• Author / Uploaded
• Linus Antonio

Citation preview

Electronic i n i in linear i FUNDAMENTALS

THROUGH ADVANCED

WAYNE TOM ASI M

lid,

MM

ELECTRONIC

COMMUNICATIONS SYSTEMS Fundamentals Through Advanced

Wayne Tomasi Mesa Community College

PRENTICE HALL, Englewood

Cliffs,

New

Jersey

07632

Library of Congress Cataloging-in-Publication Data

Tomasi, Wayne. Electronic communications systems. Includes index. 1.

I. Telecommunication systems. 62 1.38 '04 13

TK5101.T625 1988 ISBN 0-13-250804-^

Title.

87-7205

Chapters

through 12 are published

1

a

Fundamentals of Electronic Communications Systenu by Wayne Tomasi (© 1988) chapters 13 through 22 are published

/Hen County Pubk "• Wayne, Indiana

Uhm

as

Advanced Electronic Communications Systenu by Wayne Tomasi (© 1987]

Editorial/production supervision an interior design:

Kathryn Pavele

Cover design: Diane Sax Cover photo: Courtesy of Sperry Corporatio

Fumqs

Manufacturing buyer: Margaret Rizzi/Lorraine

© A

=

1988 by Prentice-Hall, Inc.

Division of Simon

Englewood

Cliffs,

All rights reserved.

reproduced,

New

No

in

&

=>

Schuster

Jersey 07632

may

part of this book

i

any form or by any mean

without permission in writing from the publish*

Printed in the United States of

10

ISBN

9

8

7

6

Ameri 5

D-13-2SDflDt

Prentice-Hall International (UK) Limited,

Lo
f)

+

sin

2x

2x sin

3* (cos

ojf)

-1-

•

•

Nx Nx

(1-12)

sin

+

3jc

(cos u>0

where X

=

N=

TTt/T

Nth harmonic and can be any value integer

From Equation

1-12

it

can be seen that a rectangular waveform has a O-Hz (dc) component

equal to

V x The narrower

or

duty cycle

(1-13)

the pulse width, the smaller the dc component. Also, from Equation

12, the amplitude of the

Nth harmonic

(sin x)lx function is

Nx Nx

2Vt sin

T

(1-14)

used to describe repetitive pulse waveforms. Sin x

simply a sinusoidal waveform whose instantaneous amplitude depends on

x is

in the

denominator, the denominator increases with

simply a damped sine wave.

(sin

A

1

is

Vm = The

V x

(sin x)lx function

is

x.

x.

is

With only

Therefore, a (sin x)lx function

shown

in

Figure 1-10.

x)/x

FIGURE

1-10

(sin x)lx function.

.

Introduction to Electronic Communications

14

T * *

Chap.

1

to

= 0.1T

t

(a)

*

*--«

1st lobe

2nd lobe

-«

2nd

1st null

>

3rd lobe

3rd null

null

I

±_L 5F

F

j_lH

J_L 20F

15F

10F

Frequency

30F

25F

(b)

FIGURE

1-11

(sin x)lx function: (a)

rectangular pulse waveform; (b) frequency

spectrum.

Figure 1-11 shows the frequency spectrum for a rectangular pulse with a pulse width-to-period ratio of 0.1. a

damped

10F Hz), there third at

It

is

30F Hz

a

can be seen that the amplitudes of the harmonics follow

The frequency whose period equals

sinusoidal shape.

0-V component.

(period

=

3/r),

A

second null occurs

and so on. All harmonics between

null frequency are considered in the first lobe of the

components between the cies

first

Hz and

in the

in the third lobe,

1

The dc component

There are 0-V components

is

=

lit),

the

a

first

second lobe, frequen-

and so on.

characteristics are true for all repetitive rectangular

2.

frequency

(period

frequency spectrum. All spectrum

and second null frequencies are

between the second and third nulls are

The following

\lt (i.e., at

20F Hz

at

waveforms:

equal to the pulse amplitude times the duty cycle. at

frequency

lit

Hz and

all

integer multiples of that

frequency. 3.

The amplitude-versus-frequency time envelope of on the shape of a damped sine wave.

EXAMPLE

1-2.

For the pulse waveform shown

in

Figure 1-12:

(a)

Determine the dc component.

(b)

Determine the peak amplitudes of the

(c) Plot the (sin x)lx function.

(d)

the spectrum

Sketch the frequency spectru

first

10 harmonics.

components take

Signal Analysis

15

t

- 0.4 ms

T = 2ms-

FIGURE Solution

(a)

From Equation

1-12

Pulse

(b)

The amplitudes of

the

first

).4

2(1)

!

Example

1-2.

is

iiM^ = o.2 v

=

2 ms

10 harmonics are determined from Equation 1-14:

ms\

"sin

( 2 ms )

N

for

component

1-13 the dc

V(0 Hz)

waveform

W

1

80(0.4

/V(3. 14)(0.4

ms/2ms)] ms/2 ms)

Amplitude

Frequency (Hz)

0.2 1

2

(c)

The

1

1.5

0.374

1000

0.303

3

1500

0.202

4

2000

0.094

5

2500

6

3000

7

3500

8

4000

9

4500

10

5000

(sin x)lx function is

0.5

500

2

shown

2.5

in

3

Frequency, F (kHz)

-0.063 -0.087 -0.076 -0.042

Figure 1-13.

3.5

4.5

FIGURE

1-13

function for

(sin x)lx

Example

1-2.

Introduction to Electronic Communications

16

Chap.

1

0.4

0.3

0.2

fS

a

0.1

J 0.5

2

1.5

1

3

2.5

JL_L 4

3.5

FIGURE

4.5

5

Frequency, F (kHz)

(d)

The frequency spectrum

is

shown

Although the frequency components

in Figure 1-14.

on

second lobe are negative,

it is

customary

the frequency spectrum.

Figure 1-15 shows the effect that reducing the duty cycle ratio)

Voltage

1-2.

in the

to plot all voltages in the positive direction

1-14

spectrum for Example

(i.e.,

has on the frequency spectrum for a nonsinusoidal waveform.

It

reducing the t/T

can be seen that

narrowing the pulse width produces a frequency spectrum with a more uniform amplitude. In fact, for infinitely

narrow pulses, the frequency spectrum comprises an

infinite

number

of frequencies of equal amplitude. Increasing the period of a rectangular waveform

while keeping the pulse width constant has the same effect on the frequency spectrum.

t/T = 0.25 sin

Jtl

x/x

— — — —i— — — — —

-1

t/T = 0.125

j

m

'

|

|

'

sin

'

I

'

1

I

Frequency

x/x

il

Frequency

|

I

•

.

.

I

I

I

t/T = 0.03125

rh sin

I

I

*

'

I

•

•

'

'

x/x

|

|

I

'

I

'

»

I

1

I

I

FIGURE Frequency

1-15

Effects of reducing the

IT ratio (either decreasing

t

increasing T).

/

or

Signal Analysis

Effects of

17

Band limiting on

Signals

Every communications channel has a limited bandwidth and therefore has a limiting effect

on signals

that are propagated through them.

channel to be equivalent to an ideal linear phase

We

can consider a communications

filter

with a

finite

bandwidth.

(a)

Time

(b)

Time

(c)

Time

(d)

Time

Time

(e)

FIGURE

1-16

Bandlimiting signals:

(a)

1-kHz square wave;

(b)

1-kHz square

wave bandlimited to 8 kHz; (c) 1-kHz square wave bandlimited to 6 kHz; (d) 1-kHz square wave bandlimited to 4 kHz; (e) 1-kHz square wave bandlimited to 2

kHz.

If a

Introduction to Electronic Communications

18

nonsinusoidal repetitive waveform passes through an ideal low-pass

filter,

Chap.

1

the harmonic

frequency components that are higher in frequency than the upper cutoff frequency for the filter are removed. Consequently, the shape of the

16a shows the time-domain waveform for the square

waveform is changed. Figure wave used in Example 1-1.

1-

If

waveform is passed through a low-pass filter with an upper cutoff frequency of 8 kHz, frequencies above the eighth harmonic (9 kHz and above) are cut off and the waveform shown in Figure l-16b results. Figures l-16c, d, and e show the waveforms produced when low-pass filters with upper cutoff frequencies of 6, 4, and 2 kHz are this

used, respectively.

can be seen from Figure 1-16 that bandlimiting a signal changes the frequency

It

its waveform and, if sufficient bandlimiting is imposed, waveform eventually comprises only the fundamental frequency. In a communications

content and thus the shape of the

system, bandlimiting reduces the information capacity of the system and, bandlimiting

is

if

excessive

imposed, a portion of the information signal can be removed from the

composite waveform.

ELECTRICAL NOISE In general terms, electrical noise

is

defined as any unwanted electrical energy present

passband of a communications

in the usable

any undesired signals that

fall into

the

band

circuit.

to 15

For instance,

kHz

audio recording

in

are audible and will interfere

with the audio information. Consequently, for audio circuits, any unwanted electrical

energy

in the

band

to 15

kHz

is

considered noise.

Essentially, noise can be divided into

two general

categories: correlated

and uncorre-

cted. Correlation implies a relationship between the signal and the noise. Uncorrected noise

noise that

is

is

present in the absence of any signal.

Correlated Noise Correlated noise signal such as

is

unwanted

electrical

distortion are both

is

present as a direct result of a

distortion.

Harmonic and intermodulation

energy that

harmonic and intermodulation

forms of nonlinear distortion; they are produced from nonlinear amplifi-

cation. Correlated noise

can not be present

Simply

no noise! Both harmonic and intermodulation

stated,

no

the shape of the

signal,

wave

in the

in a circuit unless there is

time domain and the spectral content

an input signal.

distortion in the

change

frequency

domain.

Harmonic

Harmonic distortion is the generation of unwanted multiwave when the sine wave is amplified in a nonlinear device such as a large-signal amplifier. Amplitude distortion is another name for harmonic distortion. Generally, the term "amplitude distortion" is used for analyzing a waveform in the time domain, and the term "harmonic distortion" is used for analyzing a waveform distortion.

ples of a single-frequency sine

Electrical

in the

Noise

19

frequency domain. The original input frequency

stated previously,

is

There are various degrees or orders of harmonic distortion

is

is

the

first

harmonic and, as

called the fundamental frequency. distortion.

Second-order harmonic

the ratio of the amplitude of the second harmonic to the amplitude of the

fundamental frequency. Third-order harmonic distortion

the ratio of the amplitude of

is

the third harmonic to the amplitude of the fundamental frequency, and so on. ratio

The

of the combined amplitudes of the higher harmonics to the amplitude of the funda-

mental frequency

monic

distortion

is

harmonic distortion (THD). Mathematically,

called total

total har-

is

higher

THD

x 100

(1-15)

fundamental

where

% THD = ^higher

=

percent total harmonic distortion

sum of the root-mean-square

quadratic

(rms) voltages of the higher

harmonics ^fundamental

E 4MPLE

= rms

voltage of the fundamental frequency

1-3

Determine the percent second-order, third-order, and spectrum shown

in

total

harmonic distortion for the output

Figure 1-17.

6

V,

1st

harmonic

V, 2nd harmonic

FIGURE

harmonic

Example

Frequency, F (kHz)

1-17

Harmonic

distortion for

1-3.

Solution

%

2nd-order harmonic distortion

3rd-order harmonic distortion

THD

VvT+vl V,

x 100

=

77 x 100

=7 x

100

= 33%

x 100 = - x 100= 16.7%

VFTP x

100

= 37.3%

6

Intermodulation distortion is the generation of unIntermodulation distortion. wanted cross products (sum and difference frequencies) created when two or more frequencies are amplified in a nonlinear device such as a large-signal amplifier.

As

Introduction to Electronic Communications

20

Chap.

with harmonic distortion, there are various degrees of intermodulation distortion.

1

It

would be impossible to measure all of the intermodulation components produced when two or more frequencies mix in a nonlinear device. Therefore, for comparison purposes, a common method used to measure intermodulation distortion is percent second-order intermodulation distortion. Second-order intermodulation distortion total

is

the ratio of the

amplitude of the second-order cross products to the combined amplitude of the

original input frequencies. Generally, to

measure second-order intermodulation

A

four test frequencies are used; two designated the

B band (FBI and FB2). The

designated the

distortion,

band (FA1 and FA2) and two

second-order cross products (2A-B)

are:

2FA1 - FBI, 2FA1 - FB2, 2FA2 - FBI, 2FA2 - FB2, (FA1 + FA2) - FBI, and (FA1 + FA2) — FB2. Mathematically, percent second-order intermodulation distortion (IMD)

is

2nd-order

V2nd-order cross

IMD

products

x 100

(1-16)

Voriginal

where

V2n d order = = original

EXAMPLE

quadratic

quadratic

sum of sum of

the amplitudes of the 2nd-order cross products the amplitudes of the input frequencies

1-4

Determine the percent second-order intermodulation distortion for the A-band, B-band,

and second-order intermodulation components shown

856

863

1374

in Figure 1-18.

1885

1385

1892

A-band

B-band

J

L

1896

1903

Frequency, F (Hz)

FIGURE

1-18

Intermodulation distortion for Example

1-4.

Solution r,,

-

,

.

%

2nd-order

%

2nd-order

T . ,„.

IMD =

V\ Y 2nd-order

cross products

^ Z

+

2 Z

f 62

+

6

V2 2 + 2 2 + IMD = /2

V6

% IMD

1907

2A-B cross products

= 35.4%

2

2

+

2 I

+ 62

+

2 l

x 100

1914

Noise

Electrical

21

Harmonic and intermodulation

by the same thing, nonlinear two is that harmonic distortion

distortion are caused

amplification. Essentially, the only difference between the

can occur when there

is a single input frequency, and intermodulation distortion can occur only when there are two or more input frequencies. The generation of harmonic and intermodulation components is explained in Chapter 2.

Uncorrelated Noise Uncorrected noise

is

noise that

present. Uncorrelated noise

External noise.

is

is

External noise

to enter into the circuit only if the

a signal

is

noise generated external to a circuit and allowed

frequency of the noise

filter.

Atmospheric noise. Atmospheric noise

is

naturally occurring electrical energy that

originates within the earth's atmosphere. Atmospheric noise electricity.

The source of most

commonly

is

static electricity is natural electrical

comes

as lightning. Static electricity generally its

is

falls into the passband of the There are three primary types of external noise: atmospheric noise, extraterresnoise, and man-made noise.

input trial

present regardless of whether or not there

divided into two general categories: external and internal.

called static

disturbances such

form of impulses which spread

in the

energy throughout a wide range of radio frequencies. The magnitude of the impulses

measured from naturally occurring events has been observed to frequency.

Consequently,

cant. Also, frequencies

which

tion,

distant.

manner

frequencies above 30

above 30

MHz,

is

summation of

the

to

be inversely proportional

atmospheric noise

insignifi-

80 km.

the energy from

all

sources both local

Atmospheric noise propagates through the earth's atmosphere

as radio waves. Therefore, the

on the propagation conditions

is

MHz are limited predominantly to line-of-sight propaga-

limits their interfering range to approximately

Atmospheric noise and

at

at the

magnitude of the

time and

seasonal variations. Atmospheric noise

is

is

static

dependent,

in the

same

noise received depends in part,

on diurnal and

the familiar sputtering, cracking, and so on,

heard on a radio receiver predominantly in the absence of a received signal and relatively insignificant

compared

Extraterrestrial noise. Extraterrestrial noise earth's

atmosphere and

is,

therefore,

noise originates from the Milky is

is

to the other sources of noise. is

noise that originates outside the

sometimes called deep-space noise.

Way,

Extraterrestrial

other galaxies, and the sun. Extraterrestrial noise

divided into two categories: solar and cosmic (galactic).

Solar noise

is

noise generated directly from the sun's heat. There are two

nents of solar noise: a "quiet" condition exists

when

and high-intensity sporadic disturbances caused by sun spot

flare-ups.

The sporadic disturbances come from

The magnitude of that repeats every

the disturbances caused 11

years. Also, these

compo-

a relatively constant radiation intensity

specific locations

from sun spot

activity

and solar

on the sun's surface.

activity follows a cyclic pattern

11-year periods follow a supercycle pattern

where approximately every 100 years a new maximum intensity is realized. Cosmic noise sources are continuously distributed throughout our galaxy and other

Introduction to Electronic Communications

22

galaxies. Distant stars are also suns

Chap.

1

and have high temperatures associated with them.

Consequently, they radiate noise in the same manner as our sun. Because the sources of galactic noise are located

much

Cosmic noise

relatively small.

is

farther

away than our

sun, their noise intensity

often called black body noise and

is

is

distributed fairly

evenly throughout the sky. Extraterrestrial noise contains frequencies from approximately 8

MHz

GHz, although

to 1.5

atmosphere and

frequencies below 20

MHz

seldom penetrate the earth's

are, therefore, insignificant.

Man-made noise. Man-made noise is simply noise that can be attributed to man. The sources of man-made noise include spark-producing mechanisms such as commutators in electric motors, automobile ignition systems, power switching equipment, and fluorescent lights.

Man-made

wide range of frequencies radio waves. trial

Man-made

areas and

is

noise

that are

noise

is

is

also impulsive in nature and therefore contains a in the same manner as more populated metropolitan and indus-

propagated through space

most intense

in

sometimes called industrial noise.

Internal noise.

Internal noise

is

electrical interference generated within a device.

There are three primary kinds of internally generated noise: thermal, shot, and

transit-

time.

Thermal noise. Thermal noise is a phenomenon associated with Brownian movement of electrons within a conductor. In accordance with the kinetic theory of matter, electrons within a conductor are in thermal equilibrium with the molecules and in constant

random motion. This random movement is accepted as being a confirmation of the kinetic theory of matter and was first noted by the English botanist, Robert Brown (hence the name "Brownian noise"). Brown first observed evidence for the kinetic (moving-particle) nature of matter while observing pollen grains under a microscope.

Brown noted an difficult to in the

air.

made them extremely He later noted that this same phenomenon existed for smoke particles Brownian movement of electrons was first recognized in 1927 by J. B. extraordinary agitation of the pollen grains that

examine.

Johnson of Bell Telephone Laboratories. In 1928, a quantitative theoretical treatment was furnished by H. Nyquist (also of Bell Telephone Laboratories). Electrons within a conductor carry a unit negative charge, and the mean-square velocity of an electron

is

proportional to the absolute temperature. Consequently, each flight of an electron between

Because the electron moverandom and in all directions, the average voltage produced by their movement is V dc. However, such a random movement gives rise to an ac component. This ac component has several names, which include thermal noise (because it is temperature dependent), Brownian noise (after its discoverer), Johnson noise (after the person who related Brownian particle movement to electron movement), random noise (because the direction of electron movement is totally random), resistance noise (because the magnitude of its voltage is dependent upon resistance), and white noise (because it contains all frequencies). Hence thermal noise is the random motion of free electrons

collisions with molecules constitutes a short pulse of current.

ment

is

totally

within a conductor caused by thermal agitation.

The

equipartition law of

Boltzmann and Maxwell combined with

the

works of

Electrical

Noise

23

FIGURE

19

Noise source equivalent

circuit.

Johnson and Nyquist a 1-Hz bandwidth

states that the thermal noise

power generated within

a source for

is

N = KT

(1-17)

where

N = K= = T= Thus

at

noise

power density (W/Hz)

Boltzmann's constant 23 1.38 x 10~ J/K absolute temperature (K) (room temperature

room temperature,

N = The

total noise

the available noise

1.38

power

power density

x 10" 23 J/K x 290 is

=

K=

17°C or 290 K)* is

-21 4 x 10

W/Hz

equal to the product of the bandwidth and the noise

density. Therefore, the total noise

power present

in

bandwidth (B)

N = KTB

is

(1-18)

where

KT

N— N = B=

total

noise

power in bandwidth B (W) power density (W/Hz)

noise

bandwidth of the device or system (Hz)

Figure 1-19 shows the equivalent circuit for an electrical noise source. The internal resistance of the noise source (R f )

worst case condition

(maximum

is in

series with the

rms noise voltage (VN ). For the

transfer of noise power), the load resistance (R)

is

made equal to Rj. Therefore, the noise voltage dropped across R is equal to VN/2, and the noise power (N) developed across the load resistor is equal to KTB. Therefore, VN is

determined as follows:

*0K

273°C.

Introduction to Electronic Communications

24

N=KTB

Chap.

= Q0Q

F (dB) =

A

noise figure of 6

dB

10 log 4

indicates that the

=6

power

^

dB

signal-to-noise ratio decreased by a factor

of 4 and the voltage signal-to-noise ratio decreased by a factor of 2 as the signal propagated

from the input

to the output

of the amplifier.

1

Multiplexing

When two (NF)

figure total

is

or

more

amplifiers or devices are cascaded together, the total noise

the accumulation of the individual noise figures. Mathematically, the

noise figure

is*

A

A A2

V

A A 2A 3

X

x

;

{

where

NF = Fj =

noise figure

noise figure of amplifier

1

F2 = F3 =

noise figure of amplifier 2

= A2 = A3 =

gain of amplifier

Aj

It

total

noise figure of amplifier 3 1

gain of amplifier 2

gain of amplifier 3

can be seen that the noise figure of the

each of the

first

amplifier (F

t

)

contributes the most

The noise introduced in the first stage is amplified by succeeding amplifiers. Therefore, when compared to the noise introduced

toward the overall noise

figure.

in the first stage, the noise

by a factor equal

EXAMPLE

added by each succeeding amplifier

is

effectively reduced

to the product of the gains of the preceding amplifiers.

1-7

For three cascaded amplifiers each with noise figures of 3 dB and gains of 10 dB, determine the total noise figure.

Substituting into Equation

Solution

-26 gives us

1

1

NF

F3 -

1

A A2 X

2-

1

100

=

2.11

=

3.24dB

and 10 log 2.11

An

overall noise figure of 3.24

dB

less than the

S/N

dB

indicates that the

ratio at the input :o I

A

r

S/N

ratio at the output

of

A3

is

3.24

.

MULTIPLEXING the transmission of information (either voice or data) from more than more than one destination on the same transmission medium. Transmissions occur on the same medium but not necessarily at the same time. The transmission

Multiplexing

one source

*

is

to

Noise figures and gains are given

in absolute ratios rather than in decibels.

Introduction to Electronic Communications

30

medium may be

a metallic wire pair, a coaxial cable, a

radio, or a fiber optic link.

There are several ways

common methods

although the two most

in

microwave

Chap.

1

radio, a satellite

which multiplexing can be achieved,

are frequency-division multiplexing and time-

division multiplexing.

Frequency-Division Multiplexing (FDM), multiple sources that originally occupied the same frequency band are transmitted simultaneously over a single transmission medium. Thus many relatively narrowband channels can be transmitted over a single wideband In frequency -division multiplexing

transmission system.

FDM analog and

an analog multiplexing scheme; the information entering the system

is it

remains analog throughout transmission.

An example

of

FDM

is

commercial broadcast band, which occupies a frequency spectrum from 535

kHz. Each the audio

station carries an audio intelligence signal with a

from each

would be impossible

station

to

1605

bandwidth of 5 kHz.

were transmitted with the original frequency spectrum,

to separate

one

station's transmissions

is

AM

the

If it

from another's. Instead,

each station amplitude modulates a different carrier frequency and produces a signal with a 10-kHz bandwidth. Because adjacent stations' carrier frequencies are separated

by 10 kHz, the

kHz frequency

total

commerical

AM

slots stacked next to

particular station, a receiver

is

is

divided into one hundred and seven 10in the

are frequency-division multiplexed

shows how commercial and transmitted over a

There are many other applications for

television broadcasting

frequency domain.

To

receive a

simply tuned to the frequency band associated with that

station's transmissions. Figure 1-21

(free space).

band

each other

FDM,

AM broadcast station signals single transmission

medium

such as commercial

FM

and high-volume telecommunications systems. Within any of

the commercial broadcast bands, each station's transmissions are independent of

other station's transmissions.

FIGURE stations.

1-21

and

Frequency-division-multiplexing commercial

AM

broadcast-band

all

the

Multiplexing Channel

31

1

input

Analog-to-

Sample-andhold circuit

Anti-aliasing filter

digital

converter

PCM-TDM output

TDM multiplexer

Channel 2 analog input

Analog-to-

Sample-andhold circuit

Anti-aliasing filter

digital

converter

(a)

1

PCM

TDM

frame

PCM

code

Channel

code

Channel 2

1

(b)

FIGURE

1-22

Two-channel

PCM-TDM

TDM

system: (a) block diagram; (b)

frame.

Time-Division Multiplexing With time -division multiplexing (TDM), transmissions from multiple sources occur on same facility but not at the same time. Transmissions from various sources are interleaved in the time domain. The most common type of modulation used with TDM systems is pulse code modulation (PCM). PCM is a type of digital transmission where analog signals are periodically sampled and converted to a series of binary codes, and the codes are transmitted as binary digital pulses. With a PCM-TDM system, several voice band channels are sampled, converted to PCM codes, then time-division multithe

plexed onto a single metallic cable pair. Figure l-22a shows a simplified block diagram of a two-channel system. Each channel

PCM to a

code for channel

PCM

sample

is

is 1

is

alternately

sampled and converted

being transmitted, channel 2

code. While the

PCM

taken from channel

1

is

code from channel 2

and converted to a

being transmitted, the next

code. This process continues

and samples are taken alternately from each channel, converted transmitted. 1

The multiplexer

is

takes to transmit one sample from each channel

it

total

TDM

The

PCM

to

PCM

codes, and

simply a switch with two inputs and one output. Channel

and channel 2 are alternately selected and connected

time

PCM-TDM carrier PCM code. While

being sampled and converted

is

PCM

to a

to the multiplexer output. is

called the

code for each channel occupies a fixed time

slot

frame

The

time.

{epoch) within the

frame. With a two-channel system, the time allocated for each channel

equal to one-half the total frame time.

A

sample from each channel

during each frame. Therefore, the total frame time

is

is

is

taken once

equal to the reciprocal of the

.

Introduction to Electronic Communications

32 sample

shown

rate.

Figure l-22b shows the

TDM

Chap.

1

frame allocation for the two-channel system

in Figure l-22a.

QUESTIONS 1-1. Define electronic communications. 1-2.

What

three primary

components make up a communications system?

1-3. Define modulation. 1-4. Define demodulation.

1-5. Define carrier frequency 1-6. Explain the relationships

among

the source information, the carrier, and the modulated

wave. 1-7.

What

are the three properties of an analog carrier that can be varied?

1-8.

What

organization assigns frequencies for free-space radio propagation in the United States?

1-9. Briefly describe the significance of Hartley's

1-10.

law and give the relationships between informa-

tion capacity

and bandwidth; information capacity and transmission time.

What

two primary

are the

limitations

1-11. Describe signal analysis as

it

on the performance of a communications system?

pertains to electronic communications.

1-12. Describe a time-domain display of a signal

waveform; a frequency-domain display.

1-13.

What

is

1-14.

What

is

meant by the term odd symmetry? What

1-15.

What

is

meant by the term half-wave symmetry?

meant by the term even symmetry? What

another

is is

another

name

name

for

for

even symmetry?

odd symmetry?

1-16. Describe the term duty cycle. 1-17. Describe a (sin x)lx function. 1-18. Define electrical noise. 1-19.

What

is

meant by the term correlated noise?

List

and describe two

common

forms of

correlated noise. 1-20.

What and

is

meant by the term uncorrected noise?

List several types of uncorrected noise

state their sources.

1-21. Briefly describe thermal noise. 1-22.

What

are four alternative

names

for thermal noise?

1-23. Describe the relationship between thermal noise and temperature; thermal noise and band-

width. 1-24. Define signal-to-noise ratio. 1-25. Define noise figure. 1-26.

What

is

An

What does

a signal-to-noise ratio of 100 indicate? 100

amplifier has a noise figure of 20 dB; what does this

the noise figure for a totally noiseless device?

1-27. Define multiplexing. 1-28. Describe frequency-division multiplexing. 1-29. Describe time-division multiplexing.

mean?

dB?

Chap.

Problems

1

33

PROBLEMS 1-1.

For the

train

of square waves shown below:

(a)

Determine the coefficients for the

(b)

Draw

(c)

Sketch the time-domain signal for frequency components up to the

first five

harmonics.

the frequency spectrum.

1

ms

first five

harmonics.

ms

1

+8 V

OV -8 V

1-2. For the pulse

waveform shown below:

(a)

Determine the dc component.

(b)

Determine the peak amplitudes of the

(c)

Plot the (sin x)lx function.

(d)

Sketch the frequency spectrum.

1

0.1

< 2

v

first five

harmonics.

ms

ms >

--

r^

OV

1-3.

Determine the percent second-order, third-order, and

total

harmonic distortion for the output

spectrum shown below.

8

>

6 -

>

4

?

*

1 4

8

12

Frequency, F (kHz)

1-4.

Determine the percent second-order intermodulation distortion for the A-band, B-band, and second-order intermodulation components shown below.

Introduction to Electronic Communications

34

863

856

1374

1385

1885

1892

A-band

B-band

Chap.

1

ri

1896

2A-B

1903

1914

1907

cross products

Frequency, F (Hz)

1-5.

Determine the second-order cross-product frequencies for the following A- and B-band frequencies:

1-6.

at a

A =

1356 and 1365 Hz.

temperature of 27°C with a bandwidth of 20 kHz, determine:

(c)

The noise power density (N ) in watts and dBm. The total noise power (A7) in watts and dBm. The rms noise voltage (V^) for a 50-0 internal resistance and a 50-fl load

(a)

Determine the noise power (A )

(a)

(b)

1-7.

B = 822 and 829 Hz,

For an amplifier operating

7

(b)

in watts

400°C with a bandwidth of

ture of

Determine the decrease

in noise

1

and

resistor.

dBm for an amplifier operating at a tempera-

MHz.

power

in decibels

if

the temperature decreased to

100°C. (c)

1-8.

Determine the increase

figures of 3 1-9.

in noise

power

in decibels if the

bandwidth doubled.

Determine the overall noise figure for three cascaded amplifiers, each with individual noise

dB and power

gains of 20 dB.

Determine the duty cycle for the pulse waveform shown below.

-

i

0.1

0.02

a;

ms

ms

"5.

E


20K£20K>

*i

Figure 2. Test Circuit for Single Supply Operation

Figure 3. Simplified Schematic of Frequency Control

Mechanism

1

XR-2207 SELECTION OF EXTERNAL COMPONENTS

Single Supply Operation

The

circuit should be interconnected as

shown

Figure 12

in

for single supply operation. Pin 12 should be grounded,

Pin

1

1

V+

biased from

through a

of bias voltage between to ground through a

resistive divider to a value

V+/3 and V + /2.

Pin 10

is

bypassed

3)

The

inversely

maximum For single supply operation, the dc voltage the

timing

terminals

approximately 0.6

The

logic

V

at Pin

10 and

are equal

7)

below Vg, the bias voltage

levels at the

enced to the voltage

4 through

(Pins

at Pin

and 1

1

100

fiF.

a fixed

is

in

proportional to the

Table

The minimum

1.

limited by stray capacitances and the

value by physical size and leakage current con-

Recommended

siderations.

100 pF to

values range from

The capacitor should be non-polar.

Timing Resistors (Pins

and 7)

4, 5, 6,

0.

The timing For

is

1.

binary keying terminals are refer-

at Pin

frequency

oscillator

timing capacitor, C, as indicated capacitance value

juF capacitor.

1

Timing Capacitor (Pins 2 and

and

frequency of f3 = I/R3C, the external circuit

connections can be simplified as shown

in

Figure 12B.

resistors

determine the

total timing current, lj,

available to charge the timing capacitor. Values for timing resistors can range

mum

from 2 Kft to 2M£2; however, for

temperature and power supply

stability,

ed values are 4 Kft to 200 KI2 (see Figures

To

8,

opti-

recommend10 and

11).

avoid parasitic pick up, timing resistor leads should be

kept as short as possible. For noisy environments, unused

r—OV

or

timing

deactivated

ground through

should

terminals

be

bypassed to

0.1 juF capacitors.

iio—1—O SQUARE WAVE OUT

KEYING INPUTS

Supply Voltage (Pins

The XR-2207 range of ±4 V

is

to

1

and 12)

designed to operate over a power supply

±13 V for split supplies, or 8 V to 26 V At high supply voltages, the frequency

for single supplies. 6

5

CB

=

7

sweep range

12

is

BYPASS CAP «2

>

"3

>

«4

>

y

optimum

±6 V, or 12

V

8).

Performance

single supply operation.

Binary Keying Inputs (Pins 8 and 9)

The

internal

impedance

5 K£2. Keying levels are

V*

9

reduced (see Figures 7 and

CB

o-L

cB

is

for

"one"

c

r

fin

—O— —OSQi 1

f-1/R3C CB

-

BYPASS CAP

(see

at


3 V

for

dc voltage

at Pin

10

logic levels referenced to the

Table

is

"zero" and

1).

Bias for Single Supply (Pin

1

)

For single supply operation, Pin 11 should be externally to a potential between V + /3 and V + /2 volts (see

biased

Figure 12).

The

bias current at Pin 11

is

nominally

5%

of

the total oscillation timing current, ly.

i

i i i 3

Ground

(Pin 10)

I For

split

supply operation, this pin serves as circuit ground.

For single supply operation, Pin 10 should be ac grounded through a 1 /uF bypass capacitor. During split supply operaFigure 4. Split Supply Operation

A: General

B: Fixed Frequency

tion, a qround current of 2lj flows out of where \j is the total timing current.

this terminal,

XR-2207 TYPICAL CHARACTERISTICS *T

*

TA

•

'AAAILEL COMBINATION Of ACTIVATED 1

TA

•»

C

AEMTOAS JVC

SPLIT

Figure

SUWIV VOLTAGE

5. Positive l

(VOLTtl

Supply Current,

+ (measured at pin 1)

Figure

6.

vs.

Negative Supply Current, I'

Supply Voltage*

(measured at pin 12)

vs.

Supply Voltage

1

I 5

r///////

100

KD

Ttywcal OFEHATIN °vvv^S. 'AANOE

1

%^^

w

1

NEGATIVE IU*TLV (VOLT)

Figure 7. Typical Operating Range for Split

Figure 8.

Supply Voltage

Recommended Timing Resistor Value

vs.

Power

Supply Voltage*

1

20

vs

-

c

-

•-•V

of

*n 2*

11

1

2

T.

4Kf

zE ;E

200

»T

MMMO C

?

20 Ki 1' 4 IB

...

Ta-J re OTAL •! «0»f

IKfll

I*

1 1

t

/

\

1

•HIT tUm.V VOLTAGE IVOCTSI

lEStfTANCE tOHMtl

SMOlf

Timing Resistance

\ 1

»nny^

^

Figure 9. Frequency Accuracy

i

vs.

«Om V VOLT AOE

fVOI.TR

Figure 10. Frequency Drift vs

Supply Voltage Note:

Rf

Figure 11. Normalized Frequency Drift with

Temperature

= Parallel Combination of Activated Timing Resistors

XR-2207 Frequency Control (Sweep and FM) CB -

JT"-

BYPASS CAP

The frequency

r

D. ^-O

1

is controlled by varying the drawn from the activated timing The timing current can be modulated by

of operation

total timing current,

Pins 4, 5,

6,

or 7.

Ij,

applying a control voltage,

O SQUARE WAVE OUT

through

For

\/q, to

a series resistor Rrj, as

split

supply

operation,

the activated timing pin

shown a

in

Figures 13 and 14.

negative control

voltage,

Vc, applied to the circuits of Figures 13 and 14 causes the total timing current, j, and the frequency, to increase. I

As an example, ing

in

the circuit of Figure 13, the binary key-

inputs are grounded. Therefore, only timing Pin 6

activated.

The frequency

of operation, normally

is

1

R 3C

now

is

mined CB

proportional to the control voltage, Vq, and deteras:

irtf

VCR3

1

]

R3C

R C V-

J

cB

—CMA\

—O—

f

O

WAVE —O—1—O SQUARE OUT

r

3

l I—«3C

-

-

B

i i i i -L,

Figure 12. Single Supply Operation

A: General

"•Lsu B: Fixed

Frequency CB

-

'CRT

Squarewave Output (Pin 13)

±7

»x

BYPASS CAPACITOR

J-OVcA-

I'-RCV-J

The- squarewave output at Pin 13 is an open collector stage capable of sinking up to 20 of load current. R|_ serves

mA

as a pull-up

load resistor for this output.

values for R|_ range from

1

Recommended

Figure 13. Frequency

Sweep Operation

K£2 to 100 K£2.

=

is made 5000 pF.

at Pin 14 is a triangle wave with a peak swing of approximately one-half of the total supply voltage. Pin 14 has a very low output impedance of 10 fi and is internally

then a 1000:1 frequency sweep would result for

a negative

protected against short circuits.

can be expressed

Triangle

Output

The frequency,

(Pin 14)

more The output

will increase as the control voltage

f,

negative.

R3

If

sweep voltage Vc sion gain, K,

is

=

= 2 Mfi.

V".

Re

= 2 Kft.

The voltage

to frequency conver-

controlled by the series resistance as:

Af

Bypass Capacitors

Hz/volt

The recommended value

AVC for bypass capacitors

is

1

/iF,

although, larger values are required for very low frequency operation.

C

R C CV-

Hq

and

i

XR-2207

OUTPl -

jr""

BYPASS CAPACITOR

TV CYCLE DUTY

-

CB?^

?

if

f

^

3

22o-t-0«

5

CB

•

6




IFs:

minimum bandwidth = 463 kHz - 447 kHz =

16

Figure 4-8b shows the IF bandpass characteristics for Example 4-5.

kHz

AM

Receivers

141

600

800

1200

1000

1400

1600

Frequency, F (kHz) (a)

Minimum bandwidth due

Ideal

to tracking error = 16 kHz-

minimum bandwidth = 10 kHz-

F (kHz)

447

448

449

450

451

452

453

454

455

456

458

457

459

460

461

462

463

(b)

FIGURE

4-8

Tracking error for Example 4-5:

(a)

tracking curve; (b) bandpass characteris-

tics.

Image frequency. An image frequency

RF

which,

if

is

any frequency other than the selected

allowed to enter a receiver and mix with the local oscillator will produce

a cross-product frequency that

equal to the IF. Each

is

Once an image frequency has been mixed down suppressed. If the selected

RF

carrier

same time, they both mix with

and

it

an image frequency.

cannot be

filtered out or

image frequency enter a receiver

at the

the local oscillator frequency in the mixer/converter

and produce a difference frequency equal are received

its

RF carrier has

to IF,

to the IF.

Consequently, two different stations

and demodulated simultaneously producing two audio

frequency to produce a cross product equal to the IF,

it

signals.

For a radio

must be displaced from the

by a value equal to the IF. With high-side injection, the below the local oscillator frequency. Therefore, the image frequency is the radio frequency that is the IF above the local oscillator frequency. Mathematically, for high-side injection, the image frequency is

local oscillator frequency

selected

RF

is

the IF

Amplitude Modulation Reception

142

^lo

image

RF

and since the desired

Chap. 4

+ **

(4-6a)

equals the local oscillator frequency minus the IF:

Frf +2F

Fi mage

(4-6b)

lf

Figure 4-9 shows the relative frequency spectrum for the RF, the IF, the local oscillator frequency,

and the image frequency

in a

superheterodyne receiver using high-

side injection.

From Figure 4-9

it

can be seen that the higher the IF, the farther away

frequency spectrum the image frequency

image frequency rejection, a high IF is more difficult it is to build stable amplifiers with high gain. Therefore, there off

when

in the

from the selected RF. Therefore, for better preferred. However, the higher the IF, the

is

is

a trade-

between image frequency rejection and

selecting the IF for a radio receiver

IF gain.

Image frequency

rejection ratio.

The image frequency

rejection ratio

(IFRR)

is

a

numerical measure of the ability of a preselector to reject the image frequency. For a single-tuned circuit, the ratio of

frequency

is

the

gain

its

IFRR =

RF

at the selected

IFRR. Mathematically, IFRR

V

1

to the gain at the

image

is

+

QV

(4-7a)

where

P

=

Q=

F(image)

F(RF)

F(RF)

F (image)

If there is

preselector

log

IFRR

more than one tuned

circuit in the front

and a separately tuned

filter

(4-7b)

quality factor of the tuned circuit

IFRR (dB) = 20

RF

end of the receiver (perhaps a

amplifier), the

IFRR

is

simply the product

of the two ratios.

Image-LO 2IF

LO-RF LO-RF

-«-IF (LO-RF)-

>*

(IF)-

•Frequency IF

LO

RF

FIGURE

4-9

Image frequency.

Image

AM

Receivers

143

EXAMPLE 4-6 For an

AM

broadcast-band superheterodyne receiver with an IF, RF, and local oscillator

frequency of 455, 600, and 1055 kHz, respectively: (a)

Determine the image frequency.

(b) Calculate the

Solution

(a)

IFRR

for a preselector

From Equation

Q

of 100.

4-6a,

Fimage = rF lo

=

F "if

4-

'

+ 455 kHz = 1510 kHz

1055 kHz

or from Equation 4-6b, ^image

= F rf +

2^if

= 600 kHz + (b)

From Equations 4-7a and

2(455 kHz)

= 1510 kHz

4-7b,

_ 1510 kHz _ 600 kHz

~

P

= IFRR =

=

600 kHz 2.51

-

Vl +

1510 kHz

0.397

=

2

2.113

(100 )(2.113

21 1.3 or 46.5

2 )

dB

See Figure 4-10.

Mixer/converter

LO - RF =

RF = 600kHz

^y

Image = 1510 kHz

t

1055 - 600 = 455 kHz

IF

Image - LO = IF

1510- 1055

= 455 kHz

k

Local oscillator

1055 kHz

FIGURE

4-10

Frequency conversion for Example

4-6.

Once an image frequency has been down-converted to IF, it cannot be removed. it has to be removed prior to the mixer/converter stage. Image frequency rejection is the primary purpose of the RF preselector. If the

Therefore, to reject the image frequency,

bandwidth of the preselector

is

sufficiently narrow, the

from entering the receiver. Figure 4-11

illustrates

how

image frequency

RF and IF RF carrier.

proper

prevent an image frequency from interfering with the selected

is

prevented

filtering

can

Chap. 4

Amplitude Modulation Reception

144

RF

Image

(passed)

(blocked)

Preselector

low

selectivity

(wide passband)

1 I

|

1st IF filter

medium selectivity

(medium passband)

Final IF filter

high selectivity

(narrow passband)

FIGURE

The rejection.

ratio of the

The

EXAMPLE

4-11

Image frequency

rejection.

RF to the IF is also an important consideration for image frequency RF is to the IF, the closer the RF is to the image frequency.

closer the

4-7

For a citizens' band receiver using high-side injection with an of 455 (a)

(b)

kHz

determine:

The local oscillator frequency. The image frequency.

RF

of 27

MHz

and an IF

AM

Receiver Circuits

(c)

(d)

The IFRR for a The preselector 600 kHz.

Solution

(a)

145

preselector

Q

From Equation

Flo = (b)

From Equation

of 100. as that achieved for an

RFof

4-5a,

MHz +

27

=

455 kHz

MHz +

27 455 -

From Equations 4-7a and

-

27.455

MHz

455 kHz

=

27.91

MHz

4-7b,

IFRR = (d)

same IFRR

4-6a,

^image (c)

Q

required to achieve -the

6.7 or 16.5

dB

Rearranging Equation 4-7a yields

/IFRR 2

Q~ V

p

-1

2

/211.3

2

—

V 0.0663 2

1

=

3187

From Examples 4-6 and 4-7 it can be seen that the higher the RF carrier, the more difficult it is to prevent the image frequency from entering the receiver. For the same IFRR, the higher RF carrier requires a much higher-quality filter in the preselector. This

illustrated in Figure 4-12.

is

Image

1055 kHz

27.91

MHz

I

600 kHz

455 kHz

1510 kHz

Low Q

HighQ

Frequency

-**-

RF

IF

LO

RF

Image

27

FIGURE

AM

4-12

Frequency spectrum for Example

MHz

4-7.

RECEIVER CIRCUITS RF Amplifier Circuits

An RF

amplifier

is

when used, is the first The primary purposes of an RF stage

a high-gain, low-noise, tuned amplifier that

active stage encountered

by a received

signal.

are selectivity, amplification, and sensitivity. Therefore, characteristics that are desirable in

an

1.

2.

RF

amplifier are:

Low Low

thermal noise noise figure

146

Amplitude Modulation Reception

3.

Moderate

4.

Low

5.

Moderate

6.

High image frequency

to high gain

intermodulation and harmonic distortion

Two

selectivity

rejection ratio

of the most important parameters for a receiver are amplification and noise

demodulator detects amplitude variations in its

RF

of which both are dependent on the performance of the

figure,

Chap. 4

in its input signal

AM

An

stage.

and converts them

to

changes

output signal. Consequently, amplitude variations caused by noise are converted

to erroneous fluctuations in the detector output

and the quality of the receive signal

degraded. The more gain that a signal experiences as

more pronounced

are

its

amplitude variations

demodulator input, and the

at the

noticeable are the variations caused by noise.

is

passes through a receiver, the

it

The narrower

less

the bandwidth, the less

noise propagated through the receiver and, c onsequen tly, the less noise demodulated

by the detector. From Equation 1-19 (VN

= \/4RKTB)

tional to the square root of the temperature, the if

these three parameters are minimized, the thermal noise

of an

RF

stage can be reduced

bandwidth of an is

noise voltage

,

RF

amplifier

by

artificially

is

directly propor-

bandwidth, and the resistance. Therefore, reduced. The temperature

is

cooling the front end of a receiver. The

reduced by using tuned amplifiers, and the resistance

is

reduced by using specially constructed solid-state components for the active device.

Noise figure

essentially a

is

measure of the gain of the amplifier

added by the amplifier. Therefore, the noise figure

to the noise

improved (reduced)

is

either

by

reducing the internal noise or by increasing the amplifier's gain. Intermodulation and harmonic distortion are both forms of nonlinear distortion that reduce the noise figure

more

by adding correlated noise

The IFRR

better the receiver's noise figure.

IFRR the

image frequency from entering the mixer/converter from the

is

a relative term.

efficiently radiated

wave.

RF

microwave radio

is in

excess of

broadcast band receivers

frequency, and IF

many of

is

stage. Consequently,

amplifier circuits.

simply means that the frequency

broadcast band

frequencies associated with the

fore,

amplifier combines with the

is

1

is

GHz

10.7

is

Keep

in

(1000 MHz).

MHz, which

free space as an electromagnetic

A common

is

RF

mind

high enough to be

between 535 and 1605 kHz, whereas

AM broadcast band.

RF

for

IF frequency used for

considerably higher than the

RF

simply the radiated or received

is

an intermediate frequency within a transmitter or a receiver. There-

the considerations for

neutralization, filtering,

RF

amplifiers also apply to IF amplifiers such as

and coupling.

Figure 4- 13a shows a schematic diagram for a bipolar

and L, form the coupling

circuit

from the antenna.

nonlinear distortion.

The

through T\ t which

double tuned for more selectivity.

is

moderate

stage.

by an antenna and propagated through

AM

for the

RF

RF

commonly used RF

Figure 4-13 shows several

RF

that

and the

of the preselector to reduce the receiver input bandwidth sufficiently and prevent

selectivity is all that is required

FM

RF

an

ratio of

The

to the total noise spectrum.

linear an amplifier's operation, the less nonlinear distortion produced,

collector circuit

is

Q

x

is

RF

amplifier.

class

A

Ca C h Cc ,

,

.

biased to reduce

transformer coupled to the mixer/converter

Cx

and

Cv

are

RF

bypass capaci-

v

T,

To mixer/converter

From

cd

X

Figure

in

demodulated without attenua-

=

(4 -' 3a)

2.RC

where

Fa(max) = maximum modulating frequency m = modulation coefficient RC = RC time constant For 100% modulation, the numerator

means

that all

in

Equation 4- 13a goes to 0, which essentially

modulating signal frequencies are attenuated as they are demodulated.

Typically, the modulating signal amplitude in a transmitter

such that approximately

70.7%

90%

modulation

maximum

is

limited or compressed

that

can be achieved. For

modulation, Equation 4- 13a reduces to

—

F Equation 4- 13b

mate

the

is

maximum

is

=

ii

(4 - ,3b)

commonly used when designing peak

detectors to determine an approxi-

modulating signal frequency.

AUTOMATIC GAIN CONTROL AND SQUELCH Automatic Gain Control Circuits

An

automatic gain control circuit

signal level.

The

AGC

(AGC) compensates

for

minor variations

input signals, and automatically decreases the receiver gain for strong signals can be buried in receiver noise and, consequently, tor.

Excessively strong signals can overdrive the

RF

delayed

Simple

AGC,

AGC.

receiver with simple

and forward

signals.

Weak

the audio detec-

AGC,

including direct or simple

AGC.

Figure 4-23 shows a block diagram for an

AGC. The

RF

masked from

received

weak RF

and/or IF amplifiers and produce

excessive nonlinear distortion. There are several types of

AGC,

in the

circuit automatically increases the receiver gain for

AM

superheterodyne

automatic gain control circuit monitors the received

Amplitude Modulation Reception

164

Chap. 4

Antenna

V

Preselector

and

RF

Mixer/converter

amplifier

1

IF amplifiers

Audio

Audio

detector

amplifiers

1

{

AGC correction

Local I

oscillator

voltage I

FIGURE signal level

and sends a signal back

automatically.

AGC

is

4-23

AGC

is

AM

receiver with simple

to the

RF

AGC.

and/or IF amplifiers to adjust their gain

a form of degenerative or negative feedback.

to allow a receiver to detect

The purpose of

and demodulate, equally well, signals

that are

whose output power and distance from the receiver varies. For example, an AM radio in a vehicle does not receive the same signal level from all of the transmitting stations or, for that matter, from a single station when the automobile is moving. The AGC circuit sends a voltage back to the RF and/or IF amplifiers to adjust the receiver gain and keep the IF carrier power at the input to the AM detector at a constant level. The AGC circuit is not a form of automatic volume

transmitted from different stations

IF output

D

KQ,

IF input

1

(audio

detector)

To audio

(IF

IF input

amplifiers

1^ amplifi Ca ±Z

R3

i

yC

AGC 2

^vWV=r^

C

feedback

voltage

i

FIGURE

4-24

Simple

AGC

circuit.

Automatic Gain Control and Squelch control circuit;

changes

in the

AGC

165

independent of modulation and totally unaffected by normal

is

audio modulating signal amplitude.

AGC

Figure 4-24 shows a schematic diagram for a simple

AGC

see, an

circuit is essentially a

fact,

is

that the

average dc voltage

unmodulated

at the

the carrier amplitude increases, the

decreases, the

AGC

output of a peak detector

and

carrier amplitude

AGC is

circuit

AGC detector is fed back Q When the carrier

base of

.

x

more negative, causing

it

correction

was shown

approximately equal to

is

shown

in

If

the carrier amplitude

if

Figure 4-24

is

a negative

The higher the amplitude The negative voltage from

a negative voltage.

of the input carrier, the more negative the output voltage. the

AGC

independent of modulation.

totally

voltage increases and

The

voltage decreases.

peak detector, and therefore the output

is

As you can

circuit.

very often the

taken from the output of the audio detector. In Figure 4-21

voltage

the peak

peak detector. In

to the IF stage,

where

it

controls the bias voltage on the

amplitude increases, the voltage on the base of

As

the emitter current to decrease.

a result,

r'e

Q

x

goes

increases and

the amplifier gain (rc /r'e ) decreases, causing the carrier amplitude to decrease. If the carrier amplitude decreases, the

increases,

r'e

AGC

voltage goes less negative, the emitter current

decreases, and the amplifier gain increases. Capacitor

capacitor that prevents changes in the the bias or gain of

Q

AGC

C

is

x

an audio bypass

voltage due to modulation from affecting

.

x

Delayed AGC. Simple AGC is used in most inexpensive broadcast-band receivHowever, with simple AGC, the AGC bias begins to increase as soon as the received signal level exceeds the thermal noise of the receiver. Consequently, the receiver becomes less sensitive (this is sometimes called automatic desensing). Delayed AGC prevents the AGC voltage from reaching the RF and/or IF amplifiers until the RF level exceeds ers.

a predetermined level.

AGC

delayed

voltage

Once

the carrier signal has exceeded the threshold level, the

proportional to the signal strength. Figure 4-25a shows the

is

AGC. It can be seen that with RF signal is unaffected until the AGC threshold level is exceeded, simple AGC, the RF signal is immediately affected. Delayed AGC is

response characteristics for both simple and delayed

delayed

AGC,

whereas with

the

used with more sophisticated communications receivers. Figure 4-25b shows IF gain versus

RF

input signal level for both simple and delayed

Forward AGC.

An

inherent problem with both simple and delayed

the fact that they are both forms of post-

AGC, is

AGC.

AGC

(after-the-fact compensation).

the circuit that monitors the carrier level and provides the

AGC

that

simple and delayed

Forward AGC to the front

further

is

it

may be

AGC

cannot compensate for rapid changes

similar to conventional

in the receiver.

can be compensated for

in

AGC

voltage

too late (the carrier level has already changed). Therefore,

AGC

in the carrier

except that the carrier

end of the receiver and the correction voltage

back

is

correction voltage

located after the IF amplifiers, and therefore the simple fact that the

changed indicates

AGC

With post-

is

is

fed to IF and/or

Consequently, when a signal change

is

succeeding stages. Figure 4-26 shows an

amplitude.

monitored closer

RF amplifiers

detected, the change

AM superheterodyne

Amplitude Modulation Reception

166

Chap. 4

NoAGC

Simple

AGC

Delayed

RF

AGC

input signal level (a)

60 50

1 I

N.

40 —

AGC

Delayed

30 Simple

AGC^^

20 -

10

-

I

-35

I

I

-30

RF

>l

I

-20

-25

-15

>l

I

-5

-10

FIGURE

input signal level (dBm)

(AGC):

4-25

(a)

IF gain versus

(b)

Automatic gain control

response characteristics; (b)

RF

input signal level.

Ant snna

Preselector

and

RF

amplifier

Mixer/converter

IF amplifiers

of

i


"-

+ * f\j

5.0

1.0

CARRIER FREQUENCY (MHz)

10

£ 3 _

20

i

30

X

IS5 40

z> o.

uj

3

1-

55

I

1

FREQUENCY fC fS fC t fs

II

,

as

7 a

I

NOTE:

«>

T

M

MM

1

o 1

BALANCED MODULATOR SPECTRUM

CARRIER FUNDAMENTAL MODULATING SIGNAL FUNDAMENTAL CARRIER SIOEBANOS

± nfs

fc

nfc nf C

±

number references pertain to this device when packaged numbers for plastic or ceramic packaged devices refer to the pjn

AA)

SM

«

MOTOROLA

nf s

FUNDAMENTAL CARRIER SIDEBAND HARMONICS CARRIER HARMONICS CARRIER HARMONIC SIDEBANDS

in a first

metal can. To ascertain the corresponding page of this specification sheet.

Semiconductor Products

FIGURE

pi

Inc.

6-12(b) (continued)

243

OUTLINE DIMENSIONS

G SUFFIX

METAL PACKAGE -

MILLIMETERS MAX MIN

DIM

A-

I

INCHES MIN MAX

DIM NOTE:

A

MILLIMETERS MIN MAX

INCHES MIN MAX 0.660

0.785

B

5.59

7.11

0.220

0.280

C

-

5.08

_

16.8

19.9

A

8.51

9.39

0.335

0.370

B C

7.75

8.51

0.305

0.335

4.19

4.70

0.165

0.185

D

0.381

0.584

0.015

1

D

0.407

0.533

0.016

0.021

F

0.77

1.77

0.030

|

1.02

-

0.040

0.016

0.019

-

E F

0.406

G H

0.712

0.864

0.028

0.034

J

0.737

1.14

0.029

0.045

-

0.500

-

0.250

0.500

0.483

5 84

K

12.70

L

6.35

BSC

12.70 36° BSC

M -

1.27

-

0.050

3.56

4.06

0.140

0.160

0.254

1.02

0.010

0.040

JEOEC

WWW J3E-

36° BSC

R All

MMAM 11

0.230 BSC

Q

P

DIMENSION "L" TO CENTER OF LEADS WHEN FORMED PARALLEL.

G

2.54 BSC

J

0.203

K

2.54

L

0.381

0.015

-

300 BSC

M

-

15"

_ 0.020

N

0.51

0.76

-

8.25

All

JEDEC

15°

0.030 0.325

dimensions and notes apply

dimensions and notes apply

SUFFIX

L

CERAMIC PACKAGE

NOTE:

mm

CASE 632

RADIUS OF TRUE POSITION AT SEATING PLANE AT MAXIMUM MATERIAL CONDITION.

LEADS WITHIN

R0 JA

0008 0.100

BSC

P

1

0023

0.100 BSC

-

7.62

0.200

0.070

0.18

(0.007)

TO-116

HU-—JGh—

200°C/W (Typ)

=

SEATING

R 0JA = 180°C/W (Typ)

PLANE p SUFFIX PLASTIC PACKAGE CASE 646

A A A A A A A

(MC1496

It

only)

MILLIMETERS

DIM R 0JA " 100°C/W (Typ)

V V V V V V V

2.

LEADS WITHIN 0.13 mm (0.005) RADIUS OF TRUE POSITION AT SEATING PLANE AT MAXIMUM MATERIAL CONDITION. DIMENSION "L" TO CENTER OF LEADS

MAX 18.80

6.10

6.60

0.240

0.260

4.06

4.57

0.160

0.180

0.38

0.51

0.015

0.020

1.02

1.52

0.040

0.060

2.54

BSC

PARALLEL Q

0.740

0.100 BSC

1.32

1.83

0.052

0.072

0.20

0.30

0.008

0.012

2.92

3.43

0.115

0.135

7.37

7.87

0.290

0.310 10°

-

WHEN FORMED

INCHES MIN MAX 0.715

18.16

NOTES: 1.

MIN

10°

-

0.51

1.02

0.020

0.13

0.38

0.005

0.015

0.51

0.76

0.020

0.030

0.040

THERMAL INFORMATION The maximum power consumption an

Tj(max) = Maximum Operating Junction Temperature

integrated circuit

can tolerate at a given operating ambient temperature, can be found from the equation:

J(max) -T

D T A>"
Dissipation

=

Maximum

R0j»(Typ)

;

"

V

'O'O l

allowable

at

a

s

the

Maximum

Desired Operating Ambient Temperature

= Typical

Thermal Resistance Junction to Ambient

= Total Supply Current

given

operating ambient temperature.

® MOTOROLA

Semiconductor Products

FIGURE

244

as listed

Ratings Section

6- 12(b) (continued)

Inc.

Single-Sideband Transmitters

245

DSBSC B =

B=

5kHz

SSBRC

SSBSC

10 kHz

B =

5kHz

5kHz

B =

B =

210kHz

I

\y\ 95k

5k

100k

105k

100k

100k

105k

105k

1.895M

1.9k

2M 2.1M

2.105M

BPF sum 1

Modulating signal

Amp.

input

Summer

> Balanced modulator

i

Balanced

i

modulator

i

r

-A^r\J

-Ar

Buffer

amp.

Buffer

amp.

Carrier pilot

/\ LF

^V

adiust

MF

carrier

osc.

2

SSBRC B =

carrier

osc.

100 kHz

MHz

SSBRC

5kHz

B = 4.21

MHz

B = 5

SSBRC

kHz

B =

5kHz

I

Antenna

2.1M

2.105M

17.895M

17.9M

f

22.1M 22.105M

20M

22.1

BPF sum

BPF sum

2

3

22.105M

j

22.1M 22.105M

\/

M

Linear

power amp.

Balanced

modulator

Ax Buffer

amp.

^V HF

carrier

osc.

20

MHz

FIGURE

6-13

Single-sideband transmitter:

filter

method.

"

~Z_

Single-Sideband Communications Systems

246

separated by a 200-kHz frequency band that

MHz

on 2.1025

is

kHz bandwidth.

with a 5

void of information.

with a

20-MHz

BPF 2 is centered BPF 2 is once

Therefore, the output of

again a single-sideband reduced-carrier waveform. 2.1 -MHz second IF carrier

Chap. 6

Its

spectrum comprises a reduced

and a 5-kHz upper sideband. The output of

high-frequency (HF) carrier in balanced modulator

3.

BPF

2 is mixed The output is a

double-sideband suppressed-carrier signal where the upper and lower sidebands again

each contain the original

SSBRC

4.2-MHz frequency band

that is void of information.

signal spectrum.

The sidebands

by a

are separated

sideband reduced-carrier waveform with a

BPF 3 is centered on 22.1025 the output of BPF 3 is once again a singlereduced 22.1-MHz RF carrier and a 5-kHz

upper sideband. The output waveform

amplified in the linear power amplifier and

MHz

with a 5

kHz bandwidth.

Therefore,

is

transmitted. In the transmitter just described, the original modulating signal spectrum

converted in three modulation steps to a

final carrier

MHz.

single upper sideband that extended to 22. 105

frequency of 22.1

was up-

MHz

and a

After each up-con version (frequency

from the double-sideband spectrum with BPF. The same final output spectrum can be produced with a single heterodyning process: one balanced modulator, one bandpass filter, and a single HF carrier supply. Figure 6-14 shows the block diagram and output spectrum for a single-step up-conversion transmitter. The output of the balanced modulator is a double-sideband spectrum centered translation), the desired sideband is separated

a

around a suppressed carrier frequency of 22.

1

MHz. To separate the 5-kHz upper sideband BPF with an extremely high Q is required.

from the composite spectrum, a multiple-pole

A BPF

that

meets

this criterion is in itself difficult to construct

were a multiple-channel transmitter and the

BPF must

carrier frequency

also be tunable. Constructing a tunable

BPF

in the

but suppose that

were tunable; then

this

the

MHz range with a passband

is beyond economic and engineering feasibility. The only BPF in the shown in Figure 6-13 that had to separate sidebands that were immediately adjacent was BPF 1. To construct a multiple-pole, steep-skirted BPF at 100 kHz is a relatively simple task, as only a moderate Q is required. The sidebands separated by

of only 5

kHz

transmitter

BPF

2 are 200

kHz

apart; thus a low-/)

The output of balanced modulator 2

is

side frequencies to essentially the

same

phasor except that the phase of the carrier and the modulating signal are each rotated 90° from the reference. The output of the linear summer shows the sum of the output

FIGURE

6-16

Mechanical

filter

equivalent circuit.

Single-Sideband Communications Systems

250

Modulating signal input (sin

w

Chap. 6

Phase splitter

a

t)

sin co r t

Carrier

Phase

crystal splitter

osc.

cosco r t

^cos

Balanced modulator

FIGURE

SSB

6-17

(co t c

-

wa + )t

^cos

(co

c

+

co L

(-90°)

wu

(+90°)

w

a

)t

transmitter: phase-shift method.

phasors from the two balanced modulators. The two phasors for the lower sideband are in phase and reinforce, whereas the phasors for the upper sideband are 180° out of

phase and cancel. Consequently, the upper sideband linear

Mathematical analysis. is

is

removed

at the

In Figure 6-17 the input modulating signal (sin

fed directly to balanced modulator

1

and shifted 90°

(to

oo„/)

cos (oj) and fed to balanced

2. The low-frequency carrier (sin u> c t) is also fed directly to balanced modulator and shifted 90° and fed to balanced modulator 2. The balanced modulators are product

modulator 1

output of the

summer.

modulators and their outputs are expressed mathematically as output from

balanced modulator

1

= =

x

(sin ui a t) \

cos

(co t

.

—

(sin a) r r) o) a )t

—

|

cos

(oj (

.

COS

(0) (

.

+

u> a )t

output from

balanced modulator 2

= =

(cos 5

COS

o> a t)

(to (

.

x

(cos

-

(x)

(

,)t

o) ( .r)

+

\

+

ix)

(I

)t

Single-Sideband Transmitters

251

and the output from the linear summer

— c — (w

is

i COS (0) c

+

COS

\

cos

((x)

.

(

(x)

a )t

(x)

a )t

(t>

— +

COS

(oJ c

| cos

(oj r

i

(x)

a )t

ou

J/

canceled

a )t

v

+ +

.

'

v

lower sideband (difference frequencies)

SSB Transmitter: The Third Method The

method of single-sideband generation, developed by D. K. Weaver method described previously in that it uses and summing to cancel the undesired sideband. However, it has an advan-

so-called third

1950s,

in the

phase shifting

similar to the phase shift

is

tage in that the information signal

is initially

modulated onto an audio subcarrier, thus

eliminating the need for a wideband phase shifter (which

The block diagram

for a third-method

that all of the inputs to the

and

a> (

.

+

90°).

The

SSB modulator

is

difficult to build in practice).

is

shown

in

phase shifters are single-frequencies

Figure 6-18. Notice (to,,,

co c

wa

Balanced modulator

co

to,,

+

90°,

wr

,

input audio mixes with the audio subcarrier in balanced modulators

+ 90°

1 co a

LPF

1

cj

+

90° -

CO

+

LO Q

~

w -co +

Balanced

c

CJ a o; a

+ 90 - 90°

no

3

1

t

t

i

90°

RF

90° phase

carrier

oscillator

shift

CO ' f

/

input

SSBSC

Linear

Audio i

wc

i

summing

Wa

wave

circuit 1

'

co c

Audio subcarrier

90° phase

oscillator

shifter

co c


E i

?!

k

1

r-

u

'

^_

o

o-S

'c

'o

d.

~

P

•-

1

E




= a

E co

ill

s

i

fsl

X

1

t

?5

o 1

o

2

E

«r *~

j

n o

U


-

QJ

o

>-

D

«-


a

qj

a.

E E

£8 -

i

CO

,-

X"" i

'

i

-*- --

CM

CO 4-1

LL

/

CO

-*

o

x"' •

i

-

is

(10-5) 4-nR'

Wavefront 2

FIGURE

10-3

Spherical wavefront from

an isotropic source.

Wave

402

Chap. 10

Propagation

where

Pr = R=

total

power

radiated

radius of the sphere (which

is

equal to the distance from any point on

the surface of the sphere to the source) 4tt/?

=

2

area of the sphere

and for a distance

Ra

meters from the source, the power density

is

p 2

4irR a

Equating Equations 10-3 and 10-5 gives

% 4ttR

2

377

Therefore,

%2 =

4ttR

Inverse Square

From Equation the smaller the

V30P, R

311P,

and

2

(10-6)

Law

10-3

it

can be seen that the farther the wavefront moves from the source,

power density (R a and

R c move

farther apart).

The

total

power

distributed

over the surface of the sphere remains the same. However, because the area of the sphere increases in direct proportion to the distance from the source squared radius of the sphere squared), the

power density

is

of the distance from the source. This relationship Therefore, the power density

at

(i.e., the

inversely proportional to the square is

called the inverse square law.

any point on the surface of the outer sphere

is

p

'

2

4itR 2

and the power density

at

any point on the inner sphere

is

Pr 2>,

4ir/?f

Therefore, ff 2 9>,

From Equation

10-7

it

_ Pr l4^R\ " PJ4itR 2

*! = (*i\

(10-7)

can be seen that as the distance from the source doubles,

power density decreases by a factor of 2 2 or 4. When deriving the inverse square law of radiation (Equation 10-7) it was assumed that the source radiates isotropically,

the

although

,

it

is

However, it is necessary that the velocity of propagation be uniform. Such a propagation medium is called an isotropic medium.

not necessary.

in all directions

Wave

Attenuation and Absorption

EXAMPLE

10-1

For an isotropic antenna radiating 100 (a)

(b)

403

The power density 1000 The power density 2000

W of power, determine

m from the source. m from the source.

Solution (a) Substituting into

Equation 10-5 yields

^ (b)

100

Again, substituting into Equation 10-5 gives us

100

® = 4rtM0^

^' m2

L99

we have

or substituting into Equation 10-7, 2

®o2 _ 1000 2 = 7.96 fxW/nr 2 2P,

2000

(1000

\

^J =

1.99mAV/

WAVE ATTENUATION AND ABSORPTION Attenuation The inverse square law

for radiation mathematically describes the reduction in

density with distance from the source.

As

the continuous electromagnetic field that

waves move

is

a wavefront

moves away from

power

the source,

radiated from that source spreads out. That

away from each other and, consequently, the number of waves per unit area decreases. None of the radiated power is lost or dissipated because the wavefront is moving away from the source; the wave simply spreads out or disperses over a larger area, decreasing the power density. The reduction in power density with distance is equivalent to a power loss and is commonly called wave attenuation. Because is,

the

the attenuation

is

farther

due

to the spherical spreading of the

space attenuation of the wave.

Wave

attenuation

is

wave,

it is

sometimes called the

generally expressed in terms of the

common

logarithm of the power density ratio (dB loss). Mathematically, wave attentua-

tion (y a )

is

7,= The reduction propagation

(i.e.,

a

10 log

1

(10-8)

I

power density due to the inverse-square law presumes free-space vacuum or nearly a vacuum) and is called wave attenuation. The

in

reduction in power density due to nonfree-space propagation

is

called absorption.

Wave

404

Propagation

Chap. 10

Absorption The

earth's atmosphere

is

not a vacuum. Rather,

it is

made up of atoms and molecules

Some

of various substances, such as gases, liquids, and solids. capable of absorbing electromagnetic waves.

through the earth's atmosphere, energy

Wave

molecules of the atmosphere. I

2

R power

loss.

Once absorbed,

is

As an

of these materials are

electromagnetic

from the wave

transferred

absorption by the atmosphere

the energy

is

wave propagates to the

atoms and

analogous to an

forever and causes an attenuation in

is lost

the voltage and magnetic field intensities and a corresponding reduction in

Absorption of radio frequencies in a normal atmosphere

is

power

density.

dependent on frequency

and relatively insignificant below approximately 10 GHz. Figure 10-4 shows atmospheric absorption in

GHz.

It

dB/km due

to

oxygen and water vapor

for radio frequencies above 10

can be seen that certain frequencies are affected more or

creating peaks and valleys in the curves.

Wave

depend on distance from the radiating source, but

wave propagates through

less

by absorption,

attenuation due to absorption does not rather, the total distance that the

the atmosphere. In other words, for a

homogeneous medium

(one with uniform properties throughout), the absorption experienced during the mile of propagation

is

the

same

first

as for the last mile. Also, abnormal atmospheric conditions

such as heavy rain or dense fog absorb more energy than a normal atmosphere. Atmo-

wave propagating from R to R 2 is y(R 2 — R\), where 7 is the absorption coefficient. Therefore, wave attenuation depends on the ratio R 2 /R\ and wave absorption depends on the distance between R and R 2 In a more practical situation spheric absorption

(t^)

for a

{

.

{

(i.e.,

an inhomogeneous medium), the absorption coefficient varies considerably with

location, thus creating a difficult

15

20

30

40 50 60 80 100 Frequency, F (GHz)

problem

150

200

for radio systems engineers.

FIGURE

10-4

Atmospheric absorption

of electromagnetic waves.

Optical Properties of Radio

Waves

405

OPTICAL PROPERTIES OF RADIO WAVES In the earth's atmosphere, ray-wavefront propagation

may be

from free-space

altered

behavior by optical effects such as refraction, reflection, diffraction, and interference.

Using rather unscientific terminology, refraction can be thought of as bending, reflection as bouncing, diffraction as scattering, diffraction,

and interference as colliding. Refraction,

and interference are called optical properties because they were

in the science

of optics, which

is

the behavior of light waves.

high-frequency electromagnetic waves,

it

Because

reflection,

first

light

observed

waves

also apply to radio- wave propagation. Although optical principles can be analyzed pletely

are

stands to reason that optical properties will

by application of Maxwell's equations,

this is necessarily

applications, geometric ray tracing can be substituted for analysis

com-

complex. For most

by Maxwell's equa-

tions.

Refraction Electromagnetic refraction

from one medium the velocity at

is

the change in direction of a ray as

which an electromagnetic wave propagates

the density of the

it

passes obliquely

to another with different velocities of propagation. As stated previously,

medium

in

which

ever a radio wave passes from one

it is

is

inversely proportional to

propagating. Therefore, refraction occurs when-

medium

into another

medium

of different density.

Figure 10-5 shows refraction of a wavefront at a plane boundary between two media

with different densities. For

this

example, medium

1

is

less

dense than medium 2

Normal

FIGURE

10-5

Refraction at a plane boundary between two media.

(i.e.,

Wave

406 Vj

>

ray

v 2 ).

B

can be seen that ray

It

downward

A

travels distance

more dense medium before ray B. Therefore, travels distance B-B' during the same

enters the

A and

propagates more slowly than ray

time that ray

all

A

A-A' Therefore, wavefront (A'B .

direction. Since a ray

f

)

or bent in a

is tilted

defined as being perpendicular to the wavefront

is

points, the rays in Figure 10-5

Chap. 10

Propagation

have changed direction

at the interface

media. Whenever a ray passes from a less dense to a more dense medium, bent toward the normal. (The normal

it is

at

of the two effectively

simply an imaginary line drawn perpendicular

is

whenever a ray passes from a away from the normal. The angle of incidence is the angle formed between the incident wave and the normal, and the angle of refraction is the angle formed between the refracted wave and the normal. The amount of bending or refraction that occurs at the interface of two materials of different densities is quite predictable and depends on the refractive index (also called the index of refraction) of the two materials. The refractive index is simply the ratio of to the interface at the point of incidence.) Conversely,

more dense

to a less

dense medium,

effectively bent

is

it

the velocity of propagation of a light ray in free space to the velocity of propagation

of a light ray in a given material. Mathematically, the refractive index

=

n

is

c

-

(10-9)

v

where

= = =

n c v

The

refractive index

speed of light

speed of

refractive index

most applications electromagnetic that

is

in free

light in a

is

also a function of frequency.

insignificant

wave

reacts

space

given material

and

when

it

is

However, the

variation in

therefore omitted from this discussion.

How

an

meets the interface of two transmissive materials

have different indexes of refraction can be explained with Snell's law. Snell's law

simply

states:

«! sin 0j

=

n 2 sin

(10-10)

:

and sin 0,

«2

sin 0-

where n

y

n2 0i 2

= = = =

refractive index of material

1

refractive index of material 2

angle of incidence angle of refraction

and since the refractive index of a material constant,

is

equal to the square root of

its

dielectric

Optical Properties of Radio

Waves

407

Original

wavefront

FIGURE

10-6

Wavefront refraction

in a gradient

medium.

sin 0]

(10-11) sin 6-

where e, == dielectric

e2

—

constant of

dielectric constant of

Refraction also occurs

gradient that front).

is

a wavefront propagates in a

perpendicular to the direction of propagation

medium

less

dense

its

refractive index.

at the top.

The medium

medium

the

more dense near

wave-

that has a

the

bottom

Therefore, rays traveling near the top travel faster than rays

near the bottom and consequently, the wavefront a gradual fashion as the

is

that has a density

(i.e., parallel to

Figure 10-6 shows wavefront refraction in a transmission

gradual variation in

and

when

medium 1 medium 2

wave progresses

as

tilts

downward. The

tilting

occurs in

shown.

Reflection Reflect

means

to cast or turn back,

all

when an

and reflection

wave

is

the act of reflecting. Electromagnetic

boundary of two media and some or of the incident power does not enter the second material. The waves that do not

reflection occurs

incident

strikes a

penetrate the second

medium

reflection at a plane

boundary between two media. Because

are reflected. Figure 10-7

shows electromagnetic wave all

the reflected

waves

Wave

408 Reflected

Incident

wavefront

,

wavefront

FIGURE at a plane

remain

in

medium

Chap. 10

Propagation

1,

10-7

Electromagnetic reflection

boundary of two media.

waves are equal. Conse= G r ). However, the than the incident voltage field intensity. The

the velocities of the reflected and incident

quently, the angle of reflection equals the angle of incidence (0, reflected voltage field intensity is less ratio

of the reflected to the incident voltage intensities

T (sometimes

called the coefficient of reflection)

is

called the reflection coefficient,

For a perfect conductor,

.

used to indicate both the relative amplitude of the incident and reflected

r =

fields

1.

and

T

is

also

the phase shift that occurs at the point of reflection. Mathematically, the reflection coefficient

is

F i>J® _ fV F ^r r= r

j(d r -Bd

(10-12)

where

T =

E = Er = 0, = = r t

The

reflection coefficient

incident voltage intensity reflected voltage intensity

incident phase reflected phase

ratio

total incident

of the reflected and incident power densities

power

that is not reflected is called the

(or simply the transmission coefficient).

is

Y.

The portion of

power transmission

For a perfect conductor,

T =

conservation of energy states that for a perfect reflective surface, the

power must equal

the

coefficient (T) 0.

The law of

total reflected

the total incident power. Therefore,

r+|r| 2 = For imperfect conductors, both the electric field polarization,

|T|

2

and

T

(10-13)

i

are functions of the angle of incidence,

and the dielectric constants of the two materials.

If

medium

2 is not a perfect conductor, some of the incident waves penetrate it The absorbed waves set up currents in the resistance of the material and the energy is converted to heat. The fraction of power that penetrates medium 2 is called the absorption

and are absorbed.

coefficient (or

sometimes the

coefficient

of absorption).

Optical Properties of Radio

Waves

409

Incident

/

wavefront Specular

wavefront Incident rays

Diffuse reflection-

FIGURE

When

0-8

J

the reflecting surface

Reflection

is

from a semirough surface.

not plane

(i.e.,

it

is

curved) the curvature of the

wave is different from that of the incident wave. When the wavefront of the incident wave is curved and the reflective surface is plane, the curvature of the reflected wavefront is the same as that of the incident wavefront. Reflection also occurs when the reflective surface is irregular or rough. However,

reflected

such a surface strikes is

may

destroy the shape of the wavefront.

an irregular surface,

it is

randomly scattered

in

When

many

called diffuse reflection, whereas reflection from a perfectly

specular (mirror-like) reflection. Surfaces that called semirough surfaces. lar reflection.

A

is

semirough surface

semirough surface

is

a condition is

called

between smooth and irregular are

is

shown

will not totally destroy the shape of the reflected

a reduction in the total power. will reflect as if

cosine of the angle of incidence surface irregularity and \

Such

smooth surface

Semirough surfaces cause a combination of diffuse and specu-

semirough surface

wavefront. However, there states that a

fall

an incident wavefront

directions.

is

it

The Rayleigh

criterion

were a smooth surface whenever the

greater than X/Sd,

where d

is

the depth of the

the wavelength of the incident wave. Reflection

in

from a

Figure 10-8. Mathematically, Rayleigh 's criterion

cos

6/

>

is

(10-14)

%d Diffraction Diffraction

when

it

is

defined as the modulation or redistribution of energy within a wavefront

passes near the edge of an opaque object. Diffraction

is

the

phenomenon

that

allows light or radio waves to propagate {peek) around corners. The previous discussions

of refraction and reflection assumed that the dimensions of the refracting and reflecting

Initial

Wavefront moves forward

incident

wavefront

Cancellation

Secondary wavelets (a)

Obstacle Obstacle

> Secondary point sources

Waves

\

Reflected

reflected

off obstacle

P2

rays

*

No cancellation of secondary

/

wavelet

Shadow zone (no cancellation)

Edge P3

Slot

P4

Pb

(0

(b)

FIGURE finite

410

10-9

Electromagnetic wave diffraction:

wavefront through a

slot; (c)

around an edge.

(a)

Huygens' principle for a plane wavefront;

(b)

1

Waves

Optical Properties of Radio

41

However, when a wavefront

surfaces were large in respect to a wavelength of the signal.

passes near an obstacle or discontinuity with dimensions comparable in size to a wavelength, simple geometric analysis cannot be used to explain the results and Huygens'

principle (which

is

deduced from Maxwell's equations)

Huygens' principle

states that

necessary.

is

every point on a given spherical wavefront can be

considered as a secondary point source of electromagnetic waves from which other

secondary waves (wavelets) are radiated outward. Huygens' principle Figure 10-9. Normal 10-9a. tions.

wave propagation considering an

Each secondary point source (p 1? p 2 However, the wavefront continues in

,

infinite

is

is

illustrated in

shown

energy outward

etc.) radiates its

plane

in Figure

in all direc-

original direction rather than spreading

out because cancellation of the secondary wavelets occurs in

all

directions except straight

forward. Therefore, the wavefront remains plane.

When random

a finite plane wavefront

directions

is

is

considered, as in Figure 10-9b, cancellation in

incomplete. Consequently, the wavefront spreads out or scatters.

shows

around the

This scattering effect

is

called diffraction. Figure 10-9c

edge of an obstacle.

It

can be seen that wavelet cancellation occurs only

diffraction

Diffraction occurs around the edge of the obstacle, which allows secondary

"sneak" around the corner of the obstacle into what

phenomenon can be observed when

a door

is

is

opened

partially.

waves

to

shadow zone. This

called the

into a dark

room. Light rays

diffract around the edge of the door and illuminate the area behind the door.

Interference

means to come into opposition, and interference is the act of interfering. Radio-wave interference occurs when two or more electromagnetic waves combine in such a way that system performance is degraded. Refraction, reflection, and diffraction

Interfere

are categorized as geometric optics, in

which means

that their behavior

is

terms of rays and wavefronts. Interference, on the other hand,

analyzed primarily is

subject to the

waves and occurs whenever two or more waves simultaneously occupy the same point in space. The principle of linear

principle of linear superposition of electromagnetic

superposition states that the total voltage intensity at a given point in space

is

the

sum

of the individual wave vectors. Certain types of propagation media have nonlinear properties;

however,

in

an ordinary

medium

(such as air or the earth's atmosphere), linear

superposition holds true.

Figure 10-10 shows the linear addition of two instantaneous voltage vectors whose

phase angles differ by angle

sum of

the

two vectors, but

G.

It

can be seen that the

rather, the

propagation, a phase difference

may

total voltage is not

simply the

phasor addition of the two. With free-space

exist simply because the electromagnetic polariza-

FIGURE

10-10

Linear addition of two

vectors with differing phase angles.

Wave

412

Propagation

Chap. 10

Wave A

Source

Wave B Reflection, refraction,

or diffraction changes the direction of wave B

FIGURE

Electromagnetic wave interference.

Depending on the phase angles of the two vectors, either may be more or two waves can reinforce vector because the electromagnetic or cancel.) than either less Figure 10-1 1 shows interference between two electromagnetic waves in free space. It can be seen that at point X the two waves occupy the same area of space. However, wave B has traveled a different path than wave A, and therefore their relative phase angles may be different. If the difference in distance traveled is an odd integral multiple of one-half wavelength, reinforcement takes place. If the difference is an even integral

tions of

two waves

10-11

differ.

addition or subtraction can occur. (This implies simply that the result

multiple of one-half wavelength, total cancellation occurs. in distance falls

cies

somewhere between

below VHF, the

significant problem.

the

relatively large

However, with

two and

More

likely the difference

partial cancellation occurs.

For frequen-

wavelengths prevent interference from being a

UHF

and above, wave interference can be severe.

PROPAGATION OF WAVES In radio communications systems, there are several

ways

in

which waves can be propa-

gated, depending on the type of system and the environment. Also, as previously explained, electromagnetic

atmosphere

waves

alter their path.

travel in straight lines except

when

the earth and

its

There are three ways of propagating electromagnetic waves:

ground wave, space wave (which includes both direct and ground-reflected waves), and sky wave propagation. Figure 10-12 shows the normal modes of propagation between two radio antennas. Each of these modes exists in every radio system; however, some are negligible in certain frequency ranges or over a particular type of terrain. At frequencies below 1.5 MHz, ground waves provide the best coverage. This is because ground losses increase rapidly with frequency.

waves are used

Sky waves

are used for high-frequency applications, and space

for very high frequencies

and above.

Ground-Wave Propagation

A ground wave is an electromagnetic wave that travels along the surface of the earth. Therefore, ground waves are sometimes called surface waves. Ground waves must be vertically polarized. This is

would be

because the electric

parallel to the earth's surface,

field in a horizontally polarized

wave

and such waves would be short-circuited by

'

Propagation of Waves

Transmit antenna

413

—

Earth's surface

FIGURE

10-12

the conductivity of the ground.

Normal modes of wave propagation.

With ground waves, the changing

electric field induces

voltages in the earth's surface, which cause currents to flow that are very similar to

those in a transmission line.

The

earth's surface also has resistance

and

dielectric losses.

Therefore, ground waves are attenuated as they propagate. Ground waves propagate is a good conductor, such as salt water, and poorly over dry Ground- wave losses increase rapidly with frequency. Therefore, groundwave propagation is generally limited to frequencies below 2 MHz. Figure 10-13 shows ground-wave propagation. The earth's atmosphere has a gra-

best over a surface that desert areas.

Wavefront propagation Increasing

angle of

tilt

Excessive

tilt,

wavefront dies

Wavefront perpendicular to earth's surface

FIGURE

10-13

Ground-wave propagation.

.

Wave

414

Propagation

Chap. 10

dient density (i.e., the density decreases gradually with distance from the earth's surface),

which causes the wavefront

to

tilt

wave enough power is

progressively forward. Therefore, the ground

propagates around the earth, remaining close to

its

surface,

and

if

beyond the horizon or even around the entire However, care must be taken when selecting the frequency and the terrain over which the ground wave will propagate, to ensure that the wavefront does not tilt excessively and simply turn over, lie flat on the ground, and cease to transmitted, the wavefront could propagate

earth's circumference.

propagate.

Ground-wave propagation

is

commonly used

for ship-to-ship and ship-to-shore

communications, for radio navigation, and for maritime mobile communications. Ground

waves are used at frequencies as low as 15 kHz. The disadvantages of ground- wave propagation are

as follows:

1

Ground waves

require a relatively high transmission power.

2.

Ground waves

are limited to

low and very low frequencies (LF and VLF),

ing large antennas (the reason for this 3.

Ground

is

losses vary considerably with surface material.

The advantages of ground- wave propagation 1

2.

facilitat-

explained in Chapter 11).

are as follows:

Given enough transmit power, ground waves can be used any two locations in the world.

Ground waves

are relatively unaffected

to

communicate between

by changing atmospheric conditions.

Space-Wave Propagation that travels in the lower few miles of waves include both direct and ground-reflected waves (see Figure 10-14). Direct waves are waves that travel essentially in a straight line between the transmit and receive antennas. Space- wave propagation with direct waves is commonly called line-of-sight (LOS) transmission. Therefore, space- wave propagation

Space-wave propagation includes radiated energy

the earth's atmosphere. Space

Earth's surface

FIGURE

10-14

Space-wave propagation.

Propagation of Waves

FIGURE is

415

10-15

Space waves and radio horizon.

by the curvature of the

limited

are reflected

earth. Ground-reflected waves are those waves that by the earth's surface as they propagate between the transmit and receive

antennas.

Figure 10-14 shows space- wave propagation between two antennas. seen that the

field intensity at the receive

It

can be

antenna depends on the distance between the

two antennas (attenuation and absorption) and whether the direct and ground-reflected waves are in phase (interference). The curvature of the earth presents a horizon to space wave propagation commonly called the radio horizon. Due to atmospheric refraction, the radio horizon extends beyond the optical horizon for the common standard atmosphere. The radio horizon is approximately four-thirds that of the optical horizon. Refraction is caused by the troposphere, due to changes in its density, temperature, water vapor content, and relative conductivity. The radio horizon can be lengthened simply by elevating the transmit or receive antennas (or both) above the earth's surface with towers or by placing them on top of mountains or high buildings.

Figure 10-15 shows the effect of antenna height on the radio horizon. The lineof-sight radio horizon for a single antenna

is

given as

d=V2h

(10-15)

where

d = h

=

distance to radio horizon (mi)

antenna height above sea level

(ft)

Therefore, for a transmit and receive antenna, the distance between the two antennas is

or (10-16)

where

d= d = f

total distance (mi)

radio horizon for transmit antenna (mi)

Wave

416 dr = ht hr

= =

Chap. 10

Propagation

radio horizon for receive antenna (mi)

transmit antenna height

receive antenna height

(ft)

(ft)

or

d= where d and dr are distance

4Vh + 4Vh r and h and h r are height

in kilometers

t

From Equations 10-16 and 10-17 distance can be extended simply

(10-17)

t

it

t

in meters.

can be seen that the space- wave propagation

by increasing

either the transmit or receive antenna

height, or both.

Because the conditions occurs

when

are trapped

lower atmosphere are subject to change, the

in the earth's

degree of refraction can vary with time.

A

special condition called duct propagation

the density of the lower atmosphere

between

it

is

such that electromagnetic waves

and the earth's surface. The layers of the atmosphere act

duct and an electromagnetic

wave can propagate

of the earth within this duct. Duct propagation

like a

for great distances around the curvature

is

shown

in

Figure 10-16.

Sky-Wave Propagation Electromagnetic waves that are directed above the horizon level are called sky waves. Typically, sky waves are radiated in a direction that produces a relatively large angle

with reference to the earth. Sky waves are radiated toward the sky, where they are either reflected or refracted

back

to earth

by the ionosphere. The ionosphere

region of space located approximately 50 to 400 surface.

The ionosphere

is

km

wave

the

the upper portion of the earth's atmosphere. Therefore,

absorbs large quantities of the sun's radiant energy, which ionizes the creating free electrons.

is

(30 to 250 mi) above the earth's

When

a radio

wave passes through

air

it

molecules,

the ionosphere, the electric

on the free electrons, causing them to vibrate. The vibrating electrons decrease current, which is equivalent to reducing the dielectric constant. Reducing the dielectric constant increases the velocity of propagation and causes field

of the

exerts a force

electromagnetic waves to bend

away from

regions of low electron density

(i.e.,

from

earth,

ionization increases.

the regions of high electron density toward

increasing refraction).

However, there

As

the

Upper atmosphere

Warmer

air

Trapped waves

Duct effect

FIGURE

10-16

wave moves

farther

are fewer air molecules to ionize.

Duct propagation.

Propagation of Waves mi

km

248

400

417

F, (June)

186

300

124

200 F layer

E layer

62

100

31

50

D

4

2

8

6

layer

10

14

12

20

18

16

22

24

Local time (hours of the day)

FIGURE

10-17

Ionospheric layers.

Therefore, in the upper atmosphere, there

due

to the ionosphere's

tions,

it

is

a higher percentage of ionized molecules

more

the ion density, the

nonuniform composition and

its

refraction. Also,

temperature and density varia-

Essentially, there are three layers that comprise the ionosphere

stratified.

D, E, and F

(the

is

The higher

than in the lower atmosphere.

which are shown

layers),

in

Figure 10-17.

It

can be seen that

all

three layers of the ionosphere vary in location and in ionization density with the time

of day. They also fluctuate in a cyclic pattern throughout the year, and according to

The ionosphere

the 11 -year sunspot cycle.

is

most dense during times of maximum

sunlight (i.e., during the daylight hours and in the summer).

D

layer.

The

30 and 60 mi (50

D

to

layer

is

the lowest layer of the ionosphere and

100 km) from the earth's surface. Because

it

located between

is

is

the layer farthest

from the sun, there is very little ionization in this layer. Therefore, the D layer has little effect on the direction of propagation of radio waves. However, the ions in the D layer can absorb appreciable amounts of electromagnetic energy. The amount of ionization in the D layer depends on the altitude of the sun above the horizon. Therefore, very

it

disappears at night.

HF

E

layer.

The

the earth's surface.

the

The

D

layer reflects

waves. (See Table 1-1 for VLF, LF,

two

scientists

mately 70

km

at

E

layer

The E

who

is

VLF

and

MF, and HF

LF waves and

located between 60 and 85

layer

is

discovered

absorbs

MF

mi (100

to

140 km) above

sometimes called the Kennelly-Heaviside layer it.

The E

noon, when the sun

layer has

is at its

and

frequency regions.)

its

maximum

after

density at approxi-

highest point. Like the

D

layer, the

E

Wave

418

Propagation

Chap. 10

The E layer aids MF surface-wave propagation somewhat waves during the daytime. The upper portion of the E layer and reflects considered separately and called the sporadic E layer because it seems to is sometimes rather unpredictably. The sporadic E layer is caused by solar flares and come and go sunspot activity. The sporadic E layer is a thin layer with a very high ionization density. When it appears, there generally is an unexpected improvement in long-distance transmislayer almost totally disappears at night.

HF

sion.

F

layer.

The

F

layer

is

actually

made up of two

layers: the Fj

and F 2

layers.

mi (140 to 250 km) above the earth's surface, and the F 2 layer is located 85 to 185 mi (140 to 300 km) above the earth's surface during the winter and 155 to 220 mi (250 to 350 km) in the summer. During the night, the Fj layer combines with the F 2 layer to form a single layer. The F, layer absorbs and attenuates some HF waves, although most of the waves pass through to the F 2 layer, where they are refracted back to earth. During the daytime, the Fj layer

is

located between 85 and 155

PROPAGATION TERMS AND DEFINITIONS Critical

Frequency and

Frequencies above the

Critical

UHF

Angle

range are virtually unaffected by the ionosphere because

of their extremely short wavelengths. At these frequencies, the distance between ions are appreciably large, and consequently, the electromagnetic

with

little

noticeable effect. Therefore,

it

waves pass through them

stands to reason that there must be an upper

frequency limit for sky-wave propagation. Critical frequency (Fc )

is

defined as the highest

Penetrate the ionosphere and escape the earth's atmosphere Successive reflections

Ionosphere

FIGURE

10-18

Critical angle.

Propagation Terms and Definitions

419

Specular equivalent layer height (virtual height)

Ionosphere

FIGURE

frequency that can be propagated directly upward and ionosphere.

The

critical

still

Virtual and actual height.

10-19

be returned to earth by the

frequency depends on the ionization density and therefore varies

with the time of day and the season. If the vertical angle of radiation frequencies at or above the critical frequency can

still

is

decreased,

be refracted back to the earth's

surface because they will travel a longer distance in the ionosphere and thus have a

longer time to be refracted. Therefore, critical frequency

is

used only as a point of

reference for comparison purposes. However, every frequency has a

angle at which angle

is

can be propagated and

it

called the critical angle.

The

still

critical

maximum

vertical

be refracted back by the ionosphere. This angle 6 C

is

shown

Figure 10-18.

in

Virtual Height Virtual height

is

the height

above the earth's surface from which a refracted wave

appears to have been reflected. Figure 10-19 shows a wave that has been radiated

from the earth's surface toward the ionosphere. The radiated wave

is

refracted back to

The actual maximum height that the wave reached is height h a However, path A shows the projected path that a reflected wave could have taken and still been returned to earth at the same location. The maximum height that this hypothetical reflected wave would have reached is the virtual height (h v ). earth and follows path B. .

Maximum

Usable Frequency

The maximum usable frequency (MUF) is the highest frequency sky-wave propagation between two specific points on the earth's reason, then, that there are as earth and frequencies

—an

many

infinite

values possible for

number.

MUF,

matically,

MUF

is

can be used for

surface.

It

stands to

as there are points

on

like the critical frequency, is a limiting

frequency for sky-wave propagation. However, the specific angle of incidence (the angle

MUF

that

maximum

usable frequency

is

for a

between the incident wave and the normal). Mathe-

Wave

420 critical

MUF

Propagation

Chap. 10

frequency (10-18a)

cos critical

where

is

frequency

x

sec G

(10- 18b)

the angle of incidence.

Equations 10-18 are called the secant law. The secant law assumes a

and a

flat

reflecting layer,

which of course, can never

exist. Therefore,

flat

MUF

is

earth

used

only for making preliminary calculations.

Skip Distance

The skip distance (ts ) is the minimum distance from a transmit antenna that a sky wave of given frequency (which must be greater than the critical frequency) will be returned to earth. Figure 10-20a shows several waves with different angles of incidence lonoshpere

(a)

Night hours

Earth's surface (b)

FIGURE

10-20

Skip distance:

(a) skip distance; (b)

daytime versus nighttime propagation.

.

Chap. 10

Problems

421

being radiated from the same point on earth.

wave

is

returned to earth

moves

It

can be seen that the point where the

closer to the transmitter as the angle of incidence (0)

increased. Eventually, however, the angle of incidence

is

wave

is

sufficiently high that the

penetrates through the ionosphere and totally escapes the earth's atmosphere.

Figure 10-20b shows the effect on the skip distance of the disappearance of the

D

and

E

layers during nighttime. Effectively, the ceiling

waves

raised, allowing sky

how

effect explains

formed by the ionosphere

to travel higher before being refracted

back

is

to earth. This

far-away radio stations are sometimes heard during the night that

cannot be heard during daylight hours. Figure 10-20c shows the effects of multipath

and multiple skip sky-wave propagation.

QUESTIONS 10-1. Describe an electromagnetic ray; a wavefront.

power

10-2. Describe

density; voltage intensity.

10-3. Describe a spherical wavefront. 10-4. Explain the inverse square law.

10-5. Describe

wave

attenuation.

10-6. Describe

wave

absorption.

10-7. Describe refraction. Explain Snell's law for refraction. 10-8. Describe reflection. Explain Snell's law for reflection. 10-9. Describe diffraction. Explain Huygens' principle.

10-10. Describe the composition of a good reflector. 10-11. Describe the atmospheric conditions that cause electromagnetic refraction. 10-12. Define electromagnetic

wave

interference.

10-13. Describe ground-wave propagation. List

its

advantages and disadvantages.

10-14. Describe space- wave propagation. 10-15. Explain

why

the radio horizon

is at

a greater distance than the optical horizon.

10-16. Describe the various layers of the ionosphere. 10-17. Describe sky-wave propagation.

10-18. Explain

why

atmospheric conditions vary with time of day, month of year, and so on.

10-19. Define critical frequency; critical angle. 10-20. Describe virtual height. 10-21. Define

maximum

usable frequency

10-22. Define skip distance and give the reasons that

it

varies.

PROBLEMS 10-1. Determine the

power density

an isotropic antenna.

for a radiated

power of 1000

W

at

distance of 20

km

from

Wave

422 10-2. Determine the

power density

for

Problem 10-1 for

on power density

10-3. Describe the effects

if

a point that

the distance

is

10-5. Determine the

50

ft

high; 100

maximum

m

is

km

30

from the antenna.

from a transmit antenna

10-4. Determine the radio horizon for a transmit antenna that

antenna that

Chap. 10

Propagation

is

100

is

tripled.

high and a receiving

ft

and 50 m.

usable frequency for a critical frequency of 10

MHz

and an

angle of incidence of 45°. 10-6. Determine the voltage intensity for the

same point

in

Problem 10-1.

10-7. Determine the voltage intensity for the

same point

in

Problem 10-2.

10-8. For a radiated

power

Pr —

10

kW,

determine the voltage intensity

at

a distance 20

km

from the source. 10-9. Determine the change in

power density when

the distance

from the source increases by a

factor of 4.

10-10. If the distance from the source

reduced to one-half

is

its

value, what affect does this

have on the power density? 10-11.

The power density point

is

at

a point from a source

is

0.001 |xW and the power density

0.00001 u,W; determine the attenuation

10-12. For a dielectric ratio

angle of refraction, 6 r

Ve 2/€, =0.8

at

another

in decibels.

and an angle of incidence

0,

=

26°, determine the

.

10-13. Determine the distance to the radio horizon for an antenna located 40

10-14. Determine the distance to the radio horizon for an antenna that

is

40

ft

ft

above sea

level.

above the top of

a 4000-ft mountain peak. 10-15. Determine the level for

maximum

Problem 10-13.

distance between identical antennas equally distant above sea

1

Chapter

1

ANTENNAS INTRODUCTION In essence, an antenna

is

a metallic conductor system capable of transmitting and receiv-

An antenna is used to interface a transmitter to free space An antenna couples energy from the output of a transmitter or from the earth's atmosphere to the input of a receiver. An

ing electromagnetic waves.

or free space to a receiver. to the earth's

antenna

is

atmosphere

a passive reciprocal device: passive in that

signal, at least not in the true sense of the

it

word (however,

cannot actually amplify a as

you

will see later in this

chapter, an antenna can have gain); and reciprocal in that the transmit and receive characteristics of an antenna are identical, except

where feed currents

to the antenna

element are tapered to modify the transmit pattern.

Basic Antenna Operation is best understood by looking at the voltage standing-wave on a transmission line, which are shown in Figure 11- la. The transmission line is terminated in an open circuit, which represents an abrupt discontinuity to the incident voltage wave in the form of a phase reversal. The phase reversal causes some

Basic antenna operation

patterns

of the incident voltage to be radiated, not reflected back toward the source. The radiated

energy propagates away from the antenna

in the

form of transverse electromagnetic

waves. The radiation efficiency of an open transmission efficiency

is

the ratio of radiated to reflected energy.

spread the conductors farther apart. Such an antenna poles) and

is

shown

in

line is

To is

extremely low. Radiation

radiate

more energy, simply two

called a dipole (meaning

Figure 11 -lb.

423

Antennas

424

Chap.

11

Voltage standing waves Radiated

ttitltft^iill))

Radiated

waves

(b)

(a)

Radiated

a

^/2 j Radiated

y^

waves

waves

(c)

(d)

FIGURE tors; (c)

11-1 Radiation from a transmission line: Marconi antenna; (d) Hertz antenna.

(a) transmission-line radiation; (b)

spreading conduc-

In Figure 11- lc the conductors are spread out in a straight line to a total length

of one-quarter wavelength. Such an antenna

Marconi antenna.

A

is

called a basic quarter-wave dipole or a

half- wave dipole is called a

Hertz antenna and

is

shown

in Figure

11-ld.

TERMS

AND

DEFINITIONS

Radiation Pattern

A

radiation pattern

is

a polar diagram or graph representing field strengths or power

densities at various points in space relative to an antenna. If the radiation pattern

plotted in terms of electric field strength {% is

called an absolute radiation pattern. If

respect to a reference point,

it is

= V/m) it

or power density (0*

plots field strength or

power density

called a relative radiation pattern. Figure

an absolute radiation pattern for an unspecified antenna. The pattern

= W/m 2 ),

is

1

is it

in

l-2a shows

plotted

on polar

coordinate paper with the heavy solid line representing points of equal power density

(10

(xW/m 2 ). The

maximum

circular gradients indicate distance in

radiation

from the antenna

is in

in a

a direction 90°

90° direction

is

10

2-km

steps.

It

can be seen

that

from the reference. The power density 10 km

|xW/m 2

.

In a 45° direction, the point of equal

Terms and Definitions

425 0° (Reference)

-90°

(P

= 10/uW/m 2

0° (Reference)

Major front

-90

c

5/uW

FIGURE

11-2

Radiation patterns:

(a)

absolute radiation pattern; (b) relative radiation pattern;

relative radiation pattern in decibels; (d) relative radiation pattern for

an omnidirectional antenna.

(c)

426

Antennas

Major front

90°

90°

OdB

180° (d)

FIGURE

11-2 (continued)

Chap.

11

Terms and Definitions power density there

is

5

is

km no

essentially

427

from the antenna;

only 4 km; and in a -90° direction,

at 180°,

radiation.

beam

In Figure 11 -2a the primary

is

90° direction and

in a

There can be more than one major lobe. There

lobe.

minor lobe

—180°

in a

direction. Normally,

is

called the

is

also a secondary

major

beam

or

minor lobes represent undesired radiation

or reception. Because the major lobe propagates and receives the most energy, that lobe

is

called the front lobe (i.e., the front of the antenna).

lobe are called side lobes (the 180° minor lobe

is

Lobes adjacent

exactly opposite the front lobe are called back lobes (there this pattern).

back

ratio,

ratio.

The

The

ratio

and the

of the front lobe to the back lobe

maximum

is

no back lobe shown on

simply called the front-tois

called the front-to-side

major lobe or pointing from the center of the antenna

line bisecting the

the direction of

is

of the front lobe to a side lobe

ratio

to the front

a side lobe), and lobes in a direction

radiation

is

in

called the line of shoot.

Figure 11 -2b shows a relative radiation pattern for an unspecified antenna. The

heavy solid

of equal distance from the antenna (10 km), and the power density in 1 |xW/m 2 divisions. It can be seen that maximum radiation (5 |mW/m 2 ) is in the direction of the reference (0°) and the antenna 2 radiates the least power (1 |xW/m ) in a direction 180° from the reference. Consequently, = the front-to-back ratio is 5: 1 5. Generally, relative field strength and power density are plotted in decibels (dB), where dB = 20 log (e/e max ) or 10 log (2W max ). Figure 1 12c shows a relative radiation pattern for power density in decibels. In a direction ±45° from the reference, the power density is —3 dB (half power) relative to the power density in the direction of maximum radiation (0°). Figure 1 l-2d shows a relative radiation pattern for power density for an omnidirectional antenna. As stated previously, an omnidiline represents points

circular gradients indicate

rectional antenna radiates energy equally in all directions; therefore, the radiation pattern is

no

simply a circle (actually, a sphere). Also, with an omnidirectional antenna, there are front, back, or side lobes

The tion

radiation patterns

from an actual antenna

because radiation

shown is

in

is

equal in

all

directions.

Figure 11-2 are two-dimensional. However, radia-

three-dimensional. Therefore, radiation patterns are taken

in both the horizontal (from the top) and the vertical (from the side) planes. For the

omnidirectional antenna

and

isotropic radiator

Near and Far The is at

shown

vertical planes are circular is

in

Figure 11 -2d, the radiation patterns in the horizontal

and equal because the actual radiation pattern for an

a sphere.

Fields

radiation field that

a great distance.

is

close to an antenna

The term near field

is

not the same as the radiation field that

refers to the field pattern that is close to the

antenna, and the term far field refers to the field pattern that is at great distance. During one half of a cycle, power is radiated from an antenna where some of the power is stored temporarily in the near field. During the second half of the cycle, power in the

near field

is

returned to the antenna. This action

is

similar to the

inductor stores and releases energy. Therefore, the near field induction field.

Power

is

way

in

which an

sometimes called the

that reaches the far field continues to radiate

outward and

is

Antennas

428 never returned to the antenna. Therefore, the far

field is

Chap.

11

sometimes called the radiation

more important of the two; therefore, antenna radiation patterns are generally given for the far field. The near field is defined as the area 2 within a distance D /X from the antenna where X is the wavelength and D the antenna diameter in the same units.

field.

Radiated power

is

usually the

Radiation Resistance and Antenna Efficiency

power supplied

All of the

to an antenna

and dissipated. Radiation resistance directly. Radiation resistance is

is

is

not radiated.

of

in that

an ac antenna resistance and

power radiated by the antenna to the square of the current radiation resistance

Some

somewhat "unreal"

at its

is

it is

converted to heat

cannot be measured

it

equal to the ratio of the

feed point. Mathematically,

is

i

where

R r = radiation resistance P = rms power radiated by = rms antenna current at i

the antenna the feedpoint

is the resistance which, if it replaced the antenna, would dissipate same amount of power that the antenna radiates. Antenna efficiency is the ratio of the power radiated by an antenna to the sum of power radiated and the power dissipated or the ratio of the power radiated by the

Radiation resistance exactly the

the

antenna to the

total input

power. Mathematically, antenna efficiency

is

^J^F xm

/

(

"- 2)

d

where

= antenna efficiency (%) P r = power radiated by antenna Pd = power dissipated in antenna r\

Figure 11-3 shows a simplified electrical equivalent circuit for an antenna. Some

of the input power dielectrics, is

the

is

dissipated in the dc resistances (ground resistance, corona, imperfect

eddy currents,

sum of

etc.)

and the remainder

is

radiated.

The

total

antenna power

the dissipated and radiated powers. Therefore, in terms of resistance and

current, antenna efficiency

is

n =

i2Rr ,

=

—^—

(ii-3)

Terms and Definitions

429

Dissipated

Radiated

power

power

"dc

-VW\r

-WW-

Rr

FIGURE cuit of

11-3

Simplified equivalent cir-

an antenna.

where

= — Rr = Rdc = i]

i

antenna efficiency antenna current radiation resistance

dc antenna resistance

Power Gain

Directive Gain and

The terms

and power gain are often misunderstood and, consequently,

directive gain

power density radiated in a particular direction same point by a reference antenna, assuming both antennas are radiating the same amount of power. The relative power density radiation pattern for an antenna is actually a directive gain pattern if the power density reference is taken from a standard reference antenna, which is generally an isotropic antenna. The maximum directive gain is called directivity. Mathematically, directive gain is misused. Directive gain

power density

to the

is

the ratio of the

radiated to the

3)

=

9?

(11-4)

re

I

where 2) ( i; the first represents a signal that

latter a signal that is

is

in

Bandpass Balanced

modulator

filter

BPF

Analog PSK output

i

Reference carrier oscillator

phase

180° out of phase with the reference

FIGURE

13-8

BPSK

modulator.

498

Digital

Communications

Chap. 13

T2

D1 (on)

D3and D4 (off)

+]•

•}

+

r r

(sin (a c t)

./)

or

(filtered out)

-sin 2 _

—

The highest fundamental frequency presented

c /%,

=

F —

hl -

or

The output wave from each balanced modulator

=

5

IttFj)

cos 2ir(F(

2.5)

MHz]/ -

|

cos 2

>

CM

51

+ LL Q. _l

—

.

l

LL a. _l

i

1

I

_ u

3 O o

3 o o

V)

+

+ *->

o

c

3 c

c/>

'

3

1

,L

•f-"

3

3

_c

o o

HZ*->

-IS o £ m

3

3 c

o o

O CD

^

i-

3

O

Vt

c c

+

to

a)

'55

i

-

3 CO

O U

o

"33

c c CD

zo

JO

3

3

c

a

_c

4->

o

*->

>^

u C > 3 « 8 c

Mo O

*->

Q. TJ

\

"55

O O 3 o

+

Q- -S

i

O o

/

o

'35

r 1 1 o 7; 3°

o o

1

I

+ 4-1

3

LL a.

c

m

1

3

^

eft

«°

2>

509

510

Communications

Digital

Q=

w rr +

(— sin

cos

o> t .r)

(cos

=

cos

2

input signal

—

u> c t

= U +

cos

1

o) ( i)

y

V

QPSK

carrier

(sin co ( ./)(cos a> r f)

2w

(

.0

-

+

i sin ( ( /

-0.541 and cos

inputs to the Q-channel product modulator are

a>

./. (

The output

is

Q= The outputs from

the

summer and produce

(-0.541 )(cos

(i),/)

The

u

.t (

and Q-channel product modulators are combined

I-

=

the same.

-0.541 cos

in the

linear

a modulated output of

summer output = —0.541

For the remaining

=

tribit

0.765 sin

codes (001, 010, 01

results are

shown

I/O.

in

c

sin a>

1,

./

—

(

to,./

-

100. 101.

0.541 cos

to,/

135°

1

10,

and 111), the procedi

Figure 13-28.

Output -0.541

1

1

-1.307 V +0.541

1

1

+ 1.307 V

FIGURE

13-27

and Q-channel

Truth tabic for the 2-to-4-level converters.

Eight

QAM

sin

Binary input

Q

I

1.848

1

0.765

1

1.848

1

0.765

1

1

1.848

1

1

1

1

1

\-

8QAM output amplitude phase

C

0.765

1

u>A

0.765 1.848

1

V V V V V V V V

001

-135° -135° -45° -45° +135° +135° +45° +45°

cos CJ r t

101

111

(a)

• 110

100 •

sin

000 •

001

•

u>A

• 010

-cos co c t

•011

(c)

FIGURE lation

Figure

13-28

8QAM

modulator:

(a) truth table; (b)

phasor diagram;

(c) constel-

diagram.

13-29 shows the output phase versus time relationship for an

8QAM

modulator. Note that there are two output amplitudes and only four phases are possible.

Bandwidth Considerations of In

8QAM,

same

as in

the bit rate in the

8PSK. As

I

and

8QAM Q

channels

is

one-third of the input binary rate, the

a result, the highest fundamental modulating frequency and the

522

Communications

Digital

Tribit

QIC

QIC

input

000

001

Chap. 13

8QAM output phase and amplitude

'

0.765

V

1.848

FIGURE

V

V 135° +135° +

0.765

I

-45°

8QAM 8QAM

in

required for

receiver

PAM

8PSK

V

0.765

1.848

j

+45°

V

+45

i

8QAM.

the

Figure 13-24.

and the binary

8QAM four demodulated PAM levels

the

receiver

Because there are two transmit

in

8QAM

from those achievable with 8PSK, from those

are different

the conversion factor for the analog-to-digital converters

8QAM and

I

the binary output signals

C

Q

in

8PSK. Therefore,

must also be

different. Also,

from the I-channel analog-to-digital converter

and the binary output signals from the Q-channel analog-to-digital

bits,

converter are the

shown

in

that are different

C

and

bits.

QAM 16QAM

is

groups of four (2 4

=

Like 16PSK, in

!

levels at the output of the product detectors

signals at the output of the analog-to-digital converters.

SIXTEEN

j

F^/3, the

amplitudes possible with

are the

V

+135°

i

same as with 8PSK. Therefore, same as in 8PSK.

are the is

almost identical to the

is

differences are the

with

1.848

j

Receiver

An 8QAM The

1.848

-45 b

of change

minimum bandwidth

V

0.765

Output phase and amplitude versus time relationship for

13-29

fastest output rate

8QAM

V

-135°

-135°

M

an M-ary system where 16).

As

with

8QAM,

=

16.

The

input data are acted on

both the phase and amplitude of the

transmit carrier are varied.

16QAM

Transmitter

The block diagram

for a

16QAM

transmitter

binary data are divided into four channels: the

channel

is

sjiown in Figure 13-30. The input

is I,

equal to one-fourth of the input bit rate

Q, and Q. The (///4). Four bits are I,

into the bit_splitter; then they are outputted simultaneously I,

Q, and

Q

channels.

The

I

to-4-level converters (a logic bits

and 1

Q

=

determine the magnitude (a logic

bits

determine the polarity

=

positive and a logic 1

and

=0.821 V and

the 2-to-4-level converters generate a 4-level

PAM

a logic

signal.

Two

PAM

in parallel at

each

serially clocked

with the

I,

the output of the 2-

negative).

=

The

I

and

Q

0.22 V). Consequently,

polarities

tudes are possible at the output of each 2-to-4-level converter.

±0.821 V. The

bit rate in

They

and two magni-

are

signals modulate the in-phase and quadrature carriers

±0.22 V and in the

product

Sixteen

QAM

523

2-to-4-level

Balanced

converter

modulator

F./4

Binary input -

Reference

data

oscillator

16QAM

Linear

carrier

summer

output

t

±90°

FIGURE

I

1

-0.22 V -0.821 V

1

+0.22 V +0.821 V

1

1

Balanced

converter

modu lator

13-30

16QAM

Q

Q

Output

I

2-to-4-level

transmitter block diagram.

Output

V V +0.22 V +0.821 V -0.22

-0.821

1

1 1

1

FIGURE

(b)

(a)

(a) I

Truth tables for the

13-31

and Q-channel

channel; (b)

Q channel.

modulators. Four outputs are possible for each product modulator. For the

modulator they are +0.821

Q

For the

o> r r.

cos

a) r r,

sin oj ( ./\

—0.821

sin a> ( i,

o) c t.

The

linear

product

sin oj ( ./,

(o r /,

summer combines

I

and —0.22 sin +0.22 cos a> .f, -0.821

+0.22

product modulator they are +0.821 cos

and —0.22 cos

I-

2-to-4-level converters:

c

from the

the outputs

I-

and

Q-channel product modulators and produces the 16 output conditions necessary for 16

QAM.

Figure 13-31 shows the truth table for the

EXAMPLE

I

=

amplitude and phase for the

The

Q =

and

Thus

The output

the

I = 0, Q = 0, and Q = (0000), determine 16QAM modulator shown in Figure 13-30.

0,

inputs to the I-channel 2-to-4-leveI converter are

Figure 13-31 the output are

Q =

0.

is

—0.22 V. The

is

I

=

and

1

the output

=

0.

From

inputs to the Q-channel 2-to-4-level converter

Again from Figure 13-31,

two inputs

the output

to the 1-channel product

is

-0.22 V.

modulator are —0.22

V and

sin u>j.

is

I

The two

and Q-channel 2-to-4-level converters.

13-8

For a quadbit input of

Solution

I-

=

(-0.22)(sin

(M ( .t)

= -0.22

inputs to the Q-channel product modulator are

sin co( i

-0.22 V and cos

co r /.

The ou

524

Communications

Digital

Q= The outputs from

the

I-

(-0.22)(coso)

0.22 cos

./) (

a)

j

and Q-channel product modulators are combined

summer and produce a modulated

Chap. 13

in

the

lit

output of

summer output = -0.22

=

sin

mc t —

mj

0.22 cos

0.311 sinu> r/- 135°

For the remaining quadbit codes the procedure

the same.

is

The

results are

showr

Figure 13-32.

Binary input

Q

Q

16QAM I

output

I

1 1 1

1

1

1

1.161

1

1

1

1

1

0.850 1.161

1

1 1

1

1

1

1

1

135°

V

0.850

V V

105°

1.161V

135°

V V

75°

0.850 1.161

1

175°

0.850 0.850

1

1

V V V

0.311V 0.850 V

1

1

-135° -165° -45° -15° -105° -135° -75° -45°

0.311V 0.850 V 0.311V 0.850 V 0.850 V

,

45° 15°

45°

(a)

0.850

^1.161

1101

1100»

T

•1110

11119

•1010

1011«

I

I

I

• 1001

1000« -I

• 0001

0000*

t

h-

1

-J-

•

©0010

001

•0110

0111*

1

I

I

I

• 0101 (b)

FIGURE lation

13-32

diagram.

16QAM

0100*

^ (c)

modulator:

(a) truth table; (b)

phasor diagram;

(c) constel-

QAM

Sixteen

525

Bandwidth Considerations of With 16QAM, since I,

Q, or

Q

channel

1

6QAM

the input data are divided into four channels, the bit rate in the

equal to one-fourth of the binary input data rate (F b /4). (The

is

splitter stretches the I,

I,_Q, and

Q

bits to four

times their input

bit length.)

I,

bit

Also,

Q, and Q bits are outputted simultaneously and in parallel, the 2-to4-level converters see a change in their inputs and outputs at a rate equal to one-fourth because the

I,

I,

of the input data

rate.

Figure 13-33 shows the

bit

timing relationship between the binary input data;

2-to-4-level

PAM *»

converter

Balanced

modulator

20/4 V ± sin co c t ± 0.22

V

±0.821 V sin co

Binary input data

t

—

To Q-channel 2-to-4-level

converter

Input data F

I-channel

data

FJ4

+0.821 V

+0.22 V -0.22 V

I-channel

PAM

out

-0.821 V

-0.821

FIGURE

sin oo c t

13-33

+0.22

sin oo c t

Bandwidth considerations of a

-0.821

16QAM

sin oj t c

modulator.

526 the

I,

I,

Q

Q, and

channel data; and the

fundamental frequency

in the

RAM signals. Q channel is

I

Q, or

I,

I,

rate of the binary input data (one cycle in the

amount of time

PAM

signal

I,

16QAM

modulator, there

/4, the

same

as the

Q

channel takes the same

bit rate.

one change

is

phase, amplitude, or both) for every 4 input data /7

can be seen that the highest

It

equal_ to one-eighth of the bit

Q, or

I,

Chap. 13

Also, the highest fundamental frequency of either

as 8 input bits).

equal to one-eighth of the binary input

is

With a

F

Communications

Digital

in the

output signal (either

its

Consequently, the baud equals

bits.

minimum bandwidth.

Again, the balanced modulators are product modulators and their outputs can be represented mathematically as

=

(X

2ir

Fa -—

sin a)„r)(sin a> /) (

where u>j

=

and

u>,.t

modulating signal

= 2ttFj car

and

X=±0.22

±0.821

or

Thus 6

= (x sin = - cos

2tt

2tt

2

—

\F r c

V

IttFj)

) (sin

The output frequency spectrum extends from bandwidth (FN )

\F r

cos 2tt

t

8/

Fc + Fh/S

to

Fc — F^S

and the minimum

is

2F,

For a

8/

V

(*?)-('-*)-¥ EXAMPLE

H

(

2

Fh

13-9

16QAM

modulator with an input data

frequency of 70

MHz,

determine the

rate

minimum

(Fh ) equal

to

10

Mbps and

a carrier

double-sided Nyquist frequency (FN) and

compare the results with those achieved with the BPSK. QPSK, and 8PSK Examples 13-2, 13-4, and 13-6. Use the I6QAM block diagram shown in

the baud. Also,

modulators

in

Figure 13-26 as the modulator model.

Solution bit rate

The

bit rate in the

I,

I,

Q, and

Q

channels

is

equal to one-fourth of the input

or

Fa-fw-^-J^-i^E-

2.5

Mbps

Bandwidth

Efficiency

527

Therefore, the fastest rate of change and highest fundamental frequency presented to either

balanced modulator

is

F.=

—

— Ft

or

*

7

—bQ ^

or

The output wave from

the balanced modulator (sin

rt

lixFj)

/)(sin

.25)

MHz]/ -

|

cos 2tt[(70

cos 2ir(68.75

MHz)/ -

I

cos

-

The minimum Nyquist bandwidth

1

2-rr(7

1

+ Fa )t

.

+

.25

1

.25)

MHz]/

MHz)/

is

FN = The baud equals

2irF

cos 2ir(F(

cos 2irl(70 I

is

\

cos 2ir(Fr

1.25Mhz

2

2

- Fa )t -

h I

2.5 Mbps FbQ — = -=

or

2

2

(71.25 -68.75)

MHz =

2.5

MHz

the bandwidth; thus

baud

The output spectrum

is

=

2.5

megabaud

as follows:

68.75

MHz

70

MHz

71.25

MHz

(suppressed)

FN = For the same input

16QAM

modulator

QPSK, and 25%

bit rate, is

2.5

MHz

minimum bandwidth

the

equal to one-fourth that of the

less than

required to pass the output of a

BPSK

modualtor, one-half that of

with 8PSK. For each modulation technique, the baud

is

also

reduced by the same proportions.

BANDWIDTH

EFFICIENCY

Bandwidth

compare is

efficiency (or information density as

the performance of

one

digital

is

sometimes called)

minimum bandwidth

the ratio of the transmission bit rate to the

modulation scheme. Bandwidth efficiency

and thus indicates the number of

it

is

bits that

often used to it

required for a particular

generally normalized to a 1-Hz bandwidth

can be propagated through a

each hertz of bandwidth. Mathematically, bandwidth efficiency

BW efficiency

is

modulation technique to another. In essence,

medium

for

is

transmission rate (bps)

minimum bandwidth (Hz)

(13-4)

bits/second

bits/second

bits

hertz

cycles/second

cycle

528

Communications

Digital

EXAMPLE

Chap. 13

13-10

Determine the bandwidth efficiencies for the following modulation schemes: BPSK,

QPSK,

8PSK, and 16QAM. Solution

Recall from Examples 13-2,

minimum bandwidths

13-9 the

13-6, and

13-4,

10-Mbps transmission

required to propagate a

rate with the following

modulation schemes:

Minimum bandwidth

Modulation scheme

(MHz)

BPSK QPSK

10

5

8PSK

3.33

16QAM

2.5

Substituting into Equation 13-4, the bandwidth efficiencies are determined as follows:

BW efficiency

BPSK:

= 12Mbps _

BW efficiency

QPSK:

= 10Mbps_ 5

BW efficiency

8PSK:

=

BW efficiency

16QAM:

=

The

results indicate that

16QAM

PSK

BPSK

is

the least efficient

much bandwidth

requires one-fourth as

as

2 bits

Hz

cycle

3 bps

3 bits

Hz

cycle

MHz

MHz and

BPSK

btt

2 bps

Mbps _ 4

10 2.5

1

cycle

Mbps

3.33

_

Hz

MHz

10

bps

1

MHz

10

4

bps

Hz

16QAM

for the

bits

cycle is

same

the most efficient. bit rate.

AND QAM SUMMARY The various forms of FSK, PSK, and

TABLE 13-2

DIGITAL

QAM

are

in

Table 13-2.

MODULATION SUMMARY Bandwidth

Modulation

summarized

Encoding

Baud

(Hz)

Bandwidth efficiency (bps/Hz)

bit

^Fb

Single bit

Fb

Fb Fb

Dibit

Ftt2

FtJ2

2

8PSK

Tribit

Ft/3

F„/3

3

8QAM

Tribit

Ft/3

Ft/3

3

I6PSK

Quadbit

Ft/4

Ft/4

4

16QAM

Quadbit

Ft/4

Ft/4

4

FSK BPSK

Single

QPSK

( .r)

(

+ sin wj) =

+sin 2

is

u) c t

(filtered out)

= 1(1— For a received signal of —sin output

cos

(D c t

= ( — sin

2 c t)

=Jf=

2

cos 2co

(

.r

the output of the squaring circuit

co c .r)

(

— sin

a) c .f)

= + sin 2

is

(o c t

(filtered out)

= 1(1— It

can be seen that

in

removed by

filtering,

—

\

cos

2(x) c t

both cases the output from the squaring circuit contained a

dc voltage (+i V) and a signal is

cos 2o) c t) =yg

at

twice the carrier frequency (cos 2a> c f). The dc voltage

leaving only cos 2o> c t.

vco BPSK input

—

Bandpass

*"

filter

FIGURE

13-34

~^^

c Squarer

~^

ni

.

PLL

Squaring loop carrier recovery

°"j

~^

Frequency divider

circuit for a

BPSK

"^ .

Recovered carrier

receiver.

530

Digital

A

more elaborate

combines a carrier

carrier recovery circuit

is

Communications

Chap. 13

the Costas or quadrature loop, which

and can therefore accurately recover

carrier recovery with noise suppression

from a poorer-quality received signal than can a conventional squaring loop.

Carrier recovery circuits for higher-than-binary encoding techniques are similar to

BPSK

except that circuits which raise the receive signal to the fourth, eighth, and

higher powers are used.

DIFFERENTIAL PHASE SHIFT KEYING Differential

phase

shift

keying

the binary input information

is

(DPSK)

is

an alternative form of digital modulation where

contained in the difference between two successive signal-

ing elements rather than the absolute phase.

With

DPSK

is

it

a phase-coherent carrier. Instead, a received signaling element

element time in the

is

not necessary to recover

delayed by one signaling

and then compared to the next received signaling element. The difference

slot

phase of the two signaling elements determines the logic condition of the data.

DIFFERENTIAL BPSK

DBPSK

Transmitter

Figure 13-35a shows a simplified block diagram of a differential binary phase

keying

(DBPSK)

bit prior to

An incoming information bit is XNORed with BPSK modulator (balanced modulator). For the

transmitter.

entering the

Data

s

ti: * IL

input

data

bit,

i

modulator

'

i

1-bit |

first

DBPSK

Balanced

Vi /°

shift

the preceding

>

sin co t

delay

(a)

Input data

>"1 ^1 yi

XNOR

output

0^

1/

1

1

0°

0°

/ TV

*

1

10 10

11

(reference bit)

Output phase

180°

0°

180°

0°

180°

180°

180°

(b)

FIGURE

13-35

DBPSK

modulator:

(a)

block diagram; (b) timing diagram.

0°

0°

Differential

BPSK

531

DBPSK input

Balanced

Recovered

modulator

data

Balanced modulator output 1-Bit

w c t)(+sin

co c t)

=

+| -

(-sin co c t)(-sin co c t)

=

+^ - ^

(-sin co c t)(+sin co c t)

=

-i +

(+sin

delay

^ cos 2co c t cos 2co c t

^ cos 2co c t

(a)

DBPSK

180°

input phase

180°

g

0°

0°

0°

180°

180°

0°

180°

180°

0°

(reference phase)

101110001101 \

\

Recovered stream

bit

t

I

{

*

\

\

\

\

\

(b)

FIGURE

there

is

ence

bit

the

13-36

no preceding is

XNOR

DBPSK

bit

demodulator:

which

with

output data, and the phase

In Figure

is

13-35b the

are the same, the

a logic

XNOR

first

output

1,

data bit is

a logic

The balanced modulator operates produces +sin

1

compare

to

an

Therefore,

it.

refer-

initial

co (i

output of the balanced modulator.

at the

assumed a logic ply the complement of that shown.

a logic 0.

block diagram; (b) timing sequence.

assumed. Figure 13-35b shows the relationship between the input data,

reference bit

initial

(a)

the output from the

XNORed

is

1; if

as a conventional

output and a logic

at the

If the

circuit is sim-

with the reference

BPSK

produces -sin

If

bit.

XNOR

they are different, the

same

the

XNOR

they

output

is

modulator; oj c i

at

the

output.

DBPSK

Receiver

Figure 13-36 shows the block diagram and timing sequence for a

The received element

in

generated.

phase

is

signal

is

delayed by one

the balanced modulator. If

bit time,

they are the same, a logic

If

(—

they are different, a logic

incorrectly assumed, only the

DBPSK

first

voltage)

is

demodulated

encoding can be implemented with higher-than-binary

1

The primary advantage of DPSK requires between rate as that

is

carrier recovery circuit

bit

digital

1

the simplicity with is

needed.

and 3 dB more signal-to-noise

of absolute PSK.

A

+

voltage)

is

in error.

is

Differential

modulation schemes,

which

it

al-

DBPSK.

can be implemented.

disadvantage of

ratio to

(

generated. If the reference

though the differential algorithms are much more complicated than for

With DPSK, no

receiver.

then compared with the next signaling

DPSK

achieve the same

is

that

it

bit error

532

Digital

Communications

Chap. 13

CLOCK RECOVERY As with any

digital

system, digital radio requires precise timing or clock synchronization

between the transmit and the receive clocks

at the

Because of this,

circuitry.

it

is

necessary to regenerate

receiver that are synchronous with those at the transmitter.

Figure 13-37a shows a simple circuit that

commonly used

is

to recover clocking

information from the received data. The recovered data are delayed by one-half a

time and then compared with the original data the

clock that

is

in

recovered with this method

an

is

XOR

equal

circuit.

the

to

bit

The frequency of received data rate

(Fh ). Figure 13-37b shows the relationship between the data and the recovered clock timing.

From Figure 13-37b it can be seen that as long as the receive number of transitions (1/0 sequences), the recovered clock is

substantial

the receive data

were

clock would be

to

lost.

undergo an extended period of successive

To

1

's

prevent this from occurring, the data are scrambled

Data

»

1

in 1

'

1/2-bit

delay

' i

Recovered clock (a)

PWLTUI i

I

I

I

i

i

I

i

•

I

I

I

I

I

I

Delayed data I I I

Recovered clock

I

I

i l

I

I

I

nnrnmn

i

n

(b)

FIGURE

maintained.

13-37

(a)

Clock recovery

circuit; (b) timing

If

or 0's, the recovered

transmit end and descrambled at the receive end.

Data

data contain a

diagram.

at the

8

Applications for Digital Modulation

PROBABILITY OF ERROR

AND

533

ERROR RATE

BIT

(BER)

Probability of error P(e) and bit error rate

are often used interchangeably,

although they do have slightly different meanings. P(e) expectation of the error rate for a given system.

BER

is

a theoretical (mathematical)

an empirical (historical) record -5 of a system's actual error performance. For example, if a system has a P(e) of 10 is

,

this

means

that mathematically,

=

1/100,000).

If a

was one

bit error for

every 100,000

transmitted (1/10 the past there

-5

you can expect one

Probability of error

is

system has a

every 100,000

bit error in

BER

5 of 10~

,

this

a function of the receiver carrier-to-noise ratio.

the

minimum

carrier-to-noise

required

ratio

PSK

than that required for a comparable

for

a

QAM

minimum

shown

Eh/N

for determining the

is

explained

minimum

system

carrier-to-noise ratio.

parameter often used for comparing digital system performances

are

Depending

ratio varies. In is

less

system (see Table 13-3). Also, the higher

the level of encoding used, the higher the

bit-to-noise ratio (E h /N{) ).

that in

bits transmitted.

on the M-ary used and the desired P(e), the minimum carrier-to-noise general,

means

bits

in

is

Another

the energy of the

Chapter 20, where several examples

carrier-to-noise ratio for a given

M-ary system

and desired P(e).

PERFORMANCE COMPARISON OF VARIOUS DIGITAL MODULATION SCHEMES TABLE 13-3

(BER

= 1(T 6

Modulation

)

ON

technique

BPSK QPSK

E /N

ratio

ly

()

ratio

(dB)

(dB)

3.6

10.6

13.6

10.6

1

8QAM

13.6

10.6

8PSK

18.

14

I6PSK

24.3

18.3

16QAM 32QAM 64QAM

20.5

14.5

24.4

17.4

26.6

18.8

APPLICATIONS FOR DIGITAL MODULATION

A

digitally

has

many

modulated transceiver (transmitter-receiver)

applications.

They

are used in digitally

that uses

FSK, PSK, or

modulated microwave radio and

QAM

satellite

systems (Chapter 20) with carrier frequencies from tens of megahertz to several gigahertz,

and they are also used for voice band data

between 300 and 3000 Hz.

modems

(Chapter 14) with carrier frequencies

534

Digital

Communications

Chap. 13

QUESTIONS 13-1. Explain digital transmission and digital radio. 13-2. Define information capacity. 13-3.

What

are the three

most predominant modulation schemes used

FSK

13-5. Define the following terms for

and deviation

FSK

13-7.

What

ratio.

bit rate,

is

and

(b) the

the difference

minimum bandwidth

required for an

between standard

FSK

and

MSK? What

is

13-9. Explain the relationship between bits per second and baud for a

What

FSK

system

mark and space frequencies. the advantage of

MSK?

PSK.

13-8. Define

13-10.

system.

modulation: frequency deviation, modulation index,

13-6. Explain the relationship between (a) the

and the

radio systems?

in digital

13-4. Explain the relationship between bits per second and baud for an

is

how

a constellation diagram, and

13-11. Explain the relationship between the

and the

is it

used with

BPSK

system.

PSK?

minimum bandwidth

required for a

BPSK

system

bit rate.

13-12. Explain M-ary. 13-13. Explain the relationship between bits per second and baud for a

13-14. Explain the significance of the

I

and

Q

channels

in a

QPSK

QPSK

system.

modulator.

13-15. Define dibit. 13-16. Explain the relationship between the

and the

minimum bandwidth

required for a

QPSK

system

bit rate.

13-17.

What

13-18.

What advantage does

is

a coherent demodulator?

OQPSK

have over conventional

QPSK? What

is

a disadvantage of

OQPSK? 8PSK

13-19. Explain the relationship between bits per second and baud for an 13-20. Define

system.

tribit.

13-21. Explain the relationship between the

and the

minimum bandwidth

required for an

8PSK system

bit rate.

13-22. Explain the relationship between bits per second and baud for a

16PSK system.

13-23. Define quadbit. 13-24. Define

QAM.

13-25. Explain the relationship between the

and the 13-26.

What

is

minimum bandwidth

required for a

bit rate.

the difference

between

PSK

and

QAM?

13-27. Define bandwidth efficiency.

13-28. Define carrier recovery. 13-29. Explain the differences between absolute

PSK

and

13-30.

What

is

the purpose of a clock recovery circuit?

13-31.

What

is

the difference

differential

When

between probability of error and

is

it

bit

PSK.

used? error rate?

16QAM

system

Chap. 13

Problems

535

PROBLEMS FSK

13-1. For an

modulator with space,

rest,

respectively, and an input bit rate of 10

and mark frequencies of 40, 50, and 60 MHz, Mbps, determine the output baud and minimum

bandwidth. Sketch the output spectrum. 13-2. Determine the

MHz

of 40

minimum bandwidth and baud for a BPSK modulator with a carrier frequency

and an input

QPSK

13-3. For the

—90° and sketch

to

QPSK

13-4. For the

of 500 kbps. Sketch the output spectrum.

the

new

in

ta ( J

—

8PSK modulator MHz,

frequency of 100

Figure 13-13, change the

+90°

phase-shift network

constellation diagram.

demodulator shown

input signal of sin

13-5. For an

bit rate

modulator shown

cos

Figure 13-17, determine the

in

I

and

Q

bits for

an

o) r /.

with an input data rate (F h ) equal to 20

determine the

minimum

Mbps and

a carrier

double-sided Nyquist bandwidth (F N )

and the baud. Sketch the output spectrum.

8PSK modulator shown

13-6. For the co r /

in

Figure 13-19, change the reference oscillator to cos

and sketch the new constellation diagram.

13-7. For a

16QAM

modulator with an input

frequency of 100

MHz,

determine the

bit

rate

minimum

(Fh ) equal

to

20 Mbps and a

carrier

double-sided Nyquist bandwidth (F N )

and the baud. Sketch the output spectrum.

16QAM

13-8. For the a) ( ./

modulator shown

in Figure 13-30,

change the reference

and determine the output expressions for the following

I,

I

,

Q, and

Q

oscillator to cos

input conditions:

0000, 1111, 1010, and 0101. 13-9. Determine the bandwidth efficiency for the following modulators. (a)

(b) (c)

QPSK, Fh = 10 Mbps 8PSK, F h = 21 Mbps 16QAM, Fh = 20 Mbps

13-10. For the for bit

DBPSK

the

=

1).

modulator shown

following

input

bit

in

Figure 13-35, determine the output phase sequence

sequence: 00110011010101

(assume

that

the

reference

Chapter 14

DATA COMMUNICATIONS INTRODUCTION Data communications can be defined as the transmission of digital information (usually in binary form) from a source to a destination. The original source data are in digital form and the received data are in digital form, although the data can be transmitted in analog or digital form. The source information can be binary-coded alpha/numeric characters such as ASCII or EBCDIC, microprocessor op-codes, control words, user addresses, program data, or data base information. A data communications network can be as simple as two personal computers connected together through the public telephone network, or it can comprise a complex network of one or more mainframe computers and hundreds of remote terminals. Data machines (ATMs)

communications networks are used

to

bank computers or they can be used

to interface

computer terminals (CTs) or keyboard

programs

mainframe computers. Data communimass media

displays

(KDs)

directly to application

connect automatic

in

teller

to

cations networks are used for airline and hotel reservation systems and for

and news networks such as the Associated Press (AP) or United Press International (UPI).

The

list

of applications for data communications networks goes on almost indefi-

nitely.

HISTORY OF DATA COMMUNICATIONS It

is

highly likely that data communications began long before recorded time

form of smoke signals or tom-tom drums, although

536

it

is

improbable

in

the

that these signals

Data Communications Circuits were binary coded.

If

we

limit the

electrical signals to transmit

537 scope of data communications to methods that use

binary-coded information, then data communications began

in 1 837 with the invention of the telegraph and the development of the Morse code by Samuel F. B. Morse. With telegraph, dots and dashes (analogous to binary l's and

O's) are transmitted across a wire using electromechanical induction. Various tions of these dots

and dashes were used

and punctuation. Actually, the

first

to represent binary

telegraph

Wheatstone and Sir Willaim Cooke, but

was invented

codes for in

letters,

combinanumbers,

England by Sir Charles

their contraption required six different wires

Morse secured an American patent for the telegraph and in 1844 the first telegraph line was established between Baltimore and Washington, D.C. In 1849, the first slow-speed telegraph printer was invented, but it was not until 1860 that high-speed (15 bps) printers were available. In 1850, the Western Union Telegraph Company was formed in Rochester, New York, for the purpose of carrying coded messages from one person to another. In 1874, Emile Baudot invented a telegraph multiplexer, which allowed signals from up to six different telegraph machines to be transmitted simultaneously over a single wire. The telephone was invented in 1876 by Alexander Graham Bell and, consequently, very little new evolved in telegraph until 1899, when Marconi succeeded in sending radio telegraph messages. Telegraph was the only means of sending information for a single telegraph line. In 1840,

across large spans of water until 1920,

when

the

first

commercial radio stations were

installed.

Bell Laboratories developed the

tromechanical relays. The

first

first

special-purpose computer in 1940 using elec-

general-purpose computer was an automatic sequence-

controlled calculator developed jointly by Harvard University and International Business

Machines Corporation (IBM). The UNIVAC computer, built in 1951 by Remington Rand Corporation (now Sperry Rand), was the first mass-produced electronic computer. Since 1951, the number of mainframe computers, small business computers, personal

computers, and computer terminals has increased exponentially, creating a situation

where more and more people have the need other. Consequently, the

to

exchange

digital information with

each

need for data communications has also increased exponentially.

AT&T operating tariff allowed only equipment furnished by AT&T AT&T lines. In 1968, a landmark Supreme Court decision, the decision, allowed non-Bell companies to interconnect to the vast AT&T

Until 1968, the to

be connected to

Carterfone

communications network. This decision led to competitive data

started the interconnect industry,

which has

communications offerings by a large number of independent

companies.

DATA COMMUNICATIONS CIRCUITS Figure 14-1 shows a simplified block diagram of a data communications circuit. There is

a source of digital information, a transmission

medium, and

a destination.

source and destination equipment are digital; they process information binary pulses.

The transmission medium may be

in the

Both the

form of

a digital or an analog facility

and

Data Communications

538 Transmission Source:

Chap. 14

medium

(analog or digital)

Destination:

digital

digital

equipment

equipment

FIGURE

Data communications

14-1

cuit: simplified

cir-

block diagram.

could comprise one or more of the following: metallic wire pair, coaxial cable, microwave radio, satellite radio, or an optical fiber.

Data Communications Circuit Configurations and Topologies Configurations.

Data communications

either two-point or multipoint.

A

circuits

can be generally categorized as

two-point configuration involves only two locations

or stations, whereas a multipoint configuration involves three or

more

stations.

A

two-

point circuit can involve the transfer of information between a mainframe computer

and a remote computer terminal, two mainframe computers, or two remote computer terminals.

computer or

A

multipoint circuit

(host) to

is

generally used to interconnect a single mainframe

many remote computer

more computers or computer terminals

The topology or

Topologies.

how

fies

terminals, although any combination of three constitutes a multipoint circuit.

architecture of a data

communications

the various locations within the network are interconnected.

circuit identi-

The most common

topologies used are the point to point, the star, the bus or multidrop, the ring or loop,

and to

the

point.

for data

These

mesh. Figure

are

all

multipoint

configurations

except

the

point

14-2 shows the various circuit configurations and topologies used

communications networks.

Transmission

Modes

Essentially, there are four

modes of transmission

for data

communications

circuits: sim-

plex, half duplex, full duplex, and full'/full duplex.

Simplex.

With simplex operation, data transmission is unidirectional; informaone direction. Simplex lines are also called receive-only, transmit-

tion can be sent only in

only, or one-way-only lines.

Half duplex (HDX). both directions, but not

mode, data transmission is possible in same time. Half-duplex lines are also called two-way

In the half-duplex

at the

alternate lines.

mode, transmissions arc possible in both simultaneously, but they must be between the same two stations. Full-duplex

Full duplex directions

(FDX).

lines are also called

In the full-duplex

two-way-simultaneous or simply duplex

lines.

Data Communications Circuits

Station

539

Station 2

1

(a)

Many

>

remote stations

i

i Common

communications medium

I

I

FIGURE

(b)

(0

(d)

(e)

14-2

Data network topologies: (e) mesh.

(a) point to point; (b) star; (c)

bus or

multidrop; (d) ring or loop;

Full/full

duplex (F/FDX).

directions at the

transmitting to a second station

FDX

is

In the

F/FDX mode,

transmission

is

possible in both

same two stations (i.e., one and receiving from a third station at the same

same time but not between

the

station

is

time). F/

possible only on multipoint circuits.

Two-Wire versus Four-Wire Operation Two-wire, as the name implies, involves a transmission medium wires (a signal and a reference lead) or a configuration that

only two wires. With two-wire operation, simplex,

full-,

is

that either uses

or half-duplex transmission

possible. For full-duplex operation, the signals propagating in opposite directions

occupy

different bandwidths; otherwise, they will

mix

two

equivalent to having

linearly

and

interfere with

is

must each

other.

Four-wire, as the name implies, involves a transmission wires (two are used for signals that are propagating

in

medium

that uses four

opposite directions and two are

Data Communications

540

used for reference leads) or a configuration that

With four-wire operation,

is

Chap. 14

equivalent to having four wires.

the signals propagating in opposite directions are physically

separated and therefore can occupy the same bandwidths without interfering with each other. Four- wire operation provides

more

isolation

and

is

preferred over two- wire, al-

though four-wire requires twice as many wires and, consequently, twice the cost.

A

transmitter and

its

associated receiver are equivalent to a two-wire circuit.

A

transmitter and a receiver for both directions of propagation are equivalent to a four-

wire circuit. With full-duplex transmission over a two-wire

must be divided

in half, thus

line, the available

reducing the information capacity

bandwidth

in either direction to

one-half of the half-duplex value. Consequently, full-duplex operation over two-wire

much

lines requires twice as

time to transfer the same amount of information.

DATA COMMUNICATIONS CODES Data communications codes

are used for encoding alpha/numeric characters

and symbols

(punctuation, etc.) and are consequently often called character sets, character languages, or character codes. Essentially, three types of characters are used in data communications

codes: data link control characters, which are used to facilitate the orderly flow of data

from the source

to the destination;

graphic control characters, which involve the syntax

or presentation of the data at the receive terminal; and alpha/ numeric characters, which are used to represent the various

symbols used for

letters,

numbers, and punctuation

in

the English language.

code.

The first data communications code that saw widespread usage was the Morse The Morse code used three unequal-length symbols (dot, dash, and space) to

encode alpha/numeric characters, punctuation marks, and an interrogation word.

The Morse code is inadequate for use in modern digital computer equipment because do not have the same number of symbols or take the same length of time to send, and each Morse code operator transmits code at a different rate. Also, with Morse code, there is an insufficient selection of graphic and data link control characters all

characters

to facilitate the transmission

and presentation of the data typically used

in

contemporary

computer applications.

The

three

most

common

character sets presently used for character encoding are

American Standard Code for Information Interchange (ASCII), Extended Binary-Coded Decimal Interchange Code (EBCDIC).

the Baudot code, the

and the

Baudot Code The Baudot code (sometimes called the Telex code) was the first fixed-length character code. The Baudot code was developed by a French postal engineer, Thomas Murray, in 1875 and named after Emile Baudot, an early pioneer in telegraph printing. The Baudot code

is

a 5-bit character

code

that

is

used primarily for low-speed teletype

Data Communications Codes

541

equipment such as the TWX/Telex system. With a

code there are only 2 5 or 32

5-bit

codes possible, which digits,

is insufficient to represent the 26 letters of the alphabet, the 10 and the various punctuation marks and control characters. Therefore, the Baudot

code uses figure

The

latest

Alphabet No.

TWX

shift

and

letter shift characters to

version of the Baudot code 2.

The Baudot code

is

is

still

and Telex teletype systems. The

expand

recommended by

AP

its

capabilities to

the

CCITT

used by Western Union

58 characters.

as the International

Company

for their

and UPI news services also use the Baudot

code for sending news information around the world. The most recent version of the

Baudot code

is

shown

in

Table 14-1.

TABLE

BAUDOT CODE

14-1

Character

Figure

Letter

Binary code

shift

A B

?

Bit:

4

3

1

1

1

D

$

1

E

3

1

!

1

F

&

H

#

I

8

/

1

1

C

G

2

1

1

1

1

1

1

1

1

1

1

1

1

1

i

J

K

(

L

)

M

1

1

1

1

1

1

1

1

1

1

1

1

1

•

N

,

9

1

P

Q

1

R

4

S

bel

T U V

5 7

1

1

1

1

1

1

1

1

1

1

1

1

1

1

;

1

1

1

1

w

2

1

X Y

/

1

1

6

1

1

1

I

1

1

1

1

1

"

Z

1

1

Figure shift

]

1

Letter shift

1

1

Space

1

1

Line feed (LF)

Blank

1

(null)

1

1

1

1

1

Data Communications

542 TABLE 14-2

CODE—ODD

ASCII-77

PARITY

Binary code 7

Bit:

NUL SOH STX ETX EOT

ENQ ACK

6

5

4

3

2

Binary code /

1

1

1

1

1

1

1

1

1

1

1

1

1

1

BEL

1

1

BS

HT NL VT

1

1

1

1

1

FF

1

1

SO

1

1

1

1

1

DLE DC DC2 DC3 DC4

1

1

1

SYN

1

1

1

1

ETB

1

CAN EM

!

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

&

1

1

1

1

1

1

1

%

1

1

1

1

)

* 1

+

1

1

1

1

-

05

E

1

06

F

1

1

07

G

1

1

08

H

09

I

0A

J

1

1

1

1

1

1

7

10

P

11

Q

12

R

13

S

14

T

15

16

U V

17

w

1

1

]

A

IF

—

21

1

1

1

45

46 1

47 48

1

1

1

1

1

1

1

1

1

49

4A 1

1

4B 4C

1

4D

1

4E 4F 50

1

1

1

1

51

52

1

1

53

54

1

1

1

1

1

1

1

1

1

55

56 1

57 58

1

1

1

1

1

1

1

1

1

1

1

59

5A

1

5B 5C

1

5D

1

5E 5F

1

61

60

a

22

b

23

c

24

d

25

e

1

1

26

f

1

1

27

g

28

h

29

i

62

1

1

1

1

63

64

1

1

65

1

67

66

1

1

68 1

1

2A 2B 2C 2D

in

1

2E

n

1

1

1

1

1

1

1

69

6A

1

1

k 1

43

44

1

1

J

41

42

1

1

1

IE

Hex

/

1

Z

ID

2

1

1A

[

3

1

1

\

4

1

19

1C

5

40

X Y

IB

6

1

L

20

"

(

04

C D

1

1

# $

03

18

1

RS US SP

B

N

1

GS

02

M

1

FS

A

0D

1

SUB ESC

(a

01

0E OF

1

NAK

00

1

1

1

Bit:

K

1

I

Hex

0B oc

1

1

CR SI

Chap. 14

6B 6C

6D 6E

Data Communications Codes TABLE 14-2

543

(continued) Binary code

7

Bit:

5

6

1

4

1

1

Binary code

3

2

/

1

1

1

1

1

1

1

1

2

1

1

1

1

1

1

1

1

1

3

1

4

1

5

1

1

1

1

6

1

1

1

1

1

7

1

1

1

1

8

1

1

1

1

1

1

1

1

1

1

1

1

1

1

9

;


9

1

Hex

device control 2

device control 3 device control 4 negative acknowledge

cancel substitute

escape field separator

group separator record separator unit separator

space delete

data link escape

ASCII Code In 1963, in an effort to standardize data

communications codes, the United States adopted

System model 33 teletype code as the United States of America Standard Code for Information Interchange (US ASCII), better known simply as ASCII-63. Since its adoption, ASCII has generically progressed through the 1965, 1967, and 1977 verthe Bell

sions, with the

1977 version being recommended by the

CCITT

as the International

544

Data Communications

Alphabet No.

ASCII

5.

the least significant bit

designated b 6 bit,

which

.

b7

is

is

Chap. 14

which has 2 7 or 128 codes. With ASCII, and the most significant bit (MSB) is of the ASCII code but is generally reserved for the parity

a 7-bit character set

(LSB)

not part

designated b

is

explained later in this chapter. Actually, with any character

is

set, all bits

are equally significant because the code does not represent a weighted binary number. It

is

common

with character codes to refer to bits by their order; b

the first-order bit, b 7

bit, bj is

the bit transmitted

LSB and

1977 version of the

With

is

the zero-order

serial transmission,

LSB. With ASCII, the low-order bit ASCII is probably the code most often used ASCII code is shown in Table 14-2.

first

transmitted

is

the seventh-order bit, and so on.

is

is

called the

first.

(b

)

the

is

today.

The

EBCDIC Code EBCDIC

an 8-bit character code developed by

is

and IBM-compatible equipment. With 8

EBCDIC

the

most powerful character

b 7 and the

MSB

transmitted

first

shown

in

designated b

is

set.

2

bits,

Note

8

that with

Therefore, with

.

and the low-order

bit (b

)

is

IBM

and used extensively

in

EBCDIC

EBCDIC,

transmitted

LSB

the

is

designated

the high-order bit (b 7 )

is

The EBCDIC code

is

last.

Table 14-3.

TABLE 14-3

EBCDIC CODE Binary code

Bi nary code

Bit:

/

2

3

4

5

6

NUL SOH

7

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

DLE SBA EUA

1

IC

1

1

I

2

3

4

5

6

7

1

1

1

Hex 80

01

a

02

b

1

03

c

1

04

d

1

05

e

1

06

f

1

1

07

g h

1

1

08

1

09

i

1

1

0B OC

0D 0E OF 10

1

1

NL

()

0A

1

FF

Bit:

00 1

STX ETX PT

Hex

IBM

or 256 codes are possible, making

82 1

83

84 1

85

86 1

87 88

1

1

1

1

1

89

8A 1

8B

8C

1

1

1

1

1

1

1

1

1

1

1

8D

1

8E 8F 90

1

11

J

12

k

1

1

13

1

1

1

14

in

15

n

1

1

Q

81

1

1

1

1

91

92 1

1

93 l

)4

l

>.^

Data Communications Codes TABLE 14-3

545

(continued) B nary code

Binary code Bit:

/

2

3

4

5

6

l

1

1

1

EM

7

1

1

1

1

DUP

1

1

Hex

Bit:

SF

1

1

1

ITB

1

1

1

1

J

2

3

4

5

6

16

o

1

1

17

P

1

1

18

q

19

r

7

1

1

1

1

1

ETB ESC

1

1

1

1

1

1

1

1

IB

1

ID

1

IE

1

1

IF

1

1

1

1

ENQ

1

1

22

s

1

23

t

1

24

u

25

V

1

26

w

1

1

27

X

1

1

28

y

29

z

1

1

1

1

1

1

1

2E 2F

1

1

1

1

1

1

1

1

1

EOT

1

1

1

1

1

1

1

1

RA

1

NAK SUB

1

1

1

1

1

1

SP I

1

1

1

1

1

AE AF

1

32

1

33

1

34

1

35

1

36

1

1

37

1

1

1

1

39

1

1

1

3B 3C

1

3D

1

3E

1

1

3F

1

1

1

40

{

41

A

42

B

43

C D

44

AB AC

BO

31

3A 1

A2 A3 A4 A5 A6 A7 A8 A9

AD

38

1

Al

1

1

30

SYN

9F

AA

1

2D

1

9D 9E

1

1

1

1 1

~

1

1

9B 9C

A0

21

2B 2C

1

99

9A 1

1

2A

1

97

98

20 1

Hex 96

1

1

1A 1C

FM

9

1

1

1

1

1

1

1

1

Bl

B2 B3 B4 B5 B6 B7 B8 B9

BA BB BC

BD BE BF CO CI

C2 C3 C4

Data Communications

546 TABLE 14-3

(continued) Binary code

Binary code Bit:

Chap. 14

f

2

3

4

5

6

7

1

I

1

1

1

1

1

1

$




1

6A

1

,

1

1

64

1

1

J

1

1

1

K

60

/

1

52

5E 5F

-

1

51

5D

1

1

1

1

5B 5C

7

1

}

1

I

6

1

5A

1

5

1

50

1

4

F

1

1

3

E

4E 4F

1

2

46

4D

&

1

/

45

1

1 1

Bit:

4A

1

1

Hex

1

EB EC

ED EE EF F0

71

1

1

1

72

2

1

1

1

73

3

l

1

1

1

Fl

F2 1

F3

Error Control

547

TABLE 14-3

(continued) Binary code

12

Bit.

(a

3

4

5

Binary code

6

Hex

7

10

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

I

1

Bit:

12

3

75

1

1

76

1

6

7

1

DLE =

data link escape

DUP =

duplicate

1

7A

1

7B

1

7C 7D 7E

1

1

1

1

I

I

F9

1

FA FB FC FD

1

FE FF

1

1

7F

1

F7

F8

79

1

F5

F6

1

78 1

Hex F4

I

10 111

77

1

5

10 10

74 1

4

ITB = end of intermediate transmission block

EM = ENQ = EOT =

NUL =

null

PT = program

end of medium

RA = SBA =

enquiry

end of transmission

ESC = escape ETB = end of transmission block ETX = end of text EUA = erase unprotected to address FF = form feed FM = field mark IC = insert cursor

SF =

SOH = SP =

STX = SUB = SYN = NAK =

tab

repeat to address set buffer

address

start field start

of heading

space start

of text

substitute

synchronous negative acknowledge

ERROR CONTROL A

data communications circuit can be as short as a few feet or as long as several

thousand miles, and the transmission

complex

as a

microwave,

satellite,

medium can be

as simple as a piece of wire or as

or fiber optic system. Therefore, due to the nonideal

transmission characteristics that are associated with any communications system, inevitable that errors will occur and that

it is

it

is

necessary to develop and implement proce-

dures for error control. Error control can be divided into two general categories: error detection and error correction.

Error Detection Error detection

when

is

simply the process of monitoring the received data and determining

a transmission error has occurred. Error detection techniques

bit (or bits) is in error, is

do not

identify

which

only that an error has occurred. The purpose of error detection

not to prevent errors from occurring but to prevent undetected errors from occurring.

.

Data Communications

548

How

a system reacts to transmission errors

The most common

is

Chap. 14

system dependent and varies considerably.

communications

error detection techniques used for data

circuits

redundancy, exact-count encoding, parity, vertical and longitudinal redundancy

are:

checking, and cyclic redundancy checking.

Redundancy involves transmitting each character twice. If the Redundancy. same character is not received twice in succession, a transmission error has occurred. The same concept can be used for messages. If the same sequence of characters is not received twice in succession, in exactly the same order, a transmission error has occurred.

With exact-count encoding, the number of l's in each of an exact-count encoding scheme is the ARQ

Exact-count encoding. character

An example

the same.

is

code shown

ARQ

Table 14-4. With the

in

and therefore a simple count of the number of

code, each character has three l's 's

1

received can determine

if

in

it,

a transmission

error has occurred.

Parity.

Parity

probably the simplest error detection scheme used for data

is

communications systems and ing.

With

the total

used with both vertical and horizontal redundancy check-

is

parity, a single bit (called a parity bit)

number of

number (odd

"C"

for the letter

is

There are three

the

P

made

bit is

even parity

even number (even

parity) or an

bit.

added

to

parity).

each character to force

code, not counting the parity

number of

a 0, keeping the total

P

made

bit is

a

and the

1

bit

representing the parity

bit.

If

l's at three,

total

be either an odd

For example, the ASCII code

43 hex or P1000011 binary, with the P l's in the

used, the

is

is

l's in the character, including the parity bit, to

number of

odd

parity

is

used,

an odd number. l's is four,

If

an even

number.

Taking a closer look the

number of

the

dropped, the code

bits are

and for even

parity, the

either PI

,

P

P

when

all its

output

is

parity

bit.

a

inputs are equal

1.

can be seen that the parity

all

0's or

is

by pairs of

PI

all

is is

1

—

of

both

bit is

For the

parity, the

l's are also

independent of

letter

P

"C,"

if all

bit is still

a

excluded, the code

Again, for odd parity the P

.

is

bit is a 0,

1

equivalence of equality. the

XOR

gate.

l's), the output

With an is

a 0.

A

logic gate that will determine

XOR

If all

Figure 14-3 shows two circuits that are

Essentially,

l's.

For odd

11.

P

or

bit is a

definition of parity

equal (either

it

bit is still a 1. If pairs 1,

and for even parity the P

The

at parity,

0's in the code and unaffected

gate,

if all

the inputs are

inputs are not equal, the

commonly used

to generate a

go through a comparison process eliminating

circuits

The circuit shown in Figure 14-3a uses sequential (serial) comparicircuit shown in Figure 14-3b uses combinational {parallel) comparison.

0's and pairs of l's.

son, while the

With the sequential and so on. The

parity generator b ()

result of the last

desired, the bias bit

parity

is

made

a logic

1.

is

The output of

XOR

made

is

XORed

a logic

the circuit

is

with b,, the result

is

XORed

with b :

.

compared with a bias bit. It even 0. If odd parity is desired, the bias bit is the parity bit, which is appended to the

operation

is

549

Error Control

ARQ EXACT-COUNT CODE

TABLE 14-4 ]

Bit:

2

1

Binary ;ock

3

4

5

6

1

1

1

Letter shift

1

1

Figure shift

1

1

1

1

1

1

1

1

D

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

]

1

1

1

1

8

J

(bell)

K

(

L

)

,

1

Q

1

R

4 '

T U

5

V

=

W

2

X Y

6

7

/

+

Z

1

1

1

£

I

S

1

1

I

(a-

H

P 1

1

1

G

9

1

1

1

%

1

1

1

3

F

N

1

1

(WRU)

E

M

I

1

1

1

1

1

1

7

1

1

1

I

—

B

C

1

1

A

1

1

Figure

Letter

1

1

1

7

1

1

1

Character


pL>—)0— )n>^

Parity bit

Bias bit (a)

On Parity bit

Bias bit (b)

FIGURE

14-3

Parity generators: (a) serial; (b) parallel.

odd

1,

parity; 2, even

parity.

The primary advantage of an even number of if

bits are

parity

is its

simplicity.

The disadvantage

is

that

received in error, the parity checker will not detect

it

when (i.e.,

the logic conditions of 2 bits are changed, the parity remains the same). Consequently,

over a long period of time, will detect only

parity,

assumes an equal probability

that an

50%

of the transmission errors

even or an odd number of

and horizontal redundancy checking. an error detection scheme that uses parity

is

(this

in error).

Vertical redundancy checking

Vertical

(VRC)

could be

bits

to

determine

if

a transmission

VRC is sometimes With VRC, each character has a parity bit added to it prior to transmission. It may use even or odd parity. The example shown under the topic "parity" involving the ASCII character "C" is an example of how VRC is used.

called character

error has occurred within a character. Therefore, parity.

Horizontal or longitudinal redundancy checking

scheme and

that uses parity to

all

of the other characters

in

with their respective bits from

LRC is

the

is

the result of

XORing

XORing

(HRC

or

LRC)

is

an error detection

a transmission error has occurred in a

message

message

parity.

from each character

words, b

In other

if

With LRC, each bit position has a in the message is XORed with b the message. Similarly, b,, b 2 and so on, are XORed

therefore sometimes called

is

parity bit.

from

determine

,

all

the other characters in the message.

the "characters" that

make up

of the bits within a single character. With

Essentially,

a message, whereas

LRC, only even

VRC

parity

is

used.

The

LRC

bit

sequence

is

computed

in the

transmitter prior to sending the data.

Error Control

551

then transmitted as though

it

were the

character of the message. At the receiver,

last

LRC is recomputed from the data and the recomputed LRC is compared with the LRC transmitted with the message. If they are the same, is assumed that no transmission

the

it

have occurred.

errors

Example

EXAMPLE

14-1

they are different, a transmission error must have occurred.

shows how

VRC

and

LRC

are determined.

14-1

Determine the

odd

If

parity for

VRC and LRC for the following VRC and even parity for LRC.

Character

Hex

LSB

ASCII-encoded message:

T

H

E

sp

C

A

T

LRC

54

48

45

20

43

41

54

2F

1

1

bo

I

b,

^ ^2

^

1

1

1

1

b5

The

LRC

The

LRC

is

b6

VRC

b7

2FH

VRC

bits are

1

I

1

b4

MSB

1

1

b,

1

1

1

1

I

1

1

1

each character

bit for

in the

i

o

1

or 00101111 bi nary. In

computed

Use

I

1

b2

§ w

THE CAT.

ASCII

,

this is the character

computed

is

/.

in the vertical direction,

horizontal direction. This

is

the

same scheme

and the

was

that

used with the early teletype paper tapes and keypunch cards and has subsequently been carried over to present-day data

The group of

communications applications.

make up

characters that

message

the

called a block of data. Therefore, the bit sequence for the

(i.e.,

LRC

check character (BCC) or a block check sequence (BCS). because the

LRC

has no function as a character

LRC

or data link control character); the

is

(i.e.,

it

is

is

BCS

THE CAT)

is

often

often called a block

more appropriate

is

not an alpha/numeric, graphic,

simply a sequence of

used for error

bits

detection. Historically,

LRC

detects

between 95 and

will not detect transmission errors

the

same

LRC

bit position.

is still

If

For example,

VRC

and

LRC

all

if

b4

in

two

transmission errors.

LRC

characters have an error in

different characters

are used simultaneously, the only time an error

and the happen.

VRC

is

of

is

in error,

the

valid even though multiple transmission errors have occurred.

when an even number of same bit positions in each

tected

98%

when an even number of

bits in

character are in error, which

does not identify which

would go unde-

an even number of characters were is

bit is in error in a character,

in error

highly unlikely to

and

LRC

does not

> H H

X

U u

10

'5

X

M

00

u B

X

WD v©

o X

U OS u

^ x

LL

r—-i £

V * J

vy

£

CM

"~x

""x

"~x

xM£ 1

eceive

ismit

a CO

CO

Q £

552

.


m+

+

n

=

1

+

5

1

Hamming bits are sufficient to meet the criterion of + 5 = 17 bits make up the data stream. place 5 Hamming bits into the data stream:

Equation 2-1

18; therefore, 5

Therefore, a total of 12 Arbitrarily

17 lb IS 1M 13 12 11 10 1 8 7 H

To determine a

1

H

1

1

Hamming

the logic condition of the

as a 5-bit binary

number and

XOR

Binary number

00010

6

001 10

XOR

00100 01100

12

The

17-bit

express

bits,

2

XOR J4_ XOR

00110

16

10000

XOR

10110

=

1.

3 2 1

M

all bit

positions that contain

them together.

Bit position

b, 7

b 5

HH010H010

1

01000

oino

bia

=

0,

b9

=

==

1,

Hamming code b8

=

1

,

b4

=

encoded data stream becomes

H

H

H H

H

11010100110100010

Assume stream

that during transmission, an error occurs in bit position

14.

The received

is

11000100110100010 At the receiver to determine the

them with the binary code

for

bit

position in error, extract the

each data

bit

Hamming

position that contains a

1.

bits

and

557

Synchronization

Bit position

Binary number

Hamming code

10110

2

10110

XOR

10100

6

00110

XOR

10010

12

01100

XOR

11110

16

10000

XOR Bit position 14

was received

01110

To

in error.

fix the error,

The Hamming code described here like all

FEC

in the

Hamming

lengthening the transmitted message. The purpose of

message wastes transmission time and

FEC and

simply complement

bit 14.

bits

It

cannot be

Ham-

themselves. The

codes, requires the addition of bits to the data, consequently

However,

the wasted time of retransmissions.

ARQ

binary 14

will detect only single-bit errors.

used to identify multiple-bit errors or errors

ming code,

=

in

FEC

codes

is

to reduce or eliminate

FEC

the addition of the

Obviously, a trade-off

itself.

system requirements determine which method

is

is

bits to

each

made between

best suited to a

particular system.

SYNCHRONIZATION Synchronize means to coincide or agree

in time.

In data

communications, there are

four types of synchronization that must be achieved: bit or clock synchronization,

modem

or carrier synchronization, character synchronization, and message synchronization.

clock and carrier recovery circuits discussed in Chapter 13 accomplish synchronization, and message synchronization

is

discussed

in

Chapter

bit

and

The

carrier

15.

Character Synchronization Clock synchronization ensures slot for the

occurrence of a

that the transmitter

bit.

When

and receiver agree on a precise time

a continuous string of data

significant data bit, the parity bit, zation: identifying the beginning circuits, there are

and synchronous.

and the stop

bit. In

essence, this

is

received,

is

necessary to identify which bits belong to which characters and which

it

is

bit is the least

character synchroni-

and the end of a character code. In data communications

two formats used

to achieve character synchronization:

asynchronous

Data Communications

558 Asynchronous data format. between a

start

and a stop

With asynchronous data, each character

The character code

continuing through the

MSB

The

A

logic

transmitted

5, or 2 stop

used for the

is

on a data communications

transmitted

parity bit (if used)

last bit

1.

1,

first bit

start bit

is

by a high-to-low

follows the

because an

circuit is identified

which

idle condition

time

always a logic

first

bit that

is

immediately

bits are logic l's,

which

beginning of each character. After the

at the

l's

character

start

data and parity bits are clocked into the receiver. If data are transmitted

bit is detected, the in real

is

(no data transmission)

and the

of the character code. All stop

guarantees a high-to-low transition

and

by the transmission of continuous

transition in the received data,

LSB

start bit is the

always

bits.

(these are often called idle line l's). Therefore, the start bit of the identified

is

LSB

transmitted directly after the

is

the stop bit,

is

the start bit and

beginning with the

bits are transmitted next

MSB. The

of the character. The

There can be either

I.

framed

is

Figure 14-5 shows the format used to frame a character

bit.

for asynchronous data transmission.

a logic 0.

Chap. 14

an operator types data into their computer terminal), the number

(i.e., as

of idle line l's between each character will vary. During this dead time, the receiver

simply wait for the occurrence of another

will

before clocking in the next

start bit

character.

Stop (1,

Parity

bit

1.5,2)

1

1

Start

Data bits (5-7)

bit

1/0

b6

EXAMPLE

b4

b5

MSB

b3

bit

b2

bi

b

FIGURE LSB

14-5

Asynchronous data

for-

mat.

14-4

For the following string of asynchronous ASCII-encoded data, identify each characterr(as-

sume even

parity

and 2 stop

bits).

Parity Start Parity Stop \Sto P// 1111 000100 01011 01000001011 11111111 DlDlDODlll^ DlODDDlOlll^

Start LSB

MSB

Parity

start

/£-top

DA

|

I

A

T

Synchronous data format.

With synchronous data,

rather than frame each char-

acter independently with start and stop bits, a unique synchronizing character called a

SYN

character

ASCII code, receives the character.

is

transmitted at the beginning of each message.

SYN character is SYN character, then

The character

it

used to signify the end of what kind of transmission Chapter 15.

that is

With asynchronous data, be continuously synchronized. (he

same

rate

until

it

clocks in the next 7 bits and interprets them as a

the type of protocol used and

characters are discussed in

For example, with

16H. The receiver disregards incoming data

the

a transmission varies with it

is.

Message-terminating

it

is

not necessary that the transmit and receive clocks

It

is

only necessary that they operate

and be synchronized

at

the beginning

oi'

at

each character.

approximately This

was

the

Data Communications Hardware purpose of the

start

With synchronous

559

to establish a time reference for character synchronization.

bit,

data, the transmit and receive clocks

must be synchronized because

character synchronization occurs only once at the beginning of the message.

EXAMPLE

14-5

For the following string of synchronous ASCII-encoded data, identify each character (assume

odd

parity).

LSB

MSB

i

I

DATA

11111 1011010 000001000111000001 11010001 01000000 11111111 V

-y

SYN

^

V

X —>

Y

With asynchronous data, each character has 2 or 3 bits added to each character and 1 or 2 stop bits). These bits are additional overhead and thus reduce the

start

(1

efficiency of the transmission (i.e., the ratio of information bits to total transmitted bits).

SYN

Synchronous data have two

characters (16 bits of overhead) added to each

message. Therefore, asynchronous data are more efficient for short messages, and syn-

chronous data are more

efficient for long

messages.

DATA COMMUNICATIONS HARDWARE Figure

14-6 shows the block diagram of a multipoint data communications circuit

that uses a

bus topology. This arrangement

one of the most

is

used for data communications circuits. At one station there

and

at

each of the other two stations there

hardware and associated circuitry terminals

is

that

is

a cluster of

common a

is

configurations

mainframe computer

computer terminals. The

connect the host computer to the remote computer

called a data communications link.

The

station with the

mainframe

is

called

the host or primary and the other stations are called secondaries or simply remotes.

An

arrangement such as

this

is

called a centralized network; there

is

one centrally

located station (the host) with the responsiblity of ensuring an orderly flow of data

between the remote

which

is

stations

and

At the primary and a data

modem

station there

(a data

computer terminals,

many

Data flow

is

controlled by an applications program

printers,

a mainframe computer, a line control unit (LCU),

commonly referred to simply as a modem). At modem, an LCU, and terminal equipment, such as and so on. The mainframe is the host of the network and is

is

a

where the applications program

Figure

is

modem

each secondary station there

is

itself.

stored at the primary station.

is

stored for each circuit

it

serves. For simplicity,

14-6 shows only one circuit served by the primary, although there can be different circuits served

by one mainframe computer. The primary

the capability of storing, processing, or retransmitting the data

it

station has

receives from the

secondary stations. The primary also stores software for data base management.

The

LCU

at the

primary station

is

more complicated than

the

LCUs

at the

secondary

Data Communications

560

Chap. 14

Primary station

RS-232C

Mux

Mainframe computer

serial

DTE

channel

data

processor

modem

Transmission

DCE

data

modem

modem RS-232C

RS-232C

DTE

DTE

line control

line control

unit

unit

ROP

Secondary station

ROP

CT

Secondary station 2

1

i

l

FIGURE

LCU

stations.

The

circuits,

which could

Multipoint data communicat ons circuit block diagram.

14-6

at the all

primary station directs data

have different characteristics

codes, data formats, etc.).

one data

link

The

LCU

at a

and a few terminal devices which

same character code. Generally speaking, if is called front-end processor (FEP). The an FEP. it

traffic to

and from many different

(i.e., different bit rates,

secondary station directs data

the it,

medium

DCE

data

CT

DCE

interface

front-end

a.

all

traffic

character

between

same speed and use

operate at the

LCU has software associated with LCU at the primary station is usually

the

Line Control Unit

The

LCU

interface is

has several important functions. The

between the host computer and the

connected to a different port on the

LCU

at the

circuits that

LCU. The LCU

it

primary station serves as an serves.

Each

circuit served

directs the flow of input

and

output data between the different data communications links and their respective applications

program. The

data.

The

data

LCU

in

mux

LCU

performs parallel-to-serial and serial-to-parallel conversion o(

interface channel

parallel.

between the mainframe computer and the

Data transfer between the

modem

and the

LCU

is

done

LCU

transfers

serially.

The

also houses the circuitry that performs error detection and correction. Also, data

.

Data Communications Hardware

(DLC)

link control

561

characters are inserted and deleted in the

LCU. Data

link control

characters are explained in Chapter 15.

LCU

The

when

operates on the data

it

form and

in digital

is

is

therefore called

data terminal equipment (DTE). Essentially, any piece of equipment between the main-

modem or the station equipment and its modem is classified The modem is called data communications equipment (DCE)

frame computer and the

as data terminal equipment.

because

it

interfaces the digital

LCU,

Within the

LCU's

there

is

functions. This circuit

used and a

USRT when

DTE

to the

analog transmission

line.

a single integrated circuit that performs several of the is

UART

called a

when asynchronous transmission

synchronous transmission

is

Universal asynchronous receiver/transmitter (UART).

DTE

asynchronous transmission of data between the mission means that an asynchronous data format

between the

tion transferred

DTE

UART

The

is

used for

and the DCE. Asynchronous trans-

used and there

is

is

used.

is

no clocking informa-

and the DCE. The primary functions of the

UART

are:

To perform

serial-to-parallel

2.

To perform

error detection by inserting and checking parity bits

3.

To

1

insert

and detect

start

and

parallel-to-serial conversion of data

and stop

bits

UART is divided into two sections: the transmitter and the reshows a simplified block diagram of a UART transmitter. Prior to transferring data in either direction, a control word must be programmed into the UART control register to indicate the nature of the data, such as the number of data bits; if parity is used, and if so, whether it is even or odd; and the number of Functionally, the

ceiver. Figure 14-7a

stop

bits.

the control set

word

up the data-,

UART simple. that

it

The

is

the

Essentially,

always only one

start bit

and

start it

bit

is

the

only

must be a logic

0.

for the various functions. In the parity-,

transmitter.

UART

and stop-bit steering logic

The operation of

When

the

DTE

that

is

not

UART,

word

the control

transmitter section

(TBMT)

signal to the

(TD0-TD 7)

into the transmit buffer register with the transmit data strobe signal

register

when

signal goes active (the

the shift register

is

empty and

TEOC

outputted on the transmit serial output (TSO) pin with a

clocked out, the

DTE

really quite to indicate

TBMT,

it

when

the transmit-

signal simply tells the buffer

available to receive data).

in the

used to

(TDS). The contents

shift register

have been loaded into the transmit

clock (TCP) frequency. While the data

is

is

program

an d strobes them

through the steering logic circuit, where they pick up the appropriate parity bits. After data

is

DTE

senses an active condition on

of the transmit buffer register are transferred to the transit

(TEOC)

to

circuit.

sends a parallel data character to the transmit data lines

end-of-character

there

optional;

how

UART

the

sends a transmit buffer empty

ready to receive data.

bit

Figure 14-7b shows

shift register, bit rate

The data pass start,

stop,

and

they are serially

equal to the transmit

transmit shift register are sequentially

loads the next character into the buffer register.

The process

NSP

NPB |

NDB2 NDB1 POE I

w

1

1

cs Control register

1

Parallel input

TDS

TD 7

TD 6

TD 5

data from

TD 4

TD 3

1**1

LCU TD 2

TD,

TD

1*1

+

TEOC

Transmit buffer

register

Parity

generator

Data-, parity-,

and stop-bit

steering logic

TCP

Timing

Transmit

generator

a

Start

Output

bit

circuit

shift register

out

Status word register

TBMT (a)

NPB

1

= no parity

bit

(RPE

disabled)

= parity bit

POE

1

NSB

1

= parity even = parity odd = 2 stop bits =

NDB2

1

stop bits

NDB1

Bits/word

5 6

1

7

1 1

Note:

8

1

When NDB2/NDB1

= 11 and

NSB

=

1,

1.5 stop bits

(b)

FIGURE

562

14-7

UART

*

Serial

data

Buffer empty logic circuit

SWE

TSO

transmitter: (a) simplified block diagram; (b) control word.

Data Communications Hardware continues until the in

DTE

563

has transferred

The preceding sequence

data.

all its

is

shown

Figure 14-8.

UART

A

receiver.

the

bits,

receiver

is

shown

in

data bits, and the parity-bit information for

UART receiver are determined by the same control

(i.e., the

the

UART

simplified block diagram of a

Figure 14-9. The number of stop

UART The

word that is used by the transmitter number of stop bits, and the number of data bits used for must be the same as that used for the UART transmitter).

type of parity, the receiver

UART

receiver ignores idle line l's.

When

the start bit verification circuit, the data character shift register.

If parity

used, the parity bit

is

After one complete data character

is

is

a valid start bit

serially

checked

loaded into the

is

is

check

in the parity

transferred in parallel into the buffer register and the receive data available is set in the status word word enable (SWE) and

register. if

To

U

~TJ


o\ t>

QC

-

\

V 1

1

\X

CN

w

670

/

'

.

2913/14

Combo

System

Reliability Features

Chip

671

The 2914 combo chip is powered up by pulsing the transmit frame synchronization input (FSX) and/or the receive frame synchronization input ( FSR) while a TTL high (inactive condition) is applied to the power down select pin (PDN) and all clocks and power supplies are connected. The 2914 has an internal reset on all power-ups (when ,

VBB

or

VCC

digital output

On (TSX)

are applied or temporarily interrupted). This ensures the validity of the

and thereby maintains the

the transmit channel,

PCM

are held in a high-impeda nce

power-up. After

delay

this

Due

the proper time slots. circuit requires

s tate

hook detection

on the transmit channel, the analog

to reach equilibrium. Therefore, signaling informa-

is

available almost immediately while analog input

60-ms delay.

signals are not available until after the

On

the receive channel, the signaling bit output pin

for approximately

500

jjls

after

highway.

signaling are functional and will occur in

to the auto-zeroing circuit

ms

PCM

for approximately four frames (500 |xs) after

DX, TSX, and

approximately 60

tion such as on/off

integrity of the

data output (DX) and transmit timeslot strobe

SIGR

power-up and remains inactive

is

also held low (inactive)

until

updated by reception

of a signaling frame.

TSX

and

DX

approximately 20 tion could be

are placed in the high-impedance state

(jls

after an interruption of the

caused by some kind of

and SIGR

is

held low for

master clock (CLKX). Such an interrup-

fault condition. i

Power-Down and Standby Modes To minimize power consumption, two power-down modes 2914 functions

are disabled.

Only

the

are provided in

which most

power-down, clock, and frame synchronization

buffers are enabled in these modes.

The power-down this

is

enabled by placing an external

TTL

low signal on PDN.

mode power consumption is reduced to an average of 5 mW. The standby mode for the transmit and receive channels is

by removing

FSX

In

separately controlled

and/or FSR.

Fixed-Data-Rate

Mode

In the fixed-data-rate

mode, the master transmit and receive clocks

(CLKX

and

CLKR)

perform the following functions:

1

Provide the master clock for the on-board switched capacitor

filters

2.

Provide the clock for the analog-to-digital and digital-to-analog converters

3.

Determine the input and output data

rates

between the codec and the

PCM

highway

Therefore, in the fixed-data-rate mode, the transmit and receive data rates must

be either 1.536, 1.544, or 2.048 Mbps, the same as the master clock

rate.




>

E

«J

3

u_

o

o o

Si

—& ^/\

"

E

1 1

,_ CC

CM 11

> to

"'

> >

+

t 1

'

1

,-

CM

CO CM

CM

X co O

+

1

x u. >

x >

u.

o CM

o>

o

< a o

z

CC

i

i

,

1

F

\

I

CO ,_

r^ "~

CO

in

"*

-I

IX

X Q

X

X ^ _J o

111

CO

co

|h

< X

r

*""

CO u.

cc

^ -I o

CO

Q

2

°

to

+

co CD

>

-

i

o

o

or

cc

LU CO

s

CO

O

S

Q.

V _J o

8

C?

5 cl

-1

CO

o Q

CM

CO

^

in

CO

r^

CO

O)

5

cc

>

|z

Q

-J

>

Q

o I

cc

Q

u.

O

^

cc

cc

i

J i "3d

CM

£

k

CV

1

in

-WW

1

t-

V"

ll


,

in r

'

1

O) *J

O)

oc

in

+

1

_

N

I .*

1

X ^

i

CO Analo

Analo

input

outpu

CO

| c?^ '5 2 "

o

cm en

o£

672

«>

i

l

1 DC

.2 "o

J

t

t

%

cc

3:2

X

CD i^ CO -J

^

X CO —. "- d,

n

00

's;

—

i r

i s

X

Ih N

—

00

00

(0

c

J3

2

CO

_ N

X

o.

x

CO u_

X

cc

x

LO

5"

CO-,

cc CO

x

I

iS

£

c

_t -a

qc

CO

CO

00

s

673

.

674

Chap. 17

Digital Multiplexing

Transmit and receive frame synchronizing pulses (FSX and FSR) are 8-kHz inputs set the transmit a nd rec eive

which

and nonsignaling frames. to gate th e

the line.

PC M word

TSX

is

TSX

onto the

is

sampling rates and distinguish between signaling

a time-slot strobe buffer enable output which

PCM

highway when an external buffer

is

used

is

used to drive

also used as an external gating pulse for a time-division multiplexer

(see Figure 17- 10).

Data are transmitted to the transitions of

are received after the

CLKX

PCM

DX

highway from

On

following the rising edge of FSX.

PCM

from the

highway from

DR

on the

first

eight positive

the receive channel, data

eight falling edges of

first

occurrence of FSR. Therefore, the occurrence of

on the

FSX

and

FSR must

CLKR

be synchro-

nized between codecs in a multiple-channel system to ensure that only one codec transmitting to or receiving from the

PCM

highway

is

any given time.

at

Figure 17-10 shows the block diagram and timing sequence for a single-channel

PCM

system using the 2914 combo chip

in the fixed-data-rate mode and operating MHz. In the fixed-data-rate mode, data are (This mode of operation is sometimes called the

with a master clock frequency of 1.536 inputted and outputted in short bursts.

burst mode.) With only a single channel, the total

PCM

highway

is

active only 24 of the

frame time. Additional channels can be added to the system provided that

transmissions are synchronized so that they do not occur

at the

same time

their

as transmissions

from any other channel.

From Figure 17-10 1

The

the following observations can be made:

input and output bit rates from the codec are equal to the master clock fre-

quency, 1.536 Mbps. 2.

The codec

3.

The data output (DX) and data time (125

inputs and outputs 64,000

PCM (DR)

bits

per second.

are active only

A

of the

total

frame

|xs).

To add ch anne ls FSX, FSR, and

input

TSX

to the

system shown

in

Figure

signals for each additional channel

17-10, the occurrence of the

must be synchronized so

that

they follow a timely sequence and do not allow more than one codec to transmit or receive at the

same

for a 24-channel

time. Figure 17-11

PCM-TDM

shows

the block diagram

and timing sequence

system operating with a master clock frequency of 1.536

MHz. Variable-Data-Rate

Mode

The variable-data-rate mode allows It

for a flexible data input

and output clock frequency.

provides the ability to vary the frequency of the transmit and receive

the variable data rate is still

digital

bit clocks.

mode, a master clock frequency of 1.536, 1.544, or 2.048

In

MHz

required for proper operation of the on-board bandpass filters and the analog-toand digital-to-analog converters. However, in the variable-data-rate mode, DCLKR

2913/14 and

Combo Chip

DCLKX

become

When FSX

tively.

675

high, data are transmitted onto the

DCLKX.

eight consecutive positive transitions of

from the

PCM

highway are clocked

DCLKR.

transitions of

PCM highways, respecPCM highway on the next

the data clocks for the receive and transmit

is

This

into the

mode of

Similarly, while

FSR

is

high, data

codec on the next eight consecutive negative

operation

is

sometimes called the

shift register

mode.

On

in the

l25-|xs frame as long as

high. This feature allows the

PCM

than once per frame. Signaling this

PCM word is repeated in all remaining DCLKX is pulsed and FSX is held active to be transmitted to the PCM highway more

the transmit channel, the last transmitted

time slots

mode

is

word

not allowed in the variable-data-rate

mode because

provides no means to specify a signaling frame.

Figure 17-12 shows the block diagram and timing sequence for a two-channel

PCM-TDM

system using the 2914 combo chip

master clock frequency of 1.536

MHz,

in the variable-data-rate

a sample rate of 8

mode

with a

kHz, and a transmit and

receive data rate of 128 kbps.

PCM

With a sample rate of 8 kHz, the frame time is 125 u,s. Therefore, one 8-bit word from each channel is transmitted and/or received during each 125-fxs frame.

For 16

bits to

occur 1

in

125

channel 8 bits

a

|xs,

1

128-kHz transmit and receive data clock

frame

125

bit rate

jxs

frame

2 channels

The transmit and receive enable

= 7.8125

signals

jxs

7.8125

16 bits

=-= tb

125

1

required.

is

|as

bit

28 kbps F

(jls

(FSX and FSR)

for one-half of the total frame time. Consequently, 8-kHz,

for each

50%

codec are active

duty cycle transmit

FXR) are fed directly to one codec and fed codec 180° out of phase (inverted), thereby enabling only one codec at a

and receive data enable signals (FXS and to the other

time.

To expand

to a four-channel system,

data clock rates to 256

kHz and change

simply increase the transmit and receive

the enable signals to an 8-kHz,

25%

duty

cycle pulse.

Supervisory Signaling With the 2914 combo chip, supervisory signaling can be used only rate

mode.

A

transmit signaling frame

is

identified

by making the

in the fixed-data-

FSX

and

FSR

pulses

twice their normal width. During a transmit signaling frame, the signal present on input

SIGX

PCM

word. At the receive end, the signaling

to

is

substituted into the least significant bit position (b,) of the

decoding and placed on output

frame.

SIGR

until

bit is extracted

from the

PCM

encoded word prior

updated by reception of another signaling

£

J ^

3

CO

~

0_

O

«-

o_ .E «-

CD *- to

'

1

t

-/I

I CM

t—

|X

|X

CO

to

li-

|h
+

«-

a

CTJ

-C Q. O)

•- 0.

Q. O)

o

t- o. .E

u

«

33 .—

0»

£ w

JS

•-

>

S

'—

E r.

-

—a Sf •ss S3

!_£

O

-*

E

1

w

^

* 3 CN ^«

cr

4>o|t

UU

x x * 03

"a)

c c

^^

> IT)

+

©v

r;

**

—

5 ^ M ^ S •S s —

IT)

E u ts «

u

*S

2 o" — r H £ a.

N

S = n|c

(D

678

'o ^»

a

10

U5

10

X

X

X)

^

+-

5

Q.

11

,

,

LL CQ 0_ CO CQ -J

CO 0- CO CO -1 LL.

—

.*

c

£a

'

i

£

i

i

N

X M

*

CO 00

N

—X -*

CN

3 s

00 CD

N

X -* x i 5

-S

-2

e i

8 03

"8

O

P

a

3 o s.

c C

CN

N

-8

*>

i

X

—

oo

,- -C

09

°

II

"8

c

u.

8

U

03

11Q 03

|

00 LL

-

-

-x

— XN

Jt

.*

CO

CO

CO

CO o"

ISI

d

O)

C

—

CO

i_ 0)

'£

's; oj

i

c c

(0

O

702

u

oCO j_

a>

c

0>

I

•5

±;

-

(< CO «- LO co rr lo

^

O

5 °

CO

TTTTTTTTT co^friocoi^ooiocor^ rrpPffNNN OOOOOOQQQ C0C0C0C0C0C/>C3OO

o

PI

(O

o O)

N

00 CO LO

00

^CO

ii

"8

2 (0

O u

o

o

O O)

O

O

co

P

+1

si

"D

LL

a. _i

JO

3 o E

i

i

1

°

fc

C

.

J5 7T

CO

X

O

a

Q

.9-

3 E T

"co

1

0)

*

(j

— ^ "(0^3 O

c

3r

a

< ,

UL .E

i

^_

2*2 ~

« S > »- CO C Q> Q. O o

co

1

I

k-

.O

E CO

o

~

CO

'c

3

a>

M C

CO

c

»-

IIu

JO cc

00

CO

,_

o

*-

—c

JO

«o

o

Q

-Q

719

Frequency-Division Multiplexing

720

Vestigal

SB video

FIGURE

Data

in

Data above

video (DAVID).

F (Hz)

Data

18-14

Chap. 18

Voice

in voice,

developed by Fujitsu of Japan, uses an eight-level

technique with steep

filtering.

technique which gives

it

data are transmitted in a

It

PAM-VSB

modulation

uses a highly compressed partial response encoding

a high bandwidth efficiency of nearly 5 bps/Hz

(1.544-Mbps

344-kHz bandwidth).

QUESTIONS 18-1. Describe frequency-division multiplexing. 18-2. Describe a

message channel.

18-3. Describe the formation of a group, a supergroup, and a mastergroup.

18-4. Define

baseband and composite baseband.

FDM

18-5. Describe the modulators used in

18-6. Describe the difference between an 18-7.

What

18-8. Are 18-9.

18-10.

is

a guard band?

When

is

multiplexers.

L600 and

a

U600 mastergroup.

a guard band used?

FDM-multiplexed communications systems synchronous? Explain.

Why What

are D-type supergroups used? are the

two types of

pilots

used with

FDM

systems, and what

is

the purpose of

each? 18-11.

What

are the four primary types of hybrid data networks?

18-12.

What

is

the difference

18-13. At what level

is

the

between a

104.08-kHz

DUV

and a

DAV

network?

pilot inserted?

PROBLEMS 18-1. Calculate the 12 channel carrier frequencies for the

18-2. Calculate the five group carrier frequencies for the

U600

U600

FDM system. FDM system.

18-3. Calculate the frequency range for a single channel at the output of the channel, group.

supergroup, and mastergroup combining networks for channel 3. group 4, supergroup 15.

mastergroup

2.

18-4. Determine the frequency that a 1-kHz test tone will translate to at the output of the channel.

group, supergroup, and mastergroup combining networks for channel 5, group 5. supergroup 27, mastergroup 3.

Chap. 18

Problems

721

18-5. Determine the frequency at the output of the mastergroup combining network for a group pilot

of 104.08

kHz on group

2,

supergroup 13, mastergroup

2.

18-6. Calculate the frequency range for group 4, supergroup 18, mastergroup the mastergroup

1

at the

output of

combining network.

18-7. Calculate the frequency range for supergroup 15, mastergroup 2 at the output of the master-

group combining network.

Chapter 19

MICROWAVE COMMUNICATIONS

AND SYSTEM GAIN INTRODUCTION microwave radio relay systems provide less than half of in the United States. However, at one time microwave systems carried the bulk of long-distance communications for the public telephone network, military and governmental agencies, and specialized private communications net-

Presently, terrestrial (earth) the total

message

circuit

mileage

works. There are

many

varing from 15 to

4000 miles

different types of

microwave systems

in length. Intrastate

that operate

over distances

or feeder service systems are generally

categorized as short haul because they are used for relatively short distances. Long-

haul radio systems are those used for relatively long distances, such as interstate and

backbone route applications. Microwave system capacities range from

less than 12 voice

more than 22,000. Early microwave systems carried frequency-divisionmultiplexed voice band circuits and used conventional, noncoherent frequency modulation techniques. More recently developed microwave systems carry pulse-code-modulated time-division-multiplexed voice band circuits and use more modern digital modulation band channels

to

techniques, such as phase shift keying and quadrature amplitude modulation. This chapter deals primarily with conventional

with the more modern

SIMPLIFIED

A

simplified block diagram of a is

more of

722

FDM/FM

microwave systems, and Chapter 20 deals

techniques.

MICROWAVE SYSTEM

baseband or

PCM/PSK

microwave radio system

the composite signal that modulates the

the following:

FM

is

shown

carrier

in

Figure 19-1. The

and may comprise one

.

Simplified

Microwave System

723

o

1 o c

O)

c

Baseband in

RF out

'c

Preemphasis network

BPF

!o

E o

u "55

c c CD

.C

O

(a)

J*

O

1 u

K

c c

g

Baseband out

BPF

RF

in

CO co

a 0) "53

c c CO

JC

O

(b)

FIGURE

19-1

Simplified block diagram of a microwave system: (a) transmitter;

(b) receiver.

Frequency-division-multiplexed voice band channels

1

2.

Time-division-multiplexed voice band channels

3.

Broadcast-quality composite video or picturephone

Microwave Transmitter In the

microwave transmitter (Figure 19- la), a preemphasis network precedes the FM The preemphasis network provides an artificial boost in amplitude to the higher

deviator.

baseband frequencies. This allows the lower baseband frequencies the IF carrier

assures a

An FM the

and the higher baseband frequencies

more uniform

to

to

frequency modulate

phase modulate

it.

This scheme

signal-to-noise ratio throughout the entire baseband spectrum.

deviator provides the modulation of the IF carrier which eventually becomes

main microwave

carrier. Typically,

IF carrier frequencies are between 60 and 80

MHz,

FM a

Chap. 19

Microwave Communications and System Gain

724 with 70

MHz

the

most common. Low-index frequency modulation

deviator. Typically, modulation indices are kept

narrowband

FM

between 0.5 and

1.

used

is

in the

This produces

signal at the output of the deviator. Consequently, the IF bandwidth

resembles conventional

AM

and

is

approximately equal to twice the highest baseband

frequency.

The IF and the

its

associated sidebands are up-converted to the microwave region by

AM mixer, microwave oscillator, and bandpass RF

filter.

Mixing, rather than multiplying,

frequencies because the modulation index

is

used to translate the IF frequencies to

is

unchanged by the heterodyning process. Multiplying the IF

carrier

would

also multiply

the frequency deviation and the modulation index, thus increasing the bandwidth. Typi-

MHz (1 GHz) are considered microwave frequencies. microwave systems operating with carrier frequencies up to approximately 18 GHz. The most common microwave frequencies currently being used are the 2-, 4-, 6-, 12-, and 14-GHz bands. The channel combining network provides a means of connecting more than one microwave transmitter to a single transmission line

cally,

frequencies above 1000

Presently, there are

feeding the antenna.

Microwave Receiver In the receiver (Figure

and

filtering

19- lb), the channel separation

their respective receivers.

down-convert the

FM

The bandpass

RF microwave

demodulator. The

(i.e., a

network provides the isolation

necessary to separate individual microwave channels and direct them to

FM

filter,

AM

mixer, and microwave oscillator

frequencies to IF frequencies and pass them on to the

demodulator

is

network restores the baseband signal

to

its

FM

a conventional, noncoherent

discriminator or a ratio detector). At the output of the

FM

detector

detector, a deemphasis

original amplitude versus frequency character-

istics.

MICROWAVE REPEATERS The permissible distance between

a

microwave transmitter and

its

associated microwave

receiver depends on several system variables, such as transmitter output power, receiver

noise threshold, terrain, atmospheric conditions, system capacity, reliability objectives,

and performance expectations. Typically,

this distance

is

between 15 and 40 miles.

Longhaul microwave systems span distances considerably longer than a single-hop for

most

microwave system, such

practical

when geographical

A

as the

one shown

system applications. With systems

in

this.

Figure 19-1,

Consequently, is

that are longer than

inadequate

40 miles or

obstructions, such as a mountain, block the transmission path, repeat-

microwave repeater is a receiver and a transmitter placed back to tandem with the system. A block diagram of a microwave repeater is shown in Figure 19-2. The repeater station receives a signal, amplifies and reshapes it. then retransmits the signal to the next repeater or terminal station downline from it. Basically, there are two types of microwave repeaters: baseband and IF (Figure

ers are needed.

back or

in

Diversity

725

K^h

K^h

RF

Tx IF in

RF

Rx

Rx

Microwave

Microwave

Microwave

Microwave

transmitter

receiver

transmitter

receiver

FIGURE

19-2

IF out

Microwave repeater.

19-3). IF repeaters are also called heterodyne repeaters.

With an IF repeater (Figure

RF carrier is down-converted to an IF frequency, amplified, reshaped, RF frequency, and then retransmitted. The signal is never demodulated

19-3a), the received

up-converted to an

beyond

Consequently, the baseband intelligence

IF.

a baseband repeater (Figure 19-3b), the received

is

RF

unmodified by the repeater. With

carrier

is

down-converted

to an IF

frequency, amplified, filtered, and then further demodulated to baseband. The baseband signal,

which

demodulated

is

typically frequency-division-multiplexed voice

the baseband signal to be reconfigured to tions network. carrier

which

Once

is

band channels,

even channel

to a mastergroup, supergroup, group, or

RF

further

is

This allows

meet the routing needs of the overall communica-

the baseband signal has been reconfigured,

up-converted to an

level.

carrier

it

FM

modulates an IF

and then retransmitted.

Figure 19-3c shows another baseband repeater configuration. The repeater demodulates the

With

this

RF

to baseband, amplifies

technique, the baseband

is

and reshapes

it,

then modulates the

FM

accomplishes the same thing that an IF repeater accomplishes. The difference a baseband configuration, the amplifier and equalizer act

than IF frequencies.

carrier.

not reconfigured. Essentially, this configuration is

that in

on baseband frequencies rather

The baseband frequencies are generally less than 9 MHz, whereas 60 to 80 MHz. Consequently, the filters and amplifiers

the IF frequencies are in the range

necessary for baseband repeaters are simpler to design and less expensive than the

ones required for IF repeaters. The disadvantage of a baseband configuration addition of the

FM

is

the

terminal equipment.

DIVERSITY Microwave systems use signal path

line-of-sight transmission.

There must be a

direct, line-of-sight

between the transmit and the receive antennas. Consequently,

if that

signal

path undergoes a severe degradation, a service interruption will occur. Diversity suggests that there

is

more than one transmission path or method of transmission available between microwave system, the purpose of using diversity is

a transmitter and a receiver. In a to increase the reliability of the

system by increasing

its

availability.

When

there

is

more than one transmission path or method of transmission available, the system can select the path or method that produces the highest-quality received signal. Generally,

Microwave Communications and System Gain

726

RF

Chap. 19

IF

in

Rx

Tx

amp

IF

out

Microwave

and

Microwave

receiver

equalizer

transmitter

(a)

RF

K^

in

RF out

Tx

Rx Microwave

Microwave

FM

FM

receiver

transmitter

Baseband

Baseband

Multiplexing equipment

(b)

RF

in

^h

K^

RF

in

Tx

Rx Microwave

Microwave

receiver

transmitter

FM

FM

receiver

transmitter 1

Baseband

Baseband amp and

Baseband

equalizer

(c)

FIGURE

the highest quality

is

19-3

Microwave repeaters:

(a) IF; (b)

and

(c)

baseband.

determined by evaluating the carrier-to-noise (C/AO

ratio at the

receiver input or by simply measuring the received carrier power. Although there are

many ways of achieving and polarization.

diversity, the

most

common methods

used are frequent

v.

space,

Diversity

727

Frequency Diversity Frequency diversity

same IF

RF

one

At the

that yields the better-quality IF

shows a single-channel frequency-diversity microwave

selected. Figure 19-4

is

carrier frequencies with the

signals to a given destination.

destination, both carriers are demodulated, and the signal

RF

simply modulating two different

is

intelligence, then transmitting both

system. In Figure 19-4a, the IF input signal

microwave in the

transmitters

A

and B. The

is

power

fed to a

RF outputs

splitter,

which

directs

to

it

from the two transmitters are combined

channel-combining network and fed to the transmit antenna. At the receive end

A and B RF carriers to their respective where they are down-converted to IF. The quality detector circuit

(Figure 19-4b), the channel separator directs the

microwave

receivers,

A

determines which channel, the IF switch to be further

or B,

is

the higher quality and directs that channel through

demodulated

atmospheric conditions that degrade an

Many

to baseband.

RF

of the temporary, adverse

signal are frequency selective; they

may

degrade one frequency more than another. Therefore, over a given period of time, the IF switch

many

may

switch back and forth from receiver

A

to receiver B,

and vice versa

times.

Microwave transmitter frequency

—

BPF

A

A

K

at

c

A

RF out

j5

E o o

Power IF

in'

splitter

"55

c c

B

Microwave Radio Stations

735

The RF receiver (Figure 19-8b) that

works

it

IF amplifier circuit.

essentially the same as the transmitter except However, one difference is the presence of an

is

in the opposite direction.

This IF amplifier has an automatic gain control

in the receiver.

RF

Also, very often, there are no

(AGC)

amplifiers in the receiver. Typically, a very

sensitive, low-noise-balanced demodulator is used for the receive demodulator (receive mod). This eliminates the need for an RF amplifier and improves the overall signal-to-

noise ratio.

When RF

amplifiers are required, high-quality, low-noise amplifiers

Examples of commonly used

are used.

LNAs

(LNAs)

are tunnel diodes and parametric amplifiers.

Repeater Station Figure 19-9 shows the block diagram of a microwave IF repeater station. The received

RF

signal enters the receiver through the channel separation network and bandpass

filter.

The receive mod down-converts

carrier to IF. The IF AMP/AGC and The equalizer compensates for gain versus

RF

the

equalizer circuits amplify and reshape the IF.

frequency nonlinearities and envelope delay distortion introduced

in the

system. Again,

RF for retransmission. However, in a repeater station, RF microwave carrier frequencies is slightly different

the transmod up-converts the IF to the

method used

to generate the

from the method used generator

is

in a

terminal station. In the IF repeater, only one microwave

required to supply both the transmod and the receive

carrier signal.

The microwave

mod

with an

RF

generator, shift oscillator, and shift modulator allow the

RF carrier frequency, down-convert it to IF, and then up-convert RF carrier frequency (Figure 19- 10a). It is possible for station C

repeater to receive one the IF to a different

to receive the transmissions

from both

station

A

and station B simultaneously

when

called multihop interference). This can occur only

geographical straight line in the system.

bandwidth for the system

is

To

(this is

three stations are placed in a

prevent this from occurring, the allocated

divided in half, creating a low-frequency and a high-frequency

band. Each station, in turn, alternates from a low-band to a high-band transmit carrier

frequency (Figure 19- 10b). will

be rejected

called a high/low

in the

microwave repeater system. The

RF

receives a low-band

The only time

transmission from station

If a

A

is

received by station C,

channel separation network and cause no interference. This

carrier,

that multiple carriers of the

transmission from one station

is

rules are simple: If a repeater station

retransmits a high-band

it

it

is

RF

carrier,

and vice versa.

same frequency can be received

received from another station that

is

is

when

a

three hops away.

unlikely to happen.

This

is

that

Another reason for using a high/low-frequency scheme is to prevent the power "leaks" out the back and sides of a transmit antenna from interferring with the

signal entering the input of a

antennas, no matter

how

a small percentage of their ratio for the antenna. is

neaby receive antenna. This

high their gain or

power out

may be

quite substantial

is

called ringaround. All

directive their radiation pattern, radiate

back and

sides; giving a finite front-to-back

microwave antenna amount of power that is radiated out the back of the compared to a normal received carrier power in the

Although the front-to-back

quite high, the relatively small

antenna

the

how

ratio of a typical

>|JOMjau Bumiqujoo

puueqo

n iO

aj

2

Sit

ea

o

a>

.

n

< 7x"g

£ojO c CO

E

o

;=

9L&

.2 -D 0)

O

ox O

o

Q)

CO «-

11 >jjo/wj8u uoijejedas

IrJ 736

|auueqo

Microwave Radio Stations

737

iHSKlH^Kl) F1

(a)

ilhSh^h^k ^h^h ** F1

(b)

FIGURE

19-10

(a)

Multihop interference and

(b) high/low

microwave system.

system. If the transmit and receive carrier frequencies are different,

filters in

the receiver

separation network will prevent ringaround from occurring.

A

high/low microwave repeater station (Figure 19- 10b) needs two microwave

carrier supplies for the

down- and up-converting process. Rather than use two microwave

generators, a single generator together with a shift oscillator, a shift modulator, and a

bandpass

filter

can generate the two required signals. One output from the microwave

same microwave mixed with the shift oscillator signal in the shift modulator to produce a second microwave carrier frequency. The second microwave carrier frequency is offset from the first by the shift oscillator frequency. The second microwave carrier frequency generator

generator)

is

is

fed directly into the transmod and another output (from the

is

fed into the receive modulator.

EXAMPLE

19-1

RF carrier frequency is 6180 MHz, and the transmitted RF 6000 MHz. With a 70-MHz IF frequency, a 5930-MHz microwave

In Figure 19-9 the received carrier frequency

is

generator frequency, and a shift

mod must

180-MHz

be tuned to 6110

shift oscillator

MHz.

the shift oscillator frequencies (5930

This

MHz +

is

the

180

frequency, the output

sum of

MHz =

the

61 10

filter

MHz).

This process does not reduce the number of oscillators required, but

and cheaper

it

is

simpler

microwave generator and one relatively low-frequency shift build two microwave generators. The obvious disadvantage of the

to build one

oscillator than to

of the

microwave generator and

Microwave Communications and System Gain

738 high/low scheme

is

number of channels

that the

Chap. 19

available in a given bandwidth

is

cut

in half.

Figure 19-11 shows a high/low-frequency plan with eight channels (four high-

band and four low-band). Each channel occupies a 29.7-MHz bandwidth. The west terminal transmits the low-band frequencies and receives the high-band frequencies. Channel

1

and 3 (Figure 19-1

la) are designated as

horizontally polarized channels. This

is

V

channels. This means that they

Channels 2 and 4 are designated as

are propagated with vertical polarization.

H

not a polarization diversity system. Channels

or 1

through 4 are totally independent of each other; they carry different baseband information.

The transmission of orthogonally polarized

carriers (90° out of phase) further

enhances

the isolation between the transmit and receive signals. In the west-to-east direction, the

repeater receives the low-band and transmits the high-band frequencies. After channel 1

received and down-converted to IF,

is

it is

up-converted to a different

and polarization for retransmission. The low-band channel

band channel 11, channel 2

to

RF

frequency

corresponds to the high-

1

channel 12, and so on. The east-to-west direction (Figure

19-1 lb) propagates the high- and low-band carriers in the sequence opposite to the

west-to-east system.

channel it

1

The

polarizations are also reversed. If

some of

the

power from

of the west terminal were to propagate directly to the east terminal receiver,

has a different frequency and polarization than channel 11' s transmissions. Conse-

quently,

it

would not

interfere with the reception of channel

Also, note that none of the transmit or receive channels the

same frequency and due

(no multihop interference). repeater station has both

polarization. Consequently, the interference

to ringaround

is

insignificant.

simplest form, system gain

is

the difference

to the receivers

1 1

at the

from the transmitters

SYSTEM GAIN In

its

of a transmitter and the greater than or equal to

minimum the sum of

power

input all

between the nominal output power

to a receiver.

the gains and losses incurred

propagates from a transmitter to a receiver. In essence, radio system.

System gain

is

used to predict the

parameters. Mathematically, system gain

^.v

System gain must be

it

reliability

by a signal as

of a system for given system

is

'

/

^ min

where

Gs =

system gain (dB)

= transmitter output power (dBm) C m in = minimum receiver input power for P,

a given quality objective

and where Pi

~

Cmin

—

losses

+

gains

it

represents the net loss of a

(dBm)

System Gain

739

Gains:

A,

=

Ar =

transmit antenna gain (dB) relative to an isotropic radiator

receive antenna gain (dB) relative to an isotropic radiator

Losses:

Lp =

free-space path loss between antennas (dB)

Lf = waveguide

feeder loss (dB) between the distribution network (channel combining network or channel separation network) and its respective antenna (see Table 19-1)

Lb =

coupling or branching loss (dB) in the circulators,

filters, and network between the output of a transmitter or the input to a receiver and its respective waveguide feed (see Table 19-1)

total

distribution

FM =

fade margin for a given reliability objective

Mathematically, system gain

is

Gs = P ~ C min > FM + t

where

all

values are expressed in

dB

or

Lp + Lf + L b - A - A r

(19-1)

t

dBm. Because system gain is dB values and the gains

indicative of a

net loss, the losses are represented with positive

with negative

dB

values. Figure 19-12

shows an

overall

are represented

microwave system diagram

and indicates where the respective losses and gains are incurred.

TABLE

19-1

SYSTEM GAIN PARAMETERS Branching OSS 1

Antei ina gain,

(dB)

Frequency

Feeder

loss,

Lf

A,

(GHz) Loss

Type 1.8

7.4

8.0

r

Diversity

(dB/lOOm)

Frequency

Space

5.4

5

2

S'

Gain

(m)

(dB)

1.2

25.2

coaxial

2.4

31.2

cable

3.0

33.2

3.7

34.7

Air-filled

EWP64

1.5

38.8

eliptical

2.4

43.1

waveguide

3.0

44.8

3.7

46.5

2.4

43.8

EWP69

4.7

6.5

3

3

2

2

eliptical

3.0

45.6

waveguide

3.7

47.3

4.8

49.8

'

s

-^

r—

O -* " C 0) C

II S ? > CM 5 to

0)

CC

CM

s t

So,

I"*

to -C CM to

CM

CO

CO

13'

1

r

1

{

5

O E c

O0) j- to to F £ CM to

£

oo

:

T3

d

u. a. CO

£

11

T3

•£ T5

5E

sl

I

t 0)

.2 -d

o

o

CO

8 E

is

£ o S E

5£

™ >

o oq

11 Low-ban

SI

29.

X

6029.95

BW=

(4

x

ceo

a!

68

.£ -g

III

en «• CO

o

5S Sco

'

TJ

E

8* — S

o =-

CM

5®

8".

,-

o

5co

8 E

1° is

I. 741

742

Microwave Communications and System Gain ^

A.

FM

L"

Chap. 19

XA.

M Microwave power amp

Microwave receiver

From other

To

microwave

microwave

transmitters

receivers

FIGURE

19-12

System gains and

other

losses.

Free-Space Path Loss Free-space path loss

is

defined as the loss incurred by an electromagnetic

wave

as

it

vacuum with no absorption or reflection of energy

propagates in a straight line through a

from nearby objects. The expression for free-space path

MTH 2

/4ttFD\

loss

is

given as

2

C

where

Lp —

free-space path loss

D=

distance

F= X = C=

frequency

Converting to

wavelength velocity of light in free space (3

dB

yields

Lp (dB) = 20

When

x 10 8 m/s)

the frequency

is

Lp (dB) = 20

=

4n/rD

= 20

log

given in

MHz

and the distance

47T(1Q) (1 Q) log

32.4

3

+ 20

x log

10

— + 20 4tt

log

x

+ 20

F (MHz) +

log

log

in

F+

20 log

D

km,

F (MHz) +

20 log

D

(km) (19-2)

20 log

D

(km)

System Gain

When

743

the frequency

is

given

in

GHz

and the distance

Lp (dB) = 92.4 + 20 made

Similar conversions can be

F (GHz) +

log

in

km,

D

20 log

(km)

(19-3)

using distance in miles, frequency in kHz, and so

on.

EXAMPLE

19-2

For a carrier frequency of 6

GHz

and a distance of 50 km, determine the free -space path

loss.

Solution

M

dB > = 32.4 + 20 32.4 142

+

6000 + 20 log 50

log

75.6

+

34

dB

or

VdB) =

92.4

+ 20 log 6 + 20

=

92.4

+

=

142

15.6

log

50

+ 34

dB

Fade Margin Essentially,

fade margin

is

a

"fudge factor" included

system gain equation that

in the

considers the nonideal and less predictable characteristics of radio- wave propagation,

such as multipath propagation {multipath loss) and terrain

sensitivity.

tics

cause temporary, abnormal atmospheric conditions that

loss

and are usually detrimental

to the overall

These characteris-

alter the free-space path

system performance. Fade margin also

considers system reliability objectives. Thus fade margin

is

included in the system gain

equation as a loss.

Solving the Barnett-Vignant reliability equations for a specified annual system availability for an unprotected, nondiversity

FM =

30 log

D +

system yields the following expression:

10 log (6ABF)

-

10 log

(1

-

R)

multipath

terrain

reliability

effect

sensitivity

objectives

- 70 constant

(19-4)

where

FM = D=

\

F= R = —R = A =

fade margin (dB) distance (km)

frequency (GHz) reliability

99.99% = 0.9999 one-way 400-km route

expressed as a decimal

reliability objective for a

roughness factor

(i.e.,

reliability)

Chap. 19

Microwave Communications and System Gain

744

= = = B = = = = =

EXAMPLE

4 over water or a very smooth

1 over an average terrain 0.25 over a very rough, mountainous terrain

factor to convert a worst-month probability to an annual probability to convert

1

0.5 for hot

0.

humid

areas

125 for very dry or mountainous areas

19-3

GHz. Each

air-filled

an annual availability to a worst-month basis

0.25 for average inland areas

Consider a space-diversity of 1.8

terrain

microwave radio system operating

at

an

RF

station has a 2.4-m-diameter parabolic antenna that

carrier frequency

is

fed by 100

m

of

The terrain is smooth and the area has a humid climate. The distance 40 km. A reliability objective of 99.99% is desired. Determine the

coaxial cable.

between stations

is

system gain. Solution

Substituting into Equation 19-4,

FM -

30 log 40

+

we

find that the fade

10 log (6) (4) (0.5)

= 48.06+

13.34

-(-40) -70

=

13.34

+ 40-70

48.06

+

(1 .8)

-

margin

10 log

(1

-

is

0.9999)

-

70

= 31.4dB we

Substituting into Equation 19-3,

From Table

obtain path loss

Lp =

92.4

+ 20 log

=

92.4

+

=

129.55

5.11

1.8

+

+ 20

log

40

32.04

dB

19-1,

U = ,

4 dB (2

+

2

=

4)

Lf = 10.8dB(100m+ 100m = 200 m)

A = Ar = t

31.2

dB

Substituting into Equation 19-1 gives us system gain

G = 31.4+

129.55+

10.8

+ 4-31.2-31.2=

1

13.35

dB

s

The

results indicate that for this

terrain, distribution

must be

at least 11

sytem

to

networks, transmission

3.35

dB more

than the

perform

lines,

at

99.997c

reliability

with the given

and antennas, the transmitter output power

minimum

receive signal level.

Receiver Threshold probably the most important parameter considered when evaluating the performance of a microwave communications system. The minimum wideband

Carrier-to-noise (C/N)

is

System Gain carrier

output

745

power (C min ) is

threshold

at the

input to a receiver that will produce a usable baseband

The

called the receiver threshold or, sometimes, receiver sensitivity. is

dependent on the wideband noise power present

at the input

receiver

of a receiver,

the noise introduced within the receiver, and the noise sensitivity of the baseband detector.

Before noise

C min

power

can be calculated, the input noise power must be determined. The input is

expressed mathematically as

N = KTB where

N= K=

noise

T—

equivalent noise temperature of the receiver (K) (room temperature

power

Boltzmann's constant (1.38 x 10~ 23 J/K)

= 290

K)

B = Expressed

in

noise bandwidth (Hz)

dBm,

N (dBm) = For a 1-Hz bandwidth

at

— KTB

10 log

=

^ KT

10 log

+

10 log

B

room temperature,

N=

(1.38

101og

x 10" 23 )(290)

_,

t

+1Qlogl

oooi

= -174 dBm Thus

JV(dBm) =

EXAMPLE

- 174 dBm +

(19-5)

MHz,

determine the noise power.

Substituting into Equation 19-5 yields

N=

-174 dBm + 10

= -174 dBm + minimum CIN requirement for a minimum receive carrier power is

If the

the

B

19-4

For an equivalent noise bandwidth of 10 Solution

10 log

log (10

70 dB

receiver with a

=

x 10 6 )

--104

10-MHz

dBm

noise bandwidth

is

24 dB,

Cmin = ^(dB) + AT(dB) = For a system gain of 113.35 dB, of

it

24 dB + (-104 dBm)

would require

a

= -80 dBm

minimum

transmit carrier

power (P

t)

Microwave Communications and System Gain

746

P,

= Gs + C min =

1

minimum

This indicates that a

dB + (-80 dBm) = 33.35 dBm

13.35

transmit

achieve a carrier-to-noise ratio of 24 of 10

Chap. 19

dB

dBm

power of 33.35

(2.16

with a system gain of 113.35

W)

required to

is

dB and

a bandwidth

MHz.

Carrier-to-Noise versus Signal-to-Noise Carrier-to-noise {CIN)

is

and

its

CIN

in the receiver. Essentially,

a predetection (before the

is

to-noise ratio. Signal-to-noise {SIN)

At a baseband point

FM

it

signal.

is

is

an

at

RF

power

or an IF point

FM demodulator) signalFM demodulator) ratio.

a postdetection (after the

voice band channel can be separated from

in the receiver, a single

baseband and measured independently. At an

the rest of the

(actually, not just the

associated sidebands) to the wideband noise

bandwidth of the receiver). CIN can be determined

(the noise

receiver,

wideband "carrier"

the ratio of the

carrier, but rather the carrier

RF

or IF point in the

impossible to separate a single voice band channel from the composite

For example, a typical bandwidth for a single microwave channel

MHz. The bandwidth of the composite the ratio of the

RF

signal to the

power of a

single

4 kHz. CIN

is

30

power total noise power in the 30-MHz bandwidth. SIN is voice band channel to the noise power in a 4-kHz

of a voice band channel

is

is

the ratio of the

bandwidth.

Noise Figure In

its

simplest form, noise figure (F)

device divided by the SIN ratio practical sense, noise figure

is

device divided by the SIN ratio

at the

is

the signal-to-noise ratio of an ideal noiseless

output of an amplifier or a receiver. In a more

defined as the ratio of the SIN ratio at the input to a at the output.

F = /CMA

Mathematically, noise figure

F (dB) = 10

aild

l0 S

is

1

the

or

figure

is

a ratio of ratios.

The

dB. Remember, the noise present

same gain

noise figure of a totally noiseless device

at the input to

as the signal. Consequently, only noise

decrease the signal-to-noise ratio

/C/AA

(S/AOout

(S/AOout

Thus noise

at the

an amplifier

is

added within the amplifier can

SIN

A

noise figure of 10

by a factor of

ratio

in

ratio at the output.)

Essentially, noise figure indicates the relative increase of the noise

noise to reduce the

amplified by

output and increase the noise figure. (Keep

mind, the higher the noise figure, the worse the SIN increase in signal power.

is

means

that the device

10, or the noise

power

added

power increased

to the

sufficient

tenfold in

respect to the increase in signal power.

When two

or

the total noise figure

more (NF)

amplifiers or devices are cascaded together (Figure is

cally, the total noise figure

an accumulation is

of the individual noise figures.

19-13).

Mathemati-

System Gain

747

FIGURE

NF =

F,

19-13

1

- +

+

Total noise figure.

F — 3

1

+

AA-

F — 4

AAA l

(19-6)

etc.

/1 2 /1 3

where

NF = Fj = F2 = F3 = A = A2 = ]

total noise figure

noise figure of amplifier

1

noise figure of amplifier 2

noise figure of amplifier 3

gain of amplifier

1

gain of amplifier 2

Note: Noise figures and gains are expressed as absolute values rather than dB. It

can be seen that the noise figure of the

each of the

added by each succeeding amplifier

stage

amplified by

is

the noise introduced is

effectively reduced

factor equal to the product of the gains of the preceding amplifiers.

When

precise noise calculations (0.1

more convenient is

dB

or less) are necessary,

it

is

generally

to express noise figure in terms of noise temperature or equivalent

noise temperature rather than as an absolute

(N)

amplifier (Fl) contributes the most

figure.

in the first stage, the noise

by a

first

The noise introduced in the first succeeding amplifiers. Therefore, when compared to

toward the overall noise

power (Chapter

20).

Because noise power

proportional to temperature, the noise present at the input to a device can be

expressed as a function of the device's environmental temperature (T) and noise temperature (Te ).

To

its

equivalent

convert noise figure to a term dependent on temperature

only, refer to Figure 19-14.

Let

Nd -

noise

power added by

a single amplifier

Then

N, = KTP B

FIGURE

19-14

Noise figure as a

function of temperature.

—

.

Microwave Communications and System Gain

748 where

Te

is

Chap. 19

the equivalent noise temperature. Let

Na = = A =

Nj

total

output noise power of an amplifier

total input

noise

power of an amplifier

gain of an amplifier

Therefore,

N

may be

N„ =

expressed as

AN + ANd t

and

Na = AKTB + AKT B e

Simplifying yields

N = AKB (T +

Te )

and the overall noise figure (NF) equals

N¥ =

{SIN) m

SIN;

N

(S/N) om

ASIN„

AN;

T+T„

)

(19-7)

+^

1

EXAMPLE

AKB (T + Te AKTB

19-5

In Figure 19-13, let F,

= F 2 = F3 =

3

dB and A = A 2 = A 3 = x

10 dB. Solve for the total

noise figure.

Solution

Substituting into Equation 19-6 (note: All gains and noise figures have

converted to absolute values) yields

NF = =

Fi

.

2

—

+

+

:

2-1 — —

+ -t—

:

2-1 +-

10

2. 11

An dB

overall noise figure of 3.24 less than the

The noise noise figure

S/N

in the

a gain in the total noise

power

or 10 log 2. 11

indicates that the

figure of a receiver

included

is

dB

ratio at the input to

A

100

S/N

=

3.24

dB

ratio at the output

A3

must be considered when determining

system gain equation as an equivalent is

of

is

3.24

1

equivalent to a corresponding loss

C min

.

The

loss. (Essentially,

in the signal

power.)

>

System Gain

749 G

= s

/

H

/

\

Power

112dB \ *

)—\

/

Microwave

p

amp

FM

C/N IF

receiver

r (

-min/N

NF

Baseband out S/N = 32 dB

receiver

= 6.5dB

N = -104dBm

FIGURE

EXAMPLE

19-15

System gain example.

19-6

Refer to Figure 19-15. For a system gain of 112 dB, a

total

noise figure of 6.5 dB, an

power of —104 dBm, and a minimum (S/N) out of the FM demodulator of 32 dB, determine the minimum receive carrier power and the minimum transmit power. input noise

Solution 15

dB

is

To

achieve a S/N ratio of 32

required (17

dB

dB

FM

out of the

of improvement due to

FM

demodulator, an input

C/N of

quieting). Solving for the receiver

input carrier-to-noise ratio gives

'mm

N

NF=

+

N

15

dB +

6.5

dB =

21.5

dB

Thus

mm

=

:Lmin

+

yy

ft

=

21 .5

dB + (- 104 dBm) = -82.5 dBm

P = G s + Cmin t

=

EXAMPLE

12

dB + (-82.5 dBm) =

dBm

29.5

19-7

For the system shown

and

1

in

Figure 19-16, determine the following:

G C min//V, Cmjn

1.2

m

1.2

Z. 50

50

JV,

G

s,

h-l

Space diversity

>,

km

m

•

25

m

C/N

IF

receiver

F =

Reliability objective

,

m

Microwave

Bandwidth = 6.3

s,

Pr

8GHz

Mountaineous and dry terrain

c min/N

NF

=

4.24dB

= 99.999%

MHz

FIGURE

19-16

System gain example.

FM receiver

Baseband S/N - 40 dB

Chap. 19

Microwave Communications and System Gain

750

lution

The minimum CIN

C

bjsm

receiver

is

23 d

C = h + NF

N

N

= Jubstituting into

FM

input to the

at the

23 dB

+

4.24

dB = 27.24 dB

Equation 19-5 yields

/V= - 174 dBm +

10 log

B

= - 174 dBm +

68 dB

= - 106 dBm

C =

27.24

dB + (-106 dBm) = -78.76 dBm

Substituting into Equation 19-4 gives us

FM =

=

10 log (1

32.76

Substituting into Equation 19-3,

Ln =

= From Table

+

30 log 50

92.4

10 log

[(6) (0.25) (0.

-0.99999)

125) (8)]

-70

dB

we have

dB + 20 log

92.4 dB

+

18.06

+ 20 log 50

8

dB + 33.98 dB = 144.44 dB

19-1,

4dB Lf =

0.75 (6.5 dB)

A = Ar = t

The gain of an antenna (i.e., if its

37.8

=

4.875 dB

dB

increases or decreases proportional to the square of

diameter changes by a factor of

2, its

its

diameter

gain changes by a factor of 4 or 6 dB).

Substituting into Equation 19-1 yields

G =

32 76

+

144.44

+

4.875

+4-

37.8

-

37.8

=

1

10.475

dB

P = G s + C min t

=

110.475

dB + (-78.76 dBm) = 31.715 dBm

QUESTIONS 19-1.

What

constitutes a short-haul

microwave system?

19-2. Describe the baseband signal for a 19-3.

Why

do

FDM/FM

19-4. Describe a

A

long-haul microwave system?

microwave system.

microwave systems use low-index

microwave

repeater. Contrast

FM?

baseband and IF repeaters.

.

Chap. 19

Problems

751

19-5. Define diversity. Describe the three most

commonly used

19-6. Describe a protection switching arrangement. Contrast the

diversity schemes.

two types of protection switching

arrangements. 19-7. Briefly describe the four major sections of a

microwave terminal

station.

19-8. Define ringaround.

microwave system.

19-9. Briefly describe a high/low

19-10. Define system gain. 19-11. Define the following terms: free-space path loss, branching loss, and feeder loss. 19-12. Define fade margin. Describe multipath losses, terrain, sensitivity, and reliability objectives

and how they

effect fade margin.

19-13. Define receiver threshold.

19-14. Contrast carrier-to-noise ratio and signal-to-noise ratio. 19-15. Define noise figure

PROBLEMS 19-1. Calculate the noise

4

GHz

power

at the

and a bandwidth of 30

19-2. Determine the path loss for a

the terrain

is

(assume room temperature).

3.4-GHz

19-3. Determine the fade margin for a

GHz,

input to a receiver that has a radio carrier frequency of

MHz

signal propagating

20,000 m.

60-km microwave hop. The RF

carrier frequency

very smooth and dry, and the reliability objective

19-4. Determine the noise

power

20-MHz bandwidth

for a

is

is

6

99.95%.

at the input to a receiver

with an

input noise temperature of 290°C.

minimum input C/N of 30 dB, and minimum transmit power (P ).

19-5. For a system gain of 120 dB, a

of

-115 dBm, determine

19-6. Determine the

amount of

the

an input noise power

(

loss contributed to a reliability objective of

19-7. Determine the terrain sensitivity loss for a

4-GHz

99.98%.

carrier that is propagating over a very

dry, mountainous area.

19-8.

A

RF

carrier frequency of 7.4 GHz. The baseband signal is the 1800channel FDM system described in Chapter 6 (564 to 8284 kHz). The antennas are 4.8-m-diameter parabolic dishes. The feeder lengths are 150 m at one station and 50 m at the other station. The reliability objective is 99.999%. The system propagates over an average terrain that has a very dry climate. The distance between stations is 50 km. The

frequency-diversity microwave system operates

The IF

is

minimum

at

an

a low-index frequency-modulated subcarrier.

carrier-to-noise ratio at the receiver input

is

30 dB. Determine the following:

fade margin, antenna gain, free-space path loss, total branching and feeder losses, receiver input noise power,

C min minimum ,

transmit power, and system gain.

19-9. Determine the overall noise figure for a receiver that has

noise figure of 6

dB and

two

RF

amplifiers each with a

a gain of 10 dB, a mixer down-converter with a noise figure of

10 dB, and a conversion gain of

—6

dB, and 40 dB of IF gain with a noise figure of 6

dB. 19-10.

A

microwave receiver has a

figure of

total input

4 dB. For a minimum C/N

determine the

minimum

noise ratio

power of —102

of 20

receive carrier power.

dB

at the

dBm

and an overall noise

input to the

FM

detector,

Chapter 20

SATELLITE

COMMUNICATIONS INTRODUCTION American Telephone and Telegraph Company (AT&T) released few powerful satellites of advanced design could handle more traffic than the entire AT&T long-distance communications network. The cost of these satellites was estimated to be only a fraction of the cost of equivalent terrestrial microwave facilities. Unfortunately, because AT&T was a utility, government regulations prevented them from developing the satellite systems. Smaller and much less lucrative corporations were left to develop the satellite systems, and AT&T continued to invest billions of dollars each year in conventional terrestrial microwave systems. Because of this, early developments in satellite technology were slow in coming. Throughout the years the prices of most goods and services have increased substantially; however, satellite communications services have become more affordable each year. In most instances, satellite systems offer more flexibility than submarine cables, buried underground cables, line-of-sight microwave radio, tropospheric scatter radio, In the early 1960s, the

studies indicating that a

or optical fiber systems. Essentially, a satellite

is

a radio repeater in the sky (transponder).

system consists of a transponder, a ground-based station to control

its

A

satellite

operation, and a

user network of earth stations that provide the facilities for transmission and reception

of communications

traffic

through the

ized as either bus or payload.

pay load operation. The payload the system.

Although

more and more (in

in

in

analog or digital form)

752

is

system. Satellite transmissions are categorcontrol

mechanisms that support the is conveyed through

the actual user information that

recent years

demand,

satellite

The bus includes

new

data services and television broadcasting arc

the transmission of conventional speech telephone signals

is still

the bulk of the satellite payload.

History of Satellites

753

HISTORY OF SATELLITES The simplest type of

satellite is a

from one place

signal

passive reflector, a device that simply "bounces" a

to another.

The moon

is

a natural satellite of the earth and,

consequently, in the late 1940s and early 1950s, became the In 1954, the U.S.

Navy

successfully transmitted the

first

first satellite

messages over

transponder. this earth-to-

moon-to-earth relay. In 1956, a relay service was established between Washington,

D.C., and Hawaii and,

was limited only by

until 1962, offered reliable long-distance

communications. Service

moon.

the availability of the

In 1957, Russia launched Sputnik I, the

first

active earth satellite.

An

active satellite

capable of receiving, amplifying, and retransmitting information to and from earth

is

days. Later in the same which transmitted telemetry information

stations. Sputnik I transmitted telemetry information for 21

year, the United States launched Explorer

I,

for nearly 5 months.

In

1958,

NASA

launched Score, a 150-pound conical-shaped projectory. With

an on-board tape recording, Score rebroadcasted President Eisenhower's 1958 Christmas

message. Score was the

them on magnetic

stored its

first artificial satellite

Score was a delayed repeater

tions.

tape,

satellite;

it

used for relaying

terrestrial

communica-

received transmissions from earth stations,

and rebroadcasted them

to

ground stations farther along

orbit.

In 1960,

NASA in conjunction with Bell Telephone Laboratories and the Jet Propul-

sion Laboratory launched Echo, a 100-ft-diameter plastic balloon with an coating.

Echo passively

reflected radio signals

simple and reliable but required extremely high power transmitters

The

first

transatlantic transmission using a satellite transponder

Echo. Also 3

in

at the earth stations.

was accomplished using

1960, the Department of Defense launched Courier. Courier transmitted

W of power and lasted only In 1962,

aluminum

from a large earth antenna. Echo was

17 days.

AT&T launched Telstarl, the first satellite to receive and transmit simulta-

The electronic equipment in Telstar I was damaged by radiation from the newly discovered Van Allen belts and, consequently, lasted only a few weeks. Telstar II was electronically identical to Telstar I, but it was made more radiation resistant. neously.

Telstar II

was successfully launched The

facsimile, and data transmissions.

was accomplished with Telstar

in

1963.

first

It

was used

for telephone, television,

successful transatlantic transmission of video

II.

Early satellites were both of the passive and active type. Again, a passive satellite is

one

that

simply

reflects a signal

amplify or repeat the signal. signal

back

of passive

An

back

to earth; there are

active satellite

to earth (i.e., receives, amplifies, satellites is that

is

one

no gain devices on board

to

that electronically repeats a

and retransmits the

signal).

An

advantage

they do not require sophisticated electronic equipment on

board, although they are not necessarily void of power.

Some

passive satellites require

beacon transmitter for tracking and ranging purposes. A beacon is a continuously transmitted unmodulated carrier that an earth station can lock onto and use to align its a radio

antennas or to determine the exact location of the satellites is their inefficient

satellite.

A

disadvantage of passive

use of transmitted power. With Echo, for example, only

1

754

Satellite

18

part in every 10

Communications

Chap. 20

of the earth station transmitted power was actually returned to the

earth station receiving antenna.

ORBITAL SATELLITES The

satellites

mentioned thus

far are

of the orbital or nonsynchronous type. That

is,

they rotate around the earth in a low-altitude elliptical or circular pattern with an angular velocity greater than (prograde) or less than (retrograde) that of Earth. Consequently,

they are continuously gaining or falling back on Earth and do not remain stationary to

any particular point on Earth. Thus they have to be used when available, which may be as short a period of time as 15 minutes per orbit. Another disadvantage of orbital satellites is the

stations.

orbit

need for complicated and expensive tracking equipment

Each Earth

and then lock

major advantage of

its

It is

must locate the

antenna onto the

satellite as

comes

it

and track

satellite

it

them

it

at the earth

view on each

passes overhead.

satellite

systems

the Soviet

is

Molniya system.

presently the only nonsynchronous-orbit commercial satellite system in use. Molniya

1000

km

(see Figure 20-1).

orbit reaches; the perigee

is

The apogee the

minimum

is

the

maximum

km

distance.

and perigee

at

about

distance from earth a satellite

With the Molniya system,

the apogee

reached while over the northern hemisphere and the perigee while over the southern

hemisphere. The size of the ellipse was chosen to a sidereal

day

Because of

its

(the time

it

takes the earth to

make

During

its

12-h orbit,

it

its

rotate back

unique orbital pattern, the Molniya

rotation of the earth.

period exactly one-half of

to the

satellite

spends about

1 1

is

same

constellation).

synchronous with the

h over the north hemisphere.

Eliptical

orbit

Perigee

1000

A

on board

in their respective orbits.

of the more interesting orbital

uses a highly elliptical orbit with apogee at about 40,000

is

as

into

orbital satellites is that propulsion rockets are not required

the satellites to keep

One

station

km

FIGURE

20-1

satellite orbit.

Soviet Molniya

Geostationary Satellites

755

GEOSTATIONARY SATELLITES Geostationary or geosynchronous

satellites are satellites that orbit in a circular pattern

with an angular velocity equal to that of Earth. Consequently, they remain in a fixed position in respect to a given point on Earth.

includes

all

An

obvious advantage

shadow 100% of

to all the earth stations within their

the time.

earth stations that have a line-of-sight path to

pattern of the satellite's antennas.

TABLE 20-1

CURRENT

SATELLITE

An

obvious disadvantage

Western

Operator

Frequency

they are available

and

lie

a satellite

within the radiation

they require sophisticated

is

COMMUNICATIONS SYSTEMS Characteristic

Westar

it

is

The shadow of

Intelsat

V

Intelsat

System

SBS

Satellite

Fleet-

Anik satcom U.S.

Telsat

Union

Business

Dept. of

Telegraph

Systems

Defense

C

C

Consus

Global,

and

Ku

D

Canada

Ku

UHF, X

C,

Consus

Global

Canada,

Ku

band

Coverage

northern

zonal,

U.S.

spot

Number

of

21

12

10

12

24

43

0.005-0.5

36

transponders

Transponder

BW

36-77

36

(MHz)

EIRP (dBw)

23.5-29

33

Access

TDMA

FDMA, TDMA,

Modulation

FM, QPSK

FDM/FM,

FDMA,

Multiple

40-43.7

26-28

36

TDMA

FDMA

FDMA

QPSK

FM, QPSK

FDM, FM, FM/TVD, SCPC

Fixed

Mobile

Fixed

reuse

QPSK Fixed

Service

Fixed

tele,

tele,

tele,

TTY

TVD

TVD

GHz GHz GHz

C-band: 3.4-6.425

Ku-band: 10.95-14.5 X-band: 7.25-8.4

TTY

teletype

TVD TV distribution FDMA frequency -division

TDMA

multiple access

time-division multiple access

Consus continental United States

military

tele

756

Satellite

Syncom chronous

satellite is

launched

I,

in

satellite into orbit.

Chap. 20

keep them in a fixed orbit. The orbital time same as Earth. February 1963, was the first attempt to place a geosynSyncom I was lost during orbit injection. Syncom II and

and heavy propulsion devices on board of a geosynchronous

Communications

24

to

h, the

Syncom III were successfully launched in February 1963 and August 1964, respectively. The Syncom III satellite was used to broadcast the 1964 Olympic Games from Tokyo. The Syncom projects demonstrated the feasibility of using geosynchronous satellites. Since the Syncom projects, a number of nations and private corporations have successfully launched satellites that are currently being used to provide national as well as regional

and international global communications. There are more than 80

communications systems operating

in the

common-carrier telephone and data

satellite

world today. They provide worldwide fixed

circuits;

point-to-point cable television

(CATV);

network television distribution; music broadcasting; mobile telephone service; and private networks for corporations, governmental agencies, and military applications.

A commer-

network known as Intelsat (International Telecommunications

cial global satellite

Satellite

owned and operated by a consortium of more than 100 countries. Intelsat is managed by the designated communications entities in their respective countries. The Intelsat network provides high-quality, reliable service to its member countries. Organization)

is

Table 20-1

a partial

their

is

of current international and domestic

list

satellite

systems and

primary pay load.

ORBITAL PATTERNS Once

projected, a satellite remains in orbit because the centrifugal force caused by

rotation around the earth

is

to earth the satellite rotates, the greater the gravitational pull

required to keep

it

from being pulled

it

takes approximately

the time that the satellite less

is

U

in line

and the greater the velocity

to earth. Low-altitude satellites that orbit close to

Earth (100 to 300 miles in height) travel this speed,

its

counterbalanced by the earth's gravitational pull. The closer

at

approximately 17,500 miles per hour. At

h to rotate around the entire

earth. Consequently,

of sight of a particular earth station

is

only i h or

per orbit. Medium-altitude satellites (6000 to 12,000 miles in height) have a rotation

period of 5 to 12 h and remain in line of sight of a particular earth station for 2 to 4 h

per orbit. High-altitude, geosynchronous satellites (19,000 to 25,000 miles in height) travel at

the

approximately 6879 miles per hour and have a rotation period of 24 h, exactly

same

as the earth. Consequently, they remain in a fixed position in respect to a

given earth station and have a 24-h availability time. Figure 20-2 shows a low-, medium-,

and high-altitude

geosynchronous

can be seen that three equally spaced, high-altitude rotating around the earth above the equator can cover the

satellite orbit.

satellites

entire earth except for the

It

unpopulated areas of the north and south poles.

Figure 20-3 shows the three paths that a satellite the earth.

When

the satellite rotates in an orbit

above

may

take as

the equator,

it

is

it

rotates around

called an equatorial

Orbital Patterns

757

W \

\

i / /

/

/

/

/

/

/

/

/

/

/

1/

N»' (a)

(b)

FIGURE medium

20-2

(c)

Satellite orbits: (a)

low altitude (circular orbit, 100-300 mi); (b)

altitude (elliptical orbit, 6000-12,000 mi); (c) high altitude (geosynchronous

orbit, 19,000-25,000 mi).

orbit. it is

When

the satellite rotates in an orbit that takes

called a polar orbit. It

is

Any

interesting to note that

single satellite in a polar orbit. orbit while the earth

pattern

As

is

other orbital path

is

rotating

The on a

100% of

lies

over the north and south poles,

the earth's surface can be covered with a

latitudinal axis.

on earth

it

called an inclined orbit.

satellite is rotating

a diagonal spiral around the earth

a result, every location

is

around the earth

Consequently, the

in a longitudinal

satellite's radiation

which somewhat resembles

a barber pole.

within the radiation pattern of the satellite

twice each day.

Polar

Inclined

Equatorial

Earth station

FIGURE

20-3

Satellite orbits.

.

758

Satellite

Communications

Chap. 20

SUMMARY Advantages of Geosynchronous Orbits 1.

The

satellite

remains almost stationary

quently, expensive tracking equipment 2.

There

is

no need

to switch

from one

in respect to a is

given earth station. Conse-

not required at the earth stations.

satellite to

another as they orbit overhead.

Consequently, there are no breaks in transmission because of the switching times. 3.

High-altitude geosynchronous satellites can cover a

much

larger area of the earth

than their low-altitude orbital counterparts. 4.

The

effects of

Doppler

shift are negligible.

Disadvantages of Geosynchronous Orbits 1

2.

The higher altitudes of geosynchronous satellites introduce much longer propagation times. The round-trip propagation delay between two earth stations through a geosynchronous satellite is 500 to 600 ms. Geosynchronous

higher transmit powers and more sensitive receiv-

satellites require

and greater path

ers because of the longer distances 3.

High-precision spacemanship orbit

and

to

satellites to

keep

it

there.

keep them

losses.

required to place a geosynchronous satellite into

is

Also, propulsion engines are required on board the

in their respective orbits.

LOOK ANGLES To

toward a

orient an antenna

satellite,

it is

necessary to

know

the elevation angle and

azimuth (Figure 20-4). These are called the look angles.

Angle of Elevation The angle of elevation

is

the angle

formed between the plane of a wave radiated from

an earth station antenna and the horizon, or the angle subtended antenna between the

satellite

the greater the distance a propagated

wave must pass through

As with any wave propagated through

the earth's atmosphere,

may

also be severely contaminated

too small and the distance the

wave may 5°

is

how

by noise. Consequently,

wave

is

deteriorate to a degree that

considered as the

at the earth station

and the earth's horizon. The smaller the angle of elevation,

minimum

if

the earth's atmosphere. it

suffers absorption and

the angle of elevation

within the earth's atmosphere it

is

is

too long, the

provides inadequate transmission. Generally,

acceptable angle of elevation. Figure 20-5 shows

the angle of elevation affects the signal strength of a propagated

normal atmospheric absorption, absorption due

to thick fog,

wave due

and absoiption due

to

to a

Look Angles

759 Satcom

1

135° West longitude

Equator

95.5° longitude

South

Azimuth referred to

180°

Satellite

v

Antenna

Antenna

/

Elevation -*-

North

angle

Earth

Azimuth referred to

C

Antenna from top

FIGURE

heavy

rain.

It

Antenna from

20-4

side

Azimuth and angle of elevation "look angles."

that the 14/12-GHz band (Figure 20-5b) is more severely 6/4-GHz band (Figure 20-5a). This is due to the smaller wavelengths

can be seen

affected than the

associated with the higher frequencies. Also, at elevation angles less than 5°, the attenuation increases rapidly.

Azimuth Azimuth

is

defined as the horizontal pointing angle of an antenna.

clockwise direction

in

It is

measured

in a

degrees from true north. The angle of elevation and the azimuth

both depend on the latitude of the earth station and the longitude of both the earth station

and the orbiting

the procedure

is

the earth station.

satellite.

as follows:

For a geosynchronous

From

From Table

a

satellite in

good map, determine

an equatorial orbit,

the longitude and latitude of

20-2, determine the longitude of the satellite of interest.

Calculate the difference, in degrees (AL), between the longitude of the satellite and the

(9P) SSO| J3/\AOd

P

o

in

(%) jso|

jaMod

|bu6js

(gp) sso| ja/woj

o CO 1

o

in CM 1

CM

in t-

1

1

o

in

o

T-

c

1

1

.c/

gj

1

//

_ a

/

-

o

1

v1

1

1

1

8 (%) *SO| J8MOCJ |BU6|S

760

1

1

and Frequency Allocation

Orbital Spacing

761

TABLE 20-2 LONGITUDINAL POSITION OF SEVERAL CURRENT

SYNCHRONOUS SATELLITES PARKED IN AN EQUATORIAL ARC

3

Longitude Satellite

(°W) Anik

I

104

Anik 2

109

Anik 3

114

Westar

I

Westar

II

Westar

III

91

Satcom

I

135

99 123.5

Satcom 2

Comstar Palapa

119

D2

95

I

277

Palapa 2

283

0° latitude.

longitude of the earth station. Then, from Figure 20-6, determine the azimuth and elevation angle for the antenna. Figure 20-6

is

for a

geosynchronous

satellite

in

an

equatorial orbit.

EXAMPLE An

20-1

earth station

latitude

is

located at Houston, Texas, which has a longitude of

of 29.5°N. The

satellite

of interest

is

RCA's Satcom

1,

135°W. Determine the azimuth and elevation angle for the earth Solution

First

95.5°W and a

which has a longitude of station antenna.

determine the difference between the longitude of the earth station and

the satellite.

M= Locate the intersection of

AL

135°

and the

the figure the angle of elevation

is

-95.5° = 39.5°

latitude

of the earth station on Figure 20-6. From

approximately 35-, and the azimuth

is

approximately

59° west of south.

ORBITAL SPACING

AND FREQUENCY ALLOCATION

Geosynchronous

satellites

must share a limited space and frequency spectrum within a

given arc of a geostationary orbit. Satellites operating

must be There

is

at

or near the

sufficiently separated in space to avoid interfering with

a realistic limit to the

number of

satellite

same frequency

each other (Figure 20-7).

structures that can be stationed

.

762

Satellite

Azimuth angle referenced to 180° 5

10

FIGURE

20-6

Azimuth and elevation angle

Communications

Chap. 20

(degrees)

—

for earth stations located in the northern hemi-

sphere (referred to 180°).

(parked) within a given area in space. The required spatial separation

is

dependent on

the following variables:

1

Beamwidths and sidelobe

radiation of both the earth station and satellite antennas

2.

RF

3.

Encoding or modulation technique used

carrier frequency

4.

Acceptable limits of interference

5.

Transmit carrier power

Generally, 3 to 6° of spatial separation

is

required depending on the variables stated

above.

The most common carrier frequencies used for satellite communications are the 14/12-GHz bands. The first number is the up-link (earth station-to-transponder) frequency, and the second number is the down-link (transponder-to-earth station) fre-

6/4- and

Radiation Patterns: Footprints

Satellite

A

763

Satellite

B

19,000-25,000 miles

FIGURE

20-7

satellites in

Spatial separation of

geosynchronous orbit.

quency. Different up-link and down-link frequencies are used to prevent ringaround

from occurring (Chapter

19).

The higher

the carrier frequency, the smaller the diameter

required of an antenna for a given gain.

band. Unfortunately, this band

is

Care must be taken when designing a interference with established

Most domestic

use the 6/4-GHz microwave systems. avoid interference from or satellites

also used extensively for terrestrial

microwave

satellite

network

to

links.

Certain positions in the geosynchronous orbit are in higher

For example, the mid- Atlantic position which

is

demand

than the others.

used to interconnect North America

is in exceptionally high demand. The mid-Pacific position is another. The frequencies allocated by WARC (World Administrative Radio Conference) summarized in Figure 20-8. Table 20-3 shows the bandwidths available for various

and Europe are

services in the United States.

These services include fixed-point (between earth

stations

located at fixed geographical points on earth), broadcast (wide-area coverage), mobile (ground-to-aircraft, ships, or land vehicles), and intersatellite (satellite-to-satellite crosslinks).

RADIATION PATTERNS: FOOTPRINTS The area of its

the earth covered by a satellite depends

geosynchronous

orbit, its carrier frequency,

on the location of the

and the gain of

its

satellite in

antennas. Satellite

engineers select the antenna and carrier frequency for a particular spacecraft to concentrate the limited transmitted

power on

a specific area of the earth's surface.

The geographical

764

Communications

Satellite C-band

Chap. 20

X-band

I

1

Domestic

1

mm

Domestic

m

Military

m

_i

10GHz

K-band

Ku-band

I

i

_l

+

i

ANIK

t

i

I

i

I

I

13

12

11

m

ANIK

Intelsat i

14

mn 29

I

I

I

I

I

15

16

17

18

19

20

|

35

38

Uplink

FIGURE

41

|

20-8

44

47

50

53

—

satellite

•-

56

Cross-link

frequency assignments.

TABLE 20-3 RF SATELLITE BANDWIDTHS AVAILABLE IN THE UNITED STATES Frequency band

(GHz)

Band

Bandwidth Up-link

Down-link

(MHz)

C

5.9-6.4

3.7-4.2

X

7.9-8.4

7.25-7.75

500

11.7-12.2

500

Ku Ka

V

14-14.5

V (ISL)

500

—

27-30

17-20

30-31

20-21

50-51

40-41

1000

41-43

2000

—

Q

GHz

m

Downlink

WARC

I

21

V-band

nm 32

r~n

I

Q-band

K-band

26

Ka-band i

i

—

54-58

3900

59-64

5000

59

62

Radiation Patterns: Footprints

FIGURE

20-9

Satellite

765

antenna radiation patterns ("footprints").

representation of a satellite antenna's radiation pattern

is

called a footprint (Figure 20-

The contour lines represent limits of equal receive power density. The radiation pattern from a satellite antenna may be categorized as either spot, zonal, or earth (Figure 20-10). The radiation patterns of earth coverage antennas have a beamwidth of approximately 17° and include coverage of approximately one-third of 9).

the earth's surface. Zonal coverage includes an area less than one-third of the earth's surface. Spot

beams concentrate

the radiated

power

in a

very small geographic area.

Reuse

When

an allocated frequency band

reuse of the frequency spectrum. the antenna gain) the

By

beamwidth of

is

filled,

additional capacity can be achieved

increasing the size of an antenna the antenna

is

also reduced.

Thus

(i.e.,

by

increasing

different

beams

of the same frequency can be directed to different geographical areas of the earth. This is

called frequency reuse. Another

method of frequency reuse

is

to use dual polarization.

Different information signals can be transmitted to different earth station receivers using

same band of frequencies simply by orienting their electromagnetic polarizations in an orthogonal manner (90° out of phase). Dual polarization is less effective because the earth's atmosphere has a tendency to reorient or repolarize an electromagnetic wave as it passes through. Reuse is simply another way to increase the capacity of a limited the

bandwidth.

766

Satellite

Communications

Chap. 20

Satellite

transponder

FIGURE

20-10

Beams: A,

spot; B,

zonal: C, earth.

SATELLITE SYSTEM LINK

MODELS

Essentially, a satellite system consists of three basic sections: the uplink, the satellite

transponder, and the downlink.

Uplink Model The primary component within station transmitter.

A

the uplink section of a satellite system

is

the earth

typical earth station transmitter consists of an IF modulator, an

IF-to-RF microwave up-con verter, a high-power amplifier (HPA), and some means of bandlimiting the

final

output spectrum

shows the block diagram of a

RF

carrier

an output bandpass

satellite earth station transmitter.

the input baseband signals to either an

frequency.

(i.e.,

FM,

a

PSK, or

a

filter).

Figure 20-11

The IF modulator converts

QAM

modulated intermediate

The up-con verter (mixer and bandpass filter) converts the IF to an appropriate frequency. The HPA provides adequate input sensitivity and output power

to propagate the signal to the satellite transponder.

and traveling-wave tubes.

HPAs commonly

used are klystons

Satellite

System Link Models

767 To

satellite

transponder

Up-converter

—

r Baseband

Modulator (FM, PSK,

in

FDMor PCM/TDM

or

IF

I

RF

1

BPF

HPA

BPF

Mixer

QAM)

"RF

MW

Generator 6 or 14 GHz

FIGURE

20-11

Satellite uplink

model.

Transponder

A

typical satellite transponder consists of an input bandlimiting device (BPF), an input

low-noise amplifier (LNA), a frequency translator, a low-level power amplifier, and

an output bandpass

filter.

Figure 20-12 shows a simplified block diagram of a

transponder. This transponder are IF

BPF

device used as an

to those

used

in

microwave

repeaters. In Figure

limits the total noise applied to the input of the

LNA

is

The output of

a tunnel diode.)

translator (a shift oscillator

satellite

an RF-to-RF repeater. Other transponder configurations

and baseband repeaters similar

20-12, the input

to the

is

LNA

the

is

LNA. (A common fed to a frequency

and a BPF) which converts the high-band uplink frequency

low-band downlink frequency. The low-level power amplifier, which

is

commonly

Frequency translator ~l I

Low-noise

BPF

amplifier

Low-power

RF

RF |

amplifier

BPF

Mixer

TWT

LNA

t» MW shift oscillator

\

2

.

I

From 6 or

}

GHz

I

To

earth station

14GHz

earth station

4 or 12

FIGURE

20-12

Satellite

transponder.

GHz

768 From

Satellite

Communications

Chap. 20

satellite

transponder

Down-converter "1

r Low-noise

Demodulator (FM, PSK,

IF

RFJ

amplifier

BPF

BPF

Mixer

or

LNA

QAM)

Baseband out

FDMor PCM/TDM

RF

MW generator 4or

12

GHz

L FIGURE

20-13

Satellite

downlink model.

RF signal for transmission through the downlink RF satellite channel requires a separate transponder.

a traveling-wave tube, amplifies the to the earth station receivers.

Each

Downlink Model

An

earth station receiver includes an input

verter. Figure

the

BPF

BPF, an LNA, and an RF-to-IF down-con-

20-13 shows a block diagram of a typical earth station receiver. Again,

limits the input noise

power

to the

LNA. The LNA

is

a highly sensitive, low-

noise device such as a tunnel diode amplifier or a parametric amplifier.

down-converter

is

a mixer/bandpass

filter

The RF-to-IF

combination which converts the received

signal to an IF frequency.

FIGURE

20-14

Intersatellite link.

RF

System Parameters

Satellite

769

Cross-Links Occasionally, there

This

satellites.

Figure 20-14.

is

A

an application where

is

done using

is

it

necessary to communicate between

satellite cross-links or intersatellite links (ISLs),

disadvantage of using an ISL

that both the transmitter

is

are spacebound. Consequently, both the transmitter's output

power and

shown

in

and receiver

the receiver's

input sensitivity are limited.

SATELLITE SYSTEM PARAMETERS Transmit Power and

Energy

Bit

High-power amplifiers used

in earth station transmitters

and the traveling-wave tubes

power power power is

typically used in satellite transponders are nonlinear devices; their gain (output

versus input power) characteristic curve

is

is

dependent on input signal

shown

in

Figure 20-15.

reduced by 5 dB, the output power

power compression. To reduce

is

It

level.

A

typical input/output

can be seen that as the input

reduced by only 2 dB. There

is

an obvious

amount of intermodulation distortion caused by the nonlinear amplification of the HPA, the input power must be reduced {backed off) by several dB. This allows the HPA to operate in a more linear region. The amount the input level is backed off is equivalent to a loss and is appropriately called back-off loss

To

P

power amplifier should be operated as The saturated output power is designated P (sat) or

operate as efficiently as possible, a

close as possible to saturation.

simply

the

t

.

The output power of

-12

-11

a typical satellite earth station transmitter

-8

-7

-6

-5

-4

Input back-off (dB)

FIGURE

20-15

HPA

input/output characteristic curve.

is

much

770

Communications

Satellite

Chap. 20

higher than the output power from a terrestrial microwave power amplifier. Consequently,

when to

1

dealing with satellite systems,

W)

Most modern and

QAM,

is

is

t

generally expressed in

is

keying (PSK) or quadrature

shift

With PCM-encoded, time-division-multi-

generally a

PSK

digital in nature. Also, with

carrier

(decibels in respect

rather than conventional frequency modulation (FM).

power

is

and

QAM,

several bits

may

element (baud). Consequently, a parameter

in a single transmit signaling

more meaningful than

dBW

mW).

1

systems use either phase

(QAM)

the input baseband

plexed signal which

be encoded

P

(decibels in respect to

satellite

amplitude modulation

PSK

dBm

rather than in

energy per

bit

(Eb ). Mathematically,

Eb = P Tb

Eb

is

(20-la)

t

where

Eb = P =

energy of a single

Tb =

time of a single

t

or because

Tb =

\/Fb

,

bit (J/bit)

power (W)

total carrier

where

bit (s)

Fb

is

the bit rate in bits per second.

—

(20- lb)

EXAMPLE 202 For a

total transmit

rate of

power (P ) of 1000 W, determine t

the energy per bit (Eb ) for a transmission

50 Mbps.

Solution

Tb _ J_ =

Fb

(It

appears that the units for

of:Tb time of of ,

Tb

*

50 x 10 6 bps

should be

s/bit

=

6 02 x 10"

but the per bit

s

is

implied in the definition

bit.)

Substituting into Equation 20-la yields

Eb =

1000

(Again the units appear to be (hi energy per

J/s

J/bit,

x 10~ 6

(0.02

s/bit)

but the per bit

is

= 20

implied in the definition of

bit.)

1000

J/s

on

.

50 x 10°bps Expressed as a log,

Eb = It

is

common

to express

P

t

in

P,=

10 log(20

dBW

and

Eb

10 log 1000

x 10~ 6 ) = -47 dBJ in

=

dBW/bps. Thus 30

dBW

|xJ

Eb

.

Satellite

System Parameters

Effective Isotropic Radiated Effective isotropic radiated

and

is

771

Power

power (EIRP)

is

defined as an equivalent transmit

power

expressed mathematically as

EIRP = P r A

t

where

EIRP =

Pr = A = t

effective isotropic radiated total

power

power (W) (W)

radiated from an antenna

transmit antenna gain

(W/W

or a unitless ratio)

Expressed as a log,

EIRP (dBW) = P r (dBW) + A,(dB) In respect to the transmitter output,

Thus

EIRP = P

t

-L ho -L bf +A

(20-2)

t

where

P = (

L bo = L bf = A = t

actual

power output of

back-off losses of total

HPA

the transmitter

(dBW)

(dB)

branching and feeder loss (dB)

transmit antenna gain (dB)

For an earth station transmitter with an output power o loss of 3

dB, a

total

branching and feeder loss of 3 dB, and

ID ;

a

W

VJVJ,V/VAJ

dB, determine the EIRP. Solution

Substituting into Equation 20-2 yields

EIRP = P - 1^ " Kf + (

= 40dBW- -

3

A,

dB - 3dB + 40 dB

==

W)

,

transmit antenna gain of 40

74

dBW

772

Satellite

Communications

Chap. 20

Equivalent Noise Temperature microwave systems, the noise introduced in a receiver or a component within a receiver was commonly specified by the parameter noise figure. In satellite communications systems, it is often necessary to differentiate or measure noise in increWith

terrestrial

ments as small as a tenth or a hundredth of a decibel. Noise form,

is

figure, in

inadequate for such precise calculations. Consequently,

it

is

its

standard

common

to use

environmental temperature (T) and equivalent noise temperature (Te ) when evaluating the performance of a satellite system. In Chapter 19 total noise

power was expressed

mathematically as

N = KTB Rearranging and solving for

T

gives us

N

T

KB

where

N= K= B = T=

total

noise

power (W)

Boltzmann's constant (J/K)

bandwidth (Hz) temperature of the environment (K)

Again from Chapter 19 (Equation

19-7).

NF =

1

+

^ T

where /

Te = NF = T=

equivalent noise temperature (K) noise figure (absolute value)

temperature of the environment (K)

Rearranging Equation 19-7,

we have

Te = 7XNF-

1)

Typically, equivalent noise temperatures of the receivers used in satellite transponders are about 1000 K. For earth station receivers

K. Equivalent noise temperature with the unit of

dBK,

is

generally

more

Te

values are between 20 and 1000

useful

when expressed

as follows:

r ,(dBK) (

10 log

For an equivalent noise temperature of 100 K,

7 (dBK)=

Tc

Te (dBK)

10 log 100 or 20

is

dBK

logarithmically

Satellite

System Parameters

773

Equivalent noise temperature

is

a hypothetical value that can be calculated but

cannot be measured. Equivalent noise temperature

because

or a receiver

(Te )

is

more accurate method of expressing

a

is

it

when

evaluating

its

often used rather than noise figure the noise contributed

by a device

performance. Essentially, equivalent noise temperature

the noise present at the input to a device or amplifier plus the noise

is

internally

by

by simply evaluating an equivalent input noise temperature. As you discussions,

added

that device. This allows us to analyze the noise characteristics of a device

Te

is

a very useful parameter

when

will see in subsequent

evaluating the performance of a satellite

system.

EXAMPLE 20-4 Convert noise figures of 4 and 4.01 to equivalent noise temperatures. Use 300

K

for the

environmental temperature. Solution

Substituting into Equation 20-7 yields

7;

For

NF =

= 7XNF-

4:

Te = 300(4ForNF =

300(4.01

-

1)

=

903

K

can be seen that the 3° difference in the equivalent temperatures

two noise

the difference between the is

= 900K

1)

4.01:

Te = It

I)

a

more accurate way of comparing

figures.

is

300 times

as large as

Consequently, equivalent noise temperature

the noise performances of

two receivers or devices.

Noise Density Simply

(N

stated, noise density

or the noise

power present

)

in a

is

the total noise

power normalized

to a

1-Hz bandwidth,

1-Hz bandwidth. Mathematically, noise density

Na = -

or

B

KTe

is

(20-3a)

where yV

=

noise density the per hertz

N=

total noise

B = bandwidth

(W/Hz) (NQ is

implied

is

generally expressed as simply watts;

in the definition

of

N

)

power (W) (Hz)

K=

Boltzmann's constant (J/K)

Te =

equivalent noise temperature (K)

Expressed as a log,

NG (dBW/Hz) = =

10 log

N-

io log a:

+

10 log

B

10 log t;

(20-3b) (20-3c)

774

Satellite

EXAMPLE

Communications

Chap. 20

20-5

MHz

For an equivalent noise bandwidth of 10

and a

power of 0.0276 pW,

total noise

determine the noise density and equivalent noise temperature. Substituting into Equation 20-3a,

Solution

N

N° = A,

or simply, 276

x

10" 23

W =276x10 ^, n ' „W 23

lOx.O'Hz

h;

W.

N = or simply

276 x 10~

B =

we have 16

—205.6 dBW.

10 log (276

x 10~ 23 ) = -205.6 dBW/Hz

Substituting into Equation 20-3b gives us

NQ = N (dBW) - B (dB/Hz) = -135.6 dBW - 70 (dB/Hz) = -205.6 dBW Rearranging Equation 20-3a and solving for equivalent noise temperature yields

e

K 276 x 10~ 23 J/cycle

^

=

1.38X10-*^ = 10 log 200

=

= N (dBW)-

=

-205.6

23

,

le

dBK

10 log

dBW -

^

tSMawt ° 200 K

K

(-228.6

dBWK) =

23

dBK

Carrier-to-Noise Density Ratio

CINQ is the average wideband carrier power-to-noise density ratio. The wideband carrier power is the combined power of the carrier and its associated sidebands. The noise is the thermal noise present in a normalized 1-Hz bandwidth. The carrier-to-noise density ratio may also be written as a function of noise temperature. Mathematically, C/N is

NQ Expressed as a

(20-4a)

KT

e

log,

— (dB) = C (dBW) - N

Q

(dBW)

(20-4b)

Energy of Bit-to-Noise Density Ratio E,JN

is

one of the most important and most often used parameters when evaluating

digital radio

system. The E,JN

ratio

is

a convenient

way

to

compare

a

digital systems

System Parameters

Satellite

775 modulation schemes, or encoding techniques. Mathe-

that use different transmission rates,

matically,

E^N

is

E = CIF — NIB b

b

=

NQ EfJN

is

CB (20-5)

NFb

a convenient term used for digital system calculations and performance

comparisons, but

world,

in the real

carrier power-to-noise density ratio

it

more convenient

is

and convert

it

to measure the wideband EtJNQ Rearranging Equation 20-5

to

.

yields the following expression:

Eb_C NQ ~ N EbINQ

The

ratio

X

B_

Fb

the product of the carrier-to-noise ratio

is

(ON) and

the noise

bandwidth-to-bit ratio (B/Fb ). Expressed as a log,

^(dB) = £(dB) + |-(dB)

(20-6)

The energy per bit (Eb ) will remain constant as long as the total wideband carrier power (C) and the transmission rate (bps) remain unchanged. Also, the noise density (N ) will remain constant as long as the noise temperature remains constant. The following conclusion can be made: For a given carrier power,

and noise temperature,

bit rate,

the Eij/Nq ratio will remain constant regardless of the encoding technique, modulation

scheme, or bandwidth used. Figure 20-16 graphically illustrates the relationship between an expected probability of error P{e) and the is

for the

minimum

the relationship

minimum

ON ratio required to achieve the P{e).

The

ON specified

double-sided Nyquist bandwidth. Figure 20-17 graphically illustrates

Eb/N

between an expected P(e) and the minimum

ratio required to

achieve that P(e).

A

5 5 P(e) of 10~ (1/10 ) indicates a probability that

100,000

EXAMPLE A

P(e)

bits transmitted.

is

analogous to the

coherent binary phase-shift-keyed

(a)

(b) filter,

(BPSK)

minimum

Solution

in error for

every

Determine the the

ON

(a)

ON and E/JN

ratios for a receiver

bandwidth

is

if

the noise

is

measured

bandpass

if

the noise

is

measured

at a point prior to the

bandpass iss

equal to three times the Nyquist bandwidth.

With BPSK, the minimum bandwidth the

at a point prior to the

equal to twice the Nyquist bandwidth.

is

ON

bandwidth

From Figure 20-16,

20 Mbps DS.

transmitter operates at a bit rate of

double-sided Nyquist bandwidth.

Determine the

where

be

(BER).

:

Determine the minimum theoretical

where the bandwidth (c)

filter

bit will

20-6

4 For a probability of error P(e) of 10~

equal to the

1

bit error rate

minimum

ON

is

is

equal to the bit rate, 20

MHz

8.8 dB. Substituting into Equation 20-6 gives us

776

Communications

Satellite

V

10-

K«

I

'

'

1

V

1 1

1

i '

i

1

1

'

1

i

1

1

i

1

i

U

I

|

'

Chap. 20

|

1

'

|

1

X^^^PSK 10

X^^-SPSK

X^^QPSK 10

BPSK^^ \ o

10"

5a

10"

-

-

2 Q.

10"

10"

10

-

,1,1,1,1 8

9

FIGURE

10

20-16

i

11

ill 13

1

12

i

i

1

14

i\

1

15

1,1,1,1,1, 1,1,1,1,

i

16

17

18

Minimum C/N

(dB)

Probability of error P(e) versus

C/N

19

20

21

23

22

24

25

1

26

for various digital modulation schemes.

(Bandwidth equals minimum double-sided Nyquist bandwidth.)

(dB) .V,

N

(dB)

8.8

+ ^- (dB)

dB+

20 x 1Q 6 10 log

20 x 106

= 8.8dB + 0dB = 8.8db Note: The equals the

minimum EiJN equals the minimum C/N when the receiver noise bandwidth minimum Nyquist bandwidth. The minimum E,,/N of 8.8 can be verified from

Figure 20-17.

What

Eb /N

effect

ratios?

does increasing the noise bandwidth have on the

The wideband

carrier

power

is

totally

minimum C/N and

independent of the noise bandwidth.

Similarly, an increase in the bandwidth causes a corresponding increase in the noise power.

Consequently, a decrease the noise bandwidth.

Therefore,

Eh

is

Eh

is

in

C/N

is

realized that

is

directly proportional to the increase in

dependent on the wideband carrier power and the

unaffected by an increase in the noise bandwidth.

normalized to a 1-Hz bandwidth and, consequently, the noise bandwidth.

is

N

is

bit rate

the noise

only.

power

also unaffected by an increase

in

1

Satellite

777

System Parameters

10I

I

I

I

I

I

I

i

I

I

i

I

I

I

10'

SS 16PSK

V

Q.

2

10" 2

>v8PSK

\^ 5

\ QPSK

10"

\

BPSICV

\

10-

I

I

I

I

I

I

8

FIGURE

20-17

l\

\

9

10

I

I

I

E b IN

Probability of error P(e) versus

I

12

11

1

13

\l

1

14

15

~

1

16

18

17

ratio for various digital

modulation schemes. (b) Since

Eb /N

independent of bandwidth, measuring the CIN

is

receiver where the bandwidth lutely

no

effect

on

Eh /N

used to solve for the

Eb INQ ratio, we

new

.

is

equal to twice the

Therefore,

Eb IN

at

a point in the

minimum Nyquist bandwidth

becomes

has abso-

the constant in Equation 20-6 and

is

value of CIN. Rearranging Equation 20-6 and using the calculated

have

£(dB) = ^(dB)-f-(dB)

NQ

N

Fb

x

6

40 10 -OdB-lOlog^-^

(c)

Measuring the CIN

three times the

-

8.8

dB -

=

8.8

dB

-3dB =

Eb

60 x 10 6 ,0,og

= 8.8dB=

Eb IN

ratios of 8.8, 5.8,

4.03

2o^

10 log 3

dB

and 4.03 dB indicate the CIN

at the three specified points in the

zndP(e).

where the bandwidth equals

yields the following results for CIN.

;r^-

The CIN

5.8dB

ratio at a point in the receiver

minimum bandwidth

C

measured

10 log 2

ratios that

would be

receiver to achieve the desired

minimum

778

Communications

Satellite

Because

EJNq cannot

band carrier-to-noise

E^Nq ratio,

be directly measured to determine the

measured and then substituted

ratio is

E^Nq

quently, to accurately determine the

Chap. 20 the wide-

into Equation 20-6.

Conse-

bandwidth of the receiver

ratio, the noise

must be known.

EXAMPLE 20-7 A coherent 8PSK 5 of 10~

minimum

error

ON

Determine the

(b)

where the bandwidth

filter

where the bandwidth

Determine the

(c)

(a) 8PSK minimum bandwidth

Solution

ratios for a receiver

bandwidth

double-sided Nyquist bandwidth.

filter

is

EbIN

Determine the minimum theoretical CIN and

(a)

equal to the

ON

90 Mbps. For a probability of

transmitter operates at a bit rate of

:

the noise

if

measured

is

a point prior to the bandj

at

equal to twice the Nyquist bandwidth.

is

ON

the noise

if

measured

is

a point prior to the bandpass

at

equal to three times the Nyquist bandwidth.

is

has a bandwidth efficiency of 3 bps/Hz and, consequently, requires a

of one-third the

bit rate

^ N

(

dB)=18.5dB +

= (b) Rearranging

18.5

MHz. From we obtain

or 30

18.5 dB. Substituting into Equation 20-6,

Figure 20-16, the

minimum

101og|^ 90 Mbps

dB + (-4.8 dB) =

Equation 20-6 and substituting

in

13.7 db

EbINQ yields

§«-»—«.sg =

dB - (-

13.7

£„ HPA output power; L bo backL h branching loss; A„ transmit antenna gain; Pn total radiated power = P, - L ho - L b - Lf EIRP, effective isotropic radiated power = Pr A L„, additional uplink losses due to atmosphere; L path loss; A r receive antenna gain; G/T,,, gain-to-equivap lent noise ratio; L d additional downlink losses due to atmosphere; LNA, low-noise amplifier; CITe carrier-to-equivalent noise ratio; C/7V,,, carrier-to-noise density ratio; E/JN^ energy of uplink and downlink sections.

off loss; Lf, feeder loss;

,

,

t;

;

,

,

,

,

bit-to-noise density ratio;

C/N, carrier-to-noise

ratio.

Link Equations

781

is determined by combining them in the approprimicrowave or satellite radio simply means the original and demodulated baseband signals are digital in nature. The RF portion of the radio is analog; that is, FSK, PSK, QAM, or some other higher-level modulation riding on an analog microwave carrier.

separately, then the overall performance

ate

manner. Keep

mind, a

in

digital

LINK EQUATIONS The following

used to separately analyze the uplink and the downlink

link equations are

sections of a single radio-frequency carrier satellite system. These equations consider

only the ideal gains and losses and effects of thermal noise associated with the earth station transmitter, earth station receiver,

and the

satellite

transponder.

The nonideal

aspects of the system are discussed later in this chapter.

Uplink Equation

C _ A P r(LPL u )A r _ A P r {LPL N ~ KTe K t

t

l(

G_

)

Te

where L^and L u are the additional uplink and downlink atmospheric losses, respectively. The uplink and downlink signals must pass through the earth's atmosphere, where they are partially absorbed by the moisture, oxygen, and particulates in the air. Depending on the elevation angle, the distance the

from one earth

values less than

1

.

GITe

is

RF

signal travels through the atmosphere varies

Because L u and ^represent

losses, they are decimal

the receiving antenna gain divided

by the equivalent input

station to another.

noise temperature.

Expressed as a log, C_ 101ogA,/>,. .

.

log V

.

v

EIRP

/4ttD\

- 20 -

X

v

free-space

path loss

earth

+

/

/G\ .

+

—

10 log

v

- 101ogL

\Te /

satellite

y

J

-

GlTe

additional

-

10 log v

-L ,(dB) 7

+

— (dBK -1

)

constant

- L u (dB) - AT(dBWK)

G C = A P r {LPL d)A r = AA r (LPL d y K Te KTe NQ t

)

V

/f ,

- Boltzmann's

losses

Downlink Equation

Expressed as a log

,

atmospheric

station

EIRP(dBW)

v

tt

782

—=

Communications

Satellite

lOlogVV " 201og(-2^-) + EIRP

10 log (^-\

+

free-space

10 log

Ld -

-

additional

earth station

GlTe

path loss

satellite

-

atmospheric

Chap. 20

10 log

K

Boltzmann's constant

losses

= EIRP (dBW) - LD (dB) +

LINK

- (dBK"

')

- L d (dB) - K (dBWK)

BUDGET Table 20-4

lists

the system parameters for three typical satellite

The systems and

their parameters are not necessarily for

they are hypothetical examples only. link budget.

the projected

A

link

budget

C/N and Eb /N

The system parameters

identifies the

ratios at

communication systems.

an existing or future system; are used to construct a

system parameters and

both the

satellite

is

used to determine

and earth station receivers for a

given modulation scheme and desired P(e).

EXAMPLE

20-10

Complete the

link budget for a satellite

system with the following parameters.

Uplink 1.

Earth station transmitter output power saturation,

2000

at

2.

Earth station back-off loss

3.

Earth station branching and feeder losses

4.

Earth station transmit antenna gain

(from Figure 20-19, 15

.

m

at

14

5.

Additional uplink atmospheric losses

Free-space path loss (from Figure 20-20,

7.

Satellite receiver

3dB 4dB 64 dB 0.6

dB

206.5 db

14 Ghz)

8.

Satellite

9.

Bit rate

10.

dBW

GHz)

6.

at

33

W

GITe

-5.3 dBK"

ratio

branching and feeder losses

OdB 120

Mbps

8PSK

Modulation scheme

Downlink I.

Satellite transmitter output

uration 10

at sat-

10

dBW

0.1

dB

branching and feeder losses

0.5

dB

antenna gain (from

30.8

dB

2.

Satellite back-off loss

3.

Satellite

4.

power

W

Satellite transmit

Figure 20-19, 0.37

m

at

12

GHz)

1

783

Link Budget

TABLE 20-4 SYSTEM PARAMETERS FOR THREE HYPOTHETICAL SATELLITE SYSTEMS

§>.& 2 ^

..is < » a

ffl

P.

B

4)

"5

j=

E

i

"B N =

O

a

E

^£ ~~

2£

oo

Uplink Transmitter output power (saturation,

dBW)

35

25

Earth station back-off loss (dB)

2

2

3

Earth station branching and feeder loss (dB)

3

3

4

0.6

Additional atmospheric (dB) Earth station antenna gain (dB)

Free-space path loss (dB) Satellite receive Satellite

antenna gain (dB)

Satellite equivalent noise Satellite

GITe (dBKT

temperature (K)

1

)

33

0.4

0.6

55

45

64

200

208

206.5

20

45

23.7

branching and feeder loss (dB)

1

1

1000

800

-10

16

800 -5.3

18

20

30.8

Downlink Transmitter output power (saturation, Satellite back-off loss Satellite

dBW)

0.5

(dB)

branching and feeder loss (dB)

Satellite

0.8

44

197

206

51

44

3

3

250

1000

27

14

Additional downlink atmospheric losses

0.4

Free-space path loss (dB) Earth station receive antenna gain (dB) Earth station branching and feeder loss (dB) Earth station equivalent noise temperature (K)

5.

GITe (dBKT

1

)

6. Free-space path loss (from Figure 20-20, at 12 7.

0.1

0.5

0.4

1.4

16

antenna gain (dB)

Earth station

0.2 1

Additional atmospheric loss (dB)

10

205.6

62

270 37.7

dB

205.6 dB

GHz)

Earth station receive antenna gain (15

62 dB

m, 12 GHz) 8.

Earth station branching and feeder losses

9.

Earth station equivalent noise tempera-

OdB 270

K

ture 10.

Earth station

GITe

ratio

s

-

5/s

>>

—

£

|£

E O

u

s

..88

37.7

dBK"

1

784

Satellite

I

I

I

I

I

I

Communications

Chap. 20

i

60 -

/ /-

50 -

7/f// £ a

40


II

D

?

X

i

o

18 15 10 19 10 20 10 10 16 10 17 10 10 9 10 10 10 11 10 12 10 13 10 14 10

kHz

-*

-9

E

| £

I

108

10

equal to 10 angstroms.

h 5

co

o 5 O

is

is

22-1

— Electromagnetic frequency spectrum.

21

10

22

at

s

V V

£ ?

*

>

V c

«

5

M

w




Normal

n.

Unrefracted ray

Refracted ray

dense

n

less

n,

more dense

Incident ray

FIGURE

22-7

away from

Light ray refracted

the normal.

Light Propagation

821

Angle

Critical

Figure 22-8 shows a condition in which an incident ray angle of refraction

is

note that the light ray

90° and the refracted ray is

traveling

is

at

is

an angle such that the

along the interface.

from a medium of higher

(It is

important to

refractive index to a

medium

with a lower refractive index.) Again, using Snell's law,

sin0]

With 62

= 90

=

n — sin 2

2

c

n?

sin0!=— (1) n

— n-y

sin0!=

or

n

\

\

and

-Q -6 C

sin

where 6C

is

(22-2)

x

the critical angle.

Normal

/

Un refracted

ray

dense

n2

less

n,

more dense

Reflected ray

Incident ray

FIGURE

22-8

reflection.

Critical angle

822

Fiber Optic

The ray

may

critical

angle

is

defined as the

strike the interface

minimum

when

medium

If the

dense medium.)

allowed to penetrate the

light ray is not

Chap. 22

angle of incidence at which a light

of two media and result in an angle of refraction of 90° or

greater. (This definition pertains only into a less

Communications

the light ray

is

traveling

angle of refraction

is

from a more dense 90° or greater, the

dense material. Consequently,

less

takes place at the interface, and the angle of reflection

is

total reflection

equal to the angle of incidence.

Figure 22-9 shows a comparison of the angle of refraction and the angle of reflection

when

the angle of incidence

is

less than or

more than

the critical angle.

PROPAGATION OF LIGHT THROUGH AN OPTICAL Light can be propagated

How

the light

of the

is

down an

optical fiber cable

FIBER

by

either reflection or refraction.

mode of propagation and

propagated depends on the

the index profile

fiber.

Mode

of Propagation

In fiber optics terminology, the

path for light to take

one path, of light

it

down

word mode simply means

the cable,

it

called single

is

called multimode. Figure 22-10

is

down an

shows

path. If there

mode.

single and

Normal

Refracted ray >

Angle of reflection equals

90 -

when0

1

>0

0, C

Reflected ray

Incident ray

(0,