Chapter2

ECE 4790ELECTRICAL COMMUNICATIONS

Fall 99

Dr. Bijan MobasseriECE Dept.

Villanova University

Copyright1999 by BG Mobasseri 2

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


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 [email protected]

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• signals, channels• bandwidth• signal represent.

Analog Modulation• AM and FM

Source coding• Sampling, PAM• PCM, DM, DPCM

Pulse Shaping• “best” pulse shape• interference• equalization

Digital Modulations Modem standards Spread Spectrum Wireless

SIGANLS AND CHANNELS IN COMMUNICATIONS

AN INTRODUCTION


A Block Diagram

Information source

source encoder

channelencoder

modulator

channel

demodulator

channeldecoder

source decoder

user


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

Source encoderPCM, DM, DPCM, LPC

1 0 0 1 1 0 0 1 ...


Channel encoder

The binary stream must be converted to real pulses

channelencoder

1 0 1 1 0

polar

on-off


Modulator

Signals need to be “modulated” for effective transmission

Modulator

1 0 1 1 0


Channel

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

Signals and Systems Review


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

Instantaneous power is given by

RV(t)

+

-

i(t)

p t( )=v t( )2

R=Ri t( )2


Energy

Working with normalized load, R=1Ω

p t( )=v t( )2 =i t( )2 =g t( )2

Energy is then defined as

E =lim g t( )2dt−T

T

∫T→ ∞


Average Power

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

P =limT→ ∞

12T

g t( )−T

T

∫2

dt


Energy and Power of a Sinusoid

Take• Find the energy

mt( )=Acos2πfct( )

E =limT→ ∞

g t( )2dt=−T

T

∫ limT→ ∞

A2 cos2 2πfct( )always>0

1 2 4 4 3 4 4 −T

T

∫ dt⇒ ∞


Instantaneous Power

Instantaneous power

p t( )=A2 cos2 2πfct( )

=A2

2+

A2

2cos4πfct( )

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

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1AMPLITUDE= 1, FREQ.=2

TIME(SEC)

INSTANT. POWER ORIGINAL SIGNAL


Average Power

The average power of a sinusoid is

P =limT→ ∞

12T

g t( )−T

T

∫2

dt=limT→ ∞

12T

A2 cos2 2πfct( )−T

T

∫ dt

from 2cos2 x =1+cos2x( ),then

P =limT→ ∞

12T

A2

2−T

T

∫ 1+cos4πfct( )( )dt

=limT→ ∞

12T

A2

2+

−T

T

∫ limT→ ∞

12T

A2

2−T

T

∫ cos4πfct( )dt=A2

2averages to zero

1 2 4 4 4 3 4 4 4


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

Pavg=A2

2⎛ ⎝ ⎜ ⎞

⎠ ⎟ R⇒ A= 2RPavg = 2×75×50,000

⇒ 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 satisfy0<P<infinity

Examples of power signals are sinusoidal functions


Example:energy signal

Square pulse has finite energy but zero average power

A

T

P = limT→ ∞

12T

g t( )2

−T

T∫ dt= lim

T→ ∞

12T

A2dt−T

T∫

fixed1 2 3

⇒ 0

E =A2T


Example:power signal

A sinusoid has infinite energy but finite power

A

E = A2

−∞

∞

∫ cos2 2πfct( )dt⇒ ∞

but

Pavg=limT→ ∞

12T

cos2 2πfct( )dt−T

T

∫ ⇒∞∞

⇒ finite


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?

DEFINING BANDWIDTH


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

W-W f

Spectrum


3-dB BANDWIDTH

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

W f

Spectrum


FIRST ZERO CROSSING BANDWIDTH

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

-4 -3 -2 -1 0 1 2 3 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


EQUIVALENT NOISE BANDWIDTH

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

-4 -3 -2 -1 0 1 2 3 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


RMS BANDWIDTH

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

This measures the tightness of the spectrum around its mean

Wrms =f 2 G f( )

2

−∞

∞∫

G f( )2

−∞

∞∫

⎡

⎣ ⎢

⎤

⎦ ⎥

1

2


RMS BANDWIDTH OF A SQUARE PULSE

Take a square pulse of duration 0.01 sec. Its spectrum is a sinc

0 100 200 300 400 500 600 700 8000

0.5

1

1.5

2

2.5

3

3.5

4

4.5

FREQUENCY(Hz)

SQUARE PULSE SPECTRUM (WIDTH=0.01 SEC.)


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

-60

-50

-40

-30

-20

-10

0

Frequency


Gate Function

Gate function is one of the most versatile pulse shapes in comm.

It is pulse of amplitude A and width T

T/2-T/2

A


rect function

Gate function is defined based on a function called rect

rect(t) =1 −

12

<t<12

0 otherwise

⎧ ⎨ ⎩ ⎪


Expression for gate

Based on rect we can write

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

g t( ) =ArecttT

⎛ ⎝

⎞ ⎠


Generalizing gate

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

t=to

AT


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

g t( ) =Arectt−to

T⎛ ⎝

⎞ ⎠


gate function in the Fourier Domain

The Fourier transform of a gate function is a sinc as follows

g t( ) =ArecttT

⎛ ⎝

⎞ ⎠

G f( )=ATsinc fT( )

-3 -2 -1 0 1 2 3-0.4

-0.2

0

0.2

0.4

0.6

0.8

1sinc(t)

TIME(t)

Zero crossing


Zero Crossing

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

sinc(x) =

1 x =0

0 x=±1,±2,L⎧ ⎨ ⎩


Some Numbers

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

• First zero crossing at f=1000Hz obtained by setting 10^-3f=1

g t( ) =0.5rect(1000t)

G f( )=5×10−4sinc(10−3 f)


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

This is in effect the modulated version of the original gate function

g t( ) =ArecttT

⎛ ⎝

⎞ ⎠ cos2πfct( )


Spectrum of the RF Pulse:basic rule

We resort to the following

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

g1 t( )g2 t( )⇔ G1 f( )* G2 f( )


RF Pulse Spectrum: Result

We now have to identify each term

Then, the RF pulse spectrum, G(f)

gate→ g1 t( )⇔AT2

sinc Tf( )

cosine→ g2 t( )⇔ δ f − fc( )+δ f + fc( )[ ]

G f( )=AT2

sinc Tf( )⎛ ⎝

⎞ ⎠ * δ f − fc( )+δ f + fc( )[ ]

=AT2

sinc T f − fc( )( )+AT2

sinc T f + fc( )( )


Interpretation

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

-4 0 4

RF PULSE SPECTRUM:two sincs at pulse freq

FREQUENCY (HZ)


Actual Spectrum

0 400 800 1200 1600 2000 24000

5

10

15

20

25

30

35

40

45

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

FREQUENCY (HZ)

Bandwidth=400Hz

5 msec

Baseband and Bandpass Signals and Channels


Definitions:Baseband

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

0 500 1000 1500 2000 2500 3000 3500 40000

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

FREQUENCY(HZ)

Spectrum of a baseband audio signal


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)

mt( )cos2πfct( )


Bandpass Example:AM

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

-2

-1

0

1

2

3

BANDPASSBASEBAND

mt( )cos2πfct( )


Digital Bandpass

Baseband and bandpass concepts apply equally well to digital signals

1

0

1 1 1

RF pulses


Baseband vs. Bandpass Spectrum

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

mt( )cos2πfct( )⇔12

M f −fc( )+M f +fc( )[ ]

Interpretation


Showing the Contrast

frequency

bandwidthBandwidth doubles

baseband

bandpass

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

H(f) f

Phase response

+90

-90

|H(f)|=1


More precisely

H f( ) =1e

−jπ2 f ≥0

1ejπ2 f <0

⎧

⎨ ⎪

⎩ ⎪

Equivalently

H f( ) =−jsgn(f)

sgn(f), or signum function, extracts the sign of its argument

sgn(f) =

1f >0

0f =0

−1f <0

⎧ ⎨ ⎪

⎩ ⎪


HT notation

The HT of g(t) is denoted by

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

ˆ G f( )=−jsgn(f)G f( )


Find HT of a Sinusoid

Q: what is the HT of cosine?Ans:sine

g t( ) =cos2πfct( )

we know

ˆ G f( )=−jsgn f( )G f( )=−jsgn f( )12

δ f − fc( )

pos.freq1 2 4 3 4

+δ f + fc( )

neg.freq1 2 4 3 4

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

=−j2δ f − fc( )+

j2δ f +fc( ) =

12j

δ f −fc( )−δ f +fc( )[ ]

Fourier transform of sine1 2 4 4 4 4 3 4 4 4 4


HT Properties

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

spectrum Property 2

Property 3• g and HT(g) are orthogonal, i.e.

HT ˆ g t( )( )=−g t( )

g t( )∫ ˆ g t( )dt=0


Using HT: Pre-envelope

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

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

We can re-write g(t) as follows g t( ) =mt( )cos2πfct+θ( )

g t( )=Re mt( )ej 2πfct+θ( )

because

ej 2πfct+θ( ) =cos2πfct+θ( )+jsin2πfct+θ( )


Pre-envelope is...

Compare the following two

Pre-envelope ofis

g t( )=Re mt( )ej 2πfct+θ( )

g t( )=Re g+ t( ) g+ t( )=mt( )ej 2πfct+θ( )

g t( ) =mt( )cos2πfct+θ( )

g+ t( )=mt( )ej 2πfct+θ( )


Pre-envelope in the Frequency Domain

How does pre-envelope look in the frequency domain?

We know g+ t( )=g t( )+jˆ g t( ). Fourier transform is

F g+ t( ) =G+ f( )=G f( )+j −jsgn(f)[ ]G f( )


Pre-envelope in positive and negative frequencies

Let’s evaluate G+(f) for f>0

f >0

G+ f( )=G( f)+G( f)=2G( f)

f <0

G+( f) =G( f)−G( f) =0

G(f)

G+(f)

f

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( )+jˆ g t( )


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( ) =mt( )cos2πfct+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( ) =mt( )cos2πfct+θ( )

g t( )=Re mt( )ej 2πft+θ( )

g t( )=Re g+ t( )

g+ t( )=mt( )ej 2πfct+θ( )

original signal


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-->fc=400Hz

Sampling rate=800Hz


Complex Envelope

Every bandpass signal has a lowpass equivalent or complex envelope

Take and re write asg t( ) =mt( )cos2πfct+θ( )

g t( )=Re mt( )ej 2πfct+θ( ) =Re mt( )ejθ

lowpassor complex envelop

1 2 3 ej2πfct

⎧

⎨ ⎪

⎩ ⎪

⎫

⎬ ⎪

⎭ ⎪


Complex Envelope: The Quick Way

Rewrite the signal per following model

The term in front of is the complex envelope shown by

g t( )=Re mt( )ej 2πfct+θ( ) =Re mt( )ejθ ej2πfct

ej2πfct

˜ g t( )=mt( )ejθm and theta contain all theinformation


Signal Representation Summary

Take a real-valued, baseband signal

g(t)G(f)


Pre-envelope Summary:baseband

g+(t)=g(t)+jˆ g (t)

ˆ g (t)=HT g(t)

g(t)=Re g+(t)

G(f)

G+(f)

Nothing for f<0


Pre-envelope Summary: bandpass

Baseband signal: g t( ) =mt( )cos2πfct+θ( )

G(f)

G+(f)


Complex Envelope Summary

Complex/pre envelope are related

˜ g t( )=g+(t)e−j2πfct

or

g+(t)=˜ g t( )ej2πfct

G+(f)

ˆ G f( )


RF Pulse: Complex Envelope

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

g t( ) =ArecttT

⎛ ⎝

⎞ ⎠ cos2πfct( )


Writing as Re

Rewrite g(t) as follows

g t( )=Re ArecttT

⎛ ⎝

⎞ ⎠ e

j2πfct⎧ ⎨ ⎩

⎫ ⎬ ⎭

compare

g t( )=Re ˜ g (t)ej2πfct

then

˜ g (t) =complex_envelope=ArecttT

⎛ ⎝

⎞ ⎠

Comp.Env=just a squarepulse


RF Pulse Pre-envelope

Recall

Then ˜ g t( )=g+(t)e−j2πfct

g+ t( ) =ArecttT

⎛ ⎝

⎞ ⎠ e

j2πfct


Story in the Freq. Domain

-4 0 4


FREQUENCY (HZ)

Original RF pulse spectrum

-4 0 4


FREQUENCY (HZ)

Pre-env. Spectrum(only f>0 portion)


Complex Envelope Spectrum

Complex envelope=low pass portion

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

-0.2

0

0.2

0.4

0.6

0.8

1

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

˜ Y f( ) =12

˜ H f( ) ˜ X f( )

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?

use

˜ Y f( )=12

˜ X f( ) ˜ H f( )

H(f)


What is the Complex envelope of H(f)?

It is the lowpass equivalent of H(f)

H(f)

˜ H f( )=2rectf

2B⎛ ⎝

⎞ ⎠

2B B

2

1


What is the Complex Envelope of the RF Pulse?

We found this before

g t( )=Re ArecttT

⎛ ⎝

⎞ ⎠ e

j2πfct⎧ ⎨ ⎩

⎫ ⎬ ⎭

compare

g t( )=Re ˜ g (t)ej2πfct

then

˜ g (t) =complex_envelope=ArecttT

⎛ ⎝

⎞ ⎠


Channel Output

Here is what we have• Channel complex envelope• Input complex envelope

• Output

˜ H f( )=2rectf

2B⎛ ⎝

⎞ ⎠

˜ x t( )=ArecttT

⎛ ⎝

⎞ ⎠ ⇔

˜ X ( f) =ATsinc fT( )

B=bandwidth

˜ Y f( )=12

ATsinc( fT)[ ] 2rect(f

2B)⎡

⎣ ⎤ ⎦


Interpretation

˜ Y f( )=ATsinc( fT)rect(f

2B)

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

-0.2

0

0.2

0.4

0.6

0.8

1

B

1/T

For the pulse to get through unscathed, channel bandwidthmust be larger than pulse bw

B>=1/T=bit rate


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

ThenY f( ) =AX f( )e−j2πfTd

H f( ) =Y( f)X( f)

=Ae−j2πfTd


Amplitude and Phase Response

|H(f)|

/_H(f)f

f

−(2πTd) f

Const amplitude response

Linear phase response


Complete Model

The complete transfer function is

Since this is a lowpass function, its complex envelope is the same as H(f)

H f( )=Ae−j2πfTd

˜ H f( ) =Ae−j2πfTd


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=

H f( ) =a

a+ 2πf( )2

phase_ response=

θh f( )=−tan−1 2πfa

⎛ ⎝

⎞ ⎠

a=1

RC

3-dB bandwidth=a/2pi=1/(2piRC)


Response for RC=10^-3

0 159

0.7

FREQUENCY(HZ)

AMPLITUDE RESPONSE

0 159-1.5

-1

-0.5

0

0.5

1

1.5

FREQUENCY(HZ)

PHASE RESPONSE

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 keptbelow 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

phase

f

amplitudeH f( ) = 1+kcos2πfT( )( )e−j2πftd


Pulse Dispersion

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

channel with distortionT

>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

r(t)=g(t−td)+k2

g(t−td −T)+g(t−td +T)[ ]


Nonlinear Distortion

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

Nonlinear channelg r

r

g

r=g^2


Impact of Nonlinear Dist.

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

if

r t( ) =g2 t( )

then

R f( ) =G f( )* G f( )

f

f

W

2W

G(f)

R(f)


Practice Problems

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

Technology

Chapter2