Chapter2

ECE 4790
ELECTRICAL COMMUNICATIONS
Fall 99

Dr. Bijan Mobasseri
ECE Dept.
Villanova University

Policies and procedures

3 hours of lecture per week(MWF)
2 hours of lab per week(Wed. in CEER 118)
Homework assigned and graded weekly
Lab is due the same day
2 tests and one final
Grade break down
• 15% each test
• 25% final
• 25% labs
• 20% homework

Copyright1999 by BG Mobasseri 2

Lab work

Hands-on lab work is an integrated part of
the course
There will be about 10 experiments done
using MATLAB with signal processing and
COMM toolboxes
Experiments, to the extent possible,
parallel theoretical material
Professional MATALB code is expected


Going online

Send a blank message to
ece4790_f99-subscribe@egroups.com
You will then have access to all class
notes, labs etc.
Notes/labs are in MS Office format
You can also participate in the online
discussion group and, if you wish, chat
room


Ethical standards

This course will be online and make full
use of internet. This convenience brings
with it many responsibilities
• Keep electronic class notes/material private
• Keep all passwords/accounts to yourself
• Do not exchange MATLAB code


Necessary Background

This course requires, at a minimum, the
following body of knowledge
• Signal processing
• Probability
• MATLAB


Course Outlook

Introduction Pulse Shaping
• signals, channels • “best” pulse shape
• bandwidth • interference
• signal represent. • equalization
Analog Modulation Digital Modulations
• AM and FM Modem standards
Source coding Spread Spectrum
• Sampling, PAM Wireless
• PCM, DM, DPCM


SIGANLS AND CHANNELS IN
COMMUNICATIONS

AN INTRODUCTION

A Block Diagram

Information
source user

source
source
decoder
encoder

channel
channel decoder
encoder

modulator demodulator

channel


Information source

The source can be analog, or digital to
begin with
• Voice
• Audio
• Video
• Data


Source encoder

Source encoder converts analog
information to a binary stream of 1’s and
0’s

1 0 0 1 1 0 0 1 ...
Source encoder
PCM, DM, DPCM, LPC


Channel encoder

The binary stream must be converted to
real pulses

polar

10110 channel
encoder on-off


Modulator

Signals need to be “modulated” for
effective transmission

1 01 1 0
Modulator


Channel

Channel is the “medium” through which
signals propagate. Examples are:
• Copper
• Coax
• Optical fiber
• wireless


Periodic vs. Nonperiodic

A periodic signal satisfies the condition

The smallest value of To for which this
condition is met is called a period of g(t)

g(t ) = g(t + T0 )
period


Deterministic vs. random

A deterministic signal is a signal about
which there is no uncertainty with respect
to its value at any given time
• exp(-t)
• cos(100t)


Energy and Power

Consider the following

+ i(t)
V(t) R
Instantaneous power is given by
-

2
v (t ) 2
p( t ) = = Ri(t )
R


Energy

Working with normalized load, R=1Ω

2 2 2
p(t ) = v(t ) = i(t ) = g(t )
Energy is then defined as
T
2
E = lim ∫ g(t ) dt
−T
T→∞


Average Power

The instantaneous power is a function of
time. An overall measure of signal power is
its average power

T 2
1
T→∞ 2T ∫
P = lim g( t ) dt
−T


Energy and Power of a
Sinusoid

Take m (t ) = A cos (2πfc t )
• Find the energy

T T

A2 cos2 (2πfc t )dt ⇒ ∞
2
E = lim
T→∞ ∫
−T
T →∞ ∫ 1 44 2 4 4
g(t ) dt = lim
−T
3
always>0


Instantaneous Power

Instantaneous power
AMPLITUDE= 1, FREQ.=2
1

2 2
p(t ) = A cos ( 2πfc t ) 0.8

0.6

A2 A2 0.4

= + cos( 4πfc t ) 0.2

2 2 0

-0.2 INSTANT. POWER
ORIGINAL SIGNAL
-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
TIME(SEC)


Average Power

The average power of a sinusoid is
T 2 T
1 1
T →∞ 2T ∫ T→∞ 2T ∫
P = lim g(t ) dt = lim A2 cos 2 (2πfc t )dt
−T −T

from 2 cos 2 x = 1 + cos(2x ),then
T
1 A2
P = lim
T →∞ 2T ∫ 2 (1 + cos(4πfct ))dt
−T
T T
1 A2 1 A2 A2
T→ ∞ 2T ∫ 2 T →∞ 2T ∫ 2
= lim + lim cos( 4πfc t )dt =
−T 4 4 4 2 4 4 4 3
2
−T 1
averages to zero


Average Transmitted Power

What is the peak signal amplitude in order
to transmit 50KW? Assume antenna
impedance of 75Ω.
• Note the change in Pavg for non-unit
ohm load

⎛ A2 ⎞
Pavg = ⎜ ⎟ R ⇒ A = 2RPavg = 2 × 75 × 50,000
⎝2⎠
⇒ A = 2,738volts


Energy Signals vs.
Power Signals

Do all signals have valid energy and power
levels?
• What is the energy of a sinusoid?
• What is the power of a square pulse?
In the first case, the answer is inf.
In the second case the answer is 0.


Energy Signals

A signal is classified as an energy signal if
it meets the following
0<E<infinity
Time-limited signals, such as a square
pulse, are examples of energy signals


Power Signals

A power signal must satisfy
0<P<infinity
Examples of power signals are sinusoidal
functions


Example:energy signal

Square pulse has finite energy but zero
average power

A E = A2 T

T
1 T 2 1 T 2
P = lim ∫ g(t ) dt = lim ∫ A dt ⇒ 0
T→∞ 2T −T T→∞ 2T −T
123
fixed


Example:power signal

A sinusoid has infinite energy but finite
power
A

∞

E= ∫ A2 cos 2 (2πfc t ) dt ⇒ ∞
−∞

but
T
1 ∞
cos (2πfc t )dt ⇒ ⇒ finite
T→∞ 2T ∫
2
Pavg = lim
−T
∞

SUMUP

Energy and power signals are mutually
exclusive:
• Energy signals have zero avg. power
• Power signals have infinite energy
• There are signals that are neither energy
or power?. Can you think of one?


WHAT IS BANDWIDTH?

In a nutshell, bandwidth is the “highest”
frequency contained in a signal.
We can identify at least 5 definitions for
bandwidth
• absolute
• 3-dB
• zero crossing
• equivalent noise
• RMS


ABSOLUTE BANDWIDTH

The highest frequency

Spectrum

-W W f


3-dB BANDWIDTH

The frequency where frequency response
drops to .707 of its peak

Spectrum

W f


FIRST ZERO CROSSING
BANDWIDTH

The frequency where spectrum first goes to zero is
called zero crossing bandwidth.
1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0
-4 -3 -2 -1 0 1 2 3 4


EQUIVALENT NOISE
BANDWIDTH

Bandwidth which contains the same power
as an equivalent bandlimited white noise

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0

-4 -3 -2 -1 0 1 2 3 4

RMS BANDWIDTH

RMS bandwidth is related to the second
moment of the amplitude spectrum

1
∞ 2
⎡∫−∞ f G( f ) ⎤
2 2
Wrms = ⎢ ∞ 2 ⎥
⎣ ∫−∞ G( f ) ⎦
This measures the tightness of the
spectrum around its mean


RMS BANDWIDTH OF A
SQUARE PULSE

Take a square pulse of duration 0.01 sec.
Its spectrum is a sinc
SQUARE PULSE SPECTRUM (WIDTH=0.01 SEC.)
4.5

4

3.5

3

2.5

2

1.5

1

0.5

0

0 100 200 300 400 500 600 700 800
FREQUENCY(Hz)

RMS BANDWIDTH

The RMS bandwidth can be numerically
computed using the following MATLAB
code

W rms= 35.34 Hz


Bandwidth of Real Signals

This is the spectrum a 3 sec. clip sampled
at 8KHz
0

-10

-20

-30

-40

-50

-60

-70
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Frequency


Gate Function

Gate function is one of the most versatile
pulse shapes in comm.
It is pulse of amplitude A and width T

A

-T/2 T/2


rect function

Gate function is defined based on a
function called rect

⎧ 1 1
1 − <t<
rect(t) = ⎨ 2 2
⎪
⎩0 otherwise


Expression for gate

Based on rect we can write

t
g(t ) = Arect ⎛ ⎞
⎝T ⎠
Note that for -T/2<t<T/2, the argument of
rect is inside {-1/2,1/2} therefore rect is 1
and g(t)=A


Generalizing gate

Let’ say we want a pulse with amplitude A
centered at t=to and width T

T
A

t=to


Arriving at an Expression

What we have a is a rect function shifted
to the right by to
Shift to the right of f(t) by to is written by
f(t-to)
Therefore

⎛t − to ⎞
g(t ) = Arect ⎝
T ⎠


gate function in the Fourier
Domain

The Fourier transform of a gate function is
a sinc as follows
sinc(t)
1

0.8

g(t ) = Arect ⎛t ⎞
⎝T ⎠
0.6

0.4

G( f ) = ATsinc( fT ) 0.2
Zero crossing
0

-0.2

-0.4
-3 -2 -1 0 1 2 3
TIME(t)


Zero Crossing

Zero crossings of a sinc is very significant.
ZC occurs at integer values of sinc
argument

⎧ x =0
1
sin c(x) = ⎨
⎩ x = ±1,±2,L
0


Some Numbers

What is the frequency content of a 1 msec.
square pulse of amplitude .5v?
• We have A=0.5 and T=1ms

g(t ) = 0.5rect(1000t)
G( f ) = 5 × 10 −4 sinc(10 −3 f )
• First zero crossing at f=1000Hz obtained
by setting 10^-3f=1


RF Pulse

RF(radio frequency) pulse is at the heart of
all digital communication systems.
RF pulse is a short burst of energy,
expressed by a sinusoidal function


Modeling RF Pulse

An RF pulse is a cosine wave that is
truncated on both sides
This effect can be modeled by “gating”the
cosine wave


Mathematically Speaking

Call the RF pulse g(t), then

t
g(t ) = Arect ⎛ ⎞ cos(2πfc t )
⎝T ⎠
This is in effect the modulated version of
the original gate function


Spectrum of the RF
Pulse:basic rule

We resort to the following

g1 (t )g2 (t ) ⇔ G1 ( f ) * G2 ( f )

Meaning, the Fourier transform of the
product is the convolution of individual
transforms


RF Pulse Spectrum: Result

We now have to identify each term
AT
gate → g1 (t ) ⇔ sin c(Tf )
2
cosine → g 2 (t ) ⇔ [δ ( f − fc ) + δ ( f + fc )]
Then, the RF pulse spectrum, G(f)

AT
G( f ) = ⎛ sin c(Tf )⎞ * [δ ( f − fc ) + δ ( f + fc )]
⎝ 2 ⎠
AT AT
= sin c(T ( f − fc )) + sin c(T ( f + fc ))
2 2


Interpretation

The spectrum of the RF pulse are two
sincs, one at f=- fc and the other at f=+fc

RF PULSE SPECTRUM:two sincs at pulse freq

-4 0 4
FREQUENCY (HZ)


Actual Spectrum

RF PULSE SPECTRUM, fc=1200Hz, duration= 5 msec
50

45

40

35 Bandwidth=400Hz

30

25

20
5 msec
15

10

5

0
0 400 800 1200 1600 2000 2400
FREQUENCY (HZ)


Baseband and Bandpass
Signals and Channels

Definitions:Baseband

The raw message signal is referred to as
baseband, or low freq. signal

Spectrum of a baseband audio signal
0.045

0.04

0.035

0.03

0.025

0.02

0.015

0.01

0.005

0
0 500 1000 1500 2000 2500 3000 3500 4000
FREQUENCY(HZ)


Definitions:Bandpass

When a baseband signal m(t) is modulated,
we get a bandpass signal
The bandpass signal is formed by the
following operation(modulation)

m (t )cos( 2πfc t )


Bandpass Example:AM

3
BANDPASS
BASEBAND

2

m (t )cos( 2πfc t )
1

0

-1

-2

-3
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2


Digital Bandpass

Baseband and bandpass concepts apply
equally well to digital signals

1 1 1 1

0

RF pulses


Baseband vs. Bandpass
Spectrum

Creating a bandpass signal is the same as
modulation process. We have the
following

1
m (t )cos( 2πfc t ) ⇔ [ M( f − fc ) + M ( f + fc )]
2

Interpretation


Showing the Contrast

baseband

frequency
Bandwidth doubles bandpass bandwidth


SIGNAL REPRESENTATION

How to write an expression for
signals

Introduction

We need a formalism to follow a signal as
it propagates through a channel.
To this end, we have to learn a few
concepts including Hilbert Transform,
analytic signals and complex envelope


Hilbert Transform

Hilbert transform is an operation that
affects the phase of a signal

+90

H(f) f
-90
|H(f)|=1
Phase response


More precisely

⎧ − jπ
⎪1e 2 f ≥ 0
H( f ) = ⎨
π
⎪ 2j
⎩1e f < 0
Equivalently
H ( f ) = − jsgn( f )
sgn( f ), or signum function, extracts the sign of its argument
⎧f >0
⎪
1
sgn( f ) = ⎨ f = 0
0
⎪ f <0
⎩−1

HT notation

The HT of g(t) is denoted by

HT ( g(t )) = g(t )
ˆ
In the frequency domain,

ˆ
G( f ) = − j sgn( f )G( f )


Find HT of a Sinusoid

Q: what is the HT of cosine?Ans:sine
g(t ) = cos(2πfc t )
we know
⎡ ⎛ ⎞⎤
Gˆ ( f ) = − j sgn( f )G( f ) = − j sgn( f )⎢1 ⎜δ ( f − fc ) + δ ( f + fc )⎟⎥
⎢2 ⎜14 2 43 14 2 43 ⎟⎥
⎣ ⎝ pos. freq neg. freq ⎠⎦
j j 1
= − δ ( f − fc ) + δ ( f + fc ) = [δ ( f − fc ) − δ ( f + fc )]
2 2 2j
1 4 4 4 4 2 4 4 4 43
Fourier transform of sine

HT Properties

Property 1
• g and HT(g) have the same amplitude
spectrum
Property 2

Property 3 HT ( g(t )) = −g(t )
ˆ
• g and HT(g) are orthogonal, i.e.

∫ g(t )g(t )dt = 0
ˆ


Using HT: Pre-envelope

From a real-valued signal, we can extract a
complex-valued signal by adding its HT as
follows

g+ (t ) = g(t ) + jg(t )
ˆ
g+(t) is called the pre-envelope of g(t)


Question is Why?

It turns out that it is easier to work with g+
(t) than g(t) in many comm. situations
We can always go back to g(t)

g(t ) = Re { g+ (t )}
where Re stands for " real part of"


Pre-envelope Example

Find the pre-envelope of the RF pulse
g(t ) = m (t )cos( 2πfc t + θ )
We can re-write g(t) as follows

j (2πfc t +θ )
g(t ) = Re{ m (t )e }
because
j (2πfc t +θ )
e = cos( 2πfct + θ ) + j sin( 2πfc t + θ )


Pre-envelope is...

Compare the following two

j (2πfc t +θ )
g(t ) = Re{ m (t )e }
g(t ) = Re{ g+ ( t )} g+ ( t ) = m(t )e
j ( 2πfc t +θ )

Pre-envelope of
is g(t ) = m (t )cos( 2πfc t + θ )
j ( 2πfc t +θ )
g+ ( t ) = m(t )e


Pre-envelope in the Frequency
Domain

How does pre-envelope look in the
frequency domain?

We know g+ (t ) = g(t ) + jg( t ). Fourier transform is
ˆ
F{ g+ (t )} = G+ ( f ) = G( f ) + j[ − j sgn( f )]G( f )


Pre-envelope in positive and
negative frequencies

Let’s evaluate G+(f) for f>0 G(f)

f >0
G+ ( f ) = G( f ) + G( f ) = 2G( f ) f
f <0
G+(f)
G+ ( f ) = G( f ) − G( f ) = 0

f


Interpretation

Fourier transform of Pre-envelope exists
only for positive frequencies
As such per-envelope is not a real signal. It
is complex as shown by its definition

g+ (t ) = g(t ) + jg(t )
ˆ


Corollary

To find the pre-envelope in the frequency
domain, take the original spectrum and
chop off the negative part


Example

Find the pre-envelope of a modulated
message

g(t ) = m (t )cos( 2πfc t + θ )
G(f) G+(f)

AM signal


Another Definition
for Pre-envelope

Pre-envelope is such a quantity that if you
take its real part, it will give you back your
original signal

g(t ) = m (t )cos( 2πfc t + θ ) original signal
j ( 2πfc t +θ )
g(t ) = Re{ m (t )e j (2πft +θ ) g+ ( t ) = m(t )e
}
g(t ) = Re{ g+ ( t )}


Bringing Signals Down to
Earth

Communication signals of interest are
mostly high in frequency
Simulation and handling of such signals
are very difficult and expensive
Solution: Work with their low-pass
equivalent


Tale of Two Pulses

Consider the following two pulses

Which one carries more “information”?


Lowpass Equivalent Concept

The RF pulse has no more information
content than the square pulse. They are
both sending one bit of information.
Which one is easier to work with?


Implementation issues

It takes far more samples to simulate a
bandpass signal

0.01 sec. Sampling rate=200Hz

4cycles/0.01 sec Sampling rate=800Hz
-->fc=400Hz


Complex Envelope

Every bandpass signal has a lowpass
equivalent or complex envelope
Take and re write as
g(t ) = m (t )cos( 2πfc t + θ )

⎧ ⎫
j (2πfc t +θ )
⎪ jθ j 2πfc t
⎪
g(t ) = Re{ m (t )e } = Re ⎨ m (2)e3
1 t e ⎬
⎪ lowpass ⎪
⎩ complex envelop
or ⎭


Complex Envelope:
The Quick Way

Rewrite the signal per following model

j (2πfc t +θ )
g(t ) = Re { m (t )e
} = Re { m(t )e jθ e j 2πfc t }
The term in front of j2πf t is the complex
envelope shown by e c

jθ
g( t ) = m(t )e
˜ m and theta contain all the
information


Signal Representation
Summary

Take a real-valued, baseband signal

G(f)
g(t)


Pre-envelope Summary:
baseband

G(f)

g+ (t ) = g(t) + jg(t)
ˆ
g(t) = HT { g(t)}
ˆ G+(f)
g(t) = Re{ g+ (t)} Nothing for f<0


Pre-envelope Summary:
bandpass

Baseband signal: g(t ) = m (t )cos( 2πfc t + θ )

G(f)

G+(f)


Complex Envelope Summary

Complex/pre envelope are related

− j 2πfc t G+(f)
g( t ) = g+ (t)e
˜
or
ˆ
G( f )
j 2πfc t
g+ (t) = g(t )e
˜


RF Pulse: Complex Envelope

Find the complex envelope of a T second
long RF pulse at frequency fc

t
g(t ) = Arect ⎛ ⎞ cos(2πfc t )
⎝T ⎠


Writing as Re{ }

Rewrite g(t) as follows
⎧Arect ⎛ t ⎞e j 2πfct ⎫
g(t ) = Re ⎨
⎝T ⎠ ⎬
⎩ ⎭
compare
g(t ) = Re{ g (t)e j2πfct }
˜ Comp.Env=just a square
pulse
then
t
g (t) = complex _ envelope = Arect ⎛ ⎞
˜ ⎝T ⎠


RF Pulse Pre-envelope

Recall
− j 2πfc t
Then
g( t ) = g+ (t)e
˜

g+ (t ) = Arect ⎛ t ⎞e j 2πf ct
⎝T ⎠


Story in the Freq. Domain

RF PULSE SPECTRUM:two sincs at pulse freq RF PULSE SPECTRUM:two sincs at pulse freq

-4 0 4 -4 0 4
FREQUENCY (HZ) FREQUENCY (HZ)

Original RF pulse spectrum Pre-env. Spectrum
(only f>0 portion)


Complex Envelope Spectrum

Complex envelope=low pass portion
1

0.8

0.6

0.4

0.2

0

-0.2

-0.4
-10 -8 -6 -4 -2 0 2 4 6 8 10


CHANNELS AND SIGNAL
DISTORTION

Some of the material not in the book

Signal Transmission Modeling

One of the most common tasks in
communications is transmission of RF
pulses through bandpass channels
Instead of working at high RF frequencies
at great computational cost, it is best to
work with complex envelope
representations


Channel I/O

To determine channel output, we can work
with complex envelopes

˜ 1 ˜ ˜
Y ( f ) = H ( f ) X( f )
2
where
˜
X( f ) : C.E. of input(transmitted) signal
˜
H ( f ) : C.E. of channel transfer function
˜
Y ( f ) : C.E. of output(received) signal


Passing an RF Pulse through a
Bandpass Channel

Here is the problem: what is the output of
an ideal bandpass channel in response to
an RF pulse?
H(f)

use
˜( f ) = 1 X ( f ) H( f )
Y ˜ ˜
2


What is the Complex envelope
of H(f)?

It is the lowpass equivalent of H(f)

2

˜ ( f ) = 2rect ⎛ f ⎞
H
1
⎝2B⎠

2B B

H(f)


What is the Complex Envelope
of the RF Pulse?

We found this before
⎧Arect ⎛ t ⎞e j 2πfct ⎫
g(t ) = Re ⎨
⎝T ⎠ ⎬
⎩ ⎭
compare
g(t ) = Re{ g (t)e j2πfct }
˜
then
t
g (t) = complex _ envelope = Arect ⎛ ⎞
˜ ⎝T ⎠


Channel Output

Here is what we have
f
H ( f ) = 2rect ⎛ ⎞
˜
• Channel complex envelope ⎝2B⎠

• Input complex envelope
B=bandwidth
t
x (t ) = Arect ⎛ ⎞ ⇔ X ( f ) = AT sin c( fT )
˜ ˜
• Output ⎝T ⎠

1 f
Y ( f ) = [ AT sin c( fT)] ⎡2rect( )⎤
˜
2 ⎣ 2B ⎦


Interpretation

˜ f 1

Y ( f ) = AT sin c( fT)rect( )
2B
0.8

0.6

For the pulse to get through 0.4

unscathed, channel bandwidth 0.2

must be larger than pulse bw 0

-0.2

B>=1/T=bit rate -0.4
-10 -8 -6 -4 -2 0 2 4 6 8 10

1/T
B


What Does Distortion Do?

Channel Distortion creates pulse
“dispersion”

Channel

interference


Case of No Distortion

There are two “distortions” we can live
with
• Scaling
• Delay

To


Modeling Distortion-free
Channels

The input-output relationship for a
distortion-free channel is
y(t)=Ax(t-Td)
• x(t):input
• y(t)=output
• A:scale factor
• Td: delay


Response of a Distortion-free
channel

What is channel’s frequency response?
Take FT of the I/O expression
Y ( f ) = AX ( f )e − j2πfTd
Then

Y( f )
H( f ) = = Ae − j2πfTd
X( f )


Amplitude and Phase
Response

|H(f)|
Const amplitude response

f
/_H(f)
Linear phase response
f

−(2πTd ) f


Complete Model

The complete transfer function is

H ( f ) = Ae − j 2πfTd
Since this is a lowpass function, its
complex envelope is the same as H(f)

H ( f ) = Ae − j 2πfTd
˜


Lowpass Channel

Is a first order filter an appropriate model
for a distortion-free channel?
To answer this question we have to test
the definition of the ideal channel

R

C


Amplitude and Phase
Response

amplitude _ response =
a
H( f ) = 2
a + (2πf )
3-dB bandwidth=
phase _ response =
a/2pi=1/(2piRC)
−1 ⎛ 2πf ⎞
θh ( f ) = −tan
⎝ a ⎠
1
a=
RC


Response for RC=10^-3

AMPLITUDE RESPONSE
PHASE RESPONSE
1.5

1

0.5
0.7

0

-0.5

-1

0 159 -1.5
0 159
FREQUENCY(HZ) FREQUENCY(HZ)

bandwidth=159 Hz

An “ideal” Channel?

We must have constant amplitude
response and linear phase response.
Do we?. Deviation of H(f) from the ideal is
tolerated up to .707form the peak.
The frequency at which this occurs is the
3dB bandwidth

No signal distortion if input frequencies are kept
below 3dB bandwidth or 159 Hz here


Linear Distortion

If any of the ideal channel conditions are
violated but we are still dealing with a
linear channel, we have linear distortion

amplitude
H ( f ) = { (1 + k cos (2πfT ))e− j2πftd

f
phase


Pulse Dispersion

Putting a pulse g(t) through this filter
produces 3 overlapping copies

channel with
T distortion

>T

Why?

Let g(t) and r(t) be the transmitted and
received signals. Then

R( f ) = G( f )H( f ) = G( f )[1 + kG( f )cos(2πfT )]e− j2πftd
= G( f )e− j2πftd + kG( f )cos(2πfT)e− j2πftd
Taking the inverse FT
k
r(t) = g(t − td ) + [ g(t − td − T ) + g(t − td + T )]
2


Nonlinear Distortion

This is the most serious kind where input
and output are related by a nonlinear
equation

Nonlinear
g channel r
r r=g^2

g


Impact of Nonlinear Dist.

Nonlinear channels generate new
frequencies at the output that did not exist
in the input signal. Why?
G(f)

if f
W
r(t ) = g 2 (t ) R(f)
then
R( f ) = G( f ) * G( f )
2W f


Practice Problems

For pre-envelope: 2.23
For filtering using complex envelope: 2.32


Chapter2

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