1
ASSIGNMENT OF DATA COMMUNICATION AND NETWORKING
TABLE OF CONTENTS
Amplitude modulation
2
Amplitude modulation bandwidth
2
Problems
1.1
3
1.3
6
1.5
7
1.7
9
1.9
10
Frequency Modulation
11
Frequency Modulation Bandwidth
11
Problems
2.2
2.4
2.6
2.8
2.10
Satellite Communication
Geosynchronous Satellite
Frequency Band For Satellite Communication
Cellular Telephony
Cellular Band
Receiving
Handoff
Digital
Integration With Satellite And Pcs
12
13
15
17
18
19
20
20
21
22
23
23
23
24
2
Amplitude Modulation (AM)
In am transmission, the carrier signal is modulated so that its
amplitude varies with the changing amplitudes of the modulating signal.
The frequency and phase of carrier remain same; only the amplitude changes
to follow variations in the information.
Amplitude Modulation (AM)
AMPLITUDE MODULATION BANDWIDTH (AMB)
The bandwidth of an AM signal is equal to twice the
bandwidth of the modulating signal and covers a range centered around the
carrier frequency. The bandwidth of an audio signa
is usually 5khz.therefore an am radio station needs a minimum bandwidth
of 10 kHz.
The total bandwidth required for am can be determined from the bandwidth
of the audio signal;
BWt=2*BWm
 BWm =bandwidth of the modulating signal (audio)
 BWt=total bandwidth (radio)
 fc=frequency of the carrier
(1.1) An Audio Signal
3
15Sin2π (1500t)
Amplitude modulation a carrier
60Sin2π (100000t)
(a) Sketch the audio signal
(b) Sketch the Carrier
(c) Construct the modulated wave
(d) Determine the modulation factor and percent modulation
(e) What are the frequencies of the audio signal and carrier
(f) What frequencies would show up in the spectrum analysis if the
modulation wave
Solution
Given
Audio Signal = 15Sin2π (1500t)
Carrier = 60Sin2π (100000t)
Find
(a) Sketch the audio signal
(b) Sketch the Carrier
(c) Sketch of modulated wave
(d) m, M
(e) ƒa , ƒc
(f) Frequency content of modulation wave
(a)
4
(b)
(c)
Converting modulation factor and percent modulation
M=m x 100
=0.25 x
100
M=25%
5
(d) Using the following question for modulation factor
audio amplitude
m=
B
=
Carrier amplitude
A
Since
Carrier = A Sin 2πƒc t
= 60Sin 2π(100000t) Hz
ƒc = 100000 Hz
(f) The frequency spectrum of an amplitude modulation wave consist
of
ƒc , ƒc + ƒa , And ƒc - ƒa
ƒc = 100000 HZ
ƒc + ƒa=100000 + 1500 =101500 Hz
ƒc - ƒa=100000 - 1500 =98500 Hz
The frequency content of the modulated wave is
100000 Hz
101500 Hz
98500 Hz
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(1.3) How many AM broadcast station can be accommodated in a
100KHZ bandwidth if highest frequency modulation carrier
is 5KHZ
Solution
Give:
BW = 100KHz
ƒa max = 5KHz
Find:
Number of station
Any station being modulated a 5-KHz signal will produce an
upper side frequency 5 KHz above its carrier and a lower-side
frequency 5KHz below its carrier .there by requiring a band–
-width of 10 KHz
Total BW
Number of station accommodated =
BW per Station
100 X 10
=
10 X 10 3
Number of station accommodated = 10 Station
7
(1.5)
The total power contain of an AM signal is 1000 W. Determine the
power being transmitted at the carrier frequency and at each of the
sideband when the percent modulation is 100%
Solution
Given:
Pr = 100 w
M = 100%; there fore m=1
Find:
Pc, PUSB + PLSB
From the equation of total power ,
Pr
Pc
=
m2 Pc
+
Pc
=
Pc
1000=
Pc
1000=
+
+
4
m2 Pc
+
4
m2 Pc
+
2
(1.0)2 Pc
2
0.5 Pc
=Pc +
= 1.5 Pc
1000
= Pc
1.5
Pc =666.67W
m2 Pc
4
8
This level 1000 – 666.67 = 333.33 W to be share equally between
upper and lower sideband
PUSB + PLSB = 333.33W
PUSB = PLSB
2 PLSB = 333.33
333.33
PLSB = PUSB =
2
= 166.66 w
Pc =666.67W
PLSB = PUSB= 166.66 w
9
(1.7)
The power content of the carrier of an AM wave is 5 KW
Determine the power content of each of the side band and the
total power transmitted when the carrier the modulation 75%
Solution:
Given : Pc=5 KW
M=5%, m=0.75
Find : Pr, PUSB , PLSB
Since in an AM Wave the powering each of the power band
side is equal
m2 Pc
PUSB = PLSB=
2
(0.75)(5000)
=
4
PUSB = PLSB=703.13W
The total power is the sum of the carrier power and the power
in the two side band
Pr = Pc + PUSB + PLSB
= 5000+703.13+703.13
Pr =6406.26W
10
(1.9)
Determine the percent modulation of the amplitude modulation wave
which has a power content at the carrier of 8KW and 2KW in each of
its sidebands when the carrier is modulation by a simple audio tone
Solution
Given :
Pc =8KW
PUSB = PLSB= 2KW
Find:
M
Knowing the power content of the sidebands and the carrier the
relationship of sidebands power can be used to determine the
modulation factor . Once the modulation factor is know merely
multiplying it by 1000 providing percent modulation
m2 Pc
PUSB = PLSB=
4
m2 Pc
2 X 103 =
4
(4)(2) 103
m2 =
8 103
m=1.0
M=m=100
M=100%
11
FREQUENCY MODULATION (FM)
In FM transmission the frequency of the carrier signal is
modulated to follow the changing voltage level (amplitude) of the
modulation signal. The peak amplitude and phase of the carrier signal
remain constant, but as the amplitude of the information signal changed, the
frequency of the carrier changes correspondingly.
The figure frequency modulation
Frequency modulation bandwidth (FMB)
The bandwidth of an fm signal is equal to10 time to
bandwidth of the modulation signal and like AM bandwidths, covers a range
centered around the carrier frequency. The ban- dwidth of an audio signal
broadest in stereo is almost15khzdach fm radio station, therefore needs a
minimum bandwidth of 150 kHz.
The total bandwidth required for FM can be determined from
the bandwidth of the audio signal;
BWt=10*BWm
 BWm =bandwidth of the modulating signal (audio)
 BWt=total bandwidth (radio)
 fc=frequency of the carrier
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(2.2)
Determine the frequency deviation and carrier swing for a frequency
modulation signal which has a resting frequency of 105.000 MHz and
whose upper frequency is 105.007 MHz when the modulation by the
particular wave .find the lower frequency reaches by the FM wave
Solution
Give:
ƒ0 =105.000 MHz
ƒUpper =105.007 MHz
Find:
Δƒ , cs , ƒlower
Frequency deviation is define as the maximum change in frequency
of the modulation signal away from the rest carrier frequency
Δƒ = (105.007 – 105.000) x 106
= 0.007 x 106
= 7000
Δƒ
= 7 KHz
Carrier swing can now be determine
cs = 2 Δƒ
= 2(7 x 103)
= 14 x 103
cs = 14 KHz
The low frequency reached by the modulation wave can be found by
subtracting the frequency deviation from the carrier or rest frequency.
ƒlower = ƒ0 – Δƒ
=(105.000-0.007) x 106
ƒlower = 104.993 MHz
13
(2.4)
The frequency modulation signal which is modulation by 3KHz
sine wave reach at maximum frequency of 100.02 MHz and
minimum frequency of 99.98 MHz.
(a) Determine the carrier swing
(b) Find the carrier frequency
(c) Calculate the frequency deviation of the signal
(d) what is the modulation index of signal
Solution
Given:
ƒmax =100.02 MHz
ƒmin = 99.98 MHz
Find:
(a)cs (b) ƒc (c) Δƒ (d) mƒ
(a) The carrier swing is define as the total variation in
frequency from the highest to lowest reached by the
modulation wave.
cs = ƒmax - ƒmin
= 100.02 x 106 -99.98 x 106
= 0.04 x 106
cs = 40 KHz
(b) The carrier frequency or rest frequency is midway between
the maximum frequency and minmun frequency reached by
the modulation wave
ƒmax + ƒmin
ƒc =
2
100.2 x 106 + 99.98 x 106
=
2
6
= 100 x 10
ƒc = 100.00 MHz
14
(c)Since the carrier swing is equal to twice the frequency
deviation
c.s
Δƒ =
2
40 x 10 3
=
2
Δƒ = 20 KHz
(d)Modulation index for a frequency modulation wave is define
Δƒ
mƒ =
ƒa
20 x 10 3
=
3 x 10 3
mƒ = 6667
15
(2.6)
(a) What is the frequency deviation and carrier swing necessary to
provide 75% modulation in the FM broadcast band
(b) repeat for an Fm signal serving as the audio portion of TV band
Solution:
Given:
M = 75%
Find:
ΔƒFM , c.s FM , ΔƒTV , c.sTV
(a) Frequency deviation define as
Δƒactual
M=
x100
Δƒmax
The max frequency deviation permitted in the FM bred cast 88-108
MHz by the FCC 75 KHz
ΔƒFM
75 =
x 100
3
75 x 10
75 x 75 x 103
ΔƒFM =
100
ΔƒFM = 56.25 KHz
Carrier swing related to frequency deviation by
c.s FM = 2 ΔƒFM
= 2 x 56.25 x 103
c.s FM = 112.5 KHz
(b)
Δƒactual
M=
x100
Δƒmax
The maximum frequency deviation permitted by the FCC for the audio
portion of TV signal 25KHz
Δƒtv
75 =
x 100
3
25 x 10
75 x 25 x 103
Δƒtv =
100
Δƒtv =18.75 KHz
16
c.s TV = 2 ΔƒTT
= 2 x 18.75 x 103
ΔƒTV=375 KHz
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(2.8)
The percent modulation of sound portion of the signal 80% Determine
the frequency deviation of carrier swing of the signal.
Solution:
Given:
M = 80%
Find:
Δƒ , c.s
The percent modulation of FM signal
Δƒactual
M=
x100
Δƒmax
The maximum frequency deviation for the sound portion of the TV
Specified by the FCC is 25 KHz
Δƒactua
80 =
x 100
3
25 x 10
80 x 25 x 103
Δƒactua =
100
Δƒactua = 20 KHz
Carrier Swing is related to frequency deviation by
c.s = 2 Δƒactua
= 2 x 20 x 103
c.s = 40 KHz
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(2.10)
Determine the frequency of modulation signal which is produce an
FM signal having a bandwidth of 50 KHz when the frequency
deviation of the FM signal 10 KHz
Solution:
Given:
BW = 50 KHz
Δƒ = 10 KHz
Find:
ƒa
In order to find ƒa refrence must be made to the Schwartz bandwidth
carve FIG in order to entered this carve ,determine BW/Δƒ
50 x 103
BW
=
Δƒ
10 x 103
= 5
mƒ = 2
Δƒ
=
ƒa
10 x 103
2 =
ƒa
10 x 103
ƒa =
2
ƒa = 5 KHz
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SATELLITE COMMUNICATION
Satellite transmission is much like line-of-sight microwave transmission
in which one of the stations is a satellite orbiting the earth. The
principle is the same as terrestrial microwave, with a satellite acting as a
super tall antenna and repeater ( see fig: 7.34). although in satellite
transmission signals must still travel in straight lines, the limitation
imposed on distance by the curvature of the earth are reduced. In
this way, satellite relays allow micro waves signals to span contents
and oceans with a signal bonus.
satellite micro waves can provide transmission capability to and from any
location on earth ,no matter how remote. this advantage makes highly
quality communication available to undeveloped parts of the world
without quiring a huge investment in ground based infrastructure
satellite them selves are extremely expensive ,officered ,but lazing time or
frequencies on one can be relatively cheep.
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GEOSYNCHRONOUS SATELLITES
Lines of sight propagation requires that the sending and receiving
antennas be logged on to each others location at all time (one antenna
must have the other in sight ). for this reason , a satellite that’s moves faster
or slower than the earth’s rotation is useful only for short period of time
(just as stopped clock is accurate twice a day) to ensure constant
communication , the satellite must move at the same speed as the earth
so that it seems to remain fixed above a certain spot. such satellites
are called Geosynchronous.
Such orbital speed is based on distance from the planet .only the orbit
can be geosynchronous. This orbit occurs at the equatorial plane is
approximately 22000 miles surface of the earth.
But on geosynchronous cannot cover the whole earth, on satellite in
orbital has the line-of-sight contact with the vast number so station ,but the
curvature of the equidistant from each other in geosynchronous orbit
to provide full global transmission Fig:7.35hows three satellite each 1290
degree from another in geosynchronous orbit around the equator orbit
around the equator. The view is from north poll
F R EQU ENCY BAN DS FOR SA T ELLI T E
CO M MUN ICA T ION
The frequencies reserved for the satellite communication are in the
gigahertz (GHZ) range. Each satellite send and receive over two different
bands . Transmission from the earth to the satellite is called uplink.
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Transmission from the satellite to the earth is called downlink.
table 7.2 gives the band names and frequencies for each range,
CELLULAR TELEPHONY
Cellular telephony is designed to provide stable communication
connections between two moving devices or between one mobile unit and
one stationary unit (land) unit. A service provider must be able to
locate as the caller moves out of the range of one channel and into
the range of another.
To make this tracking possible each cellular service area is divided into
small regions called cell. each cell contains an antenna and is control by a
small office called the cell office each cell office . in tern, controlled by
a switching office called a mobile telephone switching office(MTSO)the
MTSO coordinates communication center between all of the cell offices
and the telephone central office. it’s a computerized center that is
responsible for connection cells as well as recording cell information
and the billing (see fig: 7.36).
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Cell size is not fixed and can be increased or decreased depending on
the population of the area. the typical radius of a cell is 1 to 12 miles .
high density areas require more ,geographically smaller cells to meet
traffic demands than do lower density areas .once determines ,cell size is
optimize to prevent the interface of the adjacent cell signals.
The transmission power of each cell id kept low to prevent its
signal from interfering with those of other cell.
CELLULA R B EN DS
Traditional cellular transmission is analog. To minimized noise,
frequency modulation (FM) is used for communication between the mobile
telephone its self and the cell office. the
F CC has the assigned
to banned for cellular used
(see fig:7.37).
The band between 824 and 849 MHz carries that communication
that initiate from the mobile phone the band between 869 and 894 MHz
carries those communication that initiate from land phones. Carries
frequencies are spaced every 30 KHz, allowing each band to support up
to 833 carries . how ever , to carries are required for full duplex
communication , which doubles the require width of each channel to 60
KHz and leaves only 416 channels available for each band .
Each band, therefore, is divide into 416 FM channel (for a total of
832 channels). Of these , some are reserved for control and setup data
rather than voice communication .in addition , to prevent interface ,
channel are distribute among the cells in such a way that adjacent cell do
not used the same channel . this restriction means that each cell
normally has access only 40 channel.
T R ANS MI T T ING
To place a cell from a mobile phone, the caller enters a code of 7 or 10
digits (a phone number ) and presses the send button. The mobile phone
than scan the band, seeking a setup channel with a strong signal and sends
the data(phone number) to the closest cell office using that channel. The cell
23
office relays the data to the MTSO . the MTSO sends the data on to the
telephone central office. if the called party is available , a connection is
made and the result is relayed that to the MTSO.at this point ,the MTSO
assigns an unused voice channel to the cell and a connection is
established. The mobile phone automatically adjust its tuning to the
new channel and voice communication can begin.
RECIEVING
When a land phone places cell to a mobile phone, the telephone
central office sends the number to the MTSO .the MTSO searches the
location of the mobile phone by sending query signals to each cell in a
process called paging
Once the mobile phone is found, the MTSO transmits a ringing signal
and, when the mobile phone is answered assigns a voice channel to
the cell, allowing voice communication is begin
HANDOFF
It may happen that, during a conversation, the mobile phone moves from
one cell to another cell. When it does, the signal may become weak.
two solve this problem, the MTSO monitors the level of the signal every
few seconds .if the strength of the signal the diminishes, the MTSO seeks a
new cell that can accommodate the communication better. the MTSO
than changes the channel carrying the cell (hands the signal off from the
old channel to a new one). Hand off are performed so smoothly that
most of the time they are transparent to the user.
DIGITAL
Analog (FM) cellular services are based on a standard called analog
circuit switched cellular (ACSC) to transmit digital data using an ACSC
.service requires a modem with a max: speed of 9600 to 19,200 bps.
Since 1993, however, several services provider have been
moving a cellular data stranded called cellular digital packet data
(CDPD).CDPD provides low speed digital service over the existing
cellular network. It’s based on the OSI model.
To use the existing digital service, such as 56 kilo switch service,
CDPD use what is called a trisected. A trisected is a combination of 3 cell
each using 19.2 KBPS, for a total of 57.6 KBPS (which can be
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accommodate on a 56 k switched line by eliminating some over head ).
Under this scheme united states is divided into 12,30 trisected. For
every 60 trisected ,there is one router.
ITEGREATION WITH SATELLITES AND PCS
Cellular telephone is moving fast towards integrating the existing
system with satellite communication. This integration will make it
possible to have mobile communication between any two points
on the globe an other goal is to combine cellular telephony and the
personal computer communication under a scheme called mobile
personal communication to an able people to use small, mobile personal
computes to send and received data, voice, image, and the video .