ch03 part1 modified sp2014

8/13/2019 Ch03 Part1 Modified SP2014

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3.1

Chapter 3Data and Signals

Part 1

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.


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3.2

To be transmitted, data must be

transformed to electromagnetic signals.

Note


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3.3

Terms & Concepts

Data - Logical

Signal - Physical

Channel - Logical

Link Physical (carr ies one or more channels)

Bit Rate (N) (bi t per second) - Data

Bandwidth (B) (H z cycle per second) Signal and Channel

Capacity (C) (bit per second) - Channel

Signal E lement - Physical

Signal Rate (S) (Baud Rate)Data Element - Logical

Data Rate (N)


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3.4

Data & Signals

Data

Analog Digital

Signal

Analog

Digital


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3.5

3

-

1 ANALOG AND DIGITAL

Data

can

be

analog

or

digital

.

The

term

analog

data

refers

to

information

that

is

continuous

;

digital

data

refers

toinformation

that

has

discrete

states

.

Analog

data

take

on

continuous

values

.

Digital

data

take

on

discrete

values

.

Analog and Digital Data

Analog and Digital Signals

Periodic and Nonperiodic Signals

Topics discussed in this section:


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3.6

Note

Data can be analog or digital.

Analog data are continuous and take

continuous values.

Digital data have discrete states and

take discrete values.


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Signals can be analog or digital.

Analog signals can have an infinite

number of values in a range; digital

signals can have only a limited

number of values.

Note


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Figure 3.2 Comparison of analog and digi tal signals


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In data communications, we commonly

use periodic analog signals andnonperiodic digital signals.

Note


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3-2 PERIODIC ANALOG SIGNALS

Periodicanalogsignalscanbeclassif iedassimpleor

composite.Asimpleper iodicanalogsignal,asinewave,cannotbedecomposedintosimplersignals.Acomposite

periodicanalogsignal iscomposedofmultiplesine

waves.

Sine Wave

WavelengthTime and Frequency Domain

Composite Signals

Bandwidth

Topics discussed in this section:


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Figure 3.3 A sine wave


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We discuss a mathematical approach tosine waves in Appendix E online.

Note


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The power in your house can be represented by a sine

wave with a peak amplitude of 155 to 170 V. However, it

is common knowledge that the voltage of the power in

U.S. homes is 110 to 120 V. This discrepancy is due to

the fact that these areroot mean square(rms) values.

The signal is squared and then the average amplitude is

calculated. The peak value is equal to2rmsvalue.

Example 3.1


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Figure 3.4 Two signals with the same phase and f requency,but di ff erent amplitudes


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The voltage of a battery is a constant; this constant value

can be considered a sine wave, as we wil l see later. For

example, the peak value of anAAbattery is normally

1.5 V.

Example 3.2


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Frequency and period are the inverse ofeach other.

Note


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3.19

The power we use at home has a f requency of60 Hz.

The per iod of this sine wave can be determined as

follows:

Example 3.3


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3.20

Express a per iod of 100 ms in microseconds.

Example 3.4

Solution

From Table 3.1 we f ind the equivalents of 1 ms (1 ms is103s) and 1 s (1 s is 106s). We make the following

substitutions:.


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3.21

The per iod of a signal is 100 ms. What is its frequency inkilohertz?

Example 3.5

Solution

First we change 100 ms to seconds, and then we

calculate the frequency from the per iod (1 Hz = 103

kHz).


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3.22

Frequency is the rate of change with

respect to time.

Change in a short span of time

means high frequency.

Change over a long span of

time means low frequency.

Note


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3.23

If a signal does not change at all, itsfrequency is zero (for example, DC)

If a signal changes instantaneously, its

frequency is infinite (like the sharp

vertical edge of a pulse) .

Note


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3.24

Phase describes the position of thewaveform relative to time 0.

Note


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3.25

Figure 3.6 Three sine waves with the same ampl itude and f requency,but di fferent phases


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3.26

A sine wave is offset 1/6 cycle with respect to time 0.What is its phase in degrees and radians?

Example 3.6

SolutionWe know that 1 complete cycle is 360. Therefore, 1/6

cycle is


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3.27

Figure 3.7 Wavelength and period


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3.28

Figure 3.8 The time-domain and frequency-domain plots of a sine wave


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3.29

A complete sine wave in the timedomain can be represented by one

single spike in the frequency domain.

Note


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3.30

Th e f r eq u en c y d o m ain is m o re c o m p ac t an d

u s e f u l w h e n w e a r e d e a l i n g w i t h m o r e t h a n o n e

s i n e w a v e . F o r e x a m p l e , F i g u r e 3 . 8 s h o w s t h r e e

s in e w av es , eac h w i th d i ff er en t am p l it u d e an d

f r eq uen cy. A ll c an b e r ep res en ted b y t h ree

s p i k es i n t h e fr eq u en c y d o m ai n .

Example 3.7


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3.31

Figure 3.9 The time domain and frequency domain of three sine waves


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3.32

A single-frequency sine wave is notuseful in data communications;

we need to send a composite signal, a

signal made of many simple sine waves.

Note


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3.34

If the composite signal is periodic, the

decomposition gives a series of signals

with discrete frequencies;

if the composite signal is nonperiodic,

the decomposition gives a combination

of sine waves with continuousfrequencies.

Note


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3.35

Figure 3.10 shows a periodic composite signal withfrequency f. This type of signal is not typical of those

found in data communications. We can consider it to be

three alarm systems, each with a different f requency.

The analysis of this signal can give us a good

understanding of how to decompose signals.

Example 3.8


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Fi 3 11


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3.37

Figure 3.11 Decomposition of a composite per iodic signal in the time andfrequency domains


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3.38

F igure 3.12 shows a nonper iodic composite signal. I tcan be the signal created by a microphone or a telephone

set when a word or two is pronounced. I n this case, the

composite signal cannot be per iodic, because that

implies that we are repeating the same word or words

with exactly the same tone.

Example 3.9


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3.39

Figure 3.12 The time and f requency domains of a nonper iodic signal


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3.41

Figure 3.13 The bandwidth of per iodic and nonperiodic composite signals


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3.43

Figure 3.13 The bandwidth of per iodic and nonperiodic composite signals


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3.44

Figure 3.14 The bandwidth for Example 3.10

E l 3 11


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3.45

A periodic signal has a bandwidth of 20 Hz. The highestfrequency is 60 Hz. What is the lowest f requency? Draw

the spectrum if the signal contains all frequencies of the

same ampli tude.

SolutionLetfhbe the highest f requency,flthe lowest frequency,

andBthe bandwidth. Then

Example 3.11

The spectrum contains all integer frequencies. We show

this by a ser ies of spikes (see F igure 3.14).


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3.46


E l 3 12


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3.47

A nonper iodic composite signal has a bandwidth of 200kH z, with a middle frequency of 140 kH z and peak

amplitude of 20 V. The two extreme frequencies have an

amplitude of 0. Draw the frequency domain of the

signal.

Solution

The lowest f requency must be at 40 kHz and the highest

at 240 kHz. F igure 3.15 shows the frequency domain

and the bandwidth.

Example 3.12


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3.48


E l 3 13


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3.49

An example of a nonper iodic composite signal is thesignal propagated by an AM radio station. I n the United

States, each AM radio station is assigned a 10-kHz

bandwidth. The total bandwidth dedicated to AM radio

ranges from 530 to 1700 kHz. We wil l show the rationalebehind this 10-kH z bandwidth in Chapter 5.

Example 3.13

E l 3 14


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3.50

Another example of a nonper iodic composite signal isthe signal propagated by an FM radio station. In the

United States, each FM radio station is assigned a 200-

kHz bandwidth. The total bandwidth dedicated to FM

radio ranges from 88 to 108 MHz. We wil l show the

rationale behind this 200-kHz bandwidth in Chapter 5.

Example 3.14

E l 3 15


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3.51

Another example of a nonper iodic composite signal isthe signal received by an old-fashioned analog black-

and-white TV. A TV screen is made up of pixels. I f we

assume a resolution of 525 700, we have 367,500

pixels per screen. I f we scan the screen 30 times persecond, this is 367,500 30 = 11,025,000 pixels per

second. The worst-case scenario is alternating black and

white pixels. We can send 2 pixels per cycle. Therefore,

we need 11,025,000 / 2 = 5,512,500 cycles per second, orHz. The bandwidth needed is 5.5125 MHz.

Example 3.15


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3.52

L isten to Waveforms

http: //www.jhu.edu/~signals/l isten-new/listen-newindex.htm
http://www.jhu.edu/~signals/listen-new/listen-newindex.htmhttp://www.jhu.edu/~signals/listen-new/listen-newindex.htmhttp://www.jhu.edu/~signals/listen-new/listen-newindex.htm


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3.53

Four ier Seri es Demo

http://www.falstad.com/fourier/
http://www.falstad.com/fourier/http://www.falstad.com/fourier/


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Tw=1 msec, 1/Tw =1kHz, Tp=2 msec, 1/Tp=5kH z

Envelope zeros @ mul tiple integers of 2/Tw, peaks @ multiple integers of 1/Tw


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periodic

discrete spectral components

Peri od = 5 msec (peaks)

ch03 part1 modified sp2014

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