is periodic with period or .

( )X f

( ) ( )s a sk

X F F X F kF∞

=−∞

= −∑

[ ]( ) ( )s a sk

X f F X f k F∞

=−∞

= −∑

( )X F

( )aX F

1pf =( )X f

p sF F=

5



If the sampling frequency is selected such that , where is the Nyquist rate, then

Given the discrete-time signal with the spectrum ,

sF2sF B≥ 2B

( ) ( ),s aX F F X F= / 2sF F≤

( )x n( )X F

1( ),

( )

0,sa

X FFX F

=

/ 2

/ 2s

s

F F

F F

≤

>



� Assume that , we have

� Where and where is the sampling interval. The formula for reconstructing the analog signal

from its samples is called ideal interpolation formula.

2sF B≥

/2 2 / 2

/2

/2 2 ( / )

/2

1( ) ( )

1( )

sin( / )( )( )

( / )( )

ss

s

ss

s

F j Fn F j Fta F

ks

F j F t n F

Fks

ak

x t x n e e dFF

x n e dFF

T t nTx nT

T t nT

π π

π

ππ

∞−

−=−∞

∞−

−=−∞

∞

=−∞

=

=

−=

−

∑∫

∑ ∫

∑

( ) ( )ax n x nT= 1/ sT F=

( )ax t

6



� An interpolation function is

� At , the interpolation function is zero except at . Evaluated at is simply the sample

.

t kT= ( )g t nT−

k n= ( )ax t t kT=( )ax kT

sin( / )( )

( / )

T tg t

T t

ππ

=



� Sampling Theorem.

� In practice we use a finite number of samples of the signal and deal with finite-duration signals.

� When aliasing occurs due to too low a sampling rate, the effect can be described by a multiple folding of the frequency axis of the frequency variable for the analog signal.

A bandlimited continuous-time signal, with highest frequency (bandwidth) hertz, can be uniquely recovered from its samples provided that the sampling rate samples per second.

B2sF B≥

F

7



� If , the shifted replicas of overlap.

� The overlap that occurs within the fundamental frequency range

The frequency is called the folding frequency

2Fs B<( )aX F

/ 2 / 2s sF F F− ≤ ≤

/ 2sF



� Due to the folding of the frequency axis, the relationshipis piecewise linear

� Antialiasing filter ensures that frequency components of the signal above are sufficiently attenuated so that, if aliased, they cause negligible distortion on the desired signal.

sF fF=

F B≥

8



� Time-domain and frequency domain functions , ,, and relationships for sampled signals.

( )ax t ( )x n( )aX F ( )X f



� Example : Aliasing in Sinusoidal Signals

0 02 20

1 1( ) cos2

2 2j F t j F t

ax t F t e eπ ππ −= = +

Reconstructed signal

Sampling frequency range

0/ 2s sF F F< <

0ˆ ( ) cos2 ( )a sx t F F tπ= −

9



0 02 20

1 1( ) cos2

2 2j F t j F t

ax t F t e eπ ππ −= = +

Reconstructed signal

Sampling frequency range

0 3 / 2s sF F F< <

0( ) cos2 ( )a sx t F F tπ= −

In both case aliasing has occurred.



� Example : Sampling and Reconstruction of a Nonbandlimited Signal

2 2

2( ) ( ) ,

(2 )A t

a a

Ax t e X F

A Fπ− ℑ= ←→ =

+0A >

Sampling frequency 1/sF T=

( ) ( ) ( ) ,AT n nATax n x nT e e− −= = = n−∞ < < ∞

The discrete-time Fourier transform

2

2

1( ) ,

1 2 cos 2 ( / )s

aX F

a F F aπ−

=− +

ATa e−=

10



� Since is not bandlimited, aliasing cannot be avoided.( )aX F



The Spectrum of the reconstructed signal is given byˆ ( )aX F ˆ ( )ax t

( ),ˆ ( )0,a

TX FX F

=

/ 2sF F

otherwise

≤

11

Discrete-Time Processing of Continuous-Time Signals


� Many practical applications require the discrete-time processing of continuous-time signals.

� Select the bandwidth of the signal to be processed, since the bandwidth determines the minimum sample rate.

� The prefilter ensures that the bandwidth of the signal to be sampled is limited to the desired frequency range.

� The amount of signal distortion due to aliasing is negligible

� Reduce the additive noise power out of band



� Ideal A/D converter is a linear time-varying system

� Scales the analog spectrum by a factor

� Creates a periodic repetition of the scaled spectrum with period

1/sF T=

sF

( ) ( ) ( )a t nT ax n x t x nT== =

1( ) ( )a s

k

X F X F kFT

∞

=−∞

= −∑

Time domain

Frequency domain

12



� Ideal D/A converter is also a linear time-varying system with a discrete-time input and a continuous-time output.

� Scales the input spectrum by a factor

� Removes the frequency components for

�

Time domain

( ) ( ) ( )a an

y t y n g t nT∞

=−∞

= −∑

Frequency domain

?

1/sF T=

/ 2sF F>



� To obtain a frequency-domain description, evaluate the Fourier transform of the output signal

� Interpolation function of the ideal D/A converter

� Fourier transform of the output signal

,sin( / )( ) ( )

0,/ a

Tt Tga t G F

t T

ππ

ℑ = ←→ =

/ 2sF F

otherwise

≤

2

2

( ) ( ) ( )

( ) ( )

( ) ( )

j Fta

n

j FnTa

n

a

Ya F y n g t nT e dt

y n G F e

G F Y F

π

π

∞ ∞ −

−∞=−∞

∞−

=−∞

= −

=

=

∑ ∫

∑

13



� Given a continuous-time LTI system

� Where is the continuous-time system

� If is not bandlimited or it is bandlimitied but it is impossible to find such a system due to thee presence of aliasing.

2 ,sF B<

( ) ( ) ( )a a ay t h x t dtτ τ∞

−∞= −∫�

( ) ( ) ( )a a aY F H F X F=�

( )H F

( )aH F

( )ax t



� If is bandlimited and the output of system is

� To assure that the discrete-time system is

� The cascade connection of the A/D converter, LTI system, and the D/A converter is equivalent to a continuous-time LTI system.

2 ,sF B>( )ax t

( ) ( ),( ) ( ) ( ) ( )

0,a

a a

H F X FY F H F X F G F

= =

/ 2

/ 2s

s

F F

F F

≤

>

( ) ( ),a ay t y t= �

( ),( )

0,aH F

H F

=

/ 2

/ 2s

s

F F

F F

≤

>

14



� Example :Simulation of an analog integrator

Input-output relation( )

( ) ( )aa a

dy tRC y t x t

dt+ =

Frequency response( ) 1

( ) ,( ) 1 /

aa

a c

Y FH F

X F jF F= =

+

1

2Fc

RCπ=



( ) ( ) ( ) ( )AT nah n h nT A e u n−= =Sampling the continuous-time impulse response

System function1

1( ) ( )

1AT n n

ATH x A e z

e z

∞− −

− −−∞

= =−∑

( ) ( 1) ( )ATy n e y n Ax n−= − +Discrete-time system function

15



� Example : Ideal bandlimited differentiator

� If , we can define an ideal discrete-time differentiator

Ideal continuous-time differentiator( )

( ) aa

dx ty t

dt=

( )( ) 2

( )a

aa

Y FH F j F

X Fπ= =Frequency response function

Ideal bandlimited differentiator2 ,

( )0,a

j FH F

π=

c

c

F F

F F

≤

>

2s cF F=

( ) ( ) 2 ,H F Ha F j Fπ= = / 2cF F≤



� Since , we have .

� In terms of , is periodic with period

� The discrete-time impulse response is

� In a more compact form

( ) ( )a skH F H F kF= −∑ ( ) ( )ah n h nT=

2 / sF Fω π= ( )H ω 2π

2

1 cos sin( ) ( )

2j n n n n

h n H e dn T

π ω

π

π π πω ω

π π−

−= =∫

0,( ) cos

,h n n

nT

π

=

0n =

0n ≠

16



� The magnitude and phase response of the continuous-time and discrete-time ideal differentiators



� Example : Fractional delay

Continuous-time delay system for any( ) ( )a a dy t x t t= −

If is bandlimited and sampled at the Nyquist rate, obtain

( ) ( ) ( ) [( ) ] ( )a a d ay n y nT x nT t x n T x n= = − = −∆ = −∆

/dt T∆ =

0dt >

( )ax t

Where is not an integer value

17



� One way to approach this problem is by considering the frequency response

� which impulse response

� When is not an integer value, is infinite long because the sampling times fall between the zero crossings.

∆ ( )idh n

( ) jidH e ωω − ∆=

1 sin ( )( ) ( )

2 ( )j n

id

nh n H e d

n

π ω

π

πω ω

π π−

− ∆= =

−∆∫

Analog-to-Digital and Digital-to-Analog Converters


� In the previous section, the quantization error in analog-to-digital conversion and round-off errors in digital signal processing are negligible.

� Analog signal processing operations cannot be done very precisely either, since electronic components in analog systems have tolerances and they introduce noise during their operation.

� A digital system designer has better control of tolerances than an equivalent analog system designer.

18

Analog-to-Digital Converters


� The practical aspects of A/D converters and related circuitry can be found in the manufacturer’s specifications and data sheets.

� A block diagram of the basic elements of an A/D converters



� The sampling of an analog signal is performed by a sample-and-hold (S/H) circuit.

� The S/H tracks the analog input signal during the sample mode, and then holds it fixed during the hold mode to the instantaneous value of the signal.

19



� S/H allows the A/D converter to operate more slowly compared to the time actually used to acquire the sample.

� The input signal must not change by more than one-half of the quantization step during the conversion.

� Jitter : errors in the periodicity of the sampling process

� Droop : changes in the voltage held during conversion

Quantization and coding


� The conversion of the A/D converter involves the processes of quantization and coding.

� Quantization is a nonlinear and noninvertable process that maps a given amplitude at time into an amplitude taken from a finite set of values.

� Divided into intervals

� By the decision levels .

( ) ( )x n x nT≡ t nT=,kx

L

1{ ( ) },k k kI x x n x += < ≤ 1,2,...,k L=

1L + 1 2 1, ,..., Lx x x +

20



� The operation of the quantizer is defined by the relation

� In most digital signal processing operations the mapping is independent of . The quantizaiton is memoryless and is simply denoted as

ˆ( ) [ ( )] ,q kx n Q x n x≡ = ( ) kx n I∈if

n[ ]qx Q x=



� Use uniform or linear quantizers defined by

� where is the quantizer step size.

� If zero is assigned a quantization level, the quantizer is of the midtread type. If zero is assigned a decision level, the quantizer is called a midrise type.

1

1

ˆ ˆ ,

,k k

k k

x x

x x+

+

− = ∆

− = ∆

1,2,..., 1k L= −

1,k kx x +for finite

∆

21



� Define the range of the quantizer.

� The term full-scale range (FSR) is used to describe the range of an A/D converter for bipolar (positive and negative amplitudes) signals. The term full-scale (FS) is used for unipolar signals.

� The quantization error is always in the range to

�

R

( )qe n / 2−∆/ 2∆

( )2 2qe n∆ ∆

− < ≤



22



� The coding process in an A/D converter assigns a unique binary number to each quantization level.

� A word length is bits we have distinct binary numbers.

� The step size or the resolution of the A/D converter of the A/D converter is given by

� where is the range of the quantizer.

1b + 12b+

12b

R+

∆ =

R



23



� It is convenient to use the same system to represent digital signals because we can operate on them directly without any extra format conversion.

� -bit binary fraction of the form has the value

� is the most significant bit (MSB) and is the least significant bit (LSB).

� The only degradation introduced by an ideal converter is the quantization error.

( 1)b+ 0 1 2 bβ β β β⋅ ⋅ ⋅

0 1 20 1 22 2 2 2 b

bβ β β β− − −− ⋅ + ⋅ + ⋅ + ⋅⋅⋅ + ⋅

bβ0β



� A/D converters may have

� offset error

� the first transition may not occur at exactly LSB)

� scale-factor error

� the difference between the values at which the first transition and the

last transition occur is not equal to FS – 2LSB

� linear error

� the difference between the values are not all equal or or uniformly changing

1

2+

24

Analysis of Quanzation Errors


� Assume that the quantization error is random in nature.

� If the input analog signal is within the range of the quantizer, the quantization error is bounded in magnitude, and resulting error is called granular noise.

� When the input falls outside the range of the quantizer(clipping), becomes unbounded and results in overload noise.

� Only remedy is to scale the input signal so that its dynamic range falls within the range of the quantizar.

( )qe n

( )qe n



� The statsitical properties of :

� The error is uniformly distributed over the range

� The error sequence is a stationary white noise sequence. In other words, the error and the error for are uncorrelated.

� The error sequence is uncorrelated with the signal sequence .

� The signal sequence is zero mean and stationary.

( )qe n

/ 2 ( ) / 2.qe n−∆ < < ∆

( )qe n

( )x n

{ }( )qe n

{ }( )qe n

( )qe nm n≠

( )x n

25



� Signal-to-quantization-noise (power) ratio (SQNR)

� Where is the signal power and

is the power of the quantization noise. The mean value of the error is zero.

� The variance (the quantization noise power) is

1010log x

n

PSQNR

P=

2/2 /22 2 2

/2 /2

1( )

12n eP e p e de e deσ∆ ∆

−∆ −∆

∆= = = =

∆∫ ∫

2 2( )x xP E x nσ = = 2 2( )n e qP E e nσ = =



� The expression for the SQNR becomes

� If is Gaussian distributed and the range of the quantizer extends from to , then

� Each additional bit in the quantizer increases the signal-to-quantization-noise ratio by “6 dB”.

1010log 20log

6.02 16.81 20log

x x

n e

x

PSQNR

P

Rb dB

σσ

σ

= =

= + −

( )x n3 xσ− 3 xσ 6 xR σ=

6.02 1.25SQNR b dB= +

26

Digital-to-Analog Converters


� Basic operations in converting a digital signal into an analog signal.

� D/A conversion is usually performed by combining a D/A converter with a sampling-and –hold (S/H) followed by a lowpass (smoothing) filter.



� D/A converter characteristic for a 3-bit bipolar signal.

27



� settling time, a parameter which is defined as the time required for the output of the D/A converter to reach and remain within a given fraction (usually, LSB) of the final value.

� Often, the application of the input code word results in a high-amplitude transient, called a “glitch”.

� Use an S/H circuit designed to serve as a “deglitcher” to remedy this problem.

1

2±



� The interpolation function of the S/H system is a square pulse

� The frequency-domain charateristics are obtained by evaluating its Fourier transform

1,( )

0,SHg t

=

0 t T

otherwise

≤ ≤

2 2 ( /2)sin( ) ( ) j Ft F T

SH SH

FTG F g t e dt T e

FTπ ππ

π

∞ − −

−∞= =∫

28



� The S/H does not possess a sharp cutoff frequency characteristics, which passes undesirable aliased frequency components (frequencies above ) to its output.

� To remedy the problem, passing the output of the S/H through a lowpass filter, frequency above .

� The frequency response of the lowpass filter is defined by

/ 2sF

/ 2sF

2 ( / 2),( ) sin

0,a

FTe F T

H F FT

ππ

π

=

/ 2

/ 2s

s

F F

F F

≤

>


� A continuous-time bandpass signal has its frequencycontent in the two frequency bands defined by

� Bandwidth :� Center frequency :

� where

Sampling and Reconstruction of Continuous-Time Bandpass Signals

0 L HF F F< < <B

cF

2L H

c

F FF

+=

29


� Uniform or first-order sampling is the typical periodicsampling. Sampling the bandpass signal at a rateproduces a sequence with spectrum

� Where is the sampling frequency

� Integer Band Positioning :� The number is band position

Uniform or First-Order Sampling

1/SF T=

( ) ( )ax n x nT=

1( ) ( )s

k

X F Xa F kFT

∞

=−∞

= −∑

sF

HF mB=m


� Even (m=4)

� Odd (m=3)


30


� Choosing

� Note the (m=3) spectrum has the same spectralstructure as the original; the (m=4) spectrum hasbeen inverted.


2sF B=


� The original bandpass signal can be reconstructed using

� is equal to the ideal interpolation function forlowpass signals modulated by a carrier with frequency

� we can reconstruct a continuous-time bandpass signalwith spectral bands centered at

� down conversion : for we obtain the equivalentbaseband signal


( ) ( ) ( )a a an

x t x nT g t nT∞

=−∞

= −∑sin

( ) cos2a c

Btg t F t

Bt

ππ

π=where

cF( )ag t

( / 2),cF kB B= ± + 0,1,....k =0k =

31



� Arbitrary Band Positioningthe sampling frequency should that negative spectralband do not overlap with positive spectral band



� An integer k and a sampling frequency satisfy

� should be in the range

� Determine the integer k :

� The maximum value of integer k :

sF

2 2

1H L

s

F FF

k k≤ ≤

−

1

2s H

k

F F≤ ( 1) 2 2s Hk F F B− ≤ −

maxHF

kB

≤

2

( 1) 2H s

s L

F kF

k F F

≤

− ≤

sF

32



� Fit in the range from to :

� where denotes the integer part

� The minimum sampling rate required to avoid aliasing is .

� The range of acceptable uniform sampling rates is determined by

� where k is an integer number given by

0 HFmax

HFk

B =

max max2 /HF F k=

2 2

1H L

s

F FF

k k≤ ≤

−

1 HFk

B ≤ ≤



� Choosing a Sampling Frequency

2 21

1sH HFF F

k B B k B ≤ ≤ − −

33



� A practical solution is to sample at a higher sampling rate, which is equivalent to augmenting the signal band with a guard band

� Lower-order wedge and the corresponding range

H LB B B∆ = ∆ + ∆

'

'

'

L L L

H H H

F F B

F F B

B B B

= −∆

= + ∆

= + ∆

' '

' '

2 3

1H L

s

F FF

k k≤ ≤

−

''

'HF

kB

=

where



� The symmetric guard bands lead to asymmetric sampling rate to tolerance

'

'

1

2

2

L s H

H s L

kB F

kB F

−∆ = ∆

∆ = ∆

34



� Choose the practical operating point at the vertical midpoint of the wedge, the sampling rate

� The guard bands become

� where

' '

' '

2 21

2 1H L

s

F FF

k k

= + −

'

'

1

4

4

L s

H s

kB F

kB F

−∆ = ∆

∆ = ∆

2s

sL sH

FF F

∆∆ = ∆ =



� Example 6.4.1� Bandwidth kHz and kHz

� To avoid potential aliasing

25B = 10,702.5LF =

max 429HF

kB

= = max

250.0117H

s

FF

k= = kHz

'

'

2.5

2.5

' 30

10,700

10,730

L

H

L H

L L L

H H H

B

B

B B B B

F F B

F F B

∆ =

∆ =

= + ∆ + ∆ =

= − ∆ =

= − ∆ =

kHz

kHz

kHz

kHz

kHz

Minimum sampling frequency

Maximum wedge index

'max 357

HFk

B

= =

Maximum wedge index

60.1120 60.1124skHz F kHz≤ ≤

35


Interleaved or Nonuniform Second-Order Sampling� sample a continuous-time signal with sampling rate

at a time instants

� where is a fixed time offset � reconstruct continuous-time signal

� where is a reconstruction function

( )ax t1

ii

FT

= i it nT= + ∆

( ) ( ),i i a i ix nT x nT= + ∆ n−∞ < < ∞

i∆

( ) ( )( ) ( ) ( )i ia i i a i i

n

y t x nT g t nT∞

=−∞

= − −∆∑( ) ( )iag t


Interleaved or Nonuniform Second-Order Sampling� The Fourier transform of is given by

� Where is the Fourier transform of

� The Fourier transform of can be expressed in terms of as

( )i ix nT

( )iay

2 ( )( ) ( )

2( )

( ) ( )

( ) ( )

i i

i

j F nTi ia i i a

n

j Fia i

Y x nT G F e

G F X F e

π

π

∞− +∆

=−∞− ∆

=

=

∑

( )iX F

2 ( )

2( ) ( )

1( )

1( ) ( )

ii

ii

kj F

Ti a

ki i

kj

Ti ia a a

ki i

kX F X F e

T T

kY F G F X F e

T T

π

π

∞ − ∆

=−∞

∞ − ∆

=−∞

= −

= −

∑

∑

( )i ix nT2( ) ij F

aX F e π ∆

36


Interleaved or Nonuniform Second-Order Sampling

� interleaved uniformly sampled sequence

� The sum of the reconstructed signals is given by

� The Fourier transform

P.n−∞ < < ∞

( ),i ix nT

P

( )

1

( ) ( )p

ia a

i

y t y t=

=∑ 1,2,...,i p=

( ) ( )

1

2( )

( ) ( ) ( )

1( )

ii

pi i

a ai

kj

Tia

ki i

Y F G F V F

kV F X F e

T T

π

=

∞ − ∆

=−∞

=

= −

∑

∑where



� Focus on the most commonly used second-order sampling

�

1 2 2

12, 0, ,p T T

B= ∆ = ∆ = ∆ = =

37



� Split the spectrum into a “positive” or “negative” band

( )aX F

( ), 0 ( ), 0( ) , ( )

0, 0 0, 0a a

a a

X F F X F FX F X F

F F+ −≥ ≤

= = < >

(1) (2)( ) ( ) ( ) ( ) ( )ka a a a a

k k

Y F BG F X F kB BG F X F kBγ∞ ∞

=−∞ =−∞

= − + −∑ ∑

where 2j Be πγ − ∆=



� The repeated replicas of and

as four separate components.

� Each repeated copy fills the entire frequency axis without aliasing.

( )aX F kB− ( )kaX F kBγ −

38



� Determine the interpolation functions and the time offset , so that

� smallest integer

� In the region

(1) (2)( ), ( )a aG F G F

∆ ( ) ( )a aY F X F=

(1) (2)( ) ( ) 0,a aG F G F= = for andLF F< LF F B> +

2 LFm

B =

( 1)k m= ± +with andk m= ±

(1) (2)( ) ( ) ( ) ( ) ( )ka a a a a

k k


=−∞ =−∞

= − + −∑ ∑

L LF F F mB≤ ≤ − +

becomes ….



� Perfect reconstruction

� solution

(1) (2)

(1) (2)

( ) ( ) ( ) ( )

( ) ( ) ( )

a a a a

ma a a

Y F BG F BG F X F

BG F B G F X F mBγ

+ +

+

= +

+ + −

(Signal component)

(Aliasing component)

( ) ( )a aY F X F+ +=

(1) (2)

(1) (2)

( ) ( ) 1

( ) ( ) 0a a

ma a

BG F BG F

BG F B G Fγ+ =

+ =

(1) (2)1 1 1 1( ) , ( )

1 1a am mG F G F

B Bγ γ−= =− −

39



� In the region L LF mB F F B− + ≤ ≤ +

(1) (2)( ) ( ) ( ) ( ) ( )ka a a a a

k k


=−∞ =−∞

= − + −∑ ∑

becomes ….

(1) (2)

(1) 1 (2)

( ) ( ) ( ) ( )

( ) ( ) ( ( 1) )

a a a a

ma a a

Y F BG F BG F X F

BG F B G F X F m Bγ

+ +

+ +

= +

+ + − +

(Signal component)

(Aliasing component)



� Perfect reconstruction

� Solution

� The frequency range can be obtained in a similar manner.

(1) (2)

(1) 1 (2)

( ) ( ) 1

( ) ( ) 0a a

ma a

BG F BG F

BG F B G Fγ +

+ =

+ =

(1) (2)( 1) 1

1 1 1 1( ) , ( )

1 1a am mG F G F

B Bγ γ− + += =

− −

( )L LF B F F− + ≤ ≤ −

40



� The function has bandpass response

� Using the reconstruction formula

� The bandpass signal can be reconstructed perfectly from two interleaved uniformly sampled sequenceand

(2) (1) (2) (1)( ) ( ) ( ) ( )a a a aG F G F g t g t= − → = −

(1)( )aG F

( )a a a a an

n n n nx t x g t x g t

B B B B

∞

=−∞

= − + + ∆ − + + ∆

∑

( )ax t( / )ax n B=

( / ),ax n B n+ ∆ −∞ < < ∞



� Interpolation function

[ ]

[ ] [ ]

( ) ( )

cos 2 ( ) cos(2 )( )

2 sin( )

cos 2 ( ) ( 1) cos 2 ( ) ( 1)( )

2 sin[ ( 1) ]

a

L L

L L

g a t b t

mB F t mB F t mBa t

Bt mB

F B t m B mB F t m Bb t

Bt m B

π π π ππ π

π π π ππ π

= +

− − ∆ − − ∆=

∆

+ − + ∆ − − − + ∆=

+ ∆

41



� Some useful simplifications occur when is an integer for integer band positioning. The region Abecomes zero, which implies that

� We have

2 /Lm F B=

( ) 0a t =

( ) ( )ag t b t=



� For and , the interpolation function becomes

� When the result is .

� For , we can choose the time offset such that . The requirement is satisfied

� Where is the center frequency of the band.

�

0LF = 0m =

cos(2 ) cos( )( )

2 sin( )LP

Bt B Bg t

Bt B

π π ππ π− ∆ − ∆

=∆

1

2B∆ =

sin(2 )( )

2LP

Btg t

Bt

ππ

=

/ 2LF mB= ∆( 1) 1mγ ± + = −

2 1 1, 0, 1, 2,...

2 ( 1) 4 2c c

k kk

B m F F

+∆ = = + = ± ±

+

( 1)

2 2c L

B m BF F

+= + =

42



� In this requirement, the interpolation function is specified by in the range taking the inverse Fourier transform, we obtain

is the special case of the interpolation function� It is possible to sample a bandpass signal, and then to

reconstruct the discrete-time signal at a band position other than the original.

/ 2 ( 1) / 2mB F m B≤ ≤ +( ) 1/ 2QG F =

sin( ) cos 2Q c

Btg t F t

Bt

ππ

π=


Bandpass Signal Representations

� Since is real, the negative and positive frequencies in its spectrum are related by

� The signal can be completely specified by one half of the spectrum. The identity

� which can be presented another real part form

� The amplitude of the positive frequencies is doubled to compensate for the omission of the negative frequencies

( )ax t*( ) ( )a aX F X F− =

2 21 1cos 2

2 2j Fct j Fct

cF t e eπ ππ −= +

21cos 2 2

2cj F t

cF t e ππ = ℜ

43



� The extension to signals with continuous spectra is straightforward. The inverse Fourier transform of

� Change the variable in the second integral form

� equivalently

( )ax t

02 2

0( ) ( ) ( )j Ft j Ft

a a ax t X F e dF X F e dFπ π∞

−∞= +∫ ∫

2 2

0 0( ) ( ) ( )j Ft j Ft

a a ax t X F e dF X F e dFπ π∞ ∞ −= +∫ ∫

{ } { }2

0( ) 2 ( ) ( )j Ft

a a ax t X F e dF tπ ψ∞

= ℜ =ℜ∫



� is known as the analytic signal or the pre-envelope of� Using unit function

� In case we define� The inverse Fourier transform of

2

0( ) 2 ( ) j Ft

a at X F e dFπψ∞

= ∫where

( )ax t

( )aV F

2 ( ), 0( ) 2 ( ) ( )

0, 0a

a a a

X F FF X F V F

Fψ

>= =

<( ) 0,aX F ≠ (0) (0)a aXψ =

( )aV F

1( ) ( )

2 2a

jv t t

tδ

π= +

44



� By the frequency-domain convolution theorem, we obtain

� The input signal is given by

� is called the Hilbert transform , which is a convolution and does not change the domain.

1( ) 2 ( ) ( ) ( ) ( )a a a a at x t v y x t j x t

tψ

π= ∗ = + ∗

( )ax t

( )1 1ˆ ( ) ( ) a

a a

xx t x t d

t t

ττ

π π τ

∞

−∞= ∗ =

−∫ˆ ( )ax t

2

1( )

, 0( ) ( )

, 0

Q

j FtQ Q

h tt

j FH F h t e dt

j Fπ

π∞ −

−∞

=

− >= =

<∫

Impulse response

Frequency response



� In terms of the magnitude and phase

� The Hilbert transformer is an allpass quadrature filter simply shifts the phase of positive frequency components by and the phase of negative frequency components by

� Express the analytic signal using the Hilbert transform

� The Hilbert transform provides the imaginary part of its analytic signal representation

/ 2π−/ 2π

ˆ( ) ( ) ( )a a at x t jx tψ = +

/ 2, 0( ) 1, ( )

/ 2, 0Q Q

FH F H F

F

ππ− >

= = <

�

45



� Using the modulation property of the Fourier transform

� The complex lowpass signal is known as the complex envelope of

� It can be expressed in rectangular coordinates as

� quadrature representation of bandpass signals

2( ) ( ) ( ) ( )cj F tLP a LP a cx t e t X F F Fπ ψ ψ− ℑ= ←→ = +

( )LPx t

( )ax t

( ) ( ) ( )LP I Qx t x t jx t= +

( ) ( )cos 2 ( )sin 2a I c Q cx t x t F t x t F tπ π= −

In-phase component quadrature component



� Generation and Reconstruction of a bandpass signal from its in-phase and quadrature components

46



� Alternatively express the complex envelope in polar coordinates as

� The bandpass signal can be rewritten as

� The relation are

( )( ) ( ) j tLPx t A t e φ=

( ) ( )cos[2 ( )]a cx t A t F t tπ φ= +

envelope phase

2 2 1

( ) ( )cos2 , ( ) ( )sin 2

( )( ) ( ) ( ), ( ) tan

( )

I c Q c

QI Q

I

x t A t F t x t A t F t

x tA t x t x t t

x t

π π

φ −

= =

= + =


Sampling Using Bandpass Signal Representations� Since the analytic signal can be sampled at a rate

of complex samples or real samples. Also they can be obtained by sampling and its Hilbert transform at a rate.

� A complex bandpass interpolation function definrd by

� where

� The in-phase and quadrature components are lowpass signals with one-sided bandwidth , they can be represented by the sequences and , where .

( )a tψ

B 2B( )ax t

ˆ ( )ax t B

2 1,sin( ) ( )

0,c L Lj F t

a a

F F F BBtg t e G F

otherwiseBtππ

πℑ ≤ ≤ +

= ←→ =

/ 2c LF F B= +

( )Ix t ( )Qx t

B( )Ix nT ( )Qx nT

1/T B=

47


Sampling Using Bandpass Signal Representations� We can avoid the complex demodulation process

required to generate the in-phase and quadrature signals. To extract directly

� Sampling at time instants

( )Ix t

( ) ( )a n I nx t x t=

2 12 , , 0, 1, 2,...

4c n nc

nF t n or t n

Fπ π

+= = = ± ±


Sampling Using Bandpass Signal Representations� Similarly, to extract directly

� Sampling at time instants

� The quadrature approach to bandpass sampling has been widely used in radar and communication systems to generates in-phase and quadrature sequences for further processing.

( )Qx t

( ) ( )a n Q nx t x t= −

2 12 , , 0, 1, 2,...

4c n nc

nF t n or t n

Fπ π

+= = = ± ±

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Chapter 6 Sampling and Reconstruction of Signalscwlin/courses/dsp/notes/DSP... · 2010-05-12 · 1 Sampling and Reconstruction of Signals 清大電機系林嘉文 [email protected]