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EE3054
Signals and Systems
Sampling of Continuous Time Signals
Yao Wang
Polytechnic University
Some slides included are extracted from lecture presentations prepared by McClellan and Schafer
4/17/2008 © 2003, JH McClellan & RW Schafer 2
License Info for SPFirst Slides
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LECTURE OBJECTIVES
� Concept of sampling
� Sampling using periodic impulse train
� Frequency domain analysis
� Spectrum of sampled signal
� Nyquist sampling theorem
� Sampling of sinusoids
Two Processes in A/D
Conversion
� Sampling: take samples at time nT
� T: sampling period;
� fs = 1/T: sampling frequency
� Quantization: map amplitude values into a set of discrete values
� Q: quantization interval or stepsize
xc(t) x[n] = xc(nT) $[ ]x n
Quanti-
zation
Sampling
Sampling
Period
T
Quantization
Interval
Q
∞<<−∞= nnTxnx ),(][
pQ±
)]([][ˆ nTxQnx =
0 0.2 0.4 0.6 0.8 1
-1
-0.5
0
0.5
1
T=0.1Q=0.25
Analog to Digital
Conversion
A2D_plot.m
How to determine T and Q?
� T (or fs) depends on the signal frequency range� A fast varying signal should be sampled more frequently!
� Theoretically governed by the Nyquist sampling theorem
� fs > 2 fm (fm is the maximum signal frequency)
� For speech: fs >= 8 KHz; For music: fs >= 44 KHz;
� Q depends on the dynamic range of the signal amplitude and perceptual sensitivity� Q and the signal range D determine bits/sample R
� 2R=D/Q
� For speech: R = 8 bits; For music: R =16 bits;
� One can trade off T (or fs) and Q (or R)� lower R -> higher fs; higher R -> lower fs
� We only consider sampling in this class
SAMPLING x(t)
� SAMPLING PROCESS� Convert x(t) to numbers x[n]
� “n” is an integer; x[n] is a sequence of values
� Think of “n” as the storage address in memory
� UNIFORM SAMPLING at t = nTs
� IDEAL: x[n] = x(nTs)
C-to-Dx(t) x[n]
Sampling of Sinusoid
SignalsSampling above
Nyquist rate
ωs=3ωm>ωs0
Reconstructed
=original
Sampling under
Nyquist rate
ωs=1.5ωm<ωs0
Reconstructed
\= original
Aliasing: The reconstructed sinusoid has a lower frequency than the original!
Nyquist Sampling Theorem
� Theorem:� If x(t) is bandlimited, with maximum frequency fb(or
ωb =2π fb)
� and if fs =1/ Ts > 2 fb or ωs =2π / Ts >2 ωb
� Then xc(t) can be reconstructed perfectly from x[n]= x(nTs ) by using an ideal low-pass filter, with cut-off frequency at fs/2
� fs0 = 2 fb is called the Nyquist Sampling Rate
� Physical interpretation:� Must have at least two samples within each cycle!
4/17/2008 © 2003, JH McClellan & RW Schafer 10
Sampling Using Periodic Impulse
Train
x[n] = x(nTs )
FOURIER
TRANSFORM
of xs(t) ???
4/17/2008 © 2003, JH McClellan & RW Schafer 11
Periodic Impulse Train
∑∞
−∞=
−=
n
snTttp )()( δ
4/17/2008 © 2003, JH McClellan & RW Schafer 12
Impulse Train Sampling
xs (t) = x(t) δ (t − nTs )n=−∞
∞
∑ = x(t)δ (t − nTs )n=−∞
∞
∑
xs(t) = x(nTs)δ(t −nTs)n=−∞
∞
∑
4/17/2008 © 2003, JH McClellan & RW Schafer 13
Illustration of Samplingx(t)
x[n] = x(nTs )
∑∞
−∞=
−=n
sss nTtnTxtx )()()( δ
n
t
4/17/2008 © 2003, JH McClellan & RW Schafer 14
Sampling: Freq. Domain
EXPECT
FREQUENCY
SHIFTING !!!
∑∑∞
−∞=
∞
−∞=
=−=k
tjkk
n
sseanTttp
ωδ )()(
∑∞
−∞=
=k
tjkk
seaω
How is the
spectrum of xs(t)
related to that of
x(t)?
4/17/2008 © 2003, JH McClellan & RW Schafer 15
Fourier Series Representation
of Periodic Impulse Train
ωs =2π
Ts∑∑∞
−∞=
∞
−∞=
=−=k
tjkk
n
sseanTttp
ωδ )()(
s
T
T
tjk
s
kT
dtetT
as
s
s1
)(1
2/
2/
== ∫−
− ωδFourier Series
4/17/2008 © 2003, JH McClellan & RW Schafer 16
FT of Impulse Train
∑∑ ∑∞
−∞=
∞
−∞=
−=↔=−=
k
s
sn k
tjk
s
s kT
jPeT
nTttp s )(2
)(1
)()( ωωδπ
ωδ ω
s
sT
πω
2=
Frequency-Domain Analysis:
Using Fourier Series
∑ ∑∞
−∞=
=−=
n k
tjk
s
sse
TnTttp
ωδ1
)()(
)()()( tptxtxs =
xs (t) = x(t)1
Tsk=−∞
∞
∑ ejkωst =
1
Tsx(t)
k=−∞
∞
∑ ejkωst
Xs ( jω) =1
TsX( j(ω
k=−∞
∞
∑ − kωs ))
ωs =2π
Ts
Frequency-Domain Analysis:
Using Multiplication-
Convolution duality
∑∑ ∑∞
−∞=
∞
−∞=
−=↔=−=
k
s
sn k
tjk
s
s kT
jPeT
nTttp s )(2
)(1
)()( ωωδπ
ωδ ω
x(t)p(t) ⇔1
2πX( jω )∗ P( jω)
( )( )∑
∑∞
−∞=
∞
−∞=
−=
−==
k
s
s
k
s
s
s
kjXT
kjXT
jPjXjX
ωω
ωωδωπ
πωω
πω
1
)(*)(2
2
1)(*)(
2
1)(
4/17/2008 © 2003, JH McClellan & RW Schafer 19
Frequency-Domain
Representation of Sampling
Xs ( jω) =1
TsX( j(ω
k=−∞
∞
∑ − kωs ))
“Typical”
bandlimited signal
4/17/2008 © 2003, JH McClellan & RW Schafer 20
Aliasing Distortion
� If ωs < 2ωb , the copies of X(jω) overlap, and we have aliasing distortion.
“Typical”
bandlimited signal
Original signal
Sampling
impulse train
Sampled signal
ωs>2 ωm
Sampled signal
ωs<2 ωm
(Aliasing effect)
Frequency Domain
Interpretation of Sampling
The spectrum of the
sampled signal includes
the original spectrum and
its aliases (copies) shifted
to k fs , k=+/- 1,2,3,…
The reconstructed signal
from samples has the
frequency components
upto fs /2.
When fs< 2fm , aliasing
occur.
4/17/2008 © 2003, JH McClellan & RW Schafer 22
Reconstruction: Frequency-Domain
)()()(
so overlap,not do )(
of copies the,2 If
ωωω
ω
ωω
jXjHjX
jX
srr
bs
=
>
Hr ( jω )
Nyquist Sampling Theorem
� Theorem:� If x(t) is bandlimited, with maximum frequency fb(or
ωb =2π fb)
� and if fs =1/ Ts > 2 fb or ωs =2π / Ts >2 ωb
� Then xc(t) can be reconstructed perfectly from x[n]= x(nTs ) by using an ideal low-pass filter, with cut-off frequency at fs/2
� fs0 = 2 fb is called the Nyquist Sampling Rate
� Physical interpretation:� Must have at least two samples within each cycle!
Sampling of Sinusoid
Signals: Temporal domainSampling above
Nyquist rate
ωs=3ωm>ωs0
Reconstructed
=original
Sampling under
Nyquist rate
ωs=1.5ωm<ωs0
Reconstructed
\= original
Aliasing: The reconstructed sinusoid has a lower frequency than the original!
Sampling of Sinusoid:
Frequency Domain
f0-f0 fs-fs fs+f0
fs-f0-fs+f0-fs -f0
f0-f0 0
f0-f0 fs-fs fs+f0
fs-f0-fs+f0-fs -f0
fs/2-fs/2
0
Spectrum of
cos(2πf0t)
No aliasing
fs >2f0
fs -f0 >f0
Reconstructed
signal: f0
With aliasing
f0<fs <2f0 (folding)
fs -f0 <f0
Reconstructed signal: fs -f0
f0-f0
fs-fs
fs+f0f0-fs-f0+fs-fs -f0
With aliasing
fs <f0 (aliasing)
f0-fs <f0
Reconstructed signal: fs -f0
0
0
More examples with
Sinusoids
4/17/2008 © 2003, JH McClellan & RW Schafer 27
SAMPLING GUI (con2dis)
Strobe Movie
� From SP First, Chapter 4, Demo on “Strobe Movie”
How to determine the necessary
sampling frequency from a signal
waveform?
� Given the waveform, find the shortest ripple, there should be at least two samples in the shortest ripple
� The inverse of its length is approximately the highest frequency of the signal
Tmin
Fmax=1/Tmin
Need at least two
samples in this
interval, in order not
to miss the rise and
fall pattern.
Sampling with Pre-Filtering
Pre-Filter Periodic
H (f) Sampling
x(t) x’(t) xd(n)
Sampling
period T
• If fs < 2fb, aliasing will occur in sampled signal
• To prevent aliasing, pre-filter the continuous signal so that fb<fs/2
• Ideal filter is a low-pass filter with cutoff frequency at fs/2
(corresponding to sync functions in time)
•Common practical pre-filter: averaging within one sampling interval
Summary
� Sampling as multiplication with the periodic impulse train
� FT of sampled signal: original spectrum plus shifted versions (aliases) at multiples of sampling freq.
� Sampling theorem and Nyquist sampling rate
� Sampling of sinusoid signals� Can illustrate what is happening in both temporal and freq.
domain. Can determine the reconstructed signal from the sampled signal.
� Need for prefilter
� Next lecture: how to recover continuous signal from samples, ideal and practical approaches
Readings
� Textbook: Sec. 12.3.1-12.3.2, 4.1-4.3
� Oppenheim and Willsky, Signals and Systems, Chap. 7.
� Optional reading (More depth in frequency
domain interpretation)
Recommended