Digital Communication fundamentals .pdf

© Burak Kelleci - 2024
DIGITAL COMMUNICATION
Burak Kelleci

INTRODUCTION
○ Communication: Transmission of information from one point to another
point.
○ Information:
● Audio: Speech, FM, AM, DAB, HDRadio, Phone, GSM, AMPS, etc.
● Video: TV Broadcasting, Security cameras, Satellite images, etc.
● Data: WiFi, Bluetooth, RS232, PCI, CD, DVD, etc.
○ Important Parameters
● Spectral Efficiency:
○ Amount of information per given bandwidth
● Power Efficiency
○ Amount of information per used power
● Multiplexing
○ Frequency
○ Time
○ Spatial
( ) ( )
( )
t
t
f
t
a c 
 +
2
cos

HOW IS COMMUNICATION SYSTEM ORGANIZED?
○ Source of information: voice, music, pictures, videos, data
○ Transmitter: Process the information in the form that is suitable
for transmitting over the channel.
○ Channel (Transmission Medium): optical fiber, free space, twisted
cable, copper cable, CD, cassette, etc.
○ Receiver: Converts the signal transmitted over the channel back
to a form that may be understood (it may not be exactly the same
transmitted data) at the intended destination. The receiver may
also compensate the distortions introduced by the channel and
perform other functions such as synchronization of the receiver to
the transmitter.
○ Sink of information: user, computer, etc.
Information
Source
Transmitter Channel Receiver
Information
Sink

WIRELESS COMMUNICATION
TRANSMITTER
○Protocol Stack: It packages the data so that it can reliably get to
the desired destination once it crosses the radio link.
○Modulator: It transform the information upon a carrier
frequency in order to be recovered at the receiver.
○Up-Conversion & Filter: The modulated signal is converted to
the final RF frequency at which it will be transmitted.
○RF Stage: The RF signal is amplified to an appropriate power
level and then emitted via an antenna. In other words, the
modulated signal is converted to an electromagnetic wave.
Information
Source
Protocol
Stack
Modulator
Up-Conversion
& Filter
Amp
RF Stage

CHANNEL
○ Propagation Loss: Loss of a signal strength with increasing distance.
○ Frequency Selectivity: Many transmission medium conducts well over a relatively
small range of frequencies.
○ Time Varying: Some channel’s characteristics vary with time. For example, mobile
channels.
○ Nonlinear: Nonlinear elements in the channel creates distortions.
○ Shared Usage: For efficiency the communication channel is shared among
different users. This creates interference between different users.
○ Noise: The random motion of electrons creates uncertainty of the received signal.
This usually is the reason of fundamental performance limitation.
Channel
Lossy
F
r
e
q
u
e
n
c
y
S
e
l
e
c
t
i
v
e
N
onlinear
T
i
m
e
V
a
r
y
i
n
g
Noise
Noise
RX 1
RX 2
TX 2
TX 1

RECEIVER
○ RF Stage: The antenna collets RF energy in the desired band (may collect energy from unwanted as
well). The first amplifier, called low-noise amplifier, boost the signal power.
○ Down Conversion: Translate the RF signal to a frequency where the signal more easily demodulated.
○ Demodulation: Transmitted signal is recovered.
○ Synchronization: Compensates the time and frequency difference between transmitter and receiver.
○ Channel Compensation: Counteract some of the impairments that the signal encountered in the
channel. For example, equalization for frequency-selective channels, error correction for noisy
channels.
○ Protocol Stack: The receiver determines whether the detected message was intended for it or not.
Information
Source
Protocol
Stack
Demodulator
Down-Conversion
& Filter
Amp
RF Stage
Synchronization
Channel
Compensation

WHY DIGITIZE ANALOG SOURCES
○Digital systems are less sensitive to noise than
analog.
● For long transmission lengths, the signal may be
regenerated effectively error-free at different points
along the path.
○With digital systems, it is easier to integrate
different services, for example, video and
soundtrack into the same transmission scheme
○The transmission scheme is independent of the
source.
● For example, a digital transmission scheme that
transmits voice at 10kpbs can also be used to transmit
data at 10kpbs

WHY DIGITIZE ANALOG SOURCES
○Digital circuits are easy to manufacture and less
sensitive to physical effects, such temperature
○Digital signals are simpler to characterize and do not
have the same amplitude range and variability as
analog signals.
○There are techniques for adding controlled
redundancy to a digital transmission such that errors
that occur during transmission may be corrected at
the receiver. These techniques are called channel
coding.
○Channel compensation techniques, such as
equalization, are easy to implement.

THE SAMPLING PROCESS
○The sampling process is usually described in the
time domain.
○Using the sampling process an analog signal is
converted into a corresponding sequence of
samples that are usually spaced uniformly in time.
○It is important to choose the sampling rate properly
in order to define uniquely the original analog signal.

○Consider an arbitrary signal
g(t) of finite energy, which is
specified for all time.
○We sample the signal g(t)
instantaneously and at a
uniform rate once every Ts
seconds.
○We denote the samples as
g(nTs) where n takes on all
possible integer values.
○Ts: Sampling period
○fs=1/Ts: Sampling Frequency

○We refer the ideal sampled signal
○The summation is the impulse train.
○Fourier transform of the sampled signal is basically
multiplication of the continuous-time signal by
impulse train.
( ) ( ) ( )
( ) ( )



−
=

−
=
−
=
−
=
n
S
S
n
S
nT
t
nT
g
nT
t
t
g
t
g




○Fourier Transform of Impulse Train.
● Let’s rewrite the impulse train using Fourier Series
( ) ( )
( )
( ) 




−
=
−
−

−
=

−
=
=
=
=
=
−
=
k
t
f
jk
s
T
s
T
T
t
f
jk
s
k
k
t
f
jk
k
k
S
T
s
S
S
S
s
s
S
e
T
t
T
dt
e
t
T
c
e
c
kT
t
t







2
2
/
2
/
2
2
1
1
1

○Let’s use the frequency shifting property to
determine the Fourier transform
( ) ( )
0
2 0
f
f
G
t
g
e
F
t
f
j
−

− 
( ) ( )
( ) ( )
( )




−
=

−
=

−
=
−
−


=
k
s
s
k
s
s
F
T
k
t
f
jk
s
T
kf
f
f
kf
f
T
t
e
T
t
S
s
S



 
1
1
1 2
( )
f
F


1

○Let’s use multiplication property to determine the
Fourier Transform of the sampled signal.
○Note: Multiplication in time domain corresponds
convolution in the frequency domain.
( ) ( ) ( ) ( )



−
−
 

 d
f
G
G
t
g
t
g
F
1
1
2
1

○Let’s apply multiplication property
○The Fourier transform of the original signal g(t) and
fs is the sampling rate.
( ) ( ) ( ) ( )
( ) ( )
( )

 
 


−
=

−
=


−


−

−
=

−
=
−
−
−
−
−

−
n
s
s
n
s
s
n
s
s
F
n
S
nf
f
G
f
d
nf
f
G
f
d
nf
f
f
G
nT
t
t
g









( ) ( )


−
=
−

n
S
S
F
nf
f
G
f
t
g

○The process of uniformly sampling a continuous-
time signal of finite energy results in a periodic
spectrum with a period equal to the sampling rate.
○Suppose the signal g(t) is strictly band-limited with
no frequency components higher than W hertz.
The Fourier Transform of g(t)

○Let’s choose the sampling period Ts=1/2W
○There is no overlap among the repeated spectrums.
➔ Original signal can recovered.
○Nyquist Rate: Sampling Frequency = 2W
fs=2W

g(t) is sampled at lower
frequency than its
bandwidth ➔ The
spectrum is overlapped,
and it is not possible to
recover the original signal
fs<2W

g(t) is sampled at higher
frequency than its
bandwidth ➔ The
spectrum is repeated, and
it is possible to recover the
original signal
This region allows
to relax the anti-
alias filter
requirements
fs>2W

THE SAMPLING PROCESS - EXAMPLE
○Assume that voice is bandlimited to less than
3500Hz. Calculate the nyquist sampling frequency.
How many samples do we obtain per second?
○Voice signal is low-pass filtered prior sampling. The
filter significantly attenuates the frequency
components above 4KHz. Calculate the nyquist
sampling frequency for this system. How many
samples do we obtain per second?

EXAMPLE - SOLUTION
○Since the signal is bandlimited to 3500Hz, Nyquist
sampling rate is 2x3500=7000Hz
○We obtain 7000 samples per second.
○Low-pass filter removes any spectral components
above 4KHz. Therefore, nyquist sampling rate is
2x4KHz=8KHz
○We obtain 8000 samples per second.

EXAMPLE 2
○The following signal is sampled at 3KHz. Draw the
spectrum.
( ) ( )
t
KHz
t
x 
= 1
2
cos 

EXAMPLE 2 - SOLUTION
○The continuous-time spectrum will be repeated
around fs from - to +.
1 KHz
-1 KHz 4 KHz
2 KHz
-2 KHz
-4 KHz

EXAMPLE 3
○The following signal is sampled at 2KHz. Draw the
spectrum.
( ) ( )
t
KHz
t
x 
= 3
2
cos 

○The sampling rate does not satisfy Nyquist
Frequency. Therefore, there will be aliasing. The
aliasing components frequencies are calculated
using the following formula.

+

−

= to
-
from
integer
:
n
f
fs
n
f signal
sampled
Note that if the sampling starts at one of the zero crossings, the
followed samples are also at zero crossings.
Therefore, the sampled signal will be zero all the time.
1
-1 5
3
-5 (KHz)
-3

PULSE AMPLITUDE MODULATION (PAM)
○In pulse-amplitude modulation (PAM), the
amplitudes of regularly spaced pulses are varied in
proportion to the corresponding sample values of
continuous message signal.
○The pulses can be of a rectangular form or some
other appropriate shape.

PAM
○In the generation of PAM
● The message signal is sampled at every Ts seconds,
where fs=1/Ts and is chosen in accordance with the
sampling theorem.
● Each sample is lengthened to some constant value T.
○These two operation are also called sampling and
hold.
○The main important reason to increase the duration
of each sample is to avoid the use of excessive
bandwidth, since bandwidth is inversely
proportional to sample duration T.

PAM
○PAM signal can be expressed as
where Ts is the sampling period and m(nTs) is the
sample value of m(t) obtained at t=nTs.
○h(t) is the rectangular pulse of unit amplitude and
duration T, defined as follow
( ) ( ) ( )


−
=
−
=
n
S
S nT
t
h
nT
m
t
s
( )





=
=


=
otherwise
0
,
0
2
1
0
1
T
t
t
T
t
t
h

PAM
○The sampled version of m(t) is given by
○Convolving the sampled version of m(t) with the pulse
h(t) gives
( ) ( ) ( )


−
=
−
=
n
S
S nT
t
nT
m
t
m 

( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )


 



−

−
=


−

−
=


−
−
−
=
−
−
=
−
=














d
t
h
nT
nT
m
d
t
h
nT
nT
m
d
t
h
m
t
h
t
m
S
n
S
n
S
S

PAM
○Using the shifting property of the delta function
○PAM signal is basically the filtered version of m(nTs)
with filter impulse response of h(t)
○Taking Fourier Transform of both sides
( ) ( ) ( ) ( )
S
n
S nT
t
h
nT
m
t
h
t
m −
=
 

−
=

( ) ( )
  ( ) ( )
( ) ( ) ( )
( ) ( )



−
=

−
=
−
=
=






−
=

k
s
s
n
S
S
f
H
kf
f
M
f
f
H
f
M
f
S
nT
t
h
nT
m
F
t
h
t
m
F



PAM SPECTRUM

PAM
○To recover the PAM signal at the receiver, first we
need to filter the received signal to remove the
noise above 2W, W is the bandwidth of the
message signal.
○The output is equivalent to filter the continuous
time message signal through a filter with frequency
response of H(f).
○This effect causes amplitude distortion and delay of
T/2. This distortion is called aperture effect.
( ) ( ) fT
j
e
fT
T
f
H 
−
= sinc

PAM
○This distortion can be corrected by connecting an
equalizer at the output of low-pass filter. The
amplitude response of the equalizer is
( ) ( ) ( )
fT
f
fT
T
f
H 

sin
sinc
1
1
=
=

TIME DIVISION MULTIPLEXING
○PAM Signal is suitable for Time Division
Multiplexing (TDM) of multiple message signals.
○Since each sample carries only the information for a
limited time and zero otherwise, another message
signal may be sent when the PAM signal is zero.

TIME DIVISION MULTIPLEXING

PULSE DURATION (WIDTH) MODULATION
○Pulse Duration Modulation (PDM) or Pulse Length
Modulation or Pulse Width Modulation (PWM)
● The samples of the message signal are used to vary the
duration of the individual pulses.
● The modulating signal may vary the time of occurrence
of the leading edge, trailing edge or both edges.
● Since the PDM signal has only two levels, high efficient
amplifiers can be built to amplify these signals.
Theoretical efficiency can be 100%.

○Leading Edge Modulation. Information is coded at
the crossings of the first edge during a sampling
period.
○Trailing Edge Modulation: Information is coded at
the crossings of the second edge during a sampling
period.
○Both Edges Modulation: Information is coded at
both crossings during a sampling period.

GENERATION OF PWM SIGNALS
○PWM Signals are generated by comparing the
message signal with a triangular or saw-tooth signal.
Saw tooth
Generator
x(t)
PWM
Comparator

EXAMPLE
○ Determine the PWM signal of the following ramp signal, assuming
leading edge, trailing edge and both edges are modulated
○ Triangular wave is used when both edges are modulated.

EXAMPLE - SOLUTION
Both Edges are modulated

EXAMPLE - SOLUTION
Leading Edge is modulated

EXAMPLE - SOLUTION
Trailing Edge is modulated

SPECTRUM OF PWM SIGNALS
○Spectrum of a PWM signal is not easy to calculate
as FM. For example, spectrum for the following
single sinusoidal input signal
( )
( ) ( )
( )




=



=

−
=

−
=






−
−
+
−
−
−
+
+
=
1 1
0
2
2
sin
2
sin
sin
cos
2
m n
s
c
n
m
c
m
c
s
n
k
m
t
n
t
m
m
M
m
J
k
m
t
m
m
M
m
J
m
t
m
t
M
k
t
PWM













( )
t
M
k
t
V
V
t
v
s
s
SP
DC
s




cos
2
cos
+
=
+
=

SPECTRUM OF PWM SIGNALS
○The spectrum consists of infinite components like
FM spectrum.
○If the carrier wave frequency is chosen much higher
than input signal frequency, the demodulation is
done using a low-pass filter.
○PWM is also widely used in other areas such as
Class-D Audio amplifiers, motor drive, etc.

PULSE POSITION MODULATION
○In PDM, long pulses expend considerable power
while carrying no additional information.
○If only transitions are transmitted instead of long
pulses, we obtain more efficient pulse modulation
known as pulse-position modulation (PPM).
PDM (PWM)
PPM

○Let Ts donate the sample duration. Using the sample
m(nTs) of a message signal m(t) to modulate the
position of the nth pulse
where kp is the sensitivity of the pulse-position
modulator and g(t) denotes a standard pulse.
( ) ( )
( )


−
=
−
−
=
n
s
p
s nT
m
k
nT
t
g
t
s

○Clearly s(t) must be strictly non-overlapping.
○The sufficient condition for this requirement is
○The closer kp|m(t)|max is to one half the sampling
duration Ts, the narrower must be the standard
pulse g(t) in order to ensure that individual pulses of
the PPM signal s(t) do not interfere with each other.
( ) ( )
( )
2
2
0
max
max
s
s
p
s
p
s
T
nT
m
k
nT
m
k
T
t
t
g


−

=

GENERATION OF PPM
○Pulse Position Modulated signals can be produced
using a monostable circuit connected at the output
of PWM generator.
Saw tooth
Generator
x(t)
Comparator
Monostable PPM
Monostable circuit has two states, one is stable and the other is
unstable. A trigger causes the circuit to enter the unstable state.
After entering the unstable state, the circuit will return to the stable
state after a set time. Such a circuit is useful for creating a timing
period of fixed duration in response to some external event

DETECTION OF PPM WAVES
○PPM waves are first converted to PDM (PWM)
signals.
○This PWM waves are filtered to obtain the message
signal.

EXAMPLE
○ Determine the PPM signal of the following ramp signal.

EXAMPLE - SOLUTION
○You can use one of the modulated edges to
generate PPM waves

EXAMPLE 2
○24 voice signals are sampled uniformly and then time-
division multiplexed. The sampling operation uses
flat-top samples with 1s duration. The multiplexing
operation includes provision for synchronization by
adding an extra pulse of sufficient pulse and also 1s
duration. The highest frequency component of each
voice signal is 3.4KHz
○Assuming a sampling rate 8KHz, calculate the spacing
between successive pulses of the multiplexed signal.
○Repeat the calculations for Nyquist rate sampling.

○The sampling interval is 125s. Since there are 24
voice channels and 1 sync channel, the time
allocated for each channel is 125s/25=5s. The
pulse duration is 1s, so the time between pulses is
4s.
○If the channels are sampled at Nyquist rate, 6.8KHz,
the sampling period will be 147s. The allocated
time is 5.88s. The time between pulses is 4.88s.

THE QUANTIZATION PROCESS
○A continuous signal has continuous range of
amplitudes, in other words it has infinite number of
amplitude levels.
○Any human sense can detect only finite amplitude
differences.
○Therefore, the original continuous signal may be
approximated by a signal constructed of discrete
amplitudes.
○If the discrete amplitude levels have sufficiently
close spacing, the discrete-level signal will be
practically indistinguishable from the continous-
level signal.

○Amplitude quantization is defined as the process of
transforming the sample amplitude m(nTs) of a
message signal m(t) at time t=nTs into a discrete
amplitude v(nTs) taken from a finite set of possible
amplitudes.
k


○In this class, we assume that the quantization
process is memoryless and instantaneous, in other
words the transformation at time t=nTs is not
affected by earlier or later samples of message
signal.
○The signal amplitude m is specified by the index k if
it lies inside the interval
where L is the total number of amplitude levels used
in the quantizer.
○mk: decision levels or decision thresholds.
  L
k
m
m
m k
k
k ,
,
2
,
1
,
1 
=


=
 +

○The amplitudes vk are called representation levels or
reconstruction levels, and the spacing between two adjacent
representation levels is called a quantum or step-size.
○The mapping v=g(m) is the quantizer characteristic.
○Quantizers are
● uniform quantizer: representation levels are uniformly spaced.
● non-uniform quantizer: representation levels are not uniformly
spaced.
● midtread: origin lies in the middle of tread of the staircaselike
graph
● midrise: origin lies in the middle of a rising part of staircaselike
graph.

Midtread Uniform Quantizer Midtrise Uniform Quantizer

QUANTIZATION NOISE
○Quantization noise: Error between the input
signal and quantized signal.
○Let’s assume that the quantization noise is
● statistically independent from the signal
● and the message signal does not overload the
quantizer, in other words the signal levels are within
the quantizer maximum and minimum levels.
○Consider a continuous signal m in the range of
(-mmax, mmax).
○For midrise quantizer the step-size is
L
mmax
2
=


QUANTIZATION NOISE

QUANTIZATION NOISE
○The quantization noise values are bounded by
○If the step-size is sufficiently small, we can assume
that quantization error is uniformly distributed
random variable, and its effect is similar the thermal
noise.
○Its probability density is
2
/
2
/ 



− q
( )




 



−

=
otherwise
0
2
2
1
q
q
fQ

QUANTIZATION NOISE
○The mean of quantization error is zero, and its
variance is
○The quantization noise is proportional to the step-
size. Reducing the step-size reduces the
quantization noise.
( )
12
1 2
2
2
2
2
2
2
2

=

=
=




−


−
dq
q
dq
q
f
q Q
Q


QUANTIZATION NOISE
○Let R denote the number of bits per sample. The
number of levels area
○The step-size and the variance of quantization noise
L
R
L R
2
log
2
=
=
R
Q
R
m
m
2
2
max
2
max
2
3
1
2
2
−
=
=



QUANTIZATION NOISE
○The signal-to-noise ratio for a message signal of
power P
○The SNR increases exponentially with increasing the
number of bits per sample, R.
○Since R is proportional to required channel
bandwidth, there is a tradeoff between the channel
bandwidth and SNR like FM.
○Binary transmission of message provides a more
efficient method than FM for the trade-off due to
the exponential relationship.
R
Q m
P
P
SNR 2
2
max
2
2
3








=
=


QUANTIZATION NOISE
○Sinusoidal Modulating Signal
● Let’s assume a sinusoidal signal with amplitude Am is
applied to uniform quantizer.
○The average signal power on 1ohm is
○The SNR of the quantized signal is
2
2
m
A
P =
( ) dB
in
6
8
.
1
log
10
2
2
3
2
2
/
3
10
2
2
2
2
2
R
SNR
A
A
P
SNR R
R
m
m
Q
+
=
=








=
=

L R SNR(dB)
32 5 31.8
64 6 37.8
12
8
7 43.8
25
6
8 49.8

PULSE-CODE MODULATION
○In pulse-code modulation (PCM) a message signal is
represented by a sequence of coded pulses, which is
accomplished by representing the signal in discrete
form in both time and amplitude.
○The basic operations performed in a PCM
transmitter is sampling, quantizing and encoding.
○The basic operations in the receiver are
regeneration, decoding and reconstruction of the train
of quantized samples.

PULSE-CODE MODULATION

SAMPLING & QUANTIZATION
○Prior sampling a low-pass filter is used to restrict the
bandwidth half of the sampling frequency.
○This filter prevents aliasing of frequency
components above half of the sampling rate.
○The sampled signals are then quantized to pre-
determined levels.
○The quantized signal is then encoded to binary bits
for robust transmission.

QUANTIZATION
○Till now, we assumed uniform quantization.
However, some signals, such as voice, spends
different times at different levels.
○For example, voice signals are most of the time has
low amplitude levels. The loud talks are not as
common as the weak talks.
○Therefore, if step size of the quantizer changed to
favor weak talks, the overall quality of the
transmitted voice signal is improved.
○In practice, this non-uniform quantization is
performed by passing the message signal through a
compressor and then applying the compressed
signal to a uniform quantizer.

QUANTIZATION
○There are two different standard for the
compressing function.
● -law: used in North America and Japan
● A-law: used in Europe
○-law encoding is defined as
○the decoding is defined as
( )
( )
( )
1
1
1
ln
1
ln
sgn 

−
+
+

= m
m
m
v


( ) ( )
( ) 1
1
1
1
1
sgn 

−
−
+


= m
v
m
v



QUANTIZATION
○ is a positive constant and =0 corresponds to
uniform quantization.
○The slope of the compression curve or in other words
the derivative of |m| with respect to |v| is
○-law is not strictly logarithmic. For low input levels
(|m|<<1) it is linear and logarithmic for high input
levels (|m|>>1)
( ) ( )
m
v
d
m
d



+

+
= 1
1
ln

QUANTIZATION
○-law Compression and Decompression
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Normalized Input |m|
Normalized
Output
|v|
-law Compression Law
=1
=5
=100
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Normalized
Output
|v|
-law Decompression Law
=1
=5
=100

QUANTIZATION
○A-law compression law is defined as
○For A=1, A-law corresponds to uniform quantizer.
○A-law decompressing law is defined as
( ) ( )
( )
( )









+
+

+

=
1
1
ln
1
ln
1
1
ln
1
sgn
m
A
A
m
A
A
m
A
m
A
m
v
( )
( )
( )
( )
( )
( )
( )









+
+

+

= −
+
1
ln
1
1
ln
1
1
ln
1
sgn 1
ln
1
v
A
A
e
A
v
A
A
v
v
m A
v

QUANTIZATION
○A is a positive constant and A=1 corresponds to
uniform quantization.
○The slope of the compression curve or in other words
the derivative of |m| with respect to |v| is
○A-law shows also similar performance as -law. It
behaves linear for small signal levels and shows
logarithmic behavior for high signal levels.
( )
( )
( )







+

+
=
1
1
ln
1
1
ln
1
m
A
m
A
A
m
A
A
v
d
m
d

QUANTIZATION
○-law Compression and Decompression
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Normalized
Output
|v|
A-law Compression Law
A=2
A=10
A=100
0 0.2 0.4 0.6 0.8 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Normalized
Output
|v|
A-law Decompression Law
A=2
A=10
A=100

QUANTIZATION
0 0.5 1
-1
-0.5
0
0.5
1
Time (s)
Signal
Level
0 0.5 1
-1
-0.5
0
0.5
1
Time (s)
Signal
Level
0 0.5 1
-1
-0.5
0
0.5
1
Time (s)
Signal
Level
0 0.5 1
-1
-0.5
0
0.5
1
Time (s)
Signal
Level
Original
Original
Quantized
Original
Quantized
Original
Quantized

QUANTIZATION
-60 -50 -40 -30 -20 -10 0
-30
-25
-20
-15
-10
-5
0
5
10
15
20
Signal Level (dB)
SNR
(dB)
Comparasion of SNRs of -law (=100) and A-law (A-100)
Uniform Quantization
-Law Quantization
A-Law Quantization
•16-bit quantization is
used
•Below 55dB, all
methods show the same
performance, since the
signal is below the
minimum step-size
•-Law and A-Law
improves the SNR
between -55dB and -
15dB.
•For signal levels above
-15dB, uniform
quantization shows
better performance.

ENCODING
○After sampling and quantization the continuous
message signal is limited to discrete set of values,
but nor suited to transmission over a line or radio
path.
○To make this signal more robust to noise,
interference and other channel degradations, it
needs to be encoded to a more appropriate form of
signal.
○Representing each of this discrete set of values is
called a code.
○One of these discrete events in a code is called code
element or symbol.

ENCODING
○In a binary code, each symbol is assigned to one of
two distinct values.
○The main reason for using the binary code is that
the maximum advantage over the effects of noise in
a transmission medium is obtained by using a binary
code, because binary symbol withstands a relatively
high level of noise and is easy to regenerate.
○Suppose that, in a binary code each words consists
of R bits (binary digit). Therefore, R is the number of
bits per sample. This code represents 2R distinct
levels.

QUANTIZATION AND ENCODING
IN MATLAB & SIMULINK
○In Matlab round function can be used to model the
quantization.
○For example, the following code quantizes the input signal
between -2 and  where  corresponds 11 in binary. It
also encodes the signal for step-size of 0.25.
○Therefore, minimum signal level that does not overload
the quantizer is -2x0.25=-0.5V
○Maximum signal level that does not overload the
quantizer is 1x0.25=0.25V.
○The quantizer output will assign -0.5V to 00,
-0.25 to 01, 0 to 10 and 0.25 to 11

% Quantize between -0.5V and 0.25V
% with a step size of 0.25V
t=0:1e-3:1;
m=-1:2/(length(t)-1):1;
Delta=0.25;
% saturate m
ms=m;
ms(m<(-2*Delta))=-2*Delta;
ms(m>(1*Delta))=1*Delta;
% Quantize the signal
mq=round((ms+2*Delta)/Delta);
figure(1);
plot(m,mq);
xlabel('Input Signal Level');
ylabel('Output Code');
-1 -0.75 -0.5 -0.25 0 0.25 0.5 0.75 1
0
1
2
3
Input Signal Level
Output
Code
Note that this code generates midtread response

% Quantize between -0.5V and 0.25V
% with a step size of 0.25V
t=0:1e-3:1;
m=-1:2/(length(t)-1):1;
Delta=0.25;
% saturate m
ms=m;
ms(m<(-1.5*Delta))=-1.5*Delta;
ms(m>(1.5*Delta))=1.5*Delta;
% Quantize the signal
mq=round((ms+1.5*Delta)/Delta);
figure(1);
plot(m,mq);
xlabel('Input Signal Level');
ylabel('Output Code');
Note that this code generates midtrise response
-1 -0.75 -0.5 -0.25 0 0.25 0.5 0.75 1
0
1
2
3
Input Signal Level
Output
Code

○In Simulink, there is Quantizer component, but this
component has no limits and the user can only
control the step size.
○Another method is using Lookup Table to model the
quantizer.

EXAMPLE
○In the following PCM signal binary 0 is represented
by -1V and binary 1 is represented by 1V.
○The code word consists of 3 bits and step-size is 1V.
○Find the sampled version of analog signal from
which this binary signal is derived.

EXAMPLE - SOLUTION
Symbol Code
Word
Amplitu
de
1 001 1V
2 010 2V
3 011 3V
4 100 4V
5 101 5V
6 110 6V

EXAMPLE
○A PCM system that uses a uniform quantizer is
followed by a 7-bit binary encoder. The bit rate of
the system is equal to 50Mbit/s.
○What is the maximum message bandwidth for this
system?
○Determine the output signal-to-quantization noise
ratio when full-load sinusoidal modulating wave of
frequency 1MHz is applied to the input.

EXAMPLE - SOLUTION
○Sampling the message signal with bandwidth W at
Nyquist rate, and using R-bit code to represent each
sample of the message signal gives the following bit
duration.
○The output signal-to-quantization ratio is
MHz
s
Mbit
W
WR
T
WR
R
T
T
b
S
b
57
.
3
7
2
/
50
2
1
2
1
max =

=
=

=
=
( ) dB
R
SNR 8
.
43
7
6
8
.
1
6
8
.
1
log
10 10 =

+
=
+
=

LINE CODES
○Line code is the electrical representation of the
binary stream.
○Line codes often used the terminology nonreturn-
to-zero (NRZ) and return-to-zero (RZ)
○Return-to-zero implies that the pulse shape used
represent the bit always return to the 0 volts or the
neutral level before the end of the bit.
○Nonreturn-to-zero indicates that the pulse does not
necessarily return to the neutral level before the
end of the bit.

LINE CODES
○A good Line Code should have the following properties
○Timing Recovery for synchronization
● The receiver should be able to recover the timing from the
transmitted signal.
● Long series of ones and zeros could create timing problems.
○Low probability of bit error for the transmitted power
● The line code must show lowest probability of error for the
given bandwidth and transmitted power.
○Spectrum must be suitable for the channel.
● In some cases DC component can not be transmitted through
the channel and it should be avoided.
● The line code must use the minimum channel bandwidth.

LINE CODES
○Unipolar Nonreturn-to-Zero (NRZ) Signaling:
● Symbol 1: transmitting a pulse amplitude A for the duration
of the symbol.
● Symbol 0: switching off the pulse
○This line code is also referred to as on-off signaling.
○A disadvantage of this signaling is the waste of power
due to the transmitted DC

LINE CODES
○Polar Nonreturn-to-Zero(NRZ) Signaling:
● Symbol 1: transmitting a pulse amplitude +A
● Symbol 0: transmitting a pulse amplitude -A
○This code is easy to generate and is more power-
efficient than its unipolar counterpart.

LINE CODES
○Unipolar Return-to-Zero (RZ) Signaling:
● Symbol 1: A rectangular pulse of amplitude A and half-symbol
width.
● Symbol 0: transmitting no pulse
○In the power spectrum of this line code there is a delta
function at f=0, ±1/Tb, which can be used for bit-timing
recovery at the receiver.
○This line code requires 3dB more power than its polar
version for the same probability of symbol error.

LINE CODES
○Bipolar Return-to-Zero (BRZ) Signaling:
● Symbol 1: Positive and negative pulses of equal amplitude
(+A and –A) are used alternately
● Symbol 0: transmitting no pulse
○This signals has no DC component and relatively
insignificant low-frequency components if symbol 1
and symbol 0 has equal probabilities..
○This line code is also called alternate mark inversion
(AMI) signaling.

LINE CODES
○Split-Phase (Manchester Code):
● Symbol 1: a positive pulse of amplitude A followed by a
negative pulse of amplitude –A with both pulses being a
half-symbol wide.
● Symbol 0: the polarities of these two pulses are reversed.
○The manchester code suppresses the DC component
and has relatively insignificant low-frequency
components regardless of the signal statistics.

LINE CODES

POWER SPECTRUM OF DIFFERENT LINE CODES
Unipolar NRZ
Polar NRZ
R=1/Tb

Unipolar RZ
Bipolar RZ
R=1/Tb

Manchester
R=1/Tb

DIFFERENTIAL ENCODING
○In this method, the information is encoded at the
signal transitions.
○Since the information is coded at the transitions ,
this encoding requires a reference bit before
starting the encoding process.
○The original binary information is recovered, by
comparing the polarity of adjacent binary symbols.
1
−

= k
k
k d
m
d
1
−

= k
k
k d
d
m

REGENERATION
○In regenerator, the incoming noisy signal is recovered and
retransmitted again as a clean signal.
○In a regenerative repeater three basic operation is
performed.
● equalization: to compensate the effects of distortions
produced by the channel
● timing: used to sample the signal when the signal-to-noise ratio
is at its maximum.
● decision making: the extracted sample is compared with a
predetermined threshold. If threshold is exceeded, a clean new
pulse representing symbol 1 is transmitted. Otherwise, a clean
pulse representing symbol 0 is transmitted.
○Therefore, the accumulation of noise and distortion is
prevented.

DECODING AND FILTERING
○The received clean pulses are regrouped into code
words and mapped into a quantized PAM signal.
○Decoding process is basically linearly summing all the
pulses in the code word, with each pulse being weighted
by its place value (20, 21, 22, 23,…,2R-1) in the code where
R is the number of bits per sample.
○The message signal is recovered by passing the decoder
output through a low-pass filter whose cutoff frequency
is equal to the message bandwidth W.
○If there is no error occurred on the transmission path,
the recovered signal includes no noise except the
quantization noise.

MULTIPLEXING
○In PCM, usually different message signals are
multiplexed to utilize the channel efficiently.
○As the number of independent message sources is
increased, the time interval that is allocated to each
source is reduced. Therefore, the pulse widths are
reduced to accommodate the increased number of
channels.
○The narrow pulses are difficult to produce and
impairments on the channel interfere with the
proper operation.
○Therefore, the number of independent message
source is restricted depending on the channel and
system characteristics.

EXAMPLE
○T1 Carrier system is designed to accommodate 24
voice channels which is limited to a band from 300
to 3100Hz.
○Since W is 3100Hz the Nyquist rate is 6200Hz.
○The voice signals are sampled at 8KHz.
○-law with the constant =255 is used and total 255
amplitude levels are used.
○Each frame has also an extra bit for synchronization.
○Calculate the channel transmission rate.

EXAMPLE - SOLUTION
○Each frame consists of
● (24 x 8) + 1 = 193 bits
○The frame period is 1/8KHz=125s
○The duration for each bit is
● 125s/193 bits = 0.647s
○The transmission rate is
● 1/0.647s=1.544 megabits/second

DELTA MODULATION
○In Delta Modulation (DM), the message signal is
oversampled (at much higher rate than the Nyquist
rate) to increase the correlation between adjacent
samples.
○DM provides a staircase approximation to the
oversampled version of the message signal.
○The difference between the input and
approximation is quantized into only two levels, ±.

DELTA MODULATION

DELTA MODULATION
○The error between the current sample of the message
signal and previous approximated signal is
○The error signal is quantized and added to the previous
approximated signal to generate current output.
○The quantized error signal is transmitted with a rate of
fs=1/Ts, where fs is the sampling frequency.
( ) ( ) ( )
s
s
q
s
s T
nT
m
nT
m
nT
e −
−
=
( ) ( )
 
( ) ( ) ( )
s
q
s
s
q
s
q
s
s
q
nT
e
T
nT
m
nT
m
nT
e
nT
e
+
−
=

= sgn

DELTA MODULATION
○Delta Modulation transmitter consists of
comparator, quantizer and accumulator.
○The comparator compares two inputs and the
quantizer consists of a hard-limiter.
○The quantizer output is applied to an accumulator,
producing the result
( ) ( )
  ( )

 =
=
=

=
n
i
s
q
n
i
s
s
q iT
e
iT
e
nT
m
1
1
sgn

DELTA MODULATION
○In the receiver, positive and negative pulses are
passed through an accumulator to generate the
approximated message signal.
○The out-of-band quantization noise in the
approximated message signal is rejected using a
low-pass filter.

DELTA MODULATION
○The input at the quantizer is
where q(nTs) is the quantization error.
○If we assume that the quantization error small
enough the error is first backward difference of the
input signal.
○First backward difference can be seen as the digital
approximation of the derivative of the input signal.
( ) ( ) ( ) ( )
s
s
s
s
s
s T
nT
q
T
nT
m
nT
m
nT
e −
−
−
−
=
( ) ( ) ( )
s
s
s
s T
nT
m
nT
m
nT
e −
−
=

DELTA MODULATION
○Delta modulation has two types of error
● Slope Overload Distortion
● Granular Noise
○Slope overload Distortion is due to limited slope of the
delta modulation. Maximum slope of m(t) must satisfy
the following condition to prevent the slope overload
distortion.
○When the input signal does not change significantly, the
DM output changes between ±. This noise is inversely
proportional to the step size.
( )
dt
t
dm
TS
max



DELTA MODULATION
○DM requires high step size to reduce the slope
overload distortion and low step size to reduce the
granular noise.
○To obtain optimum performance an adaptive DM
modulator is needed.

EXAMPLE
○A linear delta modulator is designed to operate on
speech signals limited to 3.4KHz, The specifications
of the modulator are as follows
● Sampling rate=10 times of the Nyquist rate
● Step size =100mV
○The modulator is tested with a 1KHz sinusoidal
signal.
○Determine the maximum amplitude of this test
signal permissible to avoid slope overload.

EXAMPLE - SOLUTION
○The maximum slope of the signal is 2fA.
○Consequently, the maximum change during a
sample period is approximately 2AfTs.
○To prevent slope overload, we require
( ) ( )
V
A
A
KHz
KHz
A
AfT
mV s
08
.
1
or
092
.
0
68
/
1
2
2
100

=
=



( )
dt
t
dm
TS
max



DELTA SIGMA MODULATION
○The signal at the quantizer input is the derivative of
the message signal.
○Since the receiver must accumulate the incoming
signal, it will also accumulate the noise occurred
during the transmission.
○This drawback can be mitigated by integrating
message signal prior to applying it to the Delta
Modulator.
○The integration prior the DM improves the low-
frequency response, since the low-frequency
content is amplified by the integrator.
● Design of the receiver also is simpler than DM receiver.

○A delta modulation scheme that incorporates
integration at its input is called delta-sigma
modulation (D-M).
○Since the integration performed before the delta
modulation, this scheme should be called as sigma-
delta modulation. However, in the literature it is
usually called as delta-sigma modulation.
○Let’s consider the continuous-time form of the delta
modulator, where accumulator is replaced by an
integrator.

○Since the integration of the message signal is
cancelled by the differentiation in conventional
delta modulator, the receiver consists of a low-pass
filter.

○To simplify the design, two integrators are
combined in to a single integrator before the
quantizer.

○In a PCM system that uses 8KHz sampling rate with
an 8-bit representation requires 64KHz symbol rate.
○On the other hand, delta modulation gives the same
performance for 16KHz to 32KHz sampling
frequency for desired voice quality. Therefore, DM
provide bandwidth saving of 50% to 75% over PCM
at the expense of a more complicated
implementation.
○State-of-the-art techniques use correlated
techniques to reduce the bandwidth from 64kbps to
2.4kbps depending on the voice quality.

BASEBAND TRANSMISSION OF DIGITAL SIGNALS
○Digital data have a broad spectrum with a significant
low-frequency content.
○Baseband transmission of digital data requires the
use of low-pass channel with a bandwidth large
enough to accommodate the essential frequency
content of the data stream.
○Typically, the channel is not an ideal low-pass filter.
This causes that each received pulse is affected by
adjacent pulses. This results in error which is called
intersymbol interference (ISI)
○ISI is minimized by controlling the pulse shape of the
overall system.

BASEBAND TRANSMISSION OF DIGITAL SIGNALS
○Another noise source is the thermal noise on the
channel or the receiver.
○Noise and ISI happens in the system simultaneously.
○To understand the effect of Noise and ISI, they will
be considered separately.
○First, the detection of a known waveform in the
presence of white noise will be analyzed.
○Later effect of ISI will be analyzed.

BASEBAND PULSES
○Line codes are used to transmit any stream of binary
data.
○For example
● Unipolar NRZ
● Polar NRZ
● Unipolar RZ
● Bipolar RZ
● Manchester line code
○Each code has its advantages and disadvantages.

BASEBAND PULSES
○The power spectra of each code is calculated
assuming
● There are long sequences of random bits where 0 and 1
are equally probable.
● Since the bits are randomly distributed, the resulting line
code and its spectrum is also random.
○Assuming the channel frequency response is ideal in
other words it has little effect on the shape of the
transmitted pulse.
○In this ideal case the transmitted pulse g(t) for each
bit is only affected by the additive noise at the
receiver input.

BASEBAND PULSES
○A line coder converts the digital data to physical
waveform.
○where p(t) is the pulse shape and Tb is the bit period.
○Each line code is described by a symbol mapping
function ak and a pulse shape p(t):
Line
Coder
Digital Data ak Physical Waveform
x(t)
( ) ( )


−
=
−
=
k
b
k kT
t
p
a
t
x

POWER SPECTRA OF LINE CODES
○The Power Spectrum Density (PSD) of line codes
depends on
● the pulse shape
● the statistical properties of the data
○The PSD of output signal is
○where Sx(f) is the PSD of the output signal, P(f) is the
Fourier Transform of p(t) and S(f) is the PSD at
impulse modulator output.
Impulse
Modulator
Digital
Data ak
Pulse
Filter p(t)
( ) ( )


−
=
−
=
k
b
k kT
t
p
a
t
x
( ) ( )


−
=
−
=
k
b
k kT
t
a
t
x 

( ) ( ) ( )
f
S
f
P
f
Sx 
2
=

○Calculating the Fourier Transform of p(t) is
straightforward.
○For example, the spectrum of the rectangular pulse
p(t)
Tb/2
-Tb/2
( )
( )
( )
( )
b
b
b
b
b
fT
T
fT
fT
T
f
P
T
t
t
p
sinc
sin
otherwise
0
2
/
1
=
=


 
=



○Let’s derive the PSD of the random signal x(t). First,
the autocorrelation of x(t) is
○Since x(t) consists of impulses
( ) ( ) ( )

−

→
+
=
2
/
2
/
1
lim
T
T
T
dt
t
x
t
x
T
R 
 


( ) ( ) ( )
( ) 

−
=
+

→

−
=
=
−
=
N
N
k
n
k
k
N
n
b
b
a
a
N
n
R
nT
n
R
T
R
1
lim
where
1





○The Fourier transform of the autocorrelation gives
the PSD
( ) ( )
( ) ( ) ( )
( ) ( )
( ) b
fnT
j
n
b
f
j
b
n
b
f
j
n
b
b
f
j
e
n
R
T
d
e
nT
n
R
T
d
e
nT
n
R
T
f
S
d
e
R
f
S


















2
2
2
2
1
1
1
−

−
=


−
−

−
=


−
−

−
=


−
−



 

=
−
=
−
=
=

○For real signals R(n)=R(-n)
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( )( )
( ) ( ) ( )





+
=






+
+
=






+
−
+
=






+
+
=
=








=
−

=
−

=

=
−

=
−
−
−
=
−

−
=
b
n
b
fnT
j
fnT
j
n
b
fnT
j
n
fnT
j
n
b
fnT
j
n
fnT
j
n
b
fnT
j
n
b
fnT
n
R
R
T
e
e
n
R
R
T
e
n
R
e
n
R
R
T
e
n
R
e
n
R
R
T
e
n
R
T
f
S
b
b
b
b
b
b
b









2
cos
2
0
1
0
1
0
1
0
1
1
1
2
2
1
2
1
2
1
2
1
2
1
2

○For a given line code the autocorrelation of the bit
stream is calculated using
○where M is the number of possible values that akak+n
can take on, Ri is the ith value of akak+n, pi is the
probability that Ri occurs.
( )   
=
+ =
=
M
i
i
i
n
k
k p
R
a
a
E
n
R
1

○For polar binary,
○n=0
● ak=ak+n = -A for symbol 0 and +A for symbol 1
● probabilities are p0=p1=0.5
● R(0)=(-A*-A)*0.5+(A*A)*0.5=A2
○n0
● [akak+n ]=[-A*-A], [-A*A],[A*-A], [A*A]
● p00=0.25, p01=0.25, p10=0.25, p11=0.25
● R(n0)=0
( ) ( ) ( ) ( )
b
b
n
b T
A
fnT
n
R
R
T
f
S
2
1
2
cos
2
0
1
=






+
= 

=



○For bipolar binary
○ak’s can be 0, -A, +A
○n=0
● R(0)=(-A*-A)*0.25+(A*A)*0.25+(0*0)*0.5=A2/2
○n=1
● R(1)=(-A*A)*0.125+(A*-A)*0.125+(A*0)*0.125+(-
A*0)*0.125+(0*A)*0.125+(0*-A)*0.125+(0*0)*0.25=-
A2/4
○n2 ➔ R(n)=0
( ) ( ) ( ) ( ) ( )
( )
( ) ( )
b
b
b
b
b
b
b
n
b
fnT
T
A
fnT
T
A
fnT
T
A
fnT
n
R
R
T
f
S





2
2
2
2
1
sin
2
cos
1
2
2
cos
2
1
2
2
1
2
cos
2
0
1
=
−
=













 −
+
=






+
= 

=

○Till now, the line codes we analyzed has zero mean
value.
○Sometimes the line code has non-zero mean value
due to
● the line coding scheme, e.g., unipolar
● the probability distribution of the data
● incorrect signaling voltages owing to faults
○The non-zero mean results in that R(n) has finite
values for n in the range ±.

○For unipolar binary
○ak’s can be 0, +A
○n=0
● R(0)=(0*0)*0.5+(A*A)*0.5=A2/2
○n  1
● R(1)=(0*A)*0.25+(A*0)*0.25+(A*A)*0.25+(0*0)*0.25=A2/
4
( ) ( )














−
+
=






+
=






+
=
=





−
=

−
=
−

−
=
−
−

−
=
n b
b
b
n
fnT
j
b
n
fnT
j
b
fnT
j
n
b
T
n
f
T
T
A
e
T
A
e
T
A
e
n
R
T
f
S
b
b
b






2
1
4
1
4
4
1
4
1
1
2
2
2
2
2
2

Unipolar NRZ
( ) ( ) ( )







+
= f
T
fT
T
A
f
S
b
b
b

1
1
sinc
4
2
2
R=1/Tb
( ) 







=


+
=
b
k
T
t
t
p
A
a
rect
0
symbol
0
1
symbol

Polar NRZ
R=1/Tb
( ) ( )
b
b fT
T
A
f
S 2
2
sinc
=
( ) 







=



−
+
=
b
k
T
t
t
p
A
A
a
rect
0
symbol
1
symbol

Unipolar RZ
R=1/Tb
( ) 













−
+






= 

−
=
n b
b
b
b
T
n
f
T
fT
T
A
f
S 
1
1
2
sinc
16
2
2
( ) 







=


+
=
2
/
rect
0
symbol
0
1
symbol
b
k
T
t
t
p
A
a

Bipolar RZ
R=1/Tb
( ) ( )
b
b
b
fT
fT
T
A
f
S 
2
2
2
sin
2
sinc
4






=
( ) 







=





+
−
−
+
=
2
/
rect
A
is
mark
last
and
1
symbol
A
is
mark
last
and
1
symbol
0
symbol
0
b
k
T
t
t
p
A
A
a

Manchester
R=1/Tb
( ) 











=
2
sin
2
sinc 2
2
2 b
b
b
fT
fT
T
A
f
S

( ) 






 −
−







 +
=



−
+
=
2
/
4
/
rect
2
/
4
/
rect
0
symbol
1
symbol
b
b
b
b
k
T
T
t
T
T
t
t
p
A
A
a
p(t)

MATCHED FILTER
○Consider the receiver model with a LTI filter of
impulse response h(t).
○The filter input consists of a pulse signal g(t)
corrupted by additive noise w(t)
○where T is the observation interval.
( ) ( ) ( ) T
t
t
w
t
g
t
x 

+
= 0

MATCHED FILTER
○g(t) is the pulse signal which represents symbol 1 or
0.
○w(t) is the white noise process of zero mean and
power spectral density N0/2.
○It is assumed that the receiver has knowledge of the
pulse waveform but has no idea of the white noise.
○To detect optimally the received pulse signal, the
filter must minimize the noise and maximize the
pulse power.
○In other words, the signal power to the noise ratio
at the filter output is maximized at t=T.

MATCHED FILTER
○Since the filter is an LTI system, the filter output
○The signal to noise ratio at the filter output
where |g0(T)|2 is the instantaneous power of the
output signal and E is the statistical expectation
operator, E[n2(t)] is the average output noise power.
○For optimum performance this ration is maximized.
( ) ( ) ( )
t
n
t
g
t
y +
= 0
( )
( )
 
t
n
E
T
g
2
2
0
=


MATCHED FILTER
○Let G(f) is the Fourier transform of g(t) and H(f) is
the transfer function of the filter.
○g0(t) can be calculated using inverse Fourier
transform.
○Hence the filter output is sampled at t=T, in the
absence of noise
( ) ( ) ( )



−
= df
e
f
G
f
H
t
g ft
j 
2
0
( ) ( ) ( )
2
2
2
0 


−
= df
e
f
G
f
H
T
g fT
j 

MATCHED FILTER
○Since the noise w(t) is white, its PSD will be flat and
equal to N0/2.
○This white noise is filtered through H(f) and the
output noise density becomes
○The average noise power is
( ) ( )2
0
2
f
H
N
f
SN =
( )
  ( )
( )




−


−
=
=
df
f
H
N
df
f
S
t
n
E N
2
0
2
2

MATCHED FILTER
○The signal to noise ratio becomes
○The problem is the to find the H(f) that maximizes
this ratio for a given G(f).
○To find a solution to this optimization problem, we
apply the Schwarz’s inequality.
( ) ( )
( )




−


−
=
df
f
H
N
df
e
f
G
f
H fT
j
2
0
2
2
2



MATCHED FILTER
○Schwarz’s inequality states for two complex
functions satisfying following conditions
( ) ( )
( ) ( ) ( ) ( )
( ) ( )
x
k
x
dx
x
dx
x
dx
x
x
dx
x
dx
x
*
2
1
2
2
2
1
2
2
1
2
2
2
1
if
holds








=












−


−


−


−


−

MATCHED FILTER
○Using the Schwarz’s inequality, the numerator of the
signal to noise ratio is rewritten as follows
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( )






−


−


−


−


=
=
df
e
f
G
N
df
e
f
G
df
f
H
df
e
f
G
f
H
e
f
G
f
f
H
f
fT
j
fT
j
fT
j
fT
j
2
2
0
2
2
2
2
2
2
2
1
2
ip
relationsh
this
using








MATCHED FILTER
○The Schwarz’s inequality holds if the complex
functions are complex conjugate
where k is the scaling factor.
○Except k and e-j2fT term the transfer function of the
optimum filter is the same as the complex conjugate
of the spectrum of the input signal.
( ) ( ) fT
j
opt e
f
kG
f
H 
2
* −
=

MATCHED FILTER
○Assuming g(t) is real [G*(f)=G(f)], the impulse
response of the filter is
○The impulse response of the optimum filter is
scaled, time-reversed and delayed version of the
input signal g(t).
( ) ( ) ( )
( ) ( )
( )
t
T
kg
df
e
f
G
k
df
e
f
G
k
t
h
t
T
f
j
t
T
f
j
opt
−
=
−
=
=




−
−
−


−
−
−


2
2
*

PROPERTIES OF THE MATCHED FILTER
○The Fourier transform of the g(t) at the matched
filter output is
○The filter output at time t=T
○where E is the energy of the pulse signal g(t)
( ) ( ) ( )
( ) ( )
( ) fT
j
fT
j
opt
e
f
G
k
e
f
G
f
kG
f
G
f
H
f
G


2
2
2
*
0
−
−
=
=
=
( ) ( )
( ) kE
df
f
G
k
df
e
f
G
k
T
g fT
j
=
=
=




−


−
2
2
0
0


PROPERTIES OF THE MATCHED FILTER
○The average noise energy at the output of matched
filter is
○The peak of signal to noise ratio is
○The dependence of the signal to noise ratio on the
waveform of g(t) has been completely removed by
the matched filter.
( )
  ( )
2
2
0
2
0
2
2
E
N
k
df
f
G
N
k
t
n
E
=
= 


−
( )
0
0
2
2
max
2
2
N
E
E
N
k
kE
=
=


EXAMPLE
○Consider g(t) is a rectangular pulse of amplitude A and
duration T.
○What is the matched filter impulse response?
○Draw the matched filter output.
○What is the maximum value of output signal g0(t)?

EXAMPLE - SOLUTION
○The impulse response of the matched filter has
exactly the same waveform as the signal itself.
○The maximum value of the output signal g0(t) is
equals to kA2T, which is the energy of the input
signal scaled by the factor k. This maximum value
occurs at t=T.

PROBABILITY OF ERROR DUE TO NOISE
○Since the matched filter is the optimum detector of
a known pulse in additive noise, the probability of
error rate due to the noise can be calculated.
○Let’s consider polar NRZ signaling, where symbols 1
and 0 are represented by positive and negative
pulses by equal amplitude.
○The received signal in additive noise is
( )
( )
( )



+
−
+
+
=
sent
was
0
symbol
sent
was
1
symbol
t
w
A
t
w
A
t
x

○Let’s assume that the receiver knows the starting
and ending times of each transmitted pulse.
○The receiver samples the output of the matched
filter at t=Tb and decides whether the transmitted
symbol is 0 or 1 from the received noisy signal x(t)

○In decision device, the sampled value is compared
with threshold value .
● If the sampled signal is above  the receiver assumes
that the transmitted symbol is 1.
● If the sampled signal is below  the receiver assumes
that the transmitted symbol is 0.
● If the sampled signal is equal  the receiver makes a
guess so that the outcome does not alter the average
probability of error.
○There are two possible kinds of errors
● Symbol 1 is chosen when actually 0 was transmitted.
● Symbol 0 is chosen when actually 1 was transmitted.
○To determine the probability of error, these two
situations should be analyzed separately.

○Suppose that symbol 0 was sent, then the received
signal is
○The signal at the matched filter output sampled at
t=Tb becomes
○which represents the random variable Y
( ) ( ) b
T
t
t
w
A
t
x 

+
−
= 0
( )
( )


+
−
=
=
b
b
T
b
T
dt
t
w
T
A
dt
t
x
y
0
0
1

○Since the additive noise is Gaussian distributed, the
variance of the random variable Y is
RW(t,u) is the autocorrelation function of the white
noise w(t)
( )
 
( ) ( )
( ) ( )
 
( )
 
 
 
=
=








=
+
=
b b
b b
b b
T T
W
b
T T
b
T T
b
Y
dtdu
u
t
R
T
dtdu
u
w
t
w
E
T
dtdu
u
w
t
w
E
T
A
Y
E
0 0
0 0
0 0
2
2
,
1
1
1


○Since the power spectral density of the white noise
is N0/2, RW(t,u) is
○The variance of Y becomes
( ) ( )
u
t
N
u
t
RW −
= 
2
, 0
( )
b
T T
b
Y
T
N
dtdu
u
t
N
T
b b
2
2
1
0
0 0
0
2
=
−
=   


○The probability density function of the random
variable X which has a Gaussian distribution is
○Therefore the probability density function of the
random variable Y. given that symbol 0 has sent
( )
( )
2
2
2
2
1 X
X
x
X
X e
x
f 



−
−
=
( )
( )
b
T
N
A
y
b
Y e
T
N
y
f /
0
0
2
/
1
0
|
+
−
=


○The probability density function is plotted below/ Pe0
denotes the conditional probability of error, given
that symbol 0 has sent.
○The shaded area from  to infinity corresponds to the
range of values that are assumed as symbol 1.
○In the absence of noise, the decision is always symbol
0 for the sampled value of –A.

○The probability of error on the condition that
symbol 0 has sent is
( )
( )
( )


 +
−

=
=

=




dy
e
T
N
dy
y
f
y
P
P
b
T
N
A
y
b
Y
e
/
0
0
0
2
/
1
0
|
sent
was
0
symbol
|

○The value of he threshold  can be assigned with
the probabilities of symbols 0 and 1 denoted by p0
and p1.
○It is clear that the sum of p0 and p1 must be 1.
○If we assume that symbols 0 and 1 are occurred
with equal probabilities (p0=p1=0.5), it reasonable to
assume the threshold halfway between +A (symbol
1) and –A (symbol 0).
0
=


○The conditional probability of error when symbol 0
was sent becomes
○Let’s define a new variable z
( )

 +
−
=
0
/
0
0
0
2
/
1
dy
e
T
N
P b
T
N
A
y
b
e

b
T
N
A
y
z
2
/
0
+
=

○Let’s reformulate Pe0
○where Eb is the transmitted signal energy per bit


−
=
0
2
/
2
2
0
2
1
N
E
z
e
b
dz
e
P

b
b T
A
E 2
=

THE Q-FUNCTION
○The Q-function is used by communication engineers
to determine the area under the tails of the
Gaussian distribution.
○Error function is the twice of the area under a
normalized Gaussian from 0 to u. The
complementary error function is defined as one
minus the error function
The Q-function and complementary error function
is related to each other
( ) 

−
=
u
z
dz
e
u
2
2
erfc

( ) 





=
2
erfc
2
1 u
u
Q

○The conditional probability of error Pe0 can be
written in terms of Q-function.
○For the conditional probability of error Pe1, the
mean is +A








=
0
0
2
N
E
Q
P b
e
( )
( )
b
T
N
A
y
b
Y e
T
N
y
f /
0
0
2
/
1
1
|
−
−
=


○The probability of error Pe1 is area from - to .
○Setting the threshold to =0 and putting
○results in Pe1=Pe0
( )
( )
( )



−
−
−

−
=
=

=




dy
e
T
N
dy
y
f
y
P
P
b
T
N
A
y
b
Y
e
/
0
1
0
2
/
1
1
|
sent
was
1
symbol
|
b
T
N
A
y
z
2
/
0
−
−
=

○The average probability of symbol error Pe is
○Since Pe0=Pe1 and p0=p1=0.5
○The average probability of symbol error with binary
signaling only depends on Eb/N0, the ratio of the
transmitted signal energy per bit to the noise
spectral density.
1
1
0
0 e
e
e P
p
P
p
P +
=








=
=
=
0
1
0
2
N
E
Q
P
P
P
P
b
e
e
e
e

0 2 4 6 8 10
10
-6
10
-5
10
-4
10
-3
10
-2
10
-1
Eb
/N0
(dB)
Probability
of
Error
P
e
Matlab Code:
clear all;
close all;
EbN0_dB=0:1e-3:10;
EbN0=10.^(EbN0_dB/10);
Pe=0.5*erfc(sqrt(2*EbN0)/sqrt(2));
figure(1);
semilogy(EbN0_dB,Pe,'linewidth',2);
xlabel('E_b/N_0 (dB)');
ylabel('Probability of Error P_e');
grid;
This waterfall curve makes the digital systems superior compared to analog
systems if Eb/N0 is above few dBs.

INTERSYMBOL INTERFERENCE
○Intersymbol Interference (ISI) happens when the
channel has a frequency dependent amplitude
spectrum.
○The simplest case is band-limited channel.
● For example, a brick-wall band-limited channel passes all
frequencies |f|<W without distortion and blocks all
frequencies |f|>W.
○Consider a polar signaling where



−
+
=
sent
was
0
symbol
1
sent
was
1
symbol
1
k
a

○Once the impulse modulator output is passed
through the transmit filter with the impulse
response of the pulse g(t), the transmitted signal
becomes
○s(t) passes through the channel with impulse
response of h(t) added white Gaussian noise and
received by the receiver.
○The receiver passes the received signal through the
receive filter, which is the matched filter for the
optimum noise performance.
( ) ( )
 −
=
k
b
k kT
t
g
a
t
s

○The output of the receive filter is
sampled at t=Tb and the decision device decides the
bits.
○The receive filter output is
( ) ( ) ( )
t
n
kT
t
p
a
t
y
k
b
k +
−
= 

( ) ( )
  ( )
( )
  ( )
i
i
k
k
b
k
i
i
k
b
k
i
t
n
T
k
i
p
a
a
t
n
T
k
i
p
a
t
y
+
−
+
=
+
−
=




−
=

−
=



○The first term is the contribution of the ith
transmitted bit.
○The second term represents the residual effect of
all other transmitted bits on the decoding of ith bit.
○This residual effect due to the occurrence of pulses
before and after the sampling instant ti is called
intersymbol interference (ISI).
○The last term n(ti) represents the noise sample at the
time ti
○When the signal-to-noise ratio is high, the system is
limited by ISI rather than noise.

THE TELEPHONE CHANNEL
○The frequency response of a telephone channel is
shown below.

○The properties of this channel
● The pass-band of the channel cuts of rapidly above
3.5KHz. Therefore, we should use a line code with a
narrow spectrum, so we can maximize the data rate.
● The channel does not pass DC. It is preferable to use a
line code that has no DC component such as bipolar
signaling or Manchester code.
○These two requirements are contradictory.
● The polar NRZ has narrow spectrum but has DC
● The Manchester has no DC, but requires higher
bandwidth.

○For the data rate 1600bps, the signals are shown below.
○a is NRZ line code and b is Manchester line code.
○Since NRZ requires DC, the signal drifts to zero volt
especially when there is long strings of the symbols of the
same polarity.
○Manchester code has no DC drift but has distortion.

○For the data rate 3200bps, the DC drift in NRZ is
less evident but there is still considerable signal
distortion.
○The Manchester code has significant spectral
components outside the band limits of this channel.

EYE PATTERN
○Eye pattern is an operational too for evaluating the
effects of ISI.
○The eye pattern is defined as the synchronized
superposition of all possible realizations of the signal
of interest.
○The eye pattern derives its name from the fact that
it resembles the human eye for binary waves.
○The interior region of the eye pattern is called eye
opening.

EYE PATTERN

EYE PATTERN
○The width of the eye opening defines the time
interval over which the received signal can be
sampled without error from intersymbol
interference. The preferred time for sampling is the
instant of time at which the eye is open the widest.
○The sensitivity of the system to timing errors is
determined by the rate of closure of the eye as the
sampling time is varied.
○The height of the eye opening defines the noise
margin of the system.

EYE PATTERN
○When the effect of ISI is severe, traces from the upper
portion of the eye pattern cross traces from the lower
portion and the eye is completely closed.
● In this situation, it is impossible to avoid errors due to
the presence of intersymbol interference in the
system.
○For M-ary systems, the eye pattern contains (M-1) eye
openings stacked of vertically one on the other, where M
is the number of discrete amplitude levels used to
construct the transmitted signal.
○If the system and channel linear with random noise,
these eye openings are identical. In practice, these
openings have asymmetries due to the nonlinearities of
the system and channel.

NYQUIST’S CRITERION FOR DISTORTIONLESS
TRANSMISSION
○The extraction of the received bits involves
sampling the output y(t) at time t=iTb.
○The decoding requires that the weighted pulse
contribution akp(iTb-kTb) for k=i be free from ISI due
to the overlapping tails of all other weighted pulse
contributions represented by ki
○This requires that the overall pulse p(t) must satisfy
the following condition
○Ignoring the noise term yields to y(ti)=ai for all i
( )




=
=
−
k
i
k
i
kT
iT
p b
b
0
1

TRANSMISSION
○Transforming the condition for ISI free
communication to the frequency domain is
informative from design perspective.
○Consider the samples of p(nTb). Since sampling in
the time domain produces periodicity in the
frequency domain.
where Rb=1/Tb is the bit rate in bits per second.
○P(f) is the Fourier transform of an infinite periodic
sequence of delta functions whose areas are
weighted by respective sample value of p(t)
( ) ( )


−
=
−
=
n
b
b nR
f
P
R
f
P

TRANSMISSION
○The Fourier Transform of the sampled values
○Let m=i-k. For i=k ➔ m=0
○Applying the condition for ISI free communication
○Normalizing the pulses to p(0)=1 results in
( ) ( ) ( )
 
 


−
−

−
=
−
= dt
e
mT
t
mT
p
f
P ft
m
b
b

  2
( ) ( ) ( ) ( )
0
0 2
p
dt
e
t
p
f
P ft
=
= 


−
− 
 
( ) b
n
b T
nR
f
P =
−


−
=

TRANSMISSION
○The frequency function P(f) eliminates the ISI for
samples taken at intervals Tb if this equation is
satisfied.
○Note that P(f) refers to the overall system,
incorporating the transmit filter, the channel and the
receive filter
( ) b
n
b T
nR
f
P =
−


−
=

IDEAL NYQUIST CHANNEL
○The simplest way to satisfy ISI free communication
is using the P(f) specified in the form of rectangular
function.
○where the overall system bandwidth W is defined
by
( )






=








−
=
W
f
W
W
f
W
f
W
W
f
P
2
rect
2
1
0
2
1
b
b
T
R
W
2
1
2
=
=

○Using the inverse Fourier Transform the pulse shape
is found as
○The special value of the bit rate Rb=2W is called the
Nyquist rate, W is called Nyquist bandwidth.
( ) ( ) ( )
Wt
Wt
Wt
t
p 2
sinc
2
2
sin
=
=



○The samples do not
interfere each other
when they are sampled
at integer multiple of Tb

○Although ideal Nyquist channel achieves minimum
bandwidth and solves the ISI problem, there are two
practical difficulties
● The amplitude characteristics of P(f) must be flat from –
W to W and zero elsewhere. This is physically
unrealizable because of the abrupt transitions at the
band edges ±W.
● p(t) decreases as 1/|t| for large |t|, resulting in a slow rate
of decay. Therefore, there is no margin of error in
sampling times in the received.

○Let’s analyze the performance of ideal nyquist
channel in the presence of timing error t. To
simplify the analysis let’s choose the correct
sampling time ti equal to zero.
○The first term is the desired term and the remaining
term is ISI due to the timing error.
( ) ( )
( )
 
( )
( ) ( ) ( ) ( )




−

−

+

=

=
−

−

=
−

=

0
0
2
1
2
sin
2
sinc
1
2
2
2
sin
k
k
k
k
b
k b
b
k
k
b
k
k
t
W
a
t
W
t
W
a
t
y
WT
kT
t
W
kT
t
W
a
kT
t
p
a
t
y










RAISED COSINE SPECTRUM
○The practical difficulties of the ideal Nyquist channel
is mitigated by extending the channel bandwidth
from the minimum value W to be adjustable
between W and 2W.
○Let’s keep three terms of
and restrict the frequency band of interest to
[-W,W]
( ) W
T
nR
f
P b
n
b 2
=
=
−


−
=
( ) ( ) ( ) W
f
W
W
W
f
P
W
f
P
f
P 

−
=
+
+
−
+
2
1
2
2

○Many band-limited function can be found that
satisfies this requirement.
○One possible solution is the raised cosine spectrum.
○The frequency response consists of a flat portion
and a rolloff portion that is sinusoidal
( )
( )









−

−


















−
−
−


=
1
1
1
1
1
2
0
2
2
2
sin
1
4
1
0
2
1
f
W
f
f
W
f
f
f
W
W
f
W
f
f
W
f
P

W
f1
1−
=


○The parameter  is called the rolloff factor and it
indicates the excess bandwidth over the ideal
solution.
○The transmission bandwidth BT is defined by
2W-f1=W(1+)
○The time response p(t) or in other words the inverse
Fourier transform of P(f)
○p(t) consists of product of two factors.
● sinc(2Wt) guarantees the zero crossings as the ideal
Nyquist channel
● 1/|t|2 guarantees fast rate of decay
( ) ( )
  ( )






−
= 2
2
2
16
1
2
cos
2
sinc
t
W
Wt
Wt
t
p



○For =1 or in other words f1=0 is known as the full-
cosine rolloff characteristics.
○The frequency response simplifies to
○The time response p(t) become
( )




















+
=
W
f
W
f
W
f
W
f
P
2
0
2
0
2
cos
1
4
1 
( ) ( )
2
2
16
1
2
sinc
t
W
Wt
t
p
−
=

○This special case has two interesting properties
● At t=±Tb/2=±1/4W, p(t) = 0.5 ➔ the pulse width
measured at half amplitude is exactly equal to the bit
duration Tb
● There are zero crossings at t=±3Tb/2, ±5Tb/2, … in
addition to the usual zero crossings at the sampling
times t= =±Tb, ±2Tb,…
○These two properties are useful in extracting a
timing signal from the received signal for the
purpose of synchronization.
○However, the price is doubling the channel
bandwidth compared to ideal Nyquist channel.

BASEBAND M-ARY PAM TRANSMISSION
○Up to now for binary systems the pulses have two
possible amplitude levels.
○In a baseband M-ary PAM system, the pulse
amplitude modulator produces M possible
amplitude levels with M>2.
○In an M-ary system, the information source emits a
sequence of symbols from an alphabet that consists
of M symbols.
○Each amplitude level at the PAM modulator output
corresponds to a distinct symbol.
○The symbol duration T is also called as the signaling
rate of the system, which is expressed as symbols
per second or bauds.

○Let’s consider the following quaternary (M=4)
system.
○The symbol rate is 1/(2Tb), since each symbol
consists of two bits.

○The symbol duration T of the M-ary system is
related to the bit duration Tb of the equivalent
binary PAM system as
○For a given channel bandwidth, using M-ary PAM
system, log2M times more information is
transmitted than binary PAM system.
○The price we paid is the increased bit error rate
compared binary PAM system.
○To achieve the same probability of error as the
binary PAM system, the transmit power in <-ary
PAM system must be increased.
M
T
T b 2
log
=

○For M much larger than 2 and an average
probability of symbol error small compared to 1, the
transmitted power must be increased by a factor of
M2/log2M compared to binary PAM system.
○The M-ary PAM transmitter and receiver is similar
to the binary PAM transmitter and receiver.
○In transmitter, the M-ary pulse train is shaped by a
transmit filter and transmitted through a channel
which corrupts the signal with noise and ISI.

○The received signal is passed through a receive filter
and sampled at an appropriate rate in synchronism
with the transmitter.
○Each sample is compared with preset threshold
values and a decision is made as to which symbol
was transmitted.
○Obviously, in M-ary system there are M-1 threshold
levels which makes the system complicated.
○The raised cosine pulse shape, which is ISI-free for
binary signaling is also ISI-free for M-ary signaling.

TAPPED DELAY-LINE EQUALIZATION
○In theory, if the channel response is precisely
known, it is virtually possible to make the ISI very
small by using suitable transmit and receive filters.
○In practice, the channel response is not exactly
known. Therefore, the channel effects which
creates ISI must be compensated.
○The process to mitigate ISI issue is called
equalization.
○The filter used to perform this operation is called
equalizer.

○A tapped delay-line equalizer is well suited to
compensate the channel response.
○For symmetry, the total number of taps is chosen to
be (2N+1) with weights denoted by w-N,…,
w-1,w0,w1,…,wN.

○The impulse response of the tapped delay-line
equalizer is
where (t) is the Dirac delta function and the delay T
is chosen equal to the symbol duration.
○Let’s assume that this equalizer is connected in
cascade with a LTI system with a impulse response
of c(t).
( ) ( )

−
=
−
=
N
N
k
k kT
t
w
t
h 

○The impulse response of the equalized system is the
convolution of c(t) and h(t)
( ) ( ) ( )
( ) ( )
( ) ( )
( )



−
=
−
=
−
=
−
=
−

=
−

=

=
N
N
k
k
N
N
k
k
N
N
k
k
kT
t
c
w
kT
t
t
c
w
kT
t
w
t
c
t
h
t
c
t
p



○Let’s evaluate p(t) at the sampling times t=nT
○To eliminate ISI completely, the Nyquist criterion for
distortionless transmission must be satisfied.
( ) ( )
( )

−
=
−
=
N
N
k
k T
k
n
c
w
nT
p
( )




=
=
0
0
0
1
n
n
nT
p

○Since there are only (2N+1) adjustable coefficients,
the ideal condition can satisfied approximately as
follows.
○To simplify the notation, let’s denote the impulse
response at nT as
( )






=
=
=
N
n
n
nT
p

,
2
,
1
0
0
1
( )
nT
c
cn =

○Imposing the distortionless transmission condition
to the convolution sum results in (2N+1) equations
○In theory, the longer the equalizer is (N approaches
to infinity). the more closely will the equalized
system approach the ideal condition.






=
=
=

−
=
−
N
n
n
c
w
N
N
k
k
n
k

,
2
,
1
0
0
1

○These equations can also be written in matrix form.






















=












































−
−
−
+
+
−
+
−
−
−
−
−
−
−
−
−
−
−
+
−
0
0
1
0
0
1
0
1
0
1
1
2
1
0
1
2
1
1
0
1
1
2
1
0
1
2
1
1
0
















N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
w
w
w
w
w
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c

○In practice, the channel varies with time and the
equalizer coefficients must be changed according to
the channel changes.
○This operation is called adaptive equalization.
○Majority of the equalizers used in practical systems
are adaptive.

EFFECT OF ISI AND EQUALIZER IN
100 MBPS TRANSMISSION.
○In 100Mbps ethernet communication, the signal
attenuates over 100 meters especially with
frequency.
○The magnetic devices in close proximity to the
twisted pair ethernet cable effects the channel
characteristics.
○There is also a potential echo from the opposite end
of the cable.
○Since the system transmits and receives at the same
time, there is a possibility of crosstalk between the
transmit path and receive path.

○The transmitted signal

○The received signal

○The equalized signal

BAND-PASS TRANSMISSION OF DIGITAL SIGNALS
○In baseband pulse transmission, a data stream in the
form of a discrete pulse-amplitude modulated signal
is transmitted directly over a low-pass channel.
○In digital pass-band transmission, the incoming data
stream is modulated onto a carrier with fixed
frequency limits imposed by a band-pass channel of
interest.
○The modulation is performed by switching (keying)
the amplitude, frequency or phase of a sinusoidal
carrier in some fashion in accordance with the
incoming data.

BAND-PASS TRANSMISSION OF DIGITAL SIGNALS
○There are three basic signaling schemes
● ASK: Amplitude Shift Keying
● FSK: Frequency Shift Keying
● PSK: Phase Shift Keying
○FSK and PSK signals have constant envelope.
○This feature enables transmission of FSK and PSK
through nonlinear channels.
○The demodulation of these signals are categorized
into coherent demodulator and noncoherent
demodulator.
○In coherent demodulator the receiver is locked to
transmitter phase, whereas in noncoherent
demodulator there is no phase lock.

BAND-PASS TRANSMISSION MODEL
○In analog communication, the band-pass signals are
analyzed using their equivalent complex envelope
representation.
○Consider a general signal
a(t) is the envelope and (t) is the phase of the signal.
○We may rewrite this equation in terms of cosine and sine
○are called in-phase and quadrature components of g(t)
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( )
t
t
a
t
g
t
t
a
t
g
t
f
t
g
t
f
t
g
t
g
Q
I
c
Q
c
I




sin
cos
2
sin
2
cos
=
=
−
=
( ) ( ) ( )
 
t
t
f
t
a
t
g c 
 +
= 2
cos

○The in-phase and quadrature components, gI(t) and
gQ(t) are used to modulate the orthogonal carriers
cos(2fct) and sin(2fct) to produce the band-pass
signal g(t).

○It is assumed that the band-pass communication
channel has the following characteristics
● The channel is linear. The channel bandwidth is wide
enough to pass the band-pass signal without any
distortion.
● The received signal is perturbed by an additive
stationary Gaussian noise process of zero mean and
power spectral density N0/2.
● This idealized channel is called as additive white
Gaussian noise (AWGN) channel.
○For the situation, where the channel attenuates the
signal and adds noise
( ) ( ) ( )
( ) ( )
t
w
t
g
A
t
w
t
s
t
x
C +
=
+
=

○ The received signal is band-pass filtered with a filter
bandwidth of equal to the signal bandwidth.
○ This band-pass filtering operation converts the white noise to
narrowband noise n(t).
○ The in-phase and quadrature components of this band-pass
signal plus narrowband noise are downconverted using I-Q
down converter.

○The receiver will observe the complex representation of the
received signal [gI(t)+nI(t)]+j[gQ(t)+nQ(t)] for a duration of T
seconds and make the best estimate of the corresponding
transmitted signal gI(t)+jgQ(t) or equivalently the data symbol
0 or 1 for binary data.
○The hardware shown previous slide may be simplified
depending on the transmission strategy.
● Some methods use only in-phase signaling, so the quadrature path can
be removed.
● A noncoherent receiver recovers the symbols directly from the band-
pass signal without deriving in-phase and quadrature components.
● In modern receivers, the in-phase and quadrature oscillators are not
phase locked to the transmitter. This creates a phase rotation or even
a small frequency error. These problems are corrected by digital signal
processing algorithms.

TRANSMISSION OF BINARY PSK AND FSK
○Three basic signaling techniques are ASK, FSK and
PSK.
○ASK: Amplitude Shift Keying
● Symbol 1: transmitting a sinusoidal carrier wave of fixed
amplitude and fixed frequency for the bit duration Tb
seconds.
● Symbol 0: switching off the carrier for Tb seconds.

○FSK: Frequency Shift Keying
● Symbol 1: transmitting a carrier with fixed amplitude and
at f1 frequency.
● Symbol 0: transmitting a carrier with fixed amplitude and
at f0 frequency.
○A FSK signal can be generated by applying the
bipolar form of the input data to the voltage-
controlled oscillator.

○PSK: Phase Shift Keying
● When transmitting symbol 0 the carrier phase is shifted
by 180 degrees compared symbol 1 phase.
○A PSK signal may be generated by multiplying
bipolar input data the carrier wave.

○In binary PSK (BPSK), the in-phase component of
the complex baseband signal is modulated using the
input data. The quadrature component equals to
zero.
○For BPSK, the band-pass
signal is defined as
( ) ( )
( ) ( )
( )




+
=
−
=
=
t
f
A
t
f
A
t
s
t
f
A
t
s
c
C
c
C
c
C
2
cos
0
symbol
2
cos
1
symbol
2
cos
0
1

○For binary FSK, the complex baseband equivalent is
represented by gI(t) and gQ(t).
○Let’s define the carrier frequency fC as the midpoint
between f1 (frequency for symbol 1) and f0
(frequency for symbol 0), fC=(f1+f2)/2 and assuming
f1>f0 also define f=(f1-f0)/2.
( ) ( )
 
( )
 
( ) ( )
( ) ( )
 
( )
 
( ) ( )
0
symbol
Re
2
cos
1
symbol
Re
2
cos
2
2
0
2
2
1
ft
j
C
Q
I
t
f
f
j
C
c
C
ft
j
C
Q
I
t
f
f
j
C
c
C
e
A
t
jg
t
g
e
A
t
f
f
A
t
s
e
A
t
jg
t
g
e
A
t
f
f
A
t
s
c
c


−

−

+
=
+
=

−
=
=
+
=

+
=







COHERENT DETECTION OF
FSK AND PSK SIGNALS
○Let s0(t) and s1(t) denote the symbols 0 and 1,
respectively.
○In practice, in FSK signals the frequencies f0 and f1
are both large compared the bit rate 1/Tb, whereas
in PSK signals, fC is large compared with 1/Tb.
○The signal energy Eb within a bit interval Tb is
( ) ( )
2
2
0
2
1
0
2
0
b
c
T
T
b
T
A
dt
t
s
dt
t
s
E
b
b
=
=
= 


FSK AND PSK SIGNALS
○Assuming that the transmitted carrier phase and
frequency are known exactly, the receiver can be
built using two matched filters, one for s0(t) and the
other for s1(t).
Coherent receiver for FSK signals

FSK AND PSK SIGNALS
○For PSK, the receiver reduces two single path.
○Obviously, for these receivers, the receiver is
perfectly synchronized to the transmitter. In other
words, the receiver knows when the bit interval
starts and end.
Coherent receiver for PSK signals

FSK AND PSK SIGNALS
○To evaluate the performance of the receiver, we will
assume that the additive noise is white zero-mean
Gaussian noise with a spectral density of N0/2.
○The received signal is defined by
○The FSK receiver output l is given by
( ) ( ) ( )
( ) ( ) ( ) 1
symbol
0
symbol
1
0
t
w
t
s
t
x
t
w
t
s
t
x
+
=
+
=
( ) ( ) ( )
 
 −
=
b
T
dt
t
s
t
s
t
x
l
0
0
1

FSK AND PSK SIGNALS
○The output l is compared with a decision level of zero
volts. If l is greater than zero, the receiver chooses
symbol 1; otherwise it chooses symbol 0.
○Suppose that symbol 1 is transmitted, the output
○Since the noise w(t) has zero-mean, the random variable
L, whose value is l, has the conditional mean
( ) ( ) ( )
  ( ) ( ) ( )
 

 −
+
−
=
b
b T
T
dt
t
s
t
s
t
w
dt
t
s
t
s
t
s
l
0
0
1
0
0
1
1
  ( ) ( ) ( )
  ( )

−
=
−
=  1
1
Symbol
0
0
1
1 b
T
E
dt
t
s
t
s
t
s
L
E
b

FSK AND PSK SIGNALS
○The parameter  is the correlation coefficient of the signals
s0(t) and s1(t)
which has an absolute value which is less than or equal to
unity.
○Conditional mean of L, when symbol 0 is transmitted is
( ) ( )
( ) ( )
( ) ( )




=
=
b
b
b
b
T
b
T
T
T
dt
t
s
t
s
E
dt
t
s
dt
t
s
dt
t
s
t
s
0
1
0
0
2
1
0
2
0
0
1
0
1

  ( )

−
−
= 1
0
Symbol b
E
L
E

FSK AND PSK SIGNALS
○The variance of the random variable L is the same
regardless of whether symbol 1 or 0 was transmitted.
○where RW(t,u) is the autocorrelation function of w(t).
Since w(t) is white noise of spectral density N0/2
   
 
 
( ) ( ) ( ) ( )
  ( ) ( )
 
( ) ( )
  ( ) ( )
  ( )
 
 
−
−
=








−
−
=
−
=
b b
b b
T T
W
T T
dtdu
u
t
R
u
s
u
s
t
s
t
s
dtdu
u
s
u
s
t
s
t
s
u
w
t
w
E
L
E
L
E
L
Var
0 0
0
1
0
1
0 0
0
1
0
1
2
,
( ) ( )
u
t
N
u
t
RW −
= 
2
, 0

FSK AND PSK SIGNALS
○The variance of l is
  ( ) ( )
  ( ) ( )
  ( )
( ) ( )
 
( )


−
=
−
=
−
−
−
=

 
1
2
2
0
0
2
0
1
0
0 0
0
1
0
1
0
b
T
T T
E
N
dt
t
s
t
s
N
dtdu
u
t
u
s
u
s
t
s
t
s
N
l
Var
b
b b

FSK AND PSK SIGNALS
○Assuming that symbol 1 and 0 occur with equal
probabilities.
○The probability of error is
○For coherent PSK receiver, s0(t) = -s1(t), so the
correlation coefficient  = -1
○The probability of error for PSK is
( ) ( )
( )







 −
=

=

=
0
1
1
symbol
0
0
symbol
0
N
E
Q
l
P
l
P
P
b
e









=
0
2
N
E
Q
P b
e

FSK AND PSK SIGNALS
○For FSK signals, f0 and f1 are spaced far enough, so
s0(t) and s1(t) are orthogonal signals.
○Therefore the correlation coefficient  = 0.
○The probability of error become








=
0
N
E
Q
P b
e

CONTINUOUS PHASE FREQUENCY-SHIFT KEYING (CPFSK)
SIGNALS
○In FSK signals the phase information is only used for
synchronization of the receiver to the transmitter.
○CPFSK signals utilize the phase information to
improve the noise performance in the expense of
increased receiver complexity.
○CPFSK signal is defined as
where (t) is a continuous function of time.
( ) ( )
( )
t
t
f
A
t
s C
C 
 +
= 2
cos

SIGNALS
○The nominal carrier frequency fC is equal to the
arithmetic mean of the two frequencies f1 and f0
that are used to represent symbols 1 and 0.
○CPFSK signals for symbol 1 and 0 are as
○where 0t Tb. (0) is the value of (t) at t=0
2
0
1 f
f
fC
+
=
( )
( )
 
( )
 



+
+
=
0
symbol
0
2
cos
1
symbol
0
2
cos
0
1




t
f
A
t
f
A
t
s
C
C

SIGNALS
○Let’s rewrite the CPFSK signal as a function of fC
and phase as a linear function of time
○h is the deviation ratio of the FSK signal and
measured with respect to the bit rate 1/Tb.
( ) ( )
 
( )
( ) ( )
( ) ( )
( )
0
1
0
symbol
0
1
symbol
0
2
cos
f
f
T
h
t
T
h
t
t
T
h
t
t
t
t
f
A
t
s
b
b
b
C
C
−
=







−
=
+
=
=
+
=










SIGNALS
○When (t) is continuous function of time, the CPFSK
signal s(t) is also continuous.
○The main advantage of continuous phase is that the
spectral density of out-of band components falls of
at higher rate.
● In other words, CPFSK signal does not produce as much
interference outside the signal band as an FSK signal
with discontinuous phase.

SIGNALS

SIGNALS
○At time t=Tb, the phase difference is
○The transmission of symbol 1 increases the phase of
the CPFSK signal s(t) by h radians and the
transmission of symbol 0 reduces it by an equal
amount.
○The phase shift is an odd or even multiple of h
radians at odd or even multiples of the bit duration
Tb.
( ) ( )



−
=
−
0
symbol
1
symbol
0
h
h
Tb





SIGNALS
○Since all phase shifts are modulo 2, the case of
h=1/2 is particular interest, because the phase can
take only the two values ±/2 at odd multiples of Tb
and only two values 0 and  at even multiples of Tb.
○This special case with h=1/2 is called minimum shift
keying (MSK).

SIGNALS
0
1
1
Possible values of the
phase shift for the special
case h=1/2

SIGNALS
○Let’s write MSK signal in terms of in-phase and
quadrature components.
○The in-phase component with h=1/2 is
( ) ( )
  ( ) ( )
  ( )
t
f
t
A
t
f
t
A
t
s C
C
C
C 


 2
sin
sin
2
cos
cos −
=
( )
  ( )
( )
  ( )
  

















=







=
t
T
h
A
t
T
h
A
t
T
h
A
t
A
b
C
b
C
b
C
C







sin
0
sin
cos
0
cos
0
cos
cos


SIGNALS
○Since (0) is 0 or  depending on the past history of
the modulation process, the polarity of cos[(t)]
depends only upon (0).
○The in-phase component becomes
○where plus sign corresponds to (0)=0 and minus
sign corresponds to (0)=.
( )
  b
b
b
C
C T
t
T
t
T
A
t
A 

−









=
2
cos
cos



SIGNALS
○The quadrature component consists of a half-sine
pulse, whose polarity depends only on (Tb).
○where the plus sign corresponds to (Tb)=/2 and
the minus sign corresponds to (Tb)=-/2
( )
  b
b
C
C T
t
t
T
h
A
t
A 2
0
2
sin
sin 










=



SIGNALS
CPFSK signal
with h=1/2

COHERENT DETECTION OF CONTINUOUS PHASE
FREQUENCY-SHIFT KEYING (CPFSK) SIGNALS
○Assuming the additive noise w(t) at the receiver
input is white and Gaussian with zero-mean and
spectral density N0/2, the received MSK signal x(t) is
expressed as
○For the optimum detection of the phase states (0)
and (Tb), a pair of matched filters or correlators are
used.
( ) ( ) ( ) ( )
t
w
t
f
t
T
A
t
f
t
T
A
t
x C
b
C
C
b
C +


















= 



2
sin
2
sin
2
cos
2
cos

○The correlator in the in-phase channel compares x(t)
with the coherent reference signal
cos(t/2Tb)cos(2fCt) in the interval -TbtTb.
○If I1>0 the receiver chooses (0)=0, otherwise, it
chooses (0)=
( ) ( )

−








=
b
b
T
T
C
b
dt
t
f
t
T
t
x
I 

2
cos
2
cos
1

○The correlator in the quadrature channel compares
x(t) with the coherent reference signal
sin(t/2Tb)sin(2fCt) in the interval 0t2Tb.
○If I1>0 the receiver chooses (0)=-/2, otherwise, it
chooses (0)=/2
○The original binary sequence is reconstructed by
interleaving the phase decisions at the in-phase and
quadrature correlator outputs.
( ) ( )
 







=
b
T
C
b
dt
t
f
t
T
t
x
I
2
0
21 2
sin
2
sin 


○Since the phase information is used in the signal, the
probability of error performance of CPFSK signal is
equal to coherent PSK signal.
○This corresponds 3dB improvement over coherent
detection of FSK signals.








=
0
2
N
E
Q
P b
e

NONCOHERENT DETECTION OF FSK SIGNALS
○When the simplicity of the receiver is main concern,
the phase information is disregarded and
noncoherent detection is used.
○This simplicity is achieved in the expense of
probability of error degradation.
○The noncoherent detector of FSK signals is a pair of
matched filters followed by envelope detectors.
○The output of envelope detectors are sampled at Tb
and the sampled values are compared to determine
whether symbol 0 or 1 are transmitted.

○The calculation of probability of error rate for
noncohorent detection of FSK requires application
of the Rayleigh and Rician distribution functions.
○Evaluating the error rate gives
0
2
2
1 N
E
e
b
e
P
−
=

M-ARY DATA TRANSMISSION SYSTEMS
○Binary data transmission system: one of two
possible signals, s0(t) or s1(t) are sent during each bit
interval.
○M-ary data transmission system: one of M possible
signals, s0(t), s1(t), s2(t), … sM-1(t) are sent during each
bit interval.
○Usually, M is selected as M=2n, where n is an integer
number.
○Each of the M signals, s0(t), s1(t), s2(t), … sM-1(t), is
called a symbol.
○The transmission rate of these symbols is expressed
as bauds.

○In analog communication, the band-pass signals are
analyzed using their equivalent complex envelope
representation.
○The complex representation is
○where gI(t) and gQ(t) are in-phase and quadrature
components of g(t)
○gI(t) and gQ(t) are considered as the real and imaginary
parts of a complex number. Therefore, they are
represented geometrically on a complex plane.
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( )
t
t
a
t
g
t
t
a
t
g
t
f
t
g
t
f
t
g
t
g
Q
I
c
Q
c
I




sin
cos
2
sin
2
cos
=
=
−
=

○Binary ASK:
● Symbol 1:
transmitting a
sinusoidal carrier
wave of fixed
amplitude and fixed
frequency for the
bit duration Tb
seconds.
● Symbol 0: switching
off the carrier for Tb
seconds.
Symbol 1
Symbol 0

○Binary PSK:
● When
transmitting
symbol 0 the
carrier phase is
shifted by 180
degrees
compared
symbol 1 phase.
Symbol 1
Symbol 0

○M-ASK: similar to M-ary PAM at complex baseband.
00 01 11 10

○4-QAM and QPSK
00
01 11
10

○M-PSK
000
001
011
010
100
101
111
110

○M-QAM: This example is 16-QAM
○QAM: Quadrature Amplitude Modulation
0000
0001
0011
0010
0100
0101
0111
0110
1000
1001
1011
1010
1100
1101
1111
1110

○This geometric representation is called also signal
space representation.
○The set of points that characterize a given
modulation technique is referred as the
constellation for that modulation.
○M-FSK: It is also possible to define M-ary
representation for FSK signals, which is M signals
with M different frequencies.
○Often the frequencies are chosen integer multiple of
the symbol rate.
b
i
i
T
n
f
f
f =
−
=
 +1

○There are also, 64-QAM, 256-QAM, 2048-QAM
○The constellation does not have to be rectangular
shape.
8-QAM 16-QAM

QUADRIPHASE SHIFT KEYING (QPSK)
○In QPSK, one of four possible uniquely related pairs
of bits is transmitted during each signal interval.
( )
( )
( )
( ) 10
pair
bit
4
7
2
cos
2
00
pair
bit
4
5
2
cos
2
01
pair
bit
4
3
2
cos
2
11
pair
bit
4
2
cos
2
3
2
1
0






+
=






+
=






+
=






+
=








t
f
A
t
s
t
f
A
t
s
t
f
A
t
s
t
f
A
t
s
C
C
C
C
C
C
C
C

○QPSK transmitter:

○QPSK Receiver:

○The symbol interval of QPSK is twice of the bit
interval Tb.
○In other words, QPSK carries twice as many bits as
corresponding PSK for a give bandwidth.
○The price for improved bandwidth efficiency is
increased complexity of the receiver.
○The receiver consists of two binary detectors, one
for in-phase signal and the other is quadrature.
○QPSK systems can be seen as also two binary PSK
systems operating in parallel, with two carrier waves
in phase quadrature. In other words, quadrature
multiplexing.

○Assuming the noise is additive and Gaussian with
zero mean and spectral density N0/2, the received
signal is
○At the end of symbol interval T, the in-phase
correlator and quadrature outputs are
( ) ( ) ( ) ( ) T
t
t
w
t
f
A
t
f
A
t
x C
C
C
C 

+


= 0
2
sin
2
cos 

( ) ( )
( ) ( )


+

=
+

=
T
C
C
T
C
C
dt
t
f
t
w
T
A
I
dt
t
f
t
w
T
A
I
0
2
0
1
2
sin
2
1
2
cos
2
1



○The mean or expected value of in-phase component
is
○The variance of I1 is
○Mean and variance of Quadrature component is
  T
A
I
E C
2
1
1 
=
  ( ) ( )
4
2
cos 0
2
0
1
T
N
dt
t
f
t
w
E
I
Var
T
C =
















=  
  T
A
I
E C
2
1
2 
=  
4
0
2
T
N
I
Var =

○The probability of error at the output of first
correlator and second correlator are equal, since
their variances and mean values are the same.
○Since signal energy per symbol is








=
=
0
2
2
1
N
T
A
Q
P
P C
e
e
T
A
E C
2
=








=
=
=
0
2
1
N
E
Q
P
P
P e
e
e

○In QPSK systems, there are two bits per symbol, so
the signal energy per symbol is twice the signal
energy per bit.
○QPSK receiver must track the phase of the
transmitter for satisfactory operation. One of the
solution is four-phase Costas Loop, which tracks all
the possible phases in QPSK signal constellation.
b
E
E 2
=








=
0
2
N
E
Q
P b
e

COMPARISON OF NOISE PERFORMANCES OF VARIOUS
PSK AND FSK SYSTEMS
○Probability of error is one of the figure of merits of
the communication systems.
○Ideally, maximum number of bits per symbol is
desired to be transmitted.
○The price for increased number of bits is usually
poor noise performance, in other words higher
probability of error.
○To compare various digital communication systems,
the probability of error is expressed as a function of
signal energy per bit to mean noise power per unit
bandwidth: Eb/N0.

PSK AND FSK SYSTEMS
○Bit error probability for different schemes.
( )
( )
0
0
0
2
0
2
0
2
2
0
2
2
0
0
0
0
2
1
Detection)
nt
(Noncohore
FSK
-
M
log
1
Detection)
(Coherent
FSK
-
M
1
log
3
1
1
log
4
QAM
-
M
sin
log
2
2
PSK
-
M
2
1
DPSK
2
1
FSK
t
Noncoheren
2
QPSK
2
MSK
FSK
Coherent
2
PSK
Coherent
N
E
b
b
b
N
E
N
E
b
b
b
b
s
b
b
e
M
N
M
E
Q
M
N
E
M
M
Q
M
M
M
N
M
E
Q
e
e
N
E
Q
N
E
Q
N
E
Q
N
E
Q
−
−
−
−








−








−






−
















































PSK AND FSK SYSTEMS
-4 -2 0 2 4 6 8 10 12 14 16 18 20
10
-6
10
-5
10
-4
10
-3
10
-2
10
-1
Eb/N0 (dB)
Bit
Error
Rate
Coherent PSK
Coherent QPSK
Coherent MSK
Coherent FSK
Noncoherent FSK
DPSK

PSK AND FSK SYSTEMS
-4 -2 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40
10
-6
10
-5
10
-4
10
-3
10
-2
10
-1
Eb/N0 (dB)
Bit
Error
Rate
Coherent QPSK
8-PSK
64-PSK
16-QAM
64-QAM
256-QAM
2048-QAM
16-FSK

Digital Communication fundamentals .pdf

More Related Content

What's hot

Similar to Digital Communication fundamentals .pdf

More from ssuserca5764

Recently uploaded

Digital Communication fundamentals .pdf