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Outline

Analog and Digital Data

Analog and Digital Signals

Amplitude Modulation (AM)

Frequency Modulation (FM)

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What is a Signal ?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.

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Periodic Signals

4WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998

Aperiodic Signals

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Analog and Digital Signals

In data communication, we commonly use periodic analog signals and

aperiodic digital signals.

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Periodic signal: Sine Wave – Period(T) & Frequency(F)

T = 1 / F

Peak Amplitude

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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.

Frequency

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Frequencies

Unit Equivalent Unit Equivalent

Seconds (s) 1 s hertz (Hz) 1 Hz

Milliseconds (ms) 10–3 s kilohertz (KHz) 103 Hz

Microseconds (ms) 10–6 s megahertz (MHz) 106 Hz

Nanoseconds (ns) 10–9 s gigahertz (GHz) 109 Hz

Picoseconds (ps) 10–12 s terahertz (THz) 1012 Hz

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Example 1Example 1

Express a period of 100 ms in microseconds, and express the corresponding frequency in kilohertz.

SolutionSolution

From Table 3.1 we find the equivalent of 1 ms.We make the following substitutions:100 ms = 100 10-3 s = 100 10-3 10 s = 105 s

Now we use the inverse relationship to find the frequency, changing hertz to kilohertz100 ms = 100 10-3 s = 10-1 s f = 1/10-1 Hz = 10 10-3 KHz = 10-2 KHz

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Phase

Phase describes the position of the waveform relative to time zero.

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Example 2Example 2

A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in degrees and radians?

SolutionSolution

We know that one complete cycle is 360 degrees.

Therefore, 1/6 cycle is

(1/6) 360 = 60 degrees = 60 x 2 /360 rad = /3 rad or

1.046 rad

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ftA 2sin

where

A = Amplitude

f = frequency

= phase

Time-Domain Signal Representation

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An analog signal is best represented in the frequency domain.

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Time-Frequency Domain

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Time-Frequency Domain

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Time-Frequency Domain

18WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998

Examples

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Fourier Decomposition for Periodic Signals

k

kkk tfAts 2sin

...2sin2sin2sin 333222111 tfAtfAtfA

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1

2sin4

k

tkfk

A

...52sin5

432sin

3

42sin

4 tf

Atf

Aft

Ats

Example(1)

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Example(2)

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Example(3)

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Frequency Content of a Square Wave

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Figure 4-19

WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998

Harmonics of a Digital Signal

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Transmission Medium Imperfection

The bandwidth is a property of a medium: The bandwidth is a property of a medium: It is the difference between the highest and It is the difference between the highest and the lowest frequencies that the medium can the lowest frequencies that the medium can

satisfactorily pass.satisfactorily pass.

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Frequency Response of a Medium

Signal frequency content Frequency response of the medium

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Example 3Example 3

If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is the bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V.

SolutionSolution

B = fh  fl = 900 100 = 800 HzThe spectrum has only five spikes, at 100, 300, 500, 700, and 900

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Example 4Example 4

A signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all integral frequencies of the same amplitude.

SolutionSolution

B = fB = fhh f fll

20 = 60 20 = 60 ffll

ffll = 60 = 60 20 = 40 Hz20 = 40 Hz

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Example 5Example 5

A signal has a spectrum with frequencies between 1000 and 2000 Hz (bandwidth of 1000 Hz). A medium can pass frequencies from 3000 to 4000 Hz (a bandwidth of 1000 Hz). Can this signal faithfully pass through this medium?

SolutionSolution

The answer is definitely no. Although the signal can have The answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the range does not the same bandwidth (1000 Hz), the range does not overlap. The medium can only pass the frequencies overlap. The medium can only pass the frequencies between 3000 and 4000 Hz; the signal is totally lost.between 3000 and 4000 Hz; the signal is totally lost.

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Digital Signals

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Bit rate (R) & period of a bit (T)

1 1T or R

R T

The capacity of a data communication system can be expressed in terms of the number of data bits sent per second in time. – bits per second or (bits/sec)

This is also refers to the data rate (R) of the system – speed of transmission.

If the period of a data bit is T then R is the inverse of T.

The period of data bit is called bit interval.

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Example 6Example 6

A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval)

SolutionSolution

The bit interval is the inverse of the bit rate.

Bit interval = 1/ 2000 s = 0.000500 s = 0.000500 x 106 s = 500 s

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Transmission Impairment

Attenuation

Distortion

Noise

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Attenuation

1

2log10P

PdB

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Imagine a signal travels through a transmission medium and its power is reduced to half. This means that P2 = 1/2 P1. In this case, the attenuation (loss of power) can be calculated as

SolutionSolution

10 log10 log1010 (P2/P1) = 10 log (P2/P1) = 10 log1010 (0.5P1/P1) = 10 log (0.5P1/P1) = 10 log1010 (0.5) (0.5)

= 10(–0.3) = –3 dB = 10(–0.3) = –3 dB

Example 12Example 12

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Example 13Example 13

Imagine a signal travels through an amplifier and its power is increased ten times. This means that P2 = 10 ¥ P1. In this case, the amplification (gain of power) can be calculated as

10 log10 log1010 (P2/P1) = 10 log (P2/P1) = 10 log1010 (10P1/P1) (10P1/P1)

= 10 log= 10 log1010 (10) = 10 (1) = 10 dB (10) = 10 (1) = 10 dB

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Example 14Example 14

One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are talking about several points instead of just two (cascading). In Figure 3.22 a signal travels a long distance from point 1 to point 4. The signal is attenuated by the time it reaches point 2. Between points 2 and 3, the signal is amplified. Again, between points 3 and 4, the signal is attenuated. We can find the resultant decibel for the signal just by adding the decibel measurements between each set of points.

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dB = –3 + 7 – 3 = +1

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Distortion

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Noise

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Analog Modulation

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Amplitude Modulation

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Amplitude Modulation

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Amplitude Modulation

The total bandwidth required for AM can be determined from the bandwidth

of the audio signal: rule-of thumb: BWt = 2 x BWm.

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AM band allocation

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Example 13Example 13

We have an audio signal with a bandwidth of 4 KHz. What is the bandwidth needed if we modulate the signal using AM?

SolutionSolution

An AM signal requires twice the bandwidth of the original signal:

BW = 2 x 4 KHz = 8 KHz

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Frequency Modulation

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Frequency Modulation

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Frequency Modulation

The total bandwidth required for FM can be determined from the bandwidth

of the audio signal: rule-of thumb: BWt = 10 x BWm.

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Frequency Modulation

The bandwidth of a stereo audio signal is usually 15 KHz. Therefore, an FM station needs at least a bandwidth of

150 KHz. The FCC requires the minimum bandwidth to be at least 200

KHz (0.2 MHz).

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FM Band Allocation

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Example 14Example 14

We have an audio signal with a bandwidth of 4 MHz. What is the bandwidth needed if we modulate the signal using FM?

SolutionSolution

An FM signal requires 10 times the bandwidth of the original signal:

BW = 10 x 4 MHz = 40 MHz

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Tutorial and Examples1- Describe the three characteristics of a sine wave.

a. amplitude,b. frequency/period,c. phase

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Tutorial and Examples2- Describe a sine signal using the unit circle.

xy sin where x is in radians.

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Tutorial and Examples2- Describe a cosine signal using the unit circle.

xy cos where x is in radians.

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Tutorial and Examples3-What are the amplitude, phase, and frequency characteristics of the cosine signal that is shown below.

31002cos10

t

Amplitude= 10Frequency=100 HzPhase= Π/3 radians=60 degrees

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Tutorial and Examples4-What is the Fourier transform of the signal that is shown below.

cAtx

dtetxfX tfj

2

fAdteAfX ctfj

c

2 Dirac-Delta function

Dirac-Delta function is very useful as an approximation for tall narrow spike functions.

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Properties of Dirac-Delta functions:

0,0

0,

xfor

xforx1-

2-

1dxx

0

As red lines approaches to zero, the green line approaches to infinity.

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cAtx

t

x(t)

Ac

0

fAfX c

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Tutorial and Examples5-What is the Fourier transform of the cosine signal shown below?

tfAtx cc 2cos

tfjtfe cctfj c 2sin2cos2

tfjtfe cctfj c 2sin2cos2

2

2cos22 tfjtfj

c

cc eetf

Euler’s Equations

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xjxz sincos

xjxdx

dzcossin

xjxjxjxjdx

dzsincoscossin2

zjdx

dz

dxjdzz

1

dxjdzz

1

xjz ln xjxez xj sincos

To prove Euler’s equations

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dtetfAdtetxfX tfjcc

tfj

22 2cos

dteee

AfX tfjtfjtfj

c

cc

2

22

2

dteA

dteA

fX tffjctffjc cc 22

22

cff cff

cc

cc ff

Aff

AfX

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Hence, the Fourier transform of : tfAtx cc 2cos