chapter 16 communication

CAMBRIDGE A – LEVEL

PHYSICS

COMMUNICATION COMMUNICATION

MODULATIONI. What is modulation?I. What is modulation?

• When we tune to our favourite radiostation, we tune to a particularfrequency; e.g. 94.5 MHz, 92.9 MHz.

• These frequencies are the frequenciesof the carrier signal, a very highfrequency signal that “transports” theinformation signal.

MODULATIONI. What is modulation (cont’d)?I. What is modulation (cont’d)?

• The information signal, e.g. audio, video,media, is the information send out by thetransmitter.

• Modulation is the variation of either theamplitude or frequency of the carriersignal in synchrony with a property ofthe information signal.

MODULATIONII. Why modulation?II. Why modulation?

a. To prevent interference of differentsources. For example, if radio stationstransmit using human audible range,information from different sources wouldinterfere.

b. The carrier signal is a high frequencysignal. Modulating the information signalwith a high energy signal increases theenergy content of the modulated signal.

MODULATIONII. Why modulation?II. Why modulation?

c. The modulated signal has a highfrequency, thus a shorter wavelength.Antennae that receive the signal must

have length ��

�λ. If the frequency of

the modulated signal is low, theantennae size has to be longer, toextend of kilometres.

MODULATION

III. Two types of modulation:

• We will learn two kinds of

modulation:

a. amplitude modulation (AM),

and

b. frequency modulation (FM).

MODULATION

a. Amplitude modulation (AM):a. Amplitude modulation (AM):

• Definition: “In amplitudemodulation, the amplitude ofthe carrier signal is made to varyin synchrony with thedisplacement of the informationsignal.”

MODULATION

a. Amplitude modulation (cont’d):Fig. 3.2 , page 26,

A – Level Science

Applications

Booklet: Physics,

University of

Cambridge

International

Examinations,

Cambridge,

England, 2006.

carrier wave

information signal

MODULATION

a. Amplitude modulation (cont’d):

Fig. 3.2 , page 26, A – Level Science Applications Booklet: Physics, University of

Cambridge International Examinations, Cambridge, England, 2006.

modulated wave

MODULATIONa. Amplitude modulation (cont’d):a. Amplitude modulation (cont’d):

• A few observations:– The information signal forms an “envelope”

around the carrier signal to produce themodulated signal. This means that theinformation signal cannot have an amplitudelarger than the carrier signal.

– By finding the time for two successive “loops”in the modulated signal, we can obtain theperiod of the information signal, hence itsfrequency.

MODULATIONa. Amplitude modulation (cont’d):a. Amplitude modulation (cont’d):

• By using mathematical analysis, we canobtain the frequency spectrum of anamplitude modulated signal.

• The frequency spectrum shows us thesignal power versus frequencycomponents of the signal we haveanalysed.

MODULATIONa. Amplitude modulation (cont’d):

�


• The graph on the next slide showsthe frequency spectrum of aparticular amplitude modulatedsignal.

• The information signal is of a singlefrequency, �.

MODULATION

a. Amplitude modulation (cont’d):Figure 20.5, page

313, Chapter 20:

Communications

Systems;

Cambridge

International AS

and A Level Physics

Coursebook, Sang,

Jones, Chadha and

Woodside, 2nd

edition, Cambridge

University Press,

Cambridge,

UK,2014.

MODULATIONa. Amplitude modulation (cont’d):

� �

� � ��


• The term �� frequency of the carrierwave while �� frequency of theinformation signal (sidebandfrequencies).

• The spectrum shows that the lowestfrequency present (the lower sideband)��

MODULATION

� ��


• The spectrum shows that the highest

frequency present (the upper sideband)

��

• The range between the lower and upper

sidebands is known as the bandwidth.

The bandwidth (BW) � ��

MODULATION


• Why is the frequency spectrum

important? The receiver must be

capable of receiving all the

frequencies in the bandwidth,

otherwise some of the information

will be lost.

MODULATION


• How will the frequency spectrum look like if

the information signal has a range of

frequencies, as seen below?Fig. 3.5 , page 28,

A – Level Science

Applications

Booklet: Physics,

University of

Cambridge

International

Examinations,

Cambridge,

England, 2006.

MODULATION


• We will obtain a frequency spectrum as shown

below



←-----Bandwidth -------→

MODULATION

a. Amplitude modulation (cont’d):Questions 2 and 3,

page 311, Chapter

20:

Communications

Systems; Cambridge

International AS

and A Level Physics

Coursebook, Sang,

Jones, Chadha and

Woodside, 2nd

edition, Cambridge

University Press,

Cambridge,

UK,2014.

MODULATION

a. Amplitude modulation (cont’d):Question 6, page

314, Chapter 20:

Communications

Systems;

Cambridge

International AS

and A Level Physics

Coursebook, Sang,

Jones, Chadha and

Woodside, 2nd

edition, Cambridge

University Press,

Cambridge,

UK,2014.

MODULATION

b. Frequency modulation (FM):

• Definition: “In frequency

modulation, the frequency of the

carrier signal is made to vary in

synchrony with the displacement

of the information signal.”

MODULATION

b. Frequency modulation (cont’d)Fig. 3.2 , page 26,

A – Level Science

Applications

Booklet: Physics,

University of

Cambridge

International

Examinations,

Cambridge,

England, 2006.

carrier wave

information signal

MODULATION

b. Frequency modulation (cont’d)



MODULATIONb. Frequency modulation (cont’d):b. Frequency modulation (cont’d):

• A few observations:–The amplitude of the modulated wave

is constant.

–The frequency of the modulated waveincreases as the displacement of theinformation signal increases and ismaximum when the displacement ofthe information signal is maximum.

MODULATIONb. Frequency modulation (cont’d):b. Frequency modulation (cont’d):

• A few observations (cont’d):–For negative values of displacement of

the information signal, the frequencyof the modulated wave decreases. Thefrequency of the modulated signal isminimum when the displacement ofthe information signal has the largestnegative value.

MODULATION

b. Frequency modulation (cont’d):Questions 4 and 5,

page 312, Chapter

20:

Communications

Systems;

Cambridge

International AS

and A Level Physics

Coursebook, Sang,

Jones, Chadha and

Woodside, 2nd

edition, Cambridge

University Press,

Cambridge,

UK,2014.

MODULATION

IV. Comparing AM and FM:

• Noise and electrical interference (e.g.

external noise, lightning, etc. ) effect

the amplitude of a signal, not its

frequency. AM signals are prone to

interference due to electrical impulses

compared to FM signals.

MODULATION

IV. Comparing AM and FM (cont’d):

• AM signals have a bandwidth of 9 kHz.

This means that the maximum frequency

of the information signal is 4.5 kHz. FM

signals have a typical bandwidth of about

200 kHz. FM signals would have

frequencies of 15 kHz and higher, which

leads to better quality of sound.

MODULATIONIV. Comparing AM and FM (cont’d):IV. Comparing AM and FM (cont’d):

• AM signals have a longer wavelength. Thismeans that AM signals from a singletransmitter can travel a greater distancedue to diffraction.

• AM signals have a smaller bandwidth. Thismeans that more stations can transmitusing AM for a given frequency spectrum.

MODULATION


• The electronic circuits for AM

transmission is cheaper and less

complex compared to FM

transmission.

• The table on the next slide summarises

the relative advantages of AM and FM.

MODULATION


Table 20.1 , page 314, Chapter 20: Communications Systems; Cambridge International

AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition,

Cambridge University Press, Cambridge, UK,2014.

MODULATION

IV. Comparing AM and FM (cont’d):Questions 7, 8 and

9, page 314,

Chapter 20:

Communications

Systems; Cambridge

International AS

and A Level Physics

Coursebook, Sang,

Jones, Chadha and

Woodside, 2nd

edition, Cambridge

University Press,

Cambridge,

UK,2014.

HOMEWORKHOMEWORKModulation:

1. Question 11, Paper 4, Summer 2008.

2. Question 11, Paper 42, Winter 2009.




D I G I TA L DATA

T R A N S M I S S I O N

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T R A N S M I S S I O NWhat is the difference between digital and analogueWhat is the difference between digital and analoguesignals?

I. Analogue signals:

• Analogue signals are signals that canhave any value in between someprescribed limits.

• Examples: voltage signals produced by amicrophone

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What is the difference between digital and analogue

signals?

II. Digital signals:

• Digital signals are signals that consist of

sequence of values of 0s and 1s.

• Examples: The sequence 011110 is a

digital signal.

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D I G I TA L DATA

T R A N S M I S S I O NDecimal numbers and binary numbers:Decimal numbers and binary numbers:

• Decimal digits can have any valuebetween 0 and 9.

• We can represent numbers using thedecimal representation by using asequence of decimal digits.

• Example: 124, 302345 are all numbersrepresented using the decimal digits.

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T R A N S M I S S I O NDecimal numbers and binary numbers (cont’d):Decimal numbers and binary numbers (cont’d):

• Binary digits can only have values of 0 or1.

• We can represent numbers using thebinary representation by using asequence of decimal digits.

• Example: 10110, 111001 are all numbersrepresented using the binary digits.

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I. Conversion of a decimal number to its

binary equivalent:

• To convert a decimal number into its

binary equivalent, we perform long

division of the decimal number by 2.

• The remainders of the division

process give us the binary number.

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T R A N S M I S S I O NI. Conversion of a decimal number to its binaryI. Conversion of a decimal number to its binary

equivalent:

• The resulting binary number is read up fromthe most significant bit (MSB, the lastremainder) up to the least significant bit(LSB, the first remainder).

• We will look at an example in the next slidethat converts 156 (a decimal number) into itsbinary equivalent.

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I. Conversion of a decimal to a binary number:

Source:

http://www.wikihow.

com/Image:Convert-

from-Decimal-to-

Binary-Step-4-

Version-2.jpg

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D I G I TA L DATA


II. Conversion of a binary number to its

decimal equivalent:

• A binary number can be converted

into its decimal equivalent by

multiplying each binary digit (bit) in

the sequence by its weight, and

then sum the products.

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D I G I TA L DATA

T R A N S M I S S I O NII. Conversion of a binary number to its

II. Conversion of a binary number to itsdecimal equivalent:

• The weight is equal to � , where� � �, �, , �, ….

• The LSB has weight �2�. The bit to the leftof the LSB will have a weight �2� and soon.

• An example is shown in the next slidewhere the 6 bit binary number 110100 isconverted into its decimal equivalent.

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II. Conversion of a binary number to its

decimal equivalent:

Source: http://www.kkhsou.in/main/EVidya2/electronics/electronic/138.gif

D I G I TA L DATA


D I G I TA L DATA


Binary numbers and decimal numbers (cont’d):

Example:Questions 10 and 11,

page 317, Chapter 20:

Communications

Systems; Cambridge

International AS and A

Level Physics

Coursebook, Sang,

Jones, Chadha and

Woodside, 2nd edition,

Cambridge University

Press, Cambridge,

UK,2014.

D I G I TA L DATA


D I G I TA L DATA

T R A N S M I S S I O NRepresenting decimal numbers as their binary

� �

Representing decimal numbers as their binary equivalent:

• In digital data transmission, we need to representdecimal numbers as their binary equivalent.

• How do we find the minimum number of bits we mustuse to represent a decimal number as its binaryequivalent?

• Answer: We use the equation � � �� ,where:

1. � � minimum number of bits needed (roundedup to the nearest integer), and

2. � � the decimal value

D I G I TA L DATA


D I G I TA L DATA

T R A N S M I S S I O NRepresenting binary numbers as their decimal

� � �

Representing binary numbers as their decimal equivalent:

• In digital data transmission, we often arelimited by the number of the bits we can useto transmit data.

• We often need to find the largest decimalvalue, � that can be represented given thenumber of bits, �.

• The largest decimal value, � � � �

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Example:Example:

• For example, we need 3 bits torepresent the binary numbers 000(decimal 0) to 111 (decimal 7), and weneed at least 4 bits to represent 8(1000).

• Use the equations above to show thatthis is indeed correct.

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Comparing digital data transmission with Comparing digital data transmission with analogue data transmission:

• Data can be transmitted either in digital oranalogue format.

• Here, we will discuss two advantages of digitaldata transmission as compared to analoguedata transmission.

• Before that, we need to understand what ismeant by attenuation and noise.

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I. Attenuation:

• Definition: “Attenuation is thegradual reduction in thepower of a signal as itpropagates”.

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Comparing digital data transmission with

analogue data transmission:

I. Attenuation (cont’d):

• The amplitude of an attenuated signal is

lower than the original signal, since the

attenuated signal carries lower power

compared to the original signal.

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D I G I TA L DATA





• Attenuation is caused by:

a. the transmission medium; the

particles of the medium absorb

some of the power of the signal.

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D I G I TA L DATA




b. the distance that the signalpropagates; recall the inverse squarelaw � ! � "#⁄ �,

c. scattering of the transmitted wave,e.g. light may undergo scattering.

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D I G I TA L DATA

T R A N S M I S S I O NComparing digital data transmission with analogue Comparing digital data transmission with analogue data transmission:


• Attenuation can be overcome by:

a. adding repeaters (amplifiers of analoguesignals),

b. adding regenerators (amplifiers of digitalsignals)

along the length of the propagation medium.

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D I G I TA L DATA


II. Noise:

• Noise (or electrical noise) is theunwanted random power that addsitself to the signal.

• This electrical noise distorts thetransmitted signal.

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D I G I TA L DATA



II. Noise (cont’d):

• Noise is caused by the thermalvibrations of the particles of themedium through which the signal istransmitted. Hence, noise cannot beeliminated.

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• Sources of noise:

a. Internal sources such as the thermal

vibrations of the particles of the medium

through which the signal is transmitted.

This noise cannot be eliminated.

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D I G I TA L DATA



• Sources of noise (cont’d):b. External sources such as electrical storms,

electrostatic interference,electromagnetic interference (due toelectric currents), and radio frequencyinterference (due to radiation of noise inradio frequency and radio signals).

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D I G I TA L DATA




• Data transmitted in either analogue or digital

forms will be subject to both noise and

attenuation.

• Hence, we need to use repeaters (for

analogue signals) and regenerators (for digital

signals).

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D I G I TA L DATA




• Repeaters, however, also amplify the noise

together with the transmitted signal.



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• Repeaters, however, also amplify the noise

together with the transmitted signal.

• This causes the received signal to be a very

‘noisy’ version of the original.

• It will be hard for the receiver to recover the

original signal.

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D I G I TA L DATA




• Regenerators, need only to produce ‘high’ (a

1 bit) or ‘low’ (a 0 bit) values, hence, they do

not amplify the noise.



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D I G I TA L DATA




• Hence, we can transmit digital signals

over long distances, and by using

regenerators, we are able to recover the

original signal, without the effects of

noise.

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D I G I TA L DATA




• Another advantage of digital data

transmission is that, we can add

additional data bits for the purpose of

error correction and detection. This

minimises errors in the received signal.

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Analogue to digital conversion (ADC):

• Analogue to digital conversion (ADC) is

the conversion of an analogue signal to

its digital equivalent at the transmitter.

• Hence, we can transmit the digitised

version of an analogue signal by first

doing the ADC.

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D I G I TA L DATA



• Two processes that need to be

done during ADC are:

a. sampling, and

b. quantisation.

D I G I TA L DATA


D I G I TA L DATA

T R A N S M I S S I O NAnalogue to digital conversion (ADC):

%&


a. Sampling:

• Sampling involves obtaining values ofthe analogue signal at regular timeintervals.

• The values are known as the samples.

• The regular time intervals are known asthe sampling period, %&.

D I G I TA L DATA


D I G I TA L DATA


&�

%&


a. Sampling (cont’d):

• The sampling frequency, &, is the

number of samples obtained each

second.

• Mathematically, &�

%&

D I G I TA L DATA


D I G I TA L DATA


Analogue to digital conversion (ADC):Analogue to digital conversion (ADC):

a. Sampling (cont’d):

• Sampling produces a discrete versionof the analogue signal; the signalnow has values only at specific times.

• This is seen in the image on the nextslide.

D I G I TA L DATA


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a. Sampling (cont’d):Source:

https://cnx.org/resou

rces/89686185b0871b

5b4a5172891051a3d5

7917b326/analog_sa

mpling.jpg

D I G I TA L DATA


D I G I TA L DATA



b. Quantisation:

• Quantisation is the process in whichthe values of the samples isconverted into binary numbers (thequantised value) based on mappingvalues.

D I G I TA L DATA


D I G I TA L DATA

T R A N S M I S S I O NAnalogue to digital conversion (ADC):Analogue to digital conversion (ADC):

b. Quantisation (cont’d):

• To find the mapping values, we firstfind the number of quantisationstates given , the number of bitsby using the equation �.

• We then assign each state to itsbinary representation.

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� � 3�



• For example, if we have 3 bits (� � 3�,we will ( � 8 number of quantisationstates.

• The table in the next slide gives all thepossible states and its binaryequivalent.

D I G I TA L DATA


D I G I TA L DATA



Quantisation state Binary representation

0 000

1 001

2 010

3 011

4 100

5 101

6 110

7 111

D I G I TA L DATA


D I G I TA L DATA


�*+



• We then find the analoguequantisation size, .

• To do this, we get the largestvoltage value of the analogue signal,

�,- and also the lowest voltagevalue of the analogue signal, �*+.

D I G I TA L DATA


D I G I TA L DATA




• We then use the formula.�/01.�2�

�.

• The value of Q gives us the size ofthe range of decimal numbers thatrepresent each state.

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D I G I TA L DATA




• To do this, we start by assigning

�*+ to the lowest quantisationstate, and then incrementing byto obtain the next state and thesubsequent states.

• The next slide shows an example.

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D I G I TA L DATA


3 � 0.0 3 � 10.0



• Let us say 3�*+ � 0.0V and 3�,- � 10.0V. We also have � � 3.

• We will obtain ( � 8 and 7 �

��.�1�.�

8� 1.25.

• The table on the next slide shows thedecimal value ranges for each quantisedstate.

D I G I TA L DATA


D I G I TA L DATA

T R A N S M I S S I O NQuantisation State Binary Number Decimal value range

(V)

0 000 0.00 – 1.25

1 001 1.25 – 2.50

2 010 2.50 – 3.75

3 011 3.75 – 5.00

4 100 5.00 – 6.25

5 101 6.25 – 7.50

6 110 7.50 – 8.75

7 111 8.75 – 10.0

D I G I TA L DATA


D I G I TA L DATA



• ADC involves:

a. sampling the analogue signal, then

b. quantising the value of the sampleby finding in which range does thevalue of the sample lie in, and

c. encoding into a binaryrepresentation .

D I G I TA L DATA


D I G I TA L DATA

T R A N S M I S S I O NDigital to analogue conversion (DAC) (cont’d):Digital to analogue conversion (DAC) (cont’d):

• DAC is done at the receiver to recover theoriginal analogue signal from the receiveddigital signal at the receiver.

• We use a table, similar to the one on page73, to obtain each received binary numberto the lowest value of the correspondingdecimal range.

• The conversion table is the same fortransmitter and receiver.

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Digital to analogue conversion (DAC) (cont’d):

2 : ; < ; = 3 :;

Digital to analogue conversion (DAC) (cont’d):

• For example, using the previous example,if we receive a sequence 010111001, weget:

1. 1.25 V for 0 < ; = ;>,

2. 8.75 V for ;> < ; = 2 : ;>, and

3. 2.50 V for 2 : ;> < ; = 3 :;>respectively.

D I G I TA L DATA


D I G I TA L DATA


Digital to analogue conversion (DAC)

(cont’d):

• Note that each value is held for the

entire sampling period; i.e up till the

next sampled value.

• The analogue signal is then

reconstructed from these values.

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T R A N S M I S S I O NEfficiency factors:Efficiency factors:

• When we reconstruct the digital signal atthe receiver to recover the analoguesignal, two factors that effect thereproduction of the signal are:I. the sampling rate (sampling frequency),

and

II. the number of bits used for quantisation(the number of quantisation states)

D I G I TA L DATA


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T R A N S M I S S I O NI. Sampling rate:I. Sampling rate:

• A higher sampling rate increases thenumber of samples obtained.

• If we use an higher sampling rate, thenumber of samples obtained would belarger, hence the reproduced analoguesignal (at receiver) will be more similar tothe input analogue signal (attransmitter).

D I G I TA L DATA


D I G I TA L DATA

T R A N S M I S S I O NI. Number of quantisation bits:I. Number of quantisation bits:

• When we use too few bits forquantisation, we produce a higherquantisation error the reproducedanalogue signal.

• By increasing the number of quantisationbits, we reduce the decimal range andproduce a smoother output (less‘grainy’).

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Efficiency factors (cont’d):

• We will look at an example of ADC

and DAC of a signal with a low

sampling rate and insufficient number

of quantisation bits in the next few

slides.

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Fig. 3.8 (a) , page 31, A – Level Science Applications Booklet: Physics, University of


D I G I TA L DATA


D I G I TA L DATA



Fig. 3.8 (b)

and (c) , page

31, A – Level

Science

Applications

Booklet:

Physics,

University of

Cambridge

International

Examinations,

Cambridge,

England,

2006.

D I G I TA L DATA


D I G I TA L DATA



Fig. 3.8 (d) , page 31, A – Level Science Applications Booklet: Physics, University of


HOMEWORKHOMEWORKDigital Data Transmission

1. Question 12, Paper 41, Winter 2009 (except part

(c)).

2. Question 12, Paper 43, Winter 2010 (except part b

(ii)).


GAIN CALCULATION

Gain calculations:Gain calculations:

• The gain of a signal, G, in decibels (dB),

is given by ? � ��@�ABCAB

@2�CAB, where:

I. DEFGHFG � output power,

II. D*+HFG � input power,

III. DEFGHFG and D*+HFG must have the same

units

GAIN CALCULATION

Gain calculations (cont’d):

DEFGHFG D*+HFG


• The gain of a signal can also be quoted inunits of Bels (B).

• The gain of a signal, G, in bels (B), is given by

? � ��@�ABCAB

@2�CAB, where:

I. DEFGHFG � output power,

II. D*+HFG � input power,

III. DEFGHFG and D*+HFG must have the same units

GAIN CALCULATION

Gain calculations (cont’d):Gain calculations (cont’d):

• The gain of a signal is often quoted inlogarithmic values as the values may betoo large (e.g. 106) or too small (10-9).

• If the value of the gain is negative, itmeans the signal has been attenuated.

• If the value of the gain is positive, itmeans the signal has been amplified.

GAIN CALCULATION


• Example:

Example , page 37, A – Level Science Applications Booklet: Physics, University of


GAIN CALCULATION


• Examples:

Questions 13 and 14, page 319, Chapter 20: Communications Systems; Cambridge

International AS and A Level Physics Coursebook, Sang, Jones, Chadha and

Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

GAIN CALCULATION


• Example:

Question 16, page 319, Chapter 20: Communications Systems; Cambridge



GAIN CALCULATION


• Since signals travel along distances, it is oftenconvenient for us to specify the attenuationper unit length.

• Mathematically, attenuation per unit length

��

I: ? �

�

I: �� JKL�

@�ABCAB

@2�CAB�.

• The usual units of attenuation per unitlength: dB km-1 or dB m-1.

GAIN CALCULATION


• Example:

Example , page 37, A – Level Science Applications Booklet: Physics, University of


GAIN CALCULATION


• The signal to noise ratio (SNR) of a received

signal is SNRSNRSNRSNR � �� JKL�@&2��/�

@��2&P�.

• The minimum SNR value helps us calculate

the lowest value of Psignal that signal can

have to in order to be distinguished from

any background noise.

GAIN CALCULATION


• If the signal power goes lower than thisminimum value, the receiver would not beable to distinguish the signal from anybackground noise.

• We need to use repeaters (for analoguesignals) or regenerators (for digital signals) tohelp us restore the power of the attenuatedsignal.

GAIN CALCULATION


• Example:

Question 15, page 319, Chapter 20: Communications Systems; Cambridge



HOMEWORKHOMEWORKGain / Attenuation Calculation:


2. Question 12, Paper 42, Sumer 2010.



C O M M U N I C AT I O N

C H A N N E L S


C H A N N E L S

• A communication channel refers to the

medium, or the path, or the actual

frequency range used to transmit

information from the sender to the

receiver.

• We will look at the six different types of

communication channels.


C H A N N E L S


C H A N N E L S

We will discuss the relative advantages and• We will discuss the relative advantages anddisadvantages of these communicationchannels in terms of:

� the available bandwidth,

� noise,

� crosslinking,

� security,

� signal attenuation, and

� use of repeaters or regenerators.


C H A N N E L S


C H A N N E L S

The six different communication channels:

I. Wire - pairs,

II. Coaxial cables,

III. Radio,

IV. Microwave links,

V. Optical fibres, and

VI. Satellite links.


C H A N N E L S


C H A N N E L S

I. Wire - pairs:

• This channel consists of a pair of

insulated copper wires that connect the

transmitter to the receiver.




C H A N N E L S


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I. Wire - pairs (cont’d):I. Wire - pairs (cont’d):

• In modern communication systems, wire- pairs are used for short distance, lowfrequency communication systems.

• This communication channel is notsuitable for high frequencycommunication since signals undergohigh levels of attenuation.


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• In wire - pairs, attenuation occurs to due tothe:

a. energy loss due to the resistance, and

b. radiation emitted since these wires act asaerials.

• To overcome, the effects of attenuation,repeated amplification must be provided torestore the power levels of the signal.


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• Wire - pairs easily pick up externalinterference, degrading the originalsignal and thus increasing the amount ofnoise in the signal.

• The bandwidth of wire – pairs is onlyabout 500 kHz, thus they cannot carry alot of information.


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I. Wire - pairs (cont’d):

• Cross – linking (or cross – talk) occurs when

one wire – pair picks up another’s signal. Cross

– linking occurs when two or more wire pairs

are lined up next to each other. Cross – linking

reduces the security of this communication

channel.


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II. Coaxial cables:II. Coaxial cables:

• Coaxial cables are made up of a copperwire, covered by a polythene insulator. Acopper braid covers the polytheneinsulator, which in turn is covered by aplastic covering.

• This is shown in the diagram on the nextslide.


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II. Coaxial cables (cont’d):

Fig. 3.10 , page 32, A – Level Science Applications Booklet: Physics,

University of Cambridge International Examinations, Cambridge,

England, 2006.


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II. Coaxial cables (cont’d):II. Coaxial cables (cont’d):

• When using coaxial cables, the signal istransmitted using the inner conductor, whilethe outer conductor acts as the return wire.The outer conductor also shields the innerconductor from external interference.

• The bandwidth of coaxial cables are about 50MHz. Hence, these cables can carry moreinformation compared to wire – pairs.


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II. Coaxial cables (cont’d):II. Coaxial cables (cont’d):

• Coaxial cables are more expensive than wire– pairs but cause less attenuation to thesignal. Since attenuation is lower, repeateramplifiers (or regenerators) can be placedfurther apart.

• This cables are also less prone to externalinterference, making it more secure thanwire – pairs.


C H A N N E L S


C H A N N E L SIII. Radio waves:III. Radio waves:

• Radio waves are EM waves that have afrequency range between 30 kHz to 3GHz.

• Radio waves are produced due to theoscillations of electrons in aerials/antennae. This oscillations produceenergy that is radiated and propagate atthe speed of light.


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III. Radio waves (cont’d):

• Radio waves can be classified, based

on frequency, as either:

I. surface waves,

II. sky waves, or

III. space waves


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I. Surface waves:

• travel close to Earth’s surface,

• have frequency below 3 MHz,

• have a range of up to 1000 km since they

have long wavelengths and will diffract

around buildings or other structures.


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I. Surface waves (cont’d):

• are used in the LW (long wave) and

MW (medium wave) radio in the LF

(low frequency) band.


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II. Sky waves:

• have frequencies between 3 MHz to 30

MHz,

• due to their shorter wavelengths (relative

to surface waves), tend to travel in

straight lines (little diffraction).


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II. Sky waves (cont’d):

• can travel long distances worldwide

via multiple reflections by the Earth’s

surface and the ionosphere (a layer of

the Earth’s atmosphere).


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II. Sky waves (cont’d):

• problem: the density of the ionosphere is

not constant, hence the reflection by the

ionosphere is not reliable.

• used by SW (short wave) radio in the HF

(high frequency) band.


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III. Space waves:

• are radio waves that have frequencies

greater than 30 MHz.

• transmission is line – of – sight, i.e.

there must a clear line between

transmitter and receiver.


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III. Space waves (cont’d):

• are used for TV broadcast (in the ultra

high frequency (UHF) band) and for

FM radio broadcast (in the very high

frequency (VHF) band).


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III. Space waves (cont’d):

• The VHF and UHF bands are also used

for short range communication, e.g. in

walkie – talkies, mobile phones since

they have short wavelengths, hence

the length of aerial is short.


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• The bandwidth of the radio link increases

as the frequency of the carrier wave

increases.

• The table on the next slide summarises

the part of the EM spectrum used for

radio communication.


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Fig. 3.12, page 33, A – Level Science Applications Booklet: Physics, University of



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IV. Microwaves:

• Microwaves are radio waves in the Super

High Frequency (SHF) band.

• The SHF lies between 3 GHz to 30 GHz.

• The wavelength of microwaves are about

a few centimetres.


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IV. Microwaves (cont’d):IV. Microwaves (cont’d):

• Microwaves are used in Bluetooth, Wi –Fi communication links.

• Microwaves are commonly used for point– to – point communication.

• The diagram on the next slide shows theparabolic microwave transmitter andreceiver.


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IV. Microwaves (cont’d):

Fig. 3.13 , page 34, A – Level Science Applications Booklet: Physics,

University of Cambridge International Examinations, Cambridge,

England, 2006.


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• The transmitting element is placed at thefocal point of the parabolic mirror. Theradiated are reflected off the surface andare parallel.

• A parabolic reflector (at the receiver)focuses the parallel beam to a receivingelement.


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• The reflectors are not antennae.They function to:� to focus as much power as possible into the

parallel beams (at the transmitter), and

� collecting as much power as possible anddirecting this power to the receivingelement.


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• Parabolic dishes are most useful for shortwavelengths where the spreading of thewaves due to diffraction is limited.

• The bandwidth of the microwave linksare in the order of GHz. Hence,microwave links have a large capacity ofinformation.


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• For terrestrial use, microwave links are

limited to line – of – sight.

• To overcome this issue, we use repeaters.

We may also use a satellite to retransmit

when the transmitter and receiver do

not have a line – of – sight.


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• The beams that travel between the

transmitter and receiver are very narrow

and do not spread out. This means that it

is difficult to tap into the information

carried by the microwave beams.


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C H A N N E L S

�"EQ+

V. Satellite links:

• In a satellite communication, a transmission

tower sends a carrier wave of frequency �FH

to the satellite.

• The satellite, upon receiving the signal,

amplifies the signal, and changes the carrier

frequency to a lower frequency, �"EQ+, and

transmits this to the receiving tower.


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V. Satellite links:

• A satellite link is shown below.




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V. Satellite links (cont’d):

• The upward link is known as the uplink,

while the downward link is known as the

downlink.

• The uplink will have a higher frequency than

the downlink since the transmitting tower

will have more access to power (on Earth) as

compared to the satellite (in space).


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V. Satellite links (cont’d):V. Satellite links (cont’d):

• The uplink and downlink both havedifferent frequencies to prevent theoriginal signal send from Earth from over- swamping the signal retransmitted bythe satellite.

• Typical frequency bands used: 6/4 GHz,14/11 GHz, 30/20 GHz.


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• Satellite links are preferred to sky wavesbecause:a. the constantly changing concentration of the

ionosphere, making reflection of the sky waves notalways possible,

b. the downlink signal has more power than a signalreflected by the ionosphere, and

c. It uses higher frequencies, making the bandwidthhigher.


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• The satellites used can have twotypes of orbits:

1. polar orbits, or

2. geostationary orbits.

• We will look at both types of orbitsin a little bit of detail.


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C H A N N E L SV. Satellite links (cont’d):V. Satellite links (cont’d):

1. Polar orbits:

• Satellites in polar orbits travel frompole to pole in an orbital period ofabout 90 minutes.

• These satellites can cover the entiresurface of the Earth in 24 hours sincethe Earth also rotates below thesatellite as the satellite orbits.


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1. Polar orbits:

• The diagram below shows a satellite in polar orbit.

Fig. 3.15 , page 35, A –

Level Science

Applications Booklet:

Physics, University of

Cambridge

International

Examinations,

Cambridge, England,

2006.


C H A N N E L S



1. Polar orbits (cont’d):

• Satellites in these orbits have analtitude of about 1000 km.

• Due to their low altitude, they candetect objects of smaller detail. Theyare suitable for monitoring the Earth’ssurface, weather forecasting andespionage.


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• Due to their low altitudes also, there

is a smaller delay time between

transmission and reception of

signals.


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• Satellite dishes on the Earth need tomoved to communicate constantly withthese satellites since polar orbit satellitesare not always at the same positionrelative to the Earth.

• To maintain constant coverage, a networkof linked satellites must be used.


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2. Geostationary orbits:

• Geostationary satellites are satellitesthat have an orbital period of 24 hoursand have an altitude of 35800 km.

• Geostationary satellites have equatorialorbits; i.e. they are directly above theEarth’s equator.


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C H A N N E L SV. Satellite links (cont’d):


• The diagram below shows a geostationary satellite in its

orbit.




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• If a satellite has the same direction ofrotation as the Earth’s, then to anobserver on the Earth, that satellite willremain stationary.

• Hence, geostationary satellites appearstationary above a fixed position on theEarth’s equator.


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• These satellites are useful for TVbroadcast, e.g. MEASAT satellite.

• A network of linked geostationarysatellites can be used for trans –oceanic telephone calls.


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• Among the disadvantages of

geostationary orbits:

� high altitude means longer delay time,

� polar regions cannot be reached.


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C H A N N E L S

VI. Optical fibres:VI. Optical fibres:

• Optical fibres are very thin glass orplastic fibres that carry infra – redwaves or light waves.

• For long distances, glass and infra – redcombination is used since thiscombination as the attenuation andscattering of the waves is minimised.


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VI. Optical fibres (cont’d):

• The waves in optical fibres propagate over

long distances via total internal reflection

inside the fibre. This is shown in the figure

below.Figure 20.19, page 323, Chapter

20: Communications Systems;

Cambridge International AS and

A Level Physics Coursebook,

Sang, Jones, Chadha and


Cambridge University Press,

Cambridge, UK,2014.


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VI. Optical fibres (cont’d):VI. Optical fibres (cont’d):

• During communication, a laser orLED is caused by an electric signal toemit infra – red or light pulses offrequency of the order 1014 Hz.

• Due to this high frequency, thebandwidth can be very high.


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• A fibre optic cable is made up of

hundreds of fibres. In total, all the

fibres can carry about ten million

phone conversations at a time.


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C H A N N E L S


• Optical fibres are better than satellitesfor long distance communication sincethe time delay will be less.

• This is because the distance travelledby the signal to and from the satellite isconsiderably much greater than whenusing optical fibres.


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C H A N N E L S


• Two disadvantages of using opticalfibres:

� electrical signals must be firstconverted into infra rad or light pulses,and

� it is quite difficult to connect opticalfibre cables together.


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C H A N N E L S


• The advantages of optical fibres over copper

cables is listed below:Table from page 323,

Chapter 20:

Communications

Systems; Cambridge

International AS and A

Level Physics

Coursebook, Sang,

Jones, Chadha and


Cambridge University

Press, Cambridge,

UK,2014.

HOMEWORKHOMEWORKCommunication Channels:Communication Channels:


2. Question 9, Paper 4 (except part (b)), Winter 2008.







HOMEWORKHOMEWORKCommunication Channels (cont’d):

9. Question12, Paper 41, Winter 2012.


chapter 16 communication

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