communication systems i

CommunicationSystems I

ECE 5625/4625 Lecture NotesSpring 2007

© 2007Mark A. Wickert

Transmitter

Receiver

ChannelInputTransducer

OutputTransducer

InputMessage

MessageSignal

TransmittedSignal

OutputSignalOutput

Message

ReceivedSignal

Noise and distortion enters the system here

Chapter 1Course Introduction/Overview

Contents

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 1-3

1.2 Where are we in the Curriculum? . . . . . . . . . . . 1-4

1.3 Where are we (cont)? . . . . . . . . . . . . . . . . . . 1-5

1.4 Course Syllabus . . . . . . . . . . . . . . . . . . . . . 1-6

1.5 Instructor Policies . . . . . . . . . . . . . . . . . . . . 1-7

1.6 Communication Lab Connection . . . . . . . . . . . . 1-8

1.7 Software Tools . . . . . . . . . . . . . . . . . . . . . . 1-9

1.8 Comm I/Comm II Course Sequence . . . . . . . . . . 1-10

1.9 Course Introduction and Overview . . . . . . . . . . . 1-11

1.10 A Block Diagram . . . . . . . . . . . . . . . . . . . . . 1-12

1.11 Channel Types . . . . . . . . . . . . . . . . . . . . . . 1-13

1.11.1 Electromagnetic-wave (EM-wave) propagation . 1-131.11.2 Guided EM-wave propagation . . . . . . . . . . 1-171.11.3 Magnetic recording channel . . . . . . . . . . . 1-171.11.4 Optical channel . . . . . . . . . . . . . . . . . . 1-17

1-1

CHAPTER 1. COURSE INTRODUCTION/OVERVIEW

.

1-2 ECE 5625 Communication Systems I

1.1. INTRODUCTION

1.1 Introduction

• Where are we in the ugrad and grad curriculum?

• Course Syllabus

• Instructor policies

• Relationship to the communications lab, ECE 4670

• Software tools

• The Comm I/Comm II course sequence

• Communication systems overview

ECE 5625 Communication Systems I 1-3


1.2 Where are we in the Curriculum?

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1.3. WHERE ARE WE (CONT)?

1.3 Where are we (cont)?

ECE 3205Signals & Systems II

ECE 2205Signals & Systems I

ECE 2610Signals &Systems

ECE 3610Eng. Prob.

& Stats.

ECE 4670Comm.

Lab

ECE 4680DSPLab

ECE 5625Comm.

Systems I

ECE 5630Comm.

Systems II

ECE 5675PLL &Applic.

ECE 5610RandomSignals

ECE 6640Spread

Spectrum

ECE 5720Optical Comm.

ECE 5635Wireless Comm.

ECE 6620Detect. &

Estim. Thy.

ECE 6650Estim. &

Adapt. Fil.

ECE 5650Modern

DSP

ECE 5655Real-Time

DSP

ECE 5615Statistical

Signal Proc

Courses Offered According to Demand

You are Here!

Coding Thy,Image Proc,Sat. Comm,Radar Sys



1.4 Course Syllabus!

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1.5. INSTRUCTOR POLICIES

1.5 Instructor Policies

• Homework papers are due at the start of class

• If business travel or similar activities prevent you from attend-ing class and turning in your homework, please inform me be-forehand

• Grading is done on a straight 90, 80, 70, ... scale with curvingbelow these thresholds if needed

• Homework solutions will be placed on the course Web site inPDF format with security password required; hints pages mayalso be provided



1.6 Communication Lab Connection

• The labs are fairly tightly coupled with the lecture topics

• The communications hardware experience should enhance yourunderstanding of communications theory and analysis

• Lab topics:

– Linear System Characteristics

– Spectrum Analysis

– DSB and AM Modulation and Demodulation

– AM Superheterodyne Receivers

– Frequency Modulation and Demodulation

– Second Order Phase-Lock Loops

• Communications building blocks are dealt with for the mostpart as electronic subsystems

• The spectrum analyzer and vector network analyzer are intro-duced to extend measurement capabilities into the frequencydomain


1.7. SOFTWARE TOOLS

1.7 Software Tools

• Analysis aids

– Calculator, MATLAB, Mathematica, others

• System simulation

– MATLAB/Simulink, VisSim/Comm (used in ECE 4670),others

• Circuit simulation

– Spice type simulator, e.g. the free simulator Qucs avail-able at http://qucs.sourceforge.net/



1.8 Comm I/Comm II Course Sequence

• Communication systems I, this course, continues into a sec-ond semester when ECE 4630/5630 is offered alternate fallsemesters

• The second semester course focuses on digital communica-tions

– An introduction to random signals is provided

– Amplitude, Phase, and frequency shift-keyed modulationschemes are studied in considerable detail

– Coherent versus non-coherent modulation

– The Mobile radio channel is introduced

– Satellite communications is introduced

– Coding theory is introduced


1.9. COURSE INTRODUCTION AND OVERVIEW

1.9 Course Introduction and Overview

• “The theory of systems for the conveyance of information”

• Communication systems must deal with uncertainty (noise andinterference)

– The uncertainty aspects of noise require the use of proba-bility, random variables, and random processes

– In this first course deterministic modeling is used for themost part

• Some important dates:

1915 Transcontinental telephone line completed1918 Armstrong superheterodyne radio receiver per-

fected1938 Television broadcasting beginsWW II Radar and microwave systems developed1948 Transistor invented1956 First transoceanic telephone line completed1960 Laser demonstrated1962 First communications satellite, Telstar I1970’s Commercial relay satellites for voice and data1977 Fiber optic communication systems1980 Satellite switchboards in the sky1990’s Global positioning system (GPS) completed1990’s Cellular telephones widely used1998 Global satellite-based cellular telephone system



1.10 A Block Diagram

• A a high level communication systems are typically describedusing a block diagram

Transmitter

Receiver

ChannelInputTransducer

OutputTransducer

InputMessage

MessageSignal

TransmittedSignal

OutputSignalOutput

Message

ReceivedSignal

Noise and distortion enters the system here

• There is an information source as the input and an informationsink to receive the output

• The block diagram shown above is very general

– The source may be digital or analog

– The transmission may be at baseband or on a radio fre-quency (RF) carrier

– The channel can take on may possible forms


1.11. CHANNEL TYPES

1.11 Channel Types

1.11.1 Electromagnetic-wave (EM-wave) prop-

agation

Skip-wavepropagation

Ground wavepropagation

Line-of-sightpropagation

Earth

Ionosphere

Comm Satellite

Transiosphere (LOS)

• When you think wireless communications this is the channeltype most utilized

• The electromagnetic spectrum is a natural resource

• The above figure depicts several propagation modes

– Lower frequencies/long wavelengths tend to follow theearths surface

– Higher frequencies/short wavelengths tend to propagatein straight lines

• Reflection of radio waves by the ionosphere occurs for fre-quencies below about 100 MHz (more so at night)


1.11. CHANNEL TYPES

• Examples of public (commercial) and government (militaryapplications and the frequency bands they operate in

• There is a hierarchy of organizations that regulate how theavailable spectrum is allocated

– Worldwide there is the International TelecommunicationsUnion (ITU), which convenes regional and worldwide Ad-ministrative Radio Conferences (RARC & WARC)

– Within the United States we have the Federal Communi-cations Commission (FCC)

• Cellular telephony, wireless LAN (WLAN), and HDTV broad-casting, are examples where the FCC continues to make allo-cation changes



• At frequencies above 1–2 GHz oxygen and water vapor absorband scatter radio waves

• Satellite communications, which use the microwave frequencybands, must account for this in what is known as the link powerbudget

Water vapor and oxygen attenuation

Rainfall rate attenuation

6223 120


1.11. CHANNEL TYPES

1.11.2 Guided EM-wave propagation

• Communication using transmission lines such as twisted-pairand coax cable

1.11.3 Magnetic recording channel

• Disk drives, fixed (at one time flexible too)

• Video and audio

1.11.4 Optical channel

• Free-space

• Fiber-optic

• CD, DVD, HD-DVD, etc.



Example 1.1: Distortion in a Sat-Comm Channel

• Wideband satellite communication channels are subject to bothlinear and non-linear distortion

UplinkChannel

Transmitter

Transponder

Receiver

PSKMod

HPA(TWTA)

HPA(TWTA)

ModulationImpairments

BandpassFiltering

BandpassFiltering

BandpassFiltering

PSK Demod (bit true withfull synch)

AdaptiveEqualizer

OtherSignals

Mod

.

Mod

.

WGNNoise(off)

DataSource

RecoveredData

DownlinkChannel

WGNNoise(on)

OtherSignals

Mod

.

OtherSignals

! Spurious PM! Incidental AM! Clock jitter

! IQ amplitude imbalance! IQ phase imbalance! Waveform asymmetry

and rise/fall time

! Phase noise! Spurious PM! Incidental AM! Spurious outputs

! Phase noise! Spurious PM! Incidental AM! Spurious outputs

! BPSK! QPSK! OQPSK

Wideband Sat-Comm simulation model

• An adaptive filter can be used to estimate the channel dis-tortion, for example a technique known as decision feedbackequalization


1.11. CHANNEL TYPES

M1 TapComplex

FIRRe

z-1M1 Tap

ComplexFIR

Im 2

2

M2 TapRealFIR

M2 TapRealFIR

CM Error/LMS Update

DD Error/LMS Update

CM Error/LMS Update

DD Error/LMS Update

TapWeightUpdateSoft I/Q outputs

from demod at sample rate = 2Rs

RecoveredI Data

RecoveredQ Data

Stagger forOQPSK, omitfor QPSK

+

+

-

-

-

-

+

+AdaptMode

DecisionFeedback

DecisionFeedback

µCM, µDDµDF, !

An adaptive baseband equalizer implemented in FPGA1

• Since the distortion is both linear (bandlimiting) and nonlin-ear (amplifiers and other interference), the distortion cannot becompletely eliminated

• The following two figures show first the modulation 4-phasesignal points with and with out the equalizer, and then the biterror probability (BEP) versus received energy per bit to noisepower spectral density ratio (Eb/N0)

1Mark Wickert, Shaheen Samad, and Bryan Butler. “An Adaptive Baseband Equalizer for HighData Rate Bandlimited Channels, Proceedings 2006 International Telemetry Conference, Session5, paper 06–5-03.



!1.5 !1 !0.5 0 0.5 1 1.5!1.5

!1

!0.5

0

0.5

1

1.5

In!phase

Quadrature

!1.5 !1 !0.5 0 0.5 1 1.5!1.5

!1

!0.5

0

0.5

1

1.5

In!phaseQuadrature

Before Equalization: Rb = 300 Mbps After Equalization: Rb = 300 Mbps

OQPSK scatter plots with and without the equalizer

6 8 10 12 14 16 18 20 22 2410!7

10!6

10!5

10!4

10!3

10!2

Eb/N0 (dB)

Prob

abilit

y of B

it Erro

r

Theory EQ NO EQ

4.0 dB 8.1 dB

Semi-Analytic Simulation

300 MBPS BER Performance with a 40/0 Equalizer

BEP versus Eb/N0 in dB


Chapter 2Signal and Linear System Analysis

Contents

2.1 Signal Models . . . . . . . . . . . . . . . . . . . . . . 2-32.1.1 Deterministic and Random Signals . . . . . . . . 2-32.1.2 Periodic and Aperiodic Signals . . . . . . . . . . 2-32.1.3 Phasor Signals and Spectra . . . . . . . . . . . . 2-42.1.4 Singularity Functions . . . . . . . . . . . . . . . 2-7

2.2 Signal Classifications . . . . . . . . . . . . . . . . . . 2-112.3 Generalized Fourier Series . . . . . . . . . . . . . . . 2-142.4 Fourier Series . . . . . . . . . . . . . . . . . . . . . . 2-20

2.4.1 Complex Exponential Fourier Series . . . . . . . 2-202.4.2 Symmetry Properties of the Fourier Coefficients 2-232.4.3 Trigonometric Form . . . . . . . . . . . . . . . 2-252.4.4 Parseval’s Theorem . . . . . . . . . . . . . . . . 2-262.4.5 Line Spectra . . . . . . . . . . . . . . . . . . . 2-262.4.6 Numerical Calculation of Xn . . . . . . . . . . . 2-312.4.7 Other Fourier Series Properties . . . . . . . . . . 2-37

2.5 Fourier Transform . . . . . . . . . . . . . . . . . . . . 2-382.5.1 Amplitude and Phase Spectra . . . . . . . . . . 2-39

2-1

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

2.5.2 Symmetry Properties . . . . . . . . . . . . . . . 2-392.5.3 Energy Spectral Density . . . . . . . . . . . . . 2-402.5.4 Transform Theorems . . . . . . . . . . . . . . . 2-422.5.5 Fourier Transforms in the Limit . . . . . . . . . 2-512.5.6 Fourier Transforms of Periodic Signals . . . . . 2-532.5.7 Poisson Sum Formula . . . . . . . . . . . . . . 2-59

2.6 Power Spectral Density and Correlation . . . . . . . . 2-602.6.1 The Time Average Autocorrelation Function . . 2-612.6.2 Power Signal Case . . . . . . . . . . . . . . . . 2-622.6.3 Properties of R(τ ) . . . . . . . . . . . . . . . . 2-63

2.7 Linear Time Invariant (LTI) Systems . . . . . . . . . 2-702.7.1 Stability . . . . . . . . . . . . . . . . . . . . . . 2-722.7.2 Transfer Function . . . . . . . . . . . . . . . . . 2-722.7.3 Causality . . . . . . . . . . . . . . . . . . . . . 2-732.7.4 Properties of H( f ) . . . . . . . . . . . . . . . . 2-742.7.5 Response to Periodic Inputs . . . . . . . . . . . 2-782.7.6 Distortionless Transmission . . . . . . . . . . . 2-782.7.7 Group and Phase Delay . . . . . . . . . . . . . . 2-792.7.8 Nonlinear Distortion . . . . . . . . . . . . . . . 2-832.7.9 Ideal Filters . . . . . . . . . . . . . . . . . . . . 2-852.7.10 Realizable Filters . . . . . . . . . . . . . . . . . 2-872.7.11 Pulse Resolution, Risetime, and Bandwidth . . . 2-91

2.8 Sampling Theory . . . . . . . . . . . . . . . . . . . . . 2-972.9 The Hilbert Transform . . . . . . . . . . . . . . . . . 2-972.10 The Discrete Fourier Transform and FFT . . . . . . . 2-97


2.1. SIGNAL MODELS

2.1 Signal Models

2.1.1 Deterministic and Random Signals

• Deterministic Signals, used for this course, can be modeled ascompletely specified functions of time, e.g.,

x(t) = A(t) cos[2π f0(t)t + φ(t)]

– Note that here we have also made the amplitude, fre-quency, and phase functions of time

– To be deterministic each of these functions must be com-pletely specified functions of time

• Random Signals, used extensively in Comm Systems II, takeon random values with known probability characteristics, e.g.,

x(t) = x(t, ζi)

where ζi corresponds to an elementary outcome from a samplespace in probability theory

– The ζi create a ensemble of sample functions x(t, ζi), de-pending upon the outcome drawn from the sample space

2.1.2 Periodic and Aperiodic Signals

• A deterministic signal is periodic if we can write

x(t + nT0) = x(t)

for any integer n, with T0 being the signal fundamental period



• A signal is aperiodic otherwise, e.g.,

�(t) =�

1, |t | ≤ 1/20, otherwise

(a) periodic signal, (b) aperiodic signal, (c) random signal

2.1.3 Phasor Signals and Spectra• A complex sinusoid can be viewed as a rotating phasor

x(t) = Aej (ω0t+θ), −∞ < t < ∞

• This signal has three parameters, amplitude A, frequency ω0,and phase θ

• The fixed phasor portion is Aejθ while the rotating portion is

ejω0t


2.1. SIGNAL MODELS

• This signal is periodic with period T0 = 2π/ω0

• It also related to the real sinusoid signal A cos(ω0t + θ) viaEuler’s theorem

x(t) = Re�

x(t)�

= Re�

A cos(ω0t + θ) + j A sin(ω0t + θ)�

= A cos(ω0t + θ)

(a) obtain x(t) from x(t), (b) obtain x(t) from x(t) and x∗(t)

• We can also turn this around using the inverse Euler formula

x(t) = A cos(ω0t + θ)

= 12

x(t) + 12

x∗(t)

= Aej (ω0t+θ) + Ae

− j (ω0t+θ)

2

• The frequency spectra of a real sinusoid is the line spectra plot-ted in terms of the amplitude and phase versus frequency



• The relevant parameters are A and θ for a particular f0 =ω0/(2π)

(a) Single-sided line spectra, (b) Double-sided line spectra

• Both the single-sided and double-sided line spectra, shownabove, correspond to the real signal x(t) = A cos(2π f0t + θ)

Example 2.1: Multiple Sinusoids

• Suppose that

x(t) = 4 cos(2π(10)t + π/3) + 24 sin(2π(100)t − π/8)

• Find the two-sided amplitude and phase line spectra of x(t)

• First recall that cos(ω0t − π/2) = sin(ω0t), so

x(t) = 4 cos(2π(10)t + π/3) + 24 cos(2π(100)t − 5π/8)

• The complex sinusoid form is directly related to the two-sidedline spectra since each real sinusoid is composed of positiveand negative frequency complex sinusoids

x(t) = 2�e

j (2π(10)t+π/3) + e− j (2π(10)t+π/3)

�

+ 12�e

j (2π(100)t−5π/8) + e− j (2π(100)t−5π/8)

�


2.1. SIGNAL MODELS

f (Hz)

f (Hz)

Ampli

tude

Phase

10

12

2

100-100 -10

5!/8

-5!/8-!/3

!/3

Two-sided amplitude and phase line spectra

2.1.4 Singularity FunctionsUnit Impulse (Delta) Function

• Singularity functions, such as the delta function and unit step

• The unit impulse function, δ(t) has the operational properties�

t2

t1

δ(t − t0) dt = 1, t1 < t0 < t2

δ(t − t0) = 0, t �= t0

which implies that for x(t) continuous at t = t0, the sifting

property holds� ∞

−∞x(t)δ(t − t0) dt = x(t0)

– Alternatively the unit impulse can be defined as� ∞

−∞x(t)δ(t) dt = x(0)



• Properties:

1. δ(at) = δ(t)/|a|2. δ(−t) = δ(t)

3. Sifting property special cases

�t2

t1

x(t)δ(t − t0) dt =

x(t0), t1 < t0 < t2

0, otherwiseundefined, t0 = t1 or t0 = t2

4. Sampling property

x(t)δ(t − t0) = x(t0)δ(t − t0)

for x(t) continuous at t = t0

5. Derivative property�

t2

t1

x(t)δ(n)(t − t0) dt = (−1)nx

(n)(t0)

= (−1)nd

nx(t)

dtn

��t=t0

Note: Dealing with the derivative of a delta function re-quires care

• A test function for the unit impulse function helps our intuitionand also helps in problem solving

• Two functions of interest are

δ�(t) = 12�

�

�t

2�

�=

�12� , |t | ≤ �

0, otherwise

δ1�(t) = �

�1π t

sinπ t

�

�2


2.1. SIGNAL MODELS

Test functions for the unit impulse δ(t): (a) δ�(t), (b) δ1�(t)

• In both of the above test functions letting � → 0 results in afunction having the properties of a true delta function

Unit Step Function

• The unit step function can be defined in terms of the unit im-pulse

u(t) ≡�

t

−∞δ(τ ) dτ =

0, t < 01, t > 0undefined, t = 0

alsoδ(t) = du(t)

dt



Example 2.2: Unit Impulse 1st-Derivative

• Consider � ∞

−∞x(t)δ�(t) dt

• Using the rectangular pulse test function, δ�(t), we note that

δ�(t) = 12�

�

�t

2�

�also= 1

2�

�u(t + �) − u(t − �)

�

anddδ�(t)

dt= 1

2�

�δ(t + �) − δ(t − �)

�

• Placing the above in the integrand with x(t) we obtain, withthe aid of the sifting property, that

� ∞

−∞x(t)δ�(t) dt = lim

�→0

12�

�x(t + �) − x(t − �)

�

= lim�→0

−�x(t − �) − x(t + �)

�

2�= −x

�(0)


2.2. SIGNAL CLASSIFICATIONS

2.2 Signal Classifications• From circuits and systems we know that a real voltage or cur-

rent waveform, e(t) or i(t) respectively, measured with respec-tive a real resistance R, the instantaneous power is

P(t) = e(t)i(t) = i2(t)R W

• On a per-ohm basis, we obtain

p(t) = P(t)/R = i2(t) W/ohm

• The average energy and power can be obtain by integratingover the interval |t | ≤ T with T → ∞

E = limT →∞

�T

−T

i2(t) dt Joules/ohm

P = limT →∞

12T

�T

−T

i2(t) dt W/ohm

• In system engineering we take the above energy and powerdefinitions, and extend them to an arbitrary signal x(t), pos-sibly complex, and define the normalized energy (e.g. 1 ohmsystem) as

E�= lim

T →∞

�T

−T

|x(t)|2 dt =� ∞

−∞|x(t)|2 dt

P�= lim

T →∞1

2T

�T

−T

|x(t)|2 dt



• Signal Classes:

1. x(t) is an energy signal if and only if 0 < E < ∞ so thatP = 0

2. x(t) is a power signal if and only if 0 < P < ∞ whichimplies that E → ∞

Example 2.3: Real Exponential

• Consider x(t) = Ae−αt

u(t) where α is real

• For α > 0 the energy is given by

E =� ∞

0

�Ae

−αt�2

dt = A2e

−2αt

−2α

��∞

0

= A2

2α

• For α = 0 we just have x(t) = Au(t) and E → ∞

• For α < 0 we also have E → ∞

• In summary, we conclude that x(t) is an energy signal for α >0

• For α > 0 the power is given by

P = limT →∞

12T

A2

2α

�1 − e

−αT�

= 0

• For α = 0 we have

P = limT →∞

12T

· A2T = A

2

2


2.2. SIGNAL CLASSIFICATIONS

• For α < 0 we have P → ∞

• In summary, we conclude that x(t) is a power signal for α = 0

Example 2.4: Real Sinusoid

• Consider x(t) = A cos(ω0t + θ), −∞ < t < ∞

• The signal energy is infinite since upon squaring, and integrat-ing over one cycle, T0 = 2π/ω0, we obtain

E = limN→∞

�N T0/2

−N T0/2A

2 cos2(ω0t + θ) dt

= limN→∞

N

�T0/2

−T0/2A

2 cos2(ω0t + θ) dt

= limN→∞

NA

2

2

�T0/2

−T0/2

�1 + cos(2ω0t + 2θ) dt

= limN→∞

NA

2

2· T0 → ∞

• The signal average power is finite since the above integral isnormalized by 1/(N T0), i.e.,

P = limN→∞

1N T0

· NA

2

2· T0 = A

2

2



2.3 Generalized Fourier SeriesThe goal of generalized Fourier series is to obtain a representation ofa signal in terms of points in a signal space or abstract vector space.The coordinate vectors in this case are orthonomal functions. Thecomplex exponential Fourier series is a special case.

• Let �A be a vector in a three dimensional vector space

• Let �a1, �a2, and �a3 be linearly independent vectors in the samethree dimensional space, then

c1�a1 + c2�a2 + c3�a3 = 0 (zero vector)

only if the constants c1 = c2 = c3 = 0

• The vectors �a1, �a2, and �a3 also span the three dimensional space,that is for any vector �A there exists a set of constants c1, c2, andc3 such that

�A = c1�a1 + c2�a2 + c3�a3

• The set {�a1, �a2, �a3} forms a basis for the three dimensionalspace

• Now let {�a1, �a2, �a3} form an orthogonal basis, which impliesthat

�ai · �a j = (�ai, �a j) = ��ai, �a j� = 0, i �= j

which says the basis vectors are mutually orthogonal

• From analytic geometry (and linear algebra), we know that wecan find a representation for �A as

�A = (�a1 · �A)

|�a1|2+ (�a2 · �A)

|�a2|2+ (�a3 · �A)

|�a3|2


2.3. GENERALIZED FOURIER SERIES

which implies that

�A =3�

i=1

ci �ai

where

ci = �ai · �A|�ai |2

, i = 1, 2, 3

is the component of �A in the �ai direction

• We now extend the above concepts to a set of orthogonal func-tions {φ1(t), φ2(t), . . . ,φN(t)} defined on to ≤ t ≤ t0 + T ,where the dot product (inner product) associated with the φn’sis

�φm(t), φn(t)

�=

�t0+T

t0

φm(t)φ∗n(t) dt

= cnδmn =�

cn, n = m

0, n �= m

• The φn’s are thus orthogonal on the interval [t0, t0 + T ]

• Moving forward, let x(t) be an arbitrary function on [t0, t0+T ],and consider approximating x(t) with a linear combination ofφn’s, i.e.,

x(t) � xa(t) =N�

n=1

Xnφn(t), t0 ≤ t ≤ t0 + T,

where a denotes approximation



• A measure of the approximation error is the integral squarederror (ISE) defined as

�N =�

T

��x(t) − xa(t)��2

dt,

where�

Tdenotes integration over any T long interval

• To find the Xn’s giving the minimum �N we expand the aboveintegral into three parts (see homework problems)

�N =�

T

|x(t)|2 dt −N�

n=1

1cn

��

�

T

x(t)φ∗n(t) dt

��2

+N�

n=1

cn

��Xn − 1cn

�

T

x(t)φ∗n(t) dt

��2

– Note that the first two terms are independent of the Xn’sand the last term is nonnegative (missing steps are in texthomework problem 2.14)

• We conclude that �N is minimized for each n if each elementof the last term is made zero by setting

Xn = 1cn

�

T

x(t)φ∗n(t) dt Fourier Coefficient

• This also results in

��N

�min =

�

T

|x(t)|2 dt −N�

n=1

cn|Xn|2



• Definition: The set of of φn’s is complete if

limN→∞

(�N)min = 0

for�

T|x(t)|2 dt < ∞

– Even if though the ISE is zero when using a completeset of orthonormal functions, there may be isolated pointswhere x(t) − xa(t) �= 0

• Summary

x(t) = l.i.m.∞�

n=1

Xnφn(t)

Xn = 1cn

�

T

x(t)φ∗n(t) dt

– The notation l.i.m. stands for limit in the mean, which isa mathematical term referring to the fact that ISE is theconvergence criteria

• Parseval’s theorem: A consequence of completeness is�

T

|x(t)|2 dt =∞�

n=1

cn|Xn|2



Example 2.5: A Three Term Expansion

• Approximate the signal x(t) = cos 2π t on the interval [0, 1]using the following basis functions

0.2 0.4 0.6 0.8 1

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1

0.2 0.4 0.6 0.8 1

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1x(t) !1(t)

t t

0.2 0.4 0.6 0.8 1

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1

0.2 0.4 0.6 0.8 1

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1

!2(t) !3(t)

t t

Signal x(t) and basis functions φi(t), i = 1, 2, 3

• To begin with it should be clear that φ1(t), φ2(t), and φ3(t)are mutually orthogonal since the integrand associated with theinner product, φi(t) · φ∗

j(t) = 0, for i �= j, i, j = 1, 2, 3



• Before finding the Xn’s we need to find the cn’s

c1 =�

T

|φ1(t)|2 dt

� 1/4

0|1|2 dt = 1/4

c2 =�

T

|φ2(t)|2 dt = 1/2

c3 =�

T

|φ3(t)|2 dt = 1/4

• Now we can compute the Xn’s:

X1 = 4�

T

x(t)φ∗1(t) dt

= 4� 1/4

0cos(2π t) dt = 2

πsin(2π t)

��1/4

0= 2

π

X2 = 2� 3/4

1/4cos(2π t) dt = 1

πsin(2π t)

��3/4

1/4= −2

π

X3 = 4� 1

3/4cos(2π t) dt = 2

πsin(2π t)

��1

3/4= 2

π

t

x(t)

xa(t)

0.2 0.4 0.6 0.8 1

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1

-2/!

2/!

Functional approximation



• The integral-squared error, �N , can be computed as follows:

�N =�

T

��x(t) −3�

n=1

Xnφn(t)

��2

dt

=�

T

|x(t)|2 dt −3�

n=1

cn|Xn|2

= 12

− 14

��2π

��2

− 12

��2π

��2

− 14

��2π

��2

= 12

−��2π

��2

= 0.0947

2.4 Fourier Series

When we choose a particular set of basis functions we arrive at themore familiar Fourier series.

2.4.1 Complex Exponential Fourier Series

• A set of φn’s that is complete is

φn(t) = ejnω0t, n = 0, ±1, ±2, . . .

over the interval (t0, t0 + T0) where ω0 = 2π/T0 is the periodof the expansion interval


2.4. FOURIER SERIES

proof of orthogonality

�φm(t), φn(t)

�=

�t0+T0

t0

ejm

2π t

T0 e− jn

2π t

T0 dt =�

t0+T0

t0

ej

2πT0

(m−n)tdt

=

�t0+T0

t0dt, m = n

�t0+T0

t0

�cos[2π(m − n)t/T0]

+ j sin[2π(m − n)t/T0]�

dt, m �= n

=�

T0, m = n

0, m �= n

We also conclude that cn = T0

• Complex exponential Fourier series summary:

x(t) =∞�

n=−∞Xne

jnω0t, t0 ≤ t ≤ t0 + T0

where Xn = 1T0

�

T0

x(t)e− jnω0t

• The Fourier series expansion is unique

Example 2.6: x(t) = cos2 ω0t

• If we expand x(t) into complex exponentials we can immedi-ately determine the Fourier coefficients

x(t) = 12

+ 12

cos 2ω0t

= 12

+ 14

ej2ω0t + 1

4e

− j2ω0t



• The above implies that

Xn =

12, n = 014, n = ±20, otherwise

Time Average Operator

• The time average of signal v(t) is defined as

�v(t)� �= limT →∞

12T

�T

−T

v(t) dt

• Note that

�av1(t) + bv2(t)� = a�v1(t)� + b�v2(t)�,where a and b are arbitrary constants

• If v(t) is periodic, with period T0, then

�v(t)� = 1T0

�

T0

v(t) dt

• The Fourier coefficients can be viewed in terms of the timeaverage operator

• Let v(t) = x(t)e− jnω0t using e− jθ = cos θ − j sin θ , we find

that

Xn = �v(t)� = �x(t)e− jnω0t�= �x(t) cos nω0t� − j�x(t) sin nω0t�


2.4. FOURIER SERIES

2.4.2 Symmetry Properties of the Fourier Coef-ficients

• For x(t) real, the following coefficient symmetry propertieshold:

1. X∗n

= X−n

2. |Xn| = |X−n|3. � Xn = −� X−n

proof

X∗n

=�

1T0

�

T0

x(t)e− jnω0tdt

�∗

= 1T0

�

T0

x(t)e− j (−n)ω0tdt = X−n

since x∗(t) = x(t)

• Waveform symmetry conditions produce special results too

1. If x(−t) = x(t) (even function), then

Xn = Re�

Xn

�, i.e., Im

�Xn

�= 0

2. If x(−t) = −x(t) (odd function), then

Xn = Im�

Xn

�, i.e., Re

�Xn

�= 0

3. If x(t ± T0/2) = −x(t) (odd half-wave symmetry), then

Xn = 0 for n even



Example 2.7: Odd Half-wave Symmetry Proof

• Consider

Xn = 1T0

�t0=T0/2

t0

x(t)e− jnω0tdt + 1

T0

�t0+T0

t0+T0/2x(t �)e− jnω0t

�dt

�

• In the second integral we change variables by letting t = t� −

T0/2

Xn = 1T0

�t0+T0/2

t0

x(t)e− jnω0tdt

+ 1T0

�t+T0/2

t0

x(t − T0/2)� �� −x(t)

e− jnω0(t+T0/2)

dt

=�

1 − e− jnω0T0/2

� 1T0

�t0+T0/2

t0

x(t)e− jnω0tdt

but nω0(T0/2) = n(2π/T0)(T0/2) = nπ , thus

1 − e− jnπ =

�2, n odd0, n even

• We thus see that the even indexed Fourier coefficients are in-deed zero under odd half-wave symmetry


2.4. FOURIER SERIES

2.4.3 Trigonometric Form

• The complex exponential Fourier series can be arranged as fol-lows

x(t) =∞�

n=−∞Xne

jnω0t

= X0 +∞�

n=1

�Xne

jnω0t + X−ne− jnω0t

�

• For real x(t), we may know that |X−n| = |Xn| and � Xn =−� X−n, so

x(t) = X0 +∞�

n=1

�|Xn|e j[nω0t+ � Xn] + |Xn|e− j[nω0t+ � Xn]

�

= X0 + 2∞�

n=1

|Xn| cos�nω0t + � Xn

�

since cos(x) = (e j x + e− j x)/2

• From the trig identity cos(u + v) = cos u cos v − sin u sin v , itfollows that

x(t) = X0 +∞�

n=1

An cos(nω0t) +∞�

n=1

Bn sin(nω0t)

where

An = 2�x(t) cos(nω0t)�Bn = 2�x(t) sin(nω0t)�



2.4.4 Parseval’s Theorem

• Fourier series analysis are generally used for periodic signals,i.e., x(t) = x(t + nT0) for any integer n

• With this in mind, Parseval’s theorem becomes

P = 1T0

�

T0

|x(t)|2 dt =∞�

n=−∞|Xn|2

= X20 + 2

∞�

n=1

|Xn|2 (W)

Note: A 1 ohm system is assumed

2.4.5 Line Spectra

• Line spectra was briefly reviewed earlier for simple signals

• For any periodic signal having Fourier series representation wecan obtain both single-sided and double-sided line spectra

• The double-sided magnitude and phase line spectra is mosteasily obtained form the complex exponential Fourier series,while the single-sided magnitude and phase line spectra can beobtained from the trigonometric form

Double-sidedmag. and phase

⇐⇒∞�

n=−∞Xne

j2π(n f0)t

Single-sidedmag. and phase

⇐⇒ X0 + 2∞�

n=1

|Xn| cos[2π(n f0)t + � Xn]


2.4. FOURIER SERIES

– For the double-sided simply plot as lines |Xn| and � Xn

versus n f0 for n = 0, ±1, ±2, . . .

– For the single-sided plot |X0| and � X0 as a special casefor n = 0 at n f0 = 0 and then plot 2|Xn| and � X0 versusn f0 for n = 1, 2, . . .

Example 2.8: Cosine Squared

• Consider

x(t) = A cos2(2π f0t + θ) = A

2+ A

2cos

�2π(2 f0)t + 2θ1

�

= A

2+ A

4e

j2θ1ej2π(2 f0)t + A

4e

− j2θ1e− j2π(2 f0)t

Double-Sided

Single-Sided

-2f0-2f0

2f0

2f0 2f0

2f0f

f f

f



Example 2.9: Pulse Train

t

A

x(t)

0 ! T0 T0 + !-T0-2T0

. . .. . .

Periodic pulse train

• The pulse train signal is mathematically described by

x(t) =∞�

n=−∞A�

�t − nT0 − τ/2

τ

�

• The Fourier coefficients are

Xn = 1T0

� τ

0Ae

− j2π(n f0)t dt = A

T0· e

− j2π(n f0)t

− j2π(n f0)

��τ

0

= A

T0· 1 − e

− j2π(n f0)τ

j2π(n f0)

= Aτ

T0· e

jπ(n f0)τ − e− jπ(n f0)τ

(2 j)π(n f0)τ· e

− jπ(n f0)τ

= Aτ

T0· sin[π(n f0)τ ]

[π(n f0)τ ]· e

− jπ(n f0)τ

• To simplify further we define

sinc(x)�= sin(πx)

πx


2.4. FOURIER SERIES

• Finally,

Xn = Aτ

T0sinc(n f0τ )e− jπ(n f0)τ , n = 0, ±1, ±2, . . .

• To plot the line spectra we need to find |Xn| and � Xn

|Xn| = Aτ

T0|sinc[(n fo)τ ]|

� Xn =

−π(n f0)τ, sinc(n foτ ) > 0−π(n f0)τ + π, n f0 > 0 and sinc(n f0τ ) < 0−π(n f0)τ − π, n f0 < 0 and sinc(n f0τ ) < 0

• Plot some double-sided line spectra example using MATLAB

• First we create a helper function that takes as input a vector offrequency values n f0 and the coefficients Xn

function Line_Spectra(fk,Xk,mode)

% Line_Spectra(fk,Xk,mode1) (file Line_Spectra.m)

%

% Plot Double-Sided Line Spectra

%----------------------------------------------------

% fk = vector of real sinusoid frequencies

% Xk = magnitude and phase at each frequency in fk

% mode = ’mag’ or ’phase’ plot

%

% % Mark Wickert, January 2007

switch lower(mode) % not case sensitive

case ’mag’,’magnitude’ % two choices work

stem(fk,abs(Xk),’filled’);

grid

axis([-1.05*max(fk) 1.05*max(fk) 0 1.05*max(abs(Xk))])

ylabel(’Magnitude’)

xlabel(’Frequency (Hz)’)

case ’phase’

stem(fk,angle(Xk),’filled’);

grid

axis([-1.05*max(fk) 1.05*max(fk), ...

-1.1*max(abs(angle(Xk))) 1.1*max(abs(angle(Xk)))])

ylabel(’Phase (rad)’)



xlabel(’Frequency (Hz)’)

otherwise

error(’mode must be mag or phase’)

end

• As a specific example enter the following at the MATLAB com-mand prompt

>> n = -25:25;

>> tau = 0.125; f0 = 1; A = 1;

>> Xn = A*tau*f0*sinc(n*f0*tau).*exp(-j*pi*n*f0*tau);

>> subplot(211)

>> Line_Spectra(n*f0,Xn,’mag’)

>> subplot(212)

>> Line_Spectra(n*f0,Xn,’phase’)

!25 !20 !15 !10 !5 0 5 10 15 20 250

0.05

0.1

Magn

itude

Frequency (Hz)

!25 !20 !15 !10 !5 0 5 10 15 20 25

!2

0

2

Phas

e (ra

d)

Frequency (Hz)

1/! = 8

f0 = 1, ! = 0.125A!f0 = 0.125


2.4. FOURIER SERIES

2.4.6 Numerical Calculation of Xn

• Here we consider a purely numerical calculation of the Xk co-efficients from a single period waveform description of x(t)

• In particular, we will use MATLAB’s fast Fourier transform

(FFT) function to carry out the numerical integration

• By definition

Xk = 1T0

�

T0

x(t)e− j2πk f0tdt, k = 0, ±1, ±2, . . .

• A simple rectangular integration approximation to the aboveintegral is

Xk � 1T0

N−1�

n=0

x(nT )e− jk2π(n f0)T0/N · T0

N, k = 0, ±1, ±2, . . .

where N is the number of points used to partition the timeinterval [0, T0] and T = T0/N is the time step

• Using the fact that 2π f0T0 = 2π , we can write that

Xk � 1N

N−1�

n=0

x(nT )e− j2πkn

N , k = 0, ±1, ±2, . . .

• Note that the above must be evaluated for each Fourier coeffi-cient of interest

• Also note that the accuracy of the Xk values depends on thevalue of N



– For k small and x(t) smooth in the sense that the harmon-ics rolloff quickly, N on the order of 100 may be adequate

– For k moderate, say 5–50, N will have to become increas-ingly larger to maintain precision in the numerical integral

Calculation Using the FFT

• The FFT is a powerful digital signal processing (DSP) func-tion, which is a computationally efficient version of thediscrete

Fourier transfrom (DFT)

• For the purposes of the problem at hand, suffice it to say thatthe FFT is just an efficient algorithm for computing

X [k] =N−1�

n=0

x[n]e− j2πkn/N , k = 0, 1, 2, . . . , N − 1

• If we let x[n] = x(nT ), then it should be clear that

Xk � 1N

X [k], k = 0, 1, . . . ,N

2

• To obtain Xk for k¡0 note that

X−k � 1N

X [−k] = 1N

N−1�

n=0

x(nT )e− j2π(−k)n

N

= 1N

N−1�

n=0

x(nT )e− j2π(N−k)n

N = X [N − k]

since e− j2π Nn/N = e

− j2πn = 1


2.4. FOURIER SERIES

• In summary

Xk ��

X [k]/N , 0 ≤ k ≤ N/2X [N − k]/N , −N/2 ≤ k < 0

• To use the MATLAB function fft() to obtain the Xk we sim-ply let

X = fft(x)

where x = {x(t) : t = 0, T0/N , 2T0/N , . . . , T0(N − 1)/N }

• Remember in MATLAB that X [0] is really found in X[1], etc.

Example 2.10: Finite Rise/Fall-Time PulseTrain

t

x(t)1

1/2

0 tr T0!

!

! + tr

Pulse width = !Rise and fall time = tr

Pulse train with finite rise and fall time edges

• Shown above is one period of a finite rise and fall time pulsetrain

• We will numerically compute the Fourier series coefficients ofthis signal using the FFT

• The MATLAB function trap pulse was written to generateone period of the signal using N samples



function [xp,t] = trap_pulse(N,tau,tr)

% xp = trap_pulse(N,tau,tr)

%

% Mark Wickert, January 2007

n = 0:N-1;

t = n/N;

xp = zeros(size(t));

% Assume tr and tf are equal

T1 = tau + tr;

% Create one period of the trapezoidal pulse waveform

for k=1:N

if t(k) <= tr

xp(k) = t(k)/tr;

elseif (t(k) > tr & t(k) <= tau)

xp(k) = 1;

elseif (t(k) > tau & t(k) < T1)

xp(k) = -t(k)/tr+ 1 + tau/tr;

else

xp(k) = 0;

end

end

• We now plot the double-sided line spectra for τ = 1/8 and twovalues of rise-time tr

>> % tau = 1/8, tr = 1/20

>> N = 1024;

>> [xp,t] = trap_pulse(N,1/8,1/20);

>> Xp = fft(xp);

>> subplot(211)

>> plot(t,xp)

>> grid

>> ylabel(’x(t)’)

>> xlabel(’Time (s)’)

>> subplot(212)

>> Xp_shift = fftshift(Xp)/N;

>> f = -N/2:N/2-1;

>> Line_Spectra(f,Xp_shift,’mag’)

>> axis([-25 25 0 .15])

>> print -tiff -depsc line_spec2.eps

>> % tau = 1/8, tr = 1/10

>> xp = trap_pulse(N,1/8,1/10);

>> Xp = fft(xp);


2.4. FOURIER SERIES

>> Xp_shift = fftshift(Xp)/N;

>> f = N/2:N/2-1;

>> subplot(211)

>> plot(t,xp)

>> grid

>> ylabel(’x(t)’)

>> xlabel(’Time (s)’)

>> subplot(212)

>> Line_Spectra(f,Xp_shift,’mag’)

>> axis([-25 25 0 .15])

>> print -tiff -depsc line_spec3.eps

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

x(t)

Time (s)

!25 !20 !15 !10 !5 0 5 10 15 20 250

0.05

0.1

0.15

Magn

itude

Frequency (Hz)

1/! = 1/8

1/8

1/20

Sidelobes smaller than ideal pulse train which has zero rise time

f0 = 1, ! = 0.125, tr = 1/20

Signal x(t) and line spectrum for τ = 1/8 and tr = 1/20



0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

x(t)

Time (s)

!25 !20 !15 !10 !5 0 5 10 15 20 250

0.05

0.1

0.15

Magn

itude

Frequency (Hz)

1/! = 1/8

1/8

1/10

Sidelobes smaller than with tr = 1/20 case

f0 = 1, ! = 0.125, tr = 1/10

Signal x(t) and line spectrum for τ = 1/8 and tr = 1/10


2.4. FOURIER SERIES

2.4.7 Other Fourier Series Properties• Given x(t) has Fourier series (FS) coefficients Xn, if

y(t) = A + Bx(t)

it follows that

Yn =�

A + B X0, n = 0B Xn, n �= 0

proof:

Yn = �y(t)e− j2π(n f0)t�= A�e− j2π(n f0)t� + B�x(t)e− j2π(n f0)t�

= A

�1, n = 00, n �= 0

�+ B Xn

QED

• Likewise ify(t) = x(t − t0)

it follows thatYn = Xne

− j2π(n f0)t0

proof:Yn = �x(t − t0)e

− j2π(n f0)t�Let λ = t − t0 which implies also that t = λ + t0, so

Yn = �x(λ)e− j2π(n f0)(λ+t0)�= �x(λ)e− j2π(n f0)λ�e− j2π(n f0)t0

= Xne− j2π(n f0)t0

QED



2.5 Fourier Transform• The Fourier series provides a frequency domain representation

of a periodic signal via the Fourier coefficients and line spec-trum

• The next step is to consider the frequency domain representa-tion of aperiodic signals using the Fourier transform

• Ultimately we will be able to include periodic signals withinthe framework of the Fourier transform, using the concept oftransform in the limit

• The text establishes the Fourier transform by considering alimiting case of the expression for the Fourier series coefficientXn as T0 → ∞

• The Fourier transform (FT) and inverse Fourier transfrom (IFT)is defined as

X ( f ) =� ∞

−∞x(t)e− j2π f t

dt (FT)

x(t) =� ∞

−∞X ( f )e j2π f t

d f (IFT)

• Sufficient conditions for the existence of the Fourier transformare

1.� ∞−∞ |x(t)| dt < ∞

2. Discontinuities in x(t) be finite3. An alternate sufficient condition is that

� ∞−∞ |x(t)|2 dt <

∞, which implies that x(t) is an energy signal


2.5. FOURIER TRANSFORM

2.5.1 Amplitude and Phase Spectra• FT properties are very similar to those obtained for the Fourier

coefficients of periodic signals

• The FT, X ( f ) = F{x(t)}, is a complex function of f

X ( f ) = |X ( f )|e jθ( f ) = |X ( f )|e j � X ( f )

= Re{X ( f )} + jIm{X ( f )}

• The magnitude |X ( f )| is referred to as the amplitude spectrum

• The the angle � X ( f ) is referred to as the phase spectrum

• Note that

Re{X ( f )} =� ∞

−∞x(t) cos 2π f t dt

Im{X ( f )} =� ∞

−∞x(t) sin 2π f t dt

2.5.2 Symmetry Properties• If x(t) is real it follows that

X (− f ) =� ∞

−∞x(t)e− j2π(− f )t

dt

=�� ∞


dt

�∗= X

∗( f )

thus

|X (− f )| = |X ( f )| (even in frequency)� X (− f ) = −� X ( f ) (odd in frequency)



• Additionally,

1. For x(−t) = x(t) (even function), Im{X ( f )} = 0

2. For x(−t) = −x(t) (odd function), Re{X ( f )} = 0

2.5.3 Energy Spectral Density

• From the definition of signal energy,

E =� ∞

−∞|x(t)|2 dt

=� ∞

−∞x

∗(t)

�� ∞

−∞X ( f )e j2π f t

d f

�dt

=� ∞

−∞X ( f )

�� ∞

−∞x

∗(t)e j2π f tdt

�d f

but� ∞

−∞x

∗(t)e j2π f tdt =

�� ∞


dt

�∗= X

∗( f )

• Finally,

E =� ∞

−∞|x(t)|2 dt =

� ∞

−∞|X ( f )|2 d f

which is known as Rayleigh’s Energy Theorem

• Are the units consistent?

– Suppose x(t) has units of volts

– |X ( f )|2 has units of (volts-sec)2



– In a 1 ohm system |X ( f )|2 has units of Watts-sec/Hz =Joules/Hz

• The energy spectral density is defined as

G( f )�= |X ( f )|2 Joules/Hz

• It then follows that

E =� ∞

−∞G( f ) d f

Example 2.11: Rectangular Pulse

• Considerx(t) = A�

�t − t0

τ

�

• FT is

X ( f ) = A

�t0+τ/2

t0−τ/2e

− j2π f tdt

= A · e− j2π f t

− j2π f

��t0+τ/2

t0−τ/2

= Aτ ·�

ejπ f τ − e

− jπ f τ

( j2)π f τ

�· e

− j2π f t0

= Aτ sinc( f τ )e− j2π f t0

A�

�t − t0

τ

�F←→ Aτ sinc( f τ )e− j2π f t0

• Plot |X ( f )|, � X ( f ), and G( f )



? 3 ? 2 ? 1 1 2 3

0.20.40.60.81

? 3 ? 2 ? 1 1 2 3

? 3? 2? 1

123

? 3 ? 2 ? 1 1 2 3

0.20.40.60.81

f

f

f

0-1/! 1/! 2/!-2/!

0-1/! 1/! 2/!-2/!

-1/! 1/! 2/!-2/!

|X(f)|

G(f) = |X(f)|2

X(f)AmplitudeSpectrum

EnergySpectralDensity

PhaseSpectrum

A!

(A!)2

"

"/2

#"/2

#"t0 = !/2 slope = -"f!/2

Rectangular pulse spectra

2.5.4 Transform Theorems

• Be familiar with the FT theorems found in the table of Ap-pendix G.6 of the text

Superposition Theorem

a1x1(t) + a2x2(t)F←→ a1 X1( f ) + a2 X2( f )

proof:



Time Delay Theorem

x(t − t0)F←→ X ( f )e− j2π f t0

proof:

Frequency Translation Theorem

• In communications systems the frequency translation and mod-ulation theorems are particularly important

x(t)e j2π f0t F←→ X ( f − f0)

proof: Note that� ∞

−∞x(t)e j2π f0t

e− j2π f t

dt =� ∞

−∞x(t)e− j2π( f − f0)t dt

soF

�x(t)e j2π f0t

�= X ( f − f0)

QED

Modulation Theorem

• The modulation theorem is an extension of the frequency trans-lation theorm

x(t) cos(2π f0t)F←→ 1

2X ( f − f0) + 1

2X ( f + f0)



proof: Begin by expanding

cos(2π f0t) = 12

ej2π f0t + 1

2e

− j2π f0t

Then apply the frequency translation theorem to each term sep-arately

x(t) y(t)

cos(2!f0t)

X(f) Y(f)

f f

A/2A

f0-f0 00

signalmultiplier

A simple modulator

Duality Theorem

• Note that

F{X (t)} =� ∞

−∞X (t)e− j2π f t

dt =� ∞

−∞X (t)e j2π(− f )t

dt

which implies that

X (t)F←→ x(− f )



Example 2.12: Rectangular Spectrum

f-W W0

1X(f)

• Using duality on the above we have

X (t) = �

�t

2W

�F←→ 2W sinc(2W f ) = x(− f )

• Since sinc( ) is an even function (sinc(x) = sinc(−x)), it fol-lows that

2W sinc(2W t)F←→ �

�f

2W

�

Differentiation Theorem

• The general result isd

nx(t)

dtn

F←→ ( j2π f )nX ( f )

proof: For n = 1 we start with the integration by parts formula,�

u dv = uv�� −

�v du, and apply it to

F

�dx

dt

�=

� ∞

−∞

dx

dte

− j2π f tdt

= x(t)e− j2π f t

��∞

−∞� �� 0

+ j2π f

� ∞


dt

� �� X ( f )



alternate — From Leibnitz’s rule for differentiation of inte-grals,

d

dt

� ∞

−∞F( f, t) d f =

� ∞

−∞

∂ F( f, t)

∂ fd f

sodx(t)

dt= d

dt

� ∞

−∞X ( f )e j2π f t

d f

=� ∞

−∞X ( f )

∂ej2π f t

∂td f

=� ∞

−∞j2π f X ( f )e j2π f t

d f

⇒ dx/dtF←→ j2π f X ( f ) QED

Example 2.13: FT of Triangle Pulse

!"! 0 t

1

• Note that

!!"! "!

t t

1/!

1/!

-2/!-1/!



• Using the differentiation theorem for n = 2 we have that

F

��

�t

τ

��= 1

( j2π f )2F

�1τδ(t + τ ) − 2

τδ(t) + 1

τδ(t − τ )

�

=1τ e

j2π f τ − 2τ + 1

τ e− j2π f τ

( j2π f )2

= 2 cos(2π f τ ) − 2−τ (2π f )2

= τ4 sin2(π f τ )

4(π f τ )2 = τ sinc2( f τ )

�

�t

τ

�F←→ τ sinc2( f τ )

Convolution and Convolution Theorem

• Before discussing the convolution theorem we need to reviewconvolution

• The convolution of two signals x1(t) and x2(t) is defined as

x(t) = x1(t) ∗ x2(t) =� ∞

−∞x1(λ)x2(t − λ) dλ

= x2(t) ∗ x1(t) =� ∞

−∞x2(λ)x1(t − λ) dλ

• A special convolution case is δ(t − t0)

δ(t − t0) ∗ x(t) =� ∞

−∞δ(λ − t0)x(t − λ) dλ

= x(t − λ)��λ=t0

= x(t − t0)



Example 2.14: Rectangular Pulse Convolution

• Let x1(t) = x2(t) = �(t/τ )

• To evaluate the convolution integral we need to consider theintegrand by sketching of x1(λ) and x2(t − λ) on the λ axis fordifferent values of t

• For this example four cases are needed for t to cover the entiretime axis t ∈ (−∞, ∞)

• Case 1: When t < τ we have no overlap so the integrand iszero and x(t) is zero

!t t + "/2t - "/2 0 "/2#"/2

x2(t - !) x1(!)No overlap for t + "/2 < -"/2 or t < "

• Case 2: When −τ < t < 0 we have overlap and

x(t) =� ∞

−∞x1(λ)x2(t − λ) dλ

=�

t+τ/2

−τ/2dλ = λ

��t+τ/2

−τ/2

= t + τ/2 + τ/2 = τ + t

Overlap begins when t + !/2 = -!/2 or t = -!

"

t + !/20 !/2#!/2

x2(t - ") x1(")



• Case 3: For 0 < t < τ the leading edge of x2(t − λ) is to theright of x1(λ), but the trailing edge of the pulse is still over-lapped

x(t) =� τ/2

t−τ/2dλ = τ/2 − t + τ/2 = τ − t

!t + "/2t - "/2

0 "/2#"/2

x2(t - !)x1(!)Overlap lasts until t = "

• Case 4: For t > τ we have no overlap, and like case 1, theresult is

x(t) = 0

!t + "/2t - "/20 "/2#"/2

x2(t - !)x1(!)No overlap for t > "

• Collecting the results

x(t) =

0, t < −τ

τ + t, −τ ≤ t < 0τ − t, 0 ≤ t < τ

0, t ≥ τ

=�

τ − |t |, |t | ≤ τ

0, otherwise



• Final summary,

�

�t

τ

�∗ �

�t

τ

�= τ�

�t

τ

�

• Convolution Theorem: We now consider x1(t)∗x2(t) in termsof the FT

� ∞

−∞x1(τ )x2(t − τ ) dτ

=� ∞

−∞x1(τ )

�� ∞

−∞X2( f )e j2π f (t−τ )

d f

�dτ

=� ∞

−∞X2( f )

�� ∞

−∞x1(τ )e− j2π f τ

dτ

�e

j2π f td f

=� ∞

−∞X1( f )X2( f )e j2π f t

d f

which implies that

x1(t) ∗ x2(t)F←→ X1( f )X2( f )

Example 2.15: Revisit �(t/τ ) ∗ �(t/τ )

• Knowing that �(t/τ )∗�(t/τ ) = τ�(t/τ ) in the time domain,we can follow-up in the frequency domain by writing

F��(t/τ )

�· F

��(t/τ )

�=

�τ sinc( f τ )

�2

• We have also established the transform pair

τ�

�t

τ

�F←→ τ 2sinc2( f τ ) = τ

�τ sinc2( f τ )

�



or

�

�t

τ

�F←→ τ sinc2( f τ )

Multiplication Theorem

• Having already established the convolution theorem, it followsfrom the duality theorem or direct evaluation, that

x1(t) · x2(t)F←→ X1( f ) ∗ X2( f )

2.5.5 Fourier Transforms in the Limit

• thus far we have considered two classes of signals

1. Periodic power signals which are described by line spec-tra

2. Non-periodic (aperiodic) energy signals which are describedby continuous spectra via the FT

• We would like to have a unifying approach to spectral analysis

• To do so we must allow impulses in the frequency domain byusing limiting operations on conventional FT pairs, known asFourier transforms-in-the-limit

– Note: The corresponding time functions have infinite en-ergy, which implies that the concept of energy spectraldensity will not apply for these signals (we will introducethe concept of power spectral density for these signals)



Example 2.16: A Constant Signal

• Let x(t) = A for −∞ < t < ∞

• We can writex(t) = lim

T →∞A�(t/T )

• Note thatF

�A�(t/T )

�= AT sinc( f T )

• Using the transform-in-the-limit approach we write

F{x(t)} = limT →∞

AT sinc( f T )

? 3 ? 2 ? 1 1 2 3? 0.2

0.20.40.60.81

? 3 ? 2 ? 1 1 2 3? 0.2

0.20.40.60.81AT1 AT2

T2 >> T1

f f

Increasing T in AT sinc( f T )

• Note that since x(t) has no time variation it seems reasonablethat the spectral content ought to be confined to f = 0

• Also note that it can be shown that� ∞

−∞AT sinc( f T ) d f = A, ∀ T

• Thus we have established that

AF←→ Aδ( f )



• As a further check

F−1�

Aδ( f )�

=� ∞

−∞Aδ( f )e j2π f f t

d f = Aej2π f t

��f =0

= A

• As a result of the above example, we can obtain several moreFT-in-the-limit pairs

Aej2π f0t F←→ Aδ( f − f0)

A cos(2π f0t + θ)F←→ A

2�e

jθδ( f − f0) + e− jθδ( f + f0)

�

Aδ(t − t0)F←→ Ae

− j2π f t0

• Reciprocal Spreading Property: Compare

Aδ(t)F←→ A and A

F←→ Aδ( f )

A constant signal of infinite duration has zero spectral width,while an impulse in time has zero duration and infinite spectralwidth

2.5.6 Fourier Transforms of Periodic Signals

• For an arbitrary periodic signal with Fourier series

x(t) =∞�

n=−∞Xne

j2πn f0t



we can write

X ( f ) = F

� ∞�

n=−∞Xne

j2πn f0t

�

=∞�

n=−∞XnF

�e

j2πn f0t

�

=∞�

n=−∞Xnδ( f − n f0)

using superposition and F{Aej2π f0t} = Aδ( f − f0)

• What is the difference between line spectra and continuous

spectra? none!

• Mathematically,

LineSpectra

Convert to time domain

Convert to time domain

Sum phasors

Integrate impulses to get phasors via the inverse FT

ContinuousSpectra

• The Fourier series coefficients need to be known before the FTspectra can be obtained

• A technique that obtained the FT directly will be discussedlater



Example 2.17: Ideal Sampling Waveform

• When we discuss sampling theory it will be useful to have theFT of the periodic impulse train signal

ys(t) =∞�

m=−∞δ(t − mTs)

where Ts is the sample spacing or period

• Since this signal is periodic, it must have a Fourier series rep-resentation too

• In particular

Yn = 1Ts

�

Ts

δ(t)e− j2π(n fs)t dt = 1Ts

= fs, any n

where fs is the sampling rate in Hz

• The FT of ys(t) is given by

Ys( f ) = fs

∞�

n=−∞F

�e

j2πn f0)t�

= fs

∞�

n=−∞δ( f − n fs)

• Summary,

∞�


F←→ fs

∞�

n=−∞δ( f − n fs)



. . . . . .

t

f

ys(t)

Ys(f)0

0

Ts

fs 4fs

4Ts-Ts

-fs

. . . . . .fs

1

An impulse train in times is an impulse train in frequency

Example 2.18: Convolve Step and Exponential

• Find y(t) = Au(t) ∗ e−αt

u(t), α > 0

• For tleq0 there is no overlap so Y (t) = 0

!0t

No overlap



• For t > 0 there is always overlap

y(t) =�

t

0A · e

−α(t−λ)dλ

= Ae−αt · e

αλ

α

��t

0

= Ae−αt · e

αt − 1α

!0 t

For t > 0 there is always overlap

• Summary,

y(t) = A

α

�1 − e

−αt�

u(t)

t

y(t)

A/!

0



Direct Approach for the FT of a Periodic Signal

• The FT of a periodic signal can be found directly by expandingx(t) as follows

x(t) =� ∞�


�

∗ p(t) =∞�

m=−∞p(t − mTs)

where p(t) represents one period of x(t), having period Ts

• From the convolution theorem

X ( f ) = F

� ∞�


�

· P( f )

= fs P( f )∞�

n=−∞δ( f − n fs)

= fs

∞�

n=−∞P(n fs)δ( f − n fs)

where P( f ) = F{p(t)}

• The FT transform pair just established is

∞�


F←→∞�

n=−∞fs P(n fs)δ( f − n fs)



Example 2.19: p(t) = �(t/2) + �(t/4), T0 = 10

tT0 = 100 1-1-2 2

. . . . . .

x(t)

12

Stacked pulses periodic signal

• We begin by finding P( f ) using F{�(t/τ )} = τ sinc( f τ )

P( f ) = 2sinc(2 f ) + 4sinc(4 f )

• Plugging into the FT pair derived above with n fs = n/10,

X ( f ) = 110

∞�

n=−∞

�2sinc

�n

5

�+ 4sinc

�2n

5

��δ�

f − n

10

�

2.5.7 Poisson Sum Formula• The Poisson sum formula from mathematics can be derived

using the FT pair

e− j2π(n fs)t F←→ δ( f − n fs)

by writing

F−1

� ∞�

n=−∞fs P(n fs)δ( f − n fs)

�

= fs

∞�

n=−∞P(n fs)e

j2π(n fs)t



• From the earlier developed FT of periodic signals pair, weknow that the left side of the above is also equal to

∞�


also= fs

∞�

n=−∞P(n fs)e

j2π(n fs)t

• We can finally relate this back to the Fourier series coefficients,i.e.,

Xn = fs P(n fs)

2.6 Power Spectral Density and Corre-lation

• For energy signals we have the energy spectral density, G( f ),defined such that

E =� ∞

−∞G( f ) d f

• For power signals we can define the power spectral density

(PSD), S( f ) of x(t) such that

P =� ∞

−∞S( f ) d f = �|x(t)|2�

– Note: S( f ) is real, even and nonnegative– If x(t) is periodic S( f ) will consist of impulses at the

harmonic locations

• For x(t) = A cos(ω0t + θ), intuitively,

S( f ) = 14

A2δ( f − f0) + 1

4A

2δ( f + f0)


2.6. POWER SPECTRAL DENSITY AND CORRELATION

since�

S( f ) d f = A2/2 as expected (power on a per ohm

basis)

• To derive a general formula for the PSD we first need to con-sider the autocorrelation function

2.6.1 The Time Average Autocorrelation Func-tion

• Let φ(τ ) be the autocorrelation function of an energy signal

φ(τ ) = F−1�

G( f )�

= F−1�

X ( f )X∗( f )

�

= F−1�

X ( f )�

∗ F−1�X

∗( f )�

but x(−t)F←→ X

∗( f ) for x(t) real, so

φ(τ ) = x(t) ∗ x(−t) =� ∞

−∞x(t)x(t + τ ) dτ

or

φ(τ ) = limT →∞

�T

−T

x(t)x(t + τ ) dτ

• Observe thatG( f ) = F

�φ(τ )

�

• The autocorrelation function (ACF) gives a measure of thesimilarity of a signal at time t to that at time t + τ ; the co-herence between the signal and the delayed signal



x(t)X(f)

G(f) = |X(f)|2

!(") =

Energy spectral density and signal relationships

2.6.2 Power Signal CaseFor power signals we define the autocorrelation function as

Rx(τ ) = �x(t)x(t + τ )�

= limT →∞

12T

�T

−T

x(t)x(t + τ ) dt

if periodic= 1T0

�

T0

x(t)x(t + τ ) dt

• Note that

Rx(0) = �|x(t)|2� =� ∞

−∞Sx( f ) d f

and since for energy signals φ(τ )F←→ G( f ), a reasonable

assumption is that

Rx(τ )F←→ Sx( f )



• A formal statement of this is the Wiener-Kinchine theorem (aproof is given in text Chapter 5)

Sx( f ) =� ∞

−∞Rx(τ )e− j2π f τ

dτ

x(t) Rx(!) Sx(f)Power spectral density and signal relationships

2.6.3 Properties of R(τ )

• The following properties hold for the autocorrelation function

1. R(0) = �|x(t)|2� ≥ |R(τ )| for all values of τ

2. R(−τ ) = �x(t)x(t − τ )� = R(τ ) ⇒ an even function

3. lim|τ |→∞ R(τ ) = �x(t)�2 if x(t) is not periodic

4. If x(t) is periodic, with period T0, then R(τ ) = R(τ +T0)

5. F{R(τ )} = S( f ) ≥ 0 for all values of f

• The power spectrum and autocorrelation function are frequentlyused for systems analysis with random signals



Example 2.20: Single Sinusoid

• Consider the signal x(t) = A cos(2π f0t + θ), for all t

Rx(τ ) = 1T0

�T0

0A

2 cos(2π f0t + θ) cos(2π(t + τ ) + θ) dt

= A2

2T0

�

T0

�cos(2π f0τ ) + cos(2π(2 f0)t + 2π f0τ + 2θ)

�dt

= A2

2cos(2π f0τ )

• Note that

F�

Rx(τ )�

= Sx( f ) = A2

4�δ( f − f0) + δ( f + f0)

�

More Autocorrelation Function Properties

• Suppose that x(t) has autocorrelation function Rx(τ )

• Let y(t) = A + x(t), A = constant

Ry(τ ) = �[A + x(t)][A + x(t + τ )]�= �A

2� + �Ax(t + τ )� + �Ax(t)� + �x(t)x(t + τ )�= A

2 + 2A�x(t)�� const. terms

+Rx(τ )

• Let z(t) = x(t − t0)

Rz(τ ) = �z(t)z(t + τ )� = �x(t − t0)x(t − t0 + τ �= �x(λ)x(λ + τ )�, with λ = t − t0

= Rx(τ )



• The last result shows us that the autocorrelation function isblind to time offsets

Example 2.21: Sum of Two Sinusoids

• Consider the sum of two sinusoids

y(t) = x1(t) + x2(t)

where x1(t) = A1 cos(2π f1t +θ1) and x2(t) = A2 cos(2π f2t +θ2) and we assume that f1 �= f2

• Using the definition

Ry(τ ) = �[x1(t) + x2(t)][x1(t + τ )x2(t + τ )]�= �x1(t)x1(t + τ )� + �x2(t)x2(t + τ )�

+ �x1(t)x2(t + τ )� + �x2(t)x1(t + τ )�

• The last two terms are zero since �cos((ω1 ± ω2)t)� = 0 whenf1 �= f2 (why?), hence

Ry(τ ) = Rx1(τ ) + Rx2(τ ), for f1 �= f2

= A21

2cos(2π f1τ ) + A

22

2cos(2π f2τ )

Example 2.22: PN Sequences

• In the testing and evaluation of digital communication systemsa source of known digital data (i.e., ‘1’s and ‘0’s) is required(see text Chapter 8 p. 429–432)



• A maximal length sequence generator or pseudo-noise (PN)code is often used for this purpose

• Practical implementation of a PN code generator can be ac-complished using an N -stage shift register with appropriateexclusive-or feedback connections

• The sequence length or period of the resulting PN code is M =2N − 1 bits long

CD1 Q1

CD2 Q2

CD3 Q3

M = 23 - 1 = 7

one period = NT

t

x(t)

x(t)

ClockPeriod = T

+A

-A

Three stage PN (m-sequence) generator

• PN sequences have quite a number of properties, one being thatthe time average autocorrelation function is of the form shownbelow



!

Rx(!)

T-T

MT

MT

. . .. . .

A2

-A2/MPN sequence autocorrelation function

• The calculation of the power spectral density will be left as ahomework problem

– Hint: To find Sx( f ) = F{Rx(τ )} we use�

n

p(t − nTs)F←→ fs

�

n

P(n fs)δ( f − n fs)

where Ts = MT

– One period of Rx(τ ) is a triangle pulse with a level shift

• Suppose the logic levels are switched from ±A to positive lev-els of say v1 to v2

– Using the additional autocorrelation function propertiesthis can be done

– You need to know that a PN sequence contains one more‘1’ than ‘0’

• MATLAB for generating PN sequences from 2 to 12 stages isgiven below

function c = m_seq(m)

%function c = m_seq(m)

%

% Generate an m-sequence vector using an all-ones initialization

%



% Mark Wickert, April 2005

sr = ones(1,m);

Q = 2ˆm - 1;

c = zeros(1,Q);

switch m

case 2

taps = [1 1 1];

case 3

taps = [1 0 1 1];

case 4

taps = [1 0 0 1 1];

case 5

taps = [1 0 0 1 0 1];

case 6

taps = [1 0 0 0 0 1 1];

case 7

taps = [1 0 0 0 1 0 0 1];

case 8

taps = [1 0 0 0 1 1 1 0 1];

case 9

taps = [1 0 0 0 0 1 0 0 0 1];

case 10

taps = [1 0 0 0 0 0 0 1 0 0 1];

case 11

taps = [1 0 0 0 0 0 0 0 0 1 0 1];

case 12

taps = [1 0 0 0 0 0 1 0 1 0 0 1 1];

otherwise

disp(’Invalid length specified’)

end

for n=1:Q,

tap_xor = 0;

c(n) = sr(m);

for k=2:m,

if taps(k) == 1,

tap_xor = xor(tap_xor,xor(sr(m),sr(m-k+1)));

end

end

sr(2:end) = sr(1:end-1);

sr(1) = tap_xor;

end



R(τ ), S( f ), and Fourier Series

• For a periodic power signal, x(t), we can write

x(t) =∞�

n=−∞Xne

j2π(n f0)t

• There is an interesting linkage between the Fourier series rep-resentation of a signal, the power spectrum, and then back tothe autocorrelation function

• Using the orthogonality properties of the Fourier series expan-sion we can write

R(τ ) =�� ∞�

n=−∞Xne

j2π(n f0)t

� � ∞�

m=−∞Xme

j2π(m f0)t

�∗�

=∞�

n=−∞

∞�

m=−∞Xn X

∗m

�e

j2π(n−m) f0t�

=∞�

n=−∞|Xn|2e

j2π(n f0)t

• The power spectral density can be obtained by Fourier trans-forming both sides of the above

S( f ) =∞�

n=−∞|Xn|2δ( f − n f0)



2.7 Linear Time Invariant (LTI) Systems

x(t) y(t) = operator

Linear system block diagram

Definition

• Linearity (superposition) holds, that is for input α1x1(t)+α2x2(t),α1 and α2 constants,

y(t) = H�α1x1(t) + α2x2(t)

�

= α1H�x1(t)

�+ α2H

�x2(t)

�

= α1y1(t) + α2y2(t)

• A system is time invariant (fixed) if for y(t) = H[x(t)], adelayed input gives a correspondingly delayed output, i.e.,

y(t − t0) = H�x(t − t0)

�

Impulse Response and Superposition Integral

• The impulse response of an LTI system is denoted

h(t)�= H

�δ(t)

�

assuming the system is initially at rest

• Suppose we can write x(t) as

x(t) =N�

n=1

αnδ(t − tn)


2.7. LINEAR TIME INVARIANT (LTI) SYSTEMS

• For an LTI system with impulse response h( )

y(t) =N�

n=1

αnh(t − tn)

• To develop the superposition integral we write

x(t) =� ∞

−∞x(λ)δ(t − λ) dλ

� limN→∞

N�

n=−N

x(n�t)δ(t − n�t) �t, for �t � 1

t

x(t)

0 !t"!t 2!t 3!t 4!t 5!t 6!t

Rectangle area is approximation

. . . . . .

Impulse sequence approximation to x(t)

• If we applyH to both sides and let �t → 0 such that n�t → λwe have

y(t) � limN→∞

N�

n=−N

x(n�t)h(t − n�t) �t

=� ∞

−∞x(λ)h(t − λ) dλ = x(t) ∗ h(t)

or=� ∞

−∞x(t − σ )h(σ ) dσ = h(t) ∗ x(t)



2.7.1 Stability• In signals and systems the concept of bounded-input bounded-

output (BIBO) stability is introduced

• Satisfying this definition requires that every bounded-input (|x(t)| <∞) produces a bounded output (|y(t)| < ∞)

• For LTI systems a fundamental theorem states that a system isBIBO stable if and only if

� ∞

−∞|h(t)| dt < ∞

• Further implications of this will be discussed later

2.7.2 Transfer Function• The frequency domain result corresponding to the convolution

expression y(t) = x(t) ∗ h(t) is

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

where H( f ) is known as the transfer function or frequency

response of the system having impulse response h(t)

• It immediately follows that

h(t)F←→ H( f )

and

y(t) = F−1�

X ( f )H( f )�

=� ∞

−∞X ( f )H( f )e j2π f t

d f



2.7.3 Causality

• A system is causal if the present output relies only on past andpresent inputs, that is the output does not anticipate the input

• The fact that for LTI systems y(t) = x(t) ∗ h(t) implies thatfor a causal system we must have

h(t) = 0, t < 0

– Having h(t) nonzero for t < 0 would incorporate futurevalues of the input to form the present value of the output

• Systems that are causal have limitations are their frequencyresponse, in particular the Paley–Wiener theorem states thatfor

� ∞−∞ |h(t)|2 dt < ∞, H( f ) for a cusal system must satisfy

� ∞

−∞

| ln |H( f )||1 + f 2 d f < ∞

• In simple terms this means:

1. We cannot have |H( f )| = 0 over a finite band of frequen-cies (isolated points ok)

2. The roll-off rate of |H( f )| cannot be too great, e.g., e−k1| f |

and e−k2| f |2 are not allowed, but polynomial forms such as�

1/(1 + ( f/ fc)2N , N an integer, are acceptable

3. Practical filters such as Butterworth, Chebyshev, and el-liptical filters can come close to ideal requirements



2.7.4 Properties of H( f )

• For h(t) real it follows that

|H(− f )| = |H( f )| and � H(− f ) = −� H( f )

why?

• Input/output relationships for spectral densities are

G y( f ) = |Y ( f )|2 = |X ( f )H( f )|2 = |H( f )|2Gx( f )

Sy( f ) = |H( f )|2Sx( f ) proof in text chap. 5

Example 2.23: RC Lowpass Filter

C

Rx(t)X(f ) Y(f )

y(t)

h(t), H(f)

ic(t)

vc(t)

RC lowpass filter schematic

• To find H( f ) we may solve the circuit using AC steady-stateanalysis

Y ( jω)

X ( jω)=

1jωc

R + 1jωc

= 11 + jωRC

so

H( f ) = Y ( f )

X ( f )= 1

1 + j f/ f3, where f3 = 1/(2π RC)



• From the circuit differential equation

x(t) = ic(t)R + y(t)

but

ic(t) = cdvc(t)

dt= c

y(t)

dt

thus

RCdy(t)

dt+ y(t) = x(t)

• FT both sides using dx/dtF←→ j2π f X ( f )

j2π f RCY ( f ) + Y ( f ) = X ( f )

so again

H( f ) = Y ( f )

X ( f )= 1

1 + j f/ f3

= 1�

1 + ( f/ f3)2e

− j tan−1( f/ f3)

• The Laplace transfrom could also be used here, and perhaps ispreferred, we just need to substitute s → jω → j2π f



f

ff3-f3!/2

-!/2

1

RC lowpass frequency response

• Find the system response to

x(t) = A�

�(t − T/2)

T

�

• Finding Y ( f ) is easy since

Y ( f ) = X ( f )H( f ) = AT sinc( f T )

�1

1 + j f/ fs

�e

− jπ f t

• To find y(t) we can IFT the above, use Laplace transforms, orconvolve directly

• From the FT tables we known that

h(t) = 1RC

e−t/(RC)

u(t)



• In Example 2.18 we showed that

Au(t) ∗ e−αt

u(t) = A

α

�1 − e

−αt�u(t)

• Note that

A�

�t − T/2

T

�= A[u(t) − u(t − T )]

and here α = 1/(RC), so

y(t) = A

RCRC

�1 − e

−t/(RC)�u(t)

− A

RCRC

�1 − e

−(t−T )/(RC)�u(t − T )

0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1

-3 -2 -1 1 2 3

0.2

0.4

0.6

0.8

1

-3 -2 -1 1 2 3

0.2

0.4

0.6

0.8

1

-3 -2 -1 1 2 3

0.2

0.4

0.6

0.8

1

t/T fT

fT fT

y(t) |X(f)|, |H(f)|, |Y(f)|

|X(f)|, |H(f)|, |Y(f)| |X(f)|, |H(f)|, |Y(f)|

RC = 2T

RC = T/2RC = T/10

T/10

RC =

T/5T/2

T2T

Pulse time response and frequency spectra with A = 1



2.7.5 Response to Periodic Inputs• When the input is periodic we can write

x(t) =∞�

n=−∞Xne

j2π(n f0)t

which implies that

X ( f ) =∞�

n=−∞Xnδ( f − n f0)

• It then follows that

Y ( f ) =∞�

n=−∞Xn H(n f0)δ( f − n f0)

and

y(t) =∞�

n=−∞Xn H(n f0)e

j2π(n f0)t

=∞�

n=−∞|Xn||H(n f0|e j[2π(n f0)t+ � Xn+ � H(n f0)]

• This is a steady-state response calculation, since the analysisassumes that the periodic signal was applied to the system att = −∞

2.7.6 Distortionless Transmission• In the time domain a distortionless system is such that for any

input x(t),y(t) = H0x(t − t0)



where H0 and t0 are constants

• In the frequency domain the implies a frequency response ofthe form

H( f ) = H0e− j2π f t0,

that is the amplitude response is constant and the phase shift islinear with frequency

• Distortion types:

1. Amplitude response is not constant over a frequency band(interval) of interest ↔ amplitude distortion

2. Phase response is not linear over a frequency band of in-terest ↔ phase distortion

3. The system is non-linear, e.g., y(t) = k0+k1x(t)+k2x2(t)

↔ nonlinear distortion

2.7.7 Group and Phase Delay• The phase distortion of a linear system can be characterized

using group delay, Tg( f ),

Tg( f ) = − 12π

dθ( f )

d f

where θ( f ) is the phase response of an LTI system

• Note that for a distortionless system θ( f ) = −2π f t0, so

Tg( f ) = − 12π

d

d f− 2π f t0 = t0 s

clearly a constant group delay



• Tg( f ) is the delay that two or more frequency components un-dergo in passing through an LTI system

– If say Tg( f1) �= Tg( f2) and both of these frequencies arein a band of interest, then we know that delay distortionexists

– Having two different frequency components arrive at thesystem output at different times produces signal disper-sion

• An LTI system passing a single frequency component, x(t) =A cos(2π f1t), always appears distortionless since at a singlefrequency the output is just

y(t) = A|H( f1)| cos�2π f1t + θ( f1)

�

= A|H1( f )| cos�

2π f1

�t − −θ( f1)

2π f1

��

which is equivalent to a delay known as the phase delay

Tp( f ) = −θ( f )

2π f

• The system output now is

y(t) = A|H( f1)| cos�2π f1(t − Tp( f1))]

• Note that for a distortionless system

Tp( f ) = − 12π f

(−2π f t0) = t0



Example 2.24: Terminated Lossless Transmission Line

x(t) y(t)

Rs = R0

RL = R0R0, vp

L

y(t) = 12

x�t − L

v p

�

Lossless transmission line

• We conclude that H0 = 1/2 and t0 = L/v p

• Note that a real transmission line does have losses that intro-duces dispersion on a wideband signal



Example 2.25: Fictitious System

-20 -10 10 20

0.5

1

1.5

2

-20 -10 10 20

-1.5

-1

-0.5

0.5

1

1.5

-20 -10 10 20

0.0025

0.005

0.0075

0.01

0.0125

0.015

-20 -10 10 20

0.011

0.012

0.013

0.014

0.015

0.016

f (Hz)

f (Hz)

f (Hz)f (Hz)

H(f)|H(f)|

Tg(f) Tp(f)

No distortion on |f | < 10 Hz band

Ampl. Radians

Time Time

Amplitude, phase, group delay, phase delay

• The system in this example is artificial, but the definitions canbe observed just the same

• For signals with spectral content limited to | f | < 10 Hz thereis no distortion, amplitude or phase/group delay

• For 10 < | f | < 15 amplitude distortion is present

• For | f | > 15 both amplitude and phase distortion is present

• What about the interval 10 < | f | < 15?



2.7.8 Nonlinear Distortion

• In the time domain a nonlinear system may be written as

y(t) =∞�

n=0

anxn(t)

• Specifically consider

y(t) = a1x(t) + a2x2(t)

• Let

x(t) = A1 cos(ω1t) + A2 cos(ω2t)

• Expanding the output we have

y(t) = a1�A1 cos(ω1t) + A2 cos(ω2t)

�

+ a1�A1 cos(ω1t) + A2 cos(ω2t)

�2

= a1�A1 cos(ω1t) + A2 cos(ω2t)

�

+�

a2

2�

A21 + A

22�+ a2

2�A

21 cos(2ω1t) + A

22 cos(2ω2t)

��

+ a2 A1 A2�

cos[(ω1 + ω2)t] + cos[(ω1 − ω2)t]�

– The third line is the desired output

– The fourth line is termed harmonic distortion

– The fifth line is termed intermodulation distortion



f

f f

f

A1

A1A2

a1A1

a1A1

a1A1A2

a1A2a1a2A1

a2A12

2

a2A12

2a2A2

2

2a2(A1 + A2)

2 2

2

a2A12

2

2f1

f2

f1

f1 2f1 2f2

f1

f1 f2-f1f1+f2

f2

0

0

0

0

Non-Linear

Non-Linear

Input

Input

Output

Output

One and two tones in y(t) = a1x(t) + a2x2(t) device

• In general if y(t) = a1x(t) + a2x2(t) the multiplication theo-

rem implies that

Y ( f ) = a1 X ( f ) + a2 X ( f ) ∗ X ( f )

• In particular if X ( f ) = A�( f/(2W ))

Y ( f ) = a1 A�

�f

2W

�+ a22W A

2�

�f

2W

�



Y( f ) =

=

+f f

f

a1A 2Wa2A2

Wa2A2

-W

-W

W

W

-2W

-2W

2W

2W

a1A + Wa2A2

a1A + 2Wa2A2

Continuous spectrum in y(t) = a1x(t) + a2x2(t) device

2.7.9 Ideal Filters1. Lowpass of bandwidth B

HLP( f ) = H0�

�f

2B

�e

− j2π f t0

B B-B -B

H0slope =-2!t0

|HLP(f)| HLP(f)

f f

2. Highpass with cutoff B

HHP( f ) = H0�1 − �( f/(2B))

�e

− j2π f t0

B B-B -B

H0

slope =-2!t0

|HHP(f)|

f f

HHP(f)



3. Bandpass of bandwidth B

HBP( f ) =�Hl( f − f0) + Hl( f + f0)

�e

− j2π f t0

where Hl( f ) = H0�( f/B)

B BH0

slope = -2!t0|HBP(f)| HBP(f)

f f-f0 f0 -f0 f0

• The impulse response of the lowpass filter is

hLP(t) = F−1�

H0�( f/(2B))e− j2π f t0�

= 2B H0sinc[2B(t − t0)]

• Ideal filters are not realizable, but simplify calculations aregive useful performance upper bound results

– Note that hLP(t) �= 0 for t < 0, thus the filter is noncausaland unrealizable

• From the modulation theorem it also follows that

hBP(t) = 2hl(t − t0) cos[2π f0(t − t0)]= 2B H0sinc[B(t − t0)] cos[2π f0(t − t0)]



t t

hLP(t) hBP(t)

t0

t0

2BH0 2BH0

t0 - 12B t0 - 1

2Bt0 + 1

2B t0 + 12B

Ideal lowpass and bandpass impulse responses

2.7.10 Realizable Filters• We can approximate ideal filters with realizable filters such as

Butterworth, Chebyshev, and Bessel, to name a few

• We will only consider the lowpass case since via frequencytransformations we can obtain the others

Butterworth

• A Butterworth filter has a maximally flat (flat in the sense ofderivatives of the amplitude response at dc being zero) pass-band

• In the s-domain (s = σ+ jω) the transfer function of a lowpassdesign is

HBU(s) = ωn

c

(s − s1)(s − s2) · · · (s − sn)

where

sk = ωc exp�π

�12

+ 2k − 12n

��, k = 1, 2, . . . , n



• Note that the poles are located on a semi-circle of radius ωc =2π fc, where fc is the 3dB cuttoff frequency of the filter

• The amplitude response of a Butterworth filter is simply

|HBU( f )| 1�

1 + ( f/ fc)2n

Butterworth n = 4 lowpass filter

Chebyshev

• A Chebyshev type I filter (ripple in the passband), is is de-signed to maintain the maximum allowable attenuation in thepassband yet have maximum stopband attenuation

• The amplitude response is given by

|HC( f )| = 1�

1 + �2C2n( f )

where

Cn( f ) =�

cos(n cos−1( f/ fc)), 0 ≤ | f | ≤ fc

cosh(n cosh−1( f/ fc)), | f | > fc



• The poles are located on an ellipse as shown below

Chebyshev n = 4 lowpass filter

Bessel

• A Bessel filter is designed to maintain linear phase in the pass-band at the expense of the amplitude response

HBE( f ) = Kn

Bn( f )

where Bn( f ) is a Bessel polynomial of order n (see text) andKn is chosen so that the filter gain is unity at f = 0



Amplitude Rolloff and Group Delay Comparision

• Compare Butterworth, 0.1 dB ripple Chebyshev, and Bessel

n = 3 Amplitude response



n = 3 Group delay

Filter Construction Techniques

ConstructionType

Description of El-ements or Filter

Center Fre-quency Range

Unloaded Q

(typicalFilter Appli-cation

LC (passive) lumped elements DC–300 MHzor higher in in-tegrated form

100 Audio, video,IF and RF

Active R, C , op-amps DC–500 kHzor higher usingWB op-amps

200 Audio and lowRF

Crystal quartz crystal 1kHz – 100MHz

100,000 IF

Ceramic ceramic disks withelectrodes

10kHz – 10.7MHz

1,000 IF

Surface acousticwaves (SAW)

interdigitated fin-gers on a Piezo-electric substrate

10-800 MHz, variable IF and RF

Transmission line quarterwave stubs,open and short ckt

UHF and mi-crowave

1,000 RF

Cavity machined andplated metal

Microwave 10,000 RF

2.7.11 Pulse Resolution, Risetime, and Band-width

Problem: Given a non-bandlimited signal, what is a reasonable esti-mate of the signals transmission bandwidth?We would like to obtain a relationship to the signals time duration

• Step 1: We first consider a time domain relationship by seekinga constant T such that

T x(0) =� ∞

−∞|x(t)| dt

t

x(0)

|x(t)|

T/2-T/2 0

Make areas equal via T



• Note that� ∞

−∞x(t) dt =

� ∞


dt

��f =0

= X (0)

and � ∞

−∞|x(t)| dt ≥

� ∞

−∞x(t) dt

which impliesT x(0) ≥ X (0)

• Step 2: Find a constant W such that

2W X (0) =� ∞

−∞|X ( f )| d f

f

X(0)

|X(f)|

W-W 0

Make areas equal via W

• Note that� ∞

−∞X ( f ) d f =

� ∞

−∞X ( f )e j2π f t

d f

��t=0

= x(0)

and � ∞

−∞|X ( f )| d f ≥

� ∞

−∞X ( f ) d f

which implies that

2W X (0) ≥ x(0)



• Combining the results of Step 1 and Step 2, we have

2W X (0) ≥ x(0) ≥ 1T

X (0)

or

2W ≥ 1T

or W ≥ 12T

Example 2.26: Rectangle Pulse

• Consider the pulse x(t) = �(t/T )

• We know that X ( f ) = T sinc( f T )

-1 -0.5 0.5 1

0.2

0.4

0.6

0.8

1

-3 -2 -1 1 2 3

0.2

0.4

0.6

0.8

1

fTt /T

x(t) |X(f)|/T

f1/(2T)-1/(2T)

Lower bound for W

Pulse width versus Bandwidth, is W ≥ 1/(2T ?

• We see that for the case of the sinc( ) function the bandwidth,W , is clearly greater than the simple bound predicts



Risetime

• There is also a relationship between the risetime of a pulse-likesignal and bandwidth

• Definition: The risetime, TR, is the time required for the lead-ing edge of a pulse to go from 10% to 90% of its final value

• Given the impulse response h(t) for an LTI system, the stepresponse is just

ys(t) =� ∞

−∞h(λ)u(t − λ) dλ

=�

t

−∞h(λ) dλ

if causal=�

t

0h(λ) dλ

Example 2.27: Risetime of RC Lowpass

• The RC lowpass filter has impulse response

h(t) = 1RC

e−t/(RC)

u(t)

• The step response is

ys(t) =�1 − e

−t/(RC)�

u(t)

• The risetime can be obtained by setting ys(t1) = 0.1 and ys(t2) =0.9

0.1 =�1 − e

−t1/(RC)�

⇒ ln(0.9) = −t1

RC

0.9 =�1 − e

−t2/(RC)�

⇒ ln(0.1) = −t2

RC



• The difference t2 − t1 is the risetime

TR = t2 − t1 = RC ln(0.9/0.1) � 2.2RC = 0.35f3

where f3 is the RC lowpass 3dB frequency

Example 2.28: Risetime of Ideal Lowpass

• The risetime of an ideal lowpass filter is of interest since it isused in modeling and also to see what an ideal filter does to astep input

• The impulse response is

h(t) = F−1

��

�f

2B

��= 2Bsinc[2Bt]

• The step response then is

ys =�

t

−∞2Bsinc[2Bλ] dλ

= 1π

� 2π Bt

−∞

sin u

udu

= 12

+ 1π

Si[2π Bt]

where Si( ) is a special function known as the sine integral

• We can numerically find the risetime to be

TR � 0.44B



1 2 3 4 5

0.2

0.4

0.6

0.8

1

-2 -1 1 2 3

0.2

0.4

0.6

0.8

1

t RC t/B

Step Response of RC Lowpass Step Response of Ideal Lowpass

2.2 0.44

RC and ideal lowpass risetime comparison


2.8. SAMPLING THEORY

2.8 Sampling TheoryIntegrate with Chapter 3 material.

2.9 The Hilbert TransformIntegrate with Chapter 3 material.

2.10 The Discrete Fourier Transform andFFT

?



.


Chapter 3Analog Modulation

Contents

3.1 Linear Modulation . . . . . . . . . . . . . . . . . . . . 3-3

3.1.1 Double-Sideband Modulation (DSB) . . . . . . 3-33.1.2 Amplitude Modulation . . . . . . . . . . . . . . 3-83.1.3 Single-Sideband Modulation . . . . . . . . . . . 3-213.1.4 Vestigial-Sideband Modulation . . . . . . . . . . 3-353.1.5 Frequency Translation and Mixing . . . . . . . . 3-38

3.2 Angle Modulation . . . . . . . . . . . . . . . . . . . . 3-46

3.2.1 Narrowband Angle Modulation . . . . . . . . . 3-483.2.2 Spectrum of an Angle-Modulated Signal . . . . 3-503.2.3 Power in an Angle-Modulated Signal . . . . . . 3-563.2.4 Bandwidth of Angle-Modulated Signals . . . . . 3-563.2.5 Narrowband-to-Wideband Conversion . . . . . . 3-633.2.6 Demodulation of Angle-Modulated Signals . . . 3-63

3.3 Interference . . . . . . . . . . . . . . . . . . . . . . . 3-73

3.3.1 Interference in Linear Modulation . . . . . . . . 3-743.3.2 Interference in Angle Modulation . . . . . . . . 3-76

3.4 Feedback Demodulators . . . . . . . . . . . . . . . . . 3-81

3-1

CHAPTER 3. ANALOG MODULATION

3.4.1 Phase-Locked Loops for FM Demodulation . . . 3-813.4.2 PLL Frequency Synthesizers . . . . . . . . . . . 3-1003.4.3 Frequency-Compressive Feedback . . . . . . . . 3-1043.4.4 Coherent Carrier Recovery for DSB Demodulation 3-106

3.5 Sampling Theory . . . . . . . . . . . . . . . . . . . . . 3-110

3.6 Analog Pulse Modulation . . . . . . . . . . . . . . . . 3-115

3.6.1 Pulse-Amplitude Modulation (PAM) . . . . . . . 3-1153.6.2 Pulse-Width Modulation (PWM) . . . . . . . . . 3-1173.6.3 Pulse-Position Modulation . . . . . . . . . . . . 3-117

3.7 Delta Modulation and PCM . . . . . . . . . . . . . . . 3-118

3.7.1 Delta Modulation (DM) . . . . . . . . . . . . . 3-1183.7.2 Pulse-Code Modulation (PCM) . . . . . . . . . 3-121

3.8 Multiplexing . . . . . . . . . . . . . . . . . . . . . . . 3-124

3.8.1 Frequency-Division Multiplexing (FDM) . . . . 3-1253.8.2 Quadrature Multiplexing (QM) . . . . . . . . . . 3-1283.8.3 Time-Division Multiplexing (TDM) . . . . . . . 3-129

3.9 General Performance of Modulation Systems in Noise 3-133


3.1. LINEAR MODULATION

• We are typically interested in locating a message signal to somenew frequency location, where it can be efficiently transmitted

• The carrier of the message signal is usually sinusoidal

• A modulated carrier can be represented as

xc(t) = A(t) cos�2π fct + φ(t)

�

where A(t) is linear modulation, fc the carrier frequency, andφ(t) is phase modulation

3.1 Linear Modulation• For linear modulation schemes, we may set φ(t) = 0 without

loss of generality

xc(t) = A(t) cos(2π fct)

with A(t) placed in one-to-one correspondence with the mes-sage signal

3.1.1 Double-Sideband Modulation (DSB)• Let A(t) ∝ m(t), the message signal, thus

xc(t) = Acm(t) cos(2π fct)

• From the modulation theorem it follows that

Xc( f ) = 12

Ac M( f − fc) + 12

Ac M( f + fc)



t t

m(t) xc(t) carrier filled envelope

DSB time domain waveforms

M(0)M(f)

Xc(f) AcM(0)12

fc-fcf

f

LSB USB

DSB spectra

Coherent Demodulation

• The received signal is multiplied by the signal 2 cos(2π fct),which is synchronous with the transmitter carrier

LPFm(t) xc(t) xr(t) d(t) yD(t)

Accos[2!fct] 2cos[2!fct]

DemodulatorModulator Channel



• For an ideal channel xr(t) = xc(t), so

d(t) =�Acm(t) cos(2π fct)

�2 cos(2π fct)

= Acm(t) + Acm(t) cos(2π(2 fc)t)

where we have used the trig identity 2 cos2x = 1 + cos 2x

• The waveform and spectra of d(t) is shown below (assumingm(t) has a triangular spectrum in D( f ))

D(f)

AcM(0)

2fc-2fcf

AcM(0)12AcM(0)1

2

W-W

Lowpass modulation recovery filter

d(t)

t

Lowpass filtering will remove the double frequency carrier term

Waveform and spectrum of d(t)

• Typically the carrier frequency is much greater than the mes-sage bandwidth W , so m(t) can be recovered via lowpass fil-tering

• The scale factor Ac can be dealt with in downstream signalprocessing, e.g., an automatic gain control (AGC) amplifier



• Assuming an ideal lowpass filter, the only requirement is thatthe cutoff frequency be greater than W and less than 2 fc − W

• The difficulty with this demodulator is the need for a coherentcarrier reference

• To see how critical this is to demodulation of m(t) suppose thatthe reference signal is of the form

c(t) = 2 cos[2π fct + θ(t)]

where θ(t) is a time-varying phase error

• With the imperfect carrier reference signal

d(t) = Acm(t) cos θ(t) + Acm(t) cos[2π fct + θ(t)]yD(t) = m(t) cos θ(t)

• Suppose that θ(t) is a constant or slowly varying, then thecos θ(t) appears as a fixed or time varying attenuation factor

• Even a slowly varying attenuation can be very detrimental froma distortion standpoint

– If say θ(t) = � f t and m(t) = cos(2π fmt), then

yD(t) = 12

[cos[2π( fm − � f )t] + cos[2π( fm + � f )t]]

which is the sum of two tones

• Being able to generate a coherent local reference is also a prac-tical manner



• One scheme is to simply square the received DSB signal

x2r(t) = A

2cm

2(t) cos2(2π fct)

= 12

A2cm

2(t) + 12

A2cm

2(t) cos[2π(2 fc)t]

( )2 BPF

LPF

divideby 2

xr(t) yD(t)

xr(t)2

Acos2!fctvery narrow (tracking) band-pass filter

Carrier recovery concept using signal squaring

• Assuming that m2(t) has a nonzero DC value, then the double

frequency term will have a spectral line at 2 fc which can bedivided by two following filtering by a narrowband bandpassfilter, i.e., F{m2(t)} = kδ( f ) + · · ·

k

2fcfSp

ectru

mof m2

(t) Filter this component for coherent demod

• Note that unless m(t) has a DC component, Xc( f ) will notcontain a carrier term (read δ( f ± fc), thus DSB is also calleda suppressed carrier scheme



• Consider transmitting a small amount of unmodulated carrier

k

m(t) xc(t)

Accos2!fct

ffc-fc

AcM(0)/2use a narrowband filter (phase-locked loop) to extract the carrier in the demod.

k << 1

3.1.2 Amplitude Modulation

• Amplitude modulation (AM) can be created by simply addinga DC bias to the message signal

xc(t) =�A + m(t)

�A

�c

cos(2π fct)

= Ac

�1 + amn(t)

�cos(2π fct)

where Ac = AA�c, mn(t) is the normalized message such that

min mn(t) = −1,

mn(t) = m(t)

| min m(t)|

and a is the modulation index

a = | min m(t)|A



t

xc(t) a < 1

m(t)

A

A + m(t)

Accos[2!fct]

xc(t)

Bias term

Note that the enve-lope does not cross zero in the case of AM having a < 1

0Ac(1 - a)

A + max m(t) A + min m(t)

Generation of AM and a sample wavefrom

• Note that if m(t) is symmetrical about zero and we define d1 asthe peak-to-peak value of xc(t) and d2 as the valley-to-valleyvalue of xc(t), it follows that

a = d1 − d2

d1 + d2

proof: max m(t) = − min m(t) = | min m(t)|, so

d1 − d2

d1 + d2= 2[(A + | min m(t)|) − (A − | min m(t)|)]

2[(A + | min m(t)|) + (A − | min m(t)|)]= | min m(t)|

A= a



• The message signal can be recovered from xc(t) using a tech-nique known as envelope detection

• A diode, resistor, and capacitor is all that is needed to constructand envelope detector

eo(t)xr(t) C R

t

Recovered envelope with proper RC selectioneo(t)

0The carrier is removed if 1/fc << RC << 1/W

Envelope detector

• The circuit shown above is actually a combination of a nonlin-earity and filter (system with memory)

• A detailed analysis of this circuit is more difficult than youmight think

• A SPICE circuit simulation is relatively straight forward, but itcan be time consuming if W � fc



• The simple envelope detector fails if Ac[1 + amn(t)] < 0

– In the circuit shown above, the diode is not ideal andhence there is a turn-on voltage which further limits themaximum value of a

• The RC time constant cutoff frequency must lie between bothW and fc, hence good operation also requires that fc � W



• Digital signal processing based envelope detectors are also pos-sible

• Historically the envelope detector has provided a very low-costmeans to recover the message signal on AM carrier

• The spectrum of an AM signal is

Xc( f ) = Ac

2�δ( f − fc) + δ( f + fc)

�

� �� pure carrier spectrum

+ a Ac

2�Mn( f − fc) + Mn( f + fc)

�

� �� DSB spectrum

AM Power Efficiency

• Low-cost and easy to implement demodulators is a plus forAM, but what is the downside?

• Adding the bias term to m(t) means that a fraction of the totaltransmitted power is dedicated to a pure carrier

• The total power in xc(t) is can be written in terms of the timeaverage operator introduced in Chapter 2

�x2c(t)� = �A

2c[1 + amn(t)]2 cos2(2π fct)�

= A2c

2�[1 + 2amn(t) + a

2m

2n(t)][1 + cos(2π(2 fc)t]�

• If m(t) is slowly varying with respect to cos(2π fct), i.e.,

�m(t) cos ωct� � 0,



then

�x2c(t)� = A

2c

2�1 + 2a�mn(t)� + a

2�m2n(t)�

�

= A2c

2�1 + a

2�m2(t)��

= A2c

2��Pcarrier

+ a2A

2c

2�m2

n(t)�

� �� Psidebands

where the last line resulted from the assumption �m(t)� = 0(the DC or average value of m(t) is zero)

• Definition: AM Efficiency

E f f

�= a2�m2

n(t)�

1 + a2�m2n(t)�

also= �m2(t)�A2 + �m2(t)�

Example 3.1: Single Sinusoid AM

• An AM signal of the form

xc(t) = Ac[1 + a cos(2π fmt + π/3)] cos(2π fct)

contains a total power of 1000 W

• The modulation index is 0.8

• Find the power contained in the carrier and the sidebands, alsofind the efficiency

• The total power is

1000 = �x2c(t)� = A

2c

2+ a

2A

2c

2· �m2

n(t)�



• It should be clear that in this problem mn(t) = cos(2π fmt), so�m2

n(t)� = 1/2 and

1000 = A2c

�12

+ 14

0.64�

= 3350

A2c

• Thus we see that

A2c= 1000 · 50

33= 1515.15

and

Pcarrier = 12

A2c= 1515

2= 757.6 W

and thusPsidebands = 1000 − Pc = 242.4 W

• The efficiency is

E f f = 242.41000

= 0.242 or 24.2%

• The magnitude and phase spectra can be plotted by first ex-panding out xc(t)

xc(t) = Ac cos(2π fct) + a Ac cos(2π fmt + π/3) cos(2π fct)

= Ac cos(2π fct)

+ a Ac

2cos[2π( fc + fm)t + π/3]

+ a Ac

2cos[2π( fc − fm)t − π/3]



f

f

0

0

fc+fmfc-fm fc-fc

Ac/2

0.8Ac/4

|Xc(f)|

Xc(f)!/3

-!/3Amplitude and phase spectra for one tone AM

Example 3.2: Pulse Train with DC Offset

t

2

-1

m(t)

Tm/3 Tm

• Find mn(t) and the efficiency E

• From the definition of mn(t)

mn(t) = m(t)

| min m(t)| = m(t)

| − 1| = m(t)

• The efficiency is

E = a2�m2

n(t)�

1 + a2�m2n(t)�



• To obtain �m2n(t)� we form the time average

�m2n(t)� = 1

Tm

��Tm/3

0(2)2

dt +�

Tm

Tm/3(−1)2

dt

�

= 1Tm

�Tm

3· 4 + 2Tm

3· 1

�= 4

3+ 2

3= 7

3

thus

E = (7/3)a2

1 + (7/3)a2 = 7a2

3 + 7a2

• The best AM efficiency we can achieve with this waveform iswhen a = 1

E f f

��a=1

= 710

= 0.7 or 70%

• Suppose that the message signal is −m(t) as given here

• Now min m(t) = −2 and mn(t) = m(t)/2 and

�m2n(t)� = 1

3· (−1)2 + 2

3· (1/2)2 = 1

2

• The efficiency in this case is

E f f = (1/2)a2

1 + (1/2)a2 = a2

2 + a2

• Now when a = 1 we have E f f = 1/3 or just 33.3%

• Note that for 50% duty cycle squarewave the efficiency maxi-mum is just 50%



Example 3.3: Multiple Sinusoids

• Suppose that m(t) is a sum of multiple sinusoids (multi-toneAM)

m(t) =M�

k=1

Ak cos(2π fkt + φk)

where M is the number of sinusoids, fk values might be con-strained over some band of frequencies W , e.g., fk ≤ W , andthe phase values φk can be any value on [0, 2π ]

• To find mn(t) we need to find min m(t)

• A lower bound on min m(t) is − �M

k=1 Ak; why?

• The worst case value may not occur in practice depending uponthe phase and frequency values, so we may have to resort to anumerical search or a plot of the waveform

• Suppose that M = 3 with fk = {65, 100, 35} Hz, Ak ={2, 3.5, 4.2}, and φk = {0, π/3, −π/4} rad.

>> [m,t] = M_sinusoids(1000,[65 100 35],[2 3.5 4.2],...[0 pi/3 -pi/4], 20000);>> plot(t,m)

>> min(m)

ans = -7.2462e+00

>> -sum([2 3.5 4.2]) % worst case minimum value

ans = -9.7000e+00

>> subplot(311)>> plot(t,(1 + 0.25*m/abs(min(m))).*cos(2*pi*1000*t))>> hold



Current plot held>> plot(t,1 + 0.25*m/abs(min(m)),’r’)>> subplot(312)>> plot(t,(1 + 0.5*m/abs(min(m))).*cos(2*pi*1000*t))>> holdCurrent plot held>> plot(t,1 + 0.5*m/abs(min(m)),’r’)>> subplot(313)>> plot(t,(1 + 1.0*m/abs(min(m))).*cos(2*pi*1000*t))>> holdCurrent plot held>> plot(t,1 + 1.0*m/abs(min(m)),’r’)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05!8

!6

!4

!2

0

2

4

6

8

Time (s)

m(t)

Ampli

tude

min m(t)

Finding min m(t) graphically

• The normalization factor is approximately given by 7.246, thatis

mn(t) = m(t)

7.246• Shown below are plots of xc(t) for a = 0.25, 0.5 and 1 using

fc = 1000 Hz



0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05!2

0

2x c(t)

, a =

0.25

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05!2

0

2

x c(t), a

= 0.

5

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05!2

0

2

x c(t), a

= 1.

0

Time (s)

Modulation index comparison ( fc = 1000 Hz)

• To obtain the efficiency of multi-tone AM we first calculate�m2

n(t)� assuming unique frequencies

�m2n(t)� =

M�

k=1

A2k

2| min m(t)|2

= 22 + 3.52 + 4.22

2 × 7.2462 = 0.3227

• The maximum efficiency is just

E f f

��a=1

= 0.32271 + 0.3227

= 0.244 or 24.4%



• A remaining interest is the spectrum of xc(t)

Xc( f ) = Ac

2�δ( f − fc) + δ( f + fc)

�

+ a Ac

4

M�

k=1

Ak

�e

jφkδ( f − ( fc + fk))

+ e− jφkδ( f + ( fc + fk))

�(USB terms)

+ a Ac

4

M�

k=1

Ak

�e

jφkδ( f − ( fc − fk))

+ e− jφkδ( f + ( fc − fk))

�(LSB terms)

!1000 !800 !600 !400 !200 0 200 400 600 800 10000

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Ampli

tude S

pectr

a (|X

c(f)|)

Frequency (Hz)

Sidebands for a = 0.5

Carrier with Ac = 1

Amplitude spectra



3.1.3 Single-Sideband Modulation• In the study of DSB it was observed that the USB and LSB

spectra are related, that is the magnitude spectra about fc haseven symmetry and phase spectra about fc has odd symmetry

• The information is redundant, meaning that m(t) can be recon-structed one or the other sidebands

• Transmitting just the USB or LSB results in single-sideband

(SSB)

• For m(t) having lowpass bandwidth of W the bandwidth re-quired for DSB, centered on fc is 2W

• Since SSB operates by transmitting just one sideband, the trans-mission bandwidth is reduced to just W

M(f)

f

fc - W

fc - W fc+W

fc+Wfc

fc

fc

XDSB(f)

fXSSB(f)XSSB(f)

f

LSB USB

W

USBremoved

LSBremoved

DSB to two forms of SSB: USSB and LSSB

• The filtering required to obtain an SSB is best explained withthe aid of the Hilbert transform, so we divert from text Chapter



3 back to Chapter 2 to briefly study the basic properties of thistransform

Hilbert Transform

• The Hilbert transform is nothing more than a filter that shiftsthe phase of all frequency components by −π/2, i.e.,

H( f ) = − jsgn( f )

where

sgn( f ) =

1, f > 00, f = 0−1, f < 0

• The Hilbert transform of signal x(t) can be written in terms ofthe Fourier transform and inverse Fourier transform

x(t) = F−1� − jsgn( f )X ( f )�

= h(t) ∗ x(t)

where h(t) = F−1{H( f )}

• We can find the impulse response h(t) using the duality theo-rem and the differentiation theorem

d

d fH( f )

F←→ (− j2π t)h(−t)

where here H( f ) = − jsgn( f ), so

d

d fH( f ) = −2 jδ( f )



• Clearly,F−1{−2 jδ( f )} = −2 j

soh(t) = −2 j

− j2π t= 1

π t

and1π t

F←→ − jsgn( f )

• In the time domain the Hilbert transform is the convolutionintegral

x(t) =� ∞

−∞

x(λ)

π(t − λ)dλ =

� ∞

−∞

x(t − λ)

πλdλ

• Note that since the Hilbert transform of x(t) is a −π/2 phaseshift, the Hilbert transform of x(t) is

ˆx(t) = −x(t)

why? observe that (− jsgn( f ))2 = −1

Example 3.4: x(t) = cos ω0t

• By definition

X( f ) = − jsgn( f ) · 12�δ( f − f0) + δ( f + f0)

�

= − j12δ( f − fc) + j

12δ( f + f0)



so from ejω0t = δ( f − f0)

x(t) = − j12

ejω0t + j

12

e− jω0t

= ejω0t − e

− jω0t

2 j= sin ω0t

or�cos ω0t = sin ω0t

• It also follows that

�sin ω0t = ��cos ω0t = − cos ω0t

since ˆx(t) = −x(t)

Hilbert Transform Properties

1. The energy (power) in x(t) and x(t) are equal

The proof follows from the fact that |Y ( f )|2 = |H( f )|2|X ( f )|and | jsgn( f )|2 = 1

2. x(t) and x(t) are orthogonal, that is� ∞

−∞x(t)x(t) dt = 0 (energy signal)

limT →∞

12T

�T

−T

x(t)x(t) dt = 0 (power signal)



The proof follows for the case of energy signals by generaliz-ing Parseval’s theorem

� ∞

−∞x(t)x(t) dt =

� ∞

−∞X ( f )X

∗( f ) d f

=� ∞

−∞( jsgn( f ))� ��

odd

|X ( f )|2� �� even

d f = 0

3. Given signals m(t) and c(t) such that the corresponding spec-tra are

M( f ) = 0 for | f | > W (a lowpass signal)C( f ) = 0 for | f | < W (c(t) a highpass signal)

then�m(t)c(t) = m(t)c(t)

Example 3.5: c(t) = cos ω0t

• Suppose that M( f ) = 0 for | f | > W and f0 > W then

�m(t) cos ω0t = m(t) �cos ω0t

= m(t) sin ω0t

Analytic Signals

• Define analytic signal z(t) as

z(t) = x(t) + j x(t)

where x(t) is a real signal



• The envelope of z(t) is |z(t)| and is related to the envelopediscussed with DSB and AM signals

• The spectrum of an analytic signal have single-sideband char-acteristics

• In particular for z p(t) = x(t) + j x(t)

Z p( f ) = X ( f ) + j�

− jsgn( f )X ( f )�

= X ( f )�1 + sgn( f )

�

=�

2X ( f ), f > 00, f < 0

Note: Only positive frequencies present

• Similarly for zn(t) = x(t) − j x(t)

Zn( f ) = X ( f )�1 − sgn( f )

�

=�

0, f > 02X ( f ), f < 0



W-W

W-W

W-W

f

f

f

X(f)

Zp(f)

Zn(f)

2

2

1

The spectra of analytic signals can suppress positive or negativefrequencies

Return to SSB Development

SidebandFilter

Accos!ct

m(t)xDSB(t)

xSSB(t)LSB or USB

Basic SSB signal generation

• In simple terms, we create an SSB signal from a DSB signalusing a sideband filter

• The mathematical representation of LSSB and USSB signalsmakes use of Hilbert transform concepts and analytic signals



DSB Signal Starting Point

Formation of HL(f)-fc fc

-fc fc

f

f

f

f

sgn(f + fc)/2

-sgn(f - fc)/2

+1/2

+1/2

-1/2

-1/2HL(f) = [sgn(f + fc) - sgn(f - fc)]/21

An ideal LSSB filter

• From the frequency domain expression for the LSSB, we canultimately obtain an expression for the LSSB signal, xcLSSB(t),in the time domain

• Start with XDSB( f ) and the filter HL( f )

XcLSSB( f ) = 12

Ac

�M( f + fc) + M( f − fc)

�

· 12�sgn( f + fc) − sgn( f − fc)

�



XcLSSB( f ) = 14

Ac

�M( f + fc)sgn( f + fc)

+ M( f − fc)sgn( f − fc)�

− 14

Ac

�M( f + fc)sgn( f − fc)

+ M( f − fc)sgn( f − fc)�

= 14

Ac

�M( f + fc) + M( f − fc)

�

+ 14

Ac

�M( f + fc)sgn( f + fc)

− M( f − fc)sgn( f − fc)�

• The inverse Fourier transform of the first term is DSB, i.e.,

12

Acm(t) cos ωctF←→ 1

4Ac

�M( f + fc) + M( f − fc)

�

• The second term can be inverse transformed using

m(t)F←→ − jsgn( f ) · M( f )

soF−1�

M( f + fc)sgn( f + fc)�

= j m(t)e− jωct

since m(t)e± jωctF←→ M( f ± fc)

• Thus

14

AcF−1�M( f + fc)sgn( f + fc)− M( f − fc)sgn( f − fc)

�

= 14

Ac

�j m(t)e− jωct − j m(t)e jωct

�= 1

2m(t) sin ωct



• Finally,

xcLSSB(t) = 12

Acm(t) cos ωct + 12

Acm(t) sin ωct

• Similarly for USSB it can be shown that

xcUSSB(t) = 12

Acm(t) cos ωct − 12

Acm(t) sin ωct

• The direct implementation of SSB is very difficult due to therequirements of the filter

• By moving the phase shift frequency from fc down to DC (0Hz) the implementation is much more reasonable (this appliesto a DSP implementation as well)

• The phase shift is not perfect at low frequencies, so the modu-lation must not contain critical information at these frequencies

H(f) = -jsgn(f)

0o

-90o

m(t)xc(t)

sin!ct

cos!ctcos!ct

Carrier Osc.+

+- LSBUSB

Phase shift modulator for SSB



Demodulation

• The coherent demodulator first discussed for DSB, also worksfor SSB

LPFxr(t)d(t) yD(t)

4cos[2!fct + "(t)]1/Ac scale factorincluded

Coherent demod for SSB

• Carrying out the analysis to d(t), first we have

d(t) = 12

Ac

�m(t) cos ωct ± m(t) sin ωct

�4 cos(ωct + θ(t))

= Acm(t) cos θ(t) + Acm(t) cos[2ωct + θ(t)]∓ Acm(t) sin θ(t) ± Acm(t) sin[2ωct + θ(t)]

so

yD(t) = m(t) cos θ(t) ∓ m(t) sin θ(t)θ(t) small� m(t) ∓ m(t)θ(t)

– The m(t) sin θ(t) term represents crosstalk

• Another approach to demodulation is to use carrier reinsertion

and envelope detection

EnvelopeDetectorxr(t)

e(t)yD(t)

Kcos!ct



e(t) = xr(t) + K cos ωct

=�

12

Acm(t) + K

�cos ωct ± 1

2Acm(t) sin ωct

• To proceed with the analysis we must find the envelope of e(t),which will be the final output yD(t)

• Finding the envelope is a more general problem which will beuseful in future problem solving, so first consider the envelopeof

x(t) = a(t)��inphase

cos ωct − b(t)��quadrature

sin ωct

= Re�a(t)e jωct + jb(t)e jωct

�

= Re�

[a(t) + jb(t)]� �� R(t)=complex envelope

ejωct

�

• In a phasor diagram x(t) consists of an inphase or direct com-ponent and a quadrature component

R(t)

!(t)a(t)

b(t)

In-phase - I

Quadrature - QNote: R(t) =

R(t)ej!(t) = a(t) + jb(t)

~



where the resultant R(t) is such that

a(t) = R(t) cos θ(t)

b(t) = R(t) sin θ(t)

which implies that

x(t) = R(t)�

cos θ(t) cos ωct − sin θ(t) sin ωct�

= R(t) cos�ωct + θ(t)

�

where θ(t) = tan−1[b(t)/a(t)]

• The signal envelope is thus given by

R(t) =�

a2(t) + b2(t)

• The output of an envelope detector will be R(t) if a(t) and b(t)are slowly varying with respect to cos ωct

• In the SSB demodulator

yD(t) =��

12

Acm(t) + K

�2

+�

12

Acm(t)

�2

• If we choose k such that (Acm(t)/2 + K )2 � (Acm(t)/2)2,then

yD(t) � 12

Acm(t) + K

• Note:

– The above analysis assumed a phase coherent reference

– In speech systems the frequency and phase can be ad-justed to obtain intelligibility, but not so in data systems



– The approximation relies on the binomial expansion

(1 + x)1/2 � 1 + 12

x for |x | � 1

Example 3.6: Noncoherent Carrier Reinsertion

• Let m(t) = cos ωmt , ωm � ωc and the reinserted carrier beK cos[(ωc + �ω)t]

• Following carrier reinsertion we have

e(t) = 12

Ac cos ωmt cos ωct

∓ 12

Ac sin ωct sin ωct + K cos�(ωc + �ω)t

�

= 12

Ac cos�(ωc ± ωm)t

�+ K cos

�(ωc + �ω)t

�

• We can write e(t) as the real part of a complex envelope timesa carrier at either ωc or ωc + �ω

• In this case, since K will be large compared to Ac/2, we write

e(t) = 12

AcRe�

e± jωmt

ejωct

�

+ K Re�

1 · ej (ωc+�ω)t

�

= Re� �1

2Ace

j (±ωm−�ω)t + K

�

� �� complex envelope R(t)

ej (ωc+�ω)t

�



• Finally expanding the complex envelope into the real and imag-inary parts we can find the real envelope R(t)

yD(t) =��1

2Ac cos[±ωm + �ω)t] + K

�2

+�1

2Ac sin[(±ωm + �ω)t]

�2�1/2

� 12

Ac cos[(ωm ∓ �ω)t] + K

where the last line follows for K � Ac

• Note that the frequency error �ω causes the recovered mes-sage signal to shift up or down in frequency by �ω, but notboth at the same time as in DSB, thus the recovered speechsignal is more intelligible

3.1.4 Vestigial-Sideband Modulation

• Vestigial sideband (VSB) is derived by filtering DSB such thatone sideband is passed completely while only a vestige remainsof the other

• Why VSB?

1. Simplifies the filter design

2. Improves the low-frequency response and allows DC topass undistorted

3. Has bandwidth efficiency advantages over DSB or AM,similar to that of SSB



• A primary application of VSB is the video portion of analogtelevision (note HDTV is replacing this in the US)

• The generation of VSB starts with DSB followed by a filterthat has a 2β transition band, e.g.,

|H( f )| =

0, f < Fc − βf −( fc−β)

2β , fc − β ≤ f ≤ fc + β

1, f > fc + β

ffcfc - ! fc + !

|H(f)|

1

Ideal VSB transmitter filter amplitude response

• VSB can be demodulated using a coherent demod or using car-rier reinsertion and envelope detection

ff - f1 f + f1f - f2 f + f2fc

B/2A(1 - !)/2

A!/2

Transmitted Two-Tone Spectrum(only single-sided shown)

0

Two-tone VSB signal



• Suppose the message signal consists of two tones

m(t) = A cos ω1t + B cos ω2t

• Following the DSB modulation and VSB shaping,

xc(t) = 12

A� cos(ωc − ω1)t

+ 12

A(1 − �) cos(ωc + ω1)t + 12

B cos(ωc + ω2)t

• A coherent demod multiplies the received signal by 4 cos ωct

to produce

e(t) = A� cos ω1t + A(1 − �) cos ω1t + B cos ω2t

= A cos ω1t + B cos ω2t

which is the original message signal

• The symmetry of the VSB shaping filter has made this possible

• In the case of broadcast TV the carrier in included at the trans-mitter to insure phase coherency and easy demodulation at theTV receiver (VSB + Carrier)

– Very large video carrier power is required for typical TVstation, i.e., greater than 100,000 W

– To make matters easier still, the precise VSB filtering isnot performed at the transmitter due to the high powerrequirements, instead the TV receiver does this



0

0

-0.75

-0.75 0.75

-1.75 4.0

4.0

4.5

4.75

4.75 (f - fcv) MHz

(f - fcv) MHz

Video CarrierAudio Carrier

TransmitterOutput

ReceiverShapingFilter

12! interval

Broadcast TV transmitter spectrum and receiver shaping filter

3.1.5 Frequency Translation and Mixing• Used to translate baseband or bandpass signals to some new

center frequency

BPFatf2 ff

f1 f2Local oscillator of the form

e(t)m(t)cos!1t

Frequency translation system

• Assuming the input signal is DSB of bandwidth 2W the mixer(multiplier) output is

e(t) = m(t) cos(ω1t)

local osc (LO)� �� 2 cos(ω1 ± ω2)t

= m(t) cos(ω2t) + m(t) cos[(2ω1 ± ω2)t]



• The bandpass filter bandwidth needs to be at least 2W Hz wide

• Note that if an input of the form k(t) cos[(ω1±2ω2)t] is presentit will be converted to ω2 also, i.e.,

e(t) = k(t) cos(ω2t) + k(t) cos[(2ω1 ± 3ω2)t],

and the bandpass filter output is k(t) cos(ω2t)

• The frequencies ω1 ±2ω2 are the image frequencies of ω1 withrespect to ωLO = ω1 ± ω2

Example 3.7: AM Broadcast Superheterodyne Receiver

TunableRF-Amp

IF Filt/Amp

LocalOsc.

EnvDet

AudioAmp

Joint tuning

Automatic gaincontrol

fIF

AM Broadcast Specs: fc = 540 to 1600 kHz on 10 kHz spacings carrier stability Modulated audio flat 100 Hz to 5 kHz Typical fIF = 455 kHz

For AM BT = 2W

AM Superheterodyne receiver

• We have two choices for the local oscillator, high-side or low-

side tuning



– Low-side: 540−455 ≤ fLO ≤ 1600−455 or 85 ≤ fLO ≤1145, all frequencies in kHz

– High-side: 540 + 455 ≤ fLO ≤ 1600 + 455 or 995 ≤fLO ≤ 2055, all frequencies in kHz

• The high-side option is advantageous since the tunable oscil-lator or frequency synthesizer has the smallest frequency ratiofLO,max/ fLO,min = 2055/995 = 2.15

• Suppose the desired station is at 560 kHz, then with high-sidetuning we have fLO = 560 + 455 = 1015 kHz

• The image frequency is at fimage = fc + 2 fIF = 560 + 2 ×455 = 1470 kHz (note this is another AM radio station centerfrequency

f (kHz)

f (kHz)

f (kHz)

f (kHz)0

Input

fLO

MixerOutput

ImageOut ofmixer

455 560 1470

1470

1575(560+1015)

2485(1470+1015)

1015(560+455)

455

Desired Potential Image

This is removedwith RF BPF

IF BPF

1470-1015

1015-560

fIF fIF

BRF

BIF

Receiver frequency plan including images



Example 3.8: A Double-Conversion Receiver

TunableRF-Amp

10.7 MHzIF BPF

455 kHzIF BPF

1stLO

2ndLO

FMDemod

fc = 162.475 MHz(WX #4)

fLO1 = 173.175 MHz fLO2 = 11.155 MHz

10.7 MHz &335.65 MHz

455 kHz &21.855 MHz

Double-conversion superheterodyne receiver

• Consider a frequency modulation (FM) receiver that uses double-conversion to receive a signal con carrier frequency 162.475MHz (weather channel #4)

– Frequency modulation will be discussed in the next sec-tion

• The dual-conversion allows good image rejection by using a10.7 MHz first IF and then can provide good selectivity byusing a second IF at 455 kHz; why?

– The ratio of bandwidth to center frequency can only be sosmall in a low loss RF filter

– The second IF filter can thus have a much narrower band-width by virtue of the center frequency being much lower

• A higher first IF center frequency moves the image signal fur-ther away from the desired signal



– For high-side tuning we have fimage = fc + 2 fIF = fc +21.4 MHz

• Double-conversion receivers are more complex to implement

Mixers

• The multiplier that is used to implement frequency translationis often referred to as a mixer

• In the world of RF circuit design the term mixer is more ap-propriate, as an ideal multiplier is rarely available

• Instead active and passive circuits that approximate signal mul-tiplication are utilized

• The notion of mixing comes about from passing the sum of twosignals through a nonlinearity, e.g.,

y(t) = [a1x1(t) + a2x2(t)]2 + other terms= a

21x

21(t) + 2a1a2x1(t)x2(t) + a

22x

22(t)

• In this mixing application we are most interested in the centerterm

ydesired(t) = 2a1a2�x1(t) · x2(t)

�

• Clearly this mixer produces unwanted terms (first and third),and in general may other terms, since the nonlinearity will havemore than just a square-law input/output characteristic



• A diode or active device can be used to form mixing productsas described above, consider the dual-gate MEtal Semiconduc-tor FET (MESFET) mixer shown below

+5V

R2

C4

L3

47pF

L4

C8C7Q1NE25139

IF270nH 270nH

82pF

C3

42pF

0.01uF

10!

C8R1 C547pF 0.01uF

270!

L15 turns, 28 AWG.050 I.D.C1

0.5pFLO

5 turns, 28 AWG.050 I.D.

C2

L20.5pF

RF

G1G2

DS

Dual-Gate MESFET Active Mixer

VRF

VOUTVIN

VLO

zL

Nonlinear Device

Mixer concept

• The double-balanced mixer (DBM), which can be constructedusing a diode ring, provides better isolation between the RF,LO, and IF ports

• When properly balanced the DBM also allows even harmonicsto be suppressed in the mixing operation



• A basic transformer coupled DBM, employing a diode ring, isshown below, followed by an active version

• The DBM is suitable for use as a phase detector in phase-locked loop applications

02 91 81 71

DNG

+FI

-FI

DNG

61DNG

13

12

11

14

15

GND

GND

GND

LO2LO2

LO1LO1

4

3

2

1

RFBIAS

TAP

RFRFIN

5GND

LO SELECT

GND

6 7 8 9

VCC VCCD NG

D NG

LE SOL

01

MAX9982

IF OUT

4:1 (200:50)TRANSFORMER

41

25V

3C9

C10

C7

C6

C11C8

R3L1

R4L2

T16

5V5V

C5C4

R1

C2

C1

C3

825 MHz to 915 MHz SiGe High-Linearity Active DBM

vo(t)

mixer

D3 D4

D1D2inputLO

R G

LO source

vp(t)

inputRF

R G

RF source

R L

IF load

vi (t)

IFout

Passive Double-Balanced Mixer (DBM)



Example 3.9: Single Diode Mixer



3.2 Angle Modulation

• A general angle modulated signal is of the form

xc(t) = Ac cos[ωct + φ(t)]

• Definition: Instantaneous phase of xc(t) is

θi(t) = ωct + φ(t)

where φ(t) is the phase deviation

• Define: Intsantaneous frequency of xc(t) is

ωi(t) = dθi(t)

dt= ωc + dφ(t)

dt

where dφ(t)/dt is the frequency deviation

• There are two basic types of angle modulation

1. Phase modulation (PM)

φ(t) = kp��phase dev. const.

m(t)

which implies

xc(t) = Ac cos[ωct + kpm(t)]

– Note: the units of kp is radians per unit of m(t)

– If m(t) is a voltage, kp has units of radians/volt


3.2. ANGLE MODULATION

2. Frequency modulation (FM)

dφ(t)

dt= k f��

freq. dev. const.

m(t)

or

φ(t) = k f

�t

t0

m(α) dα + φ0

– Note: the units of k f is radians/sec per unit of m(t)

– If m(t) is a voltage, k f has units of radians/sec/volt– An alternative expression for k f is

k f = 2π fd

where fd is the frequency-deviation constant in Hz/unitof m(t)

Example 3.10: Phase and Frequency Step Modulation

• Consider m(t) = u(t) v

• We form the PM signal

xPM(t) = Ac cos�ωct + kpu(t)

�, kp = π/3 rad/v

• We form the FM signal

xFM(t) = Ac cos�ωct + 2π fd

�t

m(α) dα�, fd = 3 Hz/v



!1 1

!1

1

!1 1

!1

1

t t

!/3 phase step at t = 0 3 Hz frequency step at t = 0

Phase Modulation Frequency Modulation

fc fc fc + 3 Hzfc

Phase and frequency step modulation

3.2.1 Narrowband Angle Modulation• Begin by writing an angle modulated signal in complex form

xc(t) = Re�

Acejωct

ejφ(t)

�

• Expand ejφ(t) in a power series

xc(t) = Re�

Acejωct

�1 + jφ(t) − φ2(t)

2!− · · ·

��

• The narrowband approximation is |φ(t)| � 1, then

xc(t) � Re�

Acejωct + j Acφ(t)e jωct

�

= Ac cos(ωct) − Acφ(t) sin(ωct)



• Under the narrowband approximation we see that the signal issimilar to AM except it is carrier plus modulated quadraturecarrier

90o

!(t)

Ac sin("ct)

NBFMxc(t)

+

#

NBFM modulator block diagram

Example 3.11: Single tone narrowband FM

• Consider NBFM with m(t) = cos ωmt

φ(t) = 2π fd

�t

cos ωmt dα

= 2π fd

2π fm

sin ωmt = fd

fm

sin ωmt

• Now,

xc(t) = cos�

ωctfd

fm

sin ωmt

�

� Ac

�cos ωct − fd

fm

sin ωmt sin ωct

�

= Ac cos ωct + fd

2 fm

sin( fc + fm)t − fd

2 fm

sin( fc − fm)t

• This looks very much like AM



fc fc + fmfc - fm f

0

Single tone NBFM spectra

3.2.2 Spectrum of an Angle-Modulated Signal

• The development in this obtains the exact spectrum of an anglemodulated carrier for the case of

φ(t) = β sin ωmt

where β is the modulation index for sinusoidal angle modula-tion

• The transmitted signal is of the form

xc(t) = Ac cos�ωct + β sin ωmt

�

= AcRe�e

jωct exp jβ sin ωmt�

• Note that ejβ sin ωmt is periodic with period T = 2π/ωm, thus

we can obtain a Fourier series expansion of this signal, i.e.,

ejβ sin ωmt =

∞�

n=−∞Yne

jnωmt



• The coefficients are

Yn = ωm

2π

� π/ωm

−π/ωm

ejβ sin ωmt

e− jnωmt

dt

= ωm

2π

� π/ωm

−π/ωm

e− j (nωmt−β sin ωmt)

• Change variables in the integral by letting x = ωmt , then dx =ωmdt , t = π/ωm → x = π , and t = −π/ωm → x = −π

• With the above substitutions, we have

Yn = 12π

� π

−πe

− j (nx−β sin x)dx

= 1π

� π

0cos(nx − β sin x) dx = Jn(β)

which is a Bessel function of the first kind order n with argu-ment β

Jn(β) Properties

• Recurrence equation:

Jn+1(β) = 2n

βJn(β) − Jn−1(β)

• n – even:J−n(β) = Jn(β)

• n – odd:J−n(β) = −Jn(β)



J0(!)

J1(!) J2(!) J3(!)

!2 4 6 8 10

!0.4

!0.2

0.2

0.4

0.6

0.8

1

Bessel function of order 0–3 plotted

• The zeros of the Bessel functions are important in spectralanalysis

First five Bessel function zeros for order 0 – 5

J0(!) = 02.40483, 5.52008, 8.65373, 11.7915, 14.9309

J1(!) = 03.83171, 7.01559, 10.1735, 13.3237, 16.4706

J2(!) = 05.13562, 8.41724, 11.6198, 14.796, 17.9598

J3(!) = 06.38016, 9.76102, 13.0152, 16.2235, 19.4094

J4(!) = 07.58834, 11.0647, 14.3725, 17.616, 20.8269

J5(!) = 08.77148, 12.3386, 15.7002, 18.9801, 22.2178



Spectrum cont.

• We obtain the spectrum of xc(t) by inserting the series repre-sentation

xc(t) = AcRe

�

ejωct

∞�

n=−∞Jn(β)e jnωmt

�

= Ac

∞�

n=−∞Jn(β) cos(ωc + nωm)t

• We see that the amplitude spectrum is symmetrical about fc

due to the symmetry properties of the Bessel functions

ffc

f c + 2f

m

f c + 3f

m

f c + 4f

m

f c + 5f

m

f c + f m

f c - f m

f c - 2f

m

f c - 3f

m

f c - 5f

m

f c - 4f

m

|AcJ0(!)|

|AcJ1(!)||AcJ2(!)||AcJ-2(!)|

|AcJ3(!)||AcJ-3(!)||AcJ4(!)||AcJ-4(!)| |AcJ5(!)||AcJ-5(!)|

|AcJ-1(!)|

Ampli

tude S

pectr

um(on

e-side

d)

• For PMβ sin ωmt = kp (A sin ωmt)� ��

m(t)

⇒ β = kp A

• For FM

β sin ωmt = k f

�t

A cos ωmα dα = fd

fm

A sin ωmt

⇒ β = ( fd/ fm)A



• When β is small we have the narrowband case and as β getslarger the spectrum spreads over wider bandwidth

-5 0 5 10

0.2

0.4

0.6

0.8

1

-5 0 5 10

0.2

0.4

0.6

0.8

1

-5 0 5 10

0.2

0.4

0.6

0.8

1

-5 0 5 10

0.2

0.4

0.6

0.8

1

! = 0.2, Ac = 1(narrowband case)

! = 1, Ac = 1

! = 2.4048, Ac = 1(carrier null)

! = 3.8317, Ac = 1(1st sideband null)

! = 8, Ac = 1(spectrum becoming wide)

-5 0 5 10

0.2

0.4

0.6

0.8

1

(f - fc)/fm

(f - fc)/fm

(f - fc)/fm

(f - fc)/fm

(f - fc)/fm

Ampli

tude

Spec

trum

Ampli

tude

Spec

trum

Ampli

tude

Spec

trum

Ampli

tude

Spec

trum

Ampli

tude

Spec

trum

The amplitude spectrum relative to fc as β increases



Example 3.12: VCO FM Modulator

• Consider again single-tone FM, that is m(t) = A cos(2π fmt)

• We assume that we know fm and the modulator deviation con-stant fd

• Find A such that the spectrum of xc(t) contains no carrier com-ponent

• An FM modulator can be implemented using a voltage con-

trolled oscillator (VCO)

VCO

CenterFreq = fc

m(t)

sensitivity fd MHz/v

A VCO used as an FM modulator

• The carrier term is Ac J0(β) cos ωct

• We know that J0(β) = 0 for β = 2.4048, 5.5201, . . .

• The smallest β that will make the carrier component zero is

β = 2.4048 = fd

fm

A

which implies that we need to set

A = 2.4048 · f m

fd



• Suppose that fm = 1 kHz and fd = 2.5 MHz/v, then we wouldneed to set

A = 2.4048 · 1 × 103

2.5 × 106 = 9.6192 × 10−4

3.2.3 Power in an Angle-Modulated Signal• The average power in an angle modulated signal is

�x2c(t)� = A

2c�cos2 �

ωct + φ(t)��

= 12

A2c+ 1

2A

2c�cos

�2�ωct + φ(t)

��

• For large fc the second term is approximately zero (why?),thus

Pangle mod = �x2c(t)� = 1

2A

2c

which makes the power independent of the modulation m(t)(the assumptions must remain valid however)

3.2.4 Bandwidth of Angle-Modulated Signals• With sinusoidal angle modulation we know that the occupied

bandwidth gets larger as β increases

• There are an infinite number of sidebands, but

limn→∞

Jn(β) ≈ limn→∞

βn

2nn!= 0,

so consider the fractional power bandwidth



• Define the power ratio

Pr = Pcarrier + P±k sidebands

Ptotal=

12 A

2c

�k

n=−kJ

2n(β)

12 A2

c

= J20 (β) + 2

k�

n=1

J2n(β)

• Given an acceptable Pr implies a fractional bandwidth of

B = 2k fm (Hz)

• In the text values of Pr ≥ 0.7 and Pr ≥ 0.98 are single anddouble underlined respectively

• It turns out that for Pr ≥ 0.98 the value of k is IP[1 + β], thus

B = B98 � 2(β + 1) fm sinusoidal mod only

• For arbitrary modulation m(t), define the deviation ratio

D = peak freq. deviationbandwidth of m(t)

= fd

W

�max |m(t)|

�

• In the sinusoidal modulation bandwidth definition let β → D

and fm → W , then we obtain what is known as Carson’s rule

B = 2(D + 1)W

– Another view of Carson’s rule is to consider the maxi-mum frequency deviation � f = max |m(t)| fd , then B =2(W + � f )



• Two extremes in angle modulation are

1. Narrowband: D � 1 ⇒ B = 2W

2. Wideband: D � 1 ⇒ B = 2DW = 2� f

Example 3.13: Single Tone FM

• Consider an FM modulator for broadcasting with

xc(t) = 100 cos�2π(101.1 × 106)t + φ(t)

�

where fd = 75 kHz/v and

m(t) = cos�2π(1000)t

�v

• The β value for the transmitter is

β = fd

fm

A = 75 × 103

103 = 75

• Note that the carrier frquency is 101.1 MHz and the peak de-viation is � f = 75 kHz

• The bandwidth of the signal is thus

B � 2(1 + 75)1000 = 152 kHz

-50 0 50 100

2.55

7.510

12.515

17.5

(f - 101.1 MHz)1 kHz

Ampli

tude

Spec

trum

B = 2(! + 1)fm

76-76101.1 MHz 101.1 MHz + 76 kHz101.1 MHz - 76 kHz



• Suppose that this signal is passed through an ideal bandpassfilter of bandwidth 11 kHz centered on fc = 101.1 MHz, i.e.,

H( f ) = �

�f − fc

11000

�+ �

�f + fc

11000

�

• The carrier term and five sidebands either side of the carrierpass through this filter, resulting an output power of

Pout = A2c

2

�

J20 (75) + 2

5�

n=1

J2n(75)

�

= 241.93 W

• Note the input power is A2c/2 = 5000 W

Example 3.14: Two Tone FM

• Finding the exact spectrum of an angle modulated carrier is notalways possible

• The single-tone sinusoid case can be extended to multiple tonewith increasing complexity

• Suppose that

m(t) = A cos ω1t + B cos ω2t

• The phase deviation is given by 2π fd times the integral of thefrequency modulation, i.e.,

φ(t) = β1 sin ω1t + β2 sin ω2t

where β1 = A fd/ f1 and β2 = A fd/ f2



• The transmitted signal is of the form

xc(t) = Ac cos�ωct + β1 sin ω1t + β2 sin ω2t

�

= AcRe�e

jωcte

jβ1 sin ω1te

jβ2 sin β2t�

• We have previously seen that via Fourier series expansion

ejβ1 sin ω1t =

∞�

n=−∞Jn(β1)e

jnω1t

ejβ2 sin ω1t =

∞�

n=−∞Jn(β2)e

jnω2t

• Inserting the above Fourier series expansions into xc(t), wehave

xc(t) = AcRe

�

ejωct

∞�

n=−∞Jn(β1)e

jnω1t ·m�

m=−∞Jm(β2)e

jmω2t

�

= Ac

∞�

n=−∞

∞�

m=−∞Jn(β1)Jm(β2) cos(ωc + nω1 + mω2)t

• The nonlinear nature of angle modulation is clear, since we seenot only components at ωc + nω1 and ωc + mω2, but also at allcombinations of ωc + nω1 + mω2

• To find the bandwidth of this signal we can use Carson’s rule(the sinusoidal formula only works for one tone)

• Recall that B = 2(� f + W ), where � f is the peak frequencydeviation



• The frequency deviation is

fi(t) = 12π

d

dt

�β1 sin ω1t + β2 sin ω2t

�

= β1 f1 cos(2π f1t) + β2 f2 cos(2π f2t) Hz

• The maximum of fi(t), in this case, is β1 f1 + β2 f2

• Suppose β1 = β2 = 2 and f2 = 10 f1, then we see that W =f2 = 10 f1 and

B = 2(W +� f ) = 2�10 f1+2( f1+10 f1)

�= 2(32 f1) = 64 f1

Ampli

tude

Spec

trum

(f - fc)f1

!1 = !2 = 2, f2 = 10f1B = 2(W + "f) = 2(10f1 + 2(11)f1) = 64f1

-40 -20 0 20 40

0.050.10.150.2

0.250.3

0.35

Example 3.15: Bandlimited Noise PM and FM

• This example will utilize simulation to obtain the spectrum ofan angle modulated carrier

• The message signal in this case will be bandlimited noise hav-ing lowpass bandwidth of W Hz



• In MATLAB we can generate Gaussian amplitude distributedwhite noise using randn() and then filter this noise using ahigh-order lowpass filter (implemented as a digital filter in thiscase)

• We can then use this signal to phase or frequency modulatea carrier in terms of the peak phase deviation, derived fromknowledge of max[|φ(t)|]



3.2.5 Narrowband-to-Wideband Conversion

Narrowband FM Modulator(similar to AM)

x nFreq.

Multiplier

LO

BPF

Frequencytranslate

Narrowband FMCarrier = fc1Peak deviation = fd1Deviation ratio = D1

Wideband FMCarrier = nfc1Peak deviation = nfd1Deviation ratio = nD1

xc(t)m(t)

Ac1cos[!ct + "(t)]Ac2cos[n!ct + n"(t)]

narrowband-to-wideband conversion

• Narrowband FM can be generated using an AM-type modula-tor as discussed earlier (a VCO is not required, so the carriersource can be very stable)

• A frequency multiplier, using say a nonlinearity, can be usedto make the signal wideband FM, i.e.,

Ac1 cos[ωct + φ(t)] n×−→ Ac2 cos[nωct + nφ(t)]

so the modulator deviation constant of fd1 becomes n fd1

3.2.6 Demodulation of Angle-Modulated Signals• To demodulate FM we require a discriminator circuit, which

gives an output which is proportional to the input frequencydeviation

• For an ideal discriminator with input

xr(t) = Ac cos[ωct + φ(t)]



the output is

yD(t) = 12π

K D

dφ(t)

dt

IdealDiscriminatorxc(t) yD(t)

Ideal FM discriminator

• For FMφ(t) = 2π fd

�t

m(α) dα

soyD(t) = K D fdm(t)

slope = KD

InputFrequency

OutputSignal (voltage)

fc

Ideal discriminator I/O characteristic

• For PM signals we follow the discriminator with an integrator

yD(t)xr(t)Ideal

Discrim.

Ideal discriminator with integrator for PM demod



• For PM φ(t) = kpm(t) so

yD(t) = K Dkpm(t)

• We now consider approximating an ideal discriminator with:

EnvelopeDetectorxr(t) yD(t)

e(t)

Ideal discriminator approximation

• If xr(t) = Ac cos[ωct + φ(t)]

e(t) = dxr(t)

dt= −Ac

�ωc + dφ

dt

�sin

�ωct + φ(t)

�

• This looks like AM provided

dφ(t)

dt< ωc

which is only reasonable

• Thus

yD(t) = Ac

dφ(t)

dt= 2π Ac fdm(t) (for FM)

– Relative to an ideal discriminator, the gain constant isK D = 2π Ac

• To eliminate any amplitude variations on Ac pass xc(t) througha bandpass limiter



LimiterBPF Envelope

DetectorBandpass Limiter

xr(t) yD(t)e(t)

FM discriminator with bandpass limiter

• We can approximate the differentiator with a delay and subtractoperation

e(t) = xr(t) − xr(t − τ )

since

limτ→0

e(t)

τ= lim

τ→0

xr(t) − xr(t − τ )

τ= dxr(t)

dt,

thuse(t) � τ

dxr(t)

dt

• In a discrete-time implementation (DSP), we can perform asimilar operation, e.g.

e[n] = x[n] − x[n − 1]

Example 3.16: Complex Baseband Discriminator

• A DSP implementation in MATLAB that works with complexbaseband signals (complex envelope) is the following:

function disdata = discrim(x)% function disdata = discrimf(x)% x is the received signal in complex baseband form%% Mark Wickert



xI=real(x); % xI is the real part of the received signalxQ=imag(x); % xQ is the imaginary part of the received signalN=length(x); % N is the length of xI and xQb=[1 -1]; % filter coefficientsa=[1 0]; % for discrete derivativeder_xI=filter(b,a,xI); % derivative of xI,der_xQ=filter(b,a,xQ); % derivative of xQ% normalize by the squared envelope acts as a limiterdisdata=(xI.*der_xQ-xQ.*der_xI)./(xI.ˆ2+xQ.ˆ2);

• To understand the operation of discrim() start with a gen-eral angle modulated and obtain the complex envelope

xc(t) = Ac cos(ωct + φ(t))

= Re�

Acejφ(t)

ejωct

�

= AcRe�[cos φ(t) + j sin φ(t)]e jωct

�

• The complex envelope is

xc(t) = cos φ(t) + j sin φ(t) = xI (t) + j xQ(t)

where xI and xQ are the in-phase and quadrature signals re-spectively

• A frequency discriminator obtains dφ(t)/dt

• In terms of the I and Q signals,

φ(t) = tan−1�

xQ(t)

xI (t)

�

• The derivative of φ(t) isdφ(t)

dt= 1

1 + (xQ(t)/xI (t))2

d

dt

�xQ(t)

xI (t)

�

=xI (t)x

�Q(t) − x

�I(t)xQ(t)

x2I(t) + x

2Q(t)



• In the DSP implementation xI [n] = xI (nT ) and xQ[n] =xQ(nT ), where T is the sample period

• The derivatives, x�I(t) and x

�Q(t) are approximated by the back-

wards difference xI [n] − xI [n − 1] and xQ[n] − xQ[n − 1]respectively

• To put this code into action, consider a single tone message at1 kHz with β = 2.4048

φ(t) = 2.4048 cos(2π(1000)t)

• The complex baseband (envelope) signal is

xc(t) = ejφ(t) = e

j2.4048 cos(2π(1000)t)

• A MATLAB simulation that utilizes the function Discrim()is:

>> n = 0:5000-1;>> m = cos(2*pi*n*1000/50000); % sampling rate = 50 kHz>> xc = exp(j*2.4048*m);>> y = Discrim(xc);>> % baseband spectrum plotting tool using psd()>> bb_spec_plot(xc,2ˆ11,50);>> axis([-10 10 -30 30])>> grid>> xlabel(’Frequency (kHz)’)>> ylabel(’Spectral Density (dB)’)>> t = n/50;>> plot(t(1:200),y(1:200))>> axis([0 4 -.4 .4])>> grid>> xlabel(’Time (ms)’)>> ylabel(’Amplitude of y(t)’)



!10 !8 !6 !4 !2 0 2 4 6 8 10!30

!20

!10

0

10

20

30

Frequency (kHz)

Spec

tral D

ensit

y (dB

)

0 0.5 1 1.5 2 2.5 3 3.5 4!0.4

!0.3

!0.2

!0.1

0

0.1

0.2

0.3

0.4

Time (ms)

Ampli

tude o

f y(t)

Note: no carrier term since ! = 2.4048

Baseband FM spectrum and demodulator output wavefrom



Analog Circuit Implementations

• A simple analog circuit implementation is an RC highpass fil-ter followed by an envelope detector

Highpass Envelope Detector

ReR CeC

C R

|H(f)|1

0.707

fcf1

2!RC

Linear operating region converts FM to AM

Highpass

RC highpass filter + envelope detector discriminator (slope detector)

• For the RC highpass filter to be practical the cutoff frequencymust be reasonable

• Broadcast FM radio typically uses a 10.7 MHz IF frequency,which means the highpass filter must have cutoff above thisfrequency

• A more practical discriminator is the balanced discriminator,which offers a wider linear operating range



R

R

L1 C1

C2L2

Re

Re

Ce

Ce

yD(t)xc(t)

Bandpass Envelope Detectors

f

f

Filter

Amp

litude

Resp

onse

Filter

Amp

litude

Resp

onse

|H1(f)|

|H1(f)| - |H2(f)|

|H2(f)|

f2

f2

f1

f1

Linear region

Balanced discriminator operation (top) and a passive implementation(bottom)



FM Quadrature Detectors

C1

CpLp

xc(t)

xquad(t)

xout(t)

Tank circuit tuned to fc

Usually a lowpass filter is added here

Quadrature detector schematic

• In analog integrated circuits used for FM radio receivers andthe like, an FM demodulator known as a quadrature detector

or quadrature discriminator, is quite popular

• The input FM signal connects to one port of a multiplier (prod-uct device)

• A quadrature signal is formed by passing the input to a capaci-tor series connected to the other multiplier input and a paralleltank circuit resonant at the input carrier frequency

• The quadrature circuit receives a phase shift from the capacitorand additional phase shift from the tank circuit

• The phase shift produced by the tank circuit is time varying inproportion to the input frequency deviation

• A mathematical model for the circuit begins with the FM inputsignal

xc(t) = Ac[ωct + φ(t)]


3.3. INTERFERENCE

• The quadrature signal is

xquad(t) = K1 Ac sin�ωct + φ(t) + K2

φ(t)

dt

�

where the constants K1 and K2 are determined by circuit pa-rameters

• The multiplier output, assuming a lowpass filter removes thesum terms, is

xout(t) = 12

K1 A2c

sin�

K2dφ(t)

dt

�

• By proper choice of K2 the argument of the sin function issmall, and a small angle approximation yields

xout(t) � 12

K1K2 A2c

dφ(t)

dt= 1

2K1K2 A

2cK Dm(t)

3.3 Interference

Interference is a fact of life in communication systems. A throughunderstanding of interference requires a background in random sig-nals analysis (Chapter 6 of the text), but some basic concepts can beobtained by considering a single interference at fc + fi that lies closeto the carrier fc



3.3.1 Interference in Linear Modulation

ffc - fm fc + fm fc + fifc

Xr(f)Sin

gle-S

ided S

pectr

um12 Am

12 Am

Ac

Ai

AM carrier with single tone interference

• If a single tone carrier falls within the IF passband of the re-ceiver what problems does it cause?

• Coherent Demodulator

xr(t) =�Ac cos ωct + Am cos ωmt cos ωct

�

+ Ai cos(ωc + ωi)t

– We multiply xr(t) by 2 cos ωct and lowpass filter

yD(t) = Am cos ωmt + Ai cos ωi t� �� interference

• Envelope Detection: Here we need to find the received enve-lope relative to the strongest signal present

– Case Ac � Ai

– We will expand xr(t) in complex envelope form by firstnoting that

Ai cos(ωc + ωi)t = Ai cos ωi t cos ωct − Ai sin ωi t sin ωct


3.3. INTERFERENCE

now,

xr(t) = Re��

Ac + Am cos ωmt + Ai cos ωi t

− j Ai sin ωi t�e

jωct�

= Re�

R(t)e jωct�

so

R(t) = |R(t)|=

�(Ac + Am cos ωmt + Ai cos ωi t)

2

+ (Ai sin ωi t)2�1/2

� Ac + Am cos ωmt + Ai cos ωi t

assuming that Ac � Ai

– Finally,

yD(t) � Am cos ωmt + Ai cos ωi t� �� interference

– Case Ai >> Ac

– Now the interfering term looks like the carrier and the re-maining terms look like sidebands, LSSB sidebands rela-tive to fc + fi to be specific

– From SSB envelope detector analysis we expect

yD(t) � 12

Am cos(ωi + ωm)t + Ac cos ωi t

+ 12

Am cos(ωi − ωm)t

and we conclude that the message signal is lost!



3.3.2 Interference in Angle Modulation• Initially assume that the carrier is unmodulated

xr(t) = Ac cos ωct + Ai cos(ωc + ωi)t

• In complex envelope form we have

xr(t) = Re�(Ac + Ai cos ωi t − j Ai sin ωi t)e

jωct�

with R(t) = Ac + Ai cos ωi t − j Ai sin ωi t

• The real envelope or envelope magnitude is, R(t) = |R(t)|,

R(t) =�

(Ac + Ai cos ωi t)2 + (Ai sin ωi t)2

and the envelope phase is

φ(t) = tan−1�

Ai sin ωi t

Ac + Ai cos ωi t

�

• For future reference note that:

tan−1x = x − x

3

3+ x

5

5− x

7

7+ · · · |x |�1� x

• We can thus write that

xr(t) = R(t) cos�ωct + φ(t)

�

• If Ac � Ai

xr(t) � (Ac + Ai cos ωi t)� �� R(t)

cos�ωct + Ai

Ac

sin ωi t

� �� φ(t)

�


3.3. INTERFERENCE

• Case of PM Demodulator: The discriminator recovers dφ(t)/dt ,so the output is followed by an integrator

yD(t) = K D

Ai

Ac

sin ωi t

• Case of FM Demodulator: The discriminator output is used di-rectly to obtain dφ(t)/dt

yD(t) = 12π

K D

Ai

Ac

d

dtsin ωi t = K D

Ai

Ac

fi cos ωi t

• We thus see that the interfering tone appears directly in theoutput for both PM and FM

• For the case of FM the amplitude of the tone is proportional tothe offset frequency fi

• For fi > W , recall W is the bandwidth of the message m(t),a lowpass filter following the discriminator will remove theinterference

• When Ai is similar to Ac and larger, the above analysis nolonger holds

• In complex envelope form

xr(t) = Re��

Ac + Aiejωi t

�e

jωct�

• The phase of the complex envelope is

φ(t) = ��

Ac + Aiejωi t

�= tan−1

�Ai sin ωi t

Ac + Ai cos ωi t

�



• We now consider Ai ≈ Ac and look at plots of φ(t) and thederivative

!1 !0.5 0.5 1

!0.1

!0.05

0.05

0.1

!1 !0.5 0.5 1

!1

!0.5

0.5

1

!1 !0.5 0.5 1

!3

!2

!1

1

2

3

!1 !0.5 0.5 1

!0.6

!0.4

!0.2

0.2

0.4

0.6

!1 !0.5 0.5 1

!50

!40

!30

!20

!10

!1 !0.5 0.5 110

20

30

40

50

60

70

!(t)

!(t)

!(t)

Ai = 0.1Acfi = 1

Ai = 0.9Acfi = 1

Ai = 1.1Acfi = 1

d!(t)/dt

d!(t)/dt

d!(t)/dt

t t

t

t

t

t

Phase deviation and discriminator outputs when Ai ≈ Ac

• We see that clicks (positive or negative spikes) occur in thediscriminator output when the interference levels is near thesignal level

• When Ai � Ac the message signal is entirely lost and thediscriminator is said to be operating below threshold


3.3. INTERFERENCE

• To better see what happens when we approach threshold, applysingle tone FM to the carrier

φ(t) = ��Ace

j Am cos(ωmt) + Aiejωi t

�

• Plot the discriminator output dφ(t)/dt with Am = 5, fm = 1,fi = 3, and various values of Ai

!1 !0.5 0.5 1

!30

!20

!10

10

20

30

!1 !0.5 0.5 1

!30

!20

!10

10

20

30

!1 !0.5 0.5 1

!80

!60

!40

!20

20!1 !0.5 0.5 1

!400

!300

!200

!100

Ai = 0.005,fi = 3Am = ! = 5,fm = 1

Ai = 0.5,fi = 3Am = ! = 5,fm = 1

Ai = 0.9,fi = 3Am = ! = 5,fm = 1

Ai = 0.1,fi = 3Am = ! = 5,fm = 1

t

tt

t

d"(t)/dt d"(t)/dt

d"(t)/dt d"(t)/dt

Discriminator outputs as Ai approaches Ac with single tone FM β = 5

The Use of Preemphasis in FM

• We have seen that when Ai is small compared to Ac the inter-ference level in the case of FM demodulation is proportionalto fi

• The generalization from a single tone interferer to backgroundnoise (text Chapter 6), shows a similar behavior, that is wide



bandwidth noise entering the receiver along with the desiredFM signal creates noise in the discriminator output that hasamplitude proportional with frequency (noise power propor-tional to the square of the frequency)

• In FM radio broadcasting a preemphasis boosts the high fre-quency content of the message signal to overcome the increasednoise background level at higher frequencies, with a deem-phasis filter used at the discriminator output to gain equal-ize/flatten the end-to-end transfer function for the modulationm(t)

FMMod Discrim

C

Rr C

r

HP(f) Hd(f)

ff1 f2

|Hp(f)|

ff1

|Hd(f)|

Wf1f0

Discr

imina

tor O

utput

with

Interf

erenc

e/Nois

e No preemphasis

With preemphasis

Message Bandwidth

FM broadcast preemphasis and deemphasis filtering

• The time constant for these filters is RC = 75 µs ( f1 =1/(2π RC) = 2.1 kHz ), with a high end cutoff of about f2 =30 kHz


3.4. FEEDBACK DEMODULATORS

3.4 Feedback Demodulators

• The discriminator as described earlier first converts and FMsignal to and AM signal and then demodulates the AM

• The phase-locked loop (PLL) offer a direct way to demodulateFM and is considered a basic building block by communicationsystem engineers

3.4.1 Phase-Locked Loops for FM Demodula-tion

• The PLL has many uses and many different configurations,both analog and DSP based

• We will start with a basic configuration for demodulation ofFM

PhaseDetector

VCO

LoopFilter

LoopAmplifier

µxr(t)

xr(t)

ed(t) ev(t)

eo(t)

-eo(t)

Kd

Sinusoidal phase detectorwith invertinginput

Basic PLL block diagram



• Let

xr(t) = Ac cos�ωct + φ(t)

�

eo(t) = Av sin�ωct + θ(t)

�

– Note: Frequency error may be included in φ(t) − θ(t)

• Assume a sinusoidal phase detector with an inverting operationis included, then we can further write

ed(t) = 12

Ac Av Kd sin�φ(t) − θ(t)

�

– In the above we have assumed that the double frequencyterm is removed (e.g., by the loop filter eventually)

• Note that for the voltage controlled oscillator (VCO) we havethe following relationship

VCOKv

ev(t) !o + d"dt

but

dθ(t)

dt= Kvev(t) rad/s

⇒ θ(t) = Kv

�t

ev(α) dα

• In its present form the PLL is a nonlinear feedback controlsystem



f(t)sin( )

µ

!(t)

"(t)

ev(t)

ed(t)+

- Loop filterimpulse response

Loopnonlinearity

#(t)

Nonlinear feedback control model

• To shown tracking we first consider the loop filter to have im-pulse response δ(t) (a straight through connection or unity gainamplifier)

• The loop gain is now defined as

Kt

�= 12µAc Av Kd Kv rad/s

• The VCO output is

θ(t) = Kt

�t

sin[φ(α) − θ(α)] dα

ordθ(t)

dt= Kt sin[φ(t) − θ(t)]

• Let ψ(t) = φ(t)− θ(t) and apply an input frequency step �ω,i.e.,

dφ(t)

dt= �ω u(t)

• Now,

dθ(t)

dt= �ω − dψ(t)

dt= Kt sin ψ(t), t ≥ 0



• We can now plot dψ/dt versus ψ , which is known as a phase

plane plot

Stablelock point

d!(t)/dt

!(t)

"# > 0$

%

"#

"# + Kt

"# - Kt!ss

Phase plane plot (1st-order PLL)

ψ(t)

dt+ Kt sin ψ(t) = �ωu(t)

• At t = 0 the operating point is at B

Since dt is positive ifdψ

dt> 0 −→ dψ is positive

Since dt is positive ifdψ

dt< 0 −→ dψ is negative

therefore the steady-state operating point is at A

• The frequency error is always zero in steady-state

• The steady-state phase error is ψss

– Note that for locking to take place, the phase plane curvemust cross the dψ/dt = 0 axis



• The maximum steady-state value of �ω the loop can handle isthus Kt

• The total lock range is then

ωc − Kt ≤ ω ≤ ωc + Kt ⇒ 2Kt

– For a first-order loop the lock range and the hold-range

are identical

• For a given �ω the value of ψss can be made small by increas-ing the loop gain, i.e.,

ψss = sin−1�

�ω

Kt

�

• Thus for large Kt the in-lock operation of the loop can be mod-eled with a fully linear model since φ(t) − θ(t) is small, i.e.,

sin[φ(t) − θ(t)] � φ(t) − θ(t)

• The s-domain linear PLL model is the following

Kv/s

AcAvKd/2 F(s) µEv(s)!(s)

"(s)

+#

$(s)

Linear PLL model

• Solving for �(s) we have

�(s) = Kt

s

��(s) − �(s)

�F(s)

or �(s)

�1 + Kt

sF(s)

�= Kt

s�(s)F(s))



• Finally, the closed-loop transfer function is

H(s)�= �(s)

�(s)=

Kt

sF(s)

1 + Kt

sF(s)

= Kt F(s)

s + Kt F(s)

First-Order PLL

• Let F(s) = 1, then we have

H(s) = Kt

Kt + s

• Consider the loop response to a frequency step, that is for FM,we assume m(t) = Au(t), then

φ(t) = Ak f

�t

u(α) dα

so �(s) = Ak f

s2

• The VCO phase output is

�(s) = Ak f Kt

s2(Kt + s)

• The VCO control voltage should be closely related to the ap-plied FM message

• To see this write

Ev(s) = s

Kv�(s) = Ak f

Kv· Kt

s(s + Kt)



• Partial fraction expanding yields,

Ev(s) = Ak f

Kv

�1s

− 1s + Kt

�

thusev(t) = Ak f

Kv

�1 − e

−Kt t

�u(t)

t

A m(t)

01st-Order PLL frequency step response at VCO input Kvev(t)/k f

• In general,

�(s) = k f M(s)

s

so

Ev(s) = k f M(s)

s· s

Kv· Kt

s + Kt

= k f

Kv· Kt

s + Kt

· M(s)

• Now if the bandwidth of m(t) is W � Kt , then

Ev(s) ≈ k f

KvM(s) ⇒ ev(t) ≈ k f

Kvm(t)

• The first-order PLL has limited lock range and always has anonzero steady-state phase error when the input frequency isoffset from the quiescent VCO frequency



• Increasing the loop gain appears to help, but the loop band-width becomes large as well, which allows more noise to enterthe loop

• Spurious time constants which are always present, but not aproblem with low loop gains, are also a problem with highgain first-order PLLs

Example 3.17: First-Order PLL Simulation Example

• Tool such as MATLAB, MATLAB with Simulink, VisSim/Comm,and others provide an ideal environment for simulating PLLsat the system level

• Circuit level simulation of PLLs is very challenging due to theneed to simulate every cycle of the VCO

• The most realistic simulation method is to use the actual band-pass signals, but since the carrier frequency must be kept lowto minimize the simulation time, we have difficulties removingthe double frequency term from the phase detector output

• By simulating at baseband, using the nonlinear loop model,many PLL aspects can be modeled without worrying abouthow to remove the double frequency term

– A complex baseband simulation allows further capability,but will not be discussed at his time

• The most challenging aspect of the simulation is dealing withthe integrator found in the VCO block (Kv/s)



• We consider a discrete-time simulation where all continuous-time waveforms are replaced by their discrete-time counter-parts, i.e., x[n] = x(nT ) = x(n/ f s), where fs is the samplefrequency and T = 1/ fs is the sampling period

• The input/output relationship of an integration block can beapproximated via the trapezoidal rule

y[n] = y[n − 1] + T

2�x[n] + x[n − 1]

�

function [theta,ev,phi_error] = PLL1(phi,fs,loop_type,Kv,fn,zeta)% [theta, ev, error, t] = PLL1(phi,fs,loop_type,Kv,fn,zeta)%%% Mark Wickert, April 2007

T = 1/fs;Kv = 2*pi*Kv; % convert Kv in Hz/v to rad/s/v

if loop_type == 1% First-order loop parametersKt = 2*pi*fn; % loop natural frequency in rad/s

elseif loop_type == 2% Second-order loop parametersKt = 4*pi*zeta*fn; % loop natural frequency in rad/sa = pi*fn/zeta;

elseerror(’Loop type must be 1 or 2’);

end

% Initialize integration approximation filtersfilt_in_last = 0; filt_out_last = 0;vco_in_last = 0; vco_out = 0; vco_out_last = 0;

% Initialize working and final output vectorsn = 0:length(phi)-1;theta = zeros(size(phi));ev = zeros(size(phi));phi_error = zeros(size(phi));

% Begin the simulation loop



for k = 1:length(n)phi_error(k) = phi(k) - vco_out;% sinusoidal phase detectorpd_out = sin(phi_error(k));% Loop gaingain_out = Kt/Kv*pd_out; % apply VCO gain at VCO% Loop filterif loop_type == 2

filt_in = a*gain_out;filt_out = filt_out_last + T/2*(filt_in + filt_in_last);filt_in_last = filt_in;filt_out_last = filt_out;filt_out = filt_out + gain_out;

elsefilt_out = gain_out;

end% VCOvco_in = filt_out;vco_out = vco_out_last + T/2*(vco_in + vco_in_last);vco_in_last = vco_in;vco_out_last = vco_out;vco_out = Kv*vco_out; % apply Kv% Measured loop signalsev(k) = vco_in;theta(k) = vco_out;

end

• To simulate a frequency step we input a phase ramp

• Consider an 8 Hz frequency step turning on at 0.5 s and a -12Hz frequency step turning on at 1.5 s

φ(t) = 2π�8(t − 0.5)u(t − 0.5) − 12(t − 1.5)u(t − 1.5)

�

>> t = 0:1/1000:2.5;>> idx1 = find(t>= 0.5);>> idx2 = find(t>= 1.5);>> phi1(idx1) =2*pi* 8*(t(idx1)-0.5).*ones(size(idx1));>> phi2(idx2) = 2*pi*12*(t(idx2)-1.5).*ones(size(idx2));>> phi = phi1 - phi2;>> [theta, ev, phi_error] = PLL1(phi,1000,1,1,10,0.707);>> plot(t,phi_error); % phase error in radians



0 0.5 1 1.5 2 2.5!0.5

0

0.5

1

Time (s)

Phas

e Erro

r,!(t

) ! "(

t), (r

ad)

With Kt = 2#(10) and Kv = 2#(1) rad/s/v, we know that with the 8 Hz step ev(t) = 8, so working backwards, sin(! - ") = 8/10 = 0.8 and ! - " = 0.927 rad.

0.927

-0.412

Phase error for input within lock range

• In the above plot we see the finite rise-time due to the loop gainbeing 2π(10)

• This is a first-order lowpass step response

• The loop stays in lock since the frequency swing either side ofzero is within the ±10 Hz lock range

• Suppose now that a single positive frequency step of 12 Hz isapplied, the loop unlocks and cycle slips indefinitely; why?

>> phi = 12/8*phi1; % scale frequency step from 8 Hz to 12 Hz>> [theta, ev, phi_error] = PLL1(phi,1000,1,1,10,0.707);>> subplot(211)>> plot(t,phi_error)>> subplot(212)>> plot(t,sin(phi_error))



0 0.5 1 1.5 2 2.50

50

100

Time (s)

Phas

e Erro

r (!(t

) ! "(

t))

0 0.5 1 1.5 2 2.5!1

!0.5

0

0.5

1

Time (s)

Phas

e Erro

r sin(

!(t) !

"(t))

Cycle slips

Phase error for input exceeding lock range by 2 Hz

• By plotting the true phase detector output, sin[φ(t)−θ(t)], wesee that the error voltage is simply not large enough to pull theVCO frequency to match the input which is offset by 12 Hz

• In the phase plane plot shown earlier, this scenario correspondsto the trajectory never crossing zero



Second-Order Type II PLL

• To mitigate some of the problems of the first-order PLL, wecan include a second integrator in the open-loop transfer func-tion

• A common loop filter for building a second-order PLL is anintegrator with lead compensation

F(s) = s + a

s

• The resulting PLL is sometimes called a perfect second-order

PLL since two integrators are now in the transfer function

• In text Problem 3.52 you analyze the lead-lag loop filter

F(s) = s + a

s + �

which creates an imperfect, or finite gain integrator, second-order PLL

• Returning to the integrator with phase lead loop filter, the closed-loop transfer function is

H(s) = Kt F(s)

s + Kt F(s)= Kt(s + a)

s2 + Kts + Kta

• The transfer function from the input phase to the phase errorψ(t) is

�(s)

�(s)= �(s) − �(s)

�(s)

1 − H(s) = s2

s2 + Kts + Kta



• In standard second-order system notation we can write the de-nominator of 1 − H(s) as

s2 + Kts + Kta = s

2 + 2ζωns + ω2n

where

ωn =�

Kta = natural frequency in rad/s

ζ = 12

�Kt

a= damping factor

• For an input frequency step the steady-state phase error is zero

– Note the hold-in range is infinite, in theory, since the in-tegrator contained in the loop filter has infinite DC gain

• To verify this we can use the final value theorem

ψss = lims→0

s

��ω

s2 · s2

s2 + Kts + Kta

�

= lims→0

�ωs

s2 + Kts + Kta= 0

• In exact terms we can find ψ(t) by inverse Laplace transform-ing

�(s) = �ω

s2 + 2ζωns + ω2n

• The result for ζ < 1 is

ψ(t) = �ω

ωn

�1 − ζ 2

e−ζωnt sin

�ωn

�1 − ζ 2 t

�u(t)



Example 3.18: Second-Order PLL Simulation Example

• As a simulation example consider a loop designed with fn =10 Hz and ζ = 0.707

Kt = 2ζωn = 2 × 0.707 × 2π × 10 = 88.84

a = ωn

2ζ= 2π × 10

2 × 0.707= 44.43

• The simulation code of Example 3.17 includes the needed loopfilter via a software switch

• The integrator that is part of the loop filter is implemented us-ing the same trapezoidal formula as used in the VCO

• We input a 40 Hz frequency step and observe the VCO controlvoltage (ev(t)) as the loop first slips cycles, gradually pulls in,tracks the input signal offset by 40 Hz

• The VCO gain Kv = 1 v/Hz or 2π rad/s, so ev(t) effectivelycorresponds to the VCO frequency deviation in Hz

>> t = 0:1/1000:2.5;>> idx1 = find(t>= 0.5);>> phi(idx1) = 2*pi*40*(t(idx1)-0.5).*ones(size(idx1));>> [theta, ev, phi_error] = PLL1(phi,1000,2,1,10,0.707);>> plot(t,ev)>> axis([0.4 0.8 -10 50])



0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8!10

0

10

20

30

40

50

Time (s)

VCO

Contr

ol Vo

ltage

e v(t) (K

v = 1

Hz/v)

Cycle slipping, butpulling in to match40 Hz frequency step Cycle slipping stops,

and the loop settles

VCO control voltage for a 40 Hz frequency step

Example 3.19: Bandpass Simulation of FM Demodulation

• Baseband PLL simulations are very useful and easy to imple-ment, but sometimes a full bandpass level simulation is re-quired

• The MATLAB simulation file PLL1.m is modified to allowpassband simulation via the function file PLL2.m

• The phase detector is a multiplier followed by a lowpass filterto remove the double frequency term

function [theta, ev, phi_error] = PLL2(xr,fs,loop_type,Kv,fn,zeta)% [theta, ev, error, t] = PLL2(xr,fs,loop_type,Kv,fn,zeta)%



%% Mark Wickert, April 2007

T = 1/fs;% Set the VCO quiescent frequency in Hzfc = fs/4;% Design a lowpass filter to remove the double freq term[b,a] = butter(5,2*1/8);fstate = zeros(1,5); % LPF state vector

Kv = 2*pi*Kv; % convert Kv in Hz/v to rad/s/v

if loop_type == 1% First-order loop parametersKt = 2*pi*fn; % loop natural frequency in rad/s

elseif loop_type == 2% Second-order loop parametersKt = 4*pi*zeta*fn; % loop natural frequency in rad/sa = pi*fn/zeta;

elseerror(’Loop type musy be 1 or 2’);

end

% Initialize integration approximation filtersfilt_in_last = 0; filt_out_last = 0;vco_in_last = 0; vco_out = 0; vco_out_last = 0;

% Initialize working and final output vectorsn = 0:length(xr)-1;theta = zeros(size(xr));ev = zeros(size(xr));phi_error = zeros(size(xr));

% Begin the simulation loopfor k = 1:length(n)

% Sinusoidal phase detector (simple multiplier)phi_error(k) = 2*xr(k)*vco_out;% LPF to remove double frequency term[phi_error(k),fstate] = filter(b,a,phi_error(k),fstate);pd_out = phi_error(k);% Loop gaingain_out = Kt/Kv*pd_out; % apply VCO gain at VCO% Loop filterif loop_type == 2

filt_in = a*gain_out;filt_out = filt_out_last + T/2*(filt_in + filt_in_last);



filt_in_last = filt_in;filt_out_last = filt_out;filt_out = filt_out + gain_out;

elsefilt_out = gain_out;

end% VCOvco_in = filt_out + fc/(Kv/(2*pi)); % bias to quiescent freq.vco_out = vco_out_last + T/2*(vco_in + vco_in_last);vco_in_last = vco_in;vco_out_last = vco_out;vco_out = Kv*vco_out; % apply Kv;vco_out = sin(vco_out); % sin() for bandpass signal% Measured loop signalsev(k) = filt_out;theta(k) = vco_out;

end

• Note that the carrier frequency is fixed at fs/4 and the lowpassfilter cutoff frequency is fixed at fs/8

• The double frequency components out of the phase detectorare removed with a fifth-order Butterworth lowpass filter

• The VCO is modified to include a bias that shifts the quiescentfrequency to fc = fs/4

• The VCO output is not simply a phase deviation, but rather asinusoid with argument the VCO output phase

• We will test the PLL using a single tone FM signal

>> t = 0:1/4000:5;>> xr = cos(2*pi*1000*t+2*sin(2*pi*10*t));>> psd(xr,2ˆ14,4000)>> axis([900 1100 -40 30])>> % Process signal through PLL>> [theta, ev, phi_error] = PLL2(xr,4000,1,1,50,0.707);>> plot(t,ev)>> axis([0 1 -25 25])



900 920 940 960 980 1000 1020 1040 1060 1080 1100!40

!30

!20

!10

0

10

20

30

Frequency (Hz)

Powe

r Spe

ctrum

of x r(t)

(dB)

Single tone FM input spectrum having fm = 10 Hz and � f = 20 Hz

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1!25

!20

!15

!10

!5

0

5

10

15

20

25

Time (s)

VCO

Contr

ol Vo

ltage

e v(t) (K

v = 1

Hz/v)

Recovered modulation at VCO input, ev(t)



3.4.2 PLL Frequency Synthesizers• A frequency synthesizer is used to generate a stable, yet pro-

grammable frequency source

• A frequency synthesizer is often used to allow digital tuning ofthe local oscillator in a communications receiver

• One common frequency synthesis type is known as indirect

synthesis

• With indirect synthesis a PLL is used to create a stable fre-quency source

• The basic block diagram of an indirect frequency synthesizeris the following

PhaseDetector

LoopFilter VCO1

M

1N

freffout

frefM

foutN

Freq Div

Freq DivIndirect frequency synthesis using a PLL

• When locked the frequency error is zero, thus

fout = N

M× fref



Example 3.20: A PLL Synthesizer for Broadcast FM

• In this example the synthesizer will provide the local oscillatorsignal for frequency converting the FM broadcast band from88.1 to 107.9 MHz down to an IF of 10.7 MHz

• The minimum channel spacing should be 200 kHz

• We will choose high-side tuning for the LO, thus

88.1 + 10.7 ≤ fLO ≤ 107.9 + 10.7 MHz98.8 ≤ fLO ≤ 118.6 MHz

– The step size must be 200 kHz so the frequency must beno larger than 200 kHz

• To reduce the maximum frequency into the divide by countera frequency offset scheme will be employed

• The synthesizer with offset oscillator is the following

PhaseDetector

LoopFilter VCO1

M

1N

freffout

frefM

fmixN

Freq Div

Freq Divfoffset

fmix

DifferenceFrequency

= 200 kHz

FM broadcast band synthesizer producing fLO for fIF = 10.7 MHz



• Choose foffset < fout then fmix = fout − foffset, and for locking

fref

M= fmix

N⇒ fout = N fref

M+ foffset

– Note that Fmix = N fref/M and fout = fmix + foffset, byvirtue of the low side tuning assumption for the offset os-cillator

• Let fref/M = 200 kHz and foffset = 98.0 MHz, then

Nmax = 118.6 − 98.00.2

= 103

andNmin = 98.8 − 98.0

0.2= 4

• To program the LO such that the receiver tunes all FM stationsstep N from 4, 5, 6, . . . , 102, 103



Example 3.21: Simple PLL Frequency Multiplication

• A scheme for multiplication by three is shown below:

PhaseDetector

Loop Filt.& Ampl.

VCOCentered at 3fc

xVCO = Acos[2!(3fc)t]

Input at fc

t

f

Input Spectrum

fc 3fc0

Hard limit sinusoidal input if needed

PLL as a Frequency Multiplier



Example 3.22: Simple PLL Frequency Division

• A scheme for divide by two is shown below:

PhaseDetector

LowpassFilter

Loop Filt.& Ampl.

VCOCentered at fc/2

xLPF = Acos[2!(fc/2)t]

Keep theFundamental

Input at fc

t

VCO Output

VCO Output Spectrum

ffc/2

0

0 2T0

PLL as a Frequency Divider

3.4.3 Frequency-Compressive Feedback

BPF Discrim

VCO

xr(t) ev(t)Demod.Outputeo(t)

ed(t) x(t)

Frequency compressive feedback PLL



• If we place a discriminator inside the PLL loop a compressingaction occurs

• Assume thatxr(t) = Ac cos[ωct + φ(t)]

and

ev(t) = Av sin�(ωc − ωo)t + Kv

�t

ev(α) dα

�

• Then,

ed(t) = 12

Ac Av

� blocked by BPF� �� sin

�(2ωc − ωo)t + other terms

�

− sin[ωot + φ(t) − Kv

�t

ev(α) dα]� ��

passed by BPF

�

so

x(t) = −12

Ac Av sin�ωot + φ(t) − Kv

�t

ev(α) dα�

• Assuming an ideal discriminator

ev(t) = 12π

K D

�dφ(t)

dt− Kvev(t)

�

or

ev(t)

�1 + Kv K D

2π

�= K D

2π· dφ(t)

dt



• For FM dφ(t)/dt = 2π fdm(t), so

ev(t) = K D fd

1 + Kv K D/(2π)m(t)

which is the original modulation scaled by a constant

• The discriminator input must be

x(t) = −12

Ac Ad sin�ωot + 1

1 + Kv K D/(2π)φ(t)

�

• Assuming that Kv K D/(2π) � 1 we conclude that the discrim-inator input has been converted to a narrowband FM signal,which is justifies the name ’frequency compressive feedback’

3.4.4 Coherent Carrier Recovery for DSB De-modulation

• Recall that a DSB signal is of the form

xr(t) = m(t) cos ωct

• A PLL can be used to obtain a coherent carrier reference di-rectly from xr(t)

• Here we will consider the squaring loop and the Costas loop



( )2 LoopFilter VCO

x 2

LPF

-90o

0o

Am2(t)cos(2!ct + 2")xr(t)

m(t)cos(#)

Bsin(2!ct + 2$)

cos(!ct + $)

m(t)cos(!ct + ")

static phase error

Squaring Loop

LoopFilterVCO

LPF

LPF

0o

-90o

xr(t)sin(!ct + ")cos(!ct + ")

ksin(2#)m(t)sin(#)

m(t)cos(#)

m2(t)sin2#12

Costas Loop

• Note: For both of the above loops m2(t) must contain a DC

component

• The Costas loop or a variation of it, is often used for carrierrecovery in digital modulation

• Binary phase-shift keying (BPSK), for example, can be viewedas DSB where

m(t) =∞�

n=−∞dn p(t − nT )



where dn = ±1 represents random data bits and p(t) is a pulseshaping function, say

p(t) =�

1, 0 ≤ t ≤ T

0, otherwise

• Note that in this case m2(t) = 1, so there is a strong DC value

present

2 4 6 8 10

!1!0.5

0.51

2 4 6 8 10

!1

!0.5

0.5

1

t/T

t/T

m(t)

m(t)cos(!ct)

BPSK modulation

• Digital signal processing techniques are particularly useful forbuilding PLLs

• In the discrete-time domain, digital communication waveformsare usually processed at complex baseband following someform of I-Q demodulation



LPF A/D

A/DLPF

0o

-90o

xIF(t) cos[2!fcLt + "L]fs

fs

Sampling clock

Discrete-Time

r[n] = rI[n] + jrQ[n]

rI(t)

rQ(t)

IF to discrete-time complex baseband conversion

To Symbol Synch

[ ]y n

[ ]x n

FromMatched

Filter

NCO

LUT !1z

!1z

[ ]v n Error Generation

!2 1() 2M M Im()

pk

akLoop Filter

[ ]e ne! j"[n]

"[n]

M th-power digital PLL (DPLL) carrier phase tracking loop

![n]

"[n]

FromMatched

FilterTo Symbol Synch

( )je

( )je

M

Rect.to

Polar

[ ]x n [ ]y n

1 arg()M

L-TapMA FIR

L-TapDelay

()F #[n]

Non-Data Aided (NDA) feedforward carrier phase tracking



3.5 Sampling Theory• We now return to text Chapter 2, Section 8, for an introduc-

tion/review of sampling theory

• Consider the representation of continuous-time signal x(t) bythe sampled waveform

xδ(t) = x(t)

� ∞�

n=−∞δ(t − nTs)

�

=∞�

n=−∞x(nTs)δ(t − nTs)

x(t) x!(t)

tt0 0 Ts-Ts 2Ts 3Ts 4Ts 5Ts

Sampling

• How is Ts selected so that x(t) can be recovered from xδ(t)?

• Uniform Sampling Theorem for Lowpass Signals

GivenF{x(t)} = X ( f ) = 0, for f > W

then choose

Ts <1

2Wor fs > 2W ( fs = 1/Ts)

to reconstruct x(t) from xδ(t) and pass xδ(t) through an idealLPF with cutoff frequency W < B < fs − W

2W = Nyquist frequencyfs/2 = folding frequency



proof:

Xδ( f ) = X ( f ) ∗�

fs

∞�

n=−∞δ( f − n fs)

�

but X ( f ) ∗ δ( f − n fs) = X ( f − n fs), so

Xδ( f ) = fs

∞�

n=−∞X ( f − n fs)

-W W

-W W

0

0

f

f

X(f)X0

X!(f)

-fs fs

X0 fs

fs-W

Guard band= fs - 2W

Lowpass reconstruction filter. . . . . .

f-2fs fs-fs 2fs

X0 fs. . . . . .Aliasing

fs < 2W

fs > 2W

Spectra before and after sampling at rate fs

• As long as fs − W > W or fs > 2W there is no aliasing

(spectral overlap)



• To recover x(t) from xδ(t) all we need to do is lowpass filterthe sampled signal with an ideal lowpass filter having cutofffrequency W < fcutoff < fs − W

• In simple terms we set the lowpass bandwidth to the foldingfrequency, fs/2

• Suppose the reconstruction filter is of the form

H( f ) = H0�

�f

2B

�e

− j2π f t0

we then choose W < B < fs − W

• For input Xδ( f ), the output spectrum is

Y ( f ) = fs H0 X ( f )e− j2π f t0

and in the time domain

y(t) = fs H0x(t − t0)

• If the reconstruction filter is not ideal we then have to designthe filter in such a way that minimal desired signal energy is re-moved, yet also minimizing the contributions from the spectraltranslates either side of the n = 0 translate

• The reconstruction operation can also be viewed as interpolat-ing signal values between the available sample values

• Suppose that the reconstruction filter has impulse response h(t),



then

y(t) =∞�

n=−∞x(nTs)h(t − nTs)

= 2B H0

∞�

n=−∞x(nTs)sinc[2B(t − t0 − nTs)]

where in the last lines we invoked the ideal filter describedearlier

• Uniform Sampling Theorem for Bandpass Signals

If x(t) has a single-sided bandwidth of W Hz and

F{x(t)} = 0 for f > fu

then we may choose

fs = 2 fu

m

wherem =

�fu

W

�,

which is the greatest integer less than or equal to fu/W

Example 3.23: Bandpass signal sampling

!4 !2 2 4

0.20.40.60.81

f

X(f)

Input signal spectrum



• In the above signal spectrum we see that

W = 2, fu = 4 fu/W = 2 ⇒ m = 2

sofs = 2(4)

2= 4

will work

• The sampled signal spectrum is

Xδ( f ) = 4∞�

n=−∞X ( f − n fs)

!15 !10 !5 5 10 15

1234

f

X!(f)Recover withbandpass filter

fs 2fs 3fs-fs-2fs-3fsSpectrum after sampling


3.6. ANALOG PULSE MODULATION

3.6 Analog Pulse Modulation• The message signal m(t) is sampled at rate fs = 1/Ts

• A characteristic of the transmitted pulse is made to vary in aone-to-one correspondence with samples of the message signal

• A digital variation is to allow the pulse attribute to take onvalues from a finite set of allowable values

3.6.1 Pulse-Amplitude Modulation (PAM)• PAM produces a sequence of flat-topped pulses whose ampli-

tude varies in proportion to samples of the message signal

• Start with a message signal, m(t), that has been uniformly sam-pled

mδ(t) =∞�

n=−∞m(nTs)δ(t − nTs)

• The PAM signal is

mc(t) =∞�

n=−∞m(nTs)�

�t − (nTs + τ/2)

τ

�

m(t)

mc(t)!

0 ! Ts 2Ts 3Ts 4Tst

PAM waveform



• It is possible to create mc(t) directly from mδ(t) using a zero-

order hold filter, which has impulse response

h(t) = �

�t − τ/2

τ

�

and frequency response

H( f ) = τ sinc( f τ )e− jπ f τ

h(t)m!(t) mc(t)

• How does h(t) change the recovery operation from the case ofideal sampling?

– If τ � Ts we can get by with just a lowpass reconstruc-tion filter having cutoff frequency at fs/2 = 2/Ts

– In general, there may be a need for equalization if tau ison the order of Ts/4 to Ts/2

f-W W fs-fs

sinc() function envelope

Lowpass reconstruction filter

Lowpassmc(t) m(t)

Recovery of m(t) from mc(t)


3.6. ANALOG PULSE MODULATION

3.6.2 Pulse-Width Modulation (PWM)• A PWM waveform consists of pulses with width proportional

to the sampled analog waveform

• For bipolar m(t) signals we may choose a pulse width of Ts/2to correspond to m(t) = 0

• The biggest application for PWM is in motor control

• It is also used in class D audio power amplifiers

• A lowpass filter applied to a PWM waveform recovers themodulation m(t)

!20 !10 10 20

!1

!0.5

0.5

1PWM Signal

Analog input m(t)

t

Example PWM signal

3.6.3 Pulse-Position Modulation• With PPM the displacement in time of each pulse, with re-

spect to a reference time, is proportional to the sampled analogwaveform

• The time axis may be slotted into a discrete number of pulsepositions, then m(t) would be quantized

• Digital modulation that employs M slots, using nonoverlap-ping pulses, is a form of M-ary orthogonal communications



– PPM of this type is finding application in ultra-wideband

communications

!20 !10 10 20

!1

!0.5

0.5

1PPMSignal

Analog input m(t)

t

Example PPM signal

3.7 Delta Modulation and PCM

• This section considers two pure digital pulse modulation schemes

• Pure digital means that the output of the modulator is a binarywaveform taking on only discrete values

3.7.1 Delta Modulation (DM)

• The message signal m(t) is encoded into a binary sequencewhich corresponds to changes in m(t) relative to referencewaveform ms(t)

• DM gets its name from the fact that only the difference fromsample-to-sample is encoded

• The sampling rate in combination with the step size are the twoprimary controlling modulator design parameters


3.7. DELTA MODULATION AND PCM

1

-1+

!

m(t)

ms(t) =

xc(t)"(t)

Pulse Modulatord(t)

#0

Control the step size

Delta modulator with step size parameter δ0

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05!1

!0.50

0.51

Time (s)

m(t)

and m

s(t)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05!1

!0.50

0.51

Time (s)

x c(t)

m(t) (blue)ms(t) (red)!0 = 0.15Slope

overload

Start-up transient

Delta modulator waveforms

• The maximum slope that can be followed is δ0/Ts



• A MATLAB DM simulation function is given below

function [t_o,x,ms] = DeltaMod(m,fs,delta_0,L)% [t,x,ms] = DeltaMod(m,fs,delta_0,L)%% Mark Wickert, April 2006

n = 0:(L*length(m))-1;t_o = n/(L*fs);ms = zeros(size(m));x = zeros(size(m));ms_old = 0; % zero initial conditionfor k=1:length(m)

x(k) = sign(m(k) - ms_old);ms(k) = ms_old + x(k)*delta_0;ms_old = ms(k);

end

x = [x; zeros(L-1,length(m))];x = reshape(x,1,L*length(m));

• The message m(t) can be recovered from xc(t) by integratingand then lowpass filter to remove the stair step edges (lowpassfiltering directly is a simplification)

• Slope overload can be dealt with through an adaptive scheme

– If m(t) is nearly constant keep the step size δ0 small

– If m(t) has large variations, a larger step size is needed

• With adaptive DM the step size is controlled via a variable gainamplifier, where the gain is controlled by square-law detectingthe output of a lowpass filter acting on xc(t)



1

-1+

!

m(t)

ms(t)

xc(t)"(t)

Pulse Modulatord(t)

LPF( )2

VGA

Means to obtain a variable step size DM

3.7.2 Pulse-Code Modulation (PCM)

• Each sample of m(t) is mapped to a binary word by

1. Sampling

2. Quantizing

3. Encoding

Sampler Quantizer Encoder

Sample&

HoldAnalog to

DigitalConverter

Parallelto Serial

Converter

PCM Output

Serial Data

m(t)

m(t)

EquivalentViews

n



EncodedOutput

111110101100011010001000

Quant.Level

76543210 t

0 Ts 2Ts 3Ts 4Ts 5Ts 6Ts 7Ts

Quantizer Bits: n = 3, q = 2n = 8

Encoded Serial PCM Data: 001 100 110 111 110 100 010 010 ...

m(t)

3-Bit PCM encoding

• Assume that m(t) has bandwidth W Hz, then

– Choose fs > 2W

– Choose n bits per sample (q = 2n quantization levels)– ⇒≥ 2nW binary digits per second must be transmitted

• Each pulse has width no more than

(�τ )max = 12nW

,

so using the fact that the lowpass bandwidth of a single pulseis about 1/(2�τ ) Hz, we have that the lowpass transmissionbandwidth for PCM is approximately

B � kW n

where k is a proportionality constant

• When located on a carrier the required bandwidth is doubled



• Binary phase-shift keying (BPSK), mentioned earlier, is a pop-ular scheme for transmitting PCM using an RF carrier

• Many other digital modulation schemes are possible

• The number of quantization levels, q = log2 n, controls thequantization error, assuming m(t) lies within the full-scale rangeof the quantizer

• Increasing q reduces the quantization error, but also increasesthe transmission bandwidth

• The error between m(kTs) and the quantized value Q[m(kTs)],denoted e(n), is the quantization error

• If n = 16, for example, the ratio of signal power in the samplesof m(t), to noise power in e(n), is about 95 dB (assuming m(t)stays within the quantizer dynamic range)

Example 3.24: Compact Disk Digital Audio

• CD audio quality audio is obtained by sampling a stereo sourceat 44.1 kHz

• PCM digitizing produces 16 bits per sample per L/R audiochannel



Synch(27 bits)

SubCode Data (96 bits)Data (96 bits) Parity

(32 bits)Parity

(32 bits)

0.163 mm

One Frame of 12 Audio Samples

(8 bits)

CD recoding frame format

• The source bit rate is thus 2 × 16 × 44.1ksps = 1.4112 Msps

• Data framing and error protection bits are added to bring thetotal bit count per frame to 588 bits and a serial bit rate of4.3218 Mbps

3.8 Multiplexing• It is quite common to have multiple information sources lo-

cated at the same point within a communication system

• To simultaneously transmit these signals we need to use someform of multiplexing


3.8. MULTIPLEXING

• There is more than one form of multiplexing available to thecommunications engineer

3.8.1 Frequency-Division Multiplexing (FDM)

• With FDM the idea is to locate a group of messages on dif-ferent subcarriers and then sum then together to form a newbaseband signal which can then be modulated onto the carrier

Mod#1

RFMod

fsc1

fc

Mod#2

fsc2

Mod#N

fscN

. . .

m1(t)

m2(t)

mN(t)

xc(t)

Lower bound on the composite signal band-width

Compositebaseband

FDM transmitter

• At the receiver we first demodulate the composite signal, thenseparate into subcarrier channels using bandpass filters, thendemodulate the messages from each subcarrier



BPFfsc1

BPFfsc2

BPFfscN

RFDemod

Sub Car.Demod #1

Sub Car.Demod #2

Sub Car.Demod #N

yD1(t)

yD2(t)

yDN(t)

... ... ...

FDM receiver/demodulator

• The best spectral efficiency is obtained with SSB subcarriermodulation and no guard bands

• At one time this was the dominant means of routing calls in thepublic switched telephone network (PSTN)

• In some applications the subcarrier modulation may be combi-nations both analog and digital schemes

• The analog schemes may be combinations of amplitude mod-ulation (AM/DSM/SSB) and angle modulation (FM/PM)


3.8. MULTIPLEXING

Example 3.25: FM Stereo

x2Freq. Mult

19 kHzPilot

FMMod

l(t)

r(t)

l(t) + r(t)

l(t) - r(t)

+

++

+

+

!

+

38 kHz 19 kHzpilot

xb(t) xc(t)

fc

f (kHz)19150 3823 53

Xb(f)Other subcarrier services can occupy this region

PilotCarrier

FM stereo transmitter

FMDiscrim

LPFW = 15

kHzLPFW = 15

kHzBPF fc = 19

kHz

x 2 FreqMult

Mono output

l(t) + r(t)

l(t) - r(t)

l(t)

r(t)!

xb(t)xr(t)

Coherent demod of DSB on 38 kHz subcarrier

FM stereo receiver



3.8.2 Quadrature Multiplexing (QM)

LPF

LPFChannel

m1(t)

m2(t)

d1(t)yD1(t)

yD2(t)d2(t)

2sin!ct

2cos!ctAccos!ct

Acsin!ct

xc(t) xr(t)

QM modulation and demodulation

• With QM quadrature (sin/cos) carrier are used to send inde-pendent message sources

• The transmitted signal is

xc(t) = Ac

�m1(t) cos ωct + m2(t) sin ωct

�

• If we assume an imperfect reference at the receiver, i.e., 2 cos(ωct+θ), we have

d1(t) = Ac

�m1(t) cos θ − m2(t) sin θ

+ m1(t) cos(2ωct + θ) + m2(t) sin(2ωct + θ)� �� LPF removes these terms

�

yD1(t) = Ac

�m1(t) cos θ + m2(t) sin θ

�

• The second term in yD1(t) is termed crosstalk, and is due tothe static phase error θ


3.8. MULTIPLEXING

• Similarly

yD2(t) = Ac

�m2(t) cos θ − m1(t) sin θ

�

• Note that QM acheives a bandwidth efficiency similar to thatof SSB using adjacent two subcarriers or USSB and LSSB to-gether on the same subcarrier

3.8.3 Time-Division Multiplexing (TDM)

• Time division multiplexing can be applied to sampled analogsignals directly or accomplished at the bit level

• We assume that all sources are sample at or above the Nyquistrate

• Both schemes are similar in that the bandwidth or data rate ofthe sources being combined needs to be taken into account toproperly maintain real-time information flow from the sourceto user

• For message sources with harmonically related bandwidths wecan interleave samples such that the wideband sources are sam-pled more often

• To begin with consider equal bandwidth sources



Info. Source 1

Info. Source 2

Info. Source N

...

Channel

Info.User 1

Info. User 2

Info. User N

...

Commutators

For equal bandwidth: s1s2s3 s1s2s3 s1s2s3 s1s2s3 s1s2s3 s1s2s3 s1s2s3 ....

SynchronizationRequired

Analog TDM (equal bandwidth sources)

• Suppose that m1(t) has bandwidth 3W and sources m2(t), m3(t),and m4(t) each have bandwidth W , we could send the samplesas

s1s2s1s3s1s4s1s2s1 . . .

with the commutator rate being fs > 2W Hz

• The equivalent transmission bandwidth for multiplexed signalscan be obtained as follows

– Each channel requires greater than 2Wi samples/s

– The total number of samples, ns, over N channels in T sis thus

ns =N�

i=1

2Wi T

– An equivalent signal channel of bandwidth B would pro-duce 2BT = ns samples in T s, thus the equivalent base-


3.8. MULTIPLEXING

band signal bandwidth is

B =N�

i=1

Wi Hz

which is the same minimum bandwidth required for FDMusing SSB

– Pure digital multiplexing behaves similarly to analog mul-tiplexing, except now the number of bits per sample, whichtakes into account the sample precision, must be included

– The earlier PCM example for CD audio this was takeninto account when we said that left and right audio chan-nels each sampled at 44.1 ksps with 16-bit quantizers,multiplex up to

2 × 16 × 44, 100 = 1.4112 Msps



Example 3.26: Digital Telephone System

• The North American digital TDM hierarchy is based on singlevoice signal sampled at 8000 samples per second using an 7-bitquantizer plus one signaling bit

• The serial bit-rate per voice channel is 64 kbps

North American Digital TDM Hierarchy)

Digital No. of 64 kbpsSignal Bit Rate PCM VF Transmission

Sys. Number R (Mb/s) Channels Media UsedDS-0 0.064 1 Wire pairs

T1 DS-1 1.544 24 Wire pairsT1C DS-1C 3.152 48 Wire pairsT2 DS-2 6.312 96 Wire pairsT3 DS-3 44.736 672 Coax, radio, fiber

DS-3C 90.254 1344 Radio, fiberDS-4E 139.264 2016 Radio, fiber, coax

T4 DS-4 274.176 4032 Coax, fiberDS-432 432.00 6048 Fiber

T5 DS-5 560.160 8064 Coax, fiber

• Consider the T1 channel which contains 24 voice signals

• Eight total bits are sent per voice channel at a sampling rate of8000 Hz

• The 24 channels are multiplexed into a T1 frame with an extrabit for frame synchronization, thus there are 24 × 8 + 1 = 193bits per frame


3.9. GENERAL PERFORMANCE OF MODULATION SYSTEMS IN NOISE

• Frame period is 1/8000 = 0.125 ms, so the serial bit rate is193 × 8000 = 1.544 Mbps

• Four T1 channels are multiplexed into a T2 channel (96 voicechannels)

• Seven T2 channels are multiplexed into a T3 channel (672voice channels)

• Six T3 channels are multiplexed into a T4 channel (4032 voicechannels)

3.9 General Performance of ModulationSystems in Noise

• Regardless of the modulation scheme, the received signal xr(t),is generally perturbed by additive noise of some sort, i.e.,

xr(t) = xc(t) + n(t)

where n(t) is a noise process

• The pre- and post-detection signal-to-noise ratio (SNR) is useda figure or merit



Pre-Det.Filter

Demod/Detectorxr(t) yD(t)

Common to all systems

(SNR)T = PT

<n2(t)> (SNR)D = <m2(t)><n2(t)>

DSB, SSB, QDSB

Coherent Demod

(SNR)T

(SNR)D

FM, D = 10

FM, D = 5

FM, D = 2PCMq = 256PCMq = 64

Nonlinear modu-lation systems have a distinct threshold in noise

General modulation performance in noise


Appendix APhysical Noise Sources

ContentsA.1 Physical Noise Sources . . . . . . . . . . . . . . . . . . A-3

A.1.1 Thermal Noise . . . . . . . . . . . . . . . . . . A-4A.1.2 Nyquist’s Formula . . . . . . . . . . . . . . . . A-6A.1.3 Shot Noise . . . . . . . . . . . . . . . . . . . . A-11A.1.4 Other Noise Sources . . . . . . . . . . . . . . . A-12A.1.5 Available Power . . . . . . . . . . . . . . . . . A-13A.1.6 Frequency Dependence . . . . . . . . . . . . . . A-15A.1.7 Quantum Noise . . . . . . . . . . . . . . . . . . A-15

A.2 Characterization of Noise in Systems . . . . . . . . . A-16A.2.1 Noise Figure of a System . . . . . . . . . . . . . A-16A.2.2 Measurement of Noise Figure . . . . . . . . . . A-18A.2.3 Noise Temperature . . . . . . . . . . . . . . . . A-20A.2.4 Effective Noise Temperature . . . . . . . . . . . A-21A.2.5 Cascade of Subsystems . . . . . . . . . . . . . . A-22A.2.6 Attenuator Noise Temperature and Noise Figure A-23

A.3 Free-Space Propagation Channel . . . . . . . . . . . . A-28

A-1

APPENDIX A. PHYSICAL NOISE SOURCES

.

A-2 ECE 5625 Communication Systems I

A.1. PHYSICAL NOISE SOURCES

A.1 Physical Noise Sources

• In communication systems noise can come from both internaland external sources

• Internal noise sources include

– Active electronic devices such as amplifiers and oscilla-tors

– Passive circuitry

• Internal noise is primarily due to the random motion of chargecarriers within devices and circuits

• The focus of this chapter is modeling and analysis associatedwith internal noise sources

• External sources include

– Atmospheric, solar, and cosmic noise

– Man-made sources such as intentional or unintentionaljamming

• To analyze system performance due to external noise locationcan be very important

• Understanding the impact on system performance will requireon-site measurements

ECE 5625 Communication Systems I A-3


A.1.1 Thermal Noise

• Thermal noise is due to the random motion of charge carriers

• Nyquist’s Theorem: States that the noise voltage across a re-sistor is

v2rms = �v2

n(t)� = 4kT RB v2

where

k = Boltzmanns constant = 1.38 × 10−23 J/KT = Temperature in KelvinR = resistance in ohmsB = measurement bandwidth

R

G = 1/R

vrms = (4KTRB)1/2

irms = (4KTGB)1/2

noiseless

noiseless

Equivalent noise circuits: voltage and current



• Consider the following resistor network

R1

R1

R2

R2

R3

R3

vo

vo

v12

v22

v32

vi = 4KTRiB2

i = 1, 2, 3

Noise analysis for a resistor network

• Since the noise sources are independent, the total noise volt-age, vo can be found by summing the square of the voltage dueto each noise source (powers due to independent sources add)

v2o

= v2o1 + v2

o2 + v2o3

• The noise voltages, vo1, vo2, vo3, can be found using superpo-sition



A.1.2 Nyquist’s Formula

R, L, CNetwork vrms

Z(f)Nyquist’s formula for passive networks

• Consider a one-port R, L , C network with input impedance inthe frequency domain given by Z( f )

• Nyquist’s theorem states that

v2rms = �v2

n� = 2kT

� ∞

−∞R( f ) d f

whereR( f ) = Re

�Z( f )

�

• For a pure resistor network Nyquist’s formula reduces to

�v2n� = 2kT

�B

−B

Req d f = 4kT ReqB

• In the previous example involving three resistors

Req = R3||(R1 + R2)

Example A.1: Circuit simulation for noise characterization

• Spice and Spice-like circuit simulators, e.g. Qucs, have theability to perform noise analysis on circuit models



• The analysis is included as part of an AC simulation (in Qucsfor example it is turned off by default)

• When passive components are involved the analysis followsfrom Nyquist’s formula

• The voltage that AC noise analysis returns is of the form

vrms√Hz

=�

4kT R( f )

where the B value has been moved to the left side, making thenoise voltage a spectral density like quantity

• When active components are involved more modeling infor-mation is required

• Consider the following resistor circuit

Req

measure vrms here

T = 300oK

Pure resistor circuit



• To apply Nyquist’s formula we need Req

Req = 100K||(10 + 5 + 20)K

= 100 · 35100 + 35

K = 25.93 K

• In Nyquists formula the rms noise voltage normalized by B isvrms√

Hz=

�4kT × 25.93 × 103

= 2.072 × 10−8 v/√

Hz

assuming T = 300◦ K

• Circuit simulation results are shown below

Resistor circuit RMS noise voltage (vrms/√

Hz)

• Circuit simulation becomes particularly useful when reactiveelements are included



• To demonstrate this we modify the resistor circuit by placing a10 nf capacitor in shunt with the 5 K resistor

vrms

T = 300oK

vrms/√

Hz for a simple passive RC circuit

• The input impedance of this circuit is

Z(s) =

�R4· 1

C1s

R4+ 1C1s

+ R2 + R5

�R3

R4· 1C1s

R4+ 1C1s

+ R2 + R5 + R3

• Here the noise voltage/vrms/√

Hz takes on two limiting valuesdepending upon whether the capacitor acts as an open circuitor a short circuit

• To get the actual rms noise voltage as measured by an AC volt-meter, we need to integrate the vrms/

√Hz quantity, which can

be accomplished with a true rms measuring instrument

v2rms = 4kT

� ∞

0Re{Z( f )} d f



Example A.2: Active circuit modeling

• For Op-Amp based circuits noise model information is usuallyavailable from the data sheet1

• Circuit simulators include noise voltage and current sourcesjust for this purpose

en

inninp

NoiselessOp Amp+

!

Op Amp Noise Model

741 Noise Data

Op amp noise model with 741 data sheet noise information

• Consider an inverting amplifier with a gain of 10 using a 741op-amp

• This classic op amp, has about a 1MHz gain-bandwidth prod-uct, so with a gain of 10, the 3 dB cutoff frequency of theamplifier is at about 100 kHz

• The noise roll-off is at the same frequency1Ron Mancini, editor, Op Amps for Everyone: Design Reference, Texas Instruments Advanced

Analog Products, Literature number SLOD006, September 2000.



• The rms noise as v/√

Hz plotted below, is a function of theop amp noise model and the resistors used to configure theamplifier gain

• With relatively low impedance configured at the inputs to theop amp, the noise voltage en dominates, allowing the noisecurrents to be neglected

741

Gain = 10 so fc is at about 100 kHz

vrms

T = 300oK

Noise voltage at the op amp output

A.1.3 Shot Noise

• Due to the discrete nature of current flow in electronic devices

• Given an average current flow of Id A,

i2rms = �i2

n(t)� = 2eId B A2

where e = 1.6 × 10−19 is the charge on an electron



• Special Case: For a PN junction diode

I = Is

�exp

�eV

kT

�− 1

�A

where Is is the reverse saturation current

• Assuming Is and Is exp(eV/kT ) to be independent sources interms of noise sources, then

i2rms,tot =

�2eIs exp

�eV

kT

�+ 2eIs

�B

= 2e�I + Is

�B A2

• For I � Is the diode differential conductance is

go = d I

dV= eI

kT,

thusirms,tot � 2eI B = 2kT go B

which is half the noise due to a pure resistance

A.1.4 Other Noise Sources

• Generation-Recombination Noise: Results from generated freecarriers recombining in a semiconductor (like shot noise)

• Temperature-Fluctuation Noise: Results from fluctuating heatexchange between devices and the environment

• Flicker Noise: Has a spectral density of the form 1/ fα � 1/ f ,

also known as pink noise; the physics is not well understood



A.1.5 Available Power

R

G RL = R GL = G

(a) (b)

vrms irms

• Noise analysis is often focused around receiver circuitry wheremaximum power transfer is implemented, i.e., match the loadand sources resistances

• Under these conditions the power delivered to the load is theavailable power Pa

(a) Pa =

�12irms

�2

R= i

2rms

4R

b Pa =

�12irms

�2

G= i

2rms

4G

• For a noisy resistor

v2rms = 4kT RB,

so

Pa,R = 4kT RB

4R= kT B W



Example A.3: Fundamental Example

• Consider room temperature to be To = 290 K, then the thermalnoise power density is

Pa,R

B= 4.002 × 10−21 W/Hz

• For communication system analysis a popular measurementunit for both signal and nois epower levels, is the power ra-

tio in decibels (dB) referenced to

(i) 1 W ←→0 dBW

= 10 log10

�PWatts

1 Watt

�; PWatt = 1

(ii) 1 mW ←→0dBm

= 10 log10

�PmW

1 mW

�; PmW = 1 mW

• In dB units thermal noise power spectral density under maxi-mum power transfer is

Pwr/Hz (dBW) = 10 log10

�4.002 × 10−21

1 W

�� −204 dBW/Hz

Pwr/Hz (dBm) = 10 log10

�4.002 × 10−21

1 mW

�� −174 dBm/Hz



A.1.6 Frequency Dependence

• If frequency dependence is included, then the available powerspectral density is

Sa( f )�= Pa

B= h f

exp�

h f

kT

�− 1

W/Hz

where

h = Planck’s constant = 6.6254 × 10−34 J-sec

10 100 1000 10000 100000.

-205

-200

-195

-190

-185

-180

-175

-170

f (GHz)

Infrared

hf290 K

29 K

2.9 K

Noise

Spec

trum

(dBM

)

Thermal noise spectral density, including quantum noise

A.1.7 Quantum Noise

• To account for quantum noise the term h f must be added

• Thermal noise dominates for most applications (i.e., < 20 GHz),except in optical systems and some millimeter wave systems



A.2 Characterization of Noise in Sys-

tems

• In communication system modeling we wish to consider howthe noise introduced by each subsystem enters into the overallnoise level delivered to the demodulator

• In RF/microwave systems the concept of representing a systemas a cascade of subsystems is particularly appropriate, since allconnections between subsystems is done at a constant impedancelevel of say 50 ohms

Subsys1

Subsys2

SubsysNR0

SN)) 0

SN)) 1

SN)) 2

SN)) N

SN)) N-1

N -subsystem cascade analysis

A.2.1 Noise Figure of a System

Subsysl Rl

es,les,l-1

Rl-1, Ts

lth subsystem model

• For the lth subsystem define the noise figure, Fl , as�

S

N

�

l

= 1Fl

�S

N

�

l−1


A.2. CHARACTERIZATION OF NOISE IN SYSTEMS

• Ideally, Fl = 1, in practice Fl > 1, meaning that each subsys-tems generates some noise of its own

• In dB the noise figure (NF) is

FdB = 10 log10 Fl

• Assuming the subsystem input and put impedances (resistances)are matched, then our analysis may be done in terms of theavailable signal power and available noise power

• For the lth subsystem the available signal power at the input is

Psa,l−1 = e2s,l−1

4Rl−1

• Assuming thermal noise only, the available noise power is

Pna,l−1 = kTs B

where Ts denotes the source temperature

• Assuming that the lth subsystem (device) has power gain Ga,it follows that

Psa,l = Ga Psa,l−1

where we have also assumed the system is linear

• We can now write that�S

N

�

l

= Psa,l

Pna,l= 1

Fl

Psa,l−1

Pna,l−1= 1

Fl

�S

N

�

l−1

which implies that

Fl = Psa,l−1

Pna,l−1· Pna,l

Psa,l��Ga Psa,l−1

= Pna,l

Ga Pna,l−1� �� kTs B



• NowPna,l = Ga Pna,l−1 + Pint,l

where Pint,l is internally generated noise

• Finally we can write that

Fl = 1 + Pint,l

GakTs B

– Note that if Ga � 1 ⇒ Fl � 1, assuming that Ga isindependent of Pint,l

• As a standard, NF is normally given with Ts = T0 = 290 K, so

Fl = 1 + Pint,l

GakT0B

A.2.2 Measurement of Noise Figure

• In practice NF is measured using one or two calibrated noisesources

Method #1

• A source can be constructed using a saturated diode which pro-duces noise current

i2n

= 2eId B A2

• The current passing through the diode is adjusted until thenoise power at the output of the devide under test (DUT) isdouble the amount obtained without the diode, then we obtain

F = eId Rs

2kT0



where Rs is the diode series resistance and Id is the diode cur-rent

Method #2

• The so-called Y -factor method requires ‘hot’; and ‘cold’ cali-brated noise sources and a precision variable attenuator

NoiseSourceThot

NoiseSourceTcold

DeviceUnder TestTe, G, B

PowerMeter

CalibratedAttenuator

Y factor determination of NF

• From noise power measurements taken with the hot and coldsources we form the ratio

Ph

Pc

= Y = k(Thot + Te)BG

k(Tcold + Te)BG= Thot + Te

Tcold + Te

• Solving for Te

Te = Thot − Y Tcold

Y − 1

• The Y value is obtained by noting the attenuator setting change,�A dB, needed to maintain Pc = Ph and calculating Y =10�A/10



A.2.3 Noise Temperature

• The equivalent noise temperature of a subsystem/device, is de-fined as

Tn = Pn,max

k B

with Pn,max being the maximum noise power of the source intobandwidth B

Example A.4: Resistors in series and parallel

• Find Tn for two resistors in series

vn

R2, T2

R1, T1

R2 + R1

�v2n� = 4k B R1T1 + 4k B R2T2

and

Pna = �v2n�

4(R1 + R2)= 4k(T1 R1 + T2 R2)B

4(R1 + R2)

thereforeTn = Pna

k B= T1 R1 + T2 R2

R1 + R2

• Find Tn for two resistors in parallel

in R1 || R2 = G1 + G2R1, T1 R2, T2



�i2n� = 4k BG1T1 + 4k BG2T2

and

Pna = �i2n�

4(G1 + G2)= 4k(T1G1 + T2G2)B

4(G1 + G2)

therefore

Tn = T1G1 + T2G2

G1 + G2= T1 R2 + T2 R1

R1 + R2

A.2.4 Effective Noise Temperature

• Recall the expression for NF at stage l

Fl = 1 + Pint,l

GakT0B� �� internal noise

= 1 + Te

T0

• Note: Pint,l/(Gak B) has dimensions of temperature

• DefineTe = Pint,l

Gak B= effective noise temp.,

which is a measure of the system noisiness

• Next we use Te to determine the noise power at the output ofthe lth subsystem

• Recall that

Pna,l = Ga Pna,l−1 + Pint,l

= GakTs B + GakTe B

= Gak(Ts + Te)B



• This references all of the noise to the subsystem input by virtueof the Ga term

A.2.5 Cascade of Subsystems

• Consider two systems in cascade and the resulting output noisecontributions

1 2TsThe noise here is due to the following

1. Amplified source noise = Ga1Ga2kTs B

2. Internal noise from amplifier 1 = Ga1Ga2kTe1 B

3. Internal noise from amplifier 2 = Ga2kTe2 B

• ThusPna,2 = Ga1Ga2k

�Ts + Te1 + Te2

Ga1

�B

which implies that

Te = Te1 + Te2

Ga1

and since F = 1 + Te/T0

F = 1 + Te1

T0+ Te2

Ga1T0

= F1 +1 + Te2

T0− 1

Ga1

= F1 + F2 − 1Ga1



• In general for an arbitrary number of stages (Frii’s formula

F = F1 + F2 − 1Ga1

+ F3 − 1Ga1Ga2

+ · · ·

Te = Te1 + Te2

Ga1

+ Te3

Ga1Ga2

+ · · ·

A.2.6 Attenuator Noise Temperature and Noise

Figure

AttenLTs Pa,out =

Pa,inL

Resistive network at temperature Ts

Attenuator model

• Since the attenuator is resistive, we know that the impedancesare matched and

Pna,out = kTs B (independent of Rs or L)

• Let the equivalent temperature of the attenuator be Te, then

Pna,out = Gak(Ts + Te)B

= 1L

k(Ts + Te)B� �� looks like Pan,in

• Thus since Pan,outalso= kTs B, it follows that

1L

(Ts + Te) = Ts



orTe = (1 − L)Ts

• Now since

F = 1 + Te

T0= 1 + (L − 1)Ts

T0

with Ts = T0 (i.e., attenuator at room temperature)

Fattn = 1 + L − 1 = L

Example A.5: 6 dB attenuator

• The attenuator analysis means that a 6 dB attenuator has anoise figure of 6 dB

Example A.6: Receiver system

RFAmplifier

IFAmplifierMixer

Attn

FeedlineLoss

L = 1.5 dB G2 = 20 dB G3 = 8 dB G4 = 60 dBF1 = 1.5 dB F2 = 7 dB F3 = 10 dB F4 = 6 dB

Receiver front-end



• We need to convert from dB back to ratios to use Frii’s formula

G1 = 10−1.5/10 = 11.41 F1 = 101.5/10 = 1.41

G2 = 1020/10 = 100 F2 = 107/10 = 5.01,G3 = 108/10 = 6.3 F3 = 10G4 = 1060/10 = 106

F4 = 3.98

• The system NF is

F = 1.41 + 5.01 − 11/1.41

+ 10 − 1100/1.41

+ 3.98 − 1(100)(6.3)/1.41

= 7.19 or 8.57 dB

• The effective noise temperature is

Te = T0(F − 1) = 290(7.19 − 1)

= 1796.3 K

• To reduce the noise figure (i.e., to improve system performance)interchange the cable and RF preamp

• In practice this may mean locating an RF preamp on the backof the receive antenna, as in a satellite TV receiver

• With the system of this example,

F = 5.01 + 1.41 − 1100

+ 10 − 1100/1.41

+ 3.98 − 1(100)(6.3)/1.41

= 5.15 or 7.12 dBTe = 1202.9 K

• Note: If the first component has a high gain then its noise figuredominates in the cascade connection



• Note: The antenna noise temperature has been omitted, butcould be very important

Example A.7: Receiver system with antenna noise tempera-ture

RFAmplifier

IFAmplifierMixer

Attn

FeedlineLoss

L = 1.5 dB G2 = 20 dB G3 = 8 dB G4 = 60 dBF1 = 1.5 dB F2 = 7 dB F3 = 10 dB F4 = 6 dB

Ts = 400 K

F = 7.19 or FdB = 8.57 dB, Te = 1796.3 K

• Rework the previous example, except now we calculate avail-able noise power and signal power with additional assumptionsabout the receiving antenna

• Suppose the antenna has an effective noise temperature of Ts =400 K and the system bandwidth is B = 100 kHz

• What is the maximum available output noise power in dBm?

• Since

Pna = Gak(Ts + Te)B = (Ga)(kT0)

�Ts + Te

T0

�(B)



where

Ga,dB = −1.5 + 20 + 8 + 60 = 86.5 dBkT0 = −174 dBm/Hz, T0 = 290 K

we can write in dB that

Pna,dB = 86.5 − 174 + 10 log�

400 + 1796.3290

�+ 10 log10 105

= −28.71 dBm

• What must the received signal power at the antenna terminalsbe for a system output SNR of 20 dB?

• Let the received power be Ps or in dBm Ps,dB

10 log10

�Ga Ps

Pna

�= 20

• Solving for Ps in dbm

Ps,dB = 20 + Pna,dB − Ga, dB= 20 + (−28.71) − 86.5 = −95.21 dBm



A.3 Free-Space Propagation Channel

• A practical application of the noise analysis is in calculatingthe link budget for a free-space communications link

• This sort of analysis applies to satellite communications

RelaySatellite

Rec.

GroundStation

UserUplink

Downlink

EarthSatellite link scenario

• Consider an isotropic radiator which is an ideal omnidirec-

tional antenna

P

Power PT is radiated uniformly in all directions (a point source)

Power density at distance d from the transmitter

d

Omni antenna and received flux density

• The power density at distance d from the source (antenna) is

pt = PT

4πd2 W/m2


A.3. FREE-SPACE PROPAGATION CHANNEL

• An antenna with directivity (more power radiated in a par-ticular direction), is described by a power gain, GT , over anisotropic antenna

• For an aperture-type antenna, e.g., a parabolic dish antenna,with aperture area, AT , such that

AT � λ2

with λ the transmit wavelength, GT is given by

GT = 4π AT

λ2

• Assuming a receiver antenna with aperture area, AR, it followsthat the received power is

PR = pt AR = PT GT

4πd2 · Ar

= PT GT G Rλ2

(4πd)2

since AR = G Rλ2/(4π)

• For system analysis purposes modify the PR expression to in-clude a fudge factor called the system loss factor, L0, then wecan write

PR =�

λ

4πd

�2

� �� Free space loss

PT GT G R

L0



• In dB (actually dBW or dBm) we have

PR,dB = 10 log10 PR

= 20 log10

�λ

4πd

�

+ 10 log10 PT + 10 log10 GT� �� EIRP

+ 10 log10 G R − 10 log10 L0

where EIRP denotes the effective isotropic radiated power

Example A.8: Free-Space Propagation

• Consider a free-space link (satellite communications) where

Trans. EIRP = (28 + 10) = 38 dBWTrans. Freq = 400 MHz

• The receiver parameters are:

Rec. noise temp. = Ts + Te = 1000 KRec. ant. gain = 0 dB

Rec. system loss L0) = 3 dBRec. bandwidth = 2 kHz

Path length d = 41, 000 Km

• Find the output SNR in the 2 kHz receiver bandwidth


A.3. FREE-SPACE PROPAGATION CHANNEL

• The received signal power is

PR,dB = 20 log10

�3 × 108/4 × 108

4π × 41, 000 × 103

�+ 38 dBW + 0 − 3

= −176.74 + 38 − 3 = −141.74 dBW= −111.74 dBm

– Note: λ = c/ f = 3 × 108/400 × 106

• The receiver output noise power is

Pna,dB = 10 log10(kT0) + 10 log10

�Te

T0

�+ 10 log10 B

= −174 + 5.38 + 33= −135.62 dBm

• Hence

SNRo, dB = 10 log10

�PR

Pna

�

= −111.74 − (−135.62)

= 23.88 dB



.


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