DIGITAL COMMUNICATIONS Part II

DIGITAL COMMUNICATIONS Part II
Line Coding and Pulse Shaping 1999 Bijan Mobaseri

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 ©2000 Bijan Mobasseri

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 1 1 1 1 1 1 1 OR ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 1 1 1 1 1 DC level 0 DC ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 transmitted 1 1 It is not possible to have two neighboring pulses of the same polarity: It violates the encoding rule There must be a detection error channel 1 received ©2000 Bijan Mobasseri

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: zero DC Non zero DC 1 1 1 1 ©2000 Bijan Mobasseri

Specific line codes We will cover the following line codes Unipolar
Split phase or Manchester Bipolar or AMI (alternate mark inversion) Coded mark inversion(CMI) High density bipolar 3 (HDB3) ©2000 Bijan Mobasseri

Unipolar (on-off) Unipolar (return-to-zero, RZ) Unipolar (non RZ)
Digit 0 is represented by NO PULSE Digit 1 is represented by a half width pulse Unipolar (non RZ) Digit 1 is represented by a full width pulse 1 1 ©2000 Bijan Mobasseri

Unipolar examples The bit sequence is Unipolar RZ: Unipolar NRZ:
Unipolar RZ: Unipolar NRZ: Bit length Half-width pulses Bit length Full width pulses ©2000 Bijan Mobasseri

Unipolar properties by spectral analysis
Let the pulse rate be R. Pulse width is T=1/R Bandwidth=2xpulse rate Bandwidth=pulse rate Clock signals R R R 2R Unipolar NRZ Unipolar RZ ©2000 Bijan Mobasseri

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) ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 Clock signals 1/R R 2R BPF f ©2000 Bijan Mobasseri

Polar Encoding rule: Example: encode 1 0 1 1 0 0 0 0 1 0 1
1 is represented by a pulse 0 is represented by the negative of the same pulse Example: encode RZ NRZ ©2000 Bijan Mobasseri

Split phase or Manchester coding
Encoding rule: 1 is represented by --> 0 is represented by--> Encode Bit length ©2000 Bijan Mobasseri

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 R 2R ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

Choosing the right pulse for the job
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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

How is ISI created? There are a variety of reasons (bandlimited channels, multipath, fading). The end result is something like this transmitted ISI causes detection error 0’s may look like 1’s and vice versa received ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 H(f) ? 1/Rb Rb/2 ©2000 Bijan Mobasseri

Output in the frequency domain
In the frequency domain, input is a sinc. receive Rb Rb Rb/2 Clearly, the received pulse is severely bandlimited compared to its original shape. How does it look like in time? transmit ©2000 Bijan Mobasseri

Distorted pulse in time=pulse dispersion
As a result of bandlimiting, we witness pulse “dispersion”, i.e. spreading transmitted received Tb Tb/2 time ©2000 Bijan Mobasseri

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) ©2000 Bijan Mobasseri

Dispersion impact on a pulse train
We never send just one pulse. We send a pulse sequence such as Instead of picking up a sample from the current pulse, we are picking up samples from adjacent samples as well Sampling instants ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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) ©2000 Bijan Mobasseri

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. No ISI Nadjacent pulses Go to zero hence no ISI Tb Sampling instants ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ” ©2000 Bijan Mobasseri

Raised cosine spectrum
We can smooth the sinc spectrum by the smoothing parameter . The cost is increased bandwidth =0.5 =0 =1 Rb/2 Rb f ©2000 Bijan Mobasseri

Excess bandwidth and roll off factor
Rolling off generates increased bandwidth beyond Nyquist’s. Roll-off factor  is given: f1 =0 Excess bandwidth Rb/2 ©2000 Bijan Mobasseri

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) ©2000 Bijan Mobasseri

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 Sinc, a=0 a=1 Keeps zero crossings at t=multiples of Rb t 1/Rb ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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. BT=(1+1)Rb/2=Rb  =0  =1 Rb/ Rb ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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? ©2000 Bijan Mobasseri

HN(f)=HT(f)Hch(f)HR(f)
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) input Transmit filter Channel Receive filter ISI-free pulse HN(f) ©2000 Bijan Mobasseri

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) ©2000 Bijan Mobasseri

Nyquist’s channel response
Received pulses will have this spectrum 1 HN(f)=HT(f)HR(f) 0.5 frequency Rb Rb/2 ©2000 Bijan Mobasseri

Overall response HN(f) Receiver Transmitter
This is the equivalent model we now have HN(f) (raised cosine) Receiver Transmitter ©2000 Bijan Mobasseri

Dealing with ISI after the fact
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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 Sample time=Tb Current pulse ISI ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

What do we want? Here is what we want Equalizer p (t) po(t) po(t)
received Tb 2Tb ©2000 Bijan Mobasseri

ZF equalizer as a digital filter
One bit delay p(t) Tb Tb Tb Tb Tb input W-N W-N+1 Wo WN unknowns  ©2000 Bijan Mobasseri

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 p(0) Sample times Received pulse ©2000 Bijan Mobasseri

Example: 3-tap equalizer
Since there are 2N+1 taps, N=1 p(t) p(t+Tb) Tb Tb p(t-Tb) W-1 Wo W1  ©2000 Bijan Mobasseri

Setting up the matrix equation
The (2N+1)x(2N+1) data matrix reduces to the 3x3 matrix p(-2) p(-1) p(0) p(1) p(2) ©2000 Bijan Mobasseri

Solving for the weights
Filling in the data matrix from pulse amplitudes, we can easily solve for W’s ©2000 Bijan Mobasseri

Visual look at equalization
The matrix equation can be viewed as a sliding weighted sum Each multiply-add gives One equation W1 Wo W-1 W1 Wo W-1 W1 Wo W-1 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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? decision + 1 noise ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 Matched filter decision + 1 (SNR)1 (SNR)2 noise Want (SNR)2 higher than (SNR)1 ©2000 Bijan Mobasseri

Defining parameters ? Let’s define various symbols as follows
g(t): Transmitted pulse y(T) decision + ? y(t) t=T sample time w(t) noise Unknown filter:Matched filter ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

What is the output signal at sample time t=T?
Filter’s I/O relationship is given by Go(f)=G(f)H(f) In the time domain we have ©2000 Bijan Mobasseri

Instantaneous signal power at t=T
Evaluating g(t) at t=T and squaring it Notice t is replaced By T ©2000 Bijan Mobasseri

Matched filter output noise power
Assuming white noise coming in, Question now is what is output noise power? Sw(f) H(f) SN(f)=(No/2)|H(f)|2 No/2 f White noise spectrum ©2000 Bijan Mobasseri

Average output noise power
Average power of any random process is the area under its power spectral density function. Therefore, H(f) ©2000 Bijan Mobasseri

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) ©2000 Bijan Mobasseri

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 Filter must restrict noise passage without restricting signal Pulse spectrum Low SNR band Noise PSD ©2000 Bijan Mobasseri

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? ©2000 Bijan Mobasseri

Input pulse phase response
Input pulse has a phase spectrum profile best shown by looking at its Fourier series expansion Pulse phase spectrum π Amplitudes add incoherently f -π t ©2000 Bijan Mobasseri

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 (f)=0 f t=T ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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) conjugate Sample time ©2000 Bijan Mobasseri

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) t 1 ©2000 Bijan Mobasseri

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) ©2000 Bijan Mobasseri

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 Shifted by t=2 t ©2000 Bijan Mobasseri

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) Sample time 1 2 h(t) g(t)*h(t)= 4 2 this is g(t) flipped and shifted by 2 signal peaks at t=2for which the filter is designd for 2 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 h(t) 2A A For 0 For 1 -A ©2000 Bijan Mobasseri

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? ©2000 Bijan Mobasseri

RC filter:signal out Let’s pick the simplest of all lowpass filters
p(t) C po(t) po(t) Ap Tb Sampling point ©2000 Bijan Mobasseri

Output SNR Define output SNR as peak signal squared over noise power
Substituting Goal: find the RC that maximizes output SNR ©2000 Bijan Mobasseri

Comparison with matched filter
For matched filter output SNR was (SNR)MF=2E/No=2A2Tb/No Compare with max =(SNR)RC =(0.816)(2A2Tb/No) (SNR)RC=0.816(SNR)MF (SNR)RC=(SNR)MF-0.9 dB ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

Transversal MF It is possible to realize a MF by a transversal filter the like of equalizer r(t)=s(t)+n(t) Delay T Delay T Delay T Delay T an-1 pick the weights to maximize output SNR at sample time a1 an + ro(t)=so(t)+no(t) ©2000 Bijan Mobasseri

What happens after matched filter?
OPTIMUM DETECTION What happens after matched filter?

Baseband binary system
Data bk y(t) {ak} s(t) x(t) PAM Transmit filter channel Receive filter Decision device + 1 t=Tb w(t) noise transmitter channel receiver ©2000 Bijan Mobasseri

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 y>-->1 y<-->0 Data=x(t) + MF y decision noise ©2000 Bijan Mobasseri

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) ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

Types of bit errors Type I Type II
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 ©2000 Bijan Mobasseri

Statistics of observation
At the MF output we have y(Tb) is a random variable with the following statistics ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 f(y|0) f(y|1) y -A A ©2000 Bijan Mobasseri

Detection errors Here is the decision rule
y>-->1 is transmitted y<-->0 is transmitted Detection error occurs when a 1 is transmitted but we read y< or a 0 is sent and we read y> ©2000 Bijan Mobasseri

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  f(y|0) -A  y ©2000 Bijan Mobasseri

Pe=P(e|1)P(1)+P(e|0)P(0)
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 ©2000 Bijan Mobasseri

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 1 y<0 : decide in favor of a 0 ©2000 Bijan Mobasseri

Illustrating two BER’s
One error can be reduced at the expense of another f(y|1) f(y|0) y Pe1 Peo ©2000 Bijan Mobasseri

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” P(y>a)=Q(a) a y ©2000 Bijan Mobasseri

Area under general gaussian PDF
y is gaussian but has mean and variance other than (0,1). What then? It is a normalization problem mean m variance 2 P(y>a)=Q((a-m)/) a m ©2000 Bijan Mobasseri

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) y -a a ©2000 Bijan Mobasseri

Complimentary error function: erfc
In the textbook, instead of the Q-function, erfc is used. erfc and Q are closely related ©2000 Bijan Mobasseri

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)=0.5x =0.476 ©2000 Bijan Mobasseri

P(y>a)=Q((a-mean)/std.dev))=
Non standard gaussian If y is normal with mean 2 and variance 4, what is P(y>3) We had P(y>a)=Q((a-mean)/std.dev))= Q((3-2)/2)=Q(0.5) Find Q(0.5) ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

Eb=A2Tb=(0.001)2x(0.001)=10-9 watt-sec
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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

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 ©2000 Bijan Mobasseri

Picking the best sampling point
EYE DIAGRAM Picking the best sampling point

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? Superposition of Many pulses Which point would you choose To sample the pulse? Tb ©2000 Bijan Mobasseri

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) ©2000 Bijan Mobasseri

DIGITAL COMMUNICATIONS Part II

Similar presentations

Presentation on theme: "DIGITAL COMMUNICATIONS Part II"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

DIGITAL COMMUNICATIONS Part II

Similar presentations

Presentation on theme: "DIGITAL COMMUNICATIONS Part II"— Presentation transcript:

Similar presentations

About project

Feedback