Communication Systems 
Lecture
1
1-1
Why digital communication
1-Ease generation of digital signals if compared with analog ignals
generation as shown in fig(1.1)
!Distance 1
Original pulse
Distance 2
Distance 3
Distance 4
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
1-2
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
1-3
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.
1-4
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
1-5
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
1-6
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.
1-7
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 .
1-8
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
10
binary symbol (K=1,M=2)
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 .
1-9
Fig (1-2) Typical block diagram of a digital Comm .System
1-10
Communication Systems 
Lecture
2
2-1
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 χ ( t)  5 cos10t . For random
waveform it is not possible to write such clear mathematical
expressions .
2.Periodic and Nonperiodic Signals
A signal χ ( t) 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 χ ( t) .The period To defines
the duration of one complete cycle of χ ( t)
A signal for which there is no value of To that satisfies Equattion
(2.1) is called nonperodic signal.
2-2
3-Analog and Discrete Signals
An analog signal χ ( t) is a continuous function of time, that is
Fig (2.1)
χ ( t) 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) .
2-3
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)
g(t)
g (t)
Fig (2-1)
Analog continuous time
Digital continuous time
g (t)
g (t) Digital discrete time
t
Analog- continuous time
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 (t )
 (i) 2 R ...............(2.2)
P (t ) 
R
2-4
»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) =  2 (t ) .................(2.3)
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
T /2
Ex 

 2 (t ) d t ............................(2.4)
T / 2
and the everage power dissipated by the signal during the interval
is
T /2
1
1
P  E T 
 2 (t ) d t..............(2.5)
T
T T /2

»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 .
2-5
»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

0


nonzero but finite energy
E x  limT 

T /2
 2 (t ) d t

...........(2.6 a)
T / 2

E 

 E    for all time where
x

 2 (t ) d t ...........................(2.6 b)

» 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 (o  P   ) . For all time
x
P  limT  
1
T
T /2

 2 (t ) d t ............(2.7)
T / 2
2-6
»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 :-
where

 (t ) d t  1 .......................(2.8)
 (t ) 0 
 for t  0 ....................(2.9)
 (t )   
 for t 0 .....................(2.10)

 x(t )  (t  to ) d t  x (to ) ..............(2.11)

»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.
2-7
- (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
2
E   x (t) d t   X (f) d f.............. (2.12)
x



where X(f) is the Fourier transform of the nonperiodic signal
x (t ) . Let  ( f ) denote the squared magnitude spectrum ,
x
defined as .
ψ χ  (f)  | X(f) |
2
Joules / Hz..........(2.13)
2-8
The quantity   ( f ) is the waveform energy spectral density
(ESD ) of the signal ( t) .


The total energy of  (t) E x   ( f ) d f

x
.......(2.14)
»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
 (t ) , X ( f ) is an even function of a frequency
 Ex  2

 
( f ) d t .............(2.15)
0
2.Power Spectral Density
(PSD)
»The everage power P of a real valued power signal  (t) is
defined in Eq (2-7)
P  limT 
1
T

T /2
x 2 (t ) d t .......(2.7)
T / 2
»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
1
P  lim
x
T  T
T0 / 2 2
 (t ) d t .......(2.16)
T0 / 2

2-9
» Parseval`s theorem for a real valued periodic signal takes the
form
1
T
2

2
2
P 
 d t   C n .......(2.17)
x T  T
n  

2
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

2
G  (f)   c n δ(f  n f 0 )............(2.13)
n  
Thus the PSD of a periodic signal is a discrete function of
frequency. The average normalized power of a real valued signal
defined as
Pχ 


  G χ (f) d f  2 0
G  (f) d t ........(2.14)
»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 .
2-10
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
1
Gx ( f )  lim
X T ( f ) 2...............(2.15)
T  T
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)
P

1
T


2
(t ) dt
(2.17)
Using eq (2.13) and (2.14)

2
G  (f)   C n δ(f  n f 0 )
n  
C 1  C 1 
A
2
2-11
Cn  0
for n=0,±2, ±3,…….
2
2
A
A
G  ( f )  ( )  ( f  f )  ( )  ( f  f )
2
2
2
2
A
A


f

[
(
)
f
(
)
(
)  ( f  f )]df


2
2
P  
2
2
2

A
A
A
( )  ( ) 
2
2
2
Note Fourier series

 (t )   C n e  jn2π f 0 t
n  
where C n 
T0 /2
1
T
0 T

 (t) e
 j2π f t
dt
0 /2
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 ( ) 



χ (t) χ(t  ) dt
for   t ............(2.16)
─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.
2-12
─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 ( )  R  (τ)  symmetrical in τ about Zero
2. R (τ)  R (0) for all τ  maximum value occurs at the origin

χ
3.R

(τ)  ψ

(f)  auto correlation and energy spectral density
form a Fourier transform pair
4. R x (0) 

 χ(t) d t
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
R x    limT 
1
T

To /2
 (t)  (t  ) d t for   τ   ....(2.17)
 To /2
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
2-13
T /2
o
1
R ( ) 
x(t)  (t  τ) d t for   τ   ......(2.18)


T
0 T /2
o
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
2-14
F (χ)  P (X  χ ) ..............(2.19)
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  0  F  χ   1
2  F χ1  F χ 2  if χ1  χ 2
3  F  0
4  F ()  1
F ()

1
2
3
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
2-15
P  1     2   P X   2   P X   2   P X   
 F x  2   F x (  2 )
2
 x  d  ........( 2 . 20 )


1
The probability density function has the following probabilities:-
1) Px (  )  0
2) Px (  ) d   Fx ()  Fx()  1
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
m   E(  ) 
where E (



χP (  ) d  ......(2.21a)
) is called the expected value operator
-The mean square value of X as follows :
E ( 2 ) 



2
χ p (  ) d  ............(2.21b)
The variance of X is defined as
var (X)  δ x 2  E(x  m ) 2 
x

 (χ  m x ) 2 p  (x) d  .......(2 .22)
2-16
where (X-mx ) is the difference between x and mx
And  x 2  E  x  m

2
x 
 E  x 2  2m X  m 2 
x
x 

 E ( X 2 )  2 m E ( X )  m 2x
x


 E  X 2   2 m .mx  m 2x
x




 x 2  E  X 2   m 2x ................(2.22)


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.
2-17
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)
G  (f)  LimT  
2
1
/ XT (f ) / ..............(2.23)
T
where X (f) is the Fourier transform of (t )
2-18
-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
2-19
Fig (2.5)
2-20
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 Gaussian
noise random variable and dc signal ,that is Z=a+n
2-21
Fig (2.6)
P ( τ) 
1
e
 2π







1  Z  a  2 
 
 
2    

................(2.25)
-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)
2-22
Gn (f) = N 0 (2.26)
2
G n(f)
N0/2
R(t)
N0/2
F
0
t
0
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)
Linear system
(t)
X (f)
h(t)
H(f)
output
y (t)
Y(f)
Fig (2.8)
2-23
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)
Linear time
invariant system
Y(t)
fig (2.9)

Y (t)   (t) * h (t)    (τ) h (t  τ ) d τ ..........(22.7a)

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

Y (t)    (τ) h (t  τ ) d τ ..............(227b)
0
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)
Y (f)  X (f) H (f) ...........(2.28)
2-24
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:
G y ( f )  G x (f) H ( f ) 2............(2.30)
Distortionless Transmission
»The output of ideal distortionless transmission can be described
as y (t)  K χ (t  t o ) ............. (2.31)
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)
Y( f )  KX (f) e j 2
f t 0 .................(2.32)
2-25
Substituting the eq (2.32) for y (f)in eq(2.28)
H( f ) K e
 j2π ft 
.................(2.33a)
Y(f)=X(f) H(f)…….(2.28)
The time delay to is related to the phase shift () and the radian
frequency  = 2 π f by
( ) ( radian)
...............(2.33b)
t 
0 2 π f (radian / sec )
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
τ
1 d
...............(2.34)
2π df
-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.
2-26
Fig(2.10)
The range fL  f  fu is called passband where fL = lower cutoff
frequency and fu= upper cutoff frequency .
the filter is called bandpass filter (BPF)  0 and fu  1-When fL
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 H(f ) remains within a
specified value .
2-27
- For ideal low pass filter , the transfer function can be defined as
H( f ) K e
 j2π f t
0 .................(2.34)
where K = 1 with bandwidth H(f)=|H(f)|e-j(f)`:. Wf = fu Hz
 1....................for

H( f ) 
 0................... for

and e
 j ( f )
e
Hf
u
f f
u
 j 2ft0
The impulse response of the ideal low pass filter is show in fig
(2.11)
……….(2.35)
Thus the impulse response in fig(2.11) is noncausal. Therefore the
ideal filter is not realizable (which is described in (2.3)
2-28
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
R y (τ)  N 0f0
sin 2 π fu τ
2 π fu τ
 N 0fusinc 2 π f uτ
2-29
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
H ( f )
1
1
e  j  ( f ) ......(2.36)

1 j 2fRC
1 (2  f R C ) 2
Where Ø (f)= tan -12π f RC
W=
1
RC
rad / sec , Wf = 1/2 RC
Fig(2.12)
-The magnitude H(f ) and the phase (f)are polted in fig
2-30
(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 :-
=
N
 e
4RC

//
RC
Since
F e  a t  
2a
a 2  (2 f ) 2
if a  0
2-31
The bandwidth description
‫ــ‬Many important theorems of communication and information
theory are based on the assumption of strictly bandlimited , which
means that no signal power is allowed outside the defined band .
‫ــ‬All bandwidth criteria have in common the attempt to specify a
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  c (t ) with fc is the carrier wave frequency and T is the
pulse duration .
Fig(2.13)
a-Half power bandwidth
2-32
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.
2-33
Communication Systems 
Lecture
3
3-1
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 .
3-2
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
3-3
-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
3-4
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
1
Τ 
............(3.1 )
s
2 fm
This statement is Known as the uniform sampling theorem .
Where Ts = sampling time (which is the upper limit )
fs = sampling frequency = 1/ Ts
f s  2 f m ............(3.2 )
3-5
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)
………….(3.3)
where Ts is the sampling period and  (t) is the unit impulse .
The
............(2.4)
3-6
Fig (3.3)
3) Using the frequency convolution property of the fourier
transform
F  (t ) X  (t ) X ( f ) * X  ( f )
1
where Xs(f) =
Ts

 ( f  n f s).............(3.5)

n 
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
X (f) * δ (f  n f s )  X (f  n f s )
3-7
We can now solve for the transform X
s
(f) of the sampled
waveform
X (f)  X (f) * X
s

 1



(f)  X (f) * 
 δ(f n f s ) 
δ
 T s n   

1 
 X (f n fs ) ..............(3.6)
T n  
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.
3-8
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  p (t ) shown
in fig(3.5c) Each pulse in  p (t ) 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  s (t ) can be
expressed as
 s (t ) =  p (t )  (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
3-9
corresponding analog segment during the pulse interval .The
periodic pulse train can be expressed by a Fourier series
Fig (3.5 )
3-10

 p (t )   Cn e
j 2f s t
..........................(3.8)
n  
where f s  1
Ts
 2 fm
1
 nT 
Cn  sin c  s 
Ts
 T 
T  pulse width
Combining eq (3.7) and eq (3.8)
 s (t )   (t )


n  
Cn e
j 2  n f st
...........(3.9)
The transform Xs (f) of the sampled waveform is
Xs (f) = F [s (t) ]
…………(3.10)
For linear systems ,we can interchange the operations of
summation and Fourriers transformation .
……….(3.11)
Using the Fourier translation property of the Fourier transform:-
:. Xs ( f ) 

 Cn
n  
X ( f  nf ) ..........(3.12)
s
3-11
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 :1 
X (f) 
 X ( f  n f s ) .............(3.6)
s1
n
Ts  

X s2 ( f )   Cn X ( f  n f s ) ..............(3.12)
n
As the pulse width T0 and the pulse amplitude  (the erea
of the pulse remains unity), p (t)
3-12
δ (t)
1 T /2
 (t ) e  j 2  f s t d t
:. Cn lim 
p
T 0 T
T /2
s
s
s
Ts / 2
1
1 T /2
1
 j 2 n fs t
 j 2nf s t
(
)
(
)
Cn 
t
e
d
t
t
e
dt





Ts T
Ts  Ts / 2
Ts
s
s/2
Substitute for Cn in eq (3.12)
:. X
1
s 2(f)
Ts
In the limit T →0
1
X s2 ( f ) 
Ts

 ( f n f s )  X s ( f ) .

n 
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.
3-13
Fig (3.6)
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
3-14
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
3-15
.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.
3-16
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
3-17
Communication Systems 
Lecture
4
4-1
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)
4-2
-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 ).
.(4.1)
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 :
V
2
V p   PP
 2

2




Lq

 

 2 


L2 q 2
...................(4.2)
4
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
4-3





S 
L2 q 2 / 4
 3L2 ...................(4.3)
 
2
N 
q q / 12
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
  bit (  log 2 L) 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
4-4
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
4-5
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.
4-6
-Fig (4.4) shows the nonuniform quantizer characteristics
Fig (4.4)
Nononiform quantizer output
Input
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)
4-7
Fig(4.6)
Uniform quantizer
!!!!!!!!
-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 .
4-8
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 .
4-9
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)
4-10
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.
4-11
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) .
4-12
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 :-
4-13
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
4-14
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 .
W T (normalized bandwidth)
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
4-15
-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
L  2
...............(4.4)
4-16
-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
| e |  P V
PP
Since the maximum value for quantization error =q/2
where q is the quantile interval
/ e /
q Vpp Vpp
if L 1
 

max 2 2( L 1) 2L
Vpp
 p Vpp

2L
1
or 2  L  levels .............(4.5)
2p
where   no of bit
:.   log 2
1
(4.6)
2p
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.
4-17
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)
  log
1
 log
2p
2
1
 log 50  5.6
2
0.02
:.  = 6 bit / sample to meet the required distortion
b) Using the Nyquist sampling theorem
f  2 f  2  3000  6000 Hz  6000 sample / sec
s
m
Each sample of the PCM composed of 6 bits
:. Bit transmission rate =  fs = 6 ! ! 6000 = 3 6 k bit / sec
c) Bandwidth efficiency =
R
W
where R = Bit transmission rate
W = Bandwidth
4-18
:. Bandwidth efficiency =
36000
 3 bit / sec
12000
Convolution
The convdution of two function g (t) and w(t) dented by
g ( t)*w(t) is defined by the integral
g (t ) * w(t ) 


g ( ) w (t  ) d  ........................(4.7)

The time convolution property and its duals, the frequency
convolution property , state that if :-
g1 (t )  G1 (w) and g 2 (t )  G2 (w)
Then (time convolution )
g1 (t) * g2(t)  G1 (w) . G2(w) …………..(4.8)
and (frequency convolution )
g1 (t) . g2 (t) 
1
G1 (w) * G2 (w) …………(4.9)
2
4-19
Proof:


 j wt 
F g1(t ) * g 2 (t ) e
 g1( ) g 2 (t  )d  d t
 


 j wt 


d t d 
g1( ) g 2 (t  )


 











g1 ( ) e  j w G2 (w) d 


 G2 (w)

g1 ( ) e  j w d  G2 (w) G1 (w)

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).
4-20
Communication Systems 
Lecture
5
5-1
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.
5-2
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…..
5-3
(5.1)
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:-
eq [n]
mq [n 1]

Z 1
mq[n]
Fig(5.2)

Z 1
Fig(5.3)
5-4
-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)
5-5
»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
5-6
3-Design of the receiver is simpler.
A delta a modulation with integration at its input is called deltasigma 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 .
pulse generator
comparator
Integrator
Hard Limiter
Low pass filter
 dt
L.P.F
+ 
X
m(t)
message input
Pulse modulator
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.
5-7
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
χ  [n] 
P
 Wk χ [nK]..............(5.3)
K 1
Where p, the number of unit delay elements ,is called the
predication order .
 (n)
Z 1
 (n 1)
Z 1
 (n  2)
 (n  p  1)
Z 1
 (n p)
w1
w2
w p 1
wp



Fig (5.6)
5-8

[n]
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
Input +
m [n]
e[n]

-
Quan
-tizer
eq[n]
Encoder
+
+
^
m [n]
Predictor
DPCM
Signal

m [n]
q
Fig (4.7a) DPCM transmitter
5-9
Input Decoder
+

output
»
predictor
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]
5-10
-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.
5-11
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
5-12
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
  leg 2 21P  log 2
1
1
 log 2
 log 2 50  5.6
20.01
0.02
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
5-13
Communication Systems 
Lecture
6
6-1
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.
6-2
-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.
Note
Frequency down conversion and
Equalizing filter are optional units
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
6-3
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
 S1
Si(t)= 
S (t)
 2
0  t  T for binary
1
0  t  T for binary
0
-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) mi (also called
hard decisions)
-If error correction coding is used, the detector output consists of
estimates of channel symbols (or coded bits) ui

-The frequency down conversion block, performs frequency
translation for bandpass signals operating at some radio
frequency (R F)
6-4
-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)
6-5
-If the probability density function (Pdf)of the Gaussian random
noise n0 can be expressed as
2



no
1

1
P (n 0 ) 
exp   .........(6.5)
 2 0  
 0 2


Where 02is the noise variance .
The conditional probability density functions
Z
Z
and
P

  can be expressed as
 S1 
 S2 
!!!!!!!!! P
 1  Z a 2 
Z
1
1 
e 
P 
 .........(6.6)

2
S
 1   2   0  
2



1 Z  a2
Z 
1
 ..........(6.7)
e 
P  
 S 2   2  2   0  
a2

a1
Fig(6.2)
6-6
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
 H1  ............ (6.8)
Z(T)
H
2
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.
6-7
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 
N
where N=noise power ,W=bandwidth since the bit time
W
and bit rate Rb are reciprocal
:. Tb=1/Rb

Eb S Tb S / Rb  S  W


   ..........(6.9)
N 0 N / W N / W  N  Rb
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)

Eb  S  W
   ..........(6.10)
N0  N  R
Eq(6.10) represent that Eb/No is just version of S/N normalized
by bandwidth and bit rate
6-8
-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)
PB
Eb/N0
fig(6.3)
-Eb/No is a dimensionless ratio as shown below
E
Joule
wattS


N0 watt/ Hz watt/1/ S
watts

watts
-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
H1

Z (T)  .................(6.8)
H
2
6-9
-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).
H
p( z s1)  1 p(s2 )
p( z s2 )  H p(s1)
(6.11)
2
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
P (n ) 
0 
1
0
 1  n 
exp  02 


2
 2   0 
6-10
The likelihood ratio test described in eq (6.11)can be written as

1
Z  a1 2  H 1
 2
 
P Z / S1   0 2 
P(S 2 )

A( Z )  
2


P S1 
 1  Z  a 2  
1
P  Z / S 1 
H
2
exp   
 







2
 0 2
0  


1

exp  


2


 Z2 
 2 Za1 
a
1
 exp  

 exp( 


) exp 
2 
2
2  H1


2 0
 2 0 
 2 0   P ( S 2 ) 



2 
P
(
S
)







Z
a
2
Za
2
1
H
 exp  2  exp 

2 
2
 exp 
2
2
2
 2 
 2 
 2 


 0 
 0 
 0 


H
 Z a1  a2  a1 2  a 2 2   1 P (S 2 )
A(Z)= exp 
..........(6.12)

2
2 

P(S1)
2 0 
  0
H
2
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 .
 Z a1  a2  a1 2  a 2 2   1
P (S 2 )

In A(Z ) 
In
2
2 

P(S1 )
2


0
0

 H 2
H
Since P (S2) = P (S1) are equally likely :. In
6-11
PS 2 
0
PS1 
Z a1  a2  a 21  a 2 2


2
2 0 2

0
H1
2

a1 2  a 2 a1  a2
Z

 0 ........(6.12)
2
 H 2(a1  a2 )
2
or
Z (T )


H
1
H2
a1  a2
 0
2
p( z s2 )
p( z s1)
a2

a1
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.
6-12
-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
0
p(e/S1)=p(H2/S1)=


Z
P d z..........(6.13)
S 
 1
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) =


0
PZ / S 2 d z..........(6.14)
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
6-13
2
2
i 1
i 1
!!!!!!!!!!!!!PB=  P(e,Si)=  P(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) = 1 ]
2
:.PB= 1 P(H2/S1)+ 1 P(H1/S2)……..…(6.17)
2
2
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 .
 PZ / S

:.PB=
2
dZ .................(6.19)

Replacing the likelihood P(Z/S2)with Gaussian equivalent from
eq(8.7)

PB =


1
0
  Z a
2
1
exp 
 2  0
2
 


dZ

Where 20is the variance of the noise output of the correlator
6-14
 0  a1  a2 / 2
Let
Z  a2
, then  0 du  dz
0
U
u 
PB =

u  (a  a ) / 2
1
2
0
 u 2 
1
du
exp 
 2 
2


 a1  a2 
.................(6.20)
2

0 

PB = Q 
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
Q ( )  1
2


 u2 
exp   d u
 2 


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   3 is
Q ( ) 
1
e
 2

2
Z ............(6.21)
6-15
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)
Si (t)
AWGN

Receiving
filter
Sample
at t=T
Z (T)
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
:.
The instantaneous signal power  S 
  T
average noise power
N
 S  ai 2
   2 ......................(6.22)
 N  0
we wish to find the filter transfer function H0(f) that
6-16
S

N
 
maximizes eq (6.22) i.e maximizes 

ai (t )   H ( f ) S ( f ) e
j 2 f t
T
df

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 =
GY ( f )  Gx ( f ) | H ( f ) |
2
N0
watt / Hz
2
where G Y ( f )  0 / p power spatral density
G x ( f )  I / p power spatral density

output noise power 



 02 


 02 
G ( f ) d f  0 2
Y
G ( f ) H ( f ) 2d f
x
N0 
2
 / H ( f ) / df
2 
S
  
 N T



2
H ( f ) S ( f ) e j 2f T df

N0 / 2 

H( f )
2
............(6.23)
df
Using Schwarz`s inequality
*The equality holds if and only of f1(x)=kf*2(x)
6-17
where k is an arbitrary constant and * indicates complex
conjugate

 F1 ( x) F2 ( x)dx


2

 | f ( x)
1
|

2
dx

 |F2 (x)|dx

Let f1( x)  H ( f ) , f 2 ( x)  S ( f ) e j 2 Tf


H( f )e
j 2 fT





| H ( f ) |2 d f .


| S ( f ) |2 d f

Substituting into eq(6.23) yields

S
 
N


2
| H ( f )| df
 
T


|S ( f )|2 df


N 0 / 2 | H ( f )|2 df



2
|S ( f )|2 df
N0 

2E
S 

................(6.14)


T
N
N
0

or max 
where the energy of the input signal s(t)is

E=

/ S ( f ) / 2 df ...................(6.25)

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.
6-18
Eq (t.24)



max




S  2E 
   holds only if the
N T N 0 
Optimum filter transfer function H0(F) is employed, such that
H ( f )  H 0 ( f )  K S *( f ) e  j 2f T ............(6.26)
on
h (t) = F -1 H0 (f) = F-1 K S * (f) e-j2fT
 K s (T t )...............0  t  T

h (t )  
 0  ..................else where


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)
6-19
Communication Systems 
Lecture
7
7-1
Correlation Realization of the Matched filter
The matched flier transfer function is given by
H 0 ( f )  K S * ( f ) e  j w T F ...................(7.1)
S(f)=is the Fourier transform of the input signal
where T=symbol duration
 K S (T  t )...................0  t T
and h (t)  
........(7.2)
elsewhere
0
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

Z (t)  r (t) * h(t)  r (τ) h(t  τ) dτ ...........(7.3)
0
where r(t)= S(t)+n(t)
Substituting h(t) of Eq(7.2) into eq(7.3)and setting k=1
t
Z (t ) 
 r ( ) S T (t  )d 
0
t
Z (t ) 
 r ( ) s [T t  ]d 
0
7-2
Fig(7.1a)
r ()

T

Z(t)
0
S()
Fig (7.1 b )
When t=T
T
Z (T )   r ( )s( )d
(7.4)
0
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 .
7-3
Optimizing Error Performance
-For the binary case, the optimum decision threshold was shown
0 
a1  a2
.......….. where a1 is the signal component of Z(t)
2 0
when s1(t) is transmitted and an is the signal component of
Z(t)whenS2(t)is transmitted.
s1
s2
a2
a1
Fig(7.2)
-For minimizing the probability of error (PB), it is necessary to
choose the filter (matched filter) that maximizes the value of
 a1  a 2
 2 0
Q 



.Thus we need to determine the linear filter that
maximizes a1  a2  / 2 0 or equivalently that maximizes
(a1  a2 ) 2 2 2
7-4
Note

P 
B

0
 u2 
1
exp   d u
 2 
2


 a a 
 Q 1 2 
 2  0 


u  ( z  a2 ) 2
for Z  0
Z  a1  a2 / 2
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






S 
a1  a2 2 2 Ed
 2 .............(7.5)
 
2

N


T
where No/2 is the sided power spectral density of the noise at the
filter input
T
Ed 
 [S (t)  S (t)]dt..........(7.6)
1
2
0
7-5
Substituting Eq(7.6) in Eq(6.20)

PB 

0
 u2 
 u1 u2 
1


.......(6.20)
exp 
du Q
 2 
2  
2



 Ed 

..........(7.7)
 PB  Q
 2N 
0

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)
 
1 T
 S1 (t )S 2 (t ) d t .........(7.8)
Eb 0
T
2
From eq(7.6) Ed   [S1 (t )  S 2 (t )] dt
0
but
7-6
T
Ed  Eb  Eb  2
 S (t) S (t)d t........(7.9)
1
2
0
Substiating eq (7.8) into Eq(7.9),we obtain
Ed  2 Eb  2Eb

1 T
 S1 (t )S 2 (t )d t......(7.8)
Eb 0
Ed  2 E p 1  .............(7.10)
Substituting eq (7.10) into eq(7.7),we obtain
pB  Q
E p (1  )
N0
..............(7.11) PB  Q
2 Ed
......(7.7) !!
N0
!!!!!!!!!!!!!!!!!!!!!!
If =-1for the case S1 and S2 are antipodal signals
:. p B  Q
2Eb
..............(7.12)
N0
if =0 for the case of orthogonal signals
 Q
if  =1
Eb
N0
.......... .........( 7 . 13 )
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)
7-7
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 :
3
Eb 
2

 (t) d t
0
2
1


0

12 d t 


   
3
2 d t  12 d t  12 t  2 2 t  1t 
2
1
2


2
1
2
3
0
1
2
   
2
2
 110  3 10  6  210  3  10  6  10  3 10  6
 610 12 Joule
7-8
Since the waveforms are antipodal (=1) and detected with a
matched filter:
 2610 12
2Eb 
  Q
12

N 0 
10


PB  Q (3.46)


p B  Q



  Q 12


 
From the table of complementary error function
PB 0.0003310  4
Or, since the argument of Q() is greater than 3 we can also use
the approximate relationship in Eq(6.21).
 2 
1
PB 
exp   ..........(6.21)
 2 
 2


  (3.46) 2 
1
4

exp 
  2.89 10
2
3.46 2


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)
7-9
!!!!!!!!!!!!
Fig (7.5)
The energy difference signal, from eq(7.9) is given by
T
Ed 

0

 S1(t )
2
 S (t ) d t  A2 T
2 
.
Then the bit error performance at the output is obtained as follows


p B  Q


 A2T
Ed 

 Q
 2 N0
2 N 0 


 Q  E b 
  N 0 

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
7-10
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
7-11


P  Q
B 


2 E b(1  )
2Eb
Ed 
Q
Q
2 N 
N
N0
0
0
  1
for antipodal
-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.
7-12
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).
7-13
-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.
7-14
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
1 Rs

W
2T 2
here z  Nyquist bandwidth
»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.
7-15
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
….(7.15)
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
r
W W
W
0  r  1..............(7.16)
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.
7-16
-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
h(t )  2 W0 (Sinc 2 W0t )
cos 2  W W0 t 
..........(7.16)
2
1 4 W W0 t 
fig(7.10)
7-17
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=
1
(1+r) Rs
2
(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=
K
2
since M =4

Symbol or pulse rate R =
K=2
R 2400

1200 symbol / s
K
2
1
2
1
2
Minimum bandwidth W  1 r R = (1+1)1200=1200 H z
7-18
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:
K
M= 2 since M = 4 level :. K=2
R 2400
R  
 1200 symbol / sec .
S K
2
WDSB=(1+r)RS = (1+1) (1200) = 2400 HZ
7-19
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
H c ( f )  | H c ( f )| e
jc ( f )
...............(7.18)
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
7-20
»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).
7-21
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.
7-22
He(f)=
1
1

e  j c ( f ) ...............(7.16b)
H c ( f ) | H c ( f )|
7-23
Communication

Lecture
8
8-1
Systems
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.
‫ــ‬The distortion can be viewed as positive or negative echoes
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
8-2
…..….(8-1)
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
C N ,C N 1.........,C N
are
(2N+1)
taps
with
weights
Output samples Z(K) of the equalizer
are then found by convolving the input samples [ (k)]and tap
weights [Cn] as follows:- Z k  
N
 χ k n Cn..........(8.2)
nN
8-3




where k = -N, ……N
n = - N , ….. N
Where k=01 ,  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
Z k  
1 for k 0
0
for k  1,  2 , ... ,  N ..... 8.3
Fig (8.2)
where k = - N, ……….. N
n = - N, …….…. N
or-
8-4
0  X (0)
  
  
0  X (1)
  
: 
:
  
: 
:
  
  
:
1  
  
: 
:
  
: 
:
  
  
:
0 
  
0  X (2 N )
  
X (1)
X (0)
:
:
:
:
:
:
.
.
.
.
.
.
.
.
.
.
2 N 1
X (2 N )   C N 
X (2 N 1) C N 1 




:





:




:
  C0 
:
:
:
X (0)






  C N 1

 CN











(8.4)
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).
8-5
Solution:
………..(8.3)
2
fig(8.4a) received pulse
1-Design three tap equalizer (N=1) to find C , C 0 , C1
1
Z ( -1) =  (0) C1 +  (-1) C0 +  (-2) C1
Z (0) =  (1) C + (0) C0 +  (-1) C1
1
Z (1) =  (2) C +  (1) C0 +  (0) C1
1
Z( 1)  χ (0)
Z (0)   χ(1)
 

Z(1)  χ (2)
1
 0
1    0.2

 
0.1
1 
χ ( 1) χ( 2)  C 1 
χ (0) χ ( 1) C 0 
χ (1)
χ (0)  C 1 
0.1
1.0
 0.2
0  C 1 
0.1 C 0 
1 C1 
8-6
0 0.1
0
1 1
0.1
0  0.2 1.0
C 1 
C 1 
1
.1.0 0
 0.2 1.0 0.1
0.1  0.2 1
0 1 1.0  0.1 0.2  10.1 1.0  0 0.2  00.1 0.1  0  1


1 1.0  1  0.1 0.2  0.1 0.21  0.1 0.1  0   0.2    0.2  1 0.1
C 1   0.09606
C0 
C1 
1
0
 0.2 1
0.1 0
1
 0.2
0.1
0
0.1
1
0.1
0
1.0
0.1
 0.2 1
1
0.1
 0.2 1.0
0.1  0.2
1
0.1
 0.2 1.0
0.1
 0.2
 0.9606
0
1
0
0
0.1
1
Solution of this yields
 0.2017
C=
1
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)
8-7
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
8-8
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
8-9
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)
8-10
Waveform
Eye diagram
Tb
(a)
Tb
t
opening
(b)
width
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.
8-11
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
1
1 

L  
4
10 


1
1 

  
4
10 

C
....................(8.5)
f
where C= speed of the light=3*10 8 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
8-12
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 :-
8-13
DPSK, FSK, ASK, CPM and hybrids
Phase shift keying (PSK)
The general analytic expression for PSK is
……(8-8)
where the phase term i(t)will have M discrete value, typically
given by
2 i
 (t ) 
i
M
i  1.2............(8.9)
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
8-14
Fig (8.7)
Frequency shift keying (FSK)
The general analytical expression for FSK modulation
……..(8.10)
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.
‫ـ‬There is no requirement for the phase to be continuous for the
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…)
8-15
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
……(8.11)
Where the amplitude term
2Ei t  T 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 2E T and zero .The ASK vector corresponding to the
maximum amplitude state and a point at the origin corresponding
8-16
to the zero amplitude state. Binary ASK signaling also called (onoff 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°.
8-17
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).
8-18
fig ( 8.11) Modulator block diagram
Ex8.1
Derive the waveform amplitude coefficient 2E / T
Solution
let us consider S(t)=A Coswt
where A is the peak value of the waveform
8-19
S(t)=
2 ArmsCoswt= 2 A2 rms coswt
Assuming the signal to be a voltage or current waveform. A2rms
represent average power P normalized
to 1 Ω S  2 p Cos wt But power p 
S  2E / T Cos wt
8-20
Energy
T
Communication Systems 
Lecture
9
9-1
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.
9-2
-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.
9-3
-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
 j t  t dt  K j  Jk 0  t T ...............(9.2a)
k
0
j, k  1,...., N
T

Where the operator
 JK  
1
0
for j  k
for j  k
........................(9.2b)
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.
9-4
-Fig(9.2) shows an example with N=3
am2
am3
am1
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.
T
2
E    t  dt  K j .........................(9.2c)
j
j

0
-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)
9-5
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
S (t)  a11 ψ 1  t   a12 ψ 2  t   .......... .......... ....  a1N ψ N (t)
1
S 2 (t)  a 21 ψ 1 (t)  a 22 ψ 2 (t)  .......... .......... .  a 2N ψ (t)
N
S M (t)  a M 1ψ 1 (t)  a M 2 ψ 2 (t)  .......... .....  a MN ψ N (t)
 S i (t) 
N
 aij ψ j(t)
i 1
i  1.2.3..... ...M...... (9.3a)
N M
T
1
ai 
S (t )  j (t )dt
Kj 0 i
j

where
(9.3b)
i  1.2.3......, M
j  1........, N
0t T
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) +am3 3(t)
Thus each of the transmitted signal waveform Si(t)is completely
determined by the vector of its coefficients (am1,am2,….amN)
9-6
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.
9-7
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).
9-8

 1 (t)
1(t)
S1(t)
1
T/2
T
t
T/2
T
t
t
T/2
T
-1
-2
-3
 2  (t )
2(t)
S2(t)
2
1
1
1
t
T/2
t
T
T/2
T
(b)
1
T
T/2
T
(c)
S3(t)
T/2
t
t
-3
(a)
Fig(9.4)
9-9
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

0
T/2

S (t)S (t) dt 
1
2
T

S (t) S (t) d t  S (t)S (t)dt
1
2
1
2
T/2
0
T/2


T

(1)(2) d t  (3) (0) d t   2 (T/2-0)-T
0
T/2
T
T/2
T
0
T/2
0
T/2
 S1(t) S3(t) dt   S1(t) S3(t) dt   S1(t) S3(t)dt


T
(1) (1)dt 
0
T
(3) (3) dt  (1)[T/2 - 0]  9[T - ]  4T
2
T/2

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 /2
0
0
T
 1(t) 2 (t) dt   (1) (1) dt  T/ 2(1)(1)dt 0
9-10
T
  j (t) k (t)dt  k j jk
0


 1 J k



jk 

 J k
 0
Where
C-Using eq(9.2c) and(9.3b)
T
1
a i j
Si (t ) j (t )dt
k
j

0
i  1.2......M
j 1.2.....N
T

1 1(t)
2
K j   j (t )dt  E
T/2
-1
0
T 2

T /2

K1   (t ) dt 
0
T
K2 

1
T /2
(1) 2 dt 
0
T
2
 2 (t )dt 
0


(1) 2 dt T
0
(1) 2 dt T
0
T
1
ai j 
S (t ) (t ) dt
k
j

i
j
0
T /2
T
T

1
1
a11 
ٍ S1(t ) 1(t ) dt   (1) (1)dt  (3) (1)
K1 0
T 0

T /2

1 T
T 
  3( ) 1
2 
T 2


9-11

T
T
1
1
a12 
S1(t )  2 (t ) dt 
T
k2 0

1
a12 
k2
T

0
T /2
T
T /2
T
1
S1(t )  2 (t ) dt 
T
0 11dt T/ 231dt 2
 11dt   31dt 2
0
T /2
T
T /2
T
1
1
21dt  (0)(1)dt 1
a21
S 2 (t ) 1(t ) dt 
k1 0
T T /0
T /2


T
1
1
a22 
S 2 (t ) 2 (t ) dt 
k2 0
T

T
1
1
a31 
S3 (t ) 1(t )dt 
k1 0
T

1
a32 
k2
T

0
T /2
T
 (2)(1)dt   (0)(1)dt 1
0
T /2
T /2
T
0
T /2
T /2
 (1) (1)dt   (3) (1)dt 2
1
S3 (t ) 2 (t ) dt 
T
Using eq (9  3a)

T /2
Si (t ) 
 (1) (1)dt   (3) (1)dt 1
0
T /2
N
a
i j  j (t )
i 1,.....M
j 1
j  1,2,...N
In this case
i=1,2,3
j=1,2
 Si (t ) 
2
a
i j  j (t )
j 1
9-12
S1 (t )  a11  1 (t )  a12  2 (t )  1(t )  2 2 (t )
S 2 (t )  a21 1(t )  a22  2 (t )  1 (t )  2 (t )
S3 (t )  a31 1 (t )  a32  2(t )  2 1  2 (t )
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 1 , 2

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)
0tT
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
9-13
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).
‫ـ‬Step1can be considered as transforming 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)
9-14
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 )
9-15
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:
………..(9.6 a )
………..(9.6 b )
Fig (9.8)
9-16
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)
(9.7a)
……(9.7 b)
and n(t)=zero mean white Gaussian random process where the
phase term  is an arbitrary constant, so that the analysis by setting
=0
9-17
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
……………….(9.8)
Thus we may express the transmitted signal Si(t) in terms of 1(t)
and ai 1 as follows:
……………….(9.9)
……………….(9.10a)
……………….(9.10b)
……………….(9.10 c)
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

Z i (T )  r (t)  1 (t ) dt
0
T
Z1 (T )  S1(t )  n(t ) 1 (t )dt..........(9.11)

0
9-18
Substituting for S1(t) and 1(t) in eq(9.11)
T
(T ) { [
ZZ
1(T)
i = 
0
E1 (t ) n(t )]1(t )dt}
T
 {[ E12 (t )  n(t )1(t )]dt}
0
T
 E{[
0
2
2
E cos 2 t  n(t )
cost ]dt}
T
T
Z1(T )  E ........................(9.12a)
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).
9-19
Fig(9-11)
Fig (9-12)
9-20
-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
…...…(9.11)
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
.................(9.13 a)
.................(9.13 b)
and
where the amplitude
2 / T 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
............(9.14 a)
............(9.14 b)
9-21
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)
9-22
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)
.......................(9.13)
Si(t) =
 2i  2
Si (t )  E Cos
Cos 0t

M  T
 2i  2
sin 0t
 E sin

M  T
Si 
2E
Cos i cos 0t sin i sin 0t 
T
But r (t)  Si (t )  n (t)
9-23
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
9-24
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 .
9-25
Communication

Lecture
10
10-1
Systems
Coherent detection of FSK
FSK modulation is characterized by the information being
contained in the frequency of the carrier .Atypical set of FSK
signal waveforms was described as
2E
Cos wi t  
T
Si (t ) 
0  t  T ..........(10  1)
i  1,2,.........M
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
Si (t )  ai1 (t )  ai 2  2 (t )  .......ai j  j (t )....(10.2a)
S i 
N
aij  j (t )

j 1
i 1,2,.....M ......(10.2b)
j 1,2,....N
N M
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
10-2
T
1
a 
S (t ) (t ) d t ..............(10.3b)
ij k
i
j
0
j

where
Kj=1
for orthonormal set
Substitute for Si(t) and j(t) from equation (10-1,103a) into
(10.3b)

T
aij 
2E / T Cos i t
 2 / T Cos  t dt
j
0
0 
T
aij 



aij  



2E / T Cos i t
E
0
 2 / T Cos  td t
for i  j
j
............(10.4)
for i  j
Eq(10.4) mean that, the ith prototype (reference) signal vector is
E from the
located on the ith coordinate axis at a displacement
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
d=
 E   E 
2
2
 2E
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
10-3
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

 (t )
1(t )
0

T
2 (t )

0
fig ( 10.2)
10-4
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
…………..(10-5)
The received waveform can be characterized by
r(t) =
2E
Cos 0t i (t )   n(t ) 0  t  T .....(10  6)
T
i 1,2......M
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
j (T1) and k(T2) is independent of  that is
 k T2    j T1   k T2  j T1  i (T ).......(10.7)
-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=2i/M)
d-The detector (in the receiver), calculate the coordinates of the
incoming signal by correlating it with locally generated
waveforms ,such as
2T cos0t and
2T sin 0t , 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)
10-6
-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-1m(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.
Detectrd message
1
1
10-7
0 1
0 1 1
0
0 1
2 / T cos  t

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)= C (k 1)  m(k ) ].
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 m(k  1) 1
10-8
Then (k=2) is matched with  (k=1) again have the same value

m(k  2)  1
Then (k=3) is matched with (k=2), they are different so that

m(k  3)  0 , 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
branch and
2 / T cos1t for the I
2 / T sin1t 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 cos1t+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
10-9
output because its reference signal 2 / T 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 cos1t reference and partially correlate with
sin1t 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
10-10
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)
10-11
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)
10-12
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(2f1t+) and Cos2f2t to be used for
FSK signaling ,where f1>f2
10-13
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
 S (t)S (t)0
1
2
0
................(10-10)
Using the basic trigonometric identities, we can write eq(10-10)
as
1
2
10-14
.........(10-11)
if f1+f2  1
 Sin   1 
Cos   1.......... .(10  12 )


Combining eq(10.11) and (10-12), we can write
.........(10-13)
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 sin0 and Cos=1 occur
simultaneously when n=2k
From eq(10-13)for arbitrary , we can therefore write
10-15
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
...........(10-14)
where g(t) is known as the complex envelope as
...........(10-15)
10-16
The magnitude of the complex envelope is then
...........(10-16 a)
and its phase is
...........(10-16 b)
With respect to eq(10-14), we can called g(t) the baseband
message or data in complex form, and e
j  t
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 :-
S (t )  Re[ g (t )e
j  t
]
Quadrature implementation of a modulator
...............(10-17)
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
gk=x
10-17
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
e
j  t
and then transmitting the real part of the product.
fig(10.7)
10-18
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 (  k , k , z k ) 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)
+
+
+
+
Binary 1
0
1
1
0
Gray code 1
1
1
0
1
Binary 1
0
1
1
0
-The process of adding the current message to phase assignment k
to the prior data phase k 1 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

………..(11.2b)
D8PSKdemodulator
-Using a similar quadrature implementation demodulation consist of
reversing the process that is multiplying the received bandpass
waveform by e
 j  t
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
-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:
0.707
A 
2
Similarly, after the quadrature multiplication by sin  t in the

demodulator, and after filtering with a low pass filter we recover at
point B an ideal negative pulse as follows
B 
0.707
2
-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)
k  k 1  k 1
k  2  k  2  k 1
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 x2 y2 z 2 , x2 y2 z 2 =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 PE .The calculations for
obtaining PE 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
…….(11.2)
-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
S1 (t ) if
Z (T ) =0

decide
S 2 (t )
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).
a2
a1
Fig (11.4) Conditional probability density functions p(Z/S1) , p(Z/S2)  
0
a a
1 2
2
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 PB , 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
…….(11.4)
For equal energy antipodal signaling, such as BPSK format in
equation (11.2),the receiver components are
a1  E b when S1 (t ) is sent
and a2   Eb
when S 2 (t ) is sent
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 :
P 
B


2 Eb / N 
PB Q(
u 2
1
exp(
)du
2
2
2E B
)
N

(11.5)
(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.
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=10
11
..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
  2 
1
 for   3
Q  
exp 
 2 
 2


PB  8 10  4
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
……………(11.7)
Fig(11.5) shows bit error probability for several types of binary
systems
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
P 
B


2 Eb / N 
u 2
1
exp(
)du
2
2
(11.5)
PB Q(
2E B
)
N

(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
……(11.8)
-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)
-On-OFF keying (OOK) is an orthogonal signaling set (unipolar
pulse
signaling
eq(11.8) PB  Q
is
the
b
0
baseband
equivalent
OOK).
Thus
 E / N  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.
 Eb 
.........(11.8) for BFSK
PB  Q
 N0 


 2Eb 
........(11.7) for BPSK
PB  Q
 N0 


Thus the performance of BFSK signaling is 3dB worse than BPSK
signaling ,since for a given signal power
 E ,the distance squared
b
between orthogonal vectors is a factor of two less than the distance
squared between antipodal as shown in fig(11.6)
2(t)
2(t) S2
S2
2Eb
S1
Eb
0
1(t
Eb
Eb
1(1)
S1
2 Eb
antipodal
Eb
(b) orthogonal
(a)Binary signals vectors
Fig (11.6)