1-1

Communication Systems

Lecture 1

1-2

Why digital communication

1-Ease generation of digital signals if compared with analog ignals

generation as shown in fig(1.1)

!Distance 1 Distance 2 Distance 3 Distance 4

Original pulse some signal distortion Degraded signal Amplification

to regenerate pulse

1 2 3 4

Fig (1.1)

-The shape of the waveform is affected by two basic mechanisms;

a-All transmission lines and circuits have some nonideal frequency

transfer function, there is a distorting effect on the ideal pulse.

b-Unwanted electrical noise or other interference farther distorts the

pulse waveform .

-Circuits that perform the regeneration are called regenerative

repeaters

2-Digital circuits are less subject to distortion and interference than

analog circuits. Because binary digital cct operate in one of two

states either one (fully on) or zero (fully off) Such two state

operation facilitates signal regeneration and thus prevents noise and

1-3

other disturbance from accumulating in transmission. Analog

signals, however, are not two-state signals, they can take any

infinite variety of shaps. Once the analog signal is distorted, the

distortion cannot be removed by amplification, digital circuits are

more reliable and can be produced at a lower cost then analog ccts.

3- Digital cct . are more reliable and can be produced at a lower

cost . than analog ccts .

4-Digital hardware is more flexible implementation than analog

hardware (e.g microprocessor, digital switching and large scale

integrated (LSI) cct

5-Multiplexing of digital signals (TDM)is simpler than analog

signals(FDM).

6-Different type of digital signals (data, telegraph , telephone,

television ) can be treated as identical signals in transmission and

switching

7-Digital techniques protect themselves against interference and

jamming by using error correction codes, signal processing and

cryptography

8-Data communication always is from computer to computer or

from digital instruments or terminal to computer such digital

terminations , are best served by digital communication link

1-4

Disadvantages of Digital Communication

1-Signal processing for digital communication is more complex

than for analog communication

2-Synchronization for digital communication is more complex than

analog communication

3-Digital communication system is non graceful for degradation,

however, analog system degrade more graceful

Typical Block Diagram of a Digital Communication

Fig(1.2) shows a typical block diagram of digital communication

system (DCS) .

1-The upper blocks –format, source encode , encrypt ,channel

encode , multiplex, pulse modulate , bandpass modulate , frequency

spread , and multiple access denote signal, transformations from the

source to the transmitter(TR)

2-The lower blocks denote signal transformation from the

receiver(RC)to the sink , essentially reversing the signal processing

steps performed by the upper blocks

3-For wireless applications ,the transmitter consist of frequency up

conversion stage to radio frequency (RF),high power amplifier and

antenna. The receiver portion consist of an antenna and low-noise

amplifier(LNA).Frequency down conversion is performed in the

front end of the receiver.

1-5

4-The input information source is converted to binary digits (bits) ,

the bits are then grouped to form digital messages or message

symbols each symbol (mi, where i=1,2,….M)can be regarded as

member of a finite alphabet set containing M members. Thus, for m

=2 the message symbol Mi is binary

5-For analog system , the message waveform is typically of an

infinite set of possible waveforms.

6-For systems that use channel coding ( error correction coding),

sequence of channel symbols (code symbols). Because a message

symbol or a channel symbol can consist of a single bit or grouping

of bits, a sequence of such symbols is also described as a bit stream

as shown in fig (1.2)

7-The essential blocks for a digital communication system are;

formatting, modulation, demodulation /detection and

synchronization

a-Formatting transforms the source information into the bits from

this point in the fig(1.2) up to pulse modulation block , the

information remains in the form a bit stream .

b- Modulation is the process by which message symbols or channel

symbols (when channel coding is used) are converted to waveforms

that are compatible with the requirements of the transmission

channel . Pulse modulation is the process of conversion each

message symbols an channel symbols from binary representation to

a base band waveform. The term of a base band refers to a signal

1-6

whose spectrum extend from(or near) dc up to some finite value,

usually less than a few megahertz when pulse modulation is applied

to binary symbol the resulting binary waveform is called pulse

code modulation (PCM)waveform. When pulse modulation is

applied to nonbinary symbols, the resulting waveform is called an

M-ary pulse modulation waveform.

c-Bandpass modulation, it is required whenever the transmission

medium will not support the propagation of pulse like waveforms

The term bandpass is used to indicate that the baseband waveform

is frequency translated by a carrier wave to a frequency that is

much larger than the spectral content of the base band

additive random noise distorts the received signal during the

various points along the signal route

d-Demodulation /detection , the receiver front end and the

demodulator provides frequency down conversion for each band

pass waveform and to restores to an optimally shaped baseband

pulse in preparation for detection.

e-Equalization is used in or after the demodulator as a filtering

option to reverse any degrading effects on the signal that were

caused by the channel. Equalization becomes essential whenever

the impulse response of the channel is so poor that the received

signal is badly distorted. An equilizer is implemented to

compensate for (i-e remove or diminish) any signal distortion

caused by channel

1-7

Note

Demodulation is defined as recovery of a waveform (baseband

pulse) and detection is defined as a decision making regarding the

digital meaning of that waveform (sometimes use the terms

demodulation and detection interchangeably)

f-Source coding produces analog to digital conversion and removes

redundant (unneeded) information. The typical digital

communication system (DCS) would either used source coding or it

would use the simpler formatting transformation (for digitizing

alone)

g-Encryption is used to provide privacy for communication

channel.

h-Multiplexing and multiple access produces combine signals that

might have different characterisitics or might originate from

different sources. The main difference between the two is that

multiplexing takes place locally (e.g printed circuit board, with an

assembly or even within a facility) and multiple access takes place

remotely (e.g multiple users need to share the satellite transponder)

i-Frequency spreading is used to increase the capability of the

channel to resiste the interference(both natural and intentional ). It

is also a valuable technique used for multiple access

j. Synchronization is involved in the control of all signal processing

within the digital communication channel.

1-8

Basic Digital Communication Terms .

1-Information source; this the devise producing information to be

communicated by means of the DCS. Information sources can be

analog or discrete or discrete .The output of analog source can have

any value in a continuous range of amplitudes , whereas the output

of a discrete information source takes its value from a finite set.

Analog information can be transformed into digital source through

the use of sampling and quantization. Sampling and quantization

techniques called formatting and source coding

2-Textual message. This is a sequence of characters(e.g, ABC, ok,

..$9, 567, 273…).For digital transmission, the message will be a

sequence of digits or symbols from a finite symbol set or alphabet

3-Character. A character is a number of an alphabet or set of

symbols [e.g , A, g , $.....] characters may be mapped into a

sequence of binary digits. There are several standardized codes

used for character encoding including the American Standard code

for information interchange(ASCII)

4-Binary digit (bit); this is the fundamental information unit for all

digital systems .The term bit also is used as a unit of information

content

5-Bit stream; this is a sequence of binary digits (ones and zeros) A

bit stream is often termed a baseband signal .

1-9

6-Symbol (digital message) ; A symbol is a group of K bits

considered as a unit. We refer to this unit as a message symbol

mj(i-1,......,m)from afinit symbolset or alphabet

[e.g 1 binary symbol (K=1,M=2)

10 quaternary symbol(K=2,M=4)

011 8-ary symbol (K=3,M=8)]

7-Digital waveform ; this is a voltage or current waveform (a pulse

for baseband transmission) that represents a digital symbol

8-Data rate; this quantity in bits per second (bit/s)

Digital versus Analog Performance Criteria

A principle difference between analog and digital communication

systems has as to do with the way in which we evaluate their

performance

1-Analog systems draw their waveform from an infinite set, that is

a receiver must deal with an infinite number of possible wave

shapes .The figures of merit for the performance of analog systems

are:-

Signal to noise ratio, percent of distortion or expected mean square

error between transmitted and received waveforms.

2-Digital communication system transmits signal that represent

digits. These digits form a finite set or alphabet, and the set is

known a priori to the receiver. The figure of merit for the

performance of digital systems is :-probability of error .

1-10

Fig (1-2) Typical block diagram of a digital Comm .System

2-1

Communication Systems

Lecture 2

2-2

Classification of signals

1.Deterministic and Random Signals

A signal can be classified as deterministic , meaning that there is

no uncertainty with respect to its value at any time or as random

that there is some degree uncertainty before the signal actually

occurs .Deterministic signals or waveforms are modeled by clear

mathematical expressions , such as 5t)(χ cos10t . For random

waveform it is not possible to write such clear mathematical

expressions .

2.Periodic and Nonperiodic Signals

A signal t)(χ is called periodic in time if there exist a constant

To > 0 such that

( t) = [ t + T0 ] for -∞< t <∞ (2.1)

where t denotes time. The resultant value To that satisfies this

condition is called the periodic of t)(χ .The period To defines

the duration of one complete cycle of t)(χ

A signal for which there is no value of To that satisfies Equattion

(2.1) is called nonperodic signal.

2-3

3-Analog and Discrete Signals

An analog signal t)(χ is a continuous function of time, that is

Fig (2.1)

t)(χ is defined for all t (e.g speech).

By comparison a discrete signal χ(KT) is one that exists only at

discrete times, it is characterized by a sequence of number for

each χ (KT)

4-Analog and digital Signals

a-A signal whose amplitude can take any value in a continuous

range is an analog signal . This means that an analog signal

amplitude can take on an infinite number of values

b-A digital signal can take only a finite number of values. For a

signal to be define as digital , the number of values need not be

restricted to two (It can take any finite number as in the case

M-ary signal).

c-The terms continuous time and discrete time define the nature

of signal along the time (horizontal axis) .

2-4

The terms digital and analog define the nature of the signal

amplitude (vertical axis). Fig(2.2) shows examples of various

types of signals .It is clear that the analog signal is not necessarily

continuous time and digital signal is not necessarily discrete time

Fig (2.2)

5-Energy and power signals

»An electrical signal can be represented as a voltage (t) or

i(t)with instantaneous power P(t) across a resistor R defined by

)2.2.....(..........)()(2)( 2 Ri

R

ttP

g ( t ) g (t)

Fig (2-1)

Analog continuous time Digital continuous time

g (t) g (t) Digital discrete time

t Analog- continuous time

2-5

»In communication systems ,power is often normalized by

assuming R=1 ,although R may be another value in the actual

circuit . For normalized case , regardless of whether the signal is a

voltage or current waveform ,the normalization convention allow

us to express the instantaneous power as

P (t) = )3.2.......(..........)(2 t

where .)(t is either a voltage or a current signal

»The energy dissipated during the time interval (-T/2,T/2) by a

real signal with instantaneous power expressed by eq (2.3)can be

written as

)4.2........(....................)(2/

2/

2 tdtET

T

x

and the everage power dissipated by the signal during the interval

is

)5.2....(..........)(112/

2/

2T

T

T tdtT

ET

P

»The performance of a communication system depends on the

received signal energy ,higher energy signals are detected more

reliably (with fewer errors) than are lower energy signals

»The power is the rate at which energy is delivered .

2-6

»The power determines the voltages that must be applied to the

transmitter and the intensities of the electromagnetic field that one

must contend with radio in systems .

»We classify (t) as an energy signal if and only if ,it has

nonzero but finite energy

xE0 for all time where

)6.2.......(....................)(

2/

2/

)6.2()(lim

2

2 ...........

btdtE

T

T

atdtETx

» The periodic signal which is defined according eq.(2.1)

( t) = [ t + T0 ] ……………..…… (2.1)

-Thus the periodic signal exist for all time and thus has infinite

energy (i.e the periodic signal it is not an energy signal). Also

random signal has infinite energy it is not an energy signal.

»A signal is defined as a power signal if and only if it has finite

but nonzero power )( x

Po . For all time

)7.2..(..........)(1lim

2/

2/

2 tdtT

PT

T

T

2-7

»The energy and power classifications are mutually exclusive. An

energy signal has finit energy but zero average power , whereas

power signal has finite average power but infinite energy .

As general rule!

1-Periodic signals and random signals are classified as power

signals .

2-Determinisitic and nonperiodic signals are classified as energy

signals .

5-The unit Impulse Function or Dirac Delta Function

The Impulse function is an infinitely large amplitude pulse with

zero pulse width and unity weight (area under pulse) concentrated

at the point .the unit impulse is characterized by the following

relationships :-

)11.2....(..........)()()(

)10.2.(....................0)(

)9.2(....................00)(

)8.2...(....................1)(

oo txtdtttx

tfort

tfortwhere

tdt

»The unit impulse (t) is not a function in the usual sense. When

operation involve (t) as a unit area pulse of finie amplitude and

nonzero duration , after which the limit is considered as the pulse

duration approaches zero.

2-8

- (t) can be represented graphically as a spike located at t=t0 with

height equal to its integral area.

-Equation (2-11) is known shifting or sampling property of the

unit impulse function , the unit impulse multiplier selects a

sample of the function (t) evaluated at t=to

Spectral Density

»The spectral density of a signal characterizes the distribution of

the signals energy or power in the frequency domain .This

concept is important when considering filtering in communication

systems in order to evaluate signal and noise at the filter output .

1.Energy Spectral Density (ESD)

The total energy of a real valued energy signal (t) defined over

the interval (- to ) is described by eq (2. 6) .Using Parseral`s

theorem, we can relate the energy of such a signal expressed in

the time domain to the energy expressed in frequency domain as

(2.12).....f.........d2(f)Xtd(t)2x

xE

where X(f) is the Fourier transform of the nonperiodic signal

)(.)( fx

Lettx denote the squared magnitude spectrum ,

defined as .

..(2.13)Hz......../Joules|X(f)|(f)ψ2

χ

2-9

The quantity )( f is the waveform energy spectral density

(ESD ) of the signal ( t) .

The total energy of (t) )14.2()( .......

fdfxEx

»Energy spectral density describes the signal energy per unit

bandwidth.

»There are equal energy contribution from both positive and

negative frequency components ,since for a real signal

)(,)( fXt is an even function of a frequency

0

)15.2...(..........)(2 tdfEx

2.Power Spectral Density (PSD)

»The everage power P of a real valued power signal (t) is

defined in Eq (2-7)

)7.2.......()(1lim

2/

2/

2 tdtxT

PT

TT

»If (t)is a periodic signal with period T0 it is classified as a

power signal (since 0<P<)and the average power dissipated by

the periodic signal during the interval To (signal period) is

)16.2.......()(1lim2/0

2/0

2 tdtTTx

PT

T

2-10

» Parseval`s theorem for a real valued periodic signal takes the

form

)17.2.......(1 22

2

2

n

T

TCntd

TxP

where |Cn| terms are complex Fourier series coefficients of the

periodic signal.

»The power spectral density (PSD) function Gx(f) of the periodic

signal (t) is a real , even, and nonnegative function of frequency

that gives the distribution of the power of (t) in the frequency

domain, defined as

)3...(2.1).........fnδ(f(f)G0

2

n

cn

Thus the PSD of a periodic signal is a discrete function of

frequency. The average normalized power of a real valued signal

defined as

0

χχ.14)........(2td(f)G2fd(f)GP

»If (t) is a nonperiodic signal it cannot be expressed by Fourier

series ,and if it is a nonperiodic power signal it may not have a

Fourier transform .

2-11

If we form atruncated version (t) of the nonperiodic power

signal (t) by observing it only in the interval (-T/2,T/2),then (t)

has finite energy and has a proper Fourier transform XT(f) . The

PSD of the nonperiodic (t) can then defined in the limit as

)15.2.....(..........2)(1lim)( fXTT

fG Tx

Ex (2-1) (a) Find the average normalized power in the waveform

(t) =A cos2 π F0t using time averaging

(b) Repeat part (a) using the summation of spectral coefficients

Solution Using eq(2.17)

a)

dttTP )(1 2 (2.17)

Using eq (2.13) and (2.14)

)fnδ(f(f)G0

2

n

Cn

211

ACC

2-12

0Cn for n=0,±2, ±3,…….

)(2

)(2

)()(22

)( ffAffA

G f

2)

2()

2(

)]()2

()()2

([

222

22

AAA

dfffA

ffA

P

Note Fourier series

tdtj2π

e(t)0T

1Cwhere

tfjn2πeC

/20T

/20T

n

f

t

n

n

0)(

Autocorrelation of a Nonperiodic (Energy) Signal

» Correlation is a matching process , autocorrelation refers to the

matching of a signal with a delayed version of itself. The

autocorrelation function a real valued energy signal (t) is defined

as

R

..(2.16)..........tford)χ(t(t)χ t )(

─The autocorrelation function R(τ) provides a measure of how

closely the signal matches a copy of itself as the copy is shifted τ

units in time.

2-13

─Rx ( τ)is not a function of time , it is only function of the time

difference τ between the waveform and its shifted copy .

-The autocorrrelative function of a real valued energy signal has

the following properties:

1. R Zeroaboutτinlsymmetricaτ)(R )(

2. origintheatoccursvaluemaximumτallfor(0)(τχR R)

pairtransformFourieraform

densityspectralenergyandncorrelatioauto(f)ψ(τR ).3

4.

tdχ(t)(0)xR

value at the origin is equal to the energy of the signal

Autocorrelation of a Periodic (Power) Signal

The autocorrelation function of a real valued signal (τ) is defined

as

2.17)....τfortd(tT

1limR/2To

/2To

()()

tTx

When the power signal (t) is periodic with period To, the time

average in Eq(2.17) may be taken over period T0 , and the

autocorrelation function can be expressed as

2-14

8)......(2.1τfortdτ(t

/2oT

/2oT

x0T

1

)())( tR

The autocorrelation function of a real valued periodic signal has

properties similar to those of an energy signal.

Random signals

The main objective of a communication system is the transfer of

information over a channel. All usefull message signals appear

random, the receiver does not know which of the possible

message waveforms will be transmitted. Also the noise that added

to the message signals is random (due to random electrical

signals).

1-Random variables

The term random variable is used to define (signify a rule by

which a real number is assigned to each possible outcome of an

experiment .Let the possible outcomes of an experiment be

identified by the symbols (A) to each possible outcome.

The rule or functional relationship represented by the symbol

X ( ) is called a random variable

-The random variable may be discrete or continuous.

-The cumulative probability distribution function or distribution

function of a random variable is defined as

2-15

)) ....(2.19..........)χ(XP(χF

F() is simply the probability that an observed value will be less

than or equal to the quantity .

-The distribution function F( ) has the following properties

1)(F40F3

χχifχχF2

1χF01

2121 F

Fig (2.3)

-The density function an probability density function (Pdf) arises

from the fact that the probability of the event 1 X 2 equals

F ()

1 2 3

2-16

2

1

22

2221

)20.2........(

)(

d

FF

XPXPXPP

x

xx

The probability density function has the following probabilities:-

1)Fx()(Fxd)(Px2)0)(Px1)

The pdf is always a nonnegative function with total area =1

Note

For ease of notation, we will often omit the subscript X and write simply p ( ) - The mean value mx or expected value of a random variable X is defined by

)1......(2.2d)(Pχ)E(m a

where E ( ) is called the expected value operator

-The mean square value of X as follows :

)..(2.21..........d)(2

χ)2(E bp

The variance of X is defined as

22).......(2(x)2)xm(χ2)x

mE(xxδ 2 .(X)var dp

2-17

where (X-mx ) is the difference between x and mx

)22.2......(..........2

.22

)(2)2(

22

22

2

2

222

mXE

mmxmXE

mXEx

mXE

xmXxmxExmxEAnd

xx

xx

x

x

Thus the variance is equal to the difference between the mean

square value and the square of the mean.

Random processes

-A random process X (A,t) can be viewed as a function of two

variables; an event A and time .Fig(2-4) , shows a random

process. In the figure (2-4) ;

1-There are N sample functions of time, Xj ( t)

2-Each sample function can be considered as the output of a

different noise generator

3-For specific event Aj we have a signal time function

X(Aj, t ) = Xj (t) (i.e sample function )

4-The totality of sample function is called ensemble

5-For specific event A=Aj and a specific time t=tk ,X (Aj,tk) is

simply a number.

2-18

Fig (2-4) Random process

Power spectral density and Autocorrelation of a random process

-A random process X(t ) can generally be classified as a power

signal having a power spectral density G (f) of the form shown

in eq ( 2.15)

....(2.23)..........(fXT

1im(f)G2// )

TTL

where X (f) is the Fourier transform of (t )

2-19

-G (f) describes the distribution of a signal power in the

frequency domain as well as enables us to evaluate the signal

power that will pass through a network having known frequency

characteristics.

Fig (2.5) shows the autocorrelation and power spectral density

for binary waveform (bipolar ).The duration of each binary digit

is T second

2-20

Fig (2.5)

2-21

Noise in communication systems

-The term noise refers to unwanted electrical signals that are

always present in electrical systems .the presence of noise

superimposed on signal tends to attenuate and distorte the signal,

it limits the receiver`s ability to correct symbol decisions and

limits the rate of information transmission

-Noise arises from a sources ,both man made and natural

-we can describe thermal noise as a zero mean , Gaussian random

process. A Gaussian process n( τ ) is a random function whose

value n at any arbitrary time τ is statistically characterizedby the

Gaussian probability density function .

Where 2 = is the variance of n .

-The normalized or standardized Gaussian density function of a

zero mean process is obtained by assuming σ=l as shown in fig

(2.6)

-We will often represent a random signal as the sum of a Gaussiannoise random variable and dc signal ,that is Z=a+n

2-22

Fig (2.6)

5)......(2.2..........

2aZ

2

1

eπ2

1τ)(

P

-White Noise

The power spectral density of white noise is the same for all

frequencies of interest in most communication system .thermal

noise source has noise power per unit of bandwidth at all

frequencies from dc to 1012Hz. .Therefore , a simple model for

thermal noise assumes that its power spectral density Gn is flat for

all frequencies as shown in fig (2-7)

2-23

Gn (f) = )26.2(2

0N

Fig (2.7)

-when the noise power has such a uniform spectral density we refere to it as white noise . -the autocorrelation function of the white noise is given by the inverse fourier transformof the noise power spectral density

Signal Transmission Through Linear System

»The signal applied to the input of the system , as shown in

fig(2-8)can be described either as a time domain or frequency

domain .

1»When the input is considered in the frequency domain, we

shall define a frequency transfer function for the system .

2-When the input is considered in the time domain ,we shall

define the impulse response h(t)

output (t) h(t) y (t) X (f) H(f) Y(f)

Fig (2.8)

Linear system

G n(f) R(t) N0/2 N0/2

F t 0 0

2-24

Impulse response

-The linear time invariant system or network shown in fig(2.9) is

characterized in the time domain by an impulse response h(t),

which is the response when the input is equal to a unit impulse

δ(t)

-The response of the network to an arbitrary input signal (t)is found by the convolution of (t) with h (t) , expressed as χ(t)

Y(t) fig (2.9)

(22.7a)..........τd)τ(th(τ(t)h*(t)(t)Y )

where * denote convolution operation . If the system is assumed

causal which means that there can be no output prior to the time

,t=0, when the input is applied

0

) ....(227b)..........τd)τ(th(τ(t)Y

Frequency transfer function

1-The frequency domain output signal Y(f) is obtaind by taking

the Fourier transform of convolution in the time domain of e.g

(2.27 a)

.(2.28)..........(f)H(f)X(f)Y

Linear time invariant system

2-25

The Fourier transform of the impulse response function is called

frequency transfer function or frequency response of the network .

In general H(f) is complex and can be written as

.......(2.29)

Random Signals and Linear Systems

If the input to a time invariant linear system is a random signal ,

the output will be a random signal. That is , each sample of the

input signal yields a sample at the output . the input power

spectral density Gx(f) and also the output power spectral density

Gy(f) are related as follows:

..(2.30)..........2)(H(f)xG)(yG ff

Distortionless Transmission

»The output of ideal distortionless transmission can be described

as )(2.31.............)t(tχK(t)o

y

Where K and to are costants representes the variation in the

amplitude and the time delay of the input signal!!

»The Fourier transform for the eq (2.31)

32).......(2...........(f)KX)Y( ef tfj 02

2-26

Substituting the eq (2.32) for y (f)in eq(2.28)

)33.......(2...........eK)(H af ftj2π

The time delay to is related to the phase shift () and the radian

frequency = 2 π f by

).....(2.33..........)sec(radianfπ2

radian)()(0

t b/

From eq(2.23) is clear that phase shift must proportional to

frequency in order for the time delay of all components to be

identical. A characteristic used to measure delay distortion of a

signal is called envelope delay or group delays which is defined

as

).....(2.34..........fd

d

π2

1τ

-In practice, a signal will be distorted in passing through some

parts of a system .Phase or amplitude correction (equalization)

networks may be introduced in the system to correct for the

distortionless .

Ideal filters

-The ideal filter cannot be designed due to the problem of an

infinite capability requirement for ideal filter.

Y(f)=X(f) H(f)…….(2.28)

2-27

Fig(2.10)

The range fL f fu is called passband where fL = lower cutoff

frequency and fu= upper cutoff frequency .

1-When fL0 and futhe filter is called bandpass filter (BPF)

2-When fL= 0 and fu has finite value, the filter is called low pass

filter (LPF).

3-When fL has nonzero value and fu , the

filter is called high pass filter (HPF).

-The bandwidth of the system is defined as the interval of positive

frequencies over which the magnitude )(fH remains within a

specified value .

2-28

- For ideal low pass filter , the transfer function can be defined as

34).......(2...........0tfj2πeK)(H

f

where K = 1 with bandwidth H(f)=|H(f)|e-j(f)`:. Wf = fu Hz

0

)(

2)(

...................0

ftjefjeand

ufH

uffforfH

.for..........1.........

The impulse response of the ideal low pass filter is show in fig

(2.11)

Thus the impulse response in fig(2.11) is noncausal. Therefore the

ideal filter is not realizable (which is described in (2.3)

……….(2.35)

2-29

Fig (2.11)

Ex (2.2) White noise with power spectral density Gn(f)=No/2 is

applied to the ideal low pass filter .Find the power spectral

density and the autocorrelation function of the output signal .

Solution :

The autocorrelation is the inverse of Fourier transform

τfπ2sincfNτfπ2

τfπ2sinfN(τR uu0

u

u00y )

2-30

Realizable filters

The very simplest example of a realizable low pass filter is made

up of resistance R and capacitor C as shown in fig(2.12a) .The

transfer function can be expressed as

)36.2......()2(1

1

21

1)( )(

2

fjeCRffRCj

fH

Where Ø (f)= tan -12π f RC

W=CR

1 rad / sec , Wf = 1/2 RC

Fig(2.12)

-The magnitude )(fH and the phase (f)are polted in fig

2-31

(2.12b,c)

-The low pass filter bandwidth is defined to be its half power

point , this point is the frequency at which the output signal power

has fallen to one half of its peak value or the frequency at which

the magnitude of the output voltage has fallen to(0.707)of its peak

value.

Ex 1.3

White noise with spectral density Gn (f)=No/2 is used as an input

to low pass RC filter. Find the power spectral density Gy(f) and

the autocorrelation function of the output signal.

Solution :-

= RCeRC

N //

4

Since

0

2)2(22

aiffa

ae ta

F

2-32

The bandwidth description

Many important theorems of communication and informationــ

theory are based on the assumption of strictly bandlimited , which

means that no signal power is allowed outside the defined band .

All bandwidth criteria have in common the attempt to specify aــ

measure of width , W, of a nonnegative real-valued spectral

density defined for all frequencies |f| ∞ . Fig(2.13) shows some

of the most definitions of bandwidth for a single heterodyned

pulse )(tc

with fc is the carrier wave frequency and T is the

pulse duration .

Fig(2.13)

a-Half power bandwidth

2-33

b-Equivalent rectangular or noise equivalent bandwidth ,is

defined by the relationship WN=Px/Gx(f c )where P is the total

power over all frequencies and G(fc) is the value of Gx(f)at the

band center

c-Null to null bandwidth

d-Frctional power containment bandwidth is defined by that 99%

of the signal power is inside the occupied band

e-Bounded power spectral bandwidth defines attenuation out side

bandwidth at 35 dB or 50 dB.

3-1

Communication Systems

Lecture 3

3-2

Formatting and Baseband Modulation

-Transmit formating is transformation from source information to

digital symbols

-Formatting contains a list of topics that deals with transforming

information to digital messages:-

1-Character coding

2»Sampling

3-Quantization

4-Puls code modulation (PCM)

5- Delta modulation

6-Sigma -Delta modulation

»The formatting and transmission of baseband signals is shown

fig(3.1)

Data already in a digital format would bypass the formatting

function .Textual information is transformed into digits by use of

a coder .Analog information is formatted using three separate

processes (sampling, quantization and coding ).in all cases the

formatting step results in a sequence of binary digits. These digits

are to be transmitted through a baseband channel ,such as a pair of

wires or a coaxial cable . The binary digits firstly must be

transformed to waveforms compatable with channel .

-The conversion from a bit stream to a sequence of pulse

waveforms take place in the pulse modulator .

3-3

Fig (3.1)

Messages, Characters and Symbols

»Textual messages comprise a sequence of alphanumeric

characters . When digitally transmitted the characters are first

encoded into a sequence of bits , called a bit stream or a baseband

signal

»Groups of K bits can then be combined to form a new symbols .

A system using a symbol set size of M is referred to as an m–ary

system .

! »The value of K or M represents an impotant initial choice in

the design of any digital communication system

3-4

-Fig (3.2) shows example of messages , characters and symbol

Fig (3.2)

1-The textual message in the fig(3.2) is the word THIN

2-Using 6 bit ASCII character coding (American Standard Code

for Information Intercharge )forms a bit stream comprising

(24)bit.

3-In fig(3.2) the symbol set size M has chosen to be 8 . The bits

are therefore partitioned into groups of three [K = log28=3] bits

and forms symbol

4-The transmitter must generate 8 waveforms

Si(t) , where i=0, 1 , 2 …..7 to represent each symbol .

Formatting Analog Information

-If the information is analog , it cannot be character encoded as in

the textual data ,the information must first be transformed into a

digital format. The process of transforming an analog waveform

into a form that is compatible with a digital communication

3-5

system starts with sampling of the waveform to produce a

discrete pulse amplitude waveform .

Sampling Theorem

-The link between an analog waveform and its sampled version is

provided by the sampling process. This process can be

implemented in several ways, the most popular being the sample

and hold operation .In this operation a switch and storage device

(trasister and a capacitor ) form a sequence of samples of the

continuous input waveform . The output of the sampling process

is called pulse amplitude modulation (PAM)because the

successive output intervals can be described as as a sequence of

pulses with amplitude derived from the input waveform pulses.

The sampling theorem

-A bandlimited signal having no spectral components above fm

hertz can be determind uniquely by values sampled at uniform

intervals of

)..(3.1..........2

1Τmfs

This statement is Known as the uniform sampling theorem .

Where Ts = sampling time (which is the upper limit )

fs = sampling frequency = 1/ Ts

)..(3.2..........mf2s

f

3-6

The sampling rate fs=2fm is called the Nyquist rate. Therefore the

theorem requires that the sampling rate be rapid enough so that at

least two samples are taken during the period corresponding to the

highest frequency in the spectrum .

Let us examine case of ideal sampling with a sequence of unit

impulse functions .

1-Assume an analog signal x(t) as shown in fig(3.3a)with a

Fourier transform X(f) ,which is zero outside the interval

(-fm < f < fm) as shown in fig(3.3b) !

2-The sampling of x(t) can be viewed as the product of x(t) with a

periodic train of unit impulse functions x(t) as shown in ig(3.3c)

where Ts is the sampling period and (t) is the unit impulse .

………….(3.3)

The............(2.4)

3-7

Fig (3.3)

3) Using the frequency convolution property of the fourier

transform

F )(*)()()( fXfXtXt

where Xs(f) =

ns

sfnfT

)5.3....().........(1

Fig(3.3c) and fig(3.3d) show the impulse train X(t) and its

Fourrier transform Xδ(f).respectively .Convolution with an

impulse function simply shifts the original function as follows

)fn(fX)fn(fδ*(f)Xss

3-8

We can now solve for the transform X s (f) of the sampled

waveform

ns

nsδ

....(3.6)..........)fn(fXT

1

fnδ(fT

1*(f)X(f)X*(f)X(f)X )ss

Fig (3.3f) show the spectrum of Xs (f) . We therefore conclude

that:-

1-The spectrum Xs(f) of the sampled signal s(t) is to within a

constant factor (1/Ts) , exactly the same as that of (t)

2-The spectrum repeats itself periodically in frequency every fs

(Hertz)

3-When fs=2fm , the analog waveform can theoretically be

completely recovered by using sharp lowpass filter

4-If fs > 2 fm ,the analog waveform can be recovered easier by

using low pass filter .

5-If fs < 2 fm the spectrum of analog waveform components

overlap as shown in fig (3.4b) . Some information will be lost .

The result of under sampling ( fs < 2 fm )and the phenomenon is

called aliasing.

3-9

Fig (3.4)

Natural Sampling

- The practical method of accomplishing the sampling of a

bandlimited analog signal )(t is to multiply (t) shown in

fig(3.5a)by the pulse train or switching waveform )(tp

shown

in fig(3.5c) Each pulse in )(tp

has width T and amplitude (1/T)

The sampling frequency is designated fs , and sampling time is

designnated Ts(TS=1/fs) . The sampled data )(ts

can be

expressed as

)(ts

= )(tp

)(t ………. (3.7)

The sampling here is termed natural sampling since the top of

each pulse in the s(t) sequence has the same shape of its

3-10

corresponding analog segment during the pulse interval .The

periodic pulse train can be expressed by a Fourier series

Fig (3.5 )

3-11

n

sns

s

sn

m

ss

n

tsfjnp

tfnjeCtt

eqandeqCombiningwidthpulseT

T

Tnc

TC

fTfwhere

eCt

)9.3.(..........2)()(

)8.3()7.3(

sin1

2

1

)8.3......(....................)( 2

The transform Xs (f) of the sampled waveform is

Xs (f) = F [s (t) ]

For linear systems ,we can interchange the operations of

summation and Fourriers transformation .

Using the Fourier translation property of the Fourier transform:-

n

snffXCnfXs )12.3(..........)()(.:

…………(3.10)

……….(3.11)

3-12

Eq(3.12) and fig(3.5f) illustrate that Xs(f) is a replication of X(f)

periodically repeated in frequency every fs. In this natural

sampled case , we see that Xs(f) is weighted by the Fourier series

coefficients of the pulse train compared with a constant value in

the impulse sampled case

Ex3.1 Consider a given waveform (t) with Fourier transform

X(f) Let Xs1 (F) be the spectrum of s(t), which is the result of

sampling (t)with a unit impulse train (t).Let Xs2(f) be the

spectrum of s2(t) , the result of sampling (t)with a pulse train

p(t)with pulse width T, amplitude 1/T, and period Ts. Show that

in the limit, as T approaches zero, Xs1(f)=Xs2(f)

Solution :-

nsn

ns

s

fnfXCfsX

fnfXT

fs

X

)12.3....(..........)()(

)6.3...(..........)(1)(

2

1

As the pulse width T0 and the pulse amplitude (the erea

of the pulse remains unity), p (t) δ (t)

3-13

s

tsnfj

s

T

s

tsfnjT

sTsn

sfj

ps

T

T

sTn

Tdtet

TTtdet

TC

tdtetT

C

ss

s

1)(1)(1

)(1lim.:

2

2/

2/22/

2/

2

2/

2/

0

Substitute for Cn in eq (3.12)

:.

s

fs

TX 1

)(2

In the limit T →0

.)()(1)(2

fXsfnfT

fXs

nss

Aliasing

-Fig(3.6)shows aliasing in frequency domain .The overlapped

region , shown in fig(3.6) contains that part of the spectrum which

is aliased due to under sampling

-Higher sampling rate (fs) can eliminate the aliasing.

3-14

Why Oversampling

Oversampling is the most economic solution for the task of

transforming an analog signal to digital signal or the reverse ,

transforming a digital signal to an analog signal. This is so

because signal processing performed with high performance

analog equipment is typically much more costly than using digital

signal processing equipment to perform the same task. Consider

the task of transforming analog signals to digital signals either

using oversampling or without oversampling :-

1.Transforming analog signals to digital signals can be performed

without oversampling according the following steps:-

a)The signal passes through a high performance analog lowpass

filter to limit its bandwidth

Fig (3.6)

3-15

b)The filtered signal is sampled at the Nyquist rate for the

bandlimited. The strictly bandlimited is not realizable

c)The samples are processed by analog to digital converter that

maps the continuous valued samples to a finite list of discrete

output levels

2.Transforming analog signals to digital signals can be performed

with oversampling according the following steps ;-

a-The signal is passed through a low performance (less costly)

analog lowpass filter (prefilter)to limit its bandwidth

b-The prefilter signal is sampled with oversampling .

c- The samples are processed by analog to digital converter that

maps the continuous valued samples to a finite list of discrete

output levels.

d)The digital samples are then processed by a high performance

digital filter to reduce the bandwidth of the digital samples and

sample rate.

Sources of Corruption

The analog signal recovered from the sampled , quantized and

transmitted pulses will contain corruption from several sources:-

1-Sampling and Quantizing Effects:-

a-Quantization noise : The distortion in quantization is a

truncation error. The process of PAM signal into a quantized

PAM signal involve degradation some of the analog information

3-16

.This distortion introduced by the approximation of the analog

waveform with quantized samples is called to as quatization noise

b-Quntizer satuation : The quantizer (or analog to digital

converter ) allocates (L)level to the task of approximating the

continuous range of inputs with a finite set of output . The range

of inputs for which the difference between input and output is

small is called the operating range of the converter .If the input

exceeds this range, the difference between the input and the

output becomes large and the converter is operating in saturation .

Saturation errors being large are more effects than quantizing

noise. Generally saturation is avoided by use of automatic gain

control(AGC).

c-Timing jitter ; the sampling theorem is based on uniformly

spaced samples of the signal . If there is a slight jitter in the

position of the sample , the sampling is no longer uniform.

Timing jitter can be controlled with very good power supply

isolation and stable clock references.

2-Channel effects ;-

a-Channel noise : Thermal noise interference from other users ,

and interference from circuit switching transients can cause error

in detecting the pulses carrying the digitized samples .Channels

induced errors can degrade the reconstructed signal quality quite

quickly.

3-17

b-Intersymbol interference; the channel is always bandlimited .A

bandlimited channel spreads a pulse waveform passing through

the channel . When the channel bandwidth is much greater than

pulse width the spreading of the pulse will be slight . When the

channel bandwidth is close to the signal bandwidth, the spreading

will exceed a symbol duration and cause signal pulses to overlap.

This overlapping is called intersymbol interference (ISI). ISI

causes system degradation.

Sampling Theorem and Bandpass Signals

for a signal g(t) whose highest frequency spectral component is

fm, the sampling frequency fs must be no less than fs=2fm if the

lowest frequency spectral component of g(t) is fL=0

If fL ≠ 0 it may be that the sampling frequency need be no larger

than fs = 2 ( fm-fL)

Ex3.2 : if the spectral range of a signal extends from 10 to 10.1 M

Hz. Find the value of sampling frequency

Solution

fs= 2 ( fm – fL)

= 2 ( 10.1-10) = 0.2 M Hz

4-1

Communication Systems

Lecture 4

4-2

Signal to Noise Ratio for Quantized Pulse

Fig (4-1) shows an L level linear quantizer for an analog signal

with a peak to peak voltages ranage of vpp=vp- (Vp)= 2 Vp volts.

The quantized pulse either positive and negative as shown in

fig(4-1). The step size between quantization levels are called the

quantile interval q . When the quantization are uniformly

disterbuted over the full range , the quantizer is called uniform or

linear quantizer.

Fig (4.1)

4-3

-Each sample value of the analog waveform is approximated with

quantized pulse , this approximation will result in an error no

larger than q/2.

-The quantizer variance (mean square error) is used as a

parameter to define the uniform quantizer .

-If we assume that the quantization error, e, is uniformly

distributed over a single quantile interval q wide (i.e the analog

input takes on all values with equal probability ).

Where p(e) =1/q is the (uniform ) probability density function of

the quantization error .The variance 0 corresponds to the average

quantization noise power . The peak power of the analog signal (

normalized to 1) can be expressed as :

)2.4.........(..........4

22

2

2

22 qLqLV

V PPp

Where l is the number of quantization levels. The ratio of peak

signal power to average quantization noise power (S/N)q ,

assuming that there are no errors due to interference symbols

.(4.1)

4-4

)3.4.........(..........2312/2

4/22L

q

qL

qN

S

In the limit L , the (S/N)q approaches the PAM

format (with no quantization) and the signal to quantization noise

ratio is infinite

Pulse Code Modulation

-Pulse code modulation (PCM) can be obtained from the

quantized (PAM) signals by encoding into a digital word

-The source information is sampled and quantized to one of (L)

levels, then each quantized sample is digitally encoded into an

)log(2

Lbit codeword.

-For baseband transmission ,codeword bits will then be

transformed to pulse waveforms

-Fig(4.2) shows PCM process steps : a. Assume that the sampled

analog signal is limited in the range (-4 v to +4v).

b- The step size between quantization levels has been set at 1V.

Thus eight quantization levels are employed

c-Each quantization level is assigned by code number

(0,1,2,….7)

d-Each code number has its representation in binary ranging from

000 to 111

4-5

Fig(4.2)

-The choice of voltage levels is guided by two factors ;-

1-The quantile intervals between the levels should be equal .

2-It is convenient for the levels to be symmetrical about zero.

Uniform and Nonuniform Quantization

1-human speech is characterized by unique statistical properties

as shown in fig (4.3) the horizontal axis represents speech signal

magnitudes , normalized to the root mean square (rms) value of

such magnitude through a typical communication channel .The

y–axis is probability .

-For most voice communization channel very low speech

volumes predominate 50% of the time .Large amplitudes values

are relatively rare, only 15% of the time

4-6

Fig (4.3)

-When the step size of the quantile interval are uniform in size

,the quantization is known uniform quantization. such a system

would be wasteful for speech signals because of the low S/N

(since the quantization noise is the same in the uniform

quantization and the signal is weak )

-Nonuniform quantization can provide fine quantization of the

weak signals and coarse quantization noise of the strong

signal. Thus, in the nonuniform quantization , quantization

noise can be made proportional to signal size. Nonuniform

quantization can be used to make the SNR a constant for all

signals within the input range.

4-7

-Fig (4.4) shows the nonuniform quantizer characteristics

Fig (4.4)

Fig(4.5)

-Nonuniform quantization is achieved by first distorting the

original signal with logarithmic compression characteristic as

shown in fig(4.5) and then used uniform quantizer as shown in

fig(4.6)

Nononiform quantizer output

Input

4-8

Fig(4.6)

!!!!!!!!

-At the receiver , in inverse compression characteristic called

expansion ,is applied so that the overall transmission is not

distorted . So that the overall transmission is not distorted .The

processing pair (compression and expansion )is usually referred to

as companding .

Baseband Transmission

-PCM is used to transform the analog waveforms into binary

digits . Digits are just away to describe the message information

Thus ,we need something physical that will rerpresent or carry

the digits

-Fig(4.7) shows the representation of the binary digits with

electrical pulses in order to transmit them as information in the

PCM bit stream .

–Code word time slots are shown in fig (4.7a) where the

codeword is a 4 bit representation of each quantized pulse .

Uniform quantizer

4-9

Fig(4.7)

-Each binary one is represented by a pulse and zero is

presented by the absence of a pulse .

-At the receiver , detection is based on the received energy .Thus

there is advantage to increase the width T as wide as possible to

improve the detection .

4-10

PCM waveform types

-When pulse modulation is applied to binary symbols the

resulting binary waveform is called a pulse code modulation

(PCM) waveform .

-These PCM waveforms are called line codes if used in

telephony applications .When pulse modulation is applied to

nonbinary symbols , the resulting waveform is called an M-ary

pulse modulation waveform . The PCM waveforms fall into

four groups ;-

1-Nonreturn to zero ( NRZ)

2-Return to zero (RZ)

3-Phase encoded

4-Multilevel binary.

fig(4.8)

a. The(NRZ ) group is the most commonly used PCM waveform .

It can be divided into the following subgroups as shown in

Fig (4.8)

4-11

1-NRZ –L (L for level) is used in digitally logic circuits .A

binary is represented by one voltage level and a binary zero is

represented by another voltage level .

2-NRZ-M (M for mark) ,the one or mark is represented by a

change in level and o, or space is represented by no change

level . This is often referred to as differential encoding.NRZ-M

is used in magnetic tape recording .

3-NRZ-S (S for space) is the complement of NRZ-M.A one is

represented by no change in the level and zero is represented

by the change in the level

(b) RZ waveforms consist of following subgroups as shown in

the fig(4.9)

1- Unipolar RZ a one is represented by a half bit wide pulse

and a zero is represented by the absence of a pulse .

2-With bipdar RZ ,the ones and zeros are represented by

opposite level pulses that are one half bit.

3-RZ-AMI (AMI- for alternate mark inversion ) is a signaling

scheme used in telephone system. The ones are represented by

equal amplitude alternating pulses . The zeros are represented

by the absence of pulses.

4-12

Fig ( 4.9)

(c)The phase encoded group consist of following subgroups

shown in fig (4-10).

l-Bi--L (bi-phase-level) or Manchester coding a one is

represented by a half bit wide pulse positioned during the first

half of the bit interval , a zero is represented by a half bit wide

pulse positioned during the second half of the bit interval .

Note

The phase encoding schemes are used in magnetic recording

systems and optical communication and satellite telemetry link

2-Bi -Ø -M(bi-phase-mark).

A transition occurs at the beginning of every bit interval. A one is

represented by a second transition one half bit interval later a zero

is represented by no second transition.

3-Bi- Ø -S (bi-phase-space) .

4-13

A transition also occure at the beginning of every bit interval. A

one is represented by no second transition , a zero is represented

by a second transition one half bit interval later.

4-Delay modulation

A one is represented by a transition at the midpoint of the bit

interval ,A zero is represented by no transition , unless it is

followed by another zero. In this case , a transition is placed at the

end of the bit interval of the first zero .

Fig (4.10)

(d) Multilevel binary . Many binary waveform use three level

to encode the binary data , Bipdar RZ and RZ-AMI belong to

this group . The group also consists :-

4-14

1-Dicode NRZ, the one to zero or zero to one data transition

changes the pules polarity , without data transition , the zero

level is sent .

2-Dicode RZ,the one to zero or zero to one transition produces

a half duration polarity change ,otherwise a zero level is sent

Fig (4.11)

Why There are so Many PCM Waveforms ?

The reason for the large selection relates to the differences in

performance that characterize each waveform . There are many

parameters limits selection of the waveform for PCM :-

1-DC component

2-Self clocking

3-Error detection

4-Bandwidth compression

5-Differential encoding

6- Noise immunity

4-15

Spectral Characteristics of Various PCM Waveforms

-The most common criteria used for comparing PCM

waveforms and for selecting one waveform type from the

many available are spectral characteristics ,bit synchronization

capabilities ,error detection capabilities , interference and

noise immunity and cost and complexity of implementation .

fig (4-12) shows the spectral density of some of the most

popular PCM waveform .

Fig (4.12)

-The Fig( 4-12) plots power spectral density in (watt/Hz)

versus normalized bandwidth (WT) where W is bandwidth and

T is the duration of the pulse

W T (normalized bandwidth)

4-16

-Since the pulse or symbol rate Rs is the reciprocal of T ,

normalized bandwidth can be also be expressed as W/ Rs. This

factor (W/Rs) describes how efficiently the transmission

bandwidth is being utilized for each waveform of interest

Bits per PCM words

-The idea of binary partitioning (M=2k) is used to relate the

grouping of bits to form symbol for the purpose of signal

processing and transmission (M-ary)

-Consider the process of formatting analog information into bit

stream via sampling ,quantization and coding

-Each analog sample is transformed into a PCM word made up

of groups of bit .The PCM word size can be described by the

number of quantization levels allowed for each sample

Sampling

Analog sample

Quantization

PCM word size

Bits / analog sample

Coding

-The quantization can be described by the number of bits

required to identify that set of levels

If L= number of quantization levels in the PCM word

= number of bits needed to represent those level

)4.4.....(..........2L

4-17

-The choice of the number of levels or bits per sample,

depeneds on how much quantization distortion value in the

PCM format . It is useful to develop a general relationship

between the required number of bits per analog sample (the

PCM word size ) ,and allowable quantization distortion.

-Let the magnitude of the quantization distortion error |e| be

specified as a fraction (P) of the peak to peak analog voltage

Vpp as follows

PPVPe ||

Since the maximum value for quantization error =q/2

where q is the quantile interval

)6.4(2

1log.:

)5.4...(..........2

12

2

12)1(22max

//

2 pbitofnowhere

levelsp

Lor

VpppL

Vpp

LifL

Vpp

L

Vppqe

I t is important that we do not confuse the ideas of bits per

PCM word denoted by in Eq(4.5) with the M-level

transmission concept of K data bits per symbol.

4-18

Ex 4-1 the information in an analog waveform, with maximum

frequency fm= 3 KHz , is to be transmitted using PCM. The

quantization distortion is specified not to exceed

± 1% of the peak to peak analog signal

a-what is the minimum number of bits /sample or bits/ PCM

word

b-what is the minimum required sampling rate and what is the

resulting bit transmition rate ?

c-If the transmission bandwidth equal 12 KHz determine the

bandwidth efficieney of this system.

solution :-

a) Using Eq (4.6)

6550log02.0

1log12

log .22

p

:. = 6 bit / sample to meet the required distortion

b) Using the Nyquist sampling theorem

sec/60006000300022 sampleHzm

fs

f

Each sample of the PCM composed of 6 bits

:. Bit transmission rate = fs = 6 !! 6000 = 3 6 k bit / sec

c) Bandwidth efficiency = W

R

where R = Bit transmission rate

W = Bandwidth

4-19

:. Bandwidth efficiency = sec/312000

36000bit

Convolution

The convdution of two function g (t) and w(t) dented by

g ( t)*w(t) is defined by the integral

)7.4....(....................)()()(*)( dtwgtwtg

The time convolution property and its duals, the frequency

convolution property , state that if :-

)()()()( 2211 wGtgandwGtg

Then (time convolution )

g1 (t) * g2(t) G1 (w) . G2(w) …………..(4.8)

and (frequency convolution )

g1 (t) . g2 (t) 2

1 G1 (w) * G2 (w) …………(4.9)

4-20

Proof:

)()()()(

)()(

)()(

)()()(*)(

1212

21

21

2121

wGwGdegwG

dwGeg

dtgg

tddtggtwjetgtg

wj

wj

td

twj

F

Note

Bandwidth of the product signals: If g1(t) and g2(t) have

bandwidth B1 and B2 Hz , respectively ,the bandwidth of g1(t)

g2 (t) is B1+B2 Hz . this result follows from the application of

the width property of convolution to Eq (4.9).

5-1

Communication Systems

Lecture 5

5-2

Advantages, Limitations and Modifications of PCM

1-Advatages of PCM;

a-Robustness to channel noise and interference

b-Efficient regeneration of the coded signal along the

transmission path.

c-Auniform format for transmission of different kinds of

baseband signals ,hence their integration with other forms of

digital data in a common network

d-secure communication through the use of special modulation

schemes or encryption

2-limitations of PCM;

a-PCM involves many complex operations, today ,they can all be

implemented in a cost effective using commercially available and/

or custom made very large scale integrated (VLSI) chips.

b-Wide bandwidth requirement of PM (now wideband system

solve this limit such as satellites and fiber optics ).

The simplicity of implementation is a necessary requirement, then

use delta modulation (DM) as an alternative to PCM

Source coding deals with the task of forming efficient

descriptions of information sources.

5-3

Delta Modulation

º In delta modulation (DM) ,an incoming message signal is over

sampled (i.e at a rate much higher than the Nyquist rate)to

purposely increase the correlation between adjacent samples of

the signal . This is done to permit the use of a simple quantizing

technique for constructing the encoded signal.

-In the basic form, DM provides a staircase approximation to the

oversampled version of the message signal, as shown in fig

(5-1).The difference between the input and the approximation is

quantized into only two levels, namely , corresponding to

positive and negative differences .

1-If the approximation falls below the signal at any sampling ,it is

increased by

2-If the approximation lies above the signal it is decreased by

The staircase approximation remains within of the input

signal if the signal does not change too rapidly from sample to

sample.

Let m(t) denote the input (message )signal ,and Mq(t) denote its

staircase approximation .

:.m [ n] = m ( n Ts) n =0.1.2….. (5.1)

5-4

Fig (5.1)

Where Ts is the sampling period and M(nTs) is a sample of the

signal m(t) taken at time t=nTs . We may then formalize the basic

principles of the delta principles of delta modulation of the

following set of discrete time relations:-

][neq

1Z][nm

q

]1[ nmq

Fig(5.2)

1Z

Fig(5.3)

5-5

-The main advantage of delta modulation is simplicity as shown

in fig (5.2)and (5.3)

»The block Z-1 inside the accumulator represent a unit delay

which is equal to sampling period

e [ n] = m [ n ] – mq [n-1] ……..(5.2a)

mq [n] = mq [n-1]+eq[n] ……...(5.2 b)

when e [ n ] is an error signal representing the difference the

present sample m(n) of the I/p signal and latest approximation

Mq[n-1]

eq[n] is the quantized version of e(n)

-The quantizer consists of a hard limiter

Quantization Error of Delta Modulation

Delta modulation is subject to two types of quantization error as

shown in fig(5.4)

1-slope overload distoration

2-Granular noise

Fig (5.4)

5-6

»Slop overload distortion occurs when the step size is too

small relative to the local slop characteristics of the input

waveform m(t).

-Granular noise occurs when the step size is too large relative

to the local slop characteristics of the input waveform m(t)

-There is a need to have a large step size to accommodate a wide

dynamic range , whereas a small step size is required for the

accurate representation for relatively low level signals .It is

therefore clear that the choice of the optimum step size to satisfied

the above two requirements. Thus , we need to make the delta

modulater adaptive .

Delta -Sigma Modulation

The quantizer input in the conventional form of delta modulation

may be viewed as an approximation to the derivative of the

incoming message signal .This behavior leads to drawback of

delta modulation in that transmission disturbance such as noise in

the demodulator

-This drawback can be overcome by integrating the message

signal prior to delta modulation .The use of integration in the

manner described here has the following beneficial effects :-

1-The low frequency content of the input signal is pre-emphasized

2-Correlation between adjacent samples of the delta modulater is

increased ,which tends to improve overall system performance

5-7

3-Design of the receiver is simpler.

A delta a modulation with integration at its input is called delta-

sigma modulation (D- M).

Fig (5.5) shows the block diagram of delta –sigma modulation

system . In this diagram, the message signal m(t) is defined of a

delta sigma modulation .The message m(t) is defined in its

continuous time form ,which means that the pulse modulater now

consists of a hard limiter followed by a multiplier , to produce a

1-bit encoded signal .

Fig(5.5)

-In delta modulation , simplicity of implementation of both the

transmitter and receiver is attained y using a sampling rate far in

excess of that needed for PCM. The price paid for this benefit is a

corresponding increase in the channel bandwidth

-The receiver of D is a simple low pass filter.

dt

X

L.P.F

pulse generator comparator Integrator Hard Limiter

Low pass filter

+m(t)message input

Pulse modulator

5-8

Linear Predication

Predication will be used to forecast the future of a random signal

on the basis of its past history and its known statistical properties.

Consider a finite duration impulse response (FIR)discrete ilter

shown in fig (5.6) which involves the use of three functional

blocks:-

1-set of punit delay elements, each of which is represented by Z-1

2-set of multipliers involving the filter coefficients

w1,w2….w1,w2…..,wp

3-Set of adders used to sume the scaled versions of the delayed

inputs [n-1], [n-2],… [n-p]

to produce the output^ [n]. The filter output^ [n]

or more precisely , the linear predication of the input , is thus defined by the convolution sum

P

1K

)......(5.3K]........[nχkW[n]χ

Where p, the number of unit delay elements ,is called the predication order .

1Z

1w 1pw

pw

2w

1Z1Z

][n

)(n)1( n )2( n )1( pn

)( pn

Fig (5.6)

5-9

Differential Pulse Code Modulation (DPCM)

When a voice or video signal is sampled at a rate slightly higher

than the Nyquist rate as usually done in pulse code modulation

,the resulting sampled signal has a high degree of correlation ,

between adjacent samples

The meaning of this high correlation , is that ,the signal does not

change rapidly from one sample to the next and as a result ,the

difference between adjacent samples has a variance that is smaller

than the variance of the signal itself . When these highly

correlated samples are encoded ,as in the standard PCM system

,the resulting encoded signal contains redundant information .This

means that symbols that are not absolutely essential to the

transmission of information .By removing this redundancy before

encoding ,we obtain a more efficient coded signal , which is the

basic idea behind DPCM .

Sampled e[n] eq[n]

Input + m [n] -

+ DPCM

+ Signal

m^[n]

][nmq

Fig (4.7a) DPCM transmitter

Quan-tizer

Encoder

Predictor

5-10

Input + output

»

Fig (4.7b)DPCM receiver

-Fig (5.7)shows the block diagram of DPCM

-Suppose a baseband signal m[t] is samled at the rate fs=1/Ts to

produce the sequence [m(n)]whose samples are at Ts seconds

.The input signal to quntizer is defined by

e[n]=m[n]-m^[n]………...(5.4)

Where e[n] is the difference between the unquantized input

sample m [n] and a prediction of it denoted by m^[n] The

quantizer output may be expressed as

eq [ n ] = e [n] + q [n] ……….(5.5)

Where q[n] is the quantization error .The quatizer output eq[n] is

added to the predicted value m^[n]to produce the rediction input

mq [ n] = m^ [n] + eq [n] ………(5.6)

Substituting eq (5.5)into(5.6), we get

mq [ n] = m^ [n] + e [n]+q [n] ………(5.7)

But, from eq(5.4) m[n]=m^[n]+e[n]

Substituting m[n]in eq(5.7),we get

mq [ n] = m [n] + q [n] ……..(5.8)

Eq(5.8) represents a quantized version of the input sample m[n]

Decoder

predictor

5-11

-The receiver for reconstructing the quantized version of the input

is shown in fig (4.7b).It consists of a decoder to reconstruct the

quantized error signal.

M-ary Pulse Modulation Waveforms

-There are three basic ways to modulate information on to a

sequence of pulses , we can vary the pulse amplitude ,position or

duration which leads to names pulse amplitude modulation

(PAM),pulse position modulation (PPM)and pulse duration

modulation (PDM), respectively .PDM is sometimes called pulse

width modulation (PWM)as shown in fig(5.8)

-When information samples without any quantization are

modulated on pulses ,the resulting pulse modulation can be called

analog pulse modulation.

-When the information samples are first quantized ,yielding M

symbols (each symbol K bit, m=2k)the resulting pulse modulation

can be called digital pulse modulation and we refer to it as M-ary

pulse modulation.

-PCM waveforms with binary representation , having two

amplitude values (e.g NRZ).This PCM wavefors requiring only

two level represent the special case (M=2)of the general M-ary

PAM that requires M levels.

5-12

Fig(5.8)

»The transmission badwidth required for binary digital waveform

such as PCM may be very large.

-Multilevel signal is used to reduce the required bandwidth. Let

us consider a bit stream with data rate R bits per second. Instead

of transmitting a pulse waveform for each bit ,we might first

partition the data into K bit group, and then use (M=2k)level pulse

5-13

for transmission .With such multilevel signaling or M-ary PAM ,

each pulse waveform can now represent a K-bit symbol in a

symbol stream moving at the rate of (R/K) symbol per second

-The receiver must distinguish between the possible levels

(M=2k)of each pulse. The transmission of an M-level pulse

requires a greater amount of energy for equivalent detection

performane if compared with binary (two level) transmission.

Ex 5.1

The information in an analog waveform, with maximum

frequency fm=3 kHz, is to be transmitted over an M-ary PAM

system ,where the number of pulse levels is M=16.

.The quantization distortion is specified not to exceed 1%

of the peak to peak analog signal .what is the PAM pulse or

symbol transmission rate ?

solution

6.550log02.0

1log01.02

1log2222

12

P

leg

Therefore use =b bit/sample

fs=2fm=2! 3kHz (sample/second)

But each sample composed of b bits

:.transmission rate R=6000! b=3600 bit/sec

with M = 16 level = 2k :. k = 4 bit

6-1

Communication Systems

Lecture 6

6-2

Baseband Demodulation Detection

-The received waveforms of baseband signals are in a pulse-form

.The demodulator needed to recover the pulse waveform because

of the arriving baseband pulses are not in the same from of ideal

pulse shape.

-The filtering at the transmitter and the channel typically cause

the received pulse sequence to suffer from intersymbol

interference (ISI)and thus appear as a wide pulse (i.e the pulse

width is increased ),not quite ready for sampling and detection

-The goal of the demodulation as recovery of a waveform to an

undistorted baseband pulse with Maximum S/N and free from

any ISI can be obtained by equalization ( is a technique used to

help accomplish this goal).

-The detection is used to decision making process of selecting the

digital meaning of that waveform.

-There are two primary causes for error performance degradation

:-

1-The effect of filtering at the transmitter, channel and receiver

causes symbol smearing or intersymbol interference (ISI)

2-Electrical noise and interference produced by a variety of

sources ,such as atmospheric noise ,switching transients ,

intermodulation noise as well as interfering signals from, thermal

noise.

6-3

-Thermal noise is characterized by constant power spectral

density ,we refere to it as white noise . The spectrum of thermal

noise about 1012Hz

-Fig(6.1) shows the typical demodulation and detection

functions of a digital receiver.

Message symbol or bit if error correction not present (^mi)or

channel symbol or bit if error correction is used (

u i)

-During a given signaling interval T, a binary baseband system

will transmit one of two waveforms ,denoted g1(t) and g2(t)

.Similarly , a binary bandpass system will transmit one of two

Note Frequency down conversion and Equalizing filter are optional units

6-4

waveforms ,denoted by S1(t) and S2(t).Since the general treatment

of demodulation and detection are essentially the same for

baseband and bandpass systems,we use si(t) for a transmitted

waveform ,whther the system is baseband or bandpass. For any

binary channel, the transmitted signal over a symbol interval

(0,T)is represented by

Si(t)=

1binaryforTt01S

0binaryforTt0(t)2

S

-The received signal r(t) degraded by noise n(t) and possibly

degraded by the impulse response of the channel hc(t) is given by

r(t)=Si(t)* hc(t)+n(t) i=1,….,M…….(6.1)

Where n(t)is assumed to be a zero mean AWGN process and *

represents a convolution operation .

-For binary transmission over an ideal distortion less channel

where convolution with hc(t) produces no degradation.

:. r(t)=Si+n(t) …….………..(6.2)

-If error correction coding not present, the detector output

consists of estimates of meaasge symbols (or bits) im

(also called

hard decisions)

-If error correction coding is used, the detector output consists of

estimates of channel symbols (or coded bits) iu

-The frequency down conversion block, performs frequency

translation for bandpass signals operating at some radio

frequency (R F)

6-5

-The filtering at the transmitter and the channel typically cause

the received pulse sequence to suffer from ISI and thus not quite

ready for sampling and detection.

-The goal of the receiving filter is to recover a baseband pulse

with best possible signal to noise ratio (SNR), free of any ISI.

The optimum receiving filter for this purpose is matched filter or

correlator.

-Equalizing filter is used for those systems where channels

induced ISI can distort the signals.

-At the end of each symbol duration T, the output of the sampler,

the predetection point, yields a sample Z(T).

In step(2) ,a decision (detection)is made regarding the digital

meaning of that sample.

!!!The output step(1)is given by

Z(T)=ai(T)+n0(T)…….i=1,0………..(6.3)

where ai(T) is the desired signal component and n0(T) is the noise

component.

-If the noise component n0 (t) is a zero mean Gaussian random

variable, and thus Z(T) is Gaussian random variable with a mean

of either a1or a2 depending on whether a binary one or binary

zero was sent

Eq(6.3) can be simplified as

Z=ai+n0………...(6.4)

6-6

-If the probability density function (Pdf)of the Gaussian random

noise n0 can be expressed as

)5.6.........(2

1exp

2

1)(2

000

no

nP

Where 02is the noise variance .

The conditional probability density functions

!!!!!!!!!

21 S

ZPand

S

ZP can be expressed as

)7.6(..........2

1

2

1

)6.6.........(2

1

2

1

2

0

2

2

2

0

1

1

aZe

S

ZP

aZe

S

ZP

2a 1

a

Fig(6.2)

6-7

Fig(6.2) shows these conditional (Pdf) ,where:

-p (Z/S1) called the likelihood of S1, illustrates the probability

density function of the random variable Z(T) given that symbol

S1 was transmitted

-p(Z/S2) called the likelihood of S 2 , illustrates the Pdf of Z(T)

given that S2was transmitted

-The axis Z(T) represents the full range of possible sample output

values from step 1

-The received signal energy (not its shape)is the important

parameter in the detection process. This is why the detection

analysis for baseband signals is the same as that for band pass

signals.

-Since Z(T)is a voltage signal is proportional to the energy of the

recived symbol ,the larger the magnitude of Z(T), the more error

free will be the decision making process.

-In step 2, detection is performed by choosing the hypothesis that

results from the threshold measurement

Z(T)

2

1 )8.6(............

H

H

where H1and H2 are the two possible(binary) hypothesis .The

inequality relationship indicates that hypothesis H1is chosen if

Z(T)>and

hypothesis H2 is chosen if Z(T)<

if Z(T)=,the decision can be arbitrary one.

6-8

The basic SNR parameter for digital communication systems-The average signal power to average noise power ratio (S/N or

SNR) is used as figure of merit in analog communication

systems

-The bit energy (Eb) to the noise spectral density (No) ratio is

used as figure of merit in digital communication systems

But the bit energy Eb =S Tb

where S=signal power ,Tb=bit time and the noise power spectral

density

No W

N where N=noise power ,W=bandwidth since the bit time

and bit rate Rb are reciprocal

:. Tb=1/Rb

)9.6(........../

/

/0 b

bbb

R

WN

S

WN

RS

WN

TS

N

E

where data rate, in units of bit/sec is one of the most parameters

in digital communication (R is used instead of Rb to simplify the

notation)

)10.6(..........0 R

WN

S

N

Eb

Eq(6.10) represent that Eb/No is just version of S/N normalized

by bandwidth and bit rate

6-9

-One of the most important parameter to study the performance

of digital communication system is a plot of the bit error

probability PB versus Eb/No as shown in fig (6.3)

fig(6.3)

-Eb/No is a dimensionless ratio as shown below

swatt

swattSwatt

Swatt

Hzwatt

Joule

N

E

/1//0

-Detection of binary signals in Gaussian noise

º The decision making criterion shown in step (2) of fig(6.1)was

described by eq(6.8) as

8).......(6...........(T)Z1H

2H

PB

Eb/N0

6-10

-A popular criterion for choosing the threshold level for the

binary decision in eq (6.8)is based on minimizing the probability

of error.

-The computation for this minimum error value =0 strarts with

forming an inequality expression between the ratio of conditional

probability density functions and the signal a priori probabilities.

-Since the conditional density function p(Z/Si) is also called the

likelihood of Si the formulation (is called likelihood ratio test).

)(

)(

)(

)(

1

2

2

1

2

1

sp

sp

szp

szp

H

H

(6.11)

where

p(S1) and p(S2) are the a priori probabilities that S1(t) and S2(t)

,respectively are transmitted and H1and H2 are the two possible

hypotheses .

-The rule for minimizing the error probability states that we

should choose hypothesis H1if the ratio of likelihoods is greater

than the ratio of a priori probabilities as shown in eq(6.11)

The substitution of equations (6.6)and(6.7)in eq (6.11) to find

the optimal value for threshold 0

-The noise assumed to be independent Guassian random variable

with zero mean, variance 02 and pdf given by

20

0

02

1exp

2

1)0

(n

nP

6-11

The likelihood ratio test described in eq (6.11)can be written as

)(

)(

2

2exp

2exp

2exp

2

2exp)

2exp(

2exp

)(

2

1exp

2

1

2

1exp

2

1

/

/)(

1

2

2

1

20

22

0

22

20

2

20

12

0

21

20

2

1

2

2

1

2

0

2

0

21

0

1

1

SP

SP

ZaaZ

ZaaZ

SP

SP

H

H

aZ

aZ

SZP

SZPZA

H

H

A(Z)= exp

)12.6(..........)(

)(

2 1

2

2

1

20

222

12

0

21

SP

SP

H

HaaaaZ

where a1 is the receiver output signal when S1(t) is sent a2 is the

receiver output signal when S2(t) is sent

To simplify eq(6,12),we take the natural logarithm of both sides,

resulting in the log likelihood ratio .

)(

)(

2)(

1

2

2

1

20

222

12

0

21

SP

SPInaaaaZ

ZAInH

H

Since P (S2) = P (S1) are equally likely :. In

01

2 SP

SP

6-12

)12.6........(2)(2

2

021

21

22

21

2

1

20

22

12

2

0

21

aa

aaaa

Z

aaaaZ

H

H

or 0

21

2

1

2)( aaTZ

H

H

2a 1

a

)( 2szp )( 1szp

Fig(6.4)

where a1is the signal component of Z(T) when S1(t) is transmitted

. as is the signal component of z (T) when S2 (t) is transmitted

0 is the optimum threshold for minimizing the probability of

making an incorrect decision for this case.

-The above strategy is known as the minimum error criterion.

6-13

-for equally likelihood signals, the optimum threshold passes

through the intersection of the likelihood functions as shown in

figure(6.4).

-For M-ary instead of a binary, there would be a total of M

likelihood functions representing the M signal classes of received

signals. the maximum likelihood decision would then be to

choose the class that had the greatest likelihood of all M.

Error Probability

-For the binary decision making shown in fig(6.4), there are two

ways error can occure:-

1-An error e will occurred when S1(t) is sent ,and channel noise

results in thr receiver output signal Z(t) less than . The

probability of such an occurrence is

p(e/S1)=p(H2/S1)=

0

)13.6(..........1

zdS

ZP

2-Similary, an error occurs when S2(t)is sent ,and the channel

noise result in z(t) being greater than 0 The probability of this

occurrence is

P(e/S2)= P (H1/S2) =

02 )14.6(........../

zdSZP

The probability of an error is the sum of the probabilities of all

the ways that an error can occur. For binary case, we can express

the probability of bit error as

6-14

!!!!!!!!!!!!!PB=

2

1iP(e,Si)=

2

1iP(e/Si)P(Si)……..(6.15)

PB =P(e/S1)P(s1)+P(e/S2) P(S2)……..(6.16a)

=P(H2/S1)P(S1)+P(H1/S2)P(S2)……...(6.16b)

That is, given that signal S1(t) was transmitted ,an error result if

hypothesis H2 is chosen or given that signal S2(t) was transmitted,

an error results if hypothesis H1chosen .for the case where the a

priori probabilities are equal [i-e P(s1)=P(s2) =2

1 ]

:.PB=2

1P(H2/S1)+2

1P(H1/S2)……..…(6.17)

And because of the probability of the density functions

!!!PB=P(H2/S1)=P(H1/S2)……………..(6.18)

The probability of a bit error can be computed by integrating

p(Z/S1) between the limits - to 0 or by integrating p(z/S2)

between the limits to .

:.PB=

)19.6.......(........../ 2 dZSZP

Replacing the likelihood P(Z/S2)with Gaussian equivalent from

eq(8.7)

PB=

dZaZ

0

2

0 2

1exp

2

1

Where 20is the variance of the noise output of the correlator

6-15

2/210aa

Let dzduthenaZU

00

2 ,

PB =

u

aau

duu

02/)21(2

2exp

2

1

PB = )20.6.......(..........2 0

21

aaQ

Where Q(x) called the complementary error function is

commonly used symbol for the probability under the tail of the

Gaussian pdf .It is defined as

ud

uQ

2

2exp

21)(

Gaussian noise Q(x) can not be evaluated in closed form .It is

presented in tabular form .

-We have optimized (in the sense of minimizing PB) the threshold

level , but have not optimized the receiving filter.

-Good approximation to Q(x) if is3

)21.6..(..........2

1)(

2

ZeQ

6-16

The matched filter

-A matched filter is a linear filter designed to provide the

maximum signal to noise power ratio at its output for a given

transmitted symbol waveform.

-Let us consider signal S(t) plus AWGN n(t) is applied to a linear

time-invariant receiving filter followed by a sampler as shown in

fig(6.3)

Fig (6.5)

At t=T, the sampler output

Z(T)=ai+n0

where ai= signal component at the filter output

N0=noise component

If the variance of the output noise (average noise power) is

denoted by 02

:. TN

S

powernoiseaverage

powersignalousinstantaneThe

)22.6..(....................2

0

2

ia

N

S

we wish to find the filter transfer function H0(f) that

Si (t) AWGN Sample at t=T

Z (T)

Receiving filter

6-17

maximizes eq (6.22) i.e maximizes

TN

S

dftfjefSfHta

i2)()()(

where S(f) the Fourier transform of the input signal S(t)

This is the signal at the filter output in terms of the transfer

function H(f) and Fourier transform of the input signal

If the two sided power spectral density of the input

Noise = HzwattN /2

0

2|)(|)()( fHfGfG

xY

densityspatralpowerpIfG

densityspatralpowerpfGwhere

x

Y

/)(

/0)(

)23.6..(..........)(2/

)()(

/)(/2

2)()(

2)(

2

2

0

2

2020

20

0

dffHN

dfefSfH

N

S

dffHN

fdfHfxG

fdfYGpowernoiseoutput

Tfj

T

Using Schwarz`s inequality

*The equality holds if and only of f1(x)=kf*2(x)

6-18

where k is an arbitrary constant and * indicates complex

conjugate

dxxFdxxfdxxFxF |)(2|)(1|)()(

2

21 |2

TfjefSxffHxfLet 221

)()(,)()(

fdfSfdfHfTj

fH e 22 |)(|.|)(|2

)(

Substituting into eq(6.23) yields

)14.6......(..........

0

2max

2

0

20

22

|)(|2

|)(|2/

|)(||)(|

N

ETN

Sor

dffSN

dffHN

dffSdffH

TN

S

where the energy of the input signal s(t)is

E =

)25.6.........(..........2/)(/ dffS

Thus the max output (S/N)T= 2E/No depend on the signal energy

and the power spectral density of the noise ,not on the particular

shape of the waveform that is used.

6-19

Eq (t.24) theifonlyholdsN

E

N

S

T

0

2max

Optimum filter transfer function H0(F) is employed, such that

)26.6..(..........)()()( 2*0

TfjefSKfHfH

on h (t) = F -1 H0 (f) = F-1 K S * (f) e-j2fT

TttTsK

whereelseth

0......).........(

..................0)(

Thus the maximum response of a filter that produces the

maximum output s/n is the mirror image of the message signal

S(t),delayed by the symbol time duration T as shown in fig (6.4).

Fig(6.6)

7-1

Communication Systems

Lecture 7

7-2

Correlation Realization of the Matched filter

The matched flier transfer function is given by

)1.7.........(..........)()( *0

FTwjefSKfH

S(f)=is the Fourier transform of the input signal

where T=symbol duration

)2.7........((t)h and0..........).........(

0

TttTSK

elsewhere

The output of the matched filter Z(t) can be described in the time

domain as the convolution of a received input waveform r(t) with

the impulse response of the filter .

t

0

.(7.3)..........dττ)h(t(τrh(t)*(t)r(t)Z )

where r(t)= S(t)+n(t)

Substituting h(t) of Eq(7.2) into eq(7.3)and setting k=1

t

t

dtTsrtZ

dtTSrtZ

0

0

][)()(

)()()(

7-3

Fig(7.1a)

When t=T

T

dsrTZ0

)()()( (7.4)

Eq (7.4)described the correlation process.

Note

It is valid to implement he receiving filter with either matched

filter or a correlator.

It is important to note that the correlator output and the matched

filter output are the same only at time t=T.

Fig(7.1) shows the comparison of correlator and matched filter

output .

r () Z(t)

S()

Fig (7.1 b )

T

0

7-4

Optimizing Error Performance

-For the binary case, the optimum decision threshold was shown

.......2 0

210 aa ….. where a1 is the signal component of Z(t)

when s1(t) is transmitted and an is the signal component of

Z(t)whenS2(t)is transmitted.

2a1

a

1s2s

Fig(7.2)

-For minimizing the probability of error (PB), it is necessary to

choose the filter (matched filter) that maximizes the value of

Q

0

21

2aa .Thus we need to determine the linear filter that

maximizes 021 2/ aa or equivalently that maximizes

221 22)( aa

7-5

Note

2/

2)(

2

2

2exp

2

1

21

0

2

0

21

0

aaZZfor

azu

aaQ

udu

BP

where (a1-a2)2 is the difference of the desired signal components at

the filter output at time t=T, and the square of this difference

signal is the instantaneous power of the difference signal .

»It was shown that a matched filter achieves the maximum

possible output SNR equal to 2E/No.

»Consider that the filter is matched to the input difference signal

[S1(t)-S2(t)], thus we can write an output SNR at time t=T as

)5.7...(..........22

2221

Edaa

TN

S

where No/2 is the sided power spectral density of the noise at the

filter input

)6.7(..........)()(0

21 ][ dttStSEdT

7-6

Substituting Eq(7.6) in Eq(6.20)

)7.7(..........02

)20.6.......(2

212

2exp

2

1

0

N

dEQP

uuQdu

uP

B

B

Eq(7.7)is an important results express the optimizing the error

performance interm the energy difference signals at the matched

filter’s input.

Cross –Correlation and Optimized Probability of Error

Relationship

The time cross-correlation coefficient P is defined as a measure of

similarity between two signals S1(t) and S2(t)

T

b

tdtStSE 0

21 )8.7.........()()(1

From eq(7.6) T

dttStSEd0

221 )]()([

but

7-7

T

bbdtdtStSEEE

0

21 )9.7........()()(2

Substiating eq (7.8) into Eq(7.9),we obtain

)10.7...(..........12

)8.7......()()(1222

01

p

T

bbb

EEd

tdtStSE

EEEd

Substituting eq (7.10) into eq(7.7),we obtain

)11.7....(..........)1(

0N

EQp p

B

)7.7......(

2

0N

EdQP

B !!

!!!!!!!!!!!!!!!!!!!!!!

If =-1for the case S1 and S2 are antipodal signals

:. )12.7....(..........2

0N

EQp b

B

if =0 for the case of orthogonal signals

)13.7.........(..........0N

EQ b

if =1 for the case of correlated signals S1 (t) = S2 (t)

:. PB = 0

Ex7.1

a-Derive the relationship between the cross-correlation and

probability of error for two signals S1(t). andS2(t)

7-8

b-Consider a binary communication system that receives equally

likely signals S1(t) and S2(t) pulse AWGN as shown in fig (7.4).

Assume that the receiving filter is a matched filter and that the

noise power spectral density No=10 12 Watt/Ht, compute the bit

error probability.

Fig(7.4)

Solution :

Joule

ttttdtdtd

tdtEb

12

623623623

1

02

3

1

22

0

122

1

3

2

222

3

0

2

106

10101010210101

121121

)(

7-9

Since the waveforms are antipodal (=1) and detected with a

matched filter:

)46.3(

1210

1062212

12

0

QP

QQN

EQp

B

bB

From the table of complementary error function

41030003.0 BP

Or, since the argument of Q() is greater than 3 we can also use

the approximate relationship in Eq(6.21).

42

2

1089.22

)46.3(exp

246.3

1

)21.6(..........2

exp2

1

B

P

Error Probability Performance of Binary Signaling

1-Unipolar signaling

Fig (7.5) shows an example of baseband unipolar signaling where

S1 (t) = A 0 t T for binary 1

S2 (t) = 0 0 t T for binary 0 !!!!!!!!!!!!

where A 0 is the amplitude of symbol S1 (t)

7-10

The energy difference signal, from eq(7.9) is given by

T

TAtdtStSEd0

22 .)(2)(1

Then the bit error performance at the output is obtained as follows

00

2

0 22 N

EQ

N

TAQ

N

EQp bd

B

2-Bipolar signaling

Fig (7.6) shows an example of baseband antipodal signaling

namely, bipolar signaling

where,

S1(t)=+A 0 t T for binary 1

Fig (7.5)

7-11

S2(t)=» A 0 t T for binary o

Fig(7.6)

-In fig(7.4),the correlator output Z(T)=Z1(T)-Z2(T) and the

deicion is made using the threshold 0 = (a1+a2)/2

-For antipodal signals 0 = 0 for a1= ! a2 thus if the test statistic

Z(T) is positive, the signal detected is declared to be S1(t) and if it

is negative, it is declared to be S2(t) .

-The bit error performance at the output can be obtained as

follows

7-12

0

2

0

)1(2

02 N

EQ

N

bEQ

N

EdQ

BP b

-Fig (7.7) shows the comparison PB between bipolar and unipolar

transmissions.

fig(7.7)

Intersymbol Interference

Fig (7.8a) shows different filters are used in atypical digital

communication systems .There are various filters are used in the

transmitter receiver and channel.

1-At the transmitter, the information symbols, characterized as

impulse or voltage levels, modulate pulses that are then filtered

according to the bandwidth constraint.

1for antipodal

7-13

2-For baseband systems, the channel (such as cable),has

distributed reactances that distort the pulses .Some bandpass

systems ,such as wireless systems, are characterized by fading

channels, that behave like undesirable filters causes signal

distortion.

-When the receiving filter is used to compensated for the

distortion caused by both the transmitter and the channel, it is

often reffered to as an equalizing filter or a receiving equalizing

filter.

-Fig (7.8b) shows a convenient model for the system, collect all

the filtering effects into one equivalent system transfer function

H(f) = Ht(f) Hc(f) Hr(f) ……….(7.14)

where Ht(f) = transmitting filter transfer function.

Hc(f) = channel filter transfer function.

Hr(f)= receiving / equalizing filter transfer function.

»The characteristic H(f) then, represents the composite system

transfer function due to all the filtering at various locations in the

transmitter ,channel and receiver.

-Due to the effects of system filtering, the received pulses can

overlap one another as shown in fig(7.8b). The tail of a pulse can

"smear" into adjacent symbol intervals then interfering with the

detection process and degrading the error performance, such

interference is termed intersymbol interference (ISI).

7-14

-Even in the absence of noise, the effects of filtering and channel

induced distortion lead to (ISI).

fig(7.8)

-Nyquist showed that the theoretical minimum system bandwidth

needed in order to detect Rs symbol/S without ISI is (Rs/2)hertz.

This occure when the system transfer function H(f)is made

rectangular as shown in fig(7.9).

-For baseband system, when H(f) is such a filter with signal sided

bandwidth (1/2T) (the ideal Nyquist filter) its impulse response ,

the inverse Fourier transform of H(f)is often form h(t)=sinc(t/T).

This sinc pulse is called the ideal Nyquist pulse.

7-15

fig(7.9)

-Fig(7.9b)shows how ISI is avoided, if the sampling timing is

perfect.

-For the baseband systems, the bandwidth required to detect (1/T)

such pulses (symbols) per second is equal to1/2T or

the system bandwidth W

bandwidthNyquistzheresR

TW

22

1

»The maximum possible symbol transmission rate per hertz,

called the symbol rate packing.

-The names Nuquist filter and Nyquist pulse are often used to

describe the general class of filtering and pulse shaping that

satisfy zero ISI at the sampling point.

-It should be clear from the rectangular shaped transfer function

of the ideal Nyquist filter and the infinite length of its

corresponding pulse, that such ideal filters are not realizable, they

can only be approximately realized.

7-16

The Raised –Cosine Filter

-This filter is chosen so as to optimize the composite system

frequency transfer function H(f) ,shown in eq(7.14).This filter

belong to the Nyquist class (Zero ISI at the sampling times) is

called the raised-cosine filter .It can be expressed as

where W=Absolute bandwidth [ This is the interval between

frequencies ,outside of which spectrum is zero].

W0=1/2T represent the minimum Nyquist bandwidth for the

rectangular spectrum and the-6dB bandwidth for the raised cosine

filter.

W—W0 is the excess bandwidth, which means additional

bandwidth beyond Nyquist minimum {i.e for rectangular

.spectrum W=0} -The roll off factor is defined to be

)16.7....(..........10

rW

WWr

Thus r represents the excess bandwidth divided by W0(i.e the

fractional excess bandwidth) for a given value W0, the roll off r

specifies the required excess bandwidth as a fraction of W0 and

characterizes the steepness of the filter roll off.

….(7.15)

7-17

-Fig (7.10) shows the raised cosine filter characteristics for roll

off values of r=0, r=0.5 and r =1. The r=0 roll off is the Nyquist

minimum bandwidth case when r=1, the required excess

bandwidth is 100%

i.e twice the Nyquist minimum bandwidth.

-The impulse response for the raised cosine filter is

)16.7(..........41

2cos)2(2)(2

0

000

tWW

tWWtWSincWth

fig(7.10)

7-18

Note

-The raised cosine spectrum is not physically realizable (for the

same reason that the ideal Nyquist filter is not realizable). A

realizable filter must have an impulse response of finite duration

and a zero output prior to the pulse on time (i.e causal system).

The general relationship between required bandwidth and symbol

transmission rate involve the filter roll-off factor r is

W=2

1(1+r) Rs (7.17)

where Rs = symbol rate

Ex7.2 : Find the minimum required bandwidth for the baseband

transmission of a four level PAM pulse sequence having a data

rate of R=2400 bit / sec . If the system transfer characteristic

consist of a raised cosine spectrum with 100% excess bandwidth

(r=1).Sketch the shaped pulse and baseband raised cosine

spectrum.

Solution:

M = K2 since M =4 K=2

Symbol or pulse rate R = ssymbolK

R /12002

2400

Minimum bandwidth RrW 12

1=

2

1(1+1)1200=1200 zH

7-19

Fig(7.11) shows the baseband PAM received pulse in the time

domain .Fig (7.12b) shows the Fourier transform of h(t)the raised

cosine spectrum.

fig(7.11)

Ex.7.3(a) Discuss the operation of raised cosine filter.

(b)-Find the minimum required DSB bandwidth for transmitting

the modulated PAM sequence if the baseband transmission of

four level PAM having a data rate R=2400 bit/s if the system

transfer characteristic consist of a raised cosine spectrum with

100% excess bandwidth (r=1) .Sketch the modulated pulse and

DSB modulated raised raised cosine spectrum.

solution:

M= K2 since M = 4 level :. K=2

.sec/12002

2400 symbolK

RS

R

WDSB=(1+r)RS = (1+1) (1200) = 2400 HZ

7-20

fig(7.12)

Equalization

-Many communication channels (e.g telephone, wireless) can be

characterized as band limited linear filters with an impulse

response h(t)and frequency response

)18.7.....(..........|)(|)( )( fcjcc efHfH

where hc(t) and Hc(f) are Fourier transform pairs.

|Hc(f)| the channel’s amplitude response.

c(f) the channel phase response.

»In order to achieve ideal (nondistorting) transmission

characteristics over a channel ,within a signals bandwidth W,

1-|Hc(f) | must be constant.

2-θc(f) must be a linear function of frequency or the time delay

must be constant for all spectral range (components) of the signal

7-21

»If |Hc(f)| is not constant within W, then the effect is amplitude

distortion.

-If c(f) is not constant or a linear function of the frequency

within W , then the effect is phase distortion .

»Fading channel characterized by both type of distortions

(amplitude and phase distortions).

-Equalization refers to any signal processing or filtering technique

that is designed to eliminate or reduce (ISI).

-Equalization categories:-

1-Maximum likelihood sequence estimation (MLSE) MLSE

include measurements of hc(t) and then providing a means for

adjusting the receiver to the transmission environment, to enable

the detector to make a good estimates from the demodulated

distorted pulse sequence. With an MLSE receiver the distorted

samples are not reshaped or directly compensated in any way.

2-Equalisation with filters; uses filter to compensate the distorted

pulses. With this type, the detecter is presented with a sequence of

demodulated samples that the equalizer has modified or cleaned

up from the effects of ISI. The filters can be described as

follows;-

a-Linear devices that contain only feedforward elements

(transversal equalizers).

7-22

b-Non linear devices that contain both feedforward and feedback

elements (decision feedback equalizers)

c-Preset or adaptive; linear and non linear devices can be grouped

according to the automatic nature of their operation ,which may

be either preset or adaptive.

-Equalization with filters, the more popular

If the receiving/equalizing filter be replaced by a separate

receiving filter and equalizing filter defined by frequency transfer

functions. Hr(f) and He(f)respectively. Also,.let the overall system

transfer function H(f) be a raised cosine filter ,designated by HR

c(f). Thus we now modify the eq(7.14) as follows .

H(f) = Ht (f) Hc(f) Hr(f)…………(7.14)

HRc(f)=Ht(f) Hc(f) Hr(f) He(f) ……….(7.15)

-In practical the channels frequency transfer Hc(f) and its impulse

response hc(t)are not known with sufficient precision to allow for

a receiver deign to yield zero (ISI) for all time .Usually the

transmit and receive filters are chosen to be matched so that

HRc=(f) = Ht (f) Hr(f) ……..(7.16a)

In this way Ht(f) and Hr(f) each have frequency transfer functions

that are the square root of the raised cosine (root-raised cosine)

Then the equalizer transfer function need to compensate for

channel distortion is simply the inverse of the channel transfer

functions.

7-23

He(f)= )16.7.....(..........|)(|

1

)(

1 )( befHfH

fcj

cc

-8 1

Communication Systems

Lecture 8

-8 2

Equalizer filter types

1-Transversal equalizer

-For describing the transversal filter, consider that a single pulse

was transmitted over a system designated to have a raised cosine

transfer function HRC(f)=Ht(f)Hr(f). Also consider that the channel

induces ISI, so that the received demodulated pulse exhibits

distortion as shown in fig(8.1), such that the pulse sidelobes do

not go through zero at sample times adjacent to the mainlobe of

the pulse.

The distortion can be viewed as positive or negative echoesــ

occurring both before and after the mainlobe.

Fig (8.1)

-To achieve the desired raised cosine transfer function, the

equalizing filter should have a frequency response. He(f) is given

by

-8 3

i,e when the actual channel response Hc(f) multiplied by He(f)

yields HRc(f)=Ht(f)Hr(f). In other words we would like the

equalizing filter to generate a set of canceling echoes.

-Sampling the equalized waveform at only a few predetermined

sampling times.

-The transversal filter shown in fig (8.2),is the most popular form

of an easily adjustable equalizing filter consisting of a delay line

with T-second taps (where T is the symbol duration). In such an

equalizer, the current and past values of the received signal are

linearly weighted with equalizer coefficients or tap weights [Cn]

and are then summed to produce the output.

-If there is an infinite numbers of taps, then the tap weights could

be chosen to force the system impulse response to zero at all time

but one at the sampling times, thus making He(f) exactly to the

inverse of the channel transfer function .However ,an infinite

length filter is not realizable. Each tap is connected through a

variable gain device to a summing amplifier.

Consider that there are (2N+1) taps with weights

NNN CCC .........,, 1 Output samples Z(K) of the equalizer

are then found by convolving the input samples [ (k)]and tap

weights [Cn] as follows:- )2.8(..........

N

Nn

kZn

Cnkχ

…..….(8-1)

-8 4

where k = -N, ……N

n = - N , ….. N

Where k=01 , 2 …is a time index (Time may take on any

range of values). The index n is used two ways as a time offset,

and as a filter coefficient identifier .We can describe the relation

between the vectors Z and C and the matrix as:-

However, we can specify the value of Z (k) as (2N+1) points as

01

3.8.....,...,2,10

kfor

NkforkZ

Fig (8.2)

or-

where k = - N, ……….. N n = - N, …….…. N

-8 5

)4.8(12

)0(....)2(

:::

:::

:::

:::

:::

:::

)12(...)0()1(

)2(...)1()0(

0

0

:

:

1

:

:

0

0

1

0

1

N

C

C

C

C

C

XNX

NXXX

NXXX

N

N

N

N

Eq (8.4) represents a set of (2N + 1) simultaneous equation that

can be solved for the Cn `s. The equalizer described in Eq(8.3) is

called a zero forcing equalizer since Z(k) has N zero value on

either side . This equalizer is optimum in that it minimizes the

peak intersymbol interference. The main disadvantages of a zero

forcing equalizer is that it increases the noise power at the input to

the A/D converter.

Ex8.1

(a) Describe with aid of a block diagram the transversal equalizer

(b)Design a three tap equalizer to reduce the ISI due to the

received pulse shown in fig (8.4a).

-8 6

2

Solution:

fig(8.4a) received pulse

1-Design three tap equalizer (N=1) to find 101

,, CCC

Z ( -1) = (0) C1 + (-1) C0 + (-2) C1

Z (0) = (1) 1

C + (0) C0 + (-1) C1

Z (1) = (2) 1

C + (1) C0 + (0) C1

1

0

1

1

0

1

C

C

C

10.20.1

0.11.00.2

00.11

1

1

0

C

C

C

(0)χ(1)χ(2)χ

1)(χ(0)χχ(1)

2)χ(1)(χ(0)χ

Z(1)

(0)Z

1)Z(

………..(8.3)

-8 7

10.20.1

0.11.00.2

0.1.01

1.00.20

0.111

00.10

1C

096060

1012020010101201020101011

10101002000110120100110

1

1

.C

..........

........C

9606.0

12.01.0

1.00.12.0

01.01

101.0

1.012.0

001

0

C

2017.0

12.01.0

1.00.12.0

01.01

02.01.0

10.12.0

01.01

1

C

Solution of this yields 1

C = 0.09606, C0=O.9606 and C1=0.2017.

This tap setting produce Z(0)=1,Z(-1)=0 and Z(1)=0.

The output is sketched in fig(8.4b).

fig(8.4b)

-8 8

2-Dcision feedback equalizer (DFE)

-The basic limitation of a linear equalizer, such as the transversal

filter, is that it performs poorly on channels having spectral nulls.

These channels are used in mobile radio. A decision feedback

equalizer (DFE) is a non linear equalizer that uses previous

detctor decisions to eliminate the ISI on the pulses that are

currently being demodulated .The ISI being removed was caused

by the tails of previous pulses, in effect; the distortion on a

current pulse that was caused by previous pulses is subtracted.

Fig(8.5) shows a simplified block diagram of (DFE), where the

forward filter and the feedback filter can be a linear filter, such as

transversal filter. The nonlinearity of the DFE due to the

nonlinear characteristics of the detector that provides an input to

the feedback filter.

-The basic idea of DFE is that if the values of the symbols

previously detected are known (past decisions are assumed to be

corrected), then the ISI contributed by these symbol can be

canceled out exactly at the output of the forward filter by

subtracting past symbol value with appropriate weight.

The forward and feedback tap weights can be adjusted

simultaneously to minimize the mean square error.

The advantage of the DFE implementation is that the feedback

filter, which is additionally .working to remove ISI, operates on

-8 9

noiseless quantized levels, and thus its output is free of channel

noise.

fig(8.5)

Preset and adaptive equalizer

1-Once channel whose frequency responses are known and time

invariant, the channel characteristic can be measured and the

filter’s tap adjusted accordingly, if the weights remain fixed

during the transmission of the data, the equalization is called

preset equalization.

2-If the equalizer capable of tracking a slowly time varying

channel response, is known as adaptive equalization .It can be

-8 10

implemented to perform tap weight adjustments periodically or

continually.

Eye diagram

The ISI and other signal degradation can be studied conveniently

on an oscilloscope through what is known as the eye diagram. A

random binary pulse sequence is sent over the channel.

1-The channel output is applied to the vertical input of an

oscilloscope.

2-The time base of the scope is triggered at the same rate that of

the incoming pulses, and it yields a sweep lasting exactly Tb, the

interval of one pulse. The oscilloscope pattern thus formed looks

like a human eye and, hence, the name diagram fig(8.6)

-As an example, consider the transmission of a binary signal

bypolar rectangular pulses.

a-If the channel is ideal with infinite bandwidth ,pulses will

received without distortion. The resulting eye diagram as shown

in fig(8.6a)

b-If the channel is not distortionless or has finite bandwidth or

both ,received pulses will no longer be rectangular but will be

rounded and spread out .If the equalizer is adjusted properly to

eliminate ISI at the pulse sampling instants ,the eye diagram will

be rounded as shown in fig(8.6b)

-8 11

Fig(8.6)

Band pass Modulation and Demodulation

-Digital modulation is the process by which digital symbols are

transformed into waveforms that are compatible with the

characteristics of the channel

-In the case of baseband modulation, these waveforms usually

take the form of shaped pulses. But in the case of bandpass

modulation the shaped pulses modulate a sinusoid called a carrier

wave or simply a carrier, for a radio transmission the carrier is

converted to an electromagnetic (EM) field for propagation to the

desired distortion.

Waveform Eye diagram

Tb

(a) Tb

t opening

(b) width

-8 12

Why it is necessary to use a carrier for radio transmission of

baseband signals?

1-The transmission of EM fields through the space is

accomplished with the use of the antenna. The size of the antenna

depends on the wavelength

)5.8(....................10

1

4

1

10

1

4

1

f

CL

where C= speed of the light=3*108 m/sec

L=the length of the antenna.

f=frequency in Hz.

2-Many signals can be transmitted simultaneously by using

frequency division multiplexing such as telephony signals.

3-Modulation can be used to reduce the interference.

4-Moodulation can be used to allocate a certain frequency for

each broadcasting station or TV station in order to allowed them

transmit in the same time.

5-Modulation can be used to place signal in a frequency band

where design requirements, such as filtering and amplification,

can be easily met. This is the case when radio frequency (RF)

signals are converted to an intermediate frequency (IF) in a

receiver.

Digital bandpass modulation techniques

-The bandpass modulation can be defined as the process of

variation the amplitude, frequency or phase of an RF sinusoidal

-8 13

carrier or a combination of them accordance with the information

to be transmitted .The sinusoid carrier duration T is referred to as

a digital symbol.

-The general form of the carrier wave is

S(t)=A(t) cos (t) ……… (8.6)

where A(t) is the time varying amplitude and

(t) is the time varying angle

But (t)=0t+ (t)

:. S(t)=A(t) Cos [0t+(t)]………..(8.7)

Where 0=radian frequency of the carrier

(t) = radian phase of the carrier

When the receiver utilized (exploits) knowledge of the carrier’s

phase to detect the signals, the process is called coherent detection

-When the receiver does not utilize such phase reference

information, the process is called noncoherent detection.

-The basic bandpass modulation /demodulation types are

1-Coherent modulation /demodulation :-

Phase shift keying (PSK), frequency shift keying (FSK),

amplitude shift keying (ASK), Continuous phase modulation

(CPM) and hybrid modulation.

Some specialized techniques such as offset quadrature

PSK(OQPSK), minimum shift keying (MSK) and quadrature

amplitude modulation(QAM).

2-Noncoherent demodulation :-

-8 14

DPSK, FSK, ASK, CPM and hybrids

Phase shift keying (PSK)

The general analytic expression for PSK is

where the phase term i(t)will have M discrete value, typically

given by

)9.8..(..........2.12)( iM

iti

For binary PSK(BPSK) as shown in fig(8.7) , M=2

The parameter E=symbol energy

T=symbol time and 0 t T.

In BPSK modulation, the modulating data signal shifts the phase

of the waveform Si(t)to one of two states either zero or Π(180 ).

The signal waveforms can be represented as vectors or phasors on

a polar plot, the vector length corresponds to the signal amplitude

and the vector direction represents the signal phase

……(8-8)

-8 15

Fig (8.7)

Frequency shift keying (FSK)

The general analytical expression for FSK modulation

Where the frequency term ωi has M discrete values and the phase

term is an arbitrary constant the FSK waveform sketch in

fig(8.8) shows the typical frequency changes at the symbol

transitions At the symbol transitions ,strong change from one

frequency (tone) to another.

There is no requirement for the phase to be continuous for theـ

case MFSK.

There is need for continuous phase for special class of FSK called

continuous phase FSK (CPFSK)

-Fig(8.8) shows, M=3(to show perpendicular axes).In practice M

is power 2(2,4,8,16…)

……..(8.10)

-8 16

M-ary signals always orthogonal .But, not all FSK signaling is

orthogonal.

Fig ( 8.8)

Amplitude shift keying (ASK)

For the ASK example in fig(8.9) the general analytic expression

is

Where the amplitude term TtEi2 will have M discrete values

and the phase term an arbitrary constant.

The ASK waveform sketch in fig (8.9) can described a radar

transmission example, where the two signal amplitude

states TE2 and zero .The ASK vector corresponding to the

maximum amplitude state and a point at the origin corresponding

……(8.11)

-8 17

to the zero amplitude state. Binary ASK signaling also called (on-

off keying).

Fig (8.9)

Amplitude phase keying (APK)

For the combination of ASK and PSK(APK) example in

fig(8.10), the general analytic expression.

illustrates the indexing of both the signal amplitude term and the

phase term .The APK waveform picture in fig(8.10) shows some

typical simultaneous phase and amplitude changes at the symbol

transitions.

-For this example, M has been chosen equal to B, corresponding

to eight waveforms (8-ary). Fig(8.10)shows a hypothetical eight

vector signal set on the phase –amplitude plane .Four of the

vectors are at one amplitude and the other four vectors at a

different amplitude. Each of the vectors is separated by 45°.

-8 18

Fig (8.10)

ASK, PSK and FSK modulators

ASK, PSK and FSK can be produced by presenting an

appropriate digital baseband signal to conventional modulators.

However, appropriate modulators can be implemented more

simply by the feeding the input data directly to a switch which

can select the appropriate signal waveform form one of two signal

sources to make up the modulated signal. Modulators of this

design are shown in fig(8.11).

-8 19

fig ( 8.11) Modulator block diagram

Ex8.1

Derive the waveform amplitude coefficient TE /2

Solution

let us consider S(t)=A Coswt

where A is the peak value of the waveform

-8 20

S(t)= 2 ArmsCoswt= rmsA22 coswt

Assuming the signal to be a voltage or current waveform. A2rms

represent average power P normalized

wtCosTEST

EnergyppowerButwtCospSto

/2

2Ω1

1-9

Communication Systems

Lecture 9

2-9

Detection of signal in Gaussian noise and decision regions

-The bandpass model of the detection process is identical to the

baseband model. That is because a received bandpass waveform is

first transformed to a baseband waveform before the final detection

step takes place.

-For a linear systems, the mathematics of detection is unaffected by

a shift in frequency.

-We can define an equivalence theorem as follows performing

bandpass linear signal processing, followed by heterodyning the

signal to baseband yields the same results as heterodyning the

bandpass signal to baseband, followed by baseband linear signal

processing.

-The term heterodyning refers to a frequency conversion or mixing

process that yields a spectral shift in the signal.

-As a result of this equivalence theorem, all linear signal processing

simulations can take place at baseband (which is preferred for

simplicity), with the same results as at bandpass.

-Fig(9.1) shows the two dimensional signal space and two binary

vectors (S1+n)and (S2+n). The noise vector, n, is a zero mean

random vector, hence, the received signal vector, r, is a random

vector with means S1 or S2.

-The detectors task after receiving r to decide which of the signals

S1or S2 was actually transmitted.

3-9

-For the case where M=2 with S1and S2being equally likely and

with the noise being an additive white Gaussian noise (AWGN)

process.

-The minimum error decision rule is equivalent to choosing the

signal class such that the distance d(r1Si)=|r-Si| is minimized, where

||r-Si|| is called the magnitude of vector r-Si.

-The decision rule for the detector, stated in terms of decision

regions is as follows: When the received signal r is located in

region 1, chooses S1, when it is located in region 2, chooseS2.

-If the angle =°180, then signal set S1and S2 represents BPSK.

Fig (9-1) Two dimensional signal space

Vectorial view of signals and noise (phasor diagram)

-The geometric or vectorial view of signal waveforms that are

useful for either baseband or bandpass signals.

4-9

-We define an N-dimensional orthogonal space as a space

characterized by a set of N linearly independent functions

{j(t)}called basis function.

-Any arbitrary function in the space can be generated by a linear

combination of these basis functions. The basis functions must

satisfy the conditions

T

Nkj

aTtJkjKdtt

ktj

0 ,....,1,

)2.9.....(..........0

Where the operator

kjforbkjforJK

1

0 )2.9....(....................

The operator jk is called the Kroncher delta function when kj

constant are nonzero, the signal space is called orthogonal .When

the basis functions are normalized so that each kj=1, the space is

called an orthonormal space.

-The requirements for orthogonality;-

a-Each j(t) function of the set of basis functions must be

independent of the other member of the set.

b-Each j(t) must not interfere with any other members of the set in

the detection process.

From a geometric point of view, each j(t) is mutually

perpendicular to each other k(t) for j k.

5-9

-Fig(9.2) shows an example with N=3

am1

am2

am3

Fig (9.2)

-If j(t) correspondes to a real valued voltage or current waveform

component, associated with 1 resistive loads, the normalized

energy in joules in the load in T second.

)2.9.....(....................2

0

cKdttjj

ET

j

-One reason we focus on an orthogonal signal pace is that the

distance measurements, fundamental to the detection process, are

easily formulated in such a space.

-However, even if the signaling waveforms do not make up such an

orthogonal set, they can be transformed into a linear combination of

orthogonal waveforms.

-Any arbitrary finite set of waveforms [Si(t)], (i=1,2,..m)

6-9

where each member of the set is physically realizable and of

duration T, can be expressed as a linear combination of N

orthogonal waveforms 1(t)2(t),…N(t).

where N M

MN

S

MNMM

N

1i

(t)

N2211M

2N2221212

NN1212111(

(9.3a)...M......1.2.3.....i(t)j

ψij

aS

(t)ψa...............(t)ψa(t)ψa(t)S

(t)N

ψa.....................(t)ψa(t)ψa(t)S

(t)ψa........................tψatψat)

i

1

where

T0........,1

......,3.2.1)3.9()()(1

0

tNj

Mibdttti

SK

aj

T

ji j

The coefficient aij is the is he value j(t) component of signal Si(t).

The form of the [j(t)] is not specified, it is chosen for convenience

and will depend on the form of the signal waveforms. The set of

signal waveforms [Si(t)] ,can be viewed as a set of vectors

[Si]=[ai1,ai2….ai N].

If N=3, we may plot the vector Sm(t)

Sm(t) = am1 1(t) +am2 2(t) +am33(t)

Thus each of the transmitted signal waveform Si(t)is completely

determined by the vector of its coefficients (am1,am2,….amN)

7-9

Fig (9.3)

-A typical detection problem, conveniently viewed in terms of

signal vectors as shown in fig(9.3).Vectors Sj and Sk represent

prototype or reference signal belonging to the set of M waveforms

[Si(t)]

-The receiver knows, a priori, the location in the signal space of

each prototype vector belonging to the M-ary set.

-During the transmission, the signal is corrupted by noise, so that,

the received signal vector is corrupted version.

(e.g Si+n or SK+n)of the original one, where n represents a noise

vector.

-The noise is additive and has a Gaussian distribution, the resulting

distribution of possible received signals is a cluster or cloude of

points around Sj and Sk (maximum dense in the center and

decreased with increasing distance from the reference signal)

-Let us consider the vector r represent a signal vector arrive at the

receiver during some symbol interval .The task of the receiver is to

deside whether-r has a close to the reference signal Sj or Sk or sother

signal in the M-ary set.

8-9

Fig(9.3)

Ex(9.1)

Fig(9-4a) shows a set of three waveforms S1(t),S2(t) and S3(t)

a-Demonstrate that these waveforms [Si(t),S2(t)and S3(t)] do not

form an orthogonal set.

b-Fig (9.4b) shows a set of two waveforms 1(t) and 2(t)Verify

that these waveforms form an orthogonal set

c-Show how the nonorthogonal waveform set in part(a) can be

expressed as a liner combination of the orthogonal set in part(b)

(d) Fig (9-4 c) illustrates another set of two waveforms

1 (t)

and

2 (t).Show how the nonorthogonal set in fig (9.4a) can be

expressed as a linear combination of the set in fig(9.4c).

9-9

Fig(9.4)

S1(t) 1(t)

1 (t) 1

T/2 T t T/2 T t t T/2 T

-1

-2

-3

S2(t) 2(t) )(2 t 2

1 1 1 t t t

T/2 T T/2 T T/2 T

(b) (c)

S3(t) 1

T/2 T t

-3 (a)

10-9

Solution

a-We verify S1(t) ,S2(t)and S3(t) do not form orthogonal if they do

not meet the requirement of e.q(9..2a), that is the time integrated

value (over a symbol duration) of the cross product of any two of

three waveform is not zero. Let us verify this for S1(t) and S2

T/2

0

T

T/2

T

0

T/2

0

T

T/2

T/2

0

T

T/2

T

0

T/2

0

T

T/2

4T]-9[T0]-/2[1)(dt3)(3)((1)dt1)(

(t)dt3

S(t)1

Sdt(t)3

S(t)1

Sdt(t)3

S(t)1

S

-T0)-(T/22td(0)3)(td1)(2)(

(t)dt2

(t)S1

Std(t)2

S(t)1

Sdt(t)2

(t)S1

S

2

TT

Similarly, the integral over the interval over T of each of the cross

products S2(t)S3(t) results in nonzero values

:. The waveform set [S1(t) S2(t)andS3(t) ] is not an orthogonal set

b-Using eq (9-2a), we verify that 1(t) and 2(t) form an orthogonal

set as follows

T T T

T

dtdtdttt0 0

2/

2/

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

)(1

11-9

T

jkjjkdtt

kt

0

)()(

Where

kJ

kJkj

1

0

C-Using eq(9.2c) and(9.3b)

T T

T T T

T

jj

T

tjiji

dtdttK

TdtdtdttK

EdttK

Nj

MidttSk

aj

0 0

22

22

0

2/

0

2/

0

222

11

0

2

T)1()(

)1()1()(

)(

.....2.1

......2.1)()(1

0

T

ji dtttSjk

a0

))(1(

ji

1)2

(32

1

)1()3()1()1(1)()(12/

0 2/11

0111

TT

T

dtT

dtttSٍK

aT T

T

T

1 1(t)

T/2 T -1

12-9

T T T

T

dtdtT

dtttSk

a0

2/

0 2/

212

1221311

1)()(1

T T T

T

T T T

T

T T T

T

T T

T

T

T

T T T

T

dtdtT

dtttSk

a

dtdtT

dtttSk

a

dtdtT

dtttSk

a

dtdtT

dtttSk

a

dtdtT

dtttSk

a

0

2/

0

2/

2/

232

32

0

2/

0 2/

131

31

0

2/

0 2/

222

22

0 2/

2/

0/

121

21

0

2/

0 2/

212

12

1)1()3()1()1(1

)()(1

2)1()3()1()1(1

)()(1

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

)()(1

1)1)(0(121

)()(1

213111

)()(1

N,...2,1

,.....1)()()39(eqUsing1

j

MitatSaN

j

jjii

In this case i=1,2,3

j=1,2

2

1

)()(j

jjiitatS

13-9

)(2)()()(

)()()()()(

)(2)()()(a)(

212321313

212221212

212121111

ttatatS

tttatatS

tttattS

Thus, we express the nonorthogonal set S1(t), S2(t)and S3(t)

as a linear combination of the orthogonal basis waveforms 1(t) and

2(t).

D-Similar to part C, the nonorthogonal set S1(t), S2(t) and S3(t)

can be expressed in terms of the simple orthogonal basis

21 ,

S1(t)=- 1 (t) – 3

2 (t)

S2(t)= 2 1 (t) S3(t) = 1

(t)-3

2 (t)

Correlation receiver

-The detection of bandpass signals employ the same concepts of

baseband detection based on realization of matched filter using

correlator in the presence of AWGN.

-The received signal is the sum of the transmitted reference

(prototype) signal plus a random noise

r(t)=Si(t)+n(t) ….(9.4) 0tT

Given such received signal described by eq(9.1),the detection

process consists of two basic steps:-

a-In the first step, the received waveform r(t), is reduced to a signal

random variable Z(t) or to a set of random variables

14-9

Zi(T)[i=1,2,…M] formed at the output of the demodulate or and

sampler at time t=T, where T is the symbol duration.

b-In the second step, a symbol decision is made on the basis of

comparing Z(T) to a threshold or on the basis of choosing the

maximum Zi(T).

Step1can be considered as transformingـ the waveform into a point

in the decision space .This point can be referred to as predetection

point, the most critical point in the receiver.

-When we talk about received signal power, or received interfering

noise on Eb/N�, their values are always considered with reference

to this predetection point.

-The matched filter provides the maximum signal to noise ratio at

the filter output at time t=T. Correlator is used to realization of a

matched filter .We can define a correlation receiver comprised of M

correlators, as shown in fig(9.6), that transforms a received

waveform, r(t),to a sequence of M numbers or correlator output,

Zi(T) (i=1,2,…M).Each correlator output is characterized by the

following product integration or correlation with the received

signal:

……….(9.5)

15-9

Fig(9.6)

-In the case of binary detection, the correlation receiver can be

configured as a single matched filter or product integrated as shown

in fig (9.7a), with the reference signal being the difference between

the prototype(reference) signals S1(t)-S2(t) The output of the

correlator, Z(T) fed directly to the decision stage .Also for binary

detection, the correlation receiver can be drawn as two matched

filters or product integrators to match S1(t) and the other is matched

to S2(t) as shown in fig(9.7b)

Fig (9.7 )

16-9

The noise component n0 is a zero mean Gaussian random variable,

and thus Z(T) is a Gaussian random variable with a mean of either

a1or a2 depending on whether a binary one or binary zero was sent.

Z(T)is called the test statistic.

For the random variable Z(T) fig(9.4) shows the two conditional

probability density functions(pdf), p(Z/S1) and p(Z/S2) with mean

value of a1 and a2 respectively. These pdfs, also called the

likelihood of S1and the likelihood of S2 respectively, and are

rewritten as:

Fig (9.8)

………..(9.6 a )

………..(9.6 b )

17-9

Coherent detection of PSK

The detector shown in fig (9.9) can be used for the coherent

detection of any digital waveforms such a correlating detector is

often referred to as maximum likelihood detector consider the

following BPSK example;

Fig(9.9)

and n(t)=zero mean white Gaussian random process where the

phase term is an arbitrary constant, so that the analysis by setting

=0

……(9.7 b)

(9.7a)

18-9

E=signal energy per symbol

T=symbol duration

-For this antipodal case, only a single basis function is needed. We

can express a basis function 1(t) as

Thus we may express the transmitted signal Si(t) in terms of 1(t)

and ai 1 as follows:

Assume that S1(t) was transmitted. Then the expected values of the

product integrators in fig (9.6), with reference signal 1(t) are found

as

T

0

111

0

1

)11.9(..........)()()(S)(

)((t)r)(

dtttntTZ

dttTZT

i

……………….(9.8)

……………….(9.9)

……………….(9.10a)

……………….(9.10b)

……………….(9.10 c)

19-9

Substituting for S1(t) and 1(t) in eq(9.11)

})()]()([{)( 110

dtttntETiZT

})]()()([{1

21

0

dtttntET

}]cos2

)(2

[{ 2

0

cos dttT

tntET

ET

)12.9....(....................)(1

aETZ

Because the expected value of n(t)=0

Using the same procedure to find Z2(T) = - E (9 ,12b)

The decision stage must decide which signal was transmitted by

determining its location within single space and chooses the largest

value of Zi(T).

Coherent detection of multiple phase shift keying

Fig (9.11)shows the signal space for a multiple phase shift keying

(MPSK)signal set, the figure describes a four level (4-ary)PSK or

quadriphase shift keying (QPSK) example(M=4).

Z1(T) =

20-9

Fig(9-11)

Fig (9-12)

21-9

-At the transmitter, binary digits are collected two at a time, and for

each symbol interval, the two sequential digits instruct the

modulator as to which of four waveforms to produce. For typical

coherent M-ary PSK(MPSK)can be expressed as

where E is the received energy of such a waveform over each

symbol duration T , and 0 is the carrier frequency . If an

orthonormal signal space is assumed , we can choose a convenient

set of functions (axes) , such as

where the amplitude T/2 has been chosen to normalize the

expected output of the detector, as we done in coherent detection

PSK. Now Si(t) can be written in terms of these orthonormal

coordinates, giving

…...…(9.11)

.................(9.13 a)

.................(9.13 b)and

............(9.14 a)

............(9.14 b)

22-9

Eq(9.14) describes a set of M multiple phase waveforms in terms of

only two orthogonal carrier-wave components.

-The decision boundaries partition the signal space into M=4

regions. The decision rule for the detect or is to decide that S1(t)

was transmitted if the received signal vector falls in region 1, that

S2(t) was transmitted if the received signal vector falls in region2,

and so on:

In other words, the decision rule is to choose the ith waveform if

Zi(T) is the largest of the correlator outputs .

-There are M product correlators used for the demodulation of

MPSK signals as shown in fig(9.2)

In practice ,the implementation of an MPSK demodulator, requiring

only N=2 product integrator regardless of the size of the signal set

M as shown in fig (9.12).

Fig(9.12)

23-9

Note

The resulting value of the angle is compared with each of the

reference phase angle i .The demodulator selects the i that is

closest to the angle .

The received signal r(t) can be expressed by combining

equations(9.13) and(9.14)

ttCosT

ES

tTM

iE

tCosTM

iCosEtS

ii

i

0i0

0

0

sinsincos2

sin22

sin

22)(

But (t))(S(t)i

ntr

.......................(9.13)

Si(t) =

24-9

where i=2 i /M and n(t)is a zero mean white Gaussian noise

process.

-Notice that in fig(9.13) only two reference waveforms or basis

functions, 1(t)and 2 (t)

The upper correlator computes X

Fig (9.13)

The lower correlator computes Y

-Fig(9.13) shows that the computation of the received phase angle

can be accomplished by computing the arctan Y/X where X and

Y are in phase and quadrature components and is a noisy

estimate of the transmitted i. The resulting value of the of the

25-9

angle i is compared with the resulting value of the angle is

compared with each of the reference phase angles i. The

demodulator selects the i that is closes to .

10-1

Communication Systems

Lecture 10

10-2

Coherent detection of FSKFSK modulation is characterized by the information being

contained in the frequency of the carrier .Atypical set of FSK

signal waveforms was described as

Mi

TttwCosT

EtS ii

,.........2,1

)110(..........02

)(

Where E is the energy content of Si(t)over each symbol duration

T, and (i+1-i) is, the difference between the frequencies

typically assumed to be an integral multiple of /T. The phase

term is an arbitrary constant and can be set equal to zero

-Now Si(t) can be written in terms of the orthogonal coordinates

giving

MNNj

bMitaS

atatatatSN

j

jiji

jjiiii

,....2,1

)2.10......(,.....2,1)(

)2.10)....((.......)()()(

1

221

Assuming that the basis functions 1(t) ,2(t)….N(t)

form an orthogonal set (Kj=1),the most useful from for j(t) is

2....N....(10.3a

10-3

T

btdtjtiSjkijawhere

0

)3.10....(..........)()(1

Kj=1 for orthonormal set

Substitute for Si(t) and j(t) from equation (10-1,103a) into

(10.3b)

T

jiijdttCosTtCosTEa

0

/2/2

)4.10..(..........

0

/2/20

jiforE

jifora

tdtCosTtCosTEa

ij

T

jiij

Eq(10.4) mean that, the ith prototype (reference) signal vector is

located on the ith coordinate axis at a displacement E from the

origin of the signal space. For the general M-ary case and a given

E,the distance between any two reference signal vectors Si and Sj

is constant

Fig (10.1) shows the prototype (reference) signal and the decision

regions for a 3-ary (M=3).

-The optimum decision rule to decide that the received signal in

which region is found correspond to that transmitted signal

d= EE 22

2E

10-4

Fig (10.1)

For example ,if the transmitter had sent S2,then r would be the

sum of signal S2 plus noise na =S2+na and the decision to choose

S2is correct, however if the transmitter had actually sent S3,then r

would be the sum of signal plus noise, S3+nb and the decision to

select S2is an error.

-Fig (10.2) shows the block diagram for coherent FSK receiver

T

0

T

0

)(1 t

)(2

t

)(t

fig ( 10.2)

10-5

Noncoherent detection

1.Detection of differential PSK

The differential phase shift keying (DPSK) as the noncoherent

version of PSK. It eliminates the need for coherent reference

signal at the receiver by combining two basic operations at the

transmitter:-

-1-Differential encoding of the input binary wave

2-Phase shift keying (Hence the name differential phase shift

keying).

-With noncoherent systems, no attempt is made to determine the

actual value of the phase of the incoming signal. If the transmitted

waveform is

The received waveform can be characterized by

r(t) = Mi

TttnttCosT

Ei

......2,1

)610.....(0)()(2

0

Where α is an arbitrary constant and is typically assumed to be

random variable uniformly distributed between zero and 2 and

n(t) is AWGN process.

-If we assume that varies slowly relative to two period times

(2T), the phase difference between two successive waveforms

…………..(10-5)

10-6

j (T1) and k(T2) is independent of that is

)7.10).......((1212 TTTTTijkjk

-The basis for DPSK is as follows :-

a-The carrier phase of the previous signaling interval can be used

as a phase reference for demodulation .

b-Differential encoding is used for transmitted message sequence

at the transmitter

c-Phase shift keying is used by changing the phase according the

information (i=2i/M)

d-The detector (in the receiver), calculate the coordinates of the

incoming signal by correlating it with locally generated

waveforms ,such as tT0

cos2 and tT0

sin2 , the detector

then measures the angle between the currently signal vector and

previously received signal vector as shown in fig(10.2).

Fig (10.2)

10-7

-In general DPSK signaling performs less efficiently, than PSK,

because the errors in DPSK tend to propagate due to the

correlation between signaling waveforms. The trade off for this

performance loss is reduced system complexity.

-Fig (10.3) shows a differential encoding of binary message data

stream m(k), where k is the sample time index.

The basis for binary DPSK is as follows:-

a-The differential encoding starts with first bit of the code bit

sequence (Ck=0), chosen arbitrarily (here taken to be a one)

b-The sequence of encoded bits (Ck) can be encoded in one of two

ways:-

Ck = Ck-1m(k) ……………(10.8 a)

or Ck = Ck-1 m(k) …………(10.8 b)

where the symbol represent modulo2 addition (Exclusive–OR)

and the over bar denotes complement.

0 0 001 1 1 1 11Detectrd message

10-8

tT cos/2

Fig (10-3b) DPSK transmitters

Fig (10.3c ) DPSK receiver

-If (10.3a),the differentially encoded message was obtained by

using eq(10.8b)[C(k)= )()1( kmkC ].

The present code bit C(k) is one if the massage bit m (k) and the

prior coded bit C(k-1) are the same ,otherwise ,C(k)is zero .

C-Now, the coded bit C(k) translates into the phase shift sequence

(k),where one is characterized by a 180 phase shift and a zero is

characterized by a 0° phase shift.

a-During each system time we are matching a received symbol

with the prior symbol and looking for a correlation or

anticorrelation (180° out of phase) as shown in fi(10.3c)

The phase (k=1) is matched with (k=0), they have the same

value , hence the first bit of the decoded output is 1)1(

km

10-9

Then (k=2) is matched with (k=1) again have the same value

1)2(

km

Then (k=3) is matched with (k=2), they are different so that

0)3(

km , and so on.

2-Noncherent detection of FSK

-Fig(10-4) shows the in-phase (I) and quadrature (Q) channels

used to detect a binary FSK signal set noncoherently. The

noncoherent detection typically requires twice as many channel

branches as the coherent detection because the phase

measurements are not used during the detection process.

-Notice that the upper two channels used to detect the signal with

frequency 1, the reference signals are T/2 cos1t for the I

branch and T/2 sin1t for the Q branch.

Similarly, the lower two channels used to detect the signal with

frequency 2.

-Imaging that the received signal r(t), by chance alone ,is exactly

of the form cos1t+n(t), that is the phase exactly zero ,and thus

the signal component of the received signal exactly matches ,the

top branch reference signal with regard to frequency and phase.

In this case, the product of the top branch should yield the

maximum output. The second branch would yield a near zero

10-10

output because its reference signal T/2 sin 1t is orthogonal to

the signal component of r(t).

-In actual practice, the most likely scenario is that r(t) is of the

form cos(1t+ǿ)+n(t) that is, the incoming signal will partially

correlate with the cos1t reference and partially correlate with

sin1t reference. Hence a noncoherent quadrature receiver for

orthogonal signals requires an I and Q branch for each candidate

signal in the signaling set.

-In fig(10.4), the blocks following the product integrators perform

a squaring operation to prevent the appearance of any negative

values.

Fig (10-4) Quadrature receiver using energy detector

10-11

Another possible implementation for noncoherent FSK detection

uses bandpass filters centered at fi=i/2 with bandwidth f =1/T

followed by envelope detectors as shown in fig(10.5). An

envelope detector consists of a rectifier and a low pass filter. The

detectors matched to the signal envelopes .The phase of the

carrier is of no importance in defining the envelope.

-For binary case, the decision as to whether a one or a zero was

transmitted is made on the basis of which of two envelope

detectors has largest amplitude at the moment of measurement.

Similarly for a multiple frequency shift keying (MFSK) system,

the decision as to which of M signals was transmitted is made on

the basis of which of the M envelope detectors has the maximum

output.

Fig(10-5)

10-12

Comparison between envelope detector and energy detector

(quadrature receiver) for noncoherent detection FSK.

a-The envelope detector bock diagram looks more simple than

quadrature receiver

b-The analog filters which are used with envelope detector having

greater weight and cost.

c-Quadrature receiver can implemented digitally

d-Analog filters bank and detector can also be implemented

digitally by using FFT, but is more complex than quadrature.

Required tone spacing for noncoherent orthogonal FSK signal

Frequency shift keying (FSK)is usually implemented as

orthogonal signaling ,but not all FSK signaling is orthogonal

-Tones f1 and f2 are orthogonal if ,for transmitted tone at f1,the

sampled envelope of the receiver output filter tuned to f2 is zero

(i.e no crosstalk).A property that insures such orthogonality

between tones in an FSK signaling set states that any pair of tones

in the set must have a frequency separation that is a multiple of

1/T Hz

A tone with frequency fi that is switched on for a symbol duration

of T seconds can be described by

...........(10.9)

10-13

Fig (10-6)

The spectra of two such adjacent tones with frequency f1 for

tone1 and tone2 with frequency f2 are plotted in fig(10-.6)

from fig(10.6)

a-The minimum required spacing between (two tones)=1/T

b-The bandwidth of noncoherent detected orthogonal FSK=2/T

c-The bandwidth of coherent detected orthogonal FSK=1/T

d--The bandwidth of noncoherently detected orthogonal

MFSK=M/T

Ex(0.1)

Consider two waveforms cos(2f1t+) and Cos2f2t to be used for

FSK signaling ,where f1>f2

10-14

The symbol rate is equal to 1/T symbols/s, where T is the symbol

duration and is a constant arbitrary angle from 0to 2.

a-Prove that the minimum tone spacing for noncoherently

detected orthogonal FSK signaling is 1/T.

b. What is the minimum tone spacing for coherently detected

orthogonal FSK signaling?

Solution

a-For the two waveforms to be orthogonal, they must be

uncorrelated over symbol interval, that is

T

tStS0

21 0)()(

Using the basic trigonometric identities, we can write eq(10-10)

as

2

1

................(10-10)

10-15

if f1+f2 1

Combining eq(10.11) and (10-12), we can write

a-For noncoherent detection ≠0

For arbitrary,the terms of eq(10-13) can sum to zero

Only when sin2(f1-f2)T=0 and cos2(f1-f2)T=1

n =2k n=2 k

where n and k are integers ,then both sin0 and Cos=1 occur

simultaneously when n=2k

From eq(10-13)for arbitrary , we can therefore write

.........(10-11)

)1210.(..........1

1

Cos

Sin

.........(10-13)

10-16

Thus the minimum tone spacing for noncoherent FSK signaling

occure for k=1,in which case we write

(b)For coherent detection = 0 , we can now write eq(10-13) with

=0

Thus the minimum tone spacing for coherent FSK signal

signaling for n=1 as follows

Complex envelope

Any real bandpass waveform s(t) can be represented using

complex nation as

where g(t) is known as the complex envelope as

...........(10-14)

...........(10-15)

10-17

The magnitude of the complex envelope is then

and its phase is

With respect to eq(10-14), we can called g(t) the baseband

message or data in complex form, and tje the carrier wave in

complex for. The product of these two represents modulation, and

s(t), the real part of this product is the transmitted waveform,we

can express s(t) as follows :-

])(Re[)( tjetgtS

Quadrature implementation of a modulator

Consider an example of a baseband waveform g(t) ,described by a

sequence of ideal pulses x(t) and y(t) at discrete times

k=1,2,…..Thus g(t) , x(t) and y(t) in eq(10-15)can be written as

gk,xk and yk respectively.

Let the amplitude values xk = yk= 0.707A.

The complex envelope can then be expressed in discrete from as

...........(10-16 a)

...........(10-16 b)

...............(10-17)

gk=x

10-18

Fig(10-7)shows quadrature type modulator where we see that the

xk pulse is multiplied by cos◦t (inphase component of the carrier

wave), and the yk pulse is multiplied by sin◦t

(quadrature component of the carrier wave). The modulation

process can be described as multiplying .the complex envelope by

tje and then transmitting the real part of the product.

fig(10.7)

Communication Systems

Lecture 11

Differential 8-PSK (D8PSK) modulator

Fig (11.1) shows a quadrature implementation of a D8PSK

modulator. Because the modulation is 8-ary, we assign a 3bit

message ( kkk z,, ) to each phase k because the modulation is a

differential, at each Kth (transmission time we send a data phasor k

which can be expressed as

………..(11.1)

Fig (11.1)

1 0 1 1

1 1 1 1

1 1 1

0

0

0 0

+ + + +Binary

Binary

Gray code

-The process of adding the current message to phase assignment k

to the prior data phase 1k provides for the differential encoding

message.

-Gray code is used for quadrature implementation of a D8PSK

modulator, according the following steps:-

a-Input binary data is encoded using Gray code

b-Let the input Gray data sequence at times k=1,2,3,4 be equal to

110, 001, 110,010 respectively.

c-Table shown in fig(11.1) and eq(11.1), with starting phase at time

k=0 to be =0 are used to find the differential data phase

corresponding to

d-Taking the magnitude of the rotating ,phasor to the unity ,the in

phase (I) and the quadrature (Q) baseband pulses are -1 and 0

respectively .

e-The above pulses (from step d) are shaped with a filter such as a

root raised cosine

f-Repeat the above steps for each data symbol.

g-The transmitted waveform follows the form

……………(11.2a)

For a signaling set that can be represented on a phase –amplitude

plane ,such as (MPSK)Equation (11.2) shows that, the quadrature

implementation of the transmitted (D8PSK) transforms to simple

amplitude modulation .If we assume the data pulses have ideal

rectangular shapes (1.e without filter), for time k=2,

where and

D8PSKdemodulator

-Using a similar quadrature implementation demodulation consist of

reversing the process that is multiplying the received bandpass

waveform by tje in order to recover the baseband waveform.

Fig (11.2)

-Fig (11.2) shows the modulator and demodulator for quadrature

implementation of a D8PSK at time k=2

………..(11.2b)

-The inphase multiplication by coswot in the demodulator, we get

the signal at point A

After filtering with a low pass filter (LPF) we recover at point A an

ideal negative pulse as follows:

2

707.0A

Similarly, after the quadrature multiplication by t sin in the

demodulator, and after filtering with a low pass filter we recover at

point B an ideal negative pulse as follows

2

707.0B

-Thus, we see from the demodulated and filtered points A and B that

the differential (ideal) data pulses for I and Q channels are each

equal to -0.707

-Since the modulator/demodulator is differential, then these values

for I=Q

= -0.707 correspond the value k=2 Using eq(11.1)

122

11

kkk

kkk

eq(11.1)

If the demodulator at the earlier time k=1 had properly recovered the

signal phase to be .Using the table in fig(11.1)

!!!!!!!!!!!!!!!!!!!!

Refering back to the data encoding table in fig(11.1)we see that the

detected encoded data sequence is 222 zyx , 222 zyx =001 (Gray

code), corresponds the binary data 001

Error performance for binary systems

1-Probability of bit error for coherently detected BPSK

-An important measure of performance used for comparing digital

modulation types is the probability of error EP .The calculations for

obtaining EP can be viewed geometrically as shown in fig(11.3).

It is often convenient to specify system performance by the

probability of bit error (PB), even when decisions are made on the

basis of symbols.

-For coherent detected BPSK, the symbol error probability is the bit

error probability (because each symbol contain only one bit).

Fig (11.3)

-Assume that the signals are equally likely .Also assume that when

signal Si(t) [i=1,2] is transmitted, the received signal

r(t) = Si(t) + n (t) , where n(t) is an AWGN.

-The antipodal signals S1(t) and S2(t) can be characterized in a one

dimensional signal space as described previously

-The detector will choose the Si(t)with the largest correlate output

Zi(T)or in this case of equal energy antipodal signals, the detector

decides on the basis of

decide )(1 tS if )(TZ =0

…….(11.2)

decide )(2

tS otherwise

-Two types of errors can be made, as shown in fig(11.4):

a-The first type of error if signal S1(t)is transmitted but the detector

measures the negative value of Z(T) and chooses S2(t).

2a1

a

Fig (11.4) Conditional probability density functions p(Z/S1) , p(Z/S2)2

210

aa

b-The second type of error if signal S2(t) is transmitted but the

detector measures the positive value Z(T)and chooses S1(t).

-The probability of a bit error BP , as described previously, was

given by

…….(11.3)

Where 0 is the standard deviation of the noise out of the correlator.

The function Q(),called the complementary error function, is

defined as

For equal energy antipodal signaling, such as BPSK format in

equation (11.2),the receiver components are

sentistSwhenEaand

sentistSwhenEa

b

b

)(

)(

22

11

Where Eb is the signal energy per binary symbol For AGWN we can

replace the noise variance 02 out of the correlator with No/2, so that

we can rewrite eq(11.4) as follows :

NbE

duu

BP/2

2

)2

exp(2

1

(11.5)

)2

(

N

EQP B

B (11.6)

The result for bandpass antipodal BPSK signaling is the same as the

results that were developed earlier for the matched filter detection of

antipodal signaling and for the matched filter detection of baseband

antipodal signaling.

…….(11.4)

Ex11.1Find the bit error probability for a BPSK system with a bit

rate of 1 M bit/s .The received waveforms

are coherently detected with a matched filter, The value of A is 10m

V. Assume that the signal –sided noise power spectral density is

No= 1110 ..W/Hz and that signal power and energy per bit are

normalized relative to a 1 Ω load.

Solution:

Using eq (11.6)

Using table of complementary error function or the following eq

4

2

108

32

exp2

1

BP

forQ

Probability of bit error for coherently detected differentially

encoded binary BSK

If the carrier phase were reversed in a DPSK modulation

application, what would be the effect on the message? The only

effect would be an error in the bit during which inversion occurred

or the bit just after inversion, since the message information is

encoded in the similarity or difference between adjacent symbols.

-Sometimes messages (and their assigned waveforms) are

differentially encoded and coherently detected simply to avoid these

phase ambiguities. The probability of bit error for coherently

detected, differentially encoded PSK is given by

Fig(11.5) shows bit error probability for several types of binary

systems

……………(11.7)

Fig (11.5) shows bit error probability for several types of binary systems

Probability of bit error coherently detected binary orthogonal

FSK

-Equations (11.5) and (11.6) describes the probability of error for

coherent antipodal signals

NbE

duu

BP/2

2

)2

exp(2

1

(11.5)

)2

(

N

EQP B

B (11.6)

-Amore general treatment for binary coherent signals (not limited to

antipodal signals) yields the following equations for PB

(11.7)

Where =cos is the time cross correlation coefficient between two

signals S1(t) and S2(t),where is the angle between signals vectors

S1(t) and S2(t)

{For example, the antipodal signals, such as BPSK θ= , =-1

»For orthogonal FSK, such as binary FSK (BFSK). θ =/2

Since the S1and S2 vectors are perpendicular to each other, thus ρ=0,

and eq(11.7) can then be written as

-The result eq(11.8)for the coherent detection of orthogonal BFSK is

the same as the result that were developed earlier for the matched

filter detection of orthogonal signaling ,in general ,and for the

matched filter detection of baseband orthogonal signaling (unipolar

pulses)

……(11.8)

-On-OFF keying (OOK) is an orthogonal signaling set (unipolar

pulse signaling is the baseband equivalent OOK). Thus

eq(11.8) 0/ NEQP bB applies to the matched filter detection of

OOK, as it does to any coherent detection of orthogonal signaling

-If we compare eq(11.8) with eq(11.7), we can see that 3dB more

Eb/N0 is required for BFSK to provide the same performance as

BPSK.

BPSKforN

EQP

BFSKforN

EQP

bB

bB

)7.11........(2

)8.11.........(

0

0

Thus the performance of BFSK signaling is 3dB worse than BPSK

signaling ,since for a given signal power bE ,the distance squared

between orthogonal vectors is a factor of two less than the distance

squared between antipodal as shown in fig(11.6)

2(t) S2 2(t) S2 S1 1(t

bE2 bE

0 bE

bE

1(1) bE2

S1 antipodal

bE (b) orthogonal (a)Binary signals vectors

Fig (11.6)