Bài giảng Xử lý tín hiệu số - Chương 1: Sampling and reconstruction - Hà Hoàng Kha

Chapter 1
Sampling and Reconstruction
Click to edit Master subtitle styleHa Hoang Kha, Ph.D.
Ho Chi Minh City University of Technology
Email: hhkha@hcmut.edu.vn
Content
™ Sampling
‰ Sampling theorem
‰ Spectrum of sampling signals
™ Antialiasing prefilter
‰ Id l filea pre ter
‰ Practical prefilter
™ Analog reconstruction
‰ Ideal reconstructor
‰ Practical reconstructor
Ha H. Kha 2 Sampling and Reconstruction
Review of useful equations
™ Linear system ( )x t Linear systemh(t) ( ) ( ) ( )y t x t h t= ∗
( ) cos(2 )x t A f tπ θ= +
H(f)( )X f ( ) ( ) ( )Y f X f H f=
™ Especially 0, 
1
0 0 0( ) | ( ) | cos(2 arg( ( )))y t A H f f t H fπ θ= + +
1sin(2 ) [ ( ) ( )]FTf t j f f f fπ δ δ← → +
0 0 0cos(2 ) [ ( ) ( )]2
FTf t f f f fπ δ δ←⎯→ + + −™ Fourier transform: 
1cos( )cos( ) [cos( ) cos( )]
2
a b a b a b= + + −
0 0 02
⎯ − −
™ Trigonometric formulas: 
1sin( )sin( ) [cos( ) cos( )]
2
a b a b a b= − + − −
1sin( ) cos( ) [sin( ) sin( )]a b a b a b= + + −
3 Sampling and Reconstruction
2
Ha H. Kha
1. Introduction
™ A typical signal processing system includes 3 stages:
™ Th l i l i di i li d b A/De ana og s gna s g ta ze y an converter
™ The digitalized samples are processed by a digital signal processor.
‰ The digital processor can be programmed to perform signal processing 
operations such as filtering, spectrum estimation. Digital signal processor can be 
a general purpose computer, DSP chip or other digital hardware. 
™ The resulting output samples are converted back into analog by a 
D/A converter.
4 Sampling and ReconstructionHa H. Kha
2. Analog to digital conversion
™ Analog to digital (A/D) conversion is a three-step process.
Sampler Quantizer Coder
xQ(n)x(t) x(nT)≡x(n) 11010
t=nT
A/D converter 
xQ(n)111x(t) x(n)
011
100
101
110
n
000
001
010t n
5 Sampling and ReconstructionHa H. Kha
3. Sampling
™ Sampling is to convert a continuous time signal into a discrete time 
signal The analog signal is periodically measured at every T seconds. 
™ x(n)≡x(nT)=x(t=nT), n=.-2, -1, 0, 1, 2, 3..
™ T: sampling interval or sampling period (second); 
™ fs=1/T: sampling rate or sampling frequency (samples/second or 
6
Hz) 
Sampling and ReconstructionHa H. Kha
3. Sampling-example 1 
™ The analog signal x(t)=2cos(2πt) with t(s) is sampled at the rate fs=4 
Hz. Find the discrete-time signal x(n) ? 
Solution:
™ x(n)≡x(nT)=x(n/fs)=2cos(2πn/fs)=2cos(2πn/4)=2cos(πn/2)
n 0 1 2 3 4
x(n) 2 0 ‐2 0 2
™ Plot the signal
7 Sampling and ReconstructionHa H. Kha
3. Sampling-example 2 
™ Consider the two analog sinusoidal signals 
7( ) 2cos(2 )x t tπ= 1( ) 2cos(2 ); ( )x t t t sπ1 ,8 2 8=
These signals are sampled at the sampling frequency fs=1 Hz. 
Fi d h di i i l ?
Solution:
n t e screte-t me s gna s 
1 7 1 7
1 1 1( ) ( ) ( ) 2cos(2 ) 2cos( )8 1 4s
x n x nT x n n n
f
π π≡ = = =
12cos((2 ) ) 2cos( )
4 4
n nππ= − =
2 2 2
1 1 1 1( ) ( ) ( ) 2cos(2 ) 2cos( )
8 1 4s
x n x nT x n n n
f
π π≡ = = =
™Observation: x1(n)=x2(n) Æ based on the discrete-time signals, we 
cannot tell which of two signals are sampled ? These signals are 
8
called “alias” 
Sampling and ReconstructionHa H. Kha
3. Sampling-example 2 
f2=1/8 Hz f1=7/8 Hz
f =1 Hzs 
Fig: Illustration of aliasing 
9 Sampling and ReconstructionHa H. Kha
3. Sampling-Aliasing of Sinusoids 
at a sampling rate fs=1/T results in a discrete-
™ In general, the sampling of a continuous-time sinusoidal signal 
0( ) cos(2 )x t A f tπ θ= +
time signal x(n).
™ The sinusoids is sampled at f resulting in a( ) cos(2 )k kx t A f tπ θ= + s , 
discrete time signal xk(n). 
™ If f =f +kf k=0 ±1 ±2 then x(n)=x (n) k 0 s, , , , ., k . 
Proof: (in class)
™ Remarks: We can that the frequencies fk=f0+kfs are indistinguishable 
from the frequency f0 after sampling and hence they are aliases of f0
10 Sampling and ReconstructionHa H. Kha
4. Sampling Theorem-Sinusoids 
™ Consider the analog signal where Ω is 
the frequency (rad/s) of the analog signal, and f=Ω/2π is the 
( ) cos( ) cos(2 )x t A t A ftπ= Ω =
frequency in cycles/s or Hz. The signal is sampled at the three rate 
fs=8f, fs=4f, and fs=2f.
Fi Si id l d diff
™ Note that / sec
/
sf samples samples
f l l
= =
g: nuso samp e at erent rates 
seccyc es cyc e
™ To sample a single sinusoid properly, we must require 2sf samples
f cycle
≥
11 Sampling and ReconstructionHa H. Kha
4. Sampling Theorem
™ For accurate representation of a signal x(t) by its time samples x(nT), 
two conditions must be met:
1) The signal x(t) must be bandlimitted, i.e., its frequency spectrum must 
be limited to fmax .
Fig: Typical bandlimited spectrum
2) The sampling rate fs must be chosen at least twice the maximum 
frequency f 2f f≥ max. maxs
™ fs=2fmax is called Nyquist rate; fs/2 is called Nyquist frequency; 
[ f /2 f /2] i N i i l
12
- s , s s yqu st nterva . 
Sampling and ReconstructionHa H. Kha
4. Sampling Theorem
™ The values of fmax and fs depend on the application
Application fmax fs
Biomedical 1 KHz 2 KHz
Speech 4 KHz 8 KHz 
Audio 20 KHz 40 KHz
Video 4 MHz 8 MHz
13 Sampling and ReconstructionHa H. Kha
4. Sampling Theorem-Spectrum Replication
™ Let where ( ) ( ) ( ) ( ) ( ) ( )
n
x nT x t x t t nT x t s tδ∞
=−∞
= = − =∑ ( ) ( )
n
s t t nTδ∞
=−∞
= −∑
™ s(t) is periodic, thus, its Fourier series are given by 
2( ) sj f nts t S e π
∞
= ∑ 21 1 1( ) ( )sj f ntS t e dt t dtπδ δ−= = =∫ ∫wheren
n=−∞
n
T TT T T
21( ) sj f nts t e
T
π∞= ∑
Thus, 
n=−∞
21( ) ( ) ( ) ( ) sj nf t
n
x t x t s t x t e
T
π∞
=−∞
= = ∑
1 ∞
which results in 
( ) ( )s
n
X f X f nf
T =−∞
= −∑™ Taking the Fourier transform of yields ( )x t
™Observation: The spectrum of discrete-time signal is a sum of the 
original spectrum of analog signal and its periodic replication at the 
i l f
14
nterva s.
Sampling and ReconstructionHa H. Kha
4. Sampling Theorem-Spectrum Replication
™ fs/2 ≥ fmax
Fi T i l b dli i d
Fig: Spectrum replication caused by sampling
g: yp ca a m te spectrum
™ fs/2 < fmax
Fig: Aliasing caused by overlapping spectral replicas
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Sampling and ReconstructionHa H. Kha
5. Ideal Analog reconstruction
Fig: Ideal reconstructor as a lowpass filter 
™ An ideal reconstructor acts as a lowpass filter with cutoff frequency 
equal to the Nyquist frequency fs/2.
™ An ideal reconstructor (lowpass filter) [ / 2, / 2]( ) 0
s sT f f fH f
otherwise
∈ −⎧= ⎨⎩
( ) ( ) ( ) ( )aX f X f H f X f= =
 Then 
16 Sampling and ReconstructionHa H. Kha
5. Analog reconstruction-Example 1
™ The analog signal x(t)=cos(20πt) is sampled at the sampling 
frequency fs=40 Hz.
a) Plot the spectrum of signal x(t) ? 
b) Fi d h di i i l ( ) ? n t e screte t me s gna x n 
c) Plot the spectrum of signal x(n) ?
d) The signal x(n) is an input of the ideal reconstructor find the , 
reconstructed signal xa(t) ? 
17 Sampling and ReconstructionHa H. Kha
5. Analog reconstruction-Example 2
™ The analog signal x(t)=cos(100πt) is sampled at the sampling 
frequency fs=40 Hz.
a) Plot the spectrum of signal x(t) ? 
b) Fi d h di i i l ( ) ? n t e screte t me s gna x n 
c) Plot the spectrum of signal x(n) ?
d) The signal x(n) is an input of the ideal reconstructor find the , 
reconstructed signal xa(t) ? 
18 Sampling and ReconstructionHa H. Kha
5. Analog reconstruction
™ Remarks: xa(t) contains only the frequency components that lie in the 
Nyquist interval (NI) [ f //2 f /2] - s , s .
™ f f fsampling at fs ideal reconstructorx(t), 0 ∈ NI ------------------> x(n) ----------------------> xa(t), a= 0
sampling at f ideal reconstructor™ xk(t), fk=f0+kfs------------------> x(n) ----------------------> xa(t), fa=f0
 s 
™ Th f f f d i l ( ) i b i d b ddie requency a o reconstructe s gna xa t s o ta ne y a ng 
to or substracting from f0 (fk) enough multiples of fs until it lies 
within the Nyquist interval [-f //2 f /2] That is
mod( )a sf f f=
 s , s .. 
19 Sampling and ReconstructionHa H. Kha
5. Analog reconstruction-Example 3
™ The analog signal x(t)=10sin(4πt)+6sin(16πt) is sampled at the rate 20 
d h d lHz. Fin t e reconstructe signa xa(t) ? 
20 Sampling and ReconstructionHa H. Kha
5. Analog reconstruction-Example 4
™ Let x(t) be the sum of sinusoidal signals
x(t)=4+3cos(πt)+2cos(2πt)+cos(3πt) where t is in milliseconds .
a) Determine the minimum sampling rate that will not cause any 
aliasing effects ? 
b) To observe aliasing effects, suppose this signal is sampled at half its 
Nyquist rate. Determine the signal xa(t) that would be aliased with 
x(t) ? Plot the spectrum of signal x(n) for this sampling rate?
21 Sampling and ReconstructionHa H. Kha
6. Ideal antialiasing prefilter
™ The signals in practice may not bandlimitted, thus they must be 
f l d b l f li tere y a owpass i ter
Fi Id l ti li i p filtg: ea an a as ng re er
22 Sampling and ReconstructionHa H. Kha
6. Practical antialiasing prefilter
™ A lowpass filter: [-fpass, fpass] is the frequency range of interest for the 
l f f
™ The Nyquist frequency fs/2 is in the middle of transition region. 
app ication ( max= pass) 
™ The stopband frequency fstop and the minimum stopband attenuation 
Astop dB must be chosen appropriately to minimize the aliasing 
effects.
s pass stopf f f= +
Fig: Practical antialiasing lowpass prefilter
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Ha H. Kha
6. Practical antialiasing prefilter
™ The attenuation of the filter in decibels is defined as
10
0
( )( ) 20log ( )
( )
H fA f dB
H f
= −
where f0 is a convenient reference frequency, typically taken to be at 
DC for a lowpass filter.
™ α10 =A(10f)-A(f) (dB/decade): the increase in attenuation when f is 
changed by a factor of ten.
™ α2 =A(2f)-A(f) (dB/octave): the increase in attenuation when f is 
changed by a factor of two.
™ Analog filter with order N, |H(f)|~1/fN for large f, thus α10 =20N 
(dB/decade) and α10 =6N (dB/octave)
24 Sampling and ReconstructionHa H. Kha
6. Antialiasing prefilter-Example
™ A sound wave has the form
( ) 2 cos(10 ) 2 cos(30 ) 2 cos(50 )
2 cos(60 ) 2 cos(90 ) 2 cos(125 )
x t A t B t C t
D t E t F t
π π π
π π π
= + +
+ + +
where t is in milliseconds. What is the frequency content of this 
signal ? Which parts of it are audible and why ?
This signal is prefilter by an anlog prefilter H(f). Then, the output y(t) 
of the prefilter is sampled at a rate of 40KHz and immediately 
reconstructed by an ideal analog reconstructor, resulting into the final 
analog output ya(t), as shown below:
25 Sampling and ReconstructionHa H. Kha
6. Antialiasing prefilter-Example
Determine the output signal y(t) and ya(t) in the following cases:
a)When there is no prefilter, that is, H(f)=1 for all f.
b)When H(f) is the ideal prefilter with cutoff fs/2=20 KHz.
c)When H(f) is a practical prefilter with specifications as shown 
below:
The filter’s phase response is assumed to be ignored in this example. 
26 Sampling and ReconstructionHa H. Kha
7. Ideal and practical analog reconstructors
™ An ideal reconstructor is an ideal lowpass filter with cutoff Nyquist
f f /requency s 2.
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7. Ideal and practical analog reconstructors
™ The ideal reconstructor has the impulse response: 
h h l bl l l
sin( f t)( ) sh t
f t
π
π=w ic is not rea iza e since its impu se response is not casua s
™ It is practical to use a 
staircase reconstructor
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7. Ideal and practical analog reconstructors
Fig: Frequency response of staircase recontructor
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7. Practical reconstructors-antiimage postfilter
™ An analog lowpass postfilter whose cutoff is Nyquist frequency fs/2 
d h l lis use to remove t e surviving spectra rep icas. 
Fig: Analog anti-image postfilter
Fi S f fil
30 Sampling and Reconstruction
g: pectrum a ter post ter
Ha H. Kha
8. Homework
™ Problems: 1.2, 1.3, 1.4, 1.5, 1.9
31 Sampling and ReconstructionHa H. Kha