DIGITAL COMMUNICATIONS Part II Line Coding and Pulse Shaping

DIGITAL COMMUNICATIONS Part II

Line Coding and Pulse Shaping

©2000 Bijan Mobasseri 2

Statement of the Problem

On the way from transmitter to the receiver, we are now here…

Two questions:• How to pick pulse sequence?• How to pick pulse shape?s

Source encoder

Analog message

1 0 0 1 1...

?


Picking a pulse sequence:line coding

There are numerous ways we can convert a string of logical 1’s and 0’s to a sequence of pulses. But what are the ground rules?

Example: convert 1 0 1 1 0 1

1 1 1 10 0

1

0

1 1

0

1OR


Line coding concerns

Here are what we have to know before making a selection• Presence or absence of DC level• Spectrum at DC• Bandwidth• Built-in clock recovery • Built-in error detection• Transparency


DC level

Some line codes have a DC level some don’t. DC levels are inherently wasteful of power. Also, their DC levels will be stripped over links with transformers or capacitor coupled repeaters

1 1 1 10 0

1

0

1 1

0

1

DC level 0 DC


Spectrum at DC

This is related to the DC level. Generally we want to have the spectrum to be 0 at f=0. Finite spectrum at f=0, indicates partial presence of a DC level


Bandwidth

A most critical component. We must be able to compare the bandwidth taken up by different line codes while transmitting a fixed number of pulses per second.

BB


Clock recovery

Knowing the beginning of pulses at the receiver is critical in bit detection. Otherwise you don’t know where you are or what you are looking at. Can you tell where a pulse ends and where the next one begins?

Received pulses


Error detection

It is possible to come up with smart pulse sequences that can signal to the receiver there is something wrong

For example, let consecutive 1’s be represented by opposite polarity pulses

1 1

channel

1

1

It is not possible to have two neighboring pulses of the same polarity: It violates the encoding ruleThere must be a detection error

transmitted

received


Transparency

Data bits are often random but at times they my bunch up and form unbroken strings of 0’s or 1’s. In such cases statistics of pulse pattern may change

Look at the following two cases:

1

0

1 1

0

1

0 0 0 0

zero DC Non zero DC


Specific line codes

We will cover the following line codes• Unipolar • Polar• Split phase or Manchester• Bipolar or AMI (alternate mark inversion)• Coded mark inversion(CMI)• High density bipolar 3 (HDB3)


Unipolar (on-off)

Unipolar (return-to-zero, RZ)• Digit 0 is represented by NO PULSE• Digit 1 is represented by a half width pulse

Unipolar (non RZ)• Digit 0 is represented by NO PULSE• Digit 1 is represented by a full width pulse

10

10


Unipolar examples

The bit sequence is 1 0 1 1 0 0 0 0 1 0 1

Unipolar RZ:

Unipolar NRZ:Bit

length

Bitlength

Half-width pulses

Full width pulses


Unipolar properties by spectral analysis

Let the pulse rate be R. Pulse width is T=1/R

R 2R

Unipolar NRZ

2RR

Bandwidth=pulse rate

Bandwidth=2xpulse rate

Clock signals

Unipolar RZ

00


Could we have predicted the bandwidth?

Nyquist gave us this inequality: BT>R/2, i.e. need at least R/2 Hz to transmit R pulses per sec.

For unipolar we are transmitting R pulses per second. Unipolar spectrum gives us two answers

BT=R (NRZ)

BT=2R(RZ)


Why the difference?

Nyquist gives us the minimum possible bandwidth. Obviously, neither of the two signaling arrangements reach that goal

The reason is the choice of the pulse shape. We are working with square pulses. To meet Nyquist’s goal, we have to use sinc pulses. More on that later


Importance of clock component

If a unipolar NRZ signal is put through a narrowband bandpass filter centered at R, then a sinusoidal signal emerges that can be used for symbol timing

1/R 2RR

Clock signals

0

BPF

f


Polar

Encoding rule:• 1 is represented by a pulse• 0 is represented by the negative of the same pulse

Example: encode 1 0 1 1 0 0 0 0 1 0 1

RZ

NRZ


Polar spectrum

Identical to unipolar except for the absence of clock component at pulse rate

R 2R 2R

NRZRZ


Split phase or Manchester coding

Encoding rule:• 1 is represented by -->

• 0 is represented by-->

Encode 1 0 1 1 0 0 0 0 1 0 1 Bit length


Manchester code properties

There are some nice properties here• Zero DC because of half positive/half negative pulses• Transparent: string of 1’s and zeros will not affect DC levels

No DC at f=0 but large bandwidth and no clock

2RR


Bipolar or AMI

Bipolar or alternate mark inversion(AMI) used in PCM is defined by• O ----> no pulse• 1------> ±pulse. Sign of the pulse alternates

Example: 1 0 1 1 1

1

1 1

10

Alternating polarity


Advantages

AMI has several nice properties including null at DC. It also avoids signal “droop” over AC coupled lines

Result of a long string of1’s if polar format is used


Bipolar spectrum

Bipolar can be of RZ or NRZ variety. As usual, the NRZ has half the bandwidth of RZ

R 2RR/2f f

RZ NRZ


HDBn: G.703 standard for 2,8 and 34 Mbits/sec PCM

High density binary n line code is just like bipolar RZ with the following modification:• When the number of continuous zeros exceeds n, usually 3, they

are replaced by a special code

A string of 4 zeros is replaced by either 000V or 100V. V is a binary 1 with the sign chosen to violate the AMI rule. This will let the receiver know of the substitution


HDB3 example

Le the sequence be 10110000101. Here is the corresponding HDB3 coding

V

1 0 1 1 0 0 0 0 1 0 1

Must alternate in polarity


Coded Mark Inversion(CMI):G.703 for 140 Mbits/sec

CMI is a modified polar NRZ code. Pulses corresponding to 0’s stay put but 1’s alternate in polarity. In polar they had a fixed assignment

1 0 1 1 0 0 0 0 1 0 1


CMI spectrum

Similar 3-dB bandwidth to bipolar-NRZ but has a clock component at the pulse rate

R 2R f

Pulse Shaping

Choosing the right pulse for the job


What is the problem?

What we just went through was about picking a pulse sequence. We have not said anything about the pulse shapes that make up the sequence

Square pulses are by no means the best shapes to use


Criterion for picking pulse shapes

Whatever line coding you choose, the shape of the pulse greatly affects signal properties. The two we are most concerned with are• Bandwidth• Interference with adjacent pulses


ISI:intersymbol interference

Modern communication systems are mostly interference limited, not noise limited

ISI is a phenomenon in digital signaling that arises from the interaction of the signal and communication channel


How is ISI created?

There are a variety of reasons (bandlimited channels, multipath, fading). The end result is something like this

1 0 0 0

1 0 0 0

1 0 1 0

1 0 0 0 ISI causes detection error0’s may look like 1’s and vice versa

transmitted

received


A note on notation

We frequently talk about data rate either in terms of bits/sec(Rb) or pulses/sec(R). They are not the same because one pulse may represent several bits.

Similarly, we may talk about bit length (Tb=1/Rb) or pulse width(T=1/R)

We will use bit notations but results are applicable to pulses as well


Bandlimiting effect

Take square NRZ pulses at Rb bits/sec. The nominal pulse bandwidth is then Rb Hz.

Let’s investigate passage of this pulse through an ideal lowpass channel limited to Rb/2 Hz

1/Rb

?H(f)

Rb/2


Output in the frequency domain

In the frequency domain, input is a sinc.

RbRb

transmit

receive

Rb/2

Clearly, the received pulse is severely bandlimited compared to its original shape. How doesit look like in time?


Distorted pulse in time=pulse dispersion

As a result of bandlimiting, we witness pulse “dispersion”, i.e. spreading

transmitted

received

Tb/2

Tb

time


Notion of sampling instant

A digital receiver works on the basis of samples taken at periodic intervals.

These samples are either taken directly from the pulse (not likely) or after some preprocessing(filtering, equalization)


Dispersion impact on a pulse train

We never send just one pulse. We send a pulse sequence such as 1 1 1...

Sampling instants

Instead of picking up a sample from the current pulse, we are picking upsamples from adjacent samples as well


How to handle ISI?

ISI can be handled at two points, 1): source (by pulse shaping) and /or 2): equalization at the receiver

We will cover both but first look at ISI elimination by pulse shaping


What did Nyquist say about ISI?

He said we need not have zero ISI everywhere. In fact adjacent pulses may have substantial overlap

All we need to have is zero ISI at the sampling points where bit detection is performed


A typical scenario

Detecting bits at the receiver entails two main steps• Sampling of the received signal at the appropriate time• Decision making

Decision device

0

1


Nyquist’s first criterion for zero ISI

Nyquist defined zero ISI as the absence of ISI at sampling instants ONLY. Pulses can overlap each other at all other times with no ill effects


Pulse specification

This is what we want: transmit pulses at the rate of Rb bits/sec such that when the line is sampled at intervals 1/Rb sec., we experience NO ISI

Generally speaking, this is not possible because pulses spread out as they pass through channel


But… there is a way

Of all the candidate pulses, there is one that satisfies this seemingly difficult task. It is , you guessed it, a sinc.

To transmit Rb bits/sec. we should use pulses of the form

p(t)=sinc(Rbt)


How does a sinc work?

A sinc pulse has periodic zero crossings. If successive bits are positioned correctly, there will be no ISI at sampling instants.

Sampling instants

No ISI

Nadjacent pulsesGo to zero hence no ISI

Tb


Another distinction for sinc

We know that sinc(Rbt) has bandwidth of Rb/2 (Fourier table). Therefore, to transmit Rb bits/sec a bandwidth of Rb/2 would suffice

We learned before that this is the minimum transmission bandwidth


Nyquist’s 1st criterion for zero ISI:summary

To transmit data at Rb bits/sec use pulses of the form sinc(Rbt)

In addition to zero ISI, this pulse requires minimum possible bandwidth of Rb /2


Ideal Nyquist channel

Ideal Nyquist channel of bandwidth Rb /2 can support Rb bits/sec

Rb /2 is called the Nyquist bandwidth

Rb/2

1/Rb

Nyquist channel


Example: PCM voice channel

Remember that to transmit one voice channel, 8-bit PCM generates 64 Kb/sec. Recommend a pulse generating zero ISI

The desired pulse is p(t)=sinc(Rb t)=sinc(64000t)

Required bandwidth is BT=Rb /2=32 KHz


Problems with sinc

Sinc pulses are nice but are not realizable. Why? Because sinc is the impulse response of an ideal lowpass filter which itself cannot be built

H(f) H(f)sinc


Keeping the best of both worlds

What we want is a pulse that is 1): realizable and 2): has the same desirable ISI property of a sinc

To make a sinc realizable, we begin by smoothing the edges of its squared-off spectrum

This process is called “ roll-off ”


Raised cosine spectrum

We can smooth the sinc spectrum by the smoothing parameter . The cost is increased bandwidth

f

=0.5=0

=1

Rb/2 Rb


Excess bandwidth and roll off factor

Rolling off generates increased bandwidth beyond Nyquist’s. Roll-off factor is given:

a=1−f1 Rb /2( )

=0 f1 =Rb /2 original Nyquist's bandwidth

1 f1 =0 full roll-off⎧ ⎨ ⎩

=0

Rb/2

f1

Excess bandwidth


Bandwidth of raised cosine pulse

Raised cosine bandwidth varies according to the roll-off factor. Zero roll-off gives the Nyquist bandwidth. Full roll-off doubles it

BT=(1+)(Rb/2)


Raised-cosine pulse in the time domain

When the spectrum of sinc is smoothed out, it naturally affects the shape of the pulse in the time domain. The result is a modified sinc given by

p t( )=sincRbt( )cosπαt( )

1−4α2Rb2t2

⎡

⎣ ⎢ ⎤

⎦ ⎥

Keeps zero crossingsat t=multiples of Rb

Sinc, a=0

a=1

t1/Rb


Properties of raised-cosine

Raised cosine is very much sinc-like, i.e.• Has zero crossings at multiples of 1/Rb

• Has additional zero crossings between the old ones

Raised cosine also decays faster Side lobes are smaller than sinc All of the above properties help to diminish ISI


Special case: full raised cosine

For =1 we get a special case called full raised cosine pulse. For a bit rate of Rb, this pulse requires a bandwidth of Rb.

Rb/2 Rb

=0

BT=(1+1)Rb/2=Rb

=1


Raised-cosine pulse:full roll-off

In the time domain, raised-cosine pulse with full roll-off( =1 ) is given by

Properties:• Zero crossings at multiples of 1/2Rb (twice the sinc) . You can find

ZC’s by setting argument of sinc to ±1, ±2 etc

• Bandwidth is twice the sinc at BT=Rb

p t( )=sinc 2Rbt( )1−4Rb

2t2


Example: 64 Kbits/sec line

Compare shapes and spectra of candidate pulses below for ISI free transmission• Square pulse• Sinc pulse• full roll-off raised cosine


Square pulses:poor ISI

To transmit data at 64 KB/sec, we need pulses of 1/64000 sec (if NRZ)

Tb=1/64000 sec

BT=1/Tb=64 KHz


Other pulses

Sinc pulse corresponds to a roll-off factor of zero. Therefore,

BT=(1+ )Rb/2=32 KHz For a full roll-off raised cosine pulse, =1

BT=(1+ 1)Rb/2=64 KHz


Issue of bandwidth efficiency

The singular property of a digital signaling scheme is its bandwidth efficiency measured in bits/sec/Hz and found by the ratio =Rb/BT

measures how many bits/sec we can pump in one hertz of bandwidth


Bounds on

For a sinc we have BT=Rb/2. Therefore

=2 bits/sec/Hz For a full roll-off pulse BT=Rb. Therefore

=1 bit/sec/Hz For example, in a bandwidth of 4 KHz, the

fastest we can signal without ISI is 8 KHz using sincs


An interesting question

We know pretty well what pulse we need to work with for ISI-free transmission

The problem is we need sinc or raised cosine characteristic at the receiver

The question then is what kind of pulse we need to send in order to end up with an ISI-free pulse?


Channel model

Here is what we are looking at

The overall response must be such that the received pulse is ISI-free, e.g. sinc

HN(f)=HT(f)Hch(f)HR(f)

Transmit filter

ChannelReceive

filter

ISI-free pulseinput

HN(f)


Nyquist channel

What we need is for HN(f) to have a raised cosine response giving rise to raised cosine pulses at the receiver

Assuming perfect channel, Hch(f)=1, we can split the raised cosine response between HR(f) and HT(f)

HT f( ) =HR f( ) =cos2

πf2Rb

⎛

⎝ ⎜

⎞

⎠ ⎟ f ≤Rb

0 else

⎧

⎨ ⎪

⎩ ⎪


Nyquist’s channel response

Received pulses will have this spectrum

HN(f)=HT(f)HR(f)

frequencyRbRb/2

0.5

1


Overall response

This is the equivalent model we now have

HN(f)(raised cosine)

1 1 -1 -1 -1 1

Transmitter Receiver

anδ t−nTb( )n∑an = −1,1{ }

ann∑ hN t−nTb( )

EQUALIZATION

Dealing with ISI after the fact


Canceling ISI

We can eliminate ISI before it has happened (pulse shaping) or after it has happened (equalization)

The problem with pulse shaping is that channel characteristics, hence pulse shapes do not remain constant.

We need a mechanism to cancel ISI at the destination


Illustrating the problem

The delicate pulse shaping at the transmitter is messed up by the channel. ISI is back

Tb

ISI

Zero crossing has moved hence ISI with the adjacent pulse


What is equalization?

Equalization is a process by which a number of received pulses are processed in order to “null” ISI, i.e. drive it to zero, at the sampling point

In effect we want to create an ideal channel; flat frequency response and linear phase response. The delicate pulse shaping at the transmitter is therefore preserved


Setting the stage

Let p(t) be the current pulse sampled at t=Tb. As a result of channel, zero crossings do not coincide with sample times

Sample time=Tb

Sample time

ISI

Current pulse


Zero forcing equalizer

What we need to do is to somehow force the nonzero tails of the adjacent pulses to zero, therefore eliminating ISI at the sampling point

This can be accomplished by a weighted sum operation implemented by a digital filter


What do we want?

Here is what we want

Equalizerp (t) po(t)

po nTb( )=

1 n=0

0 n=±1,±2,K ,±N⎧ ⎨ ⎩

Tb2Tb

po(t)received


ZF equalizer as a digital filter

Tb Tb Tb

W-N W-N+1

Tb

Wo

Tb

WN

p(t)

input

po(t) = Wkp t−kTb( )k=−N

N

∑

One bit delay

unknowns


How to find the weights?

There are 2N+1 unknown weights. At the same time, po(t) must satisfy 2N+1 conditions

po nTb( )=

1 n=0

0 n=±1,±2,K ,±N⎧ ⎨ ⎩


Enforcing one condition

We want the equalized pulse to have a peak of 1 at t=0, i.e. no ISI. Setting po(0)=1

W−Np(+N)+W−N+1p(N −1)+...

+Wop(0)+W1p(−1)+...+WNp(−N) =1

p(0) Sample times

Received pulse


In matrix form...

Equalizer output can put in a compact matrix product form shown below

W−N,W−N+1,...,Wo,W1,...WN[ ]

p(N)

p(N −1)

M

p(0)

M

p(−N)

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

=1


2N+1 matrix equation

We have 2N+1 unknown weights and precisely 2N+1 equations

p(0) p(−1) L p(−2N)

p(1) p(0) L p(−2N +1)

M M M M

p(2N) p(2N −1) L p(0)

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

W−N

M

Wo

M

WN

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥

=

0

M

1

M

0

⎡

⎣

⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥

Known data


Example: 3-tap equalizer

Since there are 2N+1 taps, N=1

Tb Tb

W-1 Wo W1

p(t)

p(t+Tb) p(t-Tb)

po(t) =W−1p(t+Tb)+Wop(t)+

W1p(t−Tb)


Received pulse

Received pulse is sinc-like but its zero crossings have moved.

-0.3-0.2

1

p(0)=1p(Tb)=-0.3p(-Tb)=-0.2

Force them to zero


Setting up the matrix equation

The (2N+1)x(2N+1) data matrix reduces to the 3x3 matrix

p(0) p(−1) p(−2)

p(1) p(0) p(−1)

p(2) p(1) p(0)

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

W−1

Wo

W1

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

=

0

1

0

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

p(-2) p(-1) p(0) p(1) p(2)


Solving for the weights

Filling in the data matrix from pulse amplitudes, we can easily solve for W’s

1 −0.2 −0.05

−0.3 1 −0.2

0.1 −0.3 1

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

W−1

Wo

W1

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥ =

0

1

0

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥ ⇒ W =P−1I

in MATLAB

W=P \ I ⇒ W =

0.21

1.12

0.31

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥


Visual look at equalization

The matrix equation can be viewed as a sliding weighted sum

W1 Wo W-1

W1 Wo W-1

W1 Wo W-1

Each multiply-add givesOne equation

MATCHED FILTERING


Problem statement

The ultimate task of a receiver is detection, i.e. deciding between 1’s and 0’s. This is done by sampling the received pulse and making a decision

Matched filtering is a way to distinguish between two pulses with minimum error


Detection scenario

Here is a typical detection scenario. Received noisy pulse is sampled. Based on this sample a decision must be made; what was the transmitted bit?

+

noise

decision

0

1


Objective: lowest bit error rate (BER)

We are looking for a detector structure that over the long run provides the least number of bit errors.

Bit error is directly affected by the level of SNR at detection instant

It makes sense therefore to raise the SNR as high as possible before making a decision


Enter matched filter

Instead of sampling the received pulse directly, how about putting it through some kind of pre-processing with the objective of raising the SNR

+

noise

decision

0

1

Matchedfilter

(SNR)1 (SNR)2

Want (SNR)2 higher than (SNR)1


Defining parameters

Let’s define various symbols as follows

+ ?g(t): Transmitted pulse

w(t) noise

y(t) y(T)

t=T sample time

decision

Unknown filter:Matched filter


Filter output quantities

Filter output consists of two terms, pulse corresponding to g(t) and noise

y(t)=go(t)+n(t) Output SNR is defined at the sampling instant

t=T

η =signal powerat t =Taverage noisepower

=go(T)2

E n2 t( ){ }


What is the output signal at sample time t=T?

Filter’s I/O relationship is given byGo(f)=G(f)H(f)

In the time domain we have

go t( )= G f( )∫ H f( )ej2πftdt


Instantaneous signal power at t=T

Evaluating g(t) at t=T and squaring it

go T( )2 = H( f)G( f)ej2πfTdt−∞

∞

∫2

Notice t is replacedBy T


Matched filter output noise power

Assuming white noise coming in,

Question now is what is output noise power?

H(f)f

No/2

White noise spectrum

Sw(f)

SN(f)=(No/2)|H(f)|2


Average output noise power

Average power of any random process is the area under its power spectral density function. Therefore,

E n2(t){ }= SN( f)df =No

2H( f)2df

−∞

∞

∫−∞

∞

∫H(f)


MF output SNR

We are now in a position to write the output SNR of matched filter

Known: pulse spectrum G(f), noise PSD No/2, sampling time T

Unknown: filters response H(f)

η =

H( f)G( f)ej2πfTdf−∞

∞

∫2

No

2H( f)2 df

−∞

∞

∫


Next step: finding H(f)

Here is the premise: of all the filters you could use there is one that will generate the highest SNR at sampling instant.

The process leading to the solution is an optimization problem that in the end leads to the sought after H(f)


An intuitive solution

To maximize output SNR we must minimize output noise power. Look at the following signal and noise spectra at filter’s input

High SNR band

Low SNR band

Noise PSD

Pulse spectrum

Filter must restrict noise passage without restricting signal


Matched filter amplitude response

To pass just the signal and as little as noise as possible, filter’s amplitude response must be identical to input pulse frequency response:

|H(f)|=|G(f)| What about phase response?


Input pulse phase response

Input pulse has a phase spectrum profile best shown by looking at its Fourier series expansion

t

Amplitudes add incoherently

f

Pulse phase spectrumπ

-π


Output pulse phase response

For output signal amplitudes to add up coherently at t=T, output phase spectrum must be null, i.e. no phase difference among the components

Amplitudes add coherently

t=Tf

(f)=0


What does it take to make (f)=0?

Filter’s output phase spectrum is equal to the sum of input phase and filter’s phase spectrum. Recall

Go f( )=G( f)ejφg( f )

×H( f)ejφh( f )

=G( f ) H( f)ej φg( f )+φh( f )( )

φh( f)=−φg( f)


Characterizing Matched filter

We have established two facts• filter’s amplitude response=pulse amplitude spectrum• filter’s phase spectrum=negative pulse phase spectrum

Therefore, Matched filter is specified by

H(f)=kG*(f)exp(-j2πfT)

Sample timeconjugate


Example:define MF at t=2

Let the input pulse g(t) be

G(f)=FT((t-1))=sinc2(f)exp(-j2πf1) Using H(f) from previous slide

H(f)=kG*(f)exp(-j2πfT)=sinc2(f) exp(+j2πf1) exp(-j2πfx2)

H(f)=sinc2(f)exp(-j2πf)

t1

(t-1)


Matched filter in the time domain

MF has a more interesting form in the time domain.

Given g(t) as the input pulse, output SNR will be maximized if filter’s impulse response is given by

h(t)=kg(T-t)


MF interpretation

The impulse response of a MF is a flipped and translated pulse itself

Pulse ----->

Flipped---->

Shifted---->

t1

1

MF impulse response


Importance of sampling instant

Position of impulse response is controlled by the sampling instance. If we wanted to maximize SNR at t=2, h(t) would have been

t1 2

Shifted by t=2


MF for rectangular pulses

For a rectangular pulse centered at t=1 and width 2 design a matched filter in order to reach peak SNR at t=2

g(t)

h(t) g(t)*h(t)=2

2

1

2 4

Sample time

signal peaks at t=2for which the filter is designd for

this is g(t) flipped and shifted by 2


SNR at decision instant

We know that MF maximizes SNR at specific time T. Question is what is that SNR?

For a pulse of energy E immersed in white noise No/2, the output SNR is given by

SNR=2E/No


Interpretation

SNR is controlled by only two quantities; signal energy and noise PSD

Therefore, pulse shape has no bearing on the output SNR

Any pulse, as long as it has the same energy, will produce the same SNR


Example: SNR calculation

Let the pulse be triangular of height 10 mv and width 1.0 ms. Find SNR at the T=1 ms if noise is white with PSD 10-9 V2/Hz

10 mv

1 msec


Matched filter and sampling instant

Since sampling instant is at t=1, we should flip then shift by t=1 ms. The MF impulse response will then look like h(t)=kg(T-t)=kg(10-3-t)

Shift to t=1

T=1 msec


Finding output SNR

We must first find the pulse energy

SNR is then given by

E = g2(t)dt=0

T

∫ 20t[ ]2dt+0

0.5x10−3

∫ 0.02−20t[ ]2dt0.5x10−3

10−3

∫

=0.33×10−7 W-s

SNR=2ENo

=2×0.33×10−7

10×10−9 =6.67=8.2dB


Designing MF for two pulses

In all likelihood, we will be using at least two pulses in a digital comm system!

If they are of different shapes, how would you design a matched filter receiver?

1

0


Solution 1: filter bank

Build a filter bank, each branch matched to one pulse

h(t)=p1(T-t)

h(t)=p2(T-t)

Pick larger

0

1

p1(t)

here decision is in favor of p1 or digit 0


Solution II: use a single MF

It can be shown that when two pulses are involved, a single filter matched to the difference pulse p1(t)-p2(t), is the optimum filter

For 0

For 1

A

-A

2A

h(t)


Illustration

Let’s see what appears at the output of the single matched filter

A

-A

2A

2A2Tb

-2A2Tb

1’s and 0’s are clearlydistinguished


Nonoptimum filters

What happens if we just use a garden variety lowpass filter instead of the optimum matched filter?

Is there a loss in SNR and how much? Is the resulting hardware simplicity worth it?


RC filter:signal out

Let’s pick the simplest of all lowpass filters

R

C

H( f) =1

1+j2πfRCp(t)

po(t)

po t( )=A 1−e−

tRC

⎛

⎝ ⎜

⎞

⎠ ⎟ 0≤t≤Tb

Ap

Tb

po(t)

Ap =A 1−e−

TbRC

⎛

⎝ ⎜

⎞

⎠ ⎟

Sampling point


RC filter: noise out

Assume white noise No/2 at the input. Output PSD is then

Sno(f)=(No/2)|H(f)|2

Output noise power is the area under Sno

σno

2 =No

2df

1+(2πfRC)2=

No

4RC∫


Output SNR

Define output SNR as peak signal squared over noise power

Substituting

ρ =Ap

2

σn

2

ρ =4A2Tb

No

1−e−

TbRC

⎛

⎝ ⎜

⎞

⎠ ⎟

2

Tb RC

Goal: find the RC that maximizesoutput SNR


Solving for RC

Replace RC by x and set d/dx=0 and solve for x. The answer is

RC=Tb/1.26 The resulting output SNR is

max=(0.816)(2A2Tb/No)


Comparison with matched filter

For matched filter output SNR was (SNR)MF=2E/No=2A2Tb/No

Compare withmax =(SNR)RC =(0.816)(2A2Tb/No)

(SNR)RC=0.816(SNR)MF

(SNR)RC=(SNR)MF-0.9 dB


ISI at filter’s output

Consider the following scenario

Define ISI as the ratio of unwanted over wanted amplitudes; -11 dB with RC=Tb/1.26


Reducing ISI

One way to reduce ISI is to make pulse tails decay faster. This requires smaller time constants.

For example, for RC=Tb/2.3, ISI goes down to -20 dB.

The price we pay is that SNR is no longer at its peak value


Canceling ISI through integrate-and-dump

Integrate and dump is a procedure whereby the tail of the pulse is shorted out at the end of the bit interval or integration


Transversal MF

It is possible to realize a MF by a transversal filter the like of equalizer

DelayT

DelayT

DelayT

DelayT

anan-1 a1

+

r(t)=s(t)+n(t)

ro(t)=so(t)+no(t)

pick the weights tomaximize outputSNR at sample time

OPTIMUM DETECTION

What happens after matched filter?


Baseband binary system

PAMTransmit

filterchannel

Receivefilter

Decisiondevice

+

BinaryData

bk

{ak} s(t)

w(t)

x(t)y(t)

t=Tb

0

1

noisetransmitter channel receiver


Detection scenario

Once the received pulse is sampled (after MF or equalizer), we are left with a number

Based on this number we have to make a decision

+ MF

noise

Data=x(t)decisiony

y>-->1

y<-->0


Polar signal in noise

Assume polar signaling in noise. The received signal at MF’s input is given by

y(t)=±A+w(t)where w(t) is random white gaussian noise with

PSD of No/2


Case for bit errors

If 0(-A) is received, MF output is

Depending on noise strength, it is possible for y to swing over to positive territory causing a decision in favor of +A (1)

y t( )t=Tb= −A+w(t)[ ]

0

Tb

∫ dt= −Adesired{ +

1Tb

w(t)dt0

Tb

∫noise

1 2 4 3 4


Comments on the “observation”

y(Tb) is the defined as the “observation” upon which a decision will be made

y(Tb) however is a random variable. Therefore, laws of probability govern whether it exceeds or falls below the threshold


Types of bit errors

Type I• making a decision in favor of a 1 when 0 is transmitted

Type II• making a decision in favor of 0 when 1 is transmitted

Type I occurs when y(Tb) exceeds the pre-set threshold even though a negative pulse was transmitted

Type II occurs when y(Tb) falls short of the pre-set threshold even though a a positive pulse was transmitted


Statistics of observation

At the MF output we have

y(Tb) is a random variable with the following statistics

y t( )t=Tb= −A+w(t)[ ]

0

Tb

∫ dt= −Adesired{ +

1Tb

w(t)dt0

Tb

∫noise

1 2 4 3 4

E y Tb( ){ }=−A

var y Tb( ){ }=No2Tb


Distribution of y(Tb) when a 0 is transmitted

y(Tb) changes depending on whether a 0 or 1 is transmitted. Since noise is gaussian y(Tb) is a normal random variable with mean and variance just given

Its probability density function given that a 0 is transmitted is given by

f y0( ) =1

2πσye

−y+A( )2

2σ y2


Two density functions for y(Tb)

If a 1 is transmitted, y(Tb) is normal with mean +A. If a zero is transmitted, the mean is -A

A-A

f(y|1)f(y|0)

y


Detection errors

Here is the decision ruley>-->1 is transmittedy<-->0 is transmitted

Detection error occurs when a 1 is transmitted but we read y< or a 0 is sent and we read y>


Conditional error probability:Peo

The probability of error given that a 0 is transmitted is written as

P(e|0)=P(y>|0)=area under the curve beyond

y

f(y|0)

-A


…and Pe1

If 1 is transmitted, density function of y is normal and centered at +A

0 A

Pe1

y


Overall bit error rate

We want a single number quantifying systems bit error rate(BER). BER is obtained by averaging conditional BER’s

Pe=P(e|1)P(1)+P(e|0)P(0) P(0) and P(1) are “prior” probabilities. Meaning

in what proportion are 1’s and 0’s that are transmitted

Normally it is 50/50


Choice of threshold

Clearly threshold level on y makes a big difference in BER

It can be shown that for polar signals, the best optimum threshold is 0, i.e.

y>0 : decide in favor of 1y<0 : decide in favor of a 0


Illustrating two BER’s

One error can be reduced at the expense of another

f(y|1)f(y|0)

y

Peo

Pe1


Computing BER:Q-function

Computing BER boils down to finding areas under gaussian PDF. These numbers are tabulated for zero mean, unit variance gaussians as “Q-function”

a y

P(y>a)=Q(a)

0


Area under general gaussian PDF

y is gaussian but has mean and variance other than (0,1). What then? It is a normalization problem

ma

P(y>a)=Q((a-m)/)

mean mvariance 2


Properties of the Q-function

Due to the symmetry of gaussian PDF, there are similar symmetries on the Q-function. Examples are• P(y>-a)=1-Q(a)• P(y<-a)=Q(a)• P(-a<y<a)=1-2Q(a)

a

y

0-a


Complimentary error function:erfc

In the textbook, instead of the Q-function, erfc is used. erfc and Q are closely related

Q v( ) =12erfc

v2

⎛ ⎝

⎞ ⎠

erfc(u)=2Q 2u( )


Examples

If y is a gaussian random variable with mean 0 and variance 1, what is P(y>2)

We want Q(2). Table A7.1 shows erfc tabulation. Converting to Q

Q(2)=0.5erfc(1.4)=0.5x0.95229=0.476


Non standard gaussian

If y is normal with mean 2 and variance 4, what is P(y>3)

We hadP(y>a)=Q((a-mean)/std.dev))=

Q((3-2)/2)=Q(0.5) Find Q(0.5)


BER for polar signals

We will talk about BER’s at length in chapter 8. For now, for polar signaling, the most power efficient binary signaling:

where Eb is energy per bitBER=

12

erfcEb

No

⎛

⎝ ⎜

⎞

⎠ ⎟


Defining Eb

If a bit is transmitted via a pulse p(t) of length Tb, then

For a square pulse of height AEb=A2Tb

Eb = p2 t( )dt0

Tb

∫


Power in the bitstream

We don’t send just one pulse. It is then reasonable to ask what is the power in a bitstream running at Rb bits/sec

Power=(energy per bit)x(bits per second)=EbRb


Example

In a polar scheme running at 1000 bits/sec. pulses used have amplitudes of 1mv. What is the bitstream power?

Eb=A2Tb=(0.001)2x(0.001)=10-9 watt-sec Power is given by

P= EbRb=(10-9)x103=10-6 watts


Average Eb

In many line codings, energy per bit may not be the same for 0 and 1. Example is unipolar where one of the pulses is 0

We therefore have to work with average Eb

(Eb)avg=(1/2)Ebo+(1/2)Eb1

For unipolar,(Eb)avg=(1/2)x0+(1/2)Eb1=(1/2)Eb1

P=EavgRb


Where do we go from here?

All we have done so far has been in baseband. The next step is to take the line codes in

baseband and map them to bandpass Chapter 8 is about digital modulations, where

rubber meets the road

EYE DIAGRAM

Picking the best sampling point


Defining eye diagram

Imagine looking at a bitstream on a scope triggered by the symbol clock. Observation window is one bit long. Here is an ideal case


Problem of sampling point

In the presence of noise and ISI, the eye diagram will look something like this. Question is where in {0-Tb} do you sample?

Tb

Which point would you choose To sample the pulse?

Superposition of Many pulses


Actual examples

Let’s pass a polar NRZ bitstream through a pre-detection filter and look at its output in the presence of varying SNR.

Then, present the output in the form of eye diagram

%generating an eye pattern

%B.Mobasseri

%3/21/99

n=100;%number of bits

stdN=0.1;%noise standard deviation

%Generate bitstream, add noise then filter

p=bitstream(100,'polar_nrz');% n polar NRZ

snr=20*log10(std(p)/stdN) %show SNR

h=0.1*ones(1,10);%matched filter(output normalized)

p=p+stdN*randn(1,length(p));%add noise

y=conv(p,h);%find filter's output

y=y(6:length(y)-4);%offset pulse for proper display

z=reshape(y,10,length(y)/10);%rearrange data to simulate hold command

plot(z);grid

%Generate dynamic figure title

k=['SNR=',num2str(snr),' dB'];

title(k)


Anatomy of an eye pattern

1 2 3 4 5 6 7 8 9 10-1.5

-1

-0.5

0

0.5

1

1.5SNR=19.9974 dB

Max.eye openingDegradationIn eyeopening


Lowering SNR

1 2 3 4 5 6 7 8 9 10-1.5

-1

-0.5

0

0.5

1

1.5SNR=5.9971 dB

Eye closing


Lowering SNR still

1 2 3 4 5 6 7 8 9 10-2

-1.5

-1

-0.5

0

0.5

1

1.5

2SNR=-0.0026092 dB

Shrinking eye opening


Eye for polar RZ

1 2 3 4 5 6 7 8 9 10-1.5

-1

-0.5

0

0.5

1

1.5SNR=19.9607 dB for polar RZ


Eye for polar RZ

1 2 3 4 5 6 7 8 9 10-1.5

-1

-0.5

0

0.5

1



Eye for polar RZ

1 2 3 4 5 6 7 8 9 10-2

-1.5

-1

-0.5

0

0.5

1

1.5

2SNR=-0.023539 dB for polar RZ


Polar RZ,NRZ side-by-side

1 2 3 4 5 6 7 8 9 10-1.5

-1

-0.5

0

0.5

1


1 2 3 4 5 6 7 8 9 10-1.5

-1

-0.5

0

0.5

1

1.5SNR=19.9974 dB



1 2 3 4 5 6 7 8 9 10-1.5

-1

-0.5

0

0.5

1


1 2 3 4 5 6 7 8 9 10-1.5

-1

-0.5

0

0.5

1

1.5SNR=5.9971 dB



1 2 3 4 5 6 7 8 9 10-2

-1.5

-1

-0.5

0

0.5

1

1.5

2SNR=-0.023539 dB for polar RZ

1 2 3 4 5 6 7 8 9 10-2

-1.5

-1

-0.5

0

0.5

1

1.5

2SNR=-0.0026092 dB

Documents

DIGITAL COMMUNICATIONS Part II Line Coding and Pulse Shaping