baseband demodulation.ppt

7/27/2019 baseband demodulation.ppt

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Chapter Four:Baseband Demodulation/Detection


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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


3/45


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?


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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?


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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?


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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


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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


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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


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


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


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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

>


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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


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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


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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


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


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


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


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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


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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


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


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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


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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


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


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


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


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


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


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


30/45

The Raised Cosine Filter

sRrW )1(2

1


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


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


33/45

Performance Degradation


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


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


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


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


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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,)(


39/45


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


40/45


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


41/45


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


42/45


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


43/45

EqualizationDecision Feedback Equalizer

A nonlinear equalizer

Using previous detector decisions to eliminate the ISI


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


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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

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baseband demodulation.ppt