engn 3214 telecommunications

ENGN 3214Telecommunications

Thushara AbhayapalaDepartment of Telecommunications Engineering

Research School of Information Sciences and EngineeringAustralian National University

[email protected]://www.syseng.anu.edu.au/˜thush/engn3214.html

Semester 1, 2001

ENGN 3214 Copyright 2000, Thushara D. Abhayapala 0

Important Information

• Lecturer: Thushara Abhayapala, Room B148, RSISE Building, X58683, e-mail: [email protected] or [email protected]

• Office Hours: Wednesday 4.00-5.00 PM

• Tutor: Terence Betlehem, x8683, [email protected]

• Tutorials: Thursday 2-3PM, Friday 2-3PM

• Laboratories: Wednesday 8-10, Friday 9-11, and 11-1, Circuits andTelecoms Lab, RM103, Ian Ross Building, will be held on weeks starting19 March, 2 April, 7 May, 21 May

• Web Site: http://www.syseng.anu.edu.au/ thush/engn3214/engn3214.html

ENGN 3214 Course Overview 1

Additional Points

If there are any students with disabilities or any other difficulties: in this classwho wish to talk to me about their studies in relation to their difficulty, pleasemake an appointment. The University will try to help such students as far aspossible but also respects your privacy.

ENGN 3214 Course Overview 2

Chapter 1 — Introduction to Telecommunications

Read Chapter 1 of the textbook

We will cover

• What is Telecommunication?

• Historical perspective

• Future Telecommunication

• Communication System Model

• What are we going to learn from this course?

ENGN 3214 Chapter 1 — Introduction to Telecommunications 3

What is Telecommunication?

• Telecommunications refers to long-distance communication (the Greek telemeans ”far off”).

• At present, such communication is carried out with the aid of electronicequipment such as the Radio, Telegraph, Telephone, mobile phones,Television, and Internet.

• In earlier times, smoke signals, drums, light beacons, and various forms ofsemaphore were used.

• The transmitted information can be in the form of voice, symbols, pictures,or data.

• A telecommunications system includes a transmitter, one or more receivers,and a channel or means of communication such as the air, water, wire,cable, communications satellite, or some combination of these.


History of Telecommunication

Time chart of historical development is given in Table 1.1 of textbook. Some ofKey Events

• 1844 Morse code and Telegraph line

• 1876 Graham Bell invented the telephone.

• 1920 First radio broadcast

• 1936 BBC begins first TV broadcast

• 1942 Nyquist Sampling Theory


• 1947 Invention of transistor

• 1948 Shannon Information theory

• 1957 First earth satellite

• 1976 Personal computers developed

• 1980 Fibre Optical communication

• 1995 Internet and WWW became popular

• 1995- present: Era of Digital Communication...


Future Telecommunication

• ”In contact anytime, anywhere, with any one”

• Need wireless access to Internet, personal internet, personal multimedia

• Current wireless connection in the telecommunications chain cannotsupport the high data rate demanded by the Internet.

• Future Challenges

– Demand of high data rates– limited spectrum availability– demand for greater area coverage– increased subscriber capacity– multipath and time-varying propagation environment

Do you know these buzz words: Blue Tooth, CDMA, WCDMA, SDMA, UMTSetc.?


Digital and Analog Sources and Systems

Digital Information Source: produces a finite set of of possible messages. E.g.,English alphabet.

Analog Information Source: produces messages that are defined on acontinuum. E.g., A microphone, where output voltage is distributed over acontinuous range of values.

Analog Communication system transfers information from an analog source tothe receiver. An analog waveform is a function of time that has a continuousrange of values.

Digital Communication system transfers information from a digital source tothe intended receiver. A digital waveform has only a discrete set of amplitudevalues.

Digital communication system usually has voltage and current waveforms thathave digital values; however it may have analog waveforms.


Deterministic and Random waveforms

Deterministic Waveform

A completely specified function of time

eg., w(t) = A cos(ω0t + ψ0)

where A,ω0, and ψ0 are known constants.

Random(Stochastic) Waveform

A random waveform cannot be completely specified as a function of time andmust be modeled probabilistically. I.e., at least one of A,ω0, or ψ0 is a randomvariable (stochastic process).

eg., speech is random. Noise is also described by a random process.

Consider the letters of English alphabet and suppose each letter is transmittedby a known (deterministic) waveform. If the order of letters emitted by thesource is random, then the transmitted waveform is random.


We need random waveforms to describe communication systems. Thus, wewill study probability and statistical concepts in chapter 6.

However we only consider deterministic signals in the first half of the course.


Communication System Model

TRANSMITTERINFORMATION

RECEIVER

USER

CHANNEL

m(t)

s(t)

NOISE

s(t)

r(t)

All communication systems have 3 main subsystems: 1) Transmitter, 2)Channel, 3) Receiver.

The spectra(frequencies) of information input m(t) are concentrated about f =0 (i.e. baseband signal, e.g., Human voice 0-4kHz.)


Transmitter

• conditions m(t) for more efficient transmission

• lowpass filtering to restrict the bandwidth of m(t)

• source encoding and channel encoding so that errors can be corrected atthe receiver.

• converts the baseband signal m(t) into a frequency band that is appropriatefor the transmission medium of the channel.

s(t) = m(t) cos 2πfct

– Spectrum of s(t) is concentrated arond f = fc

– fc -Carrier frequency– This process is called modulation


Channel

• can be classified as wire and wireless channels.

• examples of wire channels: twisted-pair telephone lines, coaxial cables,waveguids, fiber-optic cables.

• examples of wireless channels: air, vaccum, seawater

• general principles of communication theory apply to all types of channels.

• channel medium attenuate the signal so that the noise of the channelcauses the delivered information m(t) to be deteriorated from that of thesource


Receiver

• takes the corrupted signal at the channel output and converts it to abaseband signal (demodulation)

• then “cleans up” the baseband signal and deliver an estimate of the sourceinformation m(t) to the user.

The goal is to design communication systems that transmit information to thereceiver with as little as deterioration as possible satisfying design constraints,of allowable transmitted energy, allowable signal bandwidth and cost.

Measure of Deterioration of the received data

m(t)

• Bit Error Rate (BER) in digital systems

• Signal to Noise Ratio (SNR) at the receiver output in analog systems.


Frequency (Spectrum) Allocation

(Read Chapter 1.7 of the textbook)

Wireless communication system often use the atmosphere for the transmissionchannel. There are various wireless systems, such as TV, radio, cellularphones etc., which uses the atmosphere as the channel.

How can all these systems share the same ‘channel’ without interfering eachother?

Different wireless systems use different frequency bands to transmit theirsignals.

Government regulation specify the modulation type, bandwidth, power, type ofinformation that user (service provider) can transmit over designated frequencybands.

Frequency allocations are given in Table 1.2: For example,30− 300MHz VHF band - VHF TV0.3− 3GHz UHF band - UHF TV, cellular telephones, GPS


Which government body regulate the frequency allocation in Australia?

What is the frequency band used for Digital Television in Australia (DTV)?


What are we going to learn from this course?

On completion of this subject students should understand:

How a basic communication system works (chapter 1-5

Effect of noise (chapter 6,7)

Simple wire and wireless communication systems (chapter 8)

Simple Network concepts


The next chapter will review signals and spectra ( Things you have learnedfrom Signals and Systems course).

Read Sections 1.1, 1.2, 1.3, 1.6, 1.7, 1.8 (NOW!)


Chapter 2 — Signals and Spectra

In this lecture we will review properties of signals (Recall Signals and Systemscourse). We will meet

• Basic signal properties

• Fourier Transform and spectra

• Convolution

ENGN 3214 Chapter 2 — Signals and Spectra 19

Basic Signal Properties

Physically realizable waveforms

• are measurable in a laboratory

• are bandlimited signals (finite frequency band)

• are continuous functions of time

• have finite power

• are non-zero for significant time interval and have only real values.

Mathematical models violate some or all of above conditions, but they areuseful to analyze systems. However, if we are careful with the mathematicalmodel, the correct result can be obtained when the answer is properlyinterpreted. E.g., see Figure 2.1 of the textbook.


Time Average Operator

This is a linear operator and defined as

< [·] >= limT→∞

1T

∫ T/2

−T/2[·] dt

Periodic Signal

w(t) is periodic with period T0 if

w(t) = w(t + T0) for all t

where T0 is the smallest positive number that satisfies this relationship.

Time Average Operator for a Periodic Signal

< [·] >=1T0

∫ T/2+a

−T/2+a[·] dt, where a is an arbitrary constant.


DC Value

Wdc = limT→∞

1T

∫ T/2

−T/2w(t) dt

For a physical waveform, we are interested the average value over a finite timeinterval of interest: say from t1 to t2,

1t2 − t1

∫ t2

t1w(t) dt


Rms Value:

Root Mean Square Value

Wrms =√

< w2(t) >

=

√

limT→∞

1T

∫ T/2

−T/2w2(t) dt

For a physical waveform, we are interested the rms over a finite time intervalT .


Average Power:

In communication systems, we need to compare the received (average) signalpower to the (average) noise power.

P =< w2(t) >= limT→∞

1T

∫ T/2

−T/2w2(t) dt

where w(t) represent a real current or voltage waveform. w(t) is called a powerwaveform if it has finite average power, i.e., 0 < P < ∞.

Total normalized energy

E = limT→∞

∫ T/2

−T/2w2(t) dt.

w(t) is an energy waveform if and only if 0 < E < ∞. w(t) can only beeither power or energy waveform but not both. There are some mathematicalwaveforms that have infinite energy and infinite power. e.g., w(t) = e−t.


Decibel

Decibels are used to measure power ratios.

decibel gain dB = 10 log10(PoutPin

)

PoutPin

= 10dB/10

Decibal Signal to Noise Ratio (SNR)

(SNR)dB = 10 log10(Psignal

Pnoise)

= 10 log10

(

√

< v2signal(t) >

√

< v2noise(t) >

)2

= 20 log10(Vrms signal

Vrms noise)


Fourier Transform (FT) and Spectra

Why we need Fourier Transform ?To find out frequencies present in a signal w(t)

What is the definition of frequency?Frequency of a sinusoidal signal with period T0 is f0 = 1/T0. I.e., s(t) =sin 2πf0t. Thus, frequency is the rate of occurrence of a sinusoidal waveform.Units of f are Hz.

All non-sinusoidal signals are made of lots of sinusoidal signals with differentfrequencies. I.e., non-sinusoidal signals have more than one frequency andFT is a tool to calculate these frequencies or spectrum.

FT of w(t) is defined as: W (f) = FT [w(t)] =∫∞−∞w(t)e−j2πft dt

(Note that ω = 2πf - radian frequency)

W (f) is also called two-sided spectrum since it has both positive and negativefrequencies. What are negative frequencies?


How do we evaluate FT?

1. Direct integration

2. Tables of FT (see Table2-1)

3. Superposition to break the problem into two or more problems

4. Differentiation or integration of w(t)

5. Using Properties of FT (see Table 2-1)

6. Numerical integration of the FT integral using Matlab etc.

7. Using Fast Fourier Transform (FFT)


Inverse Fourier Transform (IFT): w(t) =∫∞−∞W (f)ej2πft df

In Signals and Systems course we had

ω = 2πf

df =12π

dω

w(t) =12π

∫ ∞

−∞W (ω)ejωtdω.

Read section 2.2 (Today!) and refresh your knowledge on Fourier Transforms.

Example: Find the spectrum of v(t) = A cosω0t, where ω0 = 2πf0.


Parseval’s Theorem

∫ ∞

−∞|w(t)|2dt =

∫ ∞

−∞|W (f)|2df

Using Parsevals theorem we can evaluate energy of a signal either in timedomain or frequency domain description.

Energy Spectral Density

ξ(f) = |W (f)|2.


Convolution

Convolution between w1(t) and w2(t) is

w3(t) = w1(t) ∗ w2(t)

=∫ ∞

−∞w1(λ)w2(t− λ)dλ.

Example:


Recommended Problems

2.1, 2.2, 2.6, 2.13, 2.14, 2.15, 2.16, 2.17, 2.18, 2.22, 2.26, 2.27, 2.31, 2.40,2.41, 2.42


Chapter 2 — Signals and Spectra Cont....

In this lecture we will review few other properties of signals. We will meet

• Power Spectral Density (read section 2.3)

• Autocorrelation Function (read section 2.3)

• Orthogonal Functions and Series (read section 2.4)

• Fourier Series (read section 2.5)


Power Spectral Density (PSD)

The PSD is useful in describing how the power content of signals and noise isaffected by filters and other devices in communication systems.

PSD of a power signal w(t) is given by

Pw(f) = limT→∞

(

|WT (f)|2)

,

where WT (f) = FT{wT (t)} and

wT (t) =

{

w(t) if −T/2 < t < T/2,0 t elsewhere

. Also note that the normalized average power is

P =< w2(t) >=∫ ∞

−∞Pw(f)(f)df.


Autocorrelation Function

The autocorrelation function of a real (physical) waveform is

Rw(τ) ∆=< w(t)w(t + τ) >= limT→∞

1T

∫ T/2

−T/2w(t)w(t + τ)dt.

Note that

Rw(0) = limT→∞

1T

∫ T/2

−T/2w2(t)dt

= Average Power

Now we have several methods of calculating average power:

P =< w2(t) >= W 2rms =

∫∞−∞Pw(f)df = Rw(0)


Also note that the PSD and the autocorrelation function are FT pairs; i.e.,

Pw(f) = FT{Rw(τ)}.

It is easier to calculate Pw(f) by first evaluating Rw(τ) and then taking theFourier Transform.

Example: A power signal is given by w(t) = a + b cos ω0t; w0 = 2πf0. Find1) Rw(τ),2) Pw(f),3) Average Power of the signal?


Orthogonal Functions:

Functions ϕn(t) and ϕm(t) are said to be orthogonal over the interval a < t < bif

∫ b

aϕn(t)ϕ∗m(t) dt = 0, where n 6= m, (1)

where ∗ denotes complex conjugate. Further, if the functions in the set {ϕn(t)}are orthogonal, then

∫ b

aϕn(t)ϕ∗m(t) dt = Knδnm =

{

0, n 6= mKn, n = m

δnm is called the Kronecker delta function.

If Kn = 1 for all n, then the set {ϕn(t)} are called orthonormal functions.

Equation (1) is used to test pairs of functions to see if they are orthogonal.


I.e., if the integral of the product of any two functions are zero over the interval(a, b), then they are orthogonal to each other.

The zero result implies that these functions are “independent” or in“disagreement”. If the result is not zero, they are not orthogonal and the twofunctions have some “dependence” on each other.

Examples of well known orthogonal functions:

• {ejnω0t}, n = ±0,±1, . . . over 0 < t < T0 where T0 = 2πω0

• {Pn(cos θ)}, n = 0, 1, . . ., over 0 ≤ θ ≤ π, where Pn(·) are Legendrefunctions.

• The set of Bessel functions {Jn(x)} over 0 < x < a, where a > 0.

Example: Show that the functions ϕn(t) = ejnω0t, n = ±0,±1, . . . areorthogonal over the interval a < t < a + T0, where T0 = 2π

ω0


Orthogonal Series

If there is a set of functions which are orthogonal over an interval, then we canrepresent ’any other function’ in that interval using the orthogonal set.

THEOREM: w(t) can be represented over the interval (a, b) by the series

w(t) =∑

n

anϕn(t)

where the coefficients an are given by

an =1

Kn

∫ b

aw(t)ϕ∗n(t)dt,

where the set {ϕn(t)} is complete.

See page 67 for the proof. A set of function are said to be complete over aninterval, if any other function in the same interval can be represented by theset with an arbitrary small error. E.g., Fourier series, Legendre functions.


Fourier Series

The Fourier series is an orthogonal series and useful in solvingtelecommunication engineering problems.

THEOREM: A function w(t) may be represented over the interval a < t < a+T0

by the complex exponential Fourier Series

w(t) =n=∞∑

n=−∞cnejnωot

where the complex Fourier coefficients are

acn =1T0

∫ b

aw(t)ϕ∗n(t)dt,

where the set {ϕn(t)} is complete.

EXAMPLE: Problem 2.51



2.44, 2.45, 2.46, 2.47, 2.49, 2.50, 2.51, 2.53, 2.54, 2.55


Week2, Lecture 1— Signals and Spectra Cont....

In this lecture we will continue to review properties of signals. We will meet

• Quick Quiz

• Review of last week lectures

• Fourier Series (read section 2.5)


Quick Quiz

Answer the following two questions on a separate sheet of paper.

1. What is the frequency band used to broadcast DTV signals in Australia?

2. What is the relationship between Power Spectral Density Pw(f) and auto-correlation function Rw(τ) of a signal w(t)?


Review of last week lectures

• Fourier Transform (FT) : is used to find the frequencies present in awaveform w(t).

W (f) = FT [w(t)] =∫ ∞

−∞w(t)e−j2πftdt

w(t) =∫ ∞

−∞W (f)ej2πftdf

• Convolution (FT) : is useful to find the output r(t) of a system (filter), giventhe input signal w(t) and the impulse response h(t) of the system (filter).

r(t) = w(t) ∗ h(t) =∫ ∞

−∞w(λ)h(t− λ)dλ

E.g., Given a transmitted signal r(t) and the impulse response h(t) of thechannel of a communication system, we can find the received signal usingconvolution.


• Power Spectral Density ( Pw(f)): shows the distribution of power of asignal/noise over frequencies.

• Autocorrelation function ( R(τ)): indicates how much a signal waveformw(t) and it’s delayed (delay of τ ) signal waveform w(t+τ) is ’related’ to eachother.

Rw(τ) ∆=< w(t)w(t + τ) >= limT→∞

1T

∫ ∞

−∞w(t)w(t + τ)dt

If Rw(τ1) = 0, then the two signals are uncorrelated for the delay τ1.

In Communication systems, due to mutipaths, the received signal at thereceiver is a combination of the desired signal (direct path) and it’s delayedcopies. Here, we may use the autocorrelation function to find out how muchthey are correlated.

We can also use autocorrelation function to calculate the PSD

Pw(f) = FT{Rw(τ)}.


• Orthogonal Functions:

∫ b

aϕn(t)ϕ∗m(t) dt = Knδnm =

{

0, n 6= mKn, n = m

• Orthogonal Series: can be used to represent a signal over a given interval.

w(t) =∑

n

anϕn(t)

an =1

Kn

∫ b

aw(t)ϕ∗n(t)dt,


Fourier Series

The Fourier series is an orthogonal series and useful in solvingtelecommunication engineering problems.

THEOREM: A function w(t) may be represented over the interval a < t < a+T0

by the complex exponential Fourier Series

w(t) =n=∞∑

n=−∞cnejnω0t

where the complex Fourier coefficients are

cn =1T0

∫ b

aw(t)e−jnω0tdt, .

EXAMPLE 2.51:


Properties of Fourier Series

1. If w(t) is real,

cn = c∗−n

2. If w(t) is real and even [i.e., w(t) = w(−t)],

Im[cn] = 0.

3. If w(t) is real and odd [i.e., w(t) = −w(−t)],

Re[cn] = 0.

4. Parseval’s theorem

1T0

∫ a+T0

a|w(t)|2dt =

∞∑

n=∞|cn|2.


Different forms of Fourier Series (see page 70)

w(t) =n=∞∑

n=−∞cnejnω0tComplex Fourier Series

w(t) =n=∞∑

n=0

an cos(nω0t) +n=∞∑

n=0

bn sin(nω0t), Quadrature Fourier Series

If w(t) is real

w(t) = D0 +n=∞∑

n=1

an cos(nω0t + φn) Polar Fourier Series



2.52, 2.53, 2.54, 2.55, 2.56, 2.57, 2.58, 2.60, 2.62, 2.65



In this lecture we will:

• study rectangular pulse/wave and its spectrum, PSD etc

• generalize them for any periodic signal

• review Linear systems (read section 2.6)


Quick Quiz

Answers to the Quiz:

1. What is the frequency band used to broadcast DTV signals in Australia?Few channels within the VHF (300 − 300MHz) and 42 channels in the UHF(0.3 − 3GHz) band. Each DTV channel has a bandwidth of 7MHz andallocation is different from city to city and they are placed such that theydo not interfere with existing analog channels. You may read more aboutthis by following the links given in the course web page.

2. What is the relationship between Power Spectral Density Pw(f) and auto-correlation function Rw(τ) of a signal w(t)?

Pw(f) = FT{Rw(τ)}.


Rectangular Pulse∏

( tT ) and T Sa(πTf) form a Fourier Transform pair.

∏

(x) =

{

1 |x| ≤ 1/20 otherwise

Sa(x) =sin x

x

Duality Theorem

If W (f) = FT [w(t)] thenFT [W (t)] = w(−f).

Time delay Theorem

If w(t) ↔ W (f) then w(t− Td) ↔ W (f)e−j2πfTd.


Periodic waveforms have following Properties

For a periodic waveform w(t) with period T0 (ω0 = 2πf0 = 2π/T0),

w(t) =∞∑

n=−∞cnejnω0t Fourier series

W (f) =∞∑

n=−∞cnδ(f − nf0) Fourier Transform (spectrum)

Pw(f) =∞∑

n=−∞|cn|2δ(f − nf0) PSD

Rw(τ) =∞∑

n=−∞|cn|2ejnω0τ Autocorrelation


Linear Systems

Linear SystemInput waveform Output waveform

x(t) y(t)h(t) H(f)

• A system or a filter is linear when superposition holds

• A system is time invariant if

x(t) 7−→ y(t)

x(t− td) 7−→ y(t− td)


Impulse Response

• A linear system (filter) can be characterized by a linear ordinary differentialequation. E.g.,

andnydtn

+ an−1dn−1ydtn−1 + · · · = x(t)

• Impulse response is the solution to the differential equation when the forcingfunction is Dirac delta function. I.e.,if x(t) = δ(t) then y(t) = h(t).

• Causality: (see signals and system book). Output will depend on past andpresent values of the input but not future values. All physical systems needto be causal!. I.e., h(t) = 0 for t < 0.

• Convolution integral can be used to calculate output of a linear system:

y(t) = x(t) ∗ h(t) =∫ ∞

−∞x(λ)h(t− λ)dλ.


Transfer Function of a System

• H(f) -is also known as Frequency response of the system.

• h(t) ↔ H(f)

• Also Y (f) = X(f)H(f)H(f) = Y (f)

X(f)

• Transfer function is generally complex quantity and can be written in polarform:

H(f) = |H(f)|ej〈H(f)


Few Important Results

• if x(t) = A cos[2πf0t + φ] then y(t) = A|H(f0)| cos[ω0t + φ + 〈H(f0)].

• Generally if x(t) is periodic with period T0

X(f) =∞∑

n=−∞cnδ(f − nf0)

X(f) =∞∑

n=−∞cn H(nf0) δ(f − nf0)

Py(f) = |H(f0)|2Px(f)


Distortionless Transmission

Read textbook

Some systems are more sensitive to amplitude distortion (e.g. human ear) andsome sensitive to phase distortion (e.g. human eye).



You should try these problems to pass the course2.68, 2.70, 2.8

You may try these problems if you wish to learn this subject more thoroughly.I.e., to get a good grade 2.72, 2.77



In this lecture we will learn:

• Bandlimited signals (section 2.7)

• Sampling theorem (Section 2.7)

• Dimensionality Theorem ( section 2.7)

• Discrete Fourier Transform (DFT) (section 2.8)

• Definitions of Bandwidth (section 2.9)


Bandlimited signals

• A bandlimited signal has non-zero spectra only with a certain frequencyband: i.e.,

W (f) = FT [w(t)] = 0 for |f | ≥ B

where B is the absolute bandwidth of the signal.

• A time limited signal is defined as

w(t) = 0 for |t| ≥ T

• A bandlimited waveform cannot be time limited and vice versa.

• E.g., A rectangular pulse is time limited and it’s spectrum, Sa(·) is notbandlimited.


• Engineering paradox: physical waveforms are always time limited, hencethey have infinite bandwidth. But we need bandlimited signal (to transmit),so that we can efficiently use the available spectrum.

• Although, the signal may not be band limited, it may be bandlimited for allpractical purposes in the sense that the amplitude spectrum has a negligiblelevel above a certain frequency. (Read pages 86− 87)


Sampling Theorem

(Read pages 87-93)

• Sampling theorem is another application of an orthogonal series expansionand applies to digital communication systems.

• If a signal waveform is bandlimited (B Hz) then it can be completelyrepresented by and recoverable from knowledge of its samples equallyspaced (Ts) in time. Ts is the sampling time and fs = 1/Ts is the samplingfrequency.

• w(t) is bandlimited to B Hz, i.e., W (f) = 0, |f | ≥ B, then signal w(t) can bereconstructed by its samples, where the sampling frequency fs ≥ 2B.

• Nyquist frequency ∆= (fs)min = 2B.

• How do we reconstruct w(t) from its samples? (see sampling theorem onpage 87)


If W (f) = 0 for |f | ≥ B and fs = 1/Ts ≥ 2B then

w(t) =∞∑

n=−∞w(nTs)

sin[πfs(t− nTs)]πfs(t− nTs)

=∞∑

n=−∞w(nTs) Sa[πfs(t− nTs)]

where w(nTs) = w(t)|t=nTs, n = −∞, . . . ,−1, 0, 1, . . . ,∞ is the samples ofthe signal.

• How can we use sampling in a digital communication system?


Impulse Sampling

Read pages 90-92 and your notes.


Dimensionality Theorem

(page 93)

• A more general way of presenting the sampling theorem

• when BT0 is large, a real waveform may be specified by

N = 2BT0

independent piece of information.T0 waveform over T0 interval.B absolute bandwidth of the waveform.

• How can we use the dimensionality theorem?(i) to calculate number of storage locations required to represent awaveform.(ii) use to estimate bandwidth of waveforms (more in chapter 3).


Discrete Fourier Transform (DFT)

• DFT is used to compute samples of continuous Fourier transform (CFT).(Read pages 94-95 and signals and systems book)

• Fast Fourier Transforms (FFT)are fast algorithms that can be used tocalculate DFT efficiently. We can use fft and ifft functions in MATLAB tocalculate DFT and then we can interpret results to obtain CFT.


Bandwidth of Signals

• Read section 2.9

• you should learn why bandwidth of a signal is important and familiar withnumerous definitions of the bandwidth.

• What is the legal definition of bandwidth in Australia?(set by ACC (?))

• Read Example 2.18 of the textbook.


Read section 2.10 (Summary) and 2.11 (study aid examples).



You should try these problems to practise your basic skills (i.e., to pass thecourse)2.82, 2.84, 2.92

You may try these problems if you wish to learn this subject more thoroughly.I.e., to get a good grade 2.77


Week3, Lecture 1— Chapter 3

Announcements

• Course web page has a new home:http://spigot.anu.edu.au/courses/engn3214/and it is also available at the old home:http://www.syseng.anu.edu.au/ thush/engn3214/engn3214.html

• Assignment 1 will be out this week (Friday)

In this lecture we will meet:

• Weekly Quiz

• Overview of Chapter 3 (Section 3.1)

• Pulse Amplitude Modulation ( section 3.2)


Weekly Quiz

Answer the following question on a separate sheet of paper.


x(t) y(t)h(t) H(f)

Consider a linear time invariant system (LTI) with impulse response h(t):(i) Write an equation that describe output y(t) in terms of input x(t) and h(t)?(ii)Let X(f) and Y (f) be the spectrums (Fourier Transforms) of x(t) and y(t)respectively. If H(f) is the frequency response of the system, then (a) Whatis the relationship between X(f), Y (f) and H(f)? (b) What is the relationshipbetween H(f) and h(t)?(iii) If x(t) is a periodic signal, we can represent x(t) by its Fourier Series x(t) =∑∞

n=−∞ cnejω0t. Write the output y(t) in terms of Fourier series coefficients cn?


Baseband Pulse and Digital Signaling

This Chapter shows how to encode analog waveform into baseband digitalsignal. The following are the main goals of Chapter 3:

• To study how analog waveforms can be converted to digital waveforms.

• To learn how to compute the spectrum of digital signals.

• To examine how the filtering of pulse signals affect our ability to recover thedigital information at the receiver.

• To study how we can multiplex (combine) data from several sources intoone high-speed digital stream for transmission over a digital system.


Pulse Amplitude Modulation

• Convert analog signal to pulse-type signal in which the amplitude of thepulse denotes the analog information.

• Analog to PAM conversion is the first step of converting analog waveform toPulse Code Modulation (PCM) (digital signal).

• Recall that the sampling theorem provides a way to reproduce an anlogwaveform by using sample values of that waveform and sin x

x orthogonalfunctions.

• PAM signaling is another waveform that looks like pulses, yet contains theinformation that present in the analog waveforms.


There are two classes of PAM signals:

1. PAM that uses natural sampling (gating) (see Fig 3.1)

2. PAM that uses instantaneous sampling to produce flat top pulse (see Fig3.5)


PAM that uses natural sampling

If w(t) is an analog waveform bandlimited to B hertz, then the PAM signal is

ws(t) = w(t)s(t) (2)

where

s(t) =∞∑

n=−∞

∏

(t− kTs

τ) (3)

and fs = 1/Ts ≥ 2B (See Fig 3.1).

Spectrum of natural sampled PAM signal

Ws(f) = d∞∑

n=−∞

sin πndπnd

W (f − nfs) (4)

where d = τ/Ts. (see Figure 3.3).


Comments on natural sampled PAM signal

• PAM waveform is wider than that of the analog waveform

• However pulses are more practical to use in digital systems.

• (see Fig 3.2) The natural sampled PAM signal can be generated by using ananalog switch that is readily available in CMOS hardware (e.g., 4016 quadbilateral switch).

• At the receiver, the original analog waveform can be recovered from thePAM signal, ws(t) by passing ws(t) through a low-pass filter with cut-offfrequency B < fcut-off < fs − B. Note that the spectrum of the output ofthe low-pass filter would have the same shape as that of the original analogwaveform.

• Above is true only when fs ≥ 2B, otherwise spectral components inharmonic bands (of fs) would overlap (aliasing). This is another illustrationof Nyquist sampling rate requirement.


• In practice physical signals are time limited so that they cannot be absolutelybandlimited. Consequently there will be some aliasing in the PAM signal.Usually, the analog signal is pre-filtered before it is introduced to the PAMcircuits so we do not have to worry about this problem.

• Some times there may be noise due to power hum or mechanical circuitvibration that might fall in the baseband (band corresponds to n = 0)relative to other bands. For this reason, the PAM signal is multiplied bysinusoidal signal of frequency nωs before the low pass filtering. This shiftsthe frequency band of the the PAM signal that was centered about nfs tobase band (i.e., f = 0).


Instantaneous Sampling (Flat top PAM)

(see Fig. 3.5)

ws(t) =∞∑

k=−∞

w(kTs)h(t− kTs) (5)

where h(t) is the pulse shape. If h(t) is the rectangular pulse i.e.,

h(t) =∏

(tτ) (6)

we get flat top PAM signal. (See Fig 3.5).

Flat top PAM can be generate using sample and hold type of electronic circuits.

If h(t) is another form of a pulse then the resulting PAM would not be flat-topped.


Spectrum of flat-top PAM signal

Ws(f) =1Ts

H(f)∞∑

k=−∞

W (f − kfs) (7)

where H(f) = τ sin πτfπτf . (see Figure 3.6).


• Analog signal may be recovered from flat-top PAM signal by use of a low-pass filter.

• However, there is some high frequency loss due to the filter H(f), causedby the flat top pulse shape. This loss can be reduce by decreasing τ orby using some additional gain of the high frequencies in the low-pass filtertransfer function.

• To strictly remove effect of H(f), we may need a low-pass filter withfrequency response 1/H(f). Such a filter is called as equalization filter.

• The pulse width τ is alos called the aperture since τ/Ts , determines thegain of the recovered analog signal, which is small if τ is small relative toTs.

• Product detector can also be used to recover the signal, except that somepre-filtering is needed to make the spectrum flat in the appropriate band.

• Again Nyquist criterion, i.e., fs ≥ 2B is needed to avoid aliasing.


Comments on PAM signals

• Both kind of PAM signals have wider bandwidth than that of analog signal,thus it is not very good for long distance transmission (read page 37)

• PAM provides a means for converting analog signal to a PCM (pulse CodeModulation) signal.

• PAM also provides a means for breaking a signal into time slots so thatmultiple PAM signals carring information from different sources can beinterleaved to transmit all of the information over a single channel. Thisis called TDMA.



You should try these problems to practise your basic skills (i.e., to pass thecourse)3.3

You may try these problems if you wish to learn this subject more thoroughly.I.e., to get a good grade 3.1,3.2, 3.4, 3.5, 3.6


Week3, Lecture 2— Chapter 3 (Cont...Announcements

• Course web page has a new home:http://spigot.anu.edu.au/courses/engn3214/and it is also available at the old home:http://www.syseng.anu.edu.au/˜thush/engn3214/3214.html

• Assignment 1 will be out this week (Friday)-check the web page.


• Answers to Weekly Quiz

• Review of Natural sampled PAM

• Flat-top sampled PAM ( section 3.2)

• Pulse Code Modulation (PCM) (section 3.3)


Weekly Quiz

Most of you answered (i) and (ii) correctly.


x(t) y(t)h(t) H(f)

Consider a linear time invariant system (LTI) with impulse response h(t):(i) Write an equation that describe output y(t) in terms of input x(t) and h(t)?Ans: y(t) = x(t) ∗ h(t) =

∫∞−∞ x(t− λ)h(λ)dλ

(ii)Let X(f) and Y (f) be the spectrums (Fourier Transforms) of x(t) and y(t)respectively. If H(f) is the frequency response of the system, then (a) What isthe relationship between X(f), Y (f) and H(f)?Ans: Y (f) = X(f)H(f)(b) What is the relationship between H(f) and h(t)?Ans: H(f) = FT [h(t)].


(iii) If x(t) is a periodic signal, we can represent x(t) by its Fourier Series x(t) =∑∞

n=−∞ cnejnω0t. Write the output y(t) in terms of Fourier series coefficientscn?Ans:

y(t) =∞∑

n=−∞H(nf0)cnejnω0t

.


Review of Natural sampled PAM

If w(t) is an analog waveform bandlimited to B hertz, then the PAM signal is

ws(t) = w(t)s(t) (8)

where

s(t) =∞∑

n=−∞

∏

(t− kTs

τ) (9)

and fs = 1/Ts ≥ 2B (See Fig 3.1).

Spectrum of natural sampled PAM signal is

Ws(f) = d∞∑

n=−∞

sin πndπnd

W (f − nfs) (10)

where d = τ/Ts. (see Figure 3.3).


Instantaneous Sampling (Flat top PAM)

(see Fig. 3.5)

ws(t) =∞∑

k=−∞

w(kTs)h(t− kTs) (11)

where h(t) is the pulse shape. If h(t) is the rectangular pulse i.e.,

h(t) =∏

( tτ)

(12)

we get flat top PAM signal. (See Fig 3.5).

Flat top PAM can be generate using sample and hold type of electronic circuits.

If h(t) is another form of a pulse then the resulting PAM would not be flat-topped.


Spectrum of flat-top PAM signal

Ws(f) =1Ts

H(f)∞∑

k=−∞

W (f − kfs) (13)

where H(f) = τ sin πτfπτf . (see Figure 3.6).


• Analog signal may be recovered from flat-top PAM signal by use of a low-pass filter.

• However, there is some high frequency loss due to the filter H(f), causedby the flat top pulse shape. This loss can be reduce by decreasing τ orby using some additional gain of the high frequencies in the low-pass filtertransfer function.

• To strictly remove effect of H(f), we may need a low-pass filter withfrequency response 1/H(f). Such a filter is called as equalization filter.

• The pulse width τ is also called the aperture since τ/Ts , determines thegain of the recovered analog signal, which is small if τ is small relative toTs.

• Product detector can also be used to recover the signal, except that somepre-filtering is needed to make the spectrum flat in the appropriate band.

• Again Nyquist criterion, i.e., fs ≥ 2B is needed to avoid aliasing.


Comments on PAM signals

• Both kind of PAM signals have wider bandwidth than that of analog signal,thus it is not very good for long distance transmission (read page 37)

• PAM provides a means for converting analog signal to a PCM (pulse CodeModulation) signal.

• PAM also provides a means for breaking a signal into time slots so thatmultiple PAM signals carring information from different sources can beinterleaved to transmit all of the information over a single channel. Thisis called TDMA.




You may try these problems if you wish to learn this subject more thoroughly.I.e., to get a good grade 3.1,3.2, 3.4, 3.5, 3.6


Week3, Lecture 3— Chapter 3 (Cont...

In this lecture we will study:

• Pulse Code Modulation (PCM) (section 3.3)


Analog to Digital Conversion methods

• Pulse Code Modulation (PCM)

• Delta Modulation (DM)

• Differential Pulse code Modulation (DPCM)


Pulse Code Modulation (PCM)

PCM is analog to digital conversion where the information contained in theinstantaneous samples of analog signal is represented by digital words in aserial bit stream. (see Fig.3.7). PCM is used in digital Telephone systems andaudio CD recording.

1. PCM signal is generated by carrying out 3 basic operations: (i) sampling,(ii) quantizing, and (iii) encoding.

2. Sampling: generates a flat-top PAM signal.

3. Quantizing operation is illustrated in Fig 3.8a of the textbook. Input to thequantizer is a PAM signal and output is a quantized PAM signal.

• Quantizer has number of levels M (e.g., M=8),• If all steps of quantizer is equal then it is called as uniform quantizer.

There are non-uniform quantizing techniques such as µ-law and A-lawcompanding.


• Quantizing error consists of the difference between the analog signal atthe sampler input and the output of the quantizer (see Fig 3.8.c).

• Peak value of error is one half of the quantizer step size. The errors inthe recovered analog signal due to quantizer error is called as quantizingnoise or round-off errors.

4. Encoding: (refer to Fig 3.8.d and Table 3-1)

• PCM signal is obtained from the quantized PAM signal by encoding eachquantized sample value into a digital word. E.g., Table 3-1 uses gray codeto encode M = 8 quantized levels into 3 bit digital words.

• For binary code words, M = 2k unique code words are possible. Eachcode word corresponding to a certain amplitude level.

• It is up to the system designer to specify the exact code word that willrepresent a particular quantized level.

• We may use base other than 2 to represent quantized analog samplesby digital words i.e., multilevel signalling. Multilevel signals have theadvantage of possessing a much smaller bandwidth than binary signals,but the disadvantage of requiring multilevel circuitry instead of binarycircuitry.


Advantageous of PCM

Read page 137-8.

• Relatively inexpensive digital circuitry may be used extensively in thesystem.

• PCM signals can be derived from all types of analog sources (audio, video,etc.,) may be merged with data signals (e.g., from digital computer) andtransmitted over a common high-speed communication system.

• Noise performance of digital systems can be superior than that of analogsystems. In addition, coding techniques can be used in digital systems tocorrect errors.


Practical PCM circuits

• Three popular techniques are used to implement the analog-to -digitalconverter (ADC) encoding operation.

1. Counting or ramp encoders. e.g., Maxim ICL 7126 CMOS ADC.2. Serial or successive approximation encoder. e.g., Analog Devices AD570

8-bit ADC3. parallel or flash encoders. e.g., Haris CA3318 8-bit DAC

• All of the IC’s listed above have parallel digital outputs that corresponds todigital word, which represents the analog sample value. For the generationof PCM, the parallel output need to be converted to serial form for datatransmission. This is done by using serial input-output SIO chip.

• At the receiver end, the PCM signal is decoded back into an analog signalby using a digital to analog converter DAC chip (e.g., see page 142 of thetextbook). Output of a DAC is a quantized PAM signal, thus we need tolowpass filter it to obtain the analog signal.


• Semiconductor companies manufacture several hundred types of ADC andDAC circuits. Data sheets for these IC’s can be found on their web pages.

• Some of these chips are for specific applications. For example:

– Analog Devices AD 1861 is a PCM audio DAC that generate analog audiofrom from the PCM data on compact disks (CD’s).

– Motorola MC 145503 is a PCM codec for encoding and decoding datasignals for telephone analog audio applications.


Bandwidth of PCM signalsWhat is the spectrum of PCM (serial data) waveform?

• Note that the spectrum of PAM signal is a linear function of the analogsignal. This is not the case for PCM. The PCM is a nonlinear function ofthe input analog signal (due to quantizing). Thus, the spectrum of PCM isnot directly related to the spectrum of the input analog signal.

• Bandwidth of binary PCM waveforms depends on the bit rate and thewaveform pulse shape used to represent the data (see notes).

bit rateR = how many bits per seconds

= how many waveform pulses per second

• If the sampling frequency is fs and there is n bits in the PCM word (i.e., weneed n bits to represent a quantized value), then bit rate

R = nfs.


• Using the dimensionality theorem, we can show that the bandwidth of binaryencoded PCM is bounded by

BPCM ≥12R =

12nfs.

• Actual bandwidth depends on the pulse shape used. For rectangular pulses,the first null bandwidth is

BPCM = R == nfs.

• Using dimensionality theorem, a lower bound for the bandwidth of PCMsignal is

BPCM ≥ nB

where fs ≥ nB and B is the bandwidth of the analog signal.


• Thus bandwidth of PCM signal is significantly larger than the bandwidth ofthe corresponding analog signal. (read page 143)

I would expect you will read section Effect of Noise (pages 143-147).



You should try these problems to practise your basic skills (i.e., to pass thecourse)3.8, 3.10

You may try these problems if you wish to learn this subject more thoroughly.I.e., to get a good grade 3.7, 3.9, 3.11,


Week4, Lecture 1— Chapter 3 (Cont...)

Announcements

• Assignment 1 is available on the web page.

• There will be Lab sessions this week. Remember that you have to keep alab book.


• Digital Signaling (section 3.4)


Weekly Quiz

Answer the following question on a separate sheet of paper.

(i) What are the 3 basic steps involved in generating PCM signals?(ii) Name an application or a system which uses PCM signals?


Digital Signalling

How do we mathematically represent the waveform for a digital signal, such asPCM signal and how do we estimate the bandwidth of the waveform?

Voltage or current waveforms for digital signals can be expressed using anorthogonal series with a finite number of terms N ,

w(t) =N

∑

k=1

wkϕk(t), 0 < t < T0

where wk represent digital data and ϕk(t), k = 1, 2, . . . , N are N orthogonalfunctions that give the waveform its wave shape. Example (see notes):


Baud (symbol rate) D

D =NT0

=number of dimensions used

time interval

Bit Rate R

R =nT0

=number of data bits send

time interval

• For the case when wk have binary values, n = N and w(t) is said to bebinary signal.

• For binary signal symbol rate (baud) is equal to bit rate.


• At the receiver, the transmitted data wk can be recovered using,

wk =1

Kk

∫ T0

0w(t)ϕ∗k(t)dt, k = 1, 2, . . . , N,

where

Kk =∫ T0

0ϕk(t)ϕ∗k(t)dt, k = 1, 2, . . . , N,

– ϕk(t) is known to the receiver.– This is the optimum way of detecting the transmitted data, when the

received signal is corrupted by white additive noise. (matched filterdetection)


Vector Representation

We may write

w(t) =N

∑

k=1

wkϕk(t), 0 < t < T0

in vector space by

w =N

∑

j=1

wjϕj,

• w is an N dimensional vector Euclidean vector space.

• {ϕj} is an orthogonal set of N dimensional vectors that become a unitvector set if Kk = 1. (Example: see notes and Example 3.2 of the textbook)


Bandwidth Estimation

Bandwidth of w(t) is (from dimensionality theorem)

B ≥ N2T0

=D2

=symbol rate (baud)

2.

• If ϕk(t) are of sin xx type, then the lower absolute bandwidthof D/2 will be

achieved.

• for other pulse shapes (such as rectangular) bandwidth will be larger thanthis lower bound.


Binary signaling

• Can be described by the N -dimensional orthogonal series w(t) =∑N

k=1 wkϕk(t) where the series coefficients wk, take on binary values. (seeexample 3.3 of the textbook)

• Consider a digital source that produce M distinct messages. (e.g., PCMsignal (with M = 256 quantizer levels)

• Each message could be represented by n bit binary word. (e.g., M = 2n =256 = 28).


Multilevel signaling

• Here the series coefficients wk, take on L(> 2) values (see example 3.4 ofthe textbook).

• Consider a digital source that produce M distinct messages. (e.g., PCMsignal (with M = 256 quantizer levels)

• Each message could be represented by n symbols. (e.g., M = Ln = 256 =44). (see notes)

• We can reduce N ( number of dimensions in orthogonal functions) usingmultilevel signaling.

• L level signal would have 1/l bandwidth of the corresponding binary signalwhere L = 2l. This bandwidth reduction is achieved because the durationof the multilevel signal is l times that of the binary signal.


Binary line coding

Read pages 160-163.



You should try these problems to practise your basic skills (i.e., to pass thecourse)3.19, 3.20, 3.21, 3.16

You may try these problems if you wish to learn this subject more thoroughly.I.e., to get a good grade 3.14, 3.15, 3.17,



Announcements

• Assignment 1 (Due 6 April ) is available on the web page.


• Line Codes and spectra (section 3.5 )

• An Example of multilevel signaling

ENGN 3214 Chapter 3 — Baseband pulse and Digital signaling 115

Answers to the Weekly Quiz

(i) What are the 3 basic steps involved in generating PCM signals?

(0) low pass filtering, (1) sampling (PAM signal generation), (2) quantizing,and (3) encoding.

(ii) Name an application or a system which uses PCM signals?Audio CD’s, digital telephones.


Line codes and Spectra

Binary line codes

• Binary 1’s and 0’s such as in PCM signaling may be represented in variousserial bit signaling formats called line coeds (see Fig 3.15)

• Two main categories: Return-to-Zero (RZ) and Non-return-to-zero (NRZ).The waveform return to zero volt level for a portion (usually one half) of thebit interval.

• There are advantageous/disadvantageous of each line codes listed in Fig3.15 (read page 162)


Desirable properties of line codes

1. self- synchronization

2. low probability of bit errors

3. spectrum that is suitable for channel

4. transmission bandwidth

5. error detection capability

6. transparency

(read page 162)


Power Spectral Density of Line Codes

• PSD can be evaluated by using either deterministic (chapter 2) or stochastic(random) techniques (chapter 6). If the data sequence is random, we haveto use stochastic approach.

• A line code or a digital signal can be represented by

s(t) =∞∑

n=−∞anf(t− nTs)

where f(t) is the symbol pulse shape and Ts is the duration of one symbol,an is the random data. Example: see notes (from white board)

• From chapter 6, PSD of a digital signal is

Ps(f) =|F (f)|2

Ts

∞∑

n=−∞R(k)ej2πkfTs


where F (f) is the spectrum of the pulse f(t) and R(k) is the auto correlationof the data;

R(k) =I

∑

i=1

(anan+k)Pi,

where an, an+k are the voltage levels of the data pulses at the nth and(n + k)th symbol positions and Pi is the probability of having the ith anan+k

product.

• Note that the PSD of a digital signal is dependent on two things:(i) Pulse shape,(ii) statistical properties of the data.

• Example see notes (from white board)


Example of multilevel signaling

A polar NRZ line code is converted to multilevel signal for transmission over achannel. Let number of levels in multilevel signal is 32 and the pulse width ofthe polar NRZ signal is 0.1ms. Find(a) bit rate?(b) symbol rate (baud) of multilevel signal?(c) PSD of the multilevel signal? see notes (from white board), pages 176-178of textbook



Announcements



• Differential Coding (section 3.5 )

• Eye Patterns

• Bit Synchronization


Differential Coding

• When serial data are passed through many circuits along a communicationchannel, the waveform is often unintentionally inverted (i.e., datacomplemented). e.g.,

transmitted sequence 1 1 0 1 0 0

inverted sequence 0 0 1 0 1 1

• This can be occurred in a a twisted-pair transmission line channel just byswitching the two leads at a connection point when a polar line code is used.

• Differential coding is used as a solution to this problem (see Fig 3.17)


en = dn ⊕ en−1

wheredn data input to the encoderen output of the encoder

⊕

modula 2 adder (XOR).

dn en−1 en

1 0 11 1 00 0 00 1 1

• Received encoded data is decoded by

dn = en ⊕ en−1

where tilde denotes received-end data.

• Example (see Table3-4 of the textbook)


Eye Patterns

What is this?(read pages 170-171)

• The effect of channel filtering and noise can be seen by observing thereceived line code on an oscilloscope.

• An eye pattern (it looks like an eye) can be seen on an oscilloscope withmultiple sweeps of the signal, where each signal is triggered by a clocksignal and the sweep width is slightly larger than bit rate.

• You should try to see this in the labs.

• If eye is closed then there is great deal of noise or inter symbol interferenceISI (will be explained next week).

• if eye is open, then there is no or less bit errors.


• Eye patterns provides a way to asses the quality of the received line codeand the ability of the receiver to combat bit errors.

– The timing error allowed on the sampler at the receiver is given by thewidth inside the eye (eye opening.

– The sensitivity to timing error is given by the slope of the open eye,evaluate at or near the zero-crossing point.

– The noise margin of the system is given by the height of the eye opening.


Regenerative Repeaters

Repeaters amplify and “cleans up” the signal periodically when transmittingover a hardware channel over a long distance.


Bit Synchronization

• Synchronization signals are clock type signals that are necessary with areceiver to recover the transmitted message (e.g., for sampling the receivedsignal at the correct time instance.)

• These clock signals have precise frequency and phase relationship withrespect to the received input signal, and they are delayed compared to theclock signal at the transmitter, since there is a propagation delay throughthe channel.

• There are 3 types of synchronization signals in digital communicationsystems:

– bit sync - to distinguish one bit from another– frame sync - to distinguish groups of data– carrier sync - for bandpass signalling (chapter 4,5)

• how are we going to derive bit sync from the corrupted signal? The


complexity of bit synchronizer circuits depend on the sync properties of theline code used.

• For polar NRZ line code, bit synchronizer is shown in Figure 3.20. Anothermethod is given in Figure 3.21.

• Read pages 175-176 and try to understand the techniques given.


Spectral Efficiencyis defined as the number of bits per second of data that can be supported byeach hertz of bandwidth,

η =RB

(bits/s)/Hz

where R- bit rate and B - bandwidth.

• In applications in which the bandwidth is limited by physical and regulatoryconstraints, the job of the communication engineer is to choose a signalingtechnique that gives the highest spectral efficiency while achieving givencost constraints and meeting specifications for a low probability of bit errorrate at the system output.

• Maximum possible spectral efficiency is limited by the channel noise if theerror is to be small. Using Shannon’s capacity formula,

ηmax =CB

= log2(1 +SN

)


where C is the capacity of the system and S/N is the signal to noise ratio.

• Shannon does not tell us how to achieve this limit, but using error correctioncoding and multilevel signaling one can approach this limit.

• See table 3.6 for the spectral efficiencies of various line codes.

• Note that the multilevel signaling has the highest spectral efficiency.However they are costly to implement.




You may try these problems if you wish to learn this subject more thoroughly.I.e., to get a good grade3.34,



Announcements



• Intersymbol Interference (section 3.6 )


Weekly Quiz

• What is en eye pattern in a telecommunication system?

• Why do we need bit synchronization?


Intersymbol Interference (ISI)

• Absolute bandwidth of rectangular multilevel pulses is infinity. If thesepulses are filtered improperly as they pass through a communicationsystem, they will spread in time, and the pulse for each symbol may besmeared into adjacent time slots and cause ISI. (see Fig3.23)

• ISI causes errors in the recovered signal.

• see notes from white board


•

HR(f) =He(f)

H(f)HT (f)HC(f),

when He(f) is chosen to minimize ISI, HR(f) is called an equalizing filter.

• Communication engineer need to design HT (f) and HR(f) to reduce ISI.

• The channel transfer function HC(f) is dependent on channel characteristics,and may be time varying.

• For time varying channels, the equalizing filter may need to be an adaptivefilter.

• In some adaptive schemes, each communication session is preceded by atest bit (symbol) pattern that is used to adapt the filter for the maximum eyeopening (i.e., minimum ISI). Such a sequence is called learning or trainingsequence.


• Now we know, what ISI is and it’s effects on the recovered signal. How canwe design a communication system to reduce ISI?


Nyquist’s method of zero ISI

• There will be zero ISI if the equivalent impulse response

he(kTs + τ) =

{

C k = 0,0 k 6= 0,

where k is an integer, Ts is the symbol (sample) clocking period, τ is the offset in the receiver sampling clock times compared with the clock time of theinput symbol, and C is a non zero constant.


• Do we know any pulses that satisfies Nyquist’s zero ISI condition?: e.g., forτ = 0

he(t) =sin πfst

πfst.

I.e., if the transmit filter HT (f) and the receiver filter HR(f) are designed sothat overall transfer function is

He(f) =1fs

∏

(ffs

),

then there will be no ISI.

• However there are two difficulties with sin xx type of overall pulse shape:

– The overall amplitude transfer function He(f) has to be flat over −B <f < B and zero elsewhere. This is physically unrealizable (i.e., impulseresponse would be non causal and of infinite duration), and also it isdifficult to approximate steep steps at f = ±B.


– The synchronization of the clock in receiver has to be almost perfect,since the sin x

x pulse decays only as 1/x and is zero in adjacent time slotsonly when t is at exactly correct sampling time. Thus, inaccurate sync willcause ISI.

• Therefore we need to consider other pulse shapes that have a slightly widerbandwidth, The idea is to find pulse shapes that go through zero at adjacentsampling points and have envelop that decays much faster than 1/x so thatclock jitter in the sampling times does not cause appreciable ISI.

• Raised cosine-roll off Nyquist filter has most of the desirable features thatwe seek (read 183-187).



You should try these problems to practise your basic skills (i.e., to pass thecourse)3.37,3.42,3.43,3.45

You may try these problems if you wish to learn this subject more thoroughly.I.e., to get a good grade3.46, 3.47



Announcements



• Differential Pulse Code Modulation (DPCM) (section 3.7)

• Delta Modulation (section 3.8)


Differential Pulse Code Modulation (DPCM)

• When audio or video signals are sampled, it is usually found that adjacentsamples are close to the same value. I.e., there is a lot of redundancy inthe signal sample.

• The bandwidth and dynamic range of a PCM system are wasted whenredundant sample values are retransmitted.

• One way to minimize redundant transmission and reduce the bandwidth isto transmit PCM signals corresponding to the difference in adjacent samplevalues. This is called Differential Pulse Code Modulation-DPCM.

• At the receiver the present sample value is regenerated by using the pastvalues plus the update differential value that is received over the differentialsystem.


• A prediction filter or a predictor is used to estimate the present value fromthe past values.

• A prediction filter can be realized by using a tapped delay line to form atransversal filter (see Fig 3.28).

z(nTs) =K

∑

l=1

al y(nTs − lTs)

where:z(nTs) : estimated ‘present value’y(nTs − lTs) : input to the predictor (past values)al : tap gains

• Optimum tap gains are a function of the correlation properties of the audioor video signal.


• Two possible DPCM signals are given in Fig. 3.29 and 3.30.

• Note that the DPCM configuration in Fig3.30 does not accumulate noise.Why?

• For DPCM signal SNR (6dB rule):

( SN

)

dB = 6.02n + α

where −3 < α < 15 for DPCM speachn - number of quantizing bits (M = 2n).

• DPCM signal requires slower bit rates compared to PCM signal to achievesame SNR (an advantage!).


Delta modulation (DM)

• DM is a special case of DPCM, where there are only two quantizing levels.

• No encoder is necessary since M = 2 and hence DM is less expensive thanthe DPCM.

• The operation of subtractor and two-level quantizer is are implemented byusing a comparator so that the out put is ±Vc (binary).

• The cost DM is further reduced by replacing the predictor by a low costintegration circuit. (see Fig 3.31 and read section 3.8).


• The integrator is assumed to be act as a accumulator, so that the integratoroutput at time t = nTs,

zn =1Vc

n∑

i=0

δyi

whereyi = y(iTs)δ is the accumulator gain or step size.

• Read page 194 to learn more on granular noise and slope overload noise(different kinds of quantizing noise).



You should try these problems to practise your basic skills (i.e., to pass thecourse)3.49,3.50




Announcements

• Assignment 1 is (Due on 6 April ). If you need any help or hints please asknow!.

• For Q1, part 1 use properties of Fourier Transforms (Table ...) to find the FTof the Trapezoid given.


• Revisit Differential Pulse Code Modulation (DPCM) (section 3.7)

• Time Division Multiplexing (section 3.9)


Revisit DPCM

• We had few difficulties in understanding the operation of DPCM andprediction filter.

• The following questions were unanswered or partially answered! during theprevious lecture!.

1. How do you choose filter taps (coefficients ak in the prediction filter?2. Why can’t we simply send the difference between the present input and

the most recent past input rather than using a predictor?3. How do you initialize the DPCM system? i.e., what are the initial outputs

of the system?4. What are the advantageous of using DPCM instead of using PCM?5. Why a DPCM signal has less bandwidth than a corresponding PCM

signal?6. Are there any applications/devices which use DPCM?


I have copied the following text from the textbook

A detailed analysis of DPCM system is difficult and depends on the type ofinput present, the sample rate, the number of quantizing levels used, thenumber of stages in the prediction filter and the predictor gain coefficients.This type of analysis is beyond the scope of this text ....

However, we try answer our list of questions without going into too much detail.


1. How do you choose filter taps (coefficients ak in the prediction filter?

Optimum tap gains are a function of the correlation properties of the inputsignal (audio or video signal). Alternatively, we may use an adaptivealgorithm to adapt the coefficients based on the input.


2. Why can’t we simply send the difference between the present input and themost recent past input rather than using a predictor?

The equation for the predictor is

z(nTs) =K

∑

l=1

al y(nTs − lTs)

where:z(nTs) : estimated ‘present value’y(nTs − lTs) : input to the predictor (past values)al : tap gainsThe simplest case of the predictor is to estimate the present value as themost recent past value. I.e., if al = 0 for l 6= 1, then z(nTs) = y((n−1)Ts). Inthis special case, what we are sending is the difference between the presentinput and the most recent past input.So a predictor is a generalized case of the simple difference, which mightbe able to further reduce the amplitude of the DPAM.


3. How do you initialize the DPCM system? i.e., what are the initial outputs ofthe system?It is not clear from the textbook. But we can assume that the initial predictoroutput is equal to the present value, thus resulting a zero output signalinitially.

4. What are the advantageous of using DPCM instead of using PCM?DPCM uses less bandwidth.

5. Why a DPCM signal has less bandwidth than a corresponding PCM signal?We may use less number of levels (i.e., less bits per each sample value) fora DPCM signal (see notes from white board).


6. Are there any applications/devices which use DPCM?Use in some telephone systems. Specially, there are CCITT(International telegraph and Telephone Consultative Committee) standardsfor transmitting 3.2kHz bandwidth voice signals and 7kHz bandwidth audiosignals.3.2kHz bandwidth voice signals: 32kbit/s, 4 bit quantization, 8 ksamples/sencoding7kHz bandwidth audio signals: 64kbit/s, 4 bit quantization, 16 ksamples/sencoding


Time Division Multiplexing (TDM)

• TDM is the time interleaving of samples from several sources so that theinformation from these sources can be transmitted serially over a singlecommunication channel (see Fig 3.35)

• In Fig 3.35, sampling time of each channel is Ts. Then the pulse width ofthe TDM PAM is Ts

3 = 13fs

.

• Pulse width of the TDM PCM is Ts3n, where n is the number of bits used in

the PCM word.

• fs needs to satisfy the Nyquist rate for the analog source with the largestbandwidth.

• If a bandwidth of a particular source is markedly larger than the others thenthat source may be connected to several switch positions on the sampler,so that it will be sampled more often than the smaller bandwidth sources.


• AT the receiver, the demodulator (sampler) has to be synchronized with theincoming waveform so that the PAM samples corresponding to source 1, forexample, will appear on the channel 1 of the out put. - This is called framesynchronization

• Low pass filters are used to reconstruct the analog signal in each channelfrom the PAM samples.

• ISI would cause PCM from one channel to appear in another channel evenwith perfect synchronization ( bit and frame). This is called cross talk.

• Applications of TDM: public telephone system.



Announcements

• Assignment 1 is Due 6 April (Friday)!! .


• Review of TDM

• Frame Synchronization

• Example of a TDM system

• Packet Transmission system (brief overview)

• Pulse Time Modulation


Weekly Quiz

• Describe Inter Symbol Interference (ISI) with respect to a communicationsystem?


Time Division Multiplexing (TDM)

TDM is the time interleaving of samples from several sources so that theinformation from these sources can be transmitted serially over a singlecommunication channel (see Fig 3.35)


Frame Synchronization

• Frame synchronization is needed at the TDM receiver so that the receivedmultiplexed data can be sorted and directed to the appropriate outputchannel.

• The frame sync can be provided to the receiver either,

1. by sending a frame sync signal from the transmitter to the receiver over aseparate channel (straightforward but expensive).

2. by deriving the frame sync from the received TDM signal itself.

• Frame sync can be multiplexed along with the information words (see Fig3.36).

• The frame sync can be recovered at the receiver by cross correlatingthe regenerated TDM signal with the expected unique sync word s =(s1, s2, . . . , sK). (see fig 3.37)


• False sync output pulses will occur if K successive information bits happento match the bits in the sync word. For equally likely TDM data, probabilityof false sync occurring,

Pf =1

2K .

• Advantage of TDM: it can accommodate both analog and digital sources.

• Examples:

1. see Example 3.6 of the text book2. problem 3.60 (see notes from the white board)


Packet Transmission Systems

(read section 3.10)

• TDM is a synchronize transfer mode (STM). That is, a data source isassigned a specific time slot with a fixed data rate.

• In many applications, this fixed data rate assignment is not effective, sincestuff bits are inserted to match the assigned data rate when the source doesnot have any data to send. I.e., for data from bursty source, STM system isnot efficient.

• A packet transmission system partition the source data into data packets,each of which contains a destination address header.

• Then, many users can share the high-speed channel by merging theirpackets into high-speed data stream.


• Routers along the network read the header on the packets and route thepackets to the appropriate destination.

• More packets are sent from high-speed source over a given time interval,compared with only a few packets that are merged the network from lowspeed sources.

• Examples of packet transmission systems: TCP/IP and ATM technology(see Appendix C).

• A packet transmission systems are efficient when the sources have burstydata; however there is a overhead caused by the transmission of the packetheader.

• STM networks are more efficient when the source have a fixed data rate.


Pulse Time Modulation (PTM): Pulse width modulation(PWM) and Pulse position Modulation (PPM)

• PTM encodes the sample values of an analog signal into time axis of adigital signal.

• Recal that PAM, PCM, and DM techniques encode the sample values intothe amplitude characteristics of the digital signal.

• Two main types of PTM are

1. Pulse width modulation (PWM) (Pulse Duration Modulation (PDM)) - seefig. 3.43

2. Pulse position Modulation (PPM)

• In PWM, sample values of the analog waveform are used to determine thewidth of the pulse signal.


• In PPM, the analog sample value determines the position of a narrow pulserelative to the the clocking time.

• Read section 3.10 to find out how to generate PTM signals and how torecover the analog signal.

• PTM signalling is not widely used in transmitting across channels becausea relatively wide bandwidth channel is needed. However, PTM signals maybe found internally in digital communication terminal equipments.

• Main advantage of PTM signals is that they have great immunity to additivenoise compared to PAM signalling.



You should try these problems to practise your basic skills (i.e., to pass thecourse)3.57, 3.59, 3.62



Week6, Lecture 2 — Chapter 4 (Bandpass Signalling Principles

Announcements

• Assignment 1 is Due 6 April (Friday)!! .

• Don’t forget the Labs!


• Overview of the course

• Bandpass signalling principles

• Complex envelop representation of bandpass signals

• Spectrum of bandpass signals


Overview of the course

1. Deterministic signals and systems

2. Baseband signalling

3. Bandpass signalling principles (set of tools)

4. Various analog and digital modulated schemes

5. Effects of noise (random processes)

6. Overview of practical systems ( Digital TV, mobile communication systems,..etc)

7. Telecommunications and Systems Engineering ?


Bandpass signalling principles

• A bandpass communication signal is obtained by modulating a basebandanalog or digital signal onto a carrier.

• A baseband waveform has a spectral magnitude that is non-zero forfrequencies in the vicinity of f = 0 and negligible elsewhere.

• A bandpass waveform has a spectral magnitude that is non-zero forfrequencies in some band concentrated about a frequency f = ±fc, wherefc � 0 and negligible elsewhere.

• fc is the frequency of an oscillator signal in the transmitter circuit and is theassigned frequency of the transmitter, such as, for example, 850kHz for anAM broadcasting station.

• In communication problems, the information source signal, such as audiosignal from a microphone, is usually a baseband signal.


• The job of the communication engineer is to buid a system that will transferthe information in the source signal m(t) to the desired destination. Thisusually requires the use of a bandpass signal s(t), which has a bandpassspectrum that is concentrated at ±fc, where fc is selected so that s(t) willpropagate across the (wire/wireless) communication channel. (see Fig 4.1)

• Modulation is the process of imparting the source information onto abandpass signal with a carrier frequency fc by the introduction of amplitudeor phase perturbation or both. This bandpass signal is called modulatedsignal s(t), and the baseband source signal is called the modulating signal.

• Modulation can be visualized as a mapping operation that maps m(t) ontothe bandpass signal s(t). The bandpass signal will be transmitted over thechannel.


Complex envelop representation of bandpass signals

• A mathematically convenient way of representing physical bandpasswaveform is:

v(t) = Re{g(t)ejωct}

– g(t) is called the complex envelop of v(t)– fc is the associated carrier frequency, ωc = 2πfc

– This equation is a lowpass (baseband) to bandpass transformation.– ejωct factor shifts the spectrum of the baseband signal g(t) up to the

carrier frequency fc– g(t) is a complex function of time


• We can also write

v(t) = R(t) cos[ωct + θ(t)]

= x(t) cos ωct− y(t) sin ωct

where

g(t) = x(t) + jy(t) :Cartesian form

= |g(t)|ej〈g(t)

≡ R(t)ejθ(t) :Polar form

where x(t) = Re{g(t)}, y(t) = Im{g(t)},R(t) ∆= |g(t)| =

√

x2(t) + y2(t) andθ(t) ∆= 〈g(t) = tan−1 y(t)

x(t).

• The waveforms g(t), x(t), y(t), R(t), and θ(t) are all baseband waveforms,and except g(t), they are all real waveforms. R(t) is a non negative realwave form.


• Cartesian representation of g(t)

g(t) ≡ x(t) + jy(t),

-x(t) is said to be in-phase modulation associated with v(t).-y(t) is said to be quadrature modulation associated with v(t).

• Polar representation of g(t)

g(t) ≡ R(t)ejθ(t)

-R(t) is said to be amplitude modulation AM on v(t).-θ(t) is said to be phase modulation PM on v(t).


Representation of Modulated Signals

• Modulated signal is just a special application of the bandpass representation

s(t) = Re{g(t)ejωct}

where fc is the carrier frequency and wc = 2πfc.

• Complex function g(t) is a function of the modulating (information) signalm(t); i.e., g(t) = g[m(t)]

• g[·] perform mapping operation on m(t). See Fig 4.1 for examples of g(t) forvarious types of analog modulation schemes.

Scheme g(t)AM Ac[1 + m(t)]FM Acexp(jDf

∫ t−∞m(σ)dσ)

DSB-SC Acm(t)


• We will study different modulation techniques in the next chapter.

• Digitally modulated bandpass signals are obtained when m(t) is a digitalbaseband signal.


Spectrum of Bandpass Signals

If a bandpass waveform is given by


then the spectrum of v(t) is

V (f) =12[G(f − fc) + G∗(−f − fc)],

auto correlation

Rv(τ) =12

Re{Rg(τ)ejωcτ}

and power spectral density (PSD)

Pv(f) =124

[Pg(f − fc) + Pg(−f − fc)]


where G(f) = FT [g(t)], Rg(τ) is the auto correlation of g(t) and P is the PSDof g(t).Proof: see notes.




You may try these problems if you wish to learn this subject more thoroughly.I.e., to get a good grade



Announcements

• Assignment 1 is Due 6 April (Friday)!! at 4.00 PM .

• Don’t forget the Labs!


• How to find Power of Bandpass signals

• Two examples of bandpass signals

• Bandpass filtering


Bandpass Signals


V (f) =12[G(f − fc) + G∗(−f − fc)], :FT

Rv(τ) =12

Re{Rg(τ)ejωcτ} :auto correlation

Pv(f) =14[Pg(f − fc) + Pg(−f − fc)] :PSD

message: If we know the characteristics of baseband complex envelope g(t),then we can easily calculate characteristics of bandpass waveform.


Total Average Power

Pv = < v2(t) >

=∫ ∞

−∞Pv(f)df

= Rv(0)

=12

< |g(t)|2 >

Peak Envelop Power PPEP

of a bandpass signal is the average power that would be obtained if g(t) wereto be held constant at its peak value.

PPEP =12[max |g(t)|]2.

proof: see notes.


Example 1

: Find the spectrum and total average power of an Amplitude Modulated (AM)signal.

Example 2:

Problem 4.2: A double side band suppressed carrier (DSB-SC) signal s(t)with a carrier frequency fc = 3.8MHz has a complex envelop g(t) = Acm(t).Ac = 50V, and the modulation is a 1 kHx sinusoidal test tone described bym(t) = 2 sin(2π1000t). Evaluate the voltage spectrum for this DSB-SC signal.


Bandpass Filtering and Linear Distortion

(read section 4.5)Equivalent Lowpass filter

• A short cut to represent bandpass filter by a lowpass filter (see Fig 4.3)

• let h(t) be the impulse response of the bandpass filter and k(t) be thecorresponding baseband complex envelop. Then

h(t) = Re{k(t)ejωct}

H(f) =12K(f − fc) +

12K∗(−f − fc).


• Let v1(t) = Re{g1(t)ejωcT} and v2(t) = Re{g2(t)ejωcT}are the input andoutput to the bandpass filter, where g1(t) and g2(t) are the basebandcomplex envelope of the input and output respectively.

• Then the equivalent lowpass filter system has the input 12g1(t), out put 1

2g2(t)and the equivalent lowpass filter impulse response 1

2k(t). I.e.,

12g2(t) =

12g1(t) ∗

12k(t)

12G2(f) =

12G1(f)

12K(f).

• Advantageous of equivalent low pass filter

– less complicated to analyze– easy to simulate using a computer program


Linear Distortions

• A linear bandpass filter can cause variations in the phase modulation at theoutput θ(t) = 6 g2(t) as a function of the amplitude modulation of the inputcomplex envelope R1(t) = |g1(t)|. This is called AM to PM conversion.

• Similarly, a filter can cause variations in the amplitude modulation at theoutput R2(T ) of the PM of the input θ1(t). This is called PM to AMconversion.

• Because h(t) is a linear filter, g2(t) will be linear filtered version of g1(t).However, θ2(t) and R2(t) will be a nonlinear filtered version of g1(t), sinceθ2(t) and R2(t) are nonlinear functions of g2(t).

• For distortionless transmission (or to have only linear distortion), we needθ2(t) and R2(t) be linear ( and time shift) function of θ1(t) and R1(t).


• That is, for distortionless transmission of bandpass signals, the channeltransfer function H(f) = |H(f)|ejθ(f) need to satisfy:

– The amplitude response is constant,

|H(f)| = A,

where A is a positive constant.– The derivative of the phase response is a constant, i.e.,

1−2π

dθ(f)df

= Tg

where Tg is called the group delay and θ(f) = 6 H(f).– We can integrate to get

θ(f) = −2πfTg + θ0

where θ0 is a constant (see Fig 4.4 for a illustration).




• Bandpass Filtering

• Linear Distortions

• Bandpass Sampling Theorem


Recall: Representation of bandpass Signals

v(t) = Re{g(t)ejωct}= R(t) cos[ωct + θ(t)]

= x(t) cos ωct− y(t) sin ωct

R(t) = |g(t)|,θ(t) = 6 g(t)

V (f) =12[G(f − fc) + G∗(−f − fc)], :FT

Rv(τ) =12

Re{Rg(τ)ejωcτ} :auto correlation

Pv(f) =14[Pg(f − fc) + Pg(−f − fc)] :PSD


Example: Let a modulated signal,

s(t) = 100 sin(ωc + ωa)t + 500 cos ωct− 100 sin(ωc − ωa)t,

where the unmodulated carrier is 500 cos ωct.

1. Find the complex envelope for the modulated signal. What type ofmodulation is involved? What is the modulating signal?

2. Find the quadrature modulation components x(t) and y(t) for this modulatedsignal?

3. Find the magnitude and PM components R(t) and θ(t) for this modulatedsignal.


Bandpass Filtering and Linear Distortion

(read section 4.5)Equivalent Lowpass filter

• A short cut to represent bandpass filter by a lowpass filter (see Fig 4.3)

• let h(t) be the impulse response of the bandpass filter and k(t) be thecorresponding baseband complex envelop. Then

h(t) = Re{k(t)ejωct}

H(f) =12K(f − fc) +

12K∗(−f − fc).


• Let v1(t) = Re{g1(t)ejωcT} and v2(t) = Re{g2(t)ejωct}are the input andoutput to the bandpass filter, where g1(t) and g2(t) are the basebandcomplex envelope of the input and output respectively.

• Then the equivalent lowpass filter system has the input 12g1(t), out put 1

2g2(t)and the equivalent lowpass filter impulse response 1

2k(t). I.e.,

12g2(t) =

12g1(t) ∗

12k(t)

12G2(f) =

12G1(f)

12K(f).

• Advantageous of equivalent low pass filter

– less complicated to analyze– easy to simulate using a computer program


Linear Distortions

• A linear bandpass filter can cause variations in the phase modulation at theoutput θ(t) = 6 g2(t) as a function of the amplitude modulation of the inputcomplex envelope R1(t) = |g1(t)|. This is called AM to PM conversion.

• Similarly, a filter can cause variations in the amplitude modulation at theoutput R2(T ) of the PM of the input θ1(t). This is called PM to AMconversion.

• Because h(t) is a linear filter, g2(t) will be linear filtered version of g1(t).However, θ2(t) and R2(t) will be a nonlinear filtered version of g1(t), sinceθ2(t) and R2(t) are nonlinear functions of g2(t).

• For distortionless transmission (or to have only linear distortion), we needθ2(t) and R2(t) be linear ( and time shift) function of θ1(t) and R1(t).


• That is, for distortionless transmission of bandpass signals, the channeltransfer function H(f) = |H(f)|ejθ(f) need to satisfy:

– The amplitude response is constant,

|H(f)| = A,

where A is a positive constant.– The derivative of the phase response is a constant, i.e.,

1−2π

dθ(f)df

= Tg

where Tg is called the group delay and θ(f) = 6 H(f).– We can integrate to get

θ(f) = −2πfTg + θ0

where θ0 is a constant (see Fig 4.4 for a illustration).

Example: Prob 4.6:


Bandpass Sampling Theorem

• If we like to process an analog signal digitally (i.e., using a computer/dspetc.), then we have to sample the analog signal.

• If the sampling is carried out at the Nyquist rate or larger (fs ≥ 2B,where B is the highest frequency involved in the spectrum of modulatedsignal), the sampling rate can be too large to be realize. E.g., for satellitecommunication systems, fc = 6GHz, thus fs ≥ 12GHz !.

• However, for bandpass signals, the sampling rate depends only on thebandwidth of the signal not on the absolute frequencies involved.

Theorem: If a (real) bandpass waveform has a non zero spectrum only overa the frequency interval f1 < f < f2, where the transmission bandwidth BT istaken to be the absolute bandwidth BT = f2 − f1, then the waveform may bereproduced from sample values, if the sampling rate is,

fs ≥ 2BT .


E.g., If 6GHz bandpass signal has a bandwidth of 10MHz, then a sampling rateof 20MHz is sufficient instead of 12GHz!.

Bandpass Dimensionality Theorem

Assume that a bandpass waveform has a non zero spectrum only over thefrequency interval f1 < f < f2, where the transmission bandwidth BT = f2 −f1, and BT � f1. The waveform may be completely specified over a T0 secondinterval by

N = 2BTT0

independent pieces of information. N is said to be the number of dimensionsrequired to specify the waveform.


Week7, Lecture 2 — Chapter 4 (Bandpass Signalling Principles)

Announcements

• Quiz will be on 10 May (Worth 15%)

– Covers mainly Chapter 3 and 4– But you should be able to use analytical techniques from chapter 2– Questions may be modified chapter problems!


• Linear Distortions

• Bandpass Sampling Theorem


Linear Distortions (Revisit)

We have following definitions with respect to a bandpass filter H(f) =|H(f)|ejθ(f): Phase Delay

Td(f) ∆=θ(f)−2πf

(14)

Phase delay at carrier frequency fc is called the carrier time delay. The outputcarrier is delayed by Td

∆= Td(fc) relative to the input carrier.

Group Delay

Tg(f) ∆=−12π

dθ(f)df

. (15)

For a distortionless transmission, group delay needs to be a constant over thebandwidth of the bandpass signal. This is also called as the envelope timedelay, since the envelop is delayed by Tg.

Example: Prob 4.6:


Bandpass Sampling Theorem

• If we like to process an analog signal digitally (i.e., using a computer/DSPetc.), then we have to sample the analog signal.

• If the sampling is carried out at the Nyquist rate or larger (fs ≥ 2B,where B is the highest frequency involved in the spectrum of modulatedsignal), the sampling rate can be too large to be realize. E.g., for satellitecommunication systems, fc = 6GHz, thus fs ≥ 12GHz !.

• However, for bandpass signals, the sampling rate depends only on thebandwidth of the signal not on the absolute frequencies involved.

Theorem: If a (real) bandpass waveform has a non zero spectrum only overa the frequency interval f1 < f < f2, where the transmission bandwidth BT istaken to be the absolute bandwidth BT = f2 − f1, then the waveform may bereproduced from sample values, if the sampling rate is,

fs ≥ 2BT .


E.g., If 6GHz bandpass signal has a bandwidth of 10MHz, then a sampling rateof 20MHz is sufficient instead of 12GHz!.


Bandpass Dimensionality Theorem

Assume that a bandpass waveform has a non zero spectrum only over thefrequency interval f1 < f < f2, where the transmission bandwidth BT = f2 −f1, and BT � f1. The waveform may be completely specified over a T0 secondinterval by

N = 2BTT0

independent pieces of information. N is said to be the number of dimensionsrequired to specify the waveform.

Read pages 291-294 (SA4-5)



Announcements


– Covers mainly Chapter 3 and 4– But you should be able to use analytical techniques from chapter 2– Questions may be modified chapter problems!


• Bandpass Sampling

• Mixers, Up converters and Down converters


Bandpass Sampling

• Consider a bandpass signal

v(t) = x(t) cos ωct− y(t) sin ωct.

Suppose both bandpass signals x(t) and y(t) are bandlimited to B. Thenthese two signals can be sampled by fB ≥ 2B separately. (see notes fromwhite board)

• Qualitatively, if we know the carrier frequency fc, we may be able to samplev(t) using equivalent baseband rate. Then, why do we need fs ≥ 4B (i.e.fs ≥ 2BT ) not 2B???

• Even though v(t) consists of x(t) and y(t) we can’t measure them directly;what we can measure is v(t).

• Suppose we sample x(t) using the sampling rate fB, then using baseband


sampling theorem,

x(t) =∞∑

n=−∞x(

nfB

)sin πfB(t− n

fB)

πfB(t− nfB

).

Similarly,

y(t) =∞∑

n=−∞y(

nfB

)sin πfB(t− n

fB)

πfB(t− nfB

).

Thus,

v(t) =∞∑

n=−∞

[

x(nfB

) cos ωct− y(nfB

) sin ωct]sin πfB(t− n

fB)

πfB(t− nfB

).

That is, if we have samples of x(t) and y(t), we can reconstruct v(t).

• But, How can we obtain samples of x(t) and y(t)? -see notes from whiteboard.


• Note that:

cosωct = 1 7→ sinωct = 0

sin ωct = 1 7→ cos ωct = 0

• Trick: Sample v(t) using a sampling rate fB, but adjust the samplinginstance (t ∼ n

fB) such that cos ωct = 1 (then sin ωct = 0) and we have

v( nfB

) = x( nfB

).

• Similarly, v(t) can be sample when sin ωct = 1 to get v( nfB

+ t0) = y( nfB

)where t0 is a constant.

• That is, we have to sample v(t) at different points to obtain samples of x(t)and y(t). In other words, v(t) has to independently sampled to get sampling


points for x(t) and y(t). Therefore, the effective sampling rate of v(t) is

fs ≥ 2fB

fs ≥ 4B

fs ≥ 2BT .

• Another way of sampling is to down convert the signal before sampling.Read pages 291-294 (SA4-5)


Reading for Easter Holiday!!

Read sections 4.8, 4.9 and 4.10. I won’t cover them lectures.



Announcements


– Covers mainly Chapter 3 and 4– But you should be able to use analytical techniques from chapter 2


• Detector Circuits


Detector Circuits

Receivers have different detector circuits to recover the baseband modulationsignal m(t) from the received bandpass signal.


Envelop Detector

• Out put of a envelop detector is proportional to the real envelop R(t) of itsinput.

• If the input bandpass signal is

vin(t) = R(t) cos[ωct + θ(t)],

then the output of the envelop detector is

vout(t) = KR(t),

where K is a constant.

• See Fig. 4.13 a for a simple diode envelop detector (read pages 265,266).This envelop detector is a nonlinear circuit.


• The envelop detector is typically used to detect the modulation on AMsignals.

For AM: g(t) = Ac[1 + m(t)],

where m(t) is the modulation and Ac > 0. Then

vout(t) = kR(t) = k|g(t)|= KAc[1 + m(t)] if |m(t)| < 1

= KAc + KAcm(t)


Product Detector

• A product detector (Fig 4-14) is a mixer circuit that down converts the inputbandpass signal to baseband.

• Input bandpass signal vin(t) = R(t) cos[ωct+θ(t)] is multiplied by a oscillatorsignal A0 cos[ω0t + θ0] with frequency f0 and phase θ0 to get

vI(t) =12A0R(t) cos[(ωc − ω0)t + θ(t)− θ0] +

12A0R(t) cos[(ωc + ω0)t + θ(t) + θ0].

• Then vI(t) is lowpass filtered to get only the down conversion term

vout(t) =12A0R(t) cos[(ωc − ω0)t + θ(t)− θ0]


• If the oscillator has been frequency synchronized, i.e., f0 = fc, then

vout(t) =12A0R(t) cos[θ(t)− θ0]

=12A0 Re{g(t)e−jθ0}.

• If θ0 = 0, then the oscillator is phase synchronized with the in-phasecomponent (i.e., x(t)):

vout(t) =12A0 Re{g(t)}

=12A0 Re{x(t) + jy(t)}

=12A0x(t).


• If θ0 = 900, then the oscillator is phase synchronized with the quadrature-phase component (i.e., y(t)):

vout(t) =12A0 Re{g(t)ejπ/2}

=12A0y(t).


Consider the product detector output:

vout(t) =12A0R(t) cos[θ(t)− θ0]

• This equation is sensitive to AM (R(t)) and PM (θ(t)).

• If the input contain no angle modulation, i.e., θ(t) = 0 and the oscillatorphase θ0 = 0, then

vout(t) =12A0R(t),

which is the envelop detection.


• However if the input signal is only angular modulated, i.e., vin(t) =Ac cos[ωct + θ(t)], and θ0 = 900, then

vout(t) =12A0 Re{Ace−j(θ(t)−π/2}

=12A0Ac sin θ(t).

– In this case the product detector acts like a phase detector with sinusoidalcharacteristics; because the output voltage is proportional to the phasemodulated signal.

– If the phase θ(t) is small i.e. {|θ(t) � π/2,∀t}, then sin θ(t) ≈ θ(t) and

vout(t) ≈12A0Acθ(t)

– In this special case, the phase detector is directly proportional to thephase θ(t).

• The product detector is a linear time-invariant device, in contrast, envelop


detector is a non linear device. The property of being either linear or nonlinear significantly affects the results when two or more components suchas signal plus noise are applied to the input.


Frequency Modulation (FM) detector

• A device, which recovers the FM modulation signal m(t).

• Recall: for FM, R(t) = Ac and θ(t) = Df∫ t−∞m(σ)dσ.

• Output of a FM detector is proportional to instantaneous frequency of theinput (see Fig 4-15). I.e., if vin(t) = R(t) cos[ωct + θ(t)],, then

vout(t) = Kd[ωct + θ(t)]

dt= Kωc +

dθ(t)dt

,

• FM detectors are build based on 3 principles:

1. FM to AM conversion2. Phase-shift or quadrature detection3. zero crossing detection


Slope Detector

• Uses FM to AM conversion principle.

• Block diagram is shown in Fig 4-15, which has bandpass limiter,differentiator and envelop detector.

• Suppose the input is a fading signal with frequency modulation,

vin(t) = A(t) cos[ωct + θ(t)],

where

θ(t) = Df

∫ t

−∞m(σ)dσ.

A(t) represents the envelop that is fading and m(t) is the modulation (e.g.


audio) signal.

Limiter output: vI(t) = VL cos[ωct + θ(t)]

Differentiator output: vdif(t) = −VL[ωct +dθ(t)dt

] sin[ωct + θ(t)]

Envelop detector:vout(t) = | − VL[ωct +dθ(t)dt

]|

= VL[ωct +dθ(t)dt

] using ωc � dθ(t)dt .

Since

θ(t) = Df

∫ t

−∞m(σ)dσ,

dθ(t)dt

= Dfm(t),


and, therefore

vout(t) = VLωct + VLDfm(t).

• Output consist of a dc voltage VLωct and as voltage VLDfm(t) which isproportional to the modulation on the FM signal. We can block the dccomponent by using a capacitor placed in series with the output.

• An example circuit diagram of a differentiator is given in Fig 4-16 ( Recallyour circuit analysis skills!).



You should try these problems to practise your basic skills (i.e., to pass thecourse)




Announcements




• Frequency Modulation Detectors


Frequency Modulation (FM) detector

• A device, which recovers the FM modulation signal m(t).

• Recall: for FM, R(t) = Ac and θ(t) = Df∫ t−∞m(σ)dσ.

• Output of a FM detector is proportional to instantaneous frequency of theinput (see Fig 4-15). I.e., if vin(t) = R(t) cos[ωct + θ(t)],, then

vout(t) = Kd[ωct + θ(t)]

dt= Kωc +

dθ(t)dt

,

• FM detectors are build based on 3 principles:

1. FM to AM conversion2. Phase-shift or quadrature detection3. zero crossing detection


Slope Detector

• Uses FM to AM conversion principle.

• Block diagram is shown in Fig 4-15, which has bandpass limiter,differentiator and envelop detector.

• Suppose the input is a fading signal with frequency modulation,

vin(t) = A(t) cos[ωct + θ(t)],

where

θ(t) = Df

∫ t

−∞m(σ)dσ.

A(t) represents the envelop that is fading and m(t) is the modulation (e.g.


audio) signal.

Limiter output: vI(t) = VL cos[ωct + θ(t)]

Differentiator output: vdif(t) = −VL[ωct +dθ(t)dt

] sin[ωct + θ(t)]

Envelop detector:vout(t) = | − VL[ωct +dθ(t)dt

]|

= VL[ωct +dθ(t)dt

] using ωc � dθ(t)dt .

Since

θ(t) = Df

∫ t

−∞m(σ)dσ,

dθ(t)dt

= Dfm(t),


and, therefore

vout(t) = VLωct + VLDfm(t).

• Output consist of a dc voltage VLωct and as voltage VLDfm(t) which isproportional to the modulation on the FM signal. We can block the dccomponent by using a capacitor placed in series with the output.

• An example circuit diagram of a differentiator is given in Fig 4-16 ( Recallyour circuit analysis skills!).


Quadrature Detector

A quadrature signal is first obtained from the FM signal; then through the useof product detector, the quadrature signal is multiplied with the FM signal toproduce the demodulated signal. ( see block diagram and notes from the whiteboard). This principle is also used in phase locked loops (PLL).


zero crossing Detector

• Recall that for an ideal FM detector

vout(t) = Kωc + kdθ(t)dt

,

i.e., the output is directly proportional to the instantaneous frequency.Therefore, this frequency to voltage characteristics may be obtained directlyby counting the zero crossing of the input waveform.

• See Fig 4.18 for the block diagram and read pages 272-273.

• See notes from the white board



You should try these problems to practise your basic skills (i.e., to pass thecourse)4.22,4.23,4.24, 4.18




Announcements


– Covers mainly Chapter 3 and 4 (including next week lectures)– But you should be able to use analytical techniques from chapter 2


• Phase Locked Loops


Phase Locked Loops (PLL)

• PLL has numerous applications in communication systems:

1. FM detection2. generation of FM signala3. AM detection4. frequency multiplication5. frequency synthesis6. use as a building block in complex digital systems to provide bit

synchronization and data detection.

• A PLL consists of 3 basic components (see Fig 4.19)

1. a phase detector (PD)2. a low pass filter (LPF)3. a voltage controlled oscillator (VCO)

• The VCO is an oscillator that produce a periodic waveform with a frequencythat may be varied about some free running frequency f0, according to thevalue of the applied voltage v2(t).


• When the applied voltage v2(t) = 0, the frequency of the VCO output is thefree running frequency f0.

• The PD, produces an output signal vI(t) that is a function of the phasedifference between the incoming signal vin(t) and the oscillator signal v0(t).

• The filtered signal v2(t) is used to control the frequency of the VCO output.

• The PLL configuration can have two different mode of operations:

1. It can act as a narrowband tracking filter, if the LPF is a narrowband filter.In this mode, the frequency of the VCO will become that of one of theline component of the input signal spectrum. So that, the VCO outputis a periodic signal with a frequency equal to the average frequencyof this input signal component. Once the VCO acquired this frequencycomponent, the frequency of VCO will track the input signal component ifit changes slightly in frequency.

2. In another mode, the bandwidth of the LPF is wider so that VCO can trackinstantaneous frequency of the whole input signal.


• When the PLL tracks the input signal, the PLL is said to be “locked”.

• Read page 274 and understand key words; hold-in(lock) range, pull in(capture) range, and maximum locked sweep rate.

• PLL can be build using either analog circuits or digital circuits ( can beimplemented using DSP’s microprocessors etc).


Analog Phase locked Loops

Read pages 274-273 and notes from the white board.



You should try these problems to practise your basic skills (i.e., to pass thecourse)4.25, 4.26, 4.27(a), 4.29




Announcements

• Quiz will be on 10 May (Worth 15%) at 10.00 AM (duration 50 mins)



• Phase Locked Loops (cont...


Analog Phase locked Loops (APLL)

• Read pages 274-273 (see Fig 4.21) and notes from the white board.

• A multiplier is used as a PD in APLL. This PD has sinusoidal characteristics.

•

input signal vin = Ai sin[ω0t + θi(t)]

VCO signal vo = A0 sin[ω0t + θ0(t)]

where θ0(t) = kv

∫ t

−∞v2(τ)dτ,

where kv is the VCO gain constant.

• Noting that the sum frequency term of the vI(t) does not pass through LPF,

v2(t) =kmAiA0

2sin[θi(t)− θ0(t)] ∗ f(t),


where f(t) is the impulse response of the LPF.

• Overall equation describing the operation of APLL

dθe(t)dt

=dθi(t)

dt− kvkd sin θe(t) ∗ f(t),

where θe(t) = θi(t)− θ0(t) and kd∆= kmAiA0

2 .

– This is a non linear equation, and difficult to solve.– If the loop is locked, error θe(t) is small (this happen when kd is large),

and

sin θe(t) = θe(t).

– Thus, the operating equation for locked mode,

dθe(t)dt

=dθi(t)

dt− kvkdθe(t) ∗ f(t).


This equation is linear and can be converted to a block diagram as in Fig.4.22.

• Question: find the closed loop transfer function?

• A PLL system can be considered as a control system.


Hold-in Range

• If the applied signal has an initial frequency f0, the PLL will acquire a lockand the VCO will track the input signal frequency over some range, providedthat the input frequency changes slowly.

• However, the loop will remain locked only over some finite range offrequency shift. This range is called hold-in range.

• The hold in range depends the overall dc gain of the loop, which includesdc gain of the LPF.

• Maximum hold in range is given by,

∆fh =12π

kvkdF (0),

where F (0) is the dc gain of the LPF.


Example: The input to the PLL is vin(t) = A sin(ω0t + θi) and the LPF isF (s) = s+a

s+b ,. Find (a) steady state phase error, (b) maximum hold in range?



You should try these problems to practise your basic skills (i.e., to pass thecourse)4.25, 4.26, 4.27(a), 4.29



Week 9, Lecture 2 — Chapter 4 (Bandpass Signalling Principles)

Announcements

• Quiz will be on 10 May (Tomorrow) (Worth 15%) at 10.00 AM (duration 50mins)



• Generalized Transmitters and Receivers


Generalized Transmitters

• Transmitters generate the modulated signal at the carrier frequency fc fromthe modulating signal m(t).

• Modulated signal could be represented by

v(t) = Re{g(t)ejωct}= R(t) cos[ωct + θ(t)]

= x(t) cos ωc(t)− y(t) sin ωct.

where g(t) = R(t)ejωct = x(t) + jy(t).

• The complex envelop g(t) is a function of the modulating signal m(t).

• The particular relationship that is chosen for g(t) in terms of m(t) definesthe type of modulation that is used, such as AM, SSB, FM or QM.


• Two different types of generalized transmitters are given in Figures 4.27 and4.28 (read pages 281-283)


Generalized Receivers

The receiver has the job of extracting the source information from the receivedmodulated signal that may be corrupted by noise.

Tuned Radio-Frequency receiver (TRF)

• TRF receivers consists of a number of cascaded high-gain RF bandpassstages that are tuned to the carrier frequency fc followed by an appropriatedetector circuit (such as envelop detector, product detector or a FMdetector).

• TRF is not very popular.

• The ”crystal set” is an example of a single-RF stage receiver that has nogain in the RF stage.

• TRF receivers are often used to measure time-dispersive (multipath)characteristicsof radio channels.


Superheterodyne Receiver

• A popular technique- see Fig 4.29 and read pages 283-289.

• This technique consists of either down-converting or up-converting theinput signal to some conventional frequency band, called the intermediatefrequency IF band and then extracting the information (or modulation) byusing the appropriate detector.

• This basic receiver structure is used for the reception of all types ofbandpass signals, such as television, FM, AM, satellite, cellular and radarsignals.

• RF amplifier has a bandpass characteristics that passes the desired signaland provides amplification to override additional noise that is generated inthe mixer stage. It also provides some rejection of adjacent channel signalsand noise.


• The IF filter is a bandpass filter that selects either up-conversion or down-conversion component. (read pages 284-285)

• The image response is the reception of unwanted signal located at theimage frequency due to insufficient attenuation of the image signal by theRF amplifier (read the Example 4.2, page 285)

Zero-IF receivers

When fLO = fc, the superheterodyne receiver becomes zero-IF ordirect conversion receiver (read page 287 to find out advantageous anddisadvantageous).



You should try these problems to practise your basic skills (i.e., to pass thecourse)4.30, 4.31, 4.32,4.33,4.34, 4.35,4.37,4.38,4.40, 4.41



Week 10, Lecture 1 — Chapter 5 (AM,FM and Digital Modulated systems)

Announcements

• Assignment 2 will be available soon (check the web page). You needto study OFDM and QPSK modulation schemes. Read and understandsection 5-12 of the text book. OFDM is used in European/Australian DTVsystems and some of the wireless LANs.


• AM and DSB-SC


Amplitude Modulation

• Modulated signal:

s(t) = Re{g(t)ejωct}where g(t) = Ac[1 + m(t)].

Thus, s(t) = Ac[1 + m(t)] cos ωct.

• Voltage spectrum of AM

S(f) =Ac

2[δ(f − fc) + δ(f + fc) + M(f − fc) + M(f + fc)].


• AM detection methods:

1. envelop detection (pages 265-266)2. product detector (page 267)3. PLL used for coherent detection of AM (pages 279-280)

• Read section 5.1, and learn the following definitions: percentage of positivemodulation, percentage of negative modulation and modulation efficiency.

• Modulation efficiency is the percentage of total power of the modulatedsignal that conveys information.

Efficiency for AM(E) =< m2(t) >

1+ < m(t) >× 100%

• Attempt the following problems: 5.1-5.6


Double Sideband Suppressed carrier

• DSB-SC is an AM signal that has a suppressed discrete carrier.

s(t) = Acm(t) cos ωct,

where m(t) has zero dc level.

• The spectrum for DSB-SC is identical to that for AM, except that the deltafunctions at ±fc are missing:

S(f) =Ac

2[M(f − fc) + M(f + fc)].

• Modulation efficiency of DSB-SC signal is 100%, since no power is wastedin a discrete carrier.

• An envelope detector cannot be used to demodulate DSB-SC signal,instead a product detector (expensive than an envelop detector) is required.


• If m(t) signal is polar binary data then the signal is BPSK (binary phase shiftkeying).

• A QM signal can be generated by adding two DSB-SC signals, wherethere are two signals m1(t) and m2(t) modulating cosine and sine carriers,respectively.

sQM(t) = Acm1(t) cos ωc(t)−Acm2(t) sin ωc(t).


Costas Loop and Squaring Loop

• The coherence reference for product detection of DSB-SC cannot beobtained by the use of an ordinary phase-locked tracking loop, becausethere are no spectral component at ±fc.

• Since, DSB-SC has a symmetric spectrum with respect to the (suppressed)carrier frequency, we may use costas PLL or squaring loop.

• See Fig 5.3(a) Costas PLL and Fig 5.3(b) Squaring Law Loop.

• Both Loops have one major disadvantage- a 1800 phase ambiguity (readpage 311-312).



You should try these problems to practise your basic skills (i.e., to pass thecourse)5.1-5.6,5.7,5.9

You may try these problems if you wish to learn this subject more thoroughly.I.e., to get a good grade5.8


Week 10, Lecture 2 — Chapter 5 (AM,FM and Digital Modulated systems)

Announcements



• Single Sideband Modulation (SSB)

• Vestigal Sideband Modulation (VSB)

ENGN 3214 Chapter 5 — AM,FM and Digital Modulated systems 264

Single Sideband (SSB)

• An upper single sideband (USSB) signal has zero valued spectrum for |f | <fc.

• A lower single sideband (LSSB) signal has zero valued spectrum for |f | >fc.

• There are numerous ways in which the modulation m(t) may be mappedinto g(t) (see Table 4.1). We are mostly interested in SSB-AM.

• SSB-AM is the most popular type. It is used by military and radio amaturesin HF communication systems.


• An SSB (SSB-AM)signal is obtained by using the complex envelop:

g(t) = Ac[m(t)± jm(t)],

then

s(t) = Ac[m(t) cos ωct± m(t) sin ωct],

where the (−) sign used for the USSB and (+) sign used for the LSSB.

• m(t) denotes the Hilbert transform of m(t),

m(t) ∆= m(t) ∗ h(t),

where h(t) = 1πt and H(f) = FT [h(t)] corresponds to a −900 phase shift.

H(f) =−j if f > 0

j if f < 0.


• A table of Hilbert transforms are given in Section A-7.

• Spectrum of a USSB signal is given in Fig 5.4

• Spectrum of a SSB-AM complex envelope:

g(t) = Ac[m(t)± jm(t)]

G(f) = Ac[M(f)± jFT [m(t)]

= AcM(f)[1± jH(f)].

• Find G(f) for USSB and LSSB cases?

• Spectrum of a SSB-AM signal is

S(f) =12[G(f − fc) + G∗(−(f + fc))].

Substitute for G(f) and convinced yourself that SSB has indeed has onlyone sideband.


• Normalized average power of the SSB signal:

< s2(t) > =12

< g2(t) >

=12

< m2(t) + [m(t)]2 >

= Ac < m2(t) >,

which is the power of the modulating signal < m2(t) > multiplied by thepower gain factor A2

c.

• Normalized peak envelop power PEP:

12

max{|g(t)|2} =12A2

c max{m2(t) + [m(t)]2}.

• SSB-AM signal can be generated using the methods in Fig 5.5 (read pages314-315).


• Note that SSB signal has both AM and PM components:

R(t) = |g(t)| = Ac

√

m2(t) + [m(t)]2

θ(t) = 6 g(t) = tan−1 [±m(t)m(t)

]

.

• We may use superheterodyne receiver to recover the modulation signal(prove this!).

• There are some practical hints and disadvantageous of SSB given in page315.

• Advantageous: SSB has a superior detectd SNR ratio compared to that ofAM and the fact that SSB has one half the bandwidth of AM or DSB-SCsignals.


Vestigal Sideband (VSB)

• Used in American television broadcasting.

• DSB modulation technique takes too much bandwidth and SSB techniqueis too expensive too implement although it takes half the bandwidth.

• A compromise between DSB and SSB is the Vestigal Sideband (VSB).

• VSB is obtained by partial suppression of one of the sidebands of a DSBsignal (see Fig 5.6, read pages 316-318).



You should try these problems to practise your basic skills (i.e., to pass thecourse)5.10,5.12,...,5.20



Week 10, Lecture 3 — PM and FM Modulated systems)

Announcements



• Vestigal Sideband Modulation (VSB)

• Angle Modulation (FM and PM)


Vestigal Sideband (VSB)

• Used in American television broadcasting.

• DSB modulation technique takes too much bandwidth and SSB techniqueis too expensive too implement although it takes half the bandwidth.

• A compromise between DSB and SSB is the Vestigal Sideband (VSB).

• VSB is obtained by partial suppression of one of the sidebands of a DSBsignal (see Fig 5.6, read pages 316-318).

• The modulation m(t) on the VSB signal can be recovered by a receiver thatuses product detection or, if a large carrier is present by use of a envelopedetector.


• For recovery of undistorted modulation, the transfer function Hv(f) for VSBfilter must satisfy the constraint

Hv(f − fc) + Hv(f + fc) = C, for |f | < B

where C is a constant and B is the bandwidth of the modulation. (proof:see notes from the white board)


Phase Modulation and Frequency Modulation

Representation of PM and FM signals:

• PM and FM are special case of angle modulation, where the complexenvelope:

g(t) = Acejθ(t)

where the real envelope R(t) = |g(t)| = Ac.

• Note that the g(t) is a nonlinear function of the modulation m(t).

• The angle modulation signal is given by

s(t) = Ac cos[ωct + θ(t)].


• For PM, the phase is directly proportional to the modulating signal,

θ(t) = Dpm(t),

where Dp is the phase sensitivity of the modulation (radians/volt).

• For FM, the phase is proportional to the integral of m(t)

θ(t) = Df

∫ t

−∞m(σ)dσ,

where Df is the frequency deviation constant (radians/volt-second).

• Comparing the last two equations, we can see that in a PM signal, there isalso FM!. Similarly, in a FM signal there is PM present as well.

• That is, if mp(t) is the modulation signal for PM, then

mf(t) =Dp

Df

[dmp(t)dt

]


is the equivalent modulation signal for FM.

• Similarly, if mf(t) is the modulation signal for PM, then

mp(t) =Df

DD

∫ t

−∞m(σ)dσ,

is the equivalent modulation signal for PM.

• Therefore, we may use a FM modulator to modulate PM signal and also PMmodulator for FM modulation. (see Fig 5.7).

• Direct PM and Fm circuits are given in Fig 5.8.


Instantaneous Frequency fi(t)

• For a bandpass signal s(t) = R(t) cos(ωct + θ(t)) is defined as

fi(t) =12π

d[ωct + θ(t)]dt

= fc +12π

dθ(t)]dt

.

• For FM,

fi(t) = fc +12π

Dfm(t).

Note that the instantaneous frequency varies about carrier frequency fc ina manner that is directly proportional to the modulating signal (hence theterm FM).

• The instantaneous frequency should not be confused with the termfrequency as used in the spectrum of the FM signal. The spectrum is given


by the Fourier Transform of s(t) and is evaluated by looking at s(t) over theinfinite time interval (−∞ < t < ∞).

• The spectrum tells us what frequencies are present in the signal (on theaverage) over all time. On the other hand, the instantaneous frequency isthe frequency present at a particular instant of time.


The Frequency Deviation

from the carrier frequency:

fd(t)∆= fi(t)− fc =

12π

dθ(t)dt

.

Peak Frequency Deviation

∆F = max12π

dθ(t)dt

.

Peak Phase Deviation

∆θ = max[θ(t)].


Phase Modulation Index

βp∆= ∆θ.

Frequency Modulation Index

βf∆=

∆FB

.



You should try these problems to practise your basic skills (i.e., to pass thecourse)5.22, 5.23, 5.24,



Week 11, Lecture 1 — PM and FM Modulated systems)

Announcements

• Assignment 2 is now available (check the web page).


• Spectra of angular modulated signals

• Frequency Division Multiplexing (FDM)

• FM Stereo Broadcasting


Spectra of Angular Modulated Signals

• For bandpass signals

S(f) =12[G(f − fc) + G∗(−f − fc)],

where G(f) = FT [g(t)] = FT [Acejθ(t)].

• For FM and PM, g(t) is a nonlinear function of m(t), thus it is difficult to finda general formula that relate M(f) to G(f).

• See example 5.2- which calculates the spectrum of a PM/FM signal withsinusoidal modulation.

Carson’s Rule

In an angular modulated signal, 98% of the total power is contained in thebandwidth

BT = 2(β + 1)B


where β is either the phase modulation index or the frequency modulationindex and B is the bandwidth of the modulating signal.


Narrowband Angle Modulation

• When θ(t) is restricted to small values, say |θ(t)| < 0.2rad, the complexenvelope g(t) = Acejθ(t) may be approximated by the first two items of theTaylor’s series [ex = 1 + x for |x| � 1],

g(t) = Ac[1 + jθ(t)],

then

s(t) = Re{g(t)ejωct}= Ac cosωct

︸︷︷︸

−Acθ(t) sin ωct︸︷︷︸

• The the narrowband angle modulated signal consists of two parts: adiscrete carrier term Ac cos ωct and a sideband term Acθ(t) sin ωct.

• This signal is similar to AM type signal, except that the sideband term is 900

out of phase with the discrete carrier term. (see Fig. 5-12.1 and read pages328-333)


• Spectrum of the narrowband angular modulated signal,

S(f) =Ac

2{δ(f − fc) + δ(f + fc) + j[Θ(f − fc)−Θ(f + fc)]},

where

Θ(f) = FT [θ(t)] =

{

DpM(f) for PM signal,Df

j2πf for FM signal.


Frequency Division Multiplexing (FDM)

• FDM is a technique for transmitting multiple messages simultaneously overa wideband channel by first modulating the messages signal onto severalsubcarriers and forming a composite baseband signal that consists of thesum of these modulated subcarriers. This composite signal may then bemodulated onto the main carrier (see Fig 5.17).

• Any type of modulation, such as AM, DSB, SSB, PM, FM, .. can be used.

• Composite signal spectrum must consist of modulated signals that donot have overlapping spectra; otherwise, crosstalk will occur between themessage signals at the receiver output.

• The received FDM signal is first demodulated to reproduce the compositebaseband signal. Then it is passed through filters to separate into individualmodulated carriers. Then the sub-carriers are demodulated to reproducethe message signals (see FIg 5-17.c).


FM stereo Broadcasting

• Read pages 335-337 (US system) see Fig 5.18

• You shoud find out specification for Australian FM broadcasting Technicalstandard (search the WEB)



You should try these problems to practise your basic skills (i.e., to pass thecourse)5.26, 5.27, 5.28,5.30, 5.31, 5.32,5.35-5.44



Week 11, Lecture 2 — Binary Modulatation)

Announcements



• Binary modulated signals

1. On-Off Keying2. Binary Phase Shift Keying3. Frequency Shift Keying


Digitally Modulated Bandpass Signals

• Digitally modulated bandpass signals are generated using the complexenvelops (i.e., g(t)) for AM, PM, FM or QM signaling.

• For digital modulated signals, the modulating signal m(t) is a digital signalgiven by the binary or multilevel line codes (see chapter 3)


Binary Modulated Bandpass Signaling

Most common binary bandpass signaling techniques are

1. On-Off Keying (OOK) (also called Amplitude shift keying-ASK), whichconsists of keying (switching) a carrier sinusoidal on and off with a unipolarbinary signal. OOK is identical to unipolar binary modulation on a DSB-SCsignal. (see Figure 5-19)

2. Binary Phase Shift Keying BPSK, consists of shifting the phase of asinusoidal carrier 00 or 1800 with a unipolar binary signal. BPSK isequivalent to PM signaling with a digital waveform or to modulating a DSB-SC signal with a polar digital waveform.

3. Frequency Shift Keying FSK, consists of shifting the frequency of asinusoidal carrier from a one frequency (e.g., corresponding to binary 1)to another frequency (e.g., corresponding to binary 0), according to thebaseband digital signal. FSK is identical to modulating a FM carrier witha binary signal.


On-Off Keying (OOK,ASK)

• OOK signal is (identical to DSB-SC with unipolar binary modulation)

s(t) = Acm(t) cos ωct

where m(t) = 0 or 1.

• Complex envelopg(t) = Acm(t)

• PSD (power spectra density) of g(t) is proportional to that for the unipolarsignal,

Pg(f) =A2

c

2[δ(f) + Tb(

sin πfTbπfTb

)2]

where R = 1/Tb is the bit rate. The null to null Bandwidth is BT = 2R.

• OOK may be detected by using either envelop detector (non coherentdetection) or a product detector (coherent detection), because it is a form ofAM signaling.


• For optimum detection of OOK- that is to obtain the lowest bit error rate(BER) when the OOK signal is corrupted by additive white gaussian noise(AWGN)- product detection with matched filter processing is required. ( Youwill study the matched filter processing and effects of noise in final yearcommunication subjects).


Binary Phase Shift Keying (BPSK)

• BPSK signal (similar to PM signal with digital modulation)

s(t) = Ac cos[ωct + Dpm(t)]

where m(t) is a polar baseband data signal. Assume m(t) has peak valuesof ±1 and a rectangular pulse shape.

• Also, we can write

s(t) = Ac cos(Dpm(t)) cos ωct−Ac sin(Dpm(t)) sin ωct.

Since m(t) = ±1,

s(t) = Ac cos(Dp) cos ωct︸︷︷︸

−Acm(t) sin(Dp) sin ωct︸︷︷︸

.

Note that the first term is only a pilot carrier term and the second term is thedata term.


• By selecting Dp = π/2 radians, the BPSK signal becomes

s(t) = −Acm(t) sin ωct.

This is equivalent to DSB-SC signaling with a polar baseband datawaveform.

• The complex envelope for BPSK is g(t) = jAcm(t)

• The PSD of the BPSK complex envelop (using eq 3.41 ) (also see Fig 5.20b)

Pg(f) = A2cTb

(sin πfTbπfTb

)2

• Null to null bandwidth for BPSK, BT = 2R where R = 1/Tb.

• To detect BPSK signal, costas loop or squaring loop can be used (read p344). For optimum BPSK detection, matched filter processing has to beused.


Frequency Shift Keying

• FSK signal iss(t) = Ac cos[ωct + θ(t)]

where θ(t) = Df∫ t−∞m(σ)dσ, m(t) is a baseband digital signal. If m(t) is a

multilevel digital signal, then we have multilevel FSK signal.

• Similar to FM signals, it is difficult to evaluate the spectrum of FSK signal.

• Approximate transmission bandwidth is given by the carson’s rule, BT =2(β + 1)B, where B is the bandwidth of the digital modulation signal m(t).



You should try these problems to practise your basic skills (i.e., to pass thecourse)5.45, 5.46, 5.47,5.49, 5.50, 5.51, 5.54, 5.56



Week 11, Lecture 3 — Multilevel Modulated Signaling

Announcements


• Solutions to the Quiz is in the Library

• No lecture on Tuesday Week 13.


• Multilevel Modulated signals

1. M-ary phase shift keying2. Quadrature amplitude modulation


Multilevel Modulated Bandpass Signals

With multilevel signaling, digital inputs with more than two levels are allowedon the transmitter input (recall from Section 3.5)


Quadrature phase-shift keying (QPSK) and M-ary phase shift keying (MPSK)

• If the transmitter is a PM transmitter with M level digital modulation signal,MPSK is generated at the transmitter output. g(t) = ejθ(t)

• When M = 4, suppose the multilevel values at the output of the converteris −3, −1, 1, 3V. These levels might correspond to 4 different PSK values(θ(t) = θi, i = 1, 2, 3, 4)- (say 0, 90, 180, 270 respectively). This correspondsto the constellation diagram in Fig 5.30(a).

• MPSK can also be generated using two quadrature carriers modulated bythe x and y components,

g(t) = Acejθ(t)

= Ac cos θi + j cos θi

= xi + jyi, i = 1, . . . , M.

Figure 5.30(b) shows the constellation diagram for the case of M = 4.


• For rectangular data pulses the envelop of the QPSK signal is constant(i.e., there is no AM on the signal). The rectangular shaped data pulsesproduce (sin x/x)2-type power spectrum for the QPSK signal that has largeundesirable sidelobes (see Fig 5.33). These side lobes can be eliminated ifthe data pulses have raised cosine pulse shape. However, this produce AMon QPSK signal (read page 354).


Quadrature Amplitude Modulation (QAM)

• The complex envelop for QAM is

g(t) = x(t) + jy(t)

= xi + jyi, i = 1, . . . , M

= Riejθi

where (xi, yi) or (Ri, θi), i = 1, . . . , M defined the constellation.

• For example 16-QAM constellation is shown in Figure 5.32.- see notes fromwhite board.

• The waveforms of I and Q components are

x(t) =∑

n

xnh(t− nD

)


and

y(t) =∑

n

ynh(t− nD

)

where D = R/l is the symbol rate, and (xn, yn) denotes one of the permitted(xi, yi) values during the symbol time that is centered on t = nTs = n/D.h(t) is the pulse shape that is used for each symbol.

• If we use the rectangular pulse shape, then the bandwidth of the QAM islarge.

• In some applications, the timing between x(t) and y(t) components is offsetby Ts/2 (half a symbol time = 1/2D). In that case,

y(t) =∑

n

ynh(t− nD− 1

2D).

• One popular type of offset signaling is offset QPSK (OQPSK), which isidentical to offset QAM with M = 4.


• A special case of OQPSK when h(t) has a sinusoidal type of pulse shapeis called minimum-shift keying MSK.

• Offset signaling help to reduce the AM present in the modulated signal whennon-rectangular shape pulses are used.

• It is common to use root raised cosine shape pulses to transmit M-levelmodulated signals. This will reduce the bandwidth used (compared torectangular pulse) and also eliminate the inter symbol interference (ISI). Inpractice a square-root raised cosine (SRRC) filter is used at the transmitter,along with another SRRC filter at the receiver.

• If the overall pulse shape satisfies the raised cosine roll off filtercharacteristics, then the absolute bandwidth of the M-level modulatingsignal is

B =12(1 + r)D

where D = R/l and r is the roll off function of the filter characteristics. Alsonote that the transmit bandwidth BT = 2B.


• Above equations are applicable to QAM and MPSK . Thus they can be usedto compare these schemes with respect to bandwidth and symbol rate.

• Spectral Efficiency

η =data rate(bit rate)

Transmit Bandwidth=

RBT



You should try these problems to practise your basic skills (i.e., to pass thecourse)5.58, 5.60, 5.61, 5.62, 5.63, 5.65, 5.67



Week 12, Lecture 1 — Spread Spectrum Systems

Announcements


• Solutions to the Quiz is in the Library

• No lecture on Tuesday Week 13.


• Spread Spectrum Systems

1. Direct Sequences Spread Spectrum Systems2. Frequency Hopping Spread Spectrum Systems


Spread Spectrum Systems

• Spread spectrum (SS) systems have the following capabilties:

– multiple-access capability– antijam capability– interference rejection– low probability of intercept capability

• Multiple access capability is needed in cellular telephone and personalcommunications, where many uses share a same band of frequencies,because there is not enough available bandwidth to assign a personalfrequency channel to each user.

• SS can provode simultaneous use of a wide frequency band via codedivision multiple acess (CDMA) techniques (instead of band sharing - recallTDMA and FDMA).

• SS systems are used in some of the recent wireless LANs.


• A SS system must be satisfy two criteria:

1. The bandwidth of the transmitted signal s(t) needs to be much greaterthan that of the message m(t).

2. The relatively wide bandwidth of s(t) must be caused by an independentmodulation waveform c(t) called the spreading signal, (which must beknown to the receiver)

• The SS signal iss(t) = Re{g(t)ejωct}

where g(t) = gm(t)gc(t), and gm(t), gc(t) are usual type of modulationcomplex envelop functions that generate AM, PM, FM and so on. The SSsignals are classified by the type of mapping function used for gc(t).

• Most common type of SS signals:

1. Direct Sequence (DS-SS): A DSB-SC spreading modulation is used (i.e.,gc(t) = c(t), c(t) is a polar NRZ waveform).

2. Frequency Hopping (FH): gc(t) is of the FM type where there are M =2k hop frequencies determined by the k-bit words obtained from the


spreading code waveform c(t).


Direct Sequence Spread Spectrum Systems (DS-SS)

• Assume m(t) comes form a digital source and is a polar waveform having±1. Also consider the BPSK modulation where gm(t) = Acm(t).

• For DS-SS, gc(t) = c(t) - DSB-SC type modulation with c(t) being a polarNRZ waveform. Thus, SS complex envelop signal becomes

g(t) = Acm(t)c(t).

• The resulting bandpass signal s(t) = Re{g(t)ejωct} is called BPSK-DS-SS(see Fig 5.39(a)).

• The spreading code c(t) is generated bu using a pseudonoise (PN) codegenerator (see Fig 5-39(b)), where the value of c(t) is ±1.

• The pulse width of c(t) is denoted by Tc and is called a chip interval.


• The code generator uses a modula-2 addder and r shift register stages thatare clocked every Tc seconds. Also c(t) is periodic.

• Feedback taps from the stage of the shift registers and modula-2 adder arearranged so that c(t) waveform has a maximum of period of N chips, whereN = 2r − 1. This type of PN code generator is said to generate a maximumlength sequence or m-sequence wave form. (see e.g., from the notes)

• Properties of maximum length sequence:

1. In one period, the number of 1’s is always one more than the number of0’s.

2. The modula-2 sum of any m-sequence, when summed chip by chip witha shifted version of the same sequence, produces another shifted versionof the same sequences. (check this!)

3. If a window of width r is slide along the sequence for N shifts, then allpossible r-bit will appear exactly once, except for all 0 r-bit word.

4. If the 0’s and 1’s are represented by −1 and 1 v, the auto-correlation of


the sequence is

Rc(k) =

{

1 k = lN,−1/N otherwise.

and the autocorrelation of the waveform c(t) is

Rc(τ) =∞∑

l=−∞

(1 +1N

)∧

(τ − lNTc

Tc)− 1

N.

Read pages 373-374 and see Fig 5.40. (shows the capabilities of SSsystems).


• SS technique is also used in CDMA systems.

– Each user is assigned a spreading code such that the signals areorthogonal.

– Thus, multiple SS signals can be transmitted simultaneously in the samefrequency band, and yet the data on a particular SS signal can bedecoded by a receiver, provided that the receiver uses a PN code thatis identical to and synchronized with the particular SS signal that is to bedecoded.

• Read page 377- Frequency Hopping SS.


engn 3214 telecommunications

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