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3/9/2018 © 2003, JH McClellan & RW Schafer 1 Signal Processing First Chapter 4 Sampling and Aliasing
3/9/2018 © 2003, JH McClellan & RW Schafer 1
Signal Processing First
Chapter 4Sampling and Aliasing
3/9/2018 © 2003, JH McClellan & RW Schafer 2
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3/9/2018 © 2003, JH McClellan & RW Schafer 3
LECTURE OBJECTIVES
SAMPLING can cause ALIASING Sampling Theorem Sampling Rate > 2(Highest Frequency)
Spectrum for digital signals, x[n] Normalized Frequency
22ˆ s
s ffT
ALIASING
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SYSTEMS Process Signals
PROCESSING GOALS: Change x(t) into y(t) For example, more BASS
Improve x(t), e.g., image deblurring Extract Information from x(t)
SYSTEMx(t) y(t)
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System IMPLEMENTATION
DIGITAL/MICROPROCESSOR Convert x(t) to numbers stored in memory
ELECTRONICSx(t) y(t)
COMPUTER D-to-AA-to-Dx(t) y(t)y[n]x[n]
ANALOG/ELECTRONIC: Circuits: resistors, capacitors, op-amps
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4-1 SAMPLING
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]
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SAMPLING RATE, fs
SAMPLING RATE (fs) (Sampling frequency) fs =1/Ts NUMBER of SAMPLES PER SECOND
Ts = 125 microsec fs = 8000 samples/sec• UNITS ARE HERTZ: 8000 Hz
UNIFORM SAMPLING at t = nTs = n/fs IDEAL: x[n] = x(nTs)=x(n/fs)
C-to-Dx(t) x[n]=x(nTs)
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fs 2 kHz
fs 500Hz
Hz100f
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SAMPLING THEOREM
HOW OFTEN ? DEPENDS on FREQUENCY of SINUSOID ANSWERED by SHANNON/NYQUIST Theorem ALSO DEPENDS on “RECONSTRUCTION”
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NYQUIST RATE
“Nyquist Rate” Sampling fs > TWICE the HIGHEST Frequency in x(t) “Sampling above the Nyquist rate”
BANDLIMITED SIGNALS DEF: x(t) has a HIGHEST FREQUENCY
COMPONENT in its SPECTRUM NON-BANDLIMITED EXAMPLE TRIANGLE WAVE is NOT BANDLIMITED
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Reconstruction? Which One?
)4.0cos(][ nnx )4.2cos()4.0cos(
integer an is Whennn
n
Given the samples, draw a sinusoid through the values
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STORING DIGITAL SOUND
x[n] is a SAMPLED SINUSOID A list of numbers stored in memory
EXAMPLE: audio CD CD rate is 44,100 samples per second 16-bit samples Stereo uses 2 channels
Number of bytes for 1 minute is 2 X (16/8) X 60 X 44100 = 10.584 Mbytes
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sfsTnAnx
ˆ)ˆcos(][
)cos()(][)cos()(
ss nTAnTxnxtAtx
DISCRETE-TIME SINUSOID
Change x(t) into x[n] DERIVATION
))cos((][ nTAnx s
DEFINE DIGITAL FREQUENCY
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DIGITAL FREQUENCY
VARIES from 0 to 2, as f varies from 0 to the sampling frequency UNITS are radians, not rad/sec DIGITAL FREQUENCY is NORMALIZED
ss f
fT 2ˆ
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4-2 Spectrum View of Sampling
sff 2ˆ
kHz1sf
12 X1
2 X*
2–0.2
))1000/)(100(2cos(][ nAnx
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Spectrum (Digital)
ˆ 2s
ff
fs 100 Hz
12 X1
2 X*
2–2
?
x[n] is zero frequency???
))100/)(100(2cos(][ nAnx
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The Rest of the Story
Spectrum of x[n] has more than one line for each complex exponential Called ALIASING MANY SPECTRAL LINES
SPECTRUM is PERIODIC with period = 2 Because
ˆ ˆcos( ) cos(( 2 ) )A n A n
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Aliasing Derivation
Other Frequencies give the same Hz1000at sampled)400cos()(1 sfttx
)4.0cos()400cos(][ 10001 nnx n
Hz1000at sampled)2400cos()(2 sfttx
)4.2cos()2400cos(][ 10002 nnx n
)4.0cos()24.0cos()4.2cos(][2 nnnnnx
][][ 12 nxnx )1000(24002400
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Aliasing Derivation–2
Other Frequencies give the same
ss f
fT 2ˆ 2
s
s
ss
s
ff
ff
fff 22)(2ˆ :then
ˆand we want: [ ] cos( )x n A n If x (t) A cos( 2( f f s )t ) t
nfs
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Aliasing Conclusions
ADDING fs or 2fs or –fs to the FREQ of x(t) gives exactly the same x[n] The samples, x[n] = x(n/ fs ) are EXACTLY
THE SAME VALUES
GIVEN x[n], WE CAN’T DISTINGUISH fo FROM (fo + fs ) or (fo + 2fs )
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Normalized Frequency
DIGITAL FREQUENCY
ss f
fT 2ˆ 2
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SPECTRUM for x[n]
PLOT versus NORMALIZED FREQUENCY INCLUDE ALL SPECTRUM LINES ALIASES ADD MULTIPLES of 2 SUBTRACT MULTIPLES of 2
FOLDED ALIASES (to be discussed later) ALIASES of NEGATIVE FREQS
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SPECTRUM (MORE LINES)
12 X1
2 X*
2–0.2
12 X*
1.8
12 X
–1.8
))1000/)(100(2cos(][ nAnx
kHz1sf
sff 2ˆ
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SPECTRUM (ALIASING CASE)
12 X*
–0.5
12 X
–1.5
12 X
0.5 2.5–2.5
12 X1
2 X* 12 X*
1.5
))80/)(100(2cos(][ nAnxkHz80sf
sff 2ˆ
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SPECTRUM (FOLDING CASE)
ˆ 2s
ff
fs 125Hz
12 X*
0.4
12 X
–0.4 1.6–1.6
12 X1
2 X*
))125/)(100(2cos(][ nAnx
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DIGITAL FREQ AGAIN
ss f
fT 2ˆ 2
22ˆ s
s ffT FOLDED ALIAS
ALIASING
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EXAMPLE: SPECTRUM
x[n] = Acos(0.2n+) FREQS @ 0.2 and -0.2 ALIASES: {2.2, 4.2, 6.2, …} & {-1.8,-3.8,…} EX: x[n] = Acos(4.2n+)
ALIASES of NEGATIVE FREQ: {1.8,3.8,5.8,…} & {-2.2, -4.2 …}
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FOLDING (a type of ALIASING)
EXAMPLE: 3 different x(t); same x[n]
])1.0(2cos[])1.0(2cos[]2)9.0(2cos[])9.0(2cos[))900(2cos(
])1.0(2cos[])1.1(2cos[))1100(2cos(])1.0(2cos[))100(2cos(
1000
nnnnnt
nntnt
fs
)1.0(210001002ˆ
900 Hz “folds” to 100 Hz when fs=1kHz
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FOLDING DIAGRAM
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FREQUENCY DOMAINS
D-to-AA-to-Dx(t) y(t)x[n]
ff
f
y[n]
sff2ˆ
2sff 2ˆ
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4-4 Discrete-to-ContinuousConversion
Create continuous y(t) from y[n] IDEAL If you have formula for y[n]
Replace n in y[n] with fst y[n] = Acos(0.2n+) with fs = 8000 Hz y(t) = Acos(2(800)t+)
COMPUTER D-to-AA-to-Dx(t) y(t)y[n]x[n]
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D-to-A is Ambiguous!
ALIASING Given y[n], which y(t) do we pick ? ? ? Infinite number of y(t) Passing through the samples, y[n]
D-to-A reconstruction must choose one output
Reconstruct the SMOOTHEST one The LOWEST frequcney, if y[n] = sinusoid
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SPECTRUM (ALIASING CASE)
ˆ 2s
ff
fs 80Hz
12 X*
–0.5
12 X
–1.5
12 X
0.5 2.5–2.5
12 X1
2 X* 12 X*
1.5
))80/)(100(2cos(][ nAnx
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Reconstruction (D-to-A)
Convert stream of numbers to x(t) “Connect the dots” INTERPOLATION
y(t)
y[k]
kTs (k+1)Tst
INTUITIVE,conveys the idea
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SAMPLE & HOLD DEVICE
Convert y[n] to y(t) y[k] should be the value of y(t) at t = kTs
Make y(t) equal to y[k] for kTs -0.5Ts < t < kTs +0.5Ts
y(t)
y[k]
kTs (k+1)Tst
STAIR-STEPAPPROXIMATION
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SQUARE PULSE CASE
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OVER-SAMPLING CASE
EASIER TO RECONSTRUCT
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MATH MODEL for D-to-A
SQUARE PULSE:
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EXPAND the SUMMATION
SUM of SHIFTED PULSES p(t-nTs) “WEIGHTED” by y[n] CENTERED at t=nTs
SPACED by Ts RESTORES “REAL TIME”
y[n]p( t nTs ) n
y[0]p(t) y[1]p(t Ts ) y[2]p(t 2Ts )
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p(t)
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OPTIMAL PULSE ?
CALLED“BANDLIMITEDINTERPOLATION”
,2,for 0)(
for sin
)(
ss
TtT
t
TTttp
ttps
s
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4-5 Sampling Theorem
UNIFORM SAMPLING at t = nTs IDEAL: x[n] = x(nTs)
C-to-Dx(t) x[n]
