Upload
izma-izniza
View
307
Download
12
Embed Size (px)
Citation preview
7/27/2019 baseband demodulation.ppt
1/45
Chapter Four:Baseband Demodulation/Detection
7/27/2019 baseband demodulation.ppt
2/45
Baseband Demodulation/Detection
Baseband signaling
The received waveforms are already in a pulse-like form
Arriving baseband pulses are not in the form of ideal pulse
shapes due to Intersymbol interference (ISI) The task of detector
Retrieve the bit stream from the received waveform, as
error free as possible
Demodulation: a recovery of a waveform to an undistortedbaseband pulse
Detection: the decision-making process of selecting the
digital meaning of that waveform
7/27/2019 baseband demodulation.ppt
3/45
Baseband Demodulation/Detection
Receiving
filter
Equalizing
Filter
Threshold
comparison
Sample at
t=T
r(t)z(t)
z(T)
Symbol
Demodulation and sample Detection
channelIdeal0,2,1),()()(
,,2,1),()()()(
Ttitntstr
Mitnthtstr
i
ci
0
0 2,1),()()(
naz
iTnTaTz
i
i
What is a equalizing filter?
7/27/2019 baseband demodulation.ppt
4/45
Background: Performance Analysis
Signal power to average noise power ratio
Analog: SNR (S/N)
Digital: Eb/N0A normalized version of SNR
b
bbb
RW
NS
WNRS
WNST
NE
//
/0
Eb: bit energy, as signal power times the bit time
N0: noise power spectral density, as noise power N divided by bandwidth W
s-Watt
s-Watt
HzperWatt
Joule:Unit
A nature figure of merit. Why?
7/27/2019 baseband demodulation.ppt
5/45
Background: Performance Analysis
(Bit) error probability
For a binary decision-making, there are two ways errors
can occur
Case 1: when s1 is sent
the channel noise results in the receiveroutput the probability being s2 is greater
Case 2: when s2 is sent the channel noise results in the receiver
output the probability being s1 is greater
The probability of error is the sum of the probabilities of all
the ways that error can occur
Decision-making?
7/27/2019 baseband demodulation.ppt
6/45
Background: Orthogonal Signals
Signals are orthogonal
0)()(:orthogonalnotaresignalsTwo
0)()(:orthogonalaresignalsTwo
021
021
dttsts
dttsts
T
T
1/2 1
-1
-3
s1
1/2
1
2s2
1/2 1
1
f1
1/2
1
1
f2
-1
)()()(
)(2)()(
212
121
tftfts
tftfts
7/27/2019 baseband demodulation.ppt
7/45
Example:
Determine whether or not these two signals are
orthogonal over the interval )5.15.1( 22 TtT
)2cos()( 111
tfts
)2cos()( 222 tfts
212ff 0
21
21 ff 021
7/27/2019 baseband demodulation.ppt
8/45
Baseband Demodulation/Detection
Receiving
filter
Equalizing
Filter
Threshold
comparison
Sample at
t=T
r(t)z(t)
z(T)
Symbol
Demodulation and sample Detection
channelIdeal0,2,1),()()(
,,2,1),()()()(
Ttitntstr
Mitnthtstr
i
ci
0
0 2,1),()()(
naz
iTnTaTz
i
i
Desired signal component
Zero mean Gaussian random variable
7/27/2019 baseband demodulation.ppt
9/45
Background: Noise
Primary causes for error performancedegradation
The effect of filtering
Electrical noise and interferenceThermal noise modeled as AWGN
2
0
0
00 2
1
exp2
1
)(:noiserandomGaussianpdf
n
np
7/27/2019 baseband demodulation.ppt
10/45
Background: Noise
2
0
2
0
22
2
0
1
0
11
2
1exp
2
1)|(:oflikelihood
2
1exp
2
1)|(:oflikelihood
:pdfslConditiona
azszps
azszps
7/27/2019 baseband demodulation.ppt
11/45
Decision Theory
The system model
The signal source at the Tx consists of a set {si}, i=1,2M
of waveforms (or hypotheses)
The received signal r(t)=si(t)+n(t) where n(t) AWGN The waveform is reduced to a single number z(T), a
Gaussian RVz(T)=ai(T)+n0(T), where T is a symbol
duration
Receiver decision
P(s1|z) P(s2|z)
H1
H2
>
7/27/2019 baseband demodulation.ppt
13/45
Maximum Likelihood Detector
2
0
0
0
02
1exp
2
1)(:noiserandomGaussianpdf
nnp
20
2
2
2
1
2
0
21
2
0
2
2
0
2
2
2
0
2
2
0
1
2
0
2
1
2
0
2
2
0
2
0
2
0
1
0
2
1
2exp
2
2exp
2exp
2exp
2
2exp
2exp
2exp
2
1exp
2
1
2
1exp
2
1
)|(
)|(
aaaaz
zaaz
zaaz
az
az
szp
szp
H1P(z|s1) P(s2)
H2
>< A detector that minimize the error probability for
The case where the signal classes are equally likely
7/27/2019 baseband demodulation.ppt
14/45
Maximum Likelihood Detector
Error Probability
)()|(),(
)|()|()|(:2case
)|()|()|(:1case
2
1
2
1
2112
1121
0
0
i
i
i
i
iB sPsePsePP
dzszpsHPseP
dzszpsHPseP
u
aau
aaaaB
B
aaQduu
dzaz
dzszpP
sHPsHPsHPsHPP
021
210210
2/)(0
212
2/)(
2
0
2
02/)(
2
21122112
22
1exp
2
1
2
1exp
2
1)|(
)|()|()|(21)|(
21
equalareiesprobabilitprioriathewherecaseFor the
02 /)( az
Complementary error function or co-error function
7/27/2019 baseband demodulation.ppt
15/45
Q-function
Complementary function or co-error function
2exp
2
1)(3
22
1)(
22)(
2exp
2
1)(
2
2
x
xxQxif
xerfcxQ
xQxerfc
duu
xQx
7/27/2019 baseband demodulation.ppt
16/45
Example
Assume that in a binary digital communication system,
the signal component out of the receiver is ai(T)=+1 or
-1 V with equal probability. If the Gaussian noise at the
output has unit variance, find the probability of a biterror.
1587.0)1(2
)1(12 0
21
QQaaQPB
7/27/2019 baseband demodulation.ppt
17/45
Matched Filter
A linearfilterto provide the maximum signal-to-noise power
ratio
elsewhere0
0)()(
2
0
2 TttTksth
a
N
S i
T
Ttdsr
dtTsr
kdtTsr
dthrthtrtz
T
t
t
t
0
0
0
0
)(
)(
1)(
)()()()()(
Correlation
Convolution
The impulse of filter is a delayed version of the mirror image of the signal waveform
7/27/2019 baseband demodulation.ppt
18/45
Convolution vs. Correlation Matched filter convolution
Correlator correlation
)( tTh Matched tos1(t)- s2(t)
)(Tz)(tr
Correlator
)(Tz)(tr T
0)(
s1(t)- s2(t)
Matched filter output
Correlator output
7/27/2019 baseband demodulation.ppt
19/45
Matched FiltersPerformance
In general: threshold: 0=(a1+a2)/2 PB=Q[(a1-a2)/20]
Matched filter: maximize the output SNR
2/0
2
0
2
21
max
2
0
2
NEaaa
NS di
T
The signal component Average noise power
Analysis in the frequency domain
Two-sided power spectral density of the noise
T
d dttstsE 02
21 )()(
000
21
22/2
1
2 N
EQ
N
EQ
aaQP ddB
7/27/2019 baseband demodulation.ppt
20/45
Matched FiltersPerformance
TTTT
d dttstsdttsdttsdttstsE0
210
2
20
2
10
2
21 )()(2)()()()(
Energy associated with a bit,Eb
0)()(0
21 T
dttsts Orthogonal
b
T
Edttsts 0 21 )()( Perfectly correlated
b
T
Edttsts
0
21 )()( anticorrelated
T
b
dttstsE 0
21 )()(1
00
)1(
2 N
EQ
N
EQP bdB
0NEQP bB
0BP
0
2
N
E
QPb
B
7/27/2019 baseband demodulation.ppt
21/45
Examples: page 130, 3.2
Consider a binary comm. sys. That receives equally likely signals
pulse AWGN. Assume that the receiving filter is a matched filter,
and that the noise power density N0 is equal to 10-12 watt/Hz.
Compute the bit error probability.
0 1 2 3
0 1 2 32
1
-1-2
mvmv
us
us
7/27/2019 baseband demodulation.ppt
22/45
Examples: Binary Signaling
Bipolar signals
00)(
10)(
1
1
bitTtAts
bitTtAts
Correlator detector:
A
-A
)(1 Tz
)(tr T
0)(
s1(t)=A
0
2
1
)( H
H
Tz )( tsi
T
0)(
s2(t)=-A
)(2 Tz
_
+
TAETAE
N
EQ
N
EQP
db
db
B
22
00
4;
2
2
T 3T 5T
7/27/2019 baseband demodulation.ppt
23/45
Examples: Binary Signaling
Unipolar signals
000)(
10)(
1
1
bitTtts
bitTtAts
)(Tz
)(tr T
0)(
s1(t)- s2(t)=A
0
2
1
)( H
H
Tz )( tsi
Correlator detector:
;
)()(
2
1
2
1
2
1
2
2
0
2
21
2
01
00
TAdttstsETAEEE
N
EQP
N
EQP
T
dbitbitb
dB
bB
A
0
7/27/2019 baseband demodulation.ppt
24/45
Physical Meanings
-1 0 1 2 3 4 510
-10
10-8
10-6
10-4
10-2
100
SNR
BER
Unipolar
Bipolar
7/27/2019 baseband demodulation.ppt
25/45
-3 -2 -1 0 1 2 3-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Intersymbol Interference (ISI)
Due to the effects of system filtering, the received
pulses can overlap one another; The tail of a pulse
can smear into adjacent symbol intervals
Interfering with the detection process and degradingthe performance
7/27/2019 baseband demodulation.ppt
26/45
Ht(t) = Impulse response of the transmitter
Hc(t) = Impulse response of the channel
Hr(t) = Impulse response of the receiver
Transmitter
HT(f)
Receiver
HR(f)
Channel
HC(f)
+
n(t)
s(t) y(t)r(t)
t = kT
x(t)
Intersymbol Interference (ISI)
)()()()( fHfHfHfH rct
7/27/2019 baseband demodulation.ppt
27/45
Zero ISINyquist
Ideal Nyquist filter: rectangular[-1/2T, 1/2T]
Ideal Nyquist pulse: sinc-shaped pulse
Nyquist bandwidth constraint: A system withbandwidth W=1/(2T)=Rs/2 Hz can support amaximum transmission rate 2W=Rs symbols/swithout ISI
7/27/2019 baseband demodulation.ppt
28/45
Example:page 164, problem 3.8
What is the theoretical minimum system bandwidth
needed for a 10-Mbits/s signal using 16-level PAM
without ISI?
Answer:16-aryM=16 and k=4 bits/symbol
MHzRMinBW
ssymbolsMsymbolbits
sbitsM
R
s
s
25.12/
/5.2/4
/10
7/27/2019 baseband demodulation.ppt
29/45
The Raised Cosine Filter
Zero ISI at the sampling times
Equalizing filter to compensate for the distortion
caused by both the transmitter and the receiver
20
0
00
0
0
02
0
)(41
)(2cos2sin2)(
0
22
4
cos
21
)(
tWW
tWWtWcWth
Wf
WfWW
WW
WWf
WWf
fH
Excess bandwidth, additional beyond the Nyquist minimum
0
0
W
WWr
Roll-off factor
7/27/2019 baseband demodulation.ppt
30/45
The Raised Cosine Filter
sRrW )1(2
1
7/27/2019 baseband demodulation.ppt
31/45
Examples
Find the minimum required bandwidth for the baseband
transmission of a four level PAM pulse sequence having a data
rate of R=2400 bits/s if the system transfer characteristic consists
of a raised-cosine spectrum with 100% excess bandwidth (r=1).
HzRrW
ssymbolsk
RR
s
bs
1200)1(2
1:banswidthMinimum
/12002
2400:rateSymbol
7/27/2019 baseband demodulation.ppt
32/45
Performance Degradation
Due to a loss in SNR
Decreasing received signal power
Increasing noise power
Increasing interference power Intersymbol Interference (ISI)
Min BW = Rs/2
Raised Cosine filter: W=1/2(1+r) Rs
Using Nyquist filter to reduce ISI
The channel is precisely known and its characteristics do not
change with time
7/27/2019 baseband demodulation.ppt
33/45
Performance Degradation
7/27/2019 baseband demodulation.ppt
34/45
ISIChannel Characterization
Most channels can be characterized as band-limited
linear filters
Channels amplitude response
Channels phase response
ISI: amplitude and/or phase distortion Ideal channel: constant |Hc(f)| & linearc(f)
)()()(
fj
cccefHfH
7/27/2019 baseband demodulation.ppt
35/45
ISIEye Pattern
To display results from measuring a systems
response to based band signals in a prescribed way
Oscilloscopes vertical plate: connect the
receivers response to a random pulse
sequence
Oscilloscopes horizontal time base: set
equal to the symbol (pulse) duration
DA: a measure of distortion caused by ISI
JT: a measure of time jitter
MN: a measure of noise margin
ST: sensitivity to timing error
7/27/2019 baseband demodulation.ppt
36/45
ISIEye Pattern
ISI causes the eye to close
ISI distorts the position of the zero crossing, thereby
causing the system to be more sensitive to
synchronization error Moreover, noise causes a general closing of the eye
7/27/2019 baseband demodulation.ppt
37/45
Equalization
Any signal processing or filtering technique that is
designed to eliminate or reduce ISI
Two categories
Maximum-likelihood sequence estimation (MLSE) Making measurements of hc(t) and providing a means for adjusting
the receiver to the transmission environment.
Enable the detector to make good estimates
Equalization with filters Most popular approach
Using filters to compensate the distorted pulses
Transversal vs. decision feedback equalizers
7/27/2019 baseband demodulation.ppt
38/45
EqualizationTransversal Equalizer
A linear equalizer
A delay line with T-seconds taps (symbol duration)
Adjustable tap coefficient: based on channel characteristics
NNnNNkcnkxkyN
Nn
n ,...,2,...2,)(
7/27/2019 baseband demodulation.ppt
39/45
EqualizationTransversal Equalizer
Zero-forcing solution
Select weights so that the equalizer output is forced to zero
at N sample points on either side of the desired pulse
others
kky
c
c
c
Nx
NxNxNxNx
Nx
Ny
y
Ny
N
N
0
01)(
)(
)()1()1()(
)(
)2(
)0(
)2(
0
7/27/2019 baseband demodulation.ppt
40/45
EqualizationTransversal Equalizer
Example: A zero-forcing three-tap equalizer.
Given a received distorted set of pulse samples with voltage values 0.0, 0.2,
0.9, -0.3, 0.1. Weight cn to reduce the ISI. Using weights, calculate the ISI
values. Calculate the ISI at the sample times at k=2, 3.
0345.0
0071.0
0
1
00428.0
0
3448.0
9631.0
2140.0
1.000
3.01.00
9.03.01.0
2.09.03.0
02.09.0002.0
000
3448.0
9631.0
2140.0
)2(00
)1()2(0
)0()1()2(
)1()0()1(
)2()1()0(0)2()1(
00)2(
:
3448.0
9631.02140.0
9.03.01.0
2.09.03.002.09.0
0
10
)0()1()2(
)1()0()1()2()1()0(
0
10
1
0
1
1
0
1
1
0
1
x
xx
xxx
xxx
xxxxx
x
ky
ky
c
cc
c
cc
c
cc
xxx
xxxxxx
7/27/2019 baseband demodulation.ppt
41/45
EqualizationTransversal Equalizer
Minimum MSE Solution
Minimize the mean square error (MSE) of all the ISI terms
pulse the noise power
MSE: expected value of the squared difference between thedesired data symbol and the estimated data symbol
xyxx
T
xx
T
xy
xxxy
TT
RR
xxR
yxR
cRRxcxyx
1c
oration vectautocorrel:
n vectorcorrelatio-cross:where
7/27/2019 baseband demodulation.ppt
42/45
EqualizationTransversal Equalizer
Example: Consider that the tap weights of an equalizing
transversal filter are to be determined by transmitting a single
impulse as a training signal. Let the equalizer circuit be made
up of seven taps. Given a received distorted set of pulses
samples with values 0.0108, -0.0558, 0.1617, 1.0000, -0.1749,0.0227, 0.0110, use a minimum MSE solution to find the value
of the weights that will minimize the ISI.
Matlab code available on WebCT
7/27/2019 baseband demodulation.ppt
43/45
EqualizationDecision Feedback Equalizer
A nonlinear equalizer
Using previous detector decisions to eliminate the ISI
7/27/2019 baseband demodulation.ppt
44/45
Equalization -- Types
Preset equalization
The weights remain fixed during transmission of data
Done once or seldom at the start of transmission
Channel frequency responses are known and time invariant Initial training period
Adaptive equalization
Perform tap-weighted adjustment periodically or
continually
A slowly time-varying channel response
7/27/2019 baseband demodulation.ppt
45/45
Review:baseband demodulation/detection
Matched filter
Receiver filter to provide the max. SNR, thus to provide the
min. BER
Raised cosine filter A low-pass Nyquist filter
At the receiver: to eliminate ISI
The channel characteristics are known
Equalizers
At the receiver: to mitigate the effect of ISI