Lecture 21: Sampling - MIT OpenCourseWare · Courtesy of Helga Kolb, Eduardo Fernandez, and Ralph...
Transcript of Lecture 21: Sampling - MIT OpenCourseWare · Courtesy of Helga Kolb, Eduardo Fernandez, and Ralph...
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6.003: Signals and Systems
Sampling
November 22, 2011 1
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Sampling
Conversion of a continuous-time signal to discrete time.
t
x(t)
0 2 4 6 8 10n
x[n]
0 2 4 6 8 10
We have used sampling a number of times before.
Today: new insights from Fourier representations.
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Sampling
Sampling allows the use of modern digital electronics to process,
record, transmit, store, and retrieve CT signals.
• audio: MP3, CD, cell phone
• pictures: digital camera, printer
• video: DVD
• everything on the web
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Sampling
Sampling is pervasive.
Example: digital cameras record sampled images.
x
y I(x, y)
m
n I[m,n]
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Sampling
Photographs in newsprint are “half-tone” images. Each point is
black or white and the average conveys brightness.
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Sampling
Zoom in to see the binary pattern.
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Sampling
Even high-quality photographic paper records discrete images. When
AgBr crystals (0.04 − 1.5µm) are exposed to light, some of the Ag
is reduced to metal. During “development” the exposed grains are
completely reduced to metal and unexposed grains are removed.
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Images of discrete grains in photographic paper removed due to copyright restrictions.
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Sampling
Every image that we see is sampled by the retina, which contains ≈
100 million rods and 6 million cones (average spacing ≈ 3µm) which
act as discrete sensors.
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http://webvision.med.utah.edu/imageswv/sagschem.jpeg
Courtesy of Helga Kolb, Eduardo Fernandez, and Ralph Nelson. Used with permission.
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Check Yourself
Your retina is sampling this slide, which is composed of 1024×768
pixels.
Is the spatial sampling done by your rods and cones ade
quate to resolve individual pixels in this slide?
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Check Yourself
The spacing of rods and cones limits the angular resolution of your
retina to approximately
rod/cone spacing 3 × 10−6 m θeye = ≈ ≈ 10−4 radians
diameter of eye 3 cm
The angle between pixels viewed from the center of the classroom
is approximately
screen size / 1024 3 m/1024 ≈ ≈ 3 × 10−4 radiansθpixels = distance to screen 10 m
Light from a single pixel falls upon multiple rods and cones.
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Sampling
How does sampling affect the information contained in a signal?
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Sampling
We would like to sample in a way that preserves information, which
may not seem possible.
t
x(t)
Information between samples is lost. Therefore, the same samples
can represent multiple signals.
t
cos 7π3 n? cos π3n?
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Sampling and Reconstruction
To determine the effect of sampling, compare the original signal x(t)
to the signal xp(t) that is reconstructed from the samples x[n].
Uniform sampling (sampling interval T ).
tn
x[n] = x(nT )
Impulse reconstruction.
tn
xp(t) =∑n
x[n]δ(t− nT )
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Reconstruction
Impulse reconstruction maps samples x[n] (DT) to xp(t) (CT).
∞xp(t) = x[n]δ(t − nT )
n=−∞
0
∞= x(nT )δ(t − nT )
n=−∞
0
∞= x(t)δ(t − nT )
n=−∞
0
∞= x(t) δ(t − nT )
0
n=−∞ ,,≡ p(t)
- " Resulting reconstruction xp(t) is equivalent to multiplying x(t) by
impulse train.
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Sampling
Multiplication by an impulse train in time is equivalent to convolution
by an impulse train in frequency.
→ generates multiple copies of original frequency content.
ω
X(jω)
−W W
1
ω
P (jω)
−ωs ωs
2πT
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs ωs= 2πT
1T
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Check Yourself
What is the relation between the DTFT of x[n] = x(nT )
and the CTFT of xp(t) =
x[n]δ(t − nT ) for X(jω) below.
ω
X(jω)
−W W
1
1. Xp(jω) = X(e jΩ)|Ω=ω
2. Xp(jω) = X(e jΩ)|Ω= ω T
3. Xp(jω) = X(e jΩ)|Ω=ωT
4. none of the above
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Check Yourself
DTFT ∞
x[n]e n=−∞
(t)
0 X(ejΩ) =
CTFT of xp
−jΩn
∞−jωtdtXp(jω) = x[n]δ(t − nT )e
−∞ n=−∞
0 ∞
∞∞−jωtdt= x[n] δ(t − nT )e
−∞n=−∞
0
∞−jωnT = x[n]e
n=−∞
0
= X(ejΩ) Ω=ωT
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Check Yourself
Xp(jω) = X(e jΩ)
ω
X(jω)
−W W
1
Ω=ωT
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs ωs= 2πT
1T
Ω
X(ejΩ) = Xp(jω)∣∣ω=Ω
T
−2π 2π
1T
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Check Yourself
What is the relation between the DTFT of x[n] = x(nT )
and the CTFT of xp(t) = x[n]δ(t − nT ) for X(jω) below.
ω
X(jω)
−W W
1
1. Xp(jω) = X(e jΩ)|Ω=ω
2. Xp(jω) = X(e jΩ)|Ω= ω T
3. Xp(jω) = X(e jΩ)|Ω=ωT
4. none of the above
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Sampling
The high frequency copies can be removed with a low-pass filter
(also multiply by T to undo the amplitude scaling).
Impulse reconstruction followed by ideal low-pass filtering is called
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
1T
−ωs2ωs2
T
bandlimited reconstruction.
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The Sampling Theorem
If signal is bandlimited → sample without loosing information.
If x(t) is bandlimited so that
X(jω) = 0 for |ω| > ωm
then x(t) is uniquely determined by its samples x(nT ) if 2π
ωs = > 2ωm. T
The minimum sampling frequency, 2ωm, is called the “Nyquist rate.”
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−ωs2
ωs2
ωT
x[n] xr(t)Impulse
Reconstruction
xp(t) =∑x[n]δ(t− nT )
LPF
Summary
Three important ideas.
Sampling
x(t) → x[n] = x(nT )
2π Bandlimited Reconstruction ωs =
T
ωsSampling Theorem: If X(jω) = 0 ∀ |ω| > then xr(t) = x(t).2
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Check Yourself
We can hear sounds with frequency components between 20 Hz
and 20 kHz.
What is the maximum sampling interval T that can be used
to sample a signal without loss of audible information?
1. 100 µs 2. 50 µs
3. 25 µs 4. 100π µs
5. 50π µs 6. 25π µs
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Check Yourself
ωs 2π2πfm = ωm < = 2 2T 1 1
T < = = 25 µs2fm 2 × 20 kHz
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Check Yourself
We can hear sounds with frequency components between 20 Hz
and 20 kHz.
What is the maximum sampling interval T that can be used
to sample a signal without loss of audible information?
1. 100 µs 2. 50 µs
3. 25 µs 4. 100π µs
5. 50π µs 6. 25π µs
25
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CT Model of Sampling and Reconstruction
Sampling followed by bandlimited reconstruction is equivalent to
multiplying by an impulse train and then low-pass filtering.
−ωs2
ωs2
ωT
×x(t)
p(t)
xr(t)xp(t)
LPF
t
p(t) = ”sampling function”
0 T
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Aliasing
What happens if X contains frequencies |ω| > π T ?
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)
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Aliasing
What happens if X contains frequencies |ω| > π T ?
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)
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Aliasing
What happens if X contains frequencies |ω| > π T ?
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)
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Aliasing
What happens if X contains frequencies |ω| > π T ?
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)
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Aliasing
The effect of aliasing is to wrap frequencies.
ω−ωs
2ωs2
1T
ω
ω
X(jω)
Input frequency
Output frequency
ωs2
ωs2
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Aliasing
The effect of aliasing is to wrap frequencies.
ω−ωs
2ωs2
1T
ω
ω
X(jω)
Input frequency
Output frequency
ωs2
ωs2
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Aliasing
The effect of aliasing is to wrap frequencies.
ω−ωs
2ωs2
1T
ω
ω
X(jω)
Input frequency
Output frequency
ωs2
ωs2
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Aliasing
The effect of aliasing is to wrap frequencies.
ω−ωs
2ωs2
1T
ω
ω
X(jω)
Input frequency
Output frequency
ωs2
ωs2
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Check Yourself
A periodic signal, period of 0.1 ms, is sampled at 44 kHz.
To what frequency does the third harmonic alias?
1. 18 kHz
2. 16 kHz
3. 14 kHz
4. 8 kHz
5. 6 kHz
0. none of the above
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Check Yourself
Input frequency (kHz)
Output frequency (kHz)
22 44 66 88
22
44
88
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Check Yourself
Input frequency (kHz)
Output frequency (kHz)
22 44 66 88
22
44
88
Harmonic Alias
10 kHz 10 kHz
20 kHz 20 kHz
30 kHz 44 kHz-30 kHz =14 kHz
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Check Yourself
A periodic signal, period of 0.1 ms, is sampled at 44 kHz.
To what frequency does the third harmonic alias? 3
1. 18 kHz
2. 16 kHz
3. 14 kHz
4. 8 kHz
5. 6 kHz
0. none of the above
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Check Yourself
Input frequency (kHz)
Output frequency (kHz)
22 44 66 88
22
44
88
Harmonic Alias
10 kHz 10 kHz
20 kHz 20 kHz
30 kHz 44 kHz-30 kHz =14 kHz
40 kHz 44 kHz-40 kHz = 4 kHz
50 kHz 50 kHz-44 kHz = 6 kHz
60 kHz 60 kHz-44 kHz =16 kHz
70 kHz 88 kHz-70 kHz =18 kHz
80 kHz 88 kHz-80 kHz = 8 kHz
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Check Yourself
Scrambled harmonics.
ω
ω
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Aliasing
High frequency components of complex signals also wrap.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)1
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Aliasing
High frequency components of complex signals also wrap.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)1
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Aliasing
High frequency components of complex signals also wrap.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)1
43
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Aliasing
High frequency components of complex signals also wrap.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)1
44
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Aliasing
Aliasing increases as the sampling rate decreases.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)1
45
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Aliasing
Aliasing increases as the sampling rate decreases.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)1
46
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Aliasing
Aliasing increases as the sampling rate decreases.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)1
47
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Aliasing
Aliasing increases as the sampling rate decreases.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)1
48
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Aliasing Demonstration
Sampling Music
2π ωs = = 2πfs
T
• fs = 44.1 kHz
• fs = 22 kHz
• fs = 11 kHz
• fs = 5.5 kHz
• fs = 2.8 kHz
J.S. Bach, Sonata No. 1 in G minor Mvmt. IV. Presto
Nathan Milstein, violin
49
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Aliasing
Aliasing increases as the sampling rate decreases.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)1
50
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Aliasing
Aliasing increases as the sampling rate decreases.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)1
51
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Aliasing
Aliasing increases as the sampling rate decreases.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)1
52
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Aliasing
Aliasing increases as the sampling rate decreases.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)1
53
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Anti-Aliasing Filter
To avoid aliasing, remove frequency components that alias before
sampling.
−ωs2
ωs2
ω1
×−ωs
2ωs2
ωT
x(t)
p(t)
xr(t)xp(t)
ReconstructionFilter
Anti-aliasingFilter
54
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Aliasing
Aliasing increases as the sampling rate decreases.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
X(jω)1
55
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Aliasing
Aliasing increases as the sampling rate decreases.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
Anti-aliased X(jω)
56
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Aliasing
Aliasing increases as the sampling rate decreases.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
Anti-aliased X(jω)
57
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Aliasing
Aliasing increases as the sampling rate decreases.
ω
Xp(jω) = 12π (X(j · ) ∗ P (j · ))(ω)
−ωs2
ωs2
1T
ω
P (jω)
−ωs ωs
2πT
ω
Anti-aliased X(jω)
58
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Anti-Aliasing Demonstration
Sampling Music
2π ωs = = 2πfs
T
• fs = 11 kHz without anti-aliasing
• fs = 11 kHz with anti-aliasing
• fs = 5.5 kHz without anti-aliasing
• fs = 5.5 kHz with anti-aliasing
• fs = 2.8 kHz without anti-aliasing
• fs = 2.8 kHz with anti-aliasing
J.S. Bach, Sonata No. 1 in G minor Mvmt. IV. Presto
Nathan Milstein, violin
59
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Sampling: Summary
Effects of sampling are easy to visualize with Fourier representations.
Signals that are bandlimited in frequency (e.g., −W <ω <W ) can be
sampled without loss of information.
The minimum sampling frequency for sampling without loss of in
formation is called the Nyquist rate. The Nyquist rate is twice the
highest frequency contained in a bandlimited signal.
Sampling at frequencies below the Nyquist rate causes aliasing.
Aliasing can be eliminated by pre-filtering to remove frequency com
ponents that would otherwise alias.
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MIT OpenCourseWarehttp://ocw.mit.edu
6.003 Signals and SystemsFall 2011
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