1

CHAPTER

19

Communication Systemshis chapter introduces the foundations of electrical communication systems,emphasizing basic analog communications ideas. An overview of digitalcommunications concepts is provided in the last section.The subject of electrical communications is one that touches everyone’s life:telephones, TV and radio have been a part of our lives for many decades; today,

new means of communications are becoming as pervasive as the traditional ones.Computer networks, satellite weather systems, personal communication systems (pagers,cellular phones, etc.) are becoming essential parts of our everyday lives. The aim of thischapter is to present the basic mathematics of spectrum analysis, which are thefoundations of all communication systems, and the basic operation of amplitude- andfrequency-modulation systems. The explanation of these concepts is supplemented bythe use of computer-aided tools. In addition, the chapter also includes an overview ofdifferent types of commonly used communication systems.

Upon completing the chapter, you should:a. Be familiar with the most common types of communication systems in block

diagram form.b. Be capable of performing spectral analysis of simple signals using analytical and

computer-aided tools.c. Understand the principles of AM modulation and demodulation, and perform basic

calculations and numerical computations on AM signals.d. Understand the principles of FM modulation and demodulation, and perform basic

calculations and numerical computations on FM signals.

T

2

19.1 INTRODUCTION TO COMMUNICATION SYSTEMSThe modern era of communications began with the telegraph and the Morse

code (http://www.soton.ac.uk/~scp93ch/refer/alphabet.html#punctuation), and rapidlymoved towards radio and television. Table 19.1 summarizes some of the major dates inthe history of communication systems.

Table 19.1 A Brief History of CommunicationsDate Event1838 Samuel F. B. Morse demonstrates telegraph1876 Alexander Graham Bell patents the telephone1897 Marconi patents a complete wireless telegraph system1906 Lee DeForest invents the triode amplifier1915 Bell System completes a transcontinental telephone line1918 B.H. Armstrong perfects the superheterodyne receiver1937 Alec Reeves conceives pulse code modulation1938 Television broadcasting begins

WW II Radar and microwave systems are developed1948 The transistor is invented; Claude Shannon publishes

Mathematical Theory of Communications1956 First transoceanic telephone cable1960 First communications satellite, Telstar I, is launched

1962-66 High speed digital communications1965 Mariner IV transmits pictures from Mars to Earth1970 Color TV1970 Commercial relay satellite telecommunications1975 Intercontinental computer communication networks

Information, modulation and carriers

The purpose of communication systems is to communicate information; the fourmost common sources of information are: speech (or sound), video and data. Regardlessof the source, the information that is transmitted and received in a communication systemconsists of a signal, encoding the information in some appropriate fashion. Figure 19.1depicts the general layout of a communication system: an input transducer (e.g., amicrophone) converts the input message into a message signal (e.g., a time varyingvoltage) that is transmitted over a channel, and converted by a receiver into an outputsignal. An output transducer (e.g., a loudspeaker) converts the received signal into anoutput message (e.g.: sound). The transmitter performs a very important function oncommunication signals by encoding the signals in some fashion making use of a carriersignal. The information is contained in a so-called modulating signal that modulates acarrier signal. For example, in FM radio the modulating signal consists of speech andmusic, and the carrier is a sinusoidal wave of pre-determined frequency, much higherthan the modulating signal frequency. Table 19.2 summarized the frequency bandallocation and typical applications in each frequency band. More detailed informationcan be found at http://www.ntia.doc.gov/osmhome/allochrt.pdf.

There are two principal reasons for the use of a very broad spectrum of carrierfrequencies. The first is that allowing for a broad spectrum permits many simultaneoususers to broadcast information at different frequencies without interference amongdifferent transmissions; the second is that depending on the frequency of the carrier, theelectromagnetic waves that are transmitted have different propagation characteristics.Thus, different carrier frequencies are better suited for propagating over long distancesthan others. Table 19.2 summarizes the frequency spectrum allocations used today.

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Input

transducer

Transmitter Channel Receiver Output

Transducer

Carrier

Input message Message signal Transmitted signal Received signal Output signal Output message

Figure 19.1 Block diagram of a communication system

Table 19.2 Frequency bandsFrequency

BandName medium Applications

3-30 kHz Very Low Frequency(VLF)

Wire pairs Long-range navigation, sonar.

30-300 kHz Low Frequency (LF) Wire pairs Navigational aids, radio beacons.300-3000

kHzMedium Frequency

(MF)CoaxialCable

Maritime radio, directionfinding, Coast Guard,commercial AM radio.

3-30 MHz High Frequency (HF) CoaxialCable

Search and rescue, aircraftcommunications with ships,

telegraph, telephone andfacsimile.

30-300MHz

Very High Frequency(VHF)

CoaxialCable

VHF television channels, FMradio, private aircraft, air traffic

control, taxi cabs, police.0.3-3 GHz Ultra High Frequency

(UHF)CoaxialCable

Waveguide

UHF television channels,surveillance radar, satellite

communications.3-30 GHz Super High

Frequency (SHF)Waveguide Satellite communications,

airborne radar, approach radar,weather radar, land mobile.

30-300 GHz Extremely HighFrequency (EHF)

Waveguide Railroad service, radar landingsystems, experimental.

> 300 GHz Optical frequencies Opticalfiber

Wideband data, experimental.

Classification of communication systems

Communication systems can be classified into two basic families, based on thenature of the message signal: analog communication systems and digitalcommunication systems. In this chapter we shall primarily focus on analogcommunications, although it should be remarked that digital communications are takingan increasingly prominent role even in the most common applications1. Anotherclassification may be made based on the type of transmission: light wave vs. radiofrequency, or RF transmission, as is explained in the next section. A third classificationis that of carrier vs. direct baseband transmission system. This latter classification isbased on whether the signal of interest is directly transmitted (e.g., as in the case of thetelegraph), or whether the signal modulates a carrier wave, as in the case of AM and FMradio transmission.

1 An example of this phenomenon is the changeover from analog to digital systems incellular telephony. Both systems coexist at the present time, but it is reasonable toforecast that in a few years all personal communication systems will be digital.

4

Communication channels

The modulated transmitted signal can reach the receiver in a number of ways.In some cases, communication systems are hard wired. Examples of this configurationare local area computer networks, local telephone systems and local cable TV networks.Depending on the frequency range, the transmitted signal can be carried by twisted wirepair, coaxial cable, waveguides, or optical fiber. However, in most communicationssystems, after the signal had been carried over a wire or cable, it is eventually broadcastover air by an antenna, to be received by a similar antenna elsewhere. Figure 19.2 depictssome typical communication system components.

Figure 19.2 Communication system components; clockwise from top left: coaxialcables; RF connectors; detail of coaxial cable; RF cabling components; monopole

antenna; optical fiber bundle.

The range of transmission can be significant – consider that signals can bereceived from the far reaches of the solar system via radio astronomy. The mostcommon means of transmission of communication signals is via the broadcast of radiofrequency waves over the air. To understand the different types of wave propagation,we need to briefly explain the geometry of the earth’s atmosphere. With reference toFigure 19.3, the atmosphere is composed of layers, of which the troposphere and theionosphere are the most important for radio wave transmission. The troposphere (up toabout 20 km above sea level) is where the earth’s air is contained; air density,temperature and humidity decrease with increasing altitude. The propagation of radiowaves in air depends on various properties of the medium. The speed of propagation ofelectromagnetic waves and the refractive index of the medium (causing the deflection ofthe wave) increase with altitude; as a consequence, radio waves tend to bend backtowards the earth as they propagate through the troposphere.

The ionosphere is so called because of the ionization of the small amounts of airpresent at these altitudes (50 to 600 km); electromagnetic waves reaching the ionospheremay propagate through it with some losses (attenuation), or may be reflected down toearth, depending on the frequency of the transmissions. In general, frequencies above 30MHz will propagate through the ionosphere, and are therefore suitable for spacecommunications (see Web reference to radio astronomy above).

To achieve long range communications over the earth, use is made of so-calledsky waves. These are waves that are reflected by the ionosphere, and permit reachingpoints beyond the horizon. The frequencies used for these waves are below 30 MHz topermit reflection from the ionosphere. Short-wave radio makes use of the sky wave.Troposheric waves can also propagate beyond the horizon, but instead of beingreflected, as in the case of sky waves, they bend around the earth because of diffusion(scattering). Direct waves are used in line-of-sight transmission , where the transmitterand receiver are in the line of “sight” of one another. The earth’s curvature is the primarylimitation to the distance of such transmissions; however, due to reflections from the

5

ground, and to ground and surface waves, this transmission can achieve greater distancesthan one would calculate simply based on the earth’s curvature and the height of theantennas.

Coaxial cables are very commonly used for the transmission of radio-frequencywaves over short to medium distances, typically in the frequency range between afraction of a MHz to hundreds of MHz. Coaxial cable consists of a copper core,surrounded by an insulating layer, in turn surrounded by a conductive (ground) layer andby an external protective sheath. Today the most common example of the use of coaxialcables is the distribution of cable television signals from the receiving station toindividual homes.An increasingly common type of communication systems is based on light wavetransmission. Light is also electromagnetic radiation, but at much higher frequenciesthan radio waves. The main drawback in the use of light as a carrier is that it needs to beenclosed in a guide to travel over significant distances; optical fibers are used to achievesuch transmission. An optical fiber consists of a hair-thin strand of glass, the core,surrounded by a protective layer, the cladding. Snell’s law of optics ensures that, if lightenters the fiber at a sufficiently low angle of incidence, the transmission benefits fromtotal internal reflection, confining the light signal to the core with minimal losses. High-speed computer communications networks are increasingly making use of optical fibers.

Direct wave transmission with repeater

Ionosphere

Sky wave

Earth

Figure 19.3 Propagation of radio frequency waves

19.2 SPECTRAL ANALYSISSignal spectra.

You know from Chapters 4 and 6 that signals can be represented both in time-domain and in frequency-domain form. The phasor notation introduced in Chapter 4 isthe starting point of the frequency domain representation, or spectral representation ofsignals: a phasor describes a sinusoidal signal’s amplitude and phase as a function offrequency. The spectrum of a signal consists of the frequency domain representation ofthe voltage signal. For example, the signal x (t) = A1 cos( ω1t + φ1 ) only contains a single

sinusoidal frequency, ω1, and its spectrum therefore consists of a pair spectral lines atthe frequency ±ω1. Figures 19.4(a) and (b)-(c) depict the representation of a sinusoidalsignal in the time domain and in the frequency domain. Note that to completely representthe frequency domain signal one needs to consider both magnitude and phase, as wasdiscussed in Chapter 4. Note also that the spectrum of the signal exists at both positiveand negative frequencies2; this is a mathematical consequence of the definition of theFourier transform, as will be shown soon.

2 Although negative frequencies have no physical significance, the mathematical form of theFourier Series and Transform requires that we consider the spectrum of the signal at both positiveand negative frequencies.

6

-0.02 -0.015 -0.01 -0.005 0 0.005 0.01 0.015 0.02

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

time, s

x(t),

V

(a)

-200 -150 -100 -50 0 50 100 150 200

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

-200 -150 -100 -50 0 50 100 150 200

-3

-2

-1

0

1

2

3

Ang

le(X

(f)),

Rad

Figure 19.4 Time-domain (a) and frequency-domain ((b) magnitude, (c) phase)representation of a sinusoidal voltage (amplitude: 1 V-peak; phase: 0 rad)

Example 19.1 Sinusoidal signal spectrum

Problem:Generate the spectrum of a signal consisting of the addition of two unity-amplitude sinewaves for different frequencies and phases. Plot the time-domain sum and the spectrumof the sum.

Solution:

Known quantities:Sine wave amplitude, frequency and phase.

Find:Plot the time domain sum of the signals and the frequency domain spectrum.

Schematics, diagrams, circuits and given data.

ω1 = 300 rad/s; ω2 = 500 rad/s; φ1 = 0 rad; φ2 = π/4 rad/s;

Assumptions:None.

7

Analysis:The time-domain signals x1(t) and x2(t) and their sum are shown in Figure 19.5(a). Figure19.5(b) depicts the frequency spectrum of the sum signal.

-0.02 -0.015 -0.01 -0.005 0 0.005 0.01 0.015 0.02

-1.5

-1

-0.5

0

0.5

1

1.5

time, s

x(t),

V

Time-domain sum of sinusoidal signals

(a)

-400 -300 -200 -100 0 100 200 300 400

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

-400 -300 -200 -100 0 100 200 300 400

-3

-2

-1

0

1

2

3

Ang

le(X

(f)),

Rad

(b)Figure 19.5 Time-domain (a) and frequency-domain (b) representation of the sum of two

sinusoidal voltages

Comments:Note that the signal amplitude in the time domain is divided between two spectral lines ateach signal component frequency, and at the corresponding negative frequency; thus,signal power is preserved. The phase angle (at the positive frequencies) is shown to be−π/2 for the signal x1(t) because the signal is a sine wave (in Chapter 4 we defined thecosine as the reference function, with zero phase angle); thus, x2(t) has phase angle −π/2+ π/4 = −π/4.

Focus on Computer-Aided ToolsA Matlab file used to generate the figures shown in this example may be found on thebook website, http://www.mhhe.com/engcs/electrical/rizzoni. You may wish to explorefurther by changing the signal phases and frequencies, or by adding a third sinusoid.

8

Periodic signals: Fourier series

Periodic signals (see Chapter 4 for an introduction to signal waveforms) can berepresented by the infinite summation of sinusoidal signals, as was explained in the Box“Fourier Series” in Chapter 6. This section explores the idea of the Fourier seriesrepresentation of a periodic signal more formally.

Let the signal x(t) be periodic with period T, that is,x (t) = x (t + T ) = x (t + nT ) n = 0,1,2, 3... (19.1)

An example of a periodic signal is shown in Figure 19.6.

T 2T 3T t

x(t)

Figure 19.6 A periodic signal

The signal x(t) can be expressed by means of an infinite summation of sinusoidalcomponents according to one of the following three (equivalent) representations:

In each of these expressions, the period, T, is related to the fundamental frequency of thesignal, ω0, by:

ω0 = 2πf0 =2πT

rad/s (19.6)

It is straight forward to show that (19.4) is equivalent to (19.3), by expanding (19.4):

c n sin n2πT

t − θn

= c n cos θn( )sin n

2πT

t

+ c n sin θn( )cos n

2πT

t

where an = cn sin θn( ) and bn = cn cos θn( ); one can verify that the equality holds if

an2 + bn

2 = cn and bn

an

= tan θn( ). (19.7)

x (t) = a0 + an cosn2π

Tt

n =1

∞∑ + bn sin

n2πT

t

n=1

∞∑ Sine-cosine (quadrature) representation (19.3)

(19.2)

x (t) = c 0 + c n sinn2π

Tt + θn

n=1

∞∑

or

x (t) = c 0 + c n cosn2π

Tt − ψn

n=1

∞∑

Magnitude and phase representation (19.4)

x (t) = γ n ejn

2π

Tt

n =−∞

∞∑ Complex representation (19.5)

9

Similarly, one can show that (19.4) is equivalent to (19.3) if

an2 + bn

2 = cn and bn

an

= tan ψ n( ) (19.8)

Figure 19.7 is a graphical representation of the equivalence of the {an, bn} and {cn, θn}forms of the Fourier series. Equations (19.7-8) permit easy conversion between the twoforms of the Fourier series.

ncb = c sin ( θ )n n n

a = c cos (θ ) n n n

nθ

Figure 19.7 Quadrature and magnitude-phase representation of sinusoidal signals

In each of the above representations, ω0 = 2πf0 =2πT

is called the

fundamental frequency (in units of radians per second), and the frequencies 2ωo, 3ωo,

4ωo, etc. are called its harmonics.

The computation of the {an,bn} or {cn, φn} coefficients for the periodicfunction x(t) is based on the following formulas:

a0 =1

Tx (t )dt

0

T

∫ (19.9)

an =2

Tx (t ) cos n

2πT

t

dt

0

T

∫ n = 1, 2,3,... (19.10)

bn =2

Tx (t) sin n

2πT

t

dt

0

T

∫ n = 1, 2,3, ... (19.11)

Note that in the above equations it does not matter where the integration starts, providedthat it is carried out over one entire period. The cn and φn coefficients can be derivedfrom the an and bn coefficients as shown earlier. The coefficients of the complexrepresentation are computed using the following expression:

γ n =1

Tx (t) e

− jn2π

Ttdt

0

T

∫ n = 0, ±1, ±2, ... (19.12)

10

Example 19.2 Fourier series of a sawtooth waveform

Problem:

Compute the complete Fourier spectrum of the sawtooth function shown in Figure 19.9.

Solution:

Known quantities:Amplitude, period and functional form of the signal.

Find:

bn and cn coefficients as a function of n

Odd and Even functionsWhen using the quadrature form of the Fourier series equations, it is useful to recognizethat all periodic functions can be represented as the sum of odd and even functions.Even functions have mirror symmetry with respect to the y axis, while odd functionshave skew symmetry with respect to the vertical axis. Formally:

x(t) is even if x (t) = x (−t)x(t) is odd if x (t) = − x (−t) .

Figures 19.8(a) and (b) depict the appearance of an even and an odd function,respectively. You may have already noted that the basis functions, cosine and sine, usedin the quadrature representation of the Fourier series are themselves even and oddfunctions, since cos(t) = cos(−t) and sin(t) = −sin(−t). This property may actually resultin significant savigns in computation if we recognize that a given function is odd or even,since odd functions will contain only odd basis functions (sines) and even functions willcontain only even basis functions (cosines). Example 19.2 illustrates this principle.

-T Tt

x(t)

0

-T

Tt

x(t)

0

(a) Even function (b) Odd function

Figure 19.8 Examples of even and odd functions

11

Schematics, diagrams, circuits and given data.

0T/2

T0

A

-A

t

Figure 19.9 Sawtooth signal

Assumptions:The function repeats periodically.

Analysis:The function of Figure 19.9 is an odd function, since x (t) = − x (−t) . Thus we

only need compute the bn (sine) coefficients. First, we determine an expression for x(t):

x (t) = A 1 −2t

T

then we evaluate the integral of equation 19.11:

bn = 2

Tx (t) sin n

2πT

t

dt

0

T

∫

= 2

TA 1 − 2t

T

sin n

2πT

t

dt

0

T

∫ = 2

TA sin n

2πT

t

dt

0

T

∫ + 2

TA − 2t

T

sin n

2πT

t

dt

0

T

∫

= 0 − 4 A

T2 t sin n

2πT

t

dt

0

T

∫ = − 4 A

T2

1

n2 2πT

2 sin n2πT

t

− t

n2πT

cos n2πT

t

0

T

= −4 A

T 2

T 2

n 2 4π 2 sin 2nπ( )−T 2

n2πcos 2 nπ( )

=

2 A

nπ, n = 1, 2, 3, ...

Comments:Note that the final expression for the Fourier series of the sawtooth waveform is actuallyquite simple, indicating that the amplitude of the spectral components of the decreases inproportion to the harmonic number, n.

12

Example 19.3 Fourier series of pulse train

Problem:Compute the complete Fourier series of the periodic pulse train shown in Figure 19.10.

Solution:

Known quantities: Amplitude, period and functional form of the signal.

Find: bn and cn coefficients as a function of n

Schematics, diagrams, circuits and given data.: δ =τT

= 0.2

A

τ/2 t0 T

x(t)

−τ/2 Τ−τ/2

Figure 19.10 Periodic pulse train

Assumptions:The function repeats periodically.

Analysis:The function of Figure 19.10 is neither even nor odd. Since there is nothing to

be gained by using the quadrature (odd-even) representation, we shall use the complexform. The expression for x(t) is very simple in this case:

x (t) = A for − τ2

< t ≤ τ2

x (t) = 0 for τ2

< t ≤ T − τ2

To determine the complex coefficient of the Fourier series, we evaluate the integral ofequation 19.11:

γ n =1

Tx (t )e

− jn2π

Ttdt

−T

2

T

2

∫ =1

TAe

− jn2π

Ttdt

−τ

2

τ2

∫

γ n = 1T

x (t) e − jn 2 πT t dt

−δ T2

δ T2

∫

= 1T

[cos 2πnT( )t − j

−δ T2

δ T2

∫ sin 2πnT( )t]dt

γ n = 1nπ

sin nπδ( ) n = 0, ± 1, ± 2, L

We may simplify the notation by using the sinc function defined as:

sinc x( ) = sin πx( )πx

Thus we may rewrite the coefficients as:

13

γ n = 1nπδ sin nπδ( ) = 1

δ sinc nδ( )Figure 19.11(a) depicts the pulse train corresponding to the numerical values givenabove, and Figure 19.11(b) its Fourier series coefficients up to n = 1000. The envelope ofthe discrete-frequency coefficients is the sinc function defined above.

Figure 19.11 Time-domain representation and Fourier Series spectrum of periodic pulsetrain with δ = 0.2.

Comments:Note that in the complex form of the Fourier series, the coefficients range from negativeinfinity to positive infinity.

Focus on Computer-Aided ToolsA Matlab file used to generate the figures shown in this example may be found on thebook website, http://www.mhhe.com/engcs/electrical/rizzoni.

14

Non-periodic signals, Fourier transform

Practical communication signals have both periodic and non-periodiccomponents. Typically, the carrier waveform is periodic (usually a sine wave), while themodulating signal, consisting of speech, music, video or data, is non-periodic. Theanalysis of non-periodic signals uses a mathematical tool different from (but related to)the Fourier Series: the Fourier transform. The Fourier transform, also named after theFrench mathematician Fourier (see box “Fourier Series” in Chapter 6), is an integraltransform, so called because it is mathematically represented by an integral, and becauseit performs a transformation between two domains: the time domain and the frequencyor spectral domain.

The Fourier transform of a function x (t) is the function X (ω ) defined by theintegral

Conversely, if the function X (ω ) is known, the inverse Fourier transform isdefined by:

The pair x (t) and X (ω ) represent a Fourier transform pair, a relationshipusually denoted by x (t) ↔ X(ω ) . Table 19.3 provides a useful summary of commonFourier transform pairs.

Before proceeding with the Fourier transform, it will be useful to define afunction called the unit impulse or delta function. This function plays a very importantrole in Fourier analysis. The unit impulse function, δ(t), can be defined as the derivativeof the unit step function introduced in Chapter 6 (equation 6.52):

The unit impulse function has three important properties: 1) it has area equal toone; 2) it has infinite amplitude; and 3) it has zero duration, that is, its occurrence isconcentrated at one instant in time. Clearly, such a signal is a mathematical abstraction,since it is impossible to physically generate a signal that has zero duration and infiniteamplitude. Figure 19.12 shows how one can think of the delta function as the limit of asequence of rectangular pulses that are increasingly narrow and tall, such that the productof height (1/ε) and width (ε) is always equal to 1: the delta function can be thought of asthe limit of this sequence as ε approaches zero.

The delta function has one further property that is of interest in signal analysis:

x (t )δ t − t0( )dt−∞

∞∫ = x t0( ) (19.16)

that is, the delta function “samples” the function x(t) at the time of the occurrence of theimpulse.

A number of properties of the Fourier transform can assist in the computation ofFourier transforms. The most important properties are summarized in Table 19.3. SomeFourier transform pairs are listed in Table 19.4.

δ (t ) =du (t)

dt or u(t) = δ ( t' ) dt'

−∞

t

∫ Delta or unit impulse function (19.15)

X (ω ) = x ( t)e− jωt

dt−∞

∞∫ Fourier transform (19.13)

x (t) =1

2πX (ω )e

jω tdω

−∞

∞∫ Inverse Fourier transform (19.14)

t

δ(t-t )0

t0ε

ε1

Figure 19.12 Deltafunction as the limit of asequence of rectangularpulses of unit area

15

Table 19.3 Properties of Fourier transformsProperty Signal Fourier transformTime shift x (t − t0 ) e

− jωt0 X (ω )

Frequency shift e jωt0 x (t ) X (ω − ω0 )

Complex conjugate x * (t) X * (−ω )

Time reflection x (−t) X (−ω )Scaling x (at ) 1

aX

ωa

Convolution x (t) ∗ y (t) X (ω ) ⋅ Y (ω )

Multiplication x (t) ⋅ y (t ) 1

2πX(ω ) ∗Y (ω )

Differentiation d

dtx( t)

jωX (ω )

Integrationx (t )

−∞

t

∫ dt1

jωX (ω ) + πX( 0)δ (ω )

Table 19.4 Fourier transform pairsx(t) X(ωω)

1 δ (t ) 12 1 2πδ (ω )

3 u( t)(unit step)

2πδ (ω ) +1

jω4 tu (t )

(unit ramp) 2πdδ (ω )

dω−

1

ω 2

5 e−αt

u(t) α > 0 1

α + jω6 te −αt u( t) α > 0 1

α + jω( )2

7 e jω 0 t 2πδ (ω − ω0 )

8 cos ω0 t( ) π δ (ω − ω0 ) + δ (ω + ω0 )[ ]9 sin ω0t( ) − jπ δ (ω − ω0 ) − δ (ω + ω 0 )[ ]10 cos ω0 t( )u( t) π

2δ (ω − ω 0 ) + δ (ω + ω0 )[ ]+

jωω0

2 − ω 2

11 sin ω0t( )u( t) − jπ2

δ (ω − ω 0 ) − δ (ω + ω0 )[ ]+ω 0

ω02 − ω2

The importance of the Fourier transform is that it allows us to view signals inthe frequency or spectral domain. As you shall see shortly, the spectral representation ofsignals is much more convenient and effective in representing communications signals,among other reasons because it permits defining important concepts such as bandwidthand spectrum allocation. You already know that sinusoidal signals are represented bysingle frequencies in the spectrum, from the Fourier series discussion in the precedingsubsection. In the following three examples, we compute the Fourier transform of a well-know signal, a sinusoid; of a signal we have not yet analyzed: a single rectangular pulse;and of a signal that occurs very frequently in communication systems: a sine wave burst,or RF pulse. These examples will help you develop a feel for the spectral content ofdifferent types of signals.

16

Example 19.4 Fourier Transform of sine wave

Problem:Compute the Fourier transform of an arbitrary sinusoidal signal.

Solution:

Known quantities: Functional form of the signal, x(t).

Find: X(ω)

Schematics, diagrams, circuits and given data. Figure 19.13.

Assumptions: None

Analysis:The signal shown in Figure 19.13, is defined as

)sin()( 0ttx ω=

The Fourier transform for the signal is calculated using the complex representation

γ n = 1T

x (t ) e− jn 2πT t dt

− T2

T2

∫

= 1T

sin( ω0t) e− jn 2πT t dt

− T2

T2

∫

= 1T 2 j

e j 2πT t − e − j 2π

T t[ ]e− jn 2πT t dt

− T2

T2

∫

γ n = 12 j

1T

e j 2 πT ( 1−n) t − e j 2π

T ( −n −1) t[ ]dt− T

2

T2

∫

Using the relation from Table 19.3, for the defined Fourier transform pairs, the aboveintegral can be evaluated as:

γ n = 12 j 2πδ ω − ω 0( )− 2πδ ω + ω0( )[ ]

γ n = πj δ ω − ω0( )− δ ω + ω0( )[ ]

Thus, the Fourier transform of an arbitrary sinusoid consists of a pair of delta functions inthe frequency domain, as shown In Figure 19.4.

Comments:We can extend the result of this example to an arbitrary periodic signal , since we knowthat any periodic signal can be represented by the sum of an infinite number of sinusoidalfunctions. the Fourier transform, X(ω) of a periodic signal, x(t) is a train of impulsesoccurring at the harmonically related frequencies and for which the area of the impulse at

Figure 19.13 Sinusoidalsignal of frequency ω0.

Figure 19.14 Fouriertransform of sinusoid

17

the nth harmonic frequency, nω0 is 2π times the nth Fourier series coefficient an. TheFourier series coefficients for this sinusoid signal are

a1 = 12 j

,

a−1 = − 12 j

,

ak = 0, | k | ≠ 1

Hence, the Fourier transform coefficients are given by,

a1 = πj

,

a−1 = − πj

,

ak = 0 | k | ≠ 1

Example 19.5 Fourier Transform of rectangular pulse signal

Problem:Compute the Fourier transform of a square pulse signal.

Solution:

Known quantities: Functional form of the signal, x(t).

Find: X(ω)

Schematics, diagrams, circuits and given data. Figure 19.15.

Assumptions: None.

Analysis:Consider the rectangular pulse, x(t) of duration T and unity amplitude shown in Figure19.15. We define this pulse mathematically as follows:

x (t) =1 for | t |≤ T

2

0 for | t |> T2

The Fourier transform of x(t) is

X (ω ) = e − j ωt dt−T

2

T2

∫ = e − j ωt

− jω[ ]− T2

T2

= 2ω

ejω T

2 −e− j ω T

2

2 j

=2ω sin( ω T

2)

X (ω ) = T sinc (ω T2

) where , sinc ( x ) = sin( x )

x

X ( f ) = T sinc( πfT )

Figure 19.15Rectangular pulse.

Figure 19.16 FourierTransform of squarepulse (magnitude only)

18

A plot of the spectrum is shown in Figure 19.16. The figure illustrates the characteristicsof the sinc function, with zero crossings at integer multiples of π/T rad/s, and peakamplitude of 2T.

Comments:Single and repetitive square bursts occur commonly in communication systems. Theanalysis completed in this example will be useful in the following sections.

Focus on Computer-Aided ToolsA Matlab file used to generate the figures shown in this example may be found on thebook website, http://www.mhhe.com/engcs/electrical/rizzoni.

Example 19.6 Fourier Transform of sine burst (RF pulse)

Problem:Compute the Fourier transform of the sine wave burst shown in Figure 19.17.

Solution:

Known quantities: Functional form of the signal, x(t).

Find: X(ω)

Schematics, diagrams, circuits and given data: Figure 19.17.

Assumptions: None.

Analysis:The pulse signal x(t) shown in Figure 19.17 (a) consists of a sinusoidal wave of unitamplitude and frequency fc, for a duration t = -T/2 to t = T/2.The signal x(t) can bedefined mathematically as follows:

x (t) =A cos( 2πfc t ) for | t | ≤ T

2

0 for | t | > T2

The Fourier transform of x(t) is

X (ω ) = x ( t) e − j ωt dt− T

2

T2

∫ = A2

e jωc t + e − jω c t( )− T

2

T2

∫ e − jωt dt

= A2

e− j (ω−ωc ) t

− j (ω−ωc )+ e

− j (ω+ω c ) t

− j (ω +ω c )[ ]− T2

T2

= A − e− j (ω−ωc ) T

2

2 j (ω −ωc )− e

− j ( ω+ω c )T2

2 j(ω +ω c )+ e

j (ω−ω c )T2

2 j (ω −ωc )+ e

j (ω +ω c ) T2

2 j(ω +ωc )

= Asin π f + fc( )T

2π f + fc( )+ sin π f − fc( )T

2π f − fc( )

X (ω ) = AT2

sinc f + fc( )T + sinc f − fc( )T[ ]Note that we could have used the frequency shifting property of the Fourier transform toobtain the above result without carrying out the integration explicitly. The magnitudespectrum of the RF pulse is shown in Figure 19.18. It clearly illustrates the frequency-shifting property of the Fourier transform.

Figure 19.17Radio-frequency(RF) burst.

Figure 19.18Magnitude spectrumof RF burst

19

Comments:This signal is specifically referred to as a RF pulse when the frequency, fc of the sinusoidwave falls in the radio frequency range.

Focus on Computer-Aided ToolsA Matlab file used to generate the figures shown in this example may be found on thebook website, http://www.mhhe.com/engcs/electrical/rizzoni

Bandwidth

The bandwidth of a signal is the range of frequencies comprising the spectrumof the signal. Bandwidth is a very important concept in communication systems, as theallocation of radio frequency spectrum for different communication systems permits thetransmission of signal within a certain specified bandwidth. For example, standard FMradio allows a bandwidth of 200 kHz for each radio station. The most common definitionof bandwidth is that of 3-dB bandwidth, also called half-power bandwidth. The 3-dBbandwidth of a signal is defined as the frequency range between points where the signallevel is 3 dB below its maximum pass-band value. This informal definition is illustratedin Figure 19.19, where an arbitrary voltage signal is shown to have a spectrum V(f), withcenter frequency f0 and 3-dB bandwidth 2B.

You will recall from the definitions given in Chapter 6 that the 3-dB point in afrequency plot is the frequency where the amplitude has dropped to a value equal to1 / 2 , or 0.707, times the maximum value. Since signal power is proportional to thesquare of the voltage, the 3-dB bandwidth is also called the half-power bandwidth. Thus,half of the signal power is contained in the frequency band f0 − B to f0 + B ; we call 2Bthe bandwidth of the signal. Please observe that this informal definition assumes that thesignal spectrum as a band-pass shape. This is usually the case for most, if not all,communication signals.

How much bandwidth does a signal require? This depends on twofacts: i) the bandwidth of the signal itself; and ii) the type of modulation. We shallrevisit the concept of bandwidth when we explore amplitude- and frequency-modulationsystems.

f0

f −B0 0f +B

Vmax

0.707Vmax

V(f)

f

Figure 19.19 Definition of 3-dB (half-power) bandwidth

20

Example 19.7 Bandwidth of commercial AM (or TV, or FM) signals

Problem:Analyze the bandwidth of the signal from a commercial AM station and determine howmany stations can be assigned frequencies over the frequency band assigned tocommercial AM.

Solution:

Schematics, diagrams, circuits and given data:see http://www.ntia.doc.gov/osmhome/allochrt.pdf

Analysis:As can be seen in the diagram displayed in the above website, the AM band frequencyallocation goes from 535 to 1,605 kHz. Each channel is allocated a bandwidth ofapproximately 10 kHz (we shall see in the next section what this calculation includes).Thus, the total number of stations that can operate in the same region is approximatelyequal to:

N =1605 − 535

10= 107

Each AM station can operate with a total bandwidth of 10 kHz. As we shall see in thenext section, this actually corresponds to a signal bandwidth of only 5 kHz.

Comments:You are probably aware of the fact that the FCC licenses many more than 107 AMstations in the USA. This is possible because AM broadcast has a limited range, and twostations can be assigned the same frequency if there are located sufficiently far apart.You may also have noticed that at night it is possible to receive AM radio signals frommuch greater distances (for example, in Ohio one can tune in stations from as far as NewYork City and New Orleans late at night). This is a consequence of the change inionization density on the ionosphere during the night, permitting reflection of radiowaves over a longer range. The FCC regulates not only the frequency allocation, but alsothe power allocated to a given station; a station may be required to switch to a lower-power transmitter during certain times of the day.

Check Your Understanding

1. Repeat the analysis of Example 19.2 using the complex representation of the Fourierseries (equation 19.5)

2. Repeat the analysis of Example 19.3 using the quadrature representation of theFourier series (equation 19.3).

3. Compute all the coefficients of the Fourier series expansion for the signal x(t) = 1.5cos(100t).

4. Sketch the Fourier transform of a square pulse with unity amplitude and with aduration of 10 µs.

5. The spectrum of a signal can be described by the function X (ω ) =αω

ω2 + α 2. Let α

= 103, and calculate the 3-dB bandwidth of the signal. What is the center frequencyof the signal?

21

19.3 AMPLITUDE MODULATION AND DEMODULATIONThe concept of amplitude modulation (AM) was introduced in Chapter 4

(Focus on measurements: capacitive displacement transducer), where it was shown thatthe signal produced by a capacitive microphone (displacement transducer) inserted in aWheatstone bridge circuit, modulated the amplitude of a sinusoidal excitation signal. Inthat example, the pressure changes sensed by the microphone constituted the modulation,while the sinusoidal excitation provided a carrier. In Chapter 8 (Focus onmeasurements: peak detector for capacitive displacement transducer), it was shown thata diode circuit was capable of demodulating the AM signal, and of recovering the desiredinformation (pressure changes corresponding to acoustic waves, that is, sound). In thissection, the same basic principles introduced in the above mentioned examples will bediscussed more formally, as they apply to AM communication systems.

The most common manifestation of amplitude modulation in communicationsystems is commercial AM radio, or standard AM. The Federal CommunicationsCommission (FCC), a body that regulates the usage and allocation of the radio frequencyspectrum in the U.S.A., has assigned the frequency band between 540 and 1600 kHz tocommercial AM radio transmission. Each station can occupy a bandwidth of 10 kHzcentered around its carrier. As we shall see, this corresponds to an effective signalbandwidth of 5 kHz – sufficient for good reproduction of speech, and acceptablereproduction of music.

Basic principle of AM

AM signals are generated by modulating the amplitude of a carrier signal. Letthe carrier signal be a sinusoid at frequency ωc :

c (t ) = Ac cos ωc t( ) Carrier signal (19.17)and – for illustration purposes - let the modulation also be a single tone (sinusoid), at afrequency ωm << ωc :

m (t ) = Am cos ω m t( ) Modulating signal (19.18)With these definitions, we can define the basic AM signal as follows:

s( t) = Ac + m (t )[ ]cos ω c t( ) (19.19)or

s( t) = Ac 1 +Am

Accos ωm t( )

cos ω c t( ) AM signal (19.20)

The modulation index, µ, is defined to be the ratio of the modulation to carrier signalamplitudes:

µ =Am

Ac

Modulation index (19.21)

and for proper amplitude modulation should be less than 1. If Equation 19.20 isexpanded, we see that an AM signal is composed of two terms: a sinusoidal carrier wave,plus a wave that is the product of two sinusoidal terms. Using trigonometric identities,we can write the following expression:

s( t) = Ac cos ω c t( )+ µAc cos ω c t( )cos ω m t( )

= Ac cos ω c t( )+ µAc

2cos ωc − ωm( )t + µ

Ac

2cos ω c + ω m( )t

(19.22)

Equation 19.22 shows that the AM signal is really composed of three sinusoidalwaveforms: a carrier wave; a lower sideband signal, at frequency ω c − ω m( ),containing

the modulating signal; and an upper sideband signal, at frequency ω c + ωm( ), alsocontaining information (the modulation). Example 19.8 illustrates some importantproperties of an AM signal with pure sinusoidal modulation.

22

Example 19.8 Single-tone amplitude modulation

Problem:Analyze the spectrum of a single-tone modulation signal based on the WOSU AM 820radio station. Use both analytical and computational tools.

Solution:

Known quantities: Carrier frequency; modulation index.

Find: Derive expressions for and plot time-and frequency domain waveforms of the AMsignal.

Schematics, diagrams, circuits and given data.

fc = 0.82 MHz; µ = 0.5.

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x 10-4

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

Time, s

m(t)

, V)

Message signal

(a)0 1 2

x 10-5

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time, s

c(t),

V)

Carrier signal

(b)

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x 10-4

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

(c)

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

x 104

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

-fm fm

(d)-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x 10 6

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

-fc

fc

(e)7.2 7.4 7.6 7.8 8 8.2 8.4 8.6 8.8 9 9.2

x 105

0

0.2

0.4

0.6

0.8

1

(f)Figure 19.20 Time- and frequency-domain waveforms of single-tone AM signal

Assumptions: Assume unity amplitude for the carrier, Ac = 1, and a modulatingfrequency fm = 10 kHz

Analysis:Define the modulating signal m(t) and the carrier signal c(t) as

m (t ) = Am cos( 2πfm t )

c (t ) = Ac cos( 2πfc t)

23

These waveforms are plotted in Figures 19.20(a) and (b), respectively. The spectra ofthese signals are plotted in figures 19.20(d and (e).

The AM wave s(t) is given bys( t) = Ac 1 + µ cos( 2πfm t )[ ]cos( 2πfc t)

and is plotted in Figure 19.20(c). Using the Fourier transform pairs given in Table 19.3(Pair 8), the Fourier transform of s(t) can be expressed as the sum of three delta functions,centered at the carrier and at the sum (upper sideband) and difference (lower sideband)frequencies. Note also that the spectrum is repeated for negative frequencies, asexplained earlier.

S ( f ) = A c

2δ ( f − fc ) + δ ( f + fc )[ ]+ µ A c

4δ ( f − fc − fm ) + δ ( f + fc + fm )[ ]

+ µ Ac

4δ ( f − fc + fm ) + δ ( f + fc − fm )[ ]

Thus, the spectrum of an AM wave for the special case of sinusoidal modulation, consistsof delta function at ± fc, fc ± fm, and – fc ± fm, as shown in Fig. 19.20(f).

Comments:If you like, you may experiment with the value of the modulation index and see its effecton the AM wave. It is recommended that the modulation index, µ, be nearly equal to 1,but not greater. If the modulation index is greater than 1 for any t, the carrier wavebecomes over-modulated, resulting in carrier phase reversals whenever the function1 + µm( t )( )crosses zero.

Focus on Computer-Aided ToolsA Matlab file used to generate the figures shown in this example may be found on thebook website, http://www.mhhe.com/engcs/electrical/rizzoni. You may use this file toexperiment with changes in the modulation index.

The single-tone modulation example is very valuable to understand the basic propertiesof an AM signal. In the next two examples we progress to the double-tone modulation,and then to a general, non-periodic modulating signal to explore the waveforms andspectrum of more realistic AM signals.

24

Example 19.9 Double-Tone modulation

Problem:Plot the frequency spectrum of a carrier signal with unity amplitude and frequency fc = 1MHz which is amplitude modulated with a modulating signal m(t) consisting of twosinusoidal frequencies.

Solution:

Known quantities: Carrier frequency, and amplitude; modulating signal

Find: Modulation index and frequency spectrum of the AM wave with the defined carrierand modulating signal.

Schematics, diagrams, circuits and given data:

fc = 1 MHz; Ac = 1 V; m(t) = 0.5cos(2π10000t) + 0.4cos(2π80000t).

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x 10-4

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time, s

m(t)

, V)

Message signal

(a)

-18 -16 -14 -12 -10 -8 -6 -4 -2

x 10-6

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Time, s

c(t),

V)

Carrier signal

(b)

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x 10-4

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

(c)

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

x 104

-0.1

-0.05

0

0.05

0.1

0.15

0.2

-fm1 fm1

(d)-1 -0.5 0 0.5 1

x 106

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

-fc

fc

(e)

0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2

x 106

0

0.2

0.4

0.6

0.8

1

(f)

Figure 19.21 Time- and frequency-domain waveforms of double-tone AM signal

Assumptions: None.

Analysis:The modulation index for the signal is defined as

µ =AmAc

=max( m(t )) − min( m (t))

2Vc= 0.9 + 0.8626

2 = 0.8813

25

The spectrum of the AM wave in this case consists of delta functions at ± fc, fc ± fm1, fc ±fm2, -fc ± fm1, and -fc ± fm2, where fm1, fm2, are the frequencies contained in the modulatingsignal. This is seen in Figure 19.21, where all time- and frequency domain waveforms areplotted..

Comments:The frequency spectrum of the AM wave is just a shifted version of the originalmodulating signal with the shift in frequency equal to the carrier frequency. The portionof the spectrum of an AM wave lying above the carrier frequency fc is the upper side-band, whereas the symmetric portion below fc is called the lower side-band.

Focus on Computer-Aided ToolsA Matlab file used to generate the figures shown in this example may be found on thebook website, http://www.mhhe.com/engcs/electrical/rizzoni. You may use this file toexperiment with changes in the modulation index or in the signal frequencies.

Example 19.10 Non-periodic amplitude modulation

Problem:Plot the frequency spectrum of a carrier signal with unity amplitude and frequency fc =0.1 MHz which is amplitude modulated with a non-periodic modulating signal m(t)having a defined shape.

Solution:

Known quantities: Carrier frequency, and amplitude; Modulating wave m(t) defined fora certain interval of time.

Find: Frequency spectrum of the AM wave.

Schematics, diagrams, circuits and given data.

m (t ) =sin( 2πf1t) + sin( 2πf2t ) + sin( 2πf3t ) + sin( 2πf4t ) + u( t) for t ≤ T

0 otherwise

u(t ) = 1 for t ≤ T

c (t ) = Ac cos( 2πfc t)

fc = 0.1 MHz; Ac = 1; f1 = 1kHz, f2 = 2kHz, f3 = 3kHz, f4 = 4kHz.

26

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x 10-3

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

(a)

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x 10-3

-6

-4

-2

0

2

4

6Amlitude modulated signal, s(t)

Time, s

s(t),

V

(b)

-1.5 -1 -0.5 0 0.5 1 1.5

x 10 4

0

0.2

0.4

0.6

0.8

1

Frequency, Hz

Normalized spectrum of modulating signal, m(t)

|M(f)

|, V

(c)

0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2

x 105

0

0.2

0.4

0.6

0.8

1

Hz

|S(f)

|, V

Normalized spectrum of modulated signal, s(t)

fc

(d)

Figure 19.22 Time- and frequency-domain waveforms of non-periodic AM signal

Assumptions: None.

Analysis:The signal waveform and the frequency spectrum of the modulating signal and of the AMwave are shown in Figures 19.22(a-d). The spectrum of the AM wave is a shifted versionof the modulating signal spectrum around the carrier frequency.

Comments:If Bmax is the bandwidth of the modulating signal, (the highest frequency in themodulating signal), the bandwidth of the AM wave is defined as, twice the highestfrequency in the modulating signal, i.e., B = 2Bmax.

Focus on Computer-Aided ToolsA Matlab file used to generate the figures shown in this example may be found on thebook website, http://www.mhhe.com/engcs/electrical/rizzoni. You will find this fleuseful in exploring more general AM signals.

27

AM demodulation; integrated circuit receivers

Demodulation is the process of recovering the modulating signal from areceived modulated signal. With reference to Figure 19.1, one can think of thetransmitter in an AM signal as the device that imposes the modulation on a carrier, whilethe receiver extracts the modulating signal from a received AM signal. To understand thebasic principle of modulation and demodulation, we observe that amplitude modulationconsists in effect of multiplying the carrier signal times the modulating signal. Thisprocess is often called mixing, and a mixer is the device that implements this function,that is, multiplication.

Consider the AM signal of Equation (19.20), and multiply it by a second signalat the same frequency as the carrier signal:

s( t) ⋅ c( t) = Ac 1 +Am

Ac

cos ωm t( )

cos ω c t( )

⋅ cos ω c t( ) (19.23)

The resulting squared term, cos 2 ω ct( ) can be expanded to yield:

s( t) ⋅ c( t) = Ac 1 + Am

Accos ωm t( )

cos

2 ω ct( )= Ac

21 + Am

Accos ωm t( )

1 + cos 2ωc t( )( )

= Ac

2+ m (t) + Ac

2+ m( t)

cos 2ω c t( )

(19.24)

You see that the result of this mixing operation consists of two terms: a constantplus the modulation signal – what we desire to recover – and an amplitude modulatedterm at a frequency equal to twice the carrier frequency. Note that the modulation signalis back to baseband, that is low frequencies (for example, 0-5 kHz for speech and music),and that it is therefore easy to recover the modulating signal by low-pass filtering theoutput of the mixer. Figure 19.23 depicts a block diagram of a conceptual AMdemodulator as well as the spectra of the AM signal before and after mixing.

The process of demodulating AM signals is carried out today by means ofintegrated circuit receivers. Additional information may be found at the followinginternet sites:http://www.national.com/pf/LM/LM1863.htmlhttp://www.national.com/pf/LM/LM1868.html

Check Your Understanding

19.6 Use the Matlab files that accompany Examples 19.8 and 19.9 to plot only thepositive spectrum of the single- and double-tone AM signals. Determine thebandwidth of the AM signal in each case.

19.7 Determine the bandwidth of the modulating signal in Figure 19.22(c); what isthe bandwidth of the AM signal? Is this consistent with commercial FMpractice? Would the FCC allow a commercial station to broadcast such asignal?

28

19.4 FREQUENCY MODULATION AND DEMODULATIONYou are certainly familiar with the term frequency modulation because of the

great diffusion of FM radio. As its name implies, frequency modulation consists ofencoding the information contained in a modulating signal in the frequency of a carriersignal. Figure 19.24 depicts a sinusoidally-modulated FM waveform and itscorresponding magnitude spectrum. FM transmission permits significant improvementsover AM, but at the cost of an increased requirement for bandwidth. In the next sub-sections you will be introduced to the basic signal models for FM; two different cases arediscussed: narrowband FM and wideband FM. The plots of Figure 19.24 correspondto a wideband FM signal. Note the significant spread of signal frequencies relative to thecarrier frequency!

0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

time (s)

Am

plitu

de

(a)-2500 -2000 -1500 -1000 -500 0 500 1000 1500 2000 25000

0.05

0.1

0.15

0.2

0.25

(b)Figure 19.24 (a) FM signal time waveform; (b) FM signal magnitude spectrum

Basic principle of FMThe basic principle underlying FM is that the instantaneous frequency of the

carrier is modulated by the information-carrying signal. If we assume a sinusoidalcarrier, as is usually the case, say c (t ) = cos( 2πfc t ) , the modulation will cause thefrequency fc to be a function of time. In the signal of Figure 19.24, the carrier frequencyvaries sinusoidally as well. Before proceeding with the analysis of FM signals, it will beuseful to examine the relationship between the instantaneous phase and frequency of asinusoidal signal. We first define the relationship between the instantaneous phase andfrequency of a sine wave as follows:

θi = 2π fi t( )dt0

t

∫ ; fi t( ) =dθ i t( )

dt(19.23)

To illustrate this definition, consider the case of a simple sinusoidal signal,v (t ) = A cos ωt( ) ; in this signal we recognize that the phase angle (the argument of thecosine function) is θi t( ) = ωt , and therefore the instantaneous signal frequency is given

by fi t( )=dθi t( )

dt= ω . This result should not surprise you: all sinusoidal signals must

have a phase angle that increases linearly with time, so that their instantaneous frequencyis constant. In the case of an FM signal, we might have a phase angle that variessinusoidally with time, thus causing the instantaneous frequency to also vary in asinusoidal fashion; this is the simplest case of an FM signal, and is treated next.

Single tone modulationConsider the case of single tone modulation, where the modulating signal is:

m (t ) = Am cos ωm t( )= Am cos 2πfm t( ) (19.24)

29

The instantaneous frequency of the FM signal varies linearly with the modulation, that is,the carrier frequency increases and decreases with the modulating signal, as shown inequation 19.25:

f ( t) = fc + k f Am cos 2πfm t( )= fc + ∆f cos 2πfm t( ) (19.25)

In the above expression we have implicitly defined the frequency deviation, ∆f:

∆ f = k f Am frequency deviation (19.26)

The frequency deviation is a very important characteristic of an FM signal; ∆f representsthe maximum instantaneous deviation of the FM signal frequency from the carrierfrequency. ∆f is dependent on the amplitude of the modulating signal, and is independentof the modulating signal frequency. The constant kf depends on the technique used forgenerating the modulated signal.

The instantaneous phase of the FM signal is calculated using equation 19.23:

θi = 2π fi t( )dt = 2πfc0

t

∫ t +∆f

fmsin 2πfm t( ) (19.27)

Note that equation 19.27 introduces another important constant: the ratio of the frequencydeviation to the modulating frequency is called the modulation index, β:

β =∆f

fm(19.28)

The parameter β represents the maximum instantaneous deviation of the phase angle ofthe FM signal from the angle of the carrier. Now we can write the instantaneous angle ofthe FM signal as follows.

θi = 2πfc t + β sin 2πfm t( ) (19.29)

Finally, the sinusoidally modulated FM signal is defined in equation 19.30:

s FM (t) = Ac cos 2πfc t + β sin 2πfm t( )[ ] (19.30)

It is now possible to make a formal distinction between the two cases mentioned earlier,narrowband FM and wideband FM, on the basis of the modulation index.

Narrowband FMNarrowband FM corresponds to FM signals with a small modulation index (i.e.,

β << 1). We can use a trigonometric identity to expand equation 19.30:

s FM (t) = Ac cos 2πfc t( )cos β sin 2πfm t( )[ ]− Ac sin 2πfc t( )sin β sin 2πfm t( )[ ] (19.31)

and if β <<1 we can use the small-angle approximations cos β sin 2πfm t( )[ ]≈ 1 and

sin β sin 2πfm t( )[ ]≈ β sin 2πfm t( ) to write

s FM (t) ≈ Ac cos 2πfc t( )− βAc sin 2πfc t( )sin 2πfm t( ) ≈ Ac cos 2πfc t( )− 1

2βAc cos 2π fc + fm( )t[ ]− cos 2π fc − fm( )t[ ]{ } (19.34)

Note the similarity between the expression for narrowband FM and that we derivedearlier for the AM signal, repeated below for convenience:

30

s AM (t ) = Ac cos ω ct( )+ µAc

2cos ω c − ωm( )t + µ

Ac

2cos ωc + ωm( )t

(19.22)

A reasonable rule of thumb is that the approximation given in equation 19.34 holds for β< 0.3. Figure 19.25 (a)-(d) depicts the spectra of FM signals with various values of β.Note how the bandwidth increases with the value of the modulation index. Only in (a)and (b) is the signal narrowband FM.

-1 -0.5 0 0.5 1

x 10 4

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1A

mpl

itude

(a)

-1 -0.5 0 0.5 1

x 10 4

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Am

plitu

de

(b)

-1 -0.5 0 0.5 1

x 10 4

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Am

plitu

de

(c)

-1 -0.5 0 0.5 1

x 104

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Am

plitu

de

(d)Figure 19.25 Bandwidth increases with modulation index in FM signals:

(a) β = 0.1; (b) β = 0.3; (c) β = 0.6; (d) β = 1.

Example 19.11 Narrowband FM

Problem:Compare the spectrum of a narrowband FM waveform with that of an AM waveformwith the same modulating and carrier frequencies.

Solution:

Known quantities: Carrier frequency; modulation frequency; modulation index.

Find: Plot the frequency domain waveforms of the narrowband FM and AM signals.

Schematics, diagrams, circuits and given data.

fc = 1,000 Hz; fm = 100 Hz; Ac = 1 V; Am = 0.2 V; µ = 0.2; β = 0.2.

31

Assumptions: Assume sinusoidal modulation and unity amplitude for the carrier, Ac = 1.

Analysis:The files Freqmod.m and Ampmod.m are used to generate two signals; the first signal isan FM waveform with β = 0.2, the second is an AM signal with µ = 0.2. The resultingspectra are plotted in figures 19.26a and b, respectively.

0 2000 4000 6000 8000 10000 120000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Am

plitu

de

0 2000 4000 6000 8000 10000 120000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Am

plitu

de

Figure 19.26 Comparison of narrowband FM and AM signal spectra (a) FM, β = 0.2; (b)AM, µ = 0.2.

Comments: Note how the two amplitude spectra are virtually identical. Equations19.34 and 19.22 predict this result. The only difference between the two spectra is in thephase angle of the signals.

Focus on Computer-Aided ToolsA Matlab file used to generate the figures shown in this example may be found on thebook website, http://www.mhhe.com/engcs/electrical/rizzoni. You may use this file toobserve the difference in phase between the signals.

Wideband FMThe mathematical representation of wideband FM signals is far more complex

than the approximation of equation 19.34. The nonlinearity of the wideband FM signal isdescribed using Bessel functions; this analysis is beyond the scope of this chapter, andthe interested reader is referred to any one of a number of excellent textbooks in electricalcommunications.

Transmission bandwidth of FM signalsThe transmission bandwidth of a frequency-modulated signal is theoretically

infinite; however, practical approximations are possible. In the case of narrowband FMwe have already seen that we have the same transmissions bandwidth as in an AM signal:B = 2 fm . For large values of the modulation index, β, the bandwidth of the FM signal

can be experimentally observed to be close to the total frequency excursion, ±∆f, or2∆f. These observations lead to the well-known Carson’s rule, relating the approximatetransmission bandwidth to the frequency deviation and to the modulation index:

Carson’s rule (19.35)

Carson’s rule suggests that, as β becomes larger, the bandwidth approaches 2∆f, while asβ decreases, the bandwidth becomes closer to 2fm.

B = 2 ∆f + 2 fm = 2∆f 1 +1

β

32

Example 19.12 Commercial FM broadcast

Problem:Use Carson’s rule to analyze the bandwidth of a commercial FM station.

Solution:

Known quantities: Carrier frequency; modulation index.

Find: Approximate signal bandwidth

Schematics, diagrams, circuits and given data.

fc = 90.5 MHz; Ac = 1; Am = 1; fm = 10 kHz; kf = 6,000; β = 0.2.

Assumptions: Assume sinusoidal modulation.

Analysis:

For sinusoidal modulation, the frequency deviation is ∆ f = k f Am = 6 kHz.

Then, Carson’s rule predicts a bandwidth

B = 2 ∆f + 2 fm = 2∆f 1 +1

β

= 1.2 ⋅10

41 +

1

0.2

= 72 kHz

Example 19.13

Problem:Given an FM message signal, find:a. The bandwidth of the message signal, Bm.b. Bandwidth of the modulated carrier signal, Bc.c. Band of frequencies occupied by the signal, B.

Solution:

Known quantities: Modulation frequency, carrier frequency; modulation constant.

Find: Bm; Bc;

Schematics, diagrams, circuits and given data.

v m = 5 cos( 200 πt) ; fc = 100fm; k f =1000

2π Hz/V

Assumptions: Assume sinusoidal modulation.

Analysis:a. The message signal is v m = 5 cos( 200 πt) . Hence:

33

fm = 100 Hz

Am = 5

fc = 100 fm = 10 kHz

The bandwidth of the message signal is

Hz 200

2

=

=

m

mm

B

fB

b. The maximum frequency deviation is given by:∆ f = k f Am

∆ f =5000

2π Hz ≈ 795 Hz

Thus, the bandwidth of the modulated carrier signal is approximately given by:Bc = 2 ∆f + fm( )Bc = 2(795 + 100 ) = 1790 Hz

c. The carrier frequency is fc = 100 kHz . The frequency band is centered about thecarrier frequency. Therefore, the band of frequencies occupied spans from

fc −Bc

2

to fc +

Bc

2

. Hence the frequency band is

10000 −1790

2

to 10000 +

1790

2

, which equals a band from 9.9105 kHz to

10.895 kHzFigure 19.27 depicts the modulating signal and the FM spectrum of the FM signalexamined in this example.

Figure 19.27 Time-domain and spectral plots for Example 19.13

34

Example 19.14

Problem:In the United States the assigned band for FM commercial broadcast is from MHz0.88to MHz0.108 with 100 possible channels. Find the bandwidth for each channel.

Solution:

Known quantities:Band for FM commercial broadcast, number of channels.

Find: The bandwidth of each channel.

Schematics, diagrams, circuits and given data. See problem statement.

Assumptions: None.

Analysis:The bandwidth for each channel is defined as:

B =Total Band

Number of channels=

108.0 - 88

100= 200 kHz

Comments:Each commercial FM radio station has a bandwidth allocation of 200 kHz.

FM demodulationDemodulation of an FM signal is accomplished by performing a frequency-to-voltageconversion, that is, by converting the frequency modulation into a voltage signal. This isthe reverse of the modulation process, and can be realized in a number of ways. Wedescribe two basic approaches in the following sub-sections.

Frequency-to-voltage conversion

If a pulse of fixed amplitude, A, and fixed duration, τ, is generated at each zero crossingof the sensor waveform, it can be readily shown that a voltage proportional to theinstantaneous signal frequency may be obtained. Figure 19.28 depicts the functionalform of a frequency-to-voltage converter.

v (t)S zero-crossing

detector

one-shot multivibrator

τ

t i

v Z

vOS v o

τ

v (t)S

v Z

vOS

t

Ti

A

t ti-1 i

Figure 19.28 Frequency to voltage conversion

Ideally frequency to voltage (F-V) conversion could be obtained by computing thefollowing integral:

35

vost i −1

ti

∫ (t) dt =At

∆Ti

= 2 Aτfi (19.36)

yielding a voltage proportional to the frequency of νS(t) during the ith cycle of the carrierwaveform. In practice it is quite difficult to reset the integrator of Figure 19.27 at eachzero crossing, so practical F-V converters employ a low-pass filter in place of the idealintegrator.

Phase-locked demodulationAnother method for implementing F-V conversion utilizes a phase locked loop (PLL) asa frequency-to-voltage converter, or FM demodulator. The PLL, can act as an FMdemodulator once it is phase-locked to an input signal waveform. When the PLL is inlock, any change in the input signal frequency generates an error voltage at the output ofthe phase detector, which can be either analog (a mixer), or digital, consisting of a pair ofzero crossing detectors.The output voltage of the PLL, v

o(t), is the voltage which is required in order to maintain

a voltage-controlled oscillator (VCO) running at the same frequency as the input signal,and changes in the input signal frequency are matched by changes in v

o(t). In this sense,

the PLL acts as an F-V converter, with the input-output characteristic shown in Figure19.29. Note that the PLL can offer only a finite lock range.

phase detector

v S

low-pass filter

amp

voltage controlled oscillator

v o

center frequency

lock range

ω, rad/s

v o

Figure 19.29 Phase-Locked Loop F-V conversion

Integrated circuit receiversFM demodulation is performed using integrated circuit receivers; additional informationmay be found in the following websites:http://www.national.com/pf/LM/LM1868.htmlhttp://www.national.com/pf/DS/DS8906.htmlhttp://www.national.com/pf/LM/LMX3162.html

Check Your Understanding

19.8 Use the Matlab files that accompany Example 19.11 to compute and plot thephase angle of the two spectra. Are the phase spectra identical?

19.9 What is the effect of changing the amplitude of the modulating signal on thebandwidth of the FM modulated signal?

19.10 Investigate the effect of changing β on the FM modulated signal bandwidth and

analyze the spectrum of the FM signal of Example 19.11 with 5.0=β .

19.11 Find the carrier frequency for Channel 11 used by WCBE Radio, Columbus OH,per the FCC regulations for the commercial broadcast in the US (use data inExample 19.14).

36

19.5 EXAMPLES OF COMMUNICATION SYSTEMSThe objective of this chapter is to summarize some important applications of

modern communication systems. The overview given in this chapter is certainly notexhaustive, but will give you a good summary of the technology underlying some oftoday’s communication systems and of their capabilities.

Global Positioning System (GPS)

The Global Positioning System (GPS), illustrated in Figure 19.30, is rapidlysupplanting older navigation technologies based on radio communications used by theaircraft and marine industries, such as Loran-C. GPS is based on 24 satellites linked toground stations, and effectively replaces – with much greater accuracy – the century-oldsystem of navigation based on star position. You can think of the satellites as “man-madestars”. Differential GPS is capable of position measurements with accuracy of a fewcentimeters. In recent years, GPS receivers have been miniaturized, and today amateursailors, private pilots and other private users have the ability to purchase hand-held GPSunits. Among the uses of GPS we list navigation systems for cars, boats, planes andguiding agricultural machining for “precision agriculture”. The operation of GPS isexplained in five basic steps:

Table 19.5 Basic operation of GPS1. GPS operates based on triangulation of signals received from

satellites.2. Triangulation is performed by the GPS receiver by measuring

distance knowing the travel time of radio signals.3. Travel time is computed with the aid of accurate timing

references.4. The exact position of the satellites in space is used to calculate

distance.5. Delays experienced by the signals in traveling from the satellite to

the ground stations are corrected.

TriangulationThe technique of triangulation is based on distance (range) measurements from

the receiver location to the satellite. Four satellites are needed to determine exactposition of any receiver location on earth.

Measuring DistanceTo measure distance, GPS computes the time required to receive each satellite signal.The receiver and satellite both generate a synchronized signal; comparison of the signalreceived from the satellite with that in the receiver is used to calculate time of travel.Since the speed of travel of electromagnetic waves is known, distance can be calculated.

Timing AccuracySatellites carry atomic clocks on board to provide accurate timing. The clock in a low-cost GPS receiver need not be as accurate, as an additional range (distance) measurementcan be used to remove the timing error in the receiver.

Satellite PositionsThe position of the satellites is essential in calculating distance. The orbits of GPSsatellites are known, and any deviations are measured by the Deaprtment of Defense; anyerrors are transmitted to the satellites, which in turn transmit this error information to thereceivers along with their timing signals.

Figure 19.30GPS principle(Courtesy: TrimbleNavigation)

37

Correcting ErrorsSignals traveling through the ionosphere experience additional delays, which turn intotransmission errors. There are many methods for correcting errors; the technique calleddifferential GPS can eliminate almost all errors.

Further informationhttp://www.trimble.com/gps/http://www.garmin.com/http://www.cnde.iastate.edu/staff/swormley/gps/gps.htmlhttp://www.motorola.com/SPS/MCORE/gps/

Radar

The acronym RADAR stands for RAdio Direction And Ranging. Radarsystems operate by radiating electromagnetic waves (typically at microwave frequencies)and by detecting the echo returned from reflecting objects (targets). Radar technologywas developed during World War II and played a significant part in the success of theAllied Forces. While military applications are obvious, today Radar finds widespreadcivilian application in air traffic control and in tracking weather conditions, as well as inmarine navigation (see Figure 19.31 for some examples of radar technology used in thelatter application, and Figure 19.32 for an example of weather radar). In addition todetecting the position of a stationary target, radar is capable of determining the trajectoryof a moving target, thus predicting its future location. This function is, for example, veryuseful in weather radar where one wishes to predict weather conditions. The principle onwhich radar operation is based is that of the doppler shift, a concept with which you areprobably already intuitively familiar (think of the sound of a train whistle as the trainmoves by a stationary observer). The doppler shift permits distinguishing a movingtarget from a stationary one, thus allowing the radar system to discern the echo of astationary target from that of a moving target.

Further informationhttp://cirrus.sprl.umich.edu/wxnet/radsat.htmlhttp://southport.jpl.nasa.gov/education/classroom/http://www-cmpo.mit.edu/Radar_Lab/FAQ.htmlhttp://www.skywarn.ampr.org/radartut.htm

Figure 19.31 Radarantenna (top) anddisplays. (Courtesy:ProNav)

38

Figure 19.32 Weather radar from http://cirrus.sprl.umich.edu/wxnet/radsat.html

Sonar

The term SONAR stands for SOund NAvigation and Ranging. Sonar isconceptually similar to radar, in that it uses information about the reflection andtransmission of waves to determine the behavior of targets or the properties of theenvironment. The principal difference between Sonar and Radar is that the former isbased on the reflection and propagation of acoustic (sound) waves, while the latter isbased on radio frequency electromagnetic waves.

The applications of sonar are numerous, ranging from inexpensive depth findersand imaging systems used to aid the navigation of small and large sea vessels (see forexample Figure 19.33), to underwater navigation, to ocean thermal mapping. You willfind a number of interesting resources related to Sonar in the Internet sites listed below.

Further informationhttp://vision.dai.ed.ac.uk/ashley/Sonar/sonar_def.htmlhttp://www.marine-group.com/acoustic.htmhttp://oalib.njit.edu/

Figure 19.33 Sonardisplays for marinenavigation (Courtesy:ProNav)

39

Computer Networks

The distinction between communication and computer networks is increasinglyblurred today, as computers are used more and more commonly. In this brief subsectionwe describe some of the basic ideas behind today’s computer network technology.Networks are divided into two basic categories: local area networks and wide areanetworks. Generally speaking, a local area network ties computers and othercommunication systems within the same building, while wide area networks interconnectbuildings, cities and countries.

Figure 19.34 depicts the three basic network architectures: bus, ring and star.The bus architecture is characterized by a common bus shared by all elements of thenetwork; all data traffic is present on the bus, and can be read at each node. As anexample of a very common communication standard based on a bus architecture isEthernet, the computer communication protocol commonly used for high-speed datatransmission (see Chapter 15). When a computer wishes to transmit on the bus, it firstlistens to determine whether the bus is available. If the bus is available, the computer cantransmit. Should a collision occur, a protocol is followed to repeat transmission. Thestandard that rules this type of transmission is called carrier-sense multipleaccess/collision detection (CSMA/CD). The bus architecture is effective whenevernodes share in the communications approximately equally. If one or a few nodes had adisproportionate share of the traffic, then this architecture would not be very effective.

Bus architecture Ring architecture Star architecture

Figure 19.34 Communication System architectures

The ring architecture (see Figure 19.32) is characterized by each node havingtwo connections: inbound and outbound. All network traffic must pass through eachnode. To determine whether a data is to be received by a specific node, a token is passedaround the ring. When a node needs to transmit, it captures the token and sends data outin its place. When the message is received by the intended receiver, the message ismarked “read” and is sent back around the ring; when the message reaches the originalsender, the latter removes the message from the ring and replaces it with the token. Notethat a node cannot transmit until it receives a token. This method ensures that no onenode can dominate communications around the ring. Among the drawbacks of thisarchitecture is the fact that if a node is disconnected, the entire ring fails to operate.

The star architecture (see Figure 19.34) connects each node to a central hubconnection. Note that a star architecture can be used to implement both bus and ringnetworks. The hub can serve the purpose of monitoring and enforcing variousrestrictions.

Local Area NetworksLocal area networks (LANs) are physically connected networks of computers in

relative close proximity. Office and school networks are the most common examples.Connections are made through either copper wire (twisted pair or coaxial), or opticalfiber. Copper wire in the form of unshielded twisted pair (UTP) is capable of data ratesup to 100 Mbit/s (Mbps), and is most commonly used in LANs because of its low cost.Tabled 19.6 summarizes the performance of different UTP categories.

Table 19.6 Twisted pair copper cable applications

40

Category Application Maximum data rate1 Telephone 1 Mbps2 Token ring – 4 Mbps 4 Mbps3 10BaseT Ethernet 10 Mbps4 Token ring – 16 Mbps 20 Mbps5 100BaseT Ethernet 100 Mbps

Coaxial cable is also used for LANs, especially Ethernet3, and comes in thickand thin forms. Fiber optics are becoming increasingly common, through FDDI (fiberdistributed data interface) backbone networks, which carry the bulk of the traffic in alarge local area network such as corporate networks and university campuses. FDDI cansupport a 100 Mbps rate using a token ring architecture.

Wide area networksOutside of the local area, networking needs are usually handled by a

telecommunications provider, for example a telephone or cable TV company. Servicesthat are provided by telecommunications providers are based on bandwidth needs, andinclude traditional copper lines (DS-n service), Integrated Services Digital Network(ISDN), also based on copper lines, and optical fiber service (OC-n). Table 19.7summarizes the data rates achievable through OC-n. Note that the fastest fiber opticservice is nearly 10 Gbps!

Table 19.7 Optical fiber service data ratesOC-n Data rate (Mbps)OC-1 51.84OC-3 155.52

OC-12 622.08OC-48 2488.32OC-192 9953.28

Wireless Networks and Personal Communication Systems

Wireless mobile networks are becoming increasingly common in everydayapplications. With the advent of digital cellular telephony, the performance and cost ofmobile network services have improved to the point that a wireless communicationsystem may be preferable to a land-line-based one. An interesting statistic is that in a fewEuropean countries the number of wireless telephone “lines” has recently exceeded thatof fixed telephone installation. The widespread availability of such mobilecommunication capabilities has led to the concept of personal communication systems(PCS), capable of providing voice and data services in just about any setting.

Mobile wireless networks use a radio frequency (RF) carrier, and usually adopta cellular architecture. In such an architecture, a large number of antennas is used; eachantenna is at the center of a cell, as shown in Figure 19.35. A stationary user would thenreceive the signals through the antenna corresponding to the cell where s/he is located. Ifthe user is in motion, a handoff procedure is used to transfer the communication fromone cell to an adjacent cell. The new generation of digital cellular telephony is based onthe concept of microcells, with antennas spaced very closely. This technology is veryeffective in urban areas, but has limited coverage in rural areas.

A new generation of satellite-based mobile telephones has recently beenlaunched (see e.g.: Motorola Iridium); this technology permits communications to andfrom anywhere on the globe, but its cost is very high, making it suited only forspecialized applications (e.g.: explorations in remote areas).

3 See brief description of Ethernet communication protocol in Section 15.7.

Antenna

Cell coverage area

Figure 19.35Coverage pattern incellular system Navigation)

41

Further informationhttp://www.sprintpcs.com/index.htmlhttp://www.mot-sps.com/solutions/isdn/isdn-tutorial.htmlhttp://www.mot.com/SPS/WIRELESS/information/idenoverview.htmlhttp://www.mot.com/SPS/WIRELESS/information/gsmoverview.htmlhttp://www.mot.com/SPS/WIRELESS/information/gsmoverview.htmlhttp://www.mot-sps.com/cgi-bin/get?/books/apnotes/pdf/an1575*.pdfhttp://mot-sps.com/support/technical/tutorials/dspbasics.html

The Internet

The Internet is a very large network that is based on sending information packetsusing the TCP/IP protocol. Today we take the Internet for granted, but it took some 30years for the Internet to reachits present form. The earliest precursor of the Internet, theARPANet, was originally established as a project of the Defense Advanced ProjectsResearch Agency (DARPA) in 1962. In the 1980s the ARPANet was replaced by theNSFNet, administered by the National Science Foundation; NSFNet became today’sInternet in the 1990s, and the system is operated by private as well as by non-profit andcorporate Internet Service Providers. Access to the Internet is today almost ubiquitous; inmany countries a significant percentage of private homes and most businesses make useof the Internet as a means of communications on a daily basis.

42

HOMEWORK PROBLEMS

Section 19.2 Fourier series and transform19.1 Compute the Fourier series coefficients for a periodic square wave of unitamplitude, time period τ, and duty cycle η as shown in Figure P19.1 and definedmathematically as.

x (t) =1 for | t |≤ ητ−1 for | t |< τ

Figure P19.1 Square wave signal of period ττ and duty cycle ηη

Use Matlab to plot the frequency spectrum of this signal with time period,

sec 300

1=τ and duty cycle, a. %50=η , b. %30=η .

19.2 A full wave rectified sinusoidal signal of natural frequency 0ω rad/sec is shownin Figure P19.2.

a. Find the Fourier series coefficients for the full wave rectified sinusoid.b. Generate the frequency spectrum for a full wave rectified sinusoid of natural

frequency πω 2000 = rad/sec.

Figure P19.2 Full wave rectified sine wave of natural frequency ωω0 rad/sec.

19.3 A full wave rectified cosine signal of natural frequency 0ω rad/sec is shown inFigure P19.3.

a. Find the Fourier series coefficients for the full wave rectified cosine.b. Generate the frequency spectrum for a full wave rectified cosine with natural

frequency πω 1500 = rad/sec.

43

Figure P19.3 Full wave rectified cosine wave of natural frequency ωω0 rad/sec

19.4 Compute the Fourier series coefficients for the cosine-burst signal shown inFigure P19.4.

Figure P19.4 Cosine-burst signal.

19.5 The triangular pulse signal shown in Figure P19.5 is mathematically defined as:

x (t) = A 1 −| t |

T

u( t + T ) − u(t − T )( )

Figure P19.5 Triangular pulse x(t) of duration 2T.

a. Find the Fourier transform of the triangular pulse.b. Plot the frequency spectrum of a triangular pulse of period 01.0=T sec and

amplitude 5.0=A .19.6 A double exponential signal shown in Figure P19.6 is defined mathematicallyby:

x (t) =e −at , for t > 0

0, for t = 0

-e at , for t < 0

44

Figure P19.6 Double exponential signal

a. Compute the Fourier transform of the signal. (Hint: Can use the linearityproperty of Fourier transform)

b. Plot the frequency spectrum of the signal using Matlab for 8=a .19.7 Evaluate the Fourier transform of the damped sinusoidal wave shown in FigureP19.7 and having the functional form:

)()2cos(exp(-at))( tutftx cπ=where )(tu is the unit step function.

Figure P19.7 Damped Sinusoidal signal

19.8 An ideal sampling function consists of an infinite sequence of uniformly, spaceddelta functions and is mathematically defined as:

δT0( t) = δ ( t − mT0 )

m = −∞

∞∑

Figure P19.8 Dirac delta function with period T0.

a. Compute the Fourier transform of ideal sampling function shown in Figure P19.8.

45

b. Also, use Matlab to generate the time domain signal and its amplitude spectrum forsec 01.0=oT .

19.9 Download the utterance signal utter.au and use the Matlab commandauread(‘utter.au’) to load it to the workspace. Use the FFT tools in Matlab toidentify the frequency components of the signal.

19.10 A bat echolocation chirp signal is provided to you on the book website. Do afrequency analysis of the signal and explain what you observe. (Courtesy Digital SignalProcessing Group, Rice University)

Section 19.3 Amplitude modulation and demodulation19.11 Find the modulation index µ for an AM modulated signal having a carrier of

amplitude 0.1=cA and, the amplitude of the carrier at the maximum is 0.3max =A and

at the minimum is 6.0min =A .

19.12 Plot the anticipated frequency spectrum of a carrier signal with an amplitude ofunity and frequency MHz3.1=cf that is AM modulated )1( =µ with a signal, )(tm ,where

m (t ) = 0.8 sin( 2π 5000 t) + 0.4 sin( 2π10000 t) + 0.2 sin( 2π 20000 t)19.13 Plot the anticipated time domain response of a carrier signal with an amplitudeof unity and frequency MHz10=cf that is AM modulated )1( =µ with a signal, )(tm ,where

m (t ) = A 1 −| t |

T

T = 0.01 sec

Hint: The message signal is a triangular wave of time period T.19.14 An AM radio station uses a carrier signal of unity amplitude and frequency

MHz6.1=cf . The message signal is a voice signal having certain frequencycomponents and defined as:

)28902sin(1.0)22302sin(2.0)13452sin(3.0)9602sin(35.0)3402sin(4.0)( ttttttm πππππ ++++=Plot the time domain and the frequency domain AM modulated signal of modulationindex 1=µ .

19.15 A non-periodic message signal, )(tm is amplitude modulated )1( =µ by a

carrier signal of unity amplitude and frequency MHz5.0=cf . Plot the time andfrequency domain signal.

m (t ) = 0.4 sin( 2π 340 t) + 0.35 sin( 2π960 t ) + 0.3 sin( 2π1345 t) + 0.2 sin( 2π 2230 t) + 0.1sin( 2π 2890 t) + u(t )

u(t ) =1 for t ≤ 0.01

0 otherwise

19.16 Consider a modulating wave )(tm that consists of a single frequency component

and defined as:)2cos()( tfAtm mm π=

where mA is the amplitude of the modulating wave and mf is the frequency. The

sinusoidal carrier wave has amplitude cA and frequency cf . The signal is amplitude

modulated to produce the signal )(ts . Find the average power delivered to a 1-ohm

resistor by )(ts .

19.17 The carrier frequency of the W-OSU channel is 0.82 MHz. If the upper sidebandof the AM modulated signal has frequency components of amplitude 0.4 at 0.825 MHz,0.2 at 0.83 MHz, and 0.25 at 0.84 MHz,a. Find the modulating signal equation.b. Plot the spectrum of the modulating signal.c. Plot the spectrum of the AM signal including the lower sideband.

46

19.18 The AM commercial radio band in the US is authorized to operate from 525 kHzto 1.7 MHz. A carrier frequency is assigned to each station, and regulations require themto be separated by 10 kHz. Find:a. Number of channels that can be accommodated in the given frequency range.b. The maximum modulating frequency that can be transmitted without overlap.19.19 The speech signal utter.au is to be AM modulated for transmission on a AMcommercial radio band in the US. Plot the frequency spectrum of the AM signal. Use anychannel according to their separations and a modulation index 1=µ in your design.

Section 19.4 Frequency modulation and demodulation19.20 The message signal given by m (t ) = 5 cos( 750 πt), is frequency modulated by acarrier frequency 105 times the message frequency and the modulation constantisk f = 1005 . Find the bandwidth of the message signal.

19.21 If the message signal given by m (t ) = 2 cos( 360 πt), is frequency modulated by acarrier frequency 100 times the message frequency and the modulation constantk f = 1000 . Find the bandwidth of the modulated carrier signal.

19.22 For the band of frequencies occupied by the FM signal of Problem 19.2119.23 A message signal )(tm is FM modulated by a carrier of unity amplitude and

frequency MHz0.10=cf , with modulating constant 1000=fk . Plot the time and

frequency domain FM modulated signal if m (t ) = 0.8 sin( 2π 5000 t) .19.24 A packet of information is sent on an FM channel of frequency

MHz0.15=cf that uses a modulating constant 6000=fk . Plot the frequency spectrum

of the FM modulated signal.m (t ) = 0.4 sin( 2π 340 t) + 0.35 sin( 2π960 t ) + 0.3 sin( 2π1345 t)

+0.2 sin( 2π 2230 t) + 0.1sin( 2π 2890 t ) + u( t)

u(t ) =1 for t ≤ 0.001

0 otherwise

19.25 WOSU-FM uses a carrier frequency of 90.5 MHz and modulating constantk f = 66000 . The speech signal utter.au is transmitted on this channel. Plot the frequency

spectrum of the FM modulated speech signal.19.26 Consider Example 19.13. If Channel 2 is allocated for country music and themessage signal may be considered to be m (t ) = 10 cos( 2π10 3 t) . Find:a. The carrier frequency.b. The value of k f