chapter2
TRANSCRIPT
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
Copyright1999 by BG Mobasseri 3
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
Copyright1999 by BG Mobasseri 4
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
Copyright1999 by BG Mobasseri 5
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
Copyright1999 by BG Mobasseri 6
Necessary Background
This course requires, at a minimum, the following body of knowledge• Signal processing• Probability• MATLAB
Copyright1999 by BG Mobasseri 7
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
Copyright1999 by BG Mobasseri 9
A Block Diagram
Information source
source encoder
channelencoder
modulator
channel
demodulator
channeldecoder
source decoder
user
Copyright1999 by BG Mobasseri 10
Information source
The source can be analog, or digital to begin with• Voice• Audio• Video• Data
Copyright1999 by BG Mobasseri 11
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 ...
Copyright1999 by BG Mobasseri 12
Channel encoder
The binary stream must be converted to real pulses
channelencoder
1 0 1 1 0
polar
on-off
Copyright1999 by BG Mobasseri 13
Modulator
Signals need to be “modulated” for effective transmission
Modulator
1 0 1 1 0
Copyright1999 by BG Mobasseri 14
Channel
Channel is the “medium” through which signals propagate. Examples are:• Copper• Coax• Optical fiber• wireless
Signals and Systems Review
Copyright1999 by BG Mobasseri 16
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
Copyright1999 by BG Mobasseri 17
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)
Copyright1999 by BG Mobasseri 18
Energy and Power
Consider the following
Instantaneous power is given by
RV(t)
+
-
i(t)
p t( )=v t( )2
R=Ri t( )2
Copyright1999 by BG Mobasseri 19
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→ ∞
Copyright1999 by BG Mobasseri 20
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
Copyright1999 by BG Mobasseri 21
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⇒ ∞
Copyright1999 by BG Mobasseri 22
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
Copyright1999 by BG Mobasseri 23
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
Copyright1999 by BG Mobasseri 24
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
Copyright1999 by BG Mobasseri 25
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.
Copyright1999 by BG Mobasseri 26
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
Copyright1999 by BG Mobasseri 27
Power Signals
A power signal must satisfy0<P<infinity
Examples of power signals are sinusoidal functions
Copyright1999 by BG Mobasseri 28
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
Copyright1999 by BG Mobasseri 29
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
Copyright1999 by BG Mobasseri 30
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
Copyright1999 by BG Mobasseri 32
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
Copyright1999 by BG Mobasseri 33
ABSOLUTE BANDWIDTH
The highest frequency
W-W f
Spectrum
Copyright1999 by BG Mobasseri 34
3-dB BANDWIDTH
The frequency where frequency response drops to .707 of its peak
W f
Spectrum
Copyright1999 by BG Mobasseri 35
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
Copyright1999 by BG Mobasseri 36
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
Copyright1999 by BG Mobasseri 37
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
Copyright1999 by BG Mobasseri 38
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.)
Copyright1999 by BG Mobasseri 39
RMS BANDWIDTH
The RMS bandwidth can be numerically computed using the following MATLAB code
W rms= 35.34 Hz
Copyright1999 by BG Mobasseri 40
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
Copyright1999 by BG Mobasseri 41
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
Copyright1999 by BG Mobasseri 42
rect function
Gate function is defined based on a function called rect
rect(t) =1 −
12
<t<12
0 otherwise
⎧ ⎨ ⎩ ⎪
Copyright1999 by BG Mobasseri 43
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
⎛ ⎝
⎞ ⎠
Copyright1999 by BG Mobasseri 44
Generalizing gate
Let’ say we want a pulse with amplitude A centered at t=to and width T
t=to
AT
Copyright1999 by BG Mobasseri 45
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⎛ ⎝
⎞ ⎠
Copyright1999 by BG Mobasseri 46
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
Copyright1999 by BG Mobasseri 47
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⎧ ⎨ ⎩
Copyright1999 by BG Mobasseri 48
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)
Copyright1999 by BG Mobasseri 49
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
Copyright1999 by BG Mobasseri 50
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
Copyright1999 by BG Mobasseri 51
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( )
Copyright1999 by BG Mobasseri 52
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( )
Copyright1999 by BG Mobasseri 53
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( )( )
Copyright1999 by BG Mobasseri 54
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)
Copyright1999 by BG Mobasseri 55
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
Copyright1999 by BG Mobasseri 57
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
Copyright1999 by BG Mobasseri 58
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( )
Copyright1999 by BG Mobasseri 59
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( )
Copyright1999 by BG Mobasseri 60
Digital Bandpass
Baseband and bandpass concepts apply equally well to digital signals
1
0
1 1 1
RF pulses
Copyright1999 by BG Mobasseri 61
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
Copyright1999 by BG Mobasseri 62
Showing the Contrast
frequency
bandwidthBandwidth doubles
baseband
bandpass
SIGNAL REPRESENTATION
How to write an expression for signals
Copyright1999 by BG Mobasseri 64
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
Copyright1999 by BG Mobasseri 65
Hilbert Transform
Hilbert transform is an operation that affects the phase of a signal
H(f) f
Phase response
+90
-90
|H(f)|=1
Copyright1999 by BG Mobasseri 66
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
⎧ ⎨ ⎪
⎩ ⎪
Copyright1999 by BG Mobasseri 67
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( )
Copyright1999 by BG Mobasseri 68
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
Copyright1999 by BG Mobasseri 69
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
Copyright1999 by BG Mobasseri 70
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( )
Copyright1999 by BG Mobasseri 71
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"
Copyright1999 by BG Mobasseri 72
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( )
Copyright1999 by BG Mobasseri 73
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( )
Copyright1999 by BG Mobasseri 74
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( )
Copyright1999 by BG Mobasseri 75
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
Copyright1999 by BG Mobasseri 76
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( )
Copyright1999 by BG Mobasseri 77
Corollary
To find the pre-envelope in the frequency domain, take the original spectrum and chop off the negative part
Copyright1999 by BG Mobasseri 78
Example
Find the pre-envelope of a modulated message
g t( ) =mt( )cos2πfct+θ( )G(f) G+(f)
AM signal
Copyright1999 by BG Mobasseri 79
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
Copyright1999 by BG Mobasseri 80
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
Copyright1999 by BG Mobasseri 81
Tale of Two Pulses
Consider the following two pulses
Which one carries more “information”?
Copyright1999 by BG Mobasseri 82
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?
Copyright1999 by BG Mobasseri 83
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
Copyright1999 by BG Mobasseri 84
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
⎧
⎨ ⎪
⎩ ⎪
⎫
⎬ ⎪
⎭ ⎪
Copyright1999 by BG Mobasseri 85
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
Copyright1999 by BG Mobasseri 86
Signal Representation Summary
Take a real-valued, baseband signal
g(t)G(f)
Copyright1999 by BG Mobasseri 87
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
Copyright1999 by BG Mobasseri 88
Pre-envelope Summary: bandpass
Baseband signal: g t( ) =mt( )cos2πfct+θ( )
G(f)
G+(f)
Copyright1999 by BG Mobasseri 89
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( )
Copyright1999 by BG Mobasseri 90
RF Pulse: Complex Envelope
Find the complex envelope of a T second long RF pulse at frequency fc
g t( ) =ArecttT
⎛ ⎝
⎞ ⎠ cos2πfct( )
Copyright1999 by BG Mobasseri 91
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
Copyright1999 by BG Mobasseri 92
RF Pulse Pre-envelope
Recall
Then ˜ g t( )=g+(t)e−j2πfct
g+ t( ) =ArecttT
⎛ ⎝
⎞ ⎠ e
j2πfct
Copyright1999 by BG Mobasseri 93
Story in the Freq. Domain
-4 0 4
RF PULSE SPECTRUM:two sincs at pulse freq
FREQUENCY (HZ)
Original RF pulse spectrum
-4 0 4
RF PULSE SPECTRUM:two sincs at pulse freq
FREQUENCY (HZ)
Pre-env. Spectrum(only f>0 portion)
Copyright1999 by BG Mobasseri 94
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
Copyright1999 by BG Mobasseri 96
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
Copyright1999 by BG Mobasseri 97
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
Copyright1999 by BG Mobasseri 98
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)
Copyright1999 by BG Mobasseri 99
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
Copyright1999 by BG Mobasseri 100
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
⎛ ⎝
⎞ ⎠
Copyright1999 by BG Mobasseri 101
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)⎡
⎣ ⎤ ⎦
Copyright1999 by BG Mobasseri 102
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
Copyright1999 by BG Mobasseri 103
What Does Distortion Do?
Channel Distortion creates pulse “dispersion”
Channel
interference
Copyright1999 by BG Mobasseri 104
Case of No Distortion
There are two “distortions” we can live with• Scaling• Delay
To
Copyright1999 by BG Mobasseri 105
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
Copyright1999 by BG Mobasseri 106
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
Copyright1999 by BG Mobasseri 107
Amplitude and Phase Response
|H(f)|
/_H(f)f
f
−(2πTd) f
Const amplitude response
Linear phase response
Copyright1999 by BG Mobasseri 108
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
Copyright1999 by BG Mobasseri 109
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
Copyright1999 by BG Mobasseri 110
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)
Copyright1999 by BG Mobasseri 111
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
Copyright1999 by BG Mobasseri 112
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
Copyright1999 by BG Mobasseri 113
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
Copyright1999 by BG Mobasseri 114
Pulse Dispersion
Putting a pulse g(t) through this filter produces 3 overlapping copies
channel with distortionT
>T
Copyright1999 by BG Mobasseri 115
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)[ ]
Copyright1999 by BG Mobasseri 116
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
Copyright1999 by BG Mobasseri 117
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)
Copyright1999 by BG Mobasseri 118
Practice Problems
For pre-envelope: 2.23 For filtering using complex envelope: 2.32