Lectures Week 6 and 7

8/3/2019 Lectures Week 6 and 7

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Lecture slides Week 6: 6 Jan 2012


2/36

Distortion of line codes

Linear channeltime-domain effect

non-ideal magnitude, phase characteristics (or both) - dispersion fine balance of Fourier components of pulse apt to be disturbed

pulse spreads out in time inter-symbol interference (ISI)

Beyond a threshold, power is reflected back!

Low threshold for fibres ~ 5mW

Nonlinear channelfrequency-domain effect severe for high signal intensities

Stimulated Raman and Brillouin scattering in telecom fibres

cross-talk in WDM systems

Multi-path effects

transmission over data cables impedance mismatch

satellite communications

single-hop and multi-hop


3/36

How the channel can affect the signal

Special Raman fibres

extremely vigorous R&D

Can be tailored

Supercontinuum generation

Note - channel is a generic term medium between pulse generation and detection


4/36Source: G. P. Agarwals lecture

Do we have any control over the channel?


5/36

Visualizing ISI - the eye pattern

Normalized time, 1/Tb

Amplitude

time domain display of the received (PCM) line code

overlay repetitively sample received digital data signal ensure random samples

apply to oscilloscope vertical input

trigger the horizontal sweep the data rate 1/Tb - typically display one bit period

the eye

Capture one unit intervalTrigger

set trigger point sufficiently earlierin time than the acquired data -

ensure random bit values

Overlay

sync pulse

an edge in data stream

Single-Bit Eye Pattern Rendering

Also available since 2002, Continuous-Bit Eye Pattern Rendering by Le Croy


6/36

1. Ideal case

Infinite channel BW

no ISI

fully open eyes

2. Infinite BW

RZ pulsesp(t)

no ISI

fully open eyes

3. Finite channel BW

Distorting channel

Significant ISINRZ

RZ

NRZ

Illustration

Anything

unusual?

Severity of ISI Accuracy of timing extraction (time jitter)

Noise immunity how much noise can be tolerated without making a wrong decision

What information can be inferred?

S li f


7/36

Interior region of the eye pattern is called the eye opening

Sampling time - width of the eye opening defines the optimum sampling time (ISI-free)

Preferred sampling time is the instant of time at which the eye is open widest

Sensitivity to timing error is determined by the rate of closure of the eye as the sampling

time is varied

Height of eye opening, at a specified sampling time, defines the margin over noise

More noise = more closing of the eye

Salient features


8/36

Assessing signal integrity using the Eye pattern

1. Undistorted eye pattern for NRZ pulses no ISI, no significant timing jitter

2. Effect of low-pass channel


9/36

3. Effect of high-pass channel


10/36

Role of Digital Receivers

1. Reshape incoming pulses

Attenuation/ loss? Amplifier

Dispersion? Equalizer

2. Extract timing information- clock recovery

3. Symbol detection decision

4. Regenerate symbols

Absorption due to atmospheric water vapour


11/36

Equalization

Q: What is wrong with this received pulse?

Q: Didnt we take precautions?

Q: Why have things gone wrong?

Q: What can be done about it?

Received pulse

Invert the frequency response of the channel!!


12/36

Raised

cosinespectrum

Raisedcosine

spectrum

Transmitted

symbolspectrum

Transmittedsymbol

spectrum

Channel frequency

response

Channel frequency

response

Equalizer

frequencyresponse

Equalizerfrequency

response=

( ) ( ) ( ) ( )Z f B f H f E f =

0 ff

s= 1/T

( )B f

( )H f

( )E f

( )Z f

Basic idea:

Zero-forcing equalizer

Is complete equalization necessary?

Implementation


13/36

DelayTb

Delay

Tb

DelayTb

Delay

Tb

Delay

Tb

DelayTb

C-N C0 CN

C-3 C-2 C-1 C1

pr(t)

po(t)

( ) ( )N

o b n r b b

n N

p kT c p kT nT =

=

Implementation

Aim: force EQ output to have zero ISI at sampling instants

Delay

Tb

CN

Note:po(t) should satisfy Nyquist criterion

Zero forcing equalizer


14/36

Zero-forcing equalizer

Overall filter response should be

non-zero at decision time k= 0, and zero at all other sampling times k 0

0

1, 0[ ] [ ]

0, 1, 2,...

N

n r

n N

kp k c p k n

k N=

== =

=

1

1

1

1

1

[0] [ 1] ... [ 2 ] 0

[1] [0] ... [ 2 1] 0

:[ ] [ 1] ... [ ] 1

[2 1] [2 2] ... [ 1] 0

[2 ] [2 1] ... [0] 0

r N r N r N

r N r N r N

r N r N r N

r N r N r N

r N r N r N

p c p c p N c

p c p c p N c

p N c p N c p N c

p N c p N c p c

p N c p N c p c

+

+

+

+

+

+ + + =

+ + + + =

+ + + =

+ + + =

+ + + =

Infinite number of eqns!!

Specify values at (2N+1) points therefore (2N+1) equations

(k= N)

(k= 0)

(k= N)

0 r[ ] [P ][ ]p c=Toeplitz matrix

1

r 0Therefore, calculate [ ] [P ] [ ]c p=


15/36

Lecture slides Week 7: 13 Jan 2012

Detection/Decision makin


16/36

Additive Noise

Channel

input output

x(t) y(t)

Q: ISI has been removed what more is required in the detection process

Q: How does the receiver decide which bit was sent?

Q: Is it possible to knowa priori the possible outcomes?

Q: How often is a 1 mistaken for a 0? Figure of merit of a DCS

b t

A

1 bitx1(t)

b

-A

0 bit

t

x0(t)

receive

1 bit

b t

A

y1(t)

receive

0 bit

b

-A

y0(t)

Detection/Decision-making

A t i l s i


17/36

How can SNR be improved if signal and noise have overlapping PSDs?

Very importanttime-domain problem

Signalrecognition is required

Decision is based on the outcome of the process of recognition

A typical scenario

Digital 2 level PAM System


18/36

Digital 2-level PAM System

Transmitter Channel Receiver

bi

ClockTb

PAM g(t) hc(t) h(t)1

0

ak{-A,A} p(t) x(t) y(t) z(t i)

AWGN

n(t)

DecisionMaker

Threshold

Sample at

t=iTb

bits

ClockTb

matched

filter

pulse

shaper

Receiver equalizer not explicitly shown, but assumed to be there

Corrects for ISI before decision is made Often matched filter and equalizer are not distinct

1

2

( )

H

H

z T

Matched Filter


19/36

Matched Filter Detection of pulse in presence of additive noise

Receiver knowswhat pulse shape it is looking for

Channel memory assumed compensated by receiver equalizer

Additive white Gaussian noise

(AWGN) with zero mean and

varianceN0 /2

p(t)

Pulse

signal

n(t)

x(t)h(t)

y(t)

t= T, sampling instant

z(T)

Matched

filter0 0

( ) ( ) * ( ) ( ) * ( )( ) ( )

y t p t h t n t h t p t n t

= += +

Decide on the basis of the value of the test statistic z(T)

Directly proportional to received signal energy

Inversely proportional to noise

Threshold Detection


20/36

( ) ( )p T n T + Received pulses have values

Let pulse amplitude beAp

Threshold Detection

-3 -2 -1 0 1 2 3

0

0.1

0.2

0.3

0.4

x

2

21( ) exp( )

2 2n

n n

np n

= Noise has Gaussian pdf

Detection error if, 0, if

0, if

p p

p p

n n A

or

A n n A

+ > >

+ <

Ap

Or,

( |1) ( )pP P n A = < Similarly,

-Ap

Error probability given that 0

was transmitted

Error probability given that 1

was transmitted

Taking a decision


21/36

Values distributed about two mean values

Follows Gaussian distribution of the corrupting noise

Therefore, random variable with non-zero mean and variance of noise

( | 0) ( )

( )

p

p

n

P P n A

AQ

= >

=

( |1) ( )

( )

p

p

n

P P n A

AQ

= <

=

Q: Will the two error probabilities always be equal?

Q: How can P(x) be minimized?

Q: How does Q(x) vary?

-2 0 2 4 6

0

0 .1

0 .2

0 .3

0 .4

x

Likelihood ofs2 Likelihood ofs1

Taking a decision

The Q Function


22/36

he Q Funct on

Q function

Complementary errorfunction erfc

Relationship

2

/21( )2

x

y

Q y e dx

=

2

2( )y

x

erfc x e dy

=

=

22

1)(

xerfcxQ

-5 -4 -3 -2 -1 0 1 2 3 4 5

0

0.2

0.4

0.6

0.8

1

x

Q(x) = erfc (x/sqrt(2))/2

Q(x)

0 1 2 3 4 5

0

0.1

0.2

0.3

0.4

0.5

-3 -2 -1 0 1 2 3

0

0.1

0.2

0.3

0.4

x y

21( ) exp( )2

X p x x

=

Hw: Tabulate Q(x) using Matlab

x related to bit energy implications?

For a Gaussian variable given by,

Area Q(y)

Matched Filter Motivation


23/36

2

0

2

maximize , where is peak pulse SNR

| ( ) | instantaneous power =

average power{ ( )}

p T

E n T

=

Matched Filter Motivation

A matched filter attempts to -

maximize signal power at t= T

minimize noise i.e. power of 0 ( ) ( )* ( )n t n t h t =

0 ( ) ( ) * ( )p t p t h t =

Alternatively,

Pulse energy matters - pulse shape is irrelevant

p(t)

Pulse

signaln(t)

x(t)h(t)

y(t)

t= T

z(T)

Matched

filter

When?

}How?

0 bit

b-A

y0(t)

Matched Filter Derivation


24/36

2 2

0

2| ( ) | | ( )P( ) |

j fT p T H f f e df

=

Match F t r D r at on

Signal

== dffHN

dffStnE N202 |)(|

2)(})({

f

2

0N

Noise power

spectrum SN(f)

0 ( ) ( ) ( )P f H f P f =

0

2

( ) ( ) P ( )tj f

p t H f f e d f

=

0

20( ) ( ) ( ) | ( ) |

2 N N H NS f S f S f H f = =

p(t)

Pulse

signaln(t)

x(t)h(t)

y(t)

t= T

z(T)

Matched filter

0 ( ) ( ) * ( )p t p t h t =

0 ( ) ( ) * ( )n t n t h t =

AWGN Matched

filter

Noise

or,

Noise


25/36

f

2N

Noise power

spectrum SN(f)

BPower NB=

Thermal and Shot noise are practically flat up to 1THz!

( ) 2NS f kTR=

Thermal noise is due to the motion of electrons

Q: What kind of motion do electrons have?

Shot noise due to motion of charge carriers across junctions

Bandlimiting restricts the noise PSD input to a system



26/36

2 2 2 2

2

2 20

, for white Gaussian noise

| ( ) ( ) | | ( ) ( ) |

( ) | ( ) | | ( ) |2

j fT j fT

N

H f P f e df H f P f e df

NS f H f df H f df

= =

Schwartzs inequality

*

1 2( ) ( ) x k x k R =

22 2

1 2 1 2

- - -

( ) ( ) ( ) ( ) x x dx x dx x dx

Aim: To find h(t) that maximizes pulse peak SNR given by,

For functions

upper bound reached iff



27/36

*

Hence, ( ) ' ( )opth t k p T t =

*

2by Schw artz's inequality

T herefore, SN R is m aximized iff,

( )( )( )

opt

N

j fT P f H f k e k S f

=

What does this mean??

1 2

2 2 2

2

2

2

2

m ax

2

2

2

2

2

( )L et ( ) ( ) ( ) an d ( )

( )

( )P ( ) | | ( ) | | ( ) |

( ) ( ) | | ( ) |

( )( ) | ( ) |

| ( ) |,( )

N

N

-

-

N

N

N

j fT

j fT

j fT

P f e f H f S f f

S f

| H f f e d f H f d f P f d f

| H f P f e d f P fd f

S fS f H f d f

P fa n d d f S f

= =

=

=

where ' 2k k N=

The sampling instant


28/36

mp ng n nWhen should the decision be taken?

p(t)

p(-t)

p(tm-t)

p(tm-t)

p(tm-t)

t

t

t

t

t

T0

tm < T0

tm = T0

tm > T0

tm

tm

tm

Needless delay

Q: Why does not affect filtering?

Non-causal filter response - unrealizable

Optimum time instant

Matched Filter


29/36

Impulse response is hopt(t) = k p*(T - t)

Time-reversed (mirror image), and shifted version ofp(t)

Duration and shape determined by transmitted pulse shapep(t) known!

Constant k immaterial

Optimum - maximizes peak pulse SNR

2 2

max

0 0 0

max

2 22 2| ( ) | | ( ) | SNR

2Therefore, ( )

b

p

e

EP f df p t dt

N N N

EP Q Q

N

= = = =

= =

Does not

Q: Does the matched filters output necessarily resemble target signal?

Q: Does it matter?

depend on pulse shapep(t) Proportional to signal energy (energy per bit)Eb

Inversely proportional to power spectral density of noise

Matched Filter for Rectangular Pulse


30/36

g

Matched filter for causal rectangular pulse shape

Impulse response is causal rectangular pulse of same duration

Convolve input with rectangular pulse of duration Tsec and sample resultat Tsec is same as

First, integrate forTsec

Second, sample at symbol period Tsec

Third, reset integration for next time period

T

sample

Integrate and dump circuit

p(t)

Signal

n(t)

x(t) C

R

Dump switch

Sample switch

dump

Implementation of Matched Filter


31/36

Correlation receiver

0 0 0

( ) ( ) ( )

, ( ) ( ), ( ) ( ( )) ( )

r t y t h t x dx

where h t p T t and h t x p T t x p x T t

=

= = = +

t=nT

( ) ( ) ( )y t p t n t = +

( )p t( )r t

0( )r T Threshold

device Decision

0

0

0

0

( ) ( ) ( )

At the decision making instant, ( ) ( ) ( ) ( ) ( )

T

r t y t p x T t dx

r T y t p x dx y t p x dx

= +

= =

Inputs time-limitedto T0

Optimum Receiver


32/36

General case ofbinary signalling

( ) ( ) ( ), 0 for data symbol 1

( ) ( ) 0 for data symbol 0

b

b

y t p t n t t T

q t n t t T

= +

= +

p(t)H(f)

r(t)

t= Tb Threshold

devicer(Tb)

m = 0 if r(Tb) < a0

m = 1 if r(Tb) > a0

2

0

2

0

( ) ( ) ( )

( ) ( ) ( )

b

b

j fT

b

j fT

b

p T P f H f e df

q T Q f H f e df

=

=

22

( ) ( )n nS f H f df

=

-2 0 2 4 6

0

0.1

0.2

0.3

0.4

x

pr(r|0) pr(r|1)a0

Assuming a0 is the optimum detection threshold

2

0

| 2

[ ( )]1( | 0) exp2 2

b

r m

n n

r q Tp r

=

2

0

| 2

[ ( )]1( |1) exp2 2

b

r m

n n

r p Tp r

=

Conditional PDF of sampled output r(Tb)

A0A1

rq0(Tb) p0(Tb)

Notations

0.4

( |0) ( |1) P(e | m = 0) area under curve from a to


33/36

0 0 1 1

0 1

0 1

0 0 0 0

( | ) ( ) ( | ) ( )

0.5[ ( | ) ( | )], assuming equiprobable bits

1= [ ]2

( ) ( )12

e

b b

n n

P P e m P m P e m P m

P e m P e m

A A

a q T p T aQ Q

= +

= +

+

= +

Probability of error

Optimum value of a00 0

0

( ) ( )2

b bp T q T a +=

-2 0 2 4 6

0

0.1

0.2

0.3

x

pr(r|0) pr(r|1)a0

A0A1

P(e | m = 0) area under curve from a0 to

P(e | m = 1) area under curve from - to a0

q0(Tb) p0(Tb)

0 0 0 0

0

2 2

0 0 0 0

2 2

( ) ( )1 1 1= ' '2

[ ( )] [ ( ) ]1 1 1exp exp2 2 22 2

0

b be

n n n n

b b

n n nn n

a q T p T aPQ Q

a

a q T p T a

+

= =

( | 0) ( |1)Pe P e P e= =Corresponding value of Pe


34/36

2

2

0

0

( | 0) ( |1)

[ ( )]1exp2 2

( ) ( ) ( )

2

( ) ( ), where,

2

o b

n n

o b o b o b

n n

o b o b

n

a

Pe P e P e

r q T

dr

a q T p T q T Q Q

p T q T Q

=

= =

=

p g e

2

2

2

2

[ ( ) ( )] ( )

( ) ( )

b j fT

n

P f Q f H f e df

S f H f df

=

2

[ ( ) ( )]( )( )

b j fT

n

P f Q f e H f k S f

=

Therefore the optimum filter is

2

2

max

[ ( ) ( )]

( )n

P f Q f df

S f

=

Using Cauchy-Schwarz inequality

Substituting for p0(Tb) and q0(Tb)

For White Gaussian noise


35/36

2

( ) [ ( ) ( )]b j fT

H f P f Q f e

=

( ) ( ) ( )b bh t p T t q T t =

Filter transfer function

Filter matched to the pulsep(t) - q(t)

Corresponding , 2 22max

0

2 2( ) ( ) ( ) ( )

2

/ 2

bT

p q pq

P f Q f df p t q t dt N N

E E E

N

= =

+ =

Q: What do Ep and Eq represent?

0

( ) ( )bT

pqE p t q t dt =

max 2

2 2 p q pq

b

E E E P Q Q

N

+ = =

Impulse response

Bit Error Rate (BER)

Optimum value of a0 012

[ ]p qa E E=

Q: What does Epq represent? (Interference? Coherence?)

Realization


36/36

2 2

( ) ( ) ( )b b j fT j fT

H f P f e Q f e

= Equivalent optimum transfer function

p(Tb-t)

q(Tb-t)

+

-

Threshold device

0

0

( ) ( )

( ) ( )

b

b

q t if r T a

p t if r T a

r(t) r(Tb)

Parallel combination of two matched filters

Think of other combinations - simplifications

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Lectures Week 6 and 7