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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
Chapter 5
Analog Transmission
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
5.1 Modulation of Digital Data
Digital-to-Analog Conversion
Amplitude Shift Keying (ASK)
Frequency Shift Keying (FSK)
hase Shift Keying (SK)
!uadrature Amplitude "odulation #it$#aud Comparison
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Figure 5.1 Digital-to-analog modulation
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
Figure 5.2 Types of digital-to-analog modulation
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#it rate is the num%er of %its per
second& #aud rate is the num%er ofsignal units per second& #aud rate is
less than or equal to the %it rate&
Note: Note:
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
'ample 'ample
An analog signal carries 4 bits in each signal unit. If 1000
signal units are sent per second, find the baud rate and the
bit rate
Solution Solution
Baud rate = 1000 bauds per second (baud/s)Baud rate = 1000 bauds per second (baud/s)
Bit rate = 1000 x 4 = 4000 bpsBit rate = 1000 x 4 = 4000 bps
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'ample * 'ample *
The bit rate of a signal is 3000. If each signal unit carries
6 bits, what is the baud rate?
Solution Solution
Baud rate = 3000 / 6 = 500 baud/sBaud rate = 3000 / 6 = 500 baud/s
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
Figure 5.3 ASK
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Figure 5.4 +elationship %et,een %aud rate and %and,idth in ASK
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
'ample 'ample
Find the minimum bandwidth for an ASK signal
transmitting at 2000 bps. The transmission mode is half-
duplex.
Solution Solution
In !" the baud rate and bit rate are the sa#e$ %he baud
rate is there&ore '000$ n !" sina re*uires a#ini#u# band+idth e*ua to its baud rate$ %here&ore, the
#ini#u# band+idth is '000 -.$
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
'ample . 'ample .
Given a bandwidth of 5000 Hz for an ASK signal, what
are the baud rate and bit rate?
Solution SolutionIn !" the baud rate is the sa#e as the band+idth,
+hich #eans the baud rate is 5000$ But because the baud
rate and the bit rate are aso the sa#e &or !", the bitrate is 5000 bps$
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
'ample / 'ample /
Given a bandwidth of 10,000 Hz (1000 to 11,000 Hz),
draw the full-duplex ASK diagram of the system. Find
the carriers and the bandwidths in each direction. Assume
there is no gap between the bands in the two directions.
Solution Solution
or &udupex !", the band+idth &or each direction is
B = 10000 / ' = 5000 -.
%he carrier &re*uencies can be chosen at the #idde o&
each band (see i$ 5$5)$
&c (&or+ard) = 1000 2 5000/' = 3500 -.
&c (bac+ard) = 11000 5000/' = 500 -.
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
Figure 5.5 Solution to 'ample /
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Figure 5.6 FSK
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Figure 5.7 +elationship %et,een %aud rate and %and,idth in FSK
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'ample 0 'ample 0
Find the minimum bandwidth for an FSK signal
transmitting at 2000 bps. Transmission is in half-duplex
mode, and the carriers are separated by 3000 Hz.
Solution Solution
or !"
B = baud rate 2 & B = baud rate 2 & c1c1 −− & & c0c0
B = bit rate 2 &c1B = bit rate 2 &c1−−
&c0 = '000 2 3000 = 5000 -.&c0 = '000 2 3000 = 5000 -.
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
'ample 1 'ample 1
Find the maximum bit rates for an FSK signal if the
bandwidth of the medium is 12,000 Hz and the difference
between the two carriers is 2000 Hz. Transmission is in
full-duplex mode.
Solution Solution
Because the trans#ission is &u dupex, on 6000 -. is
aocated &or each direction$
B = baud rate 2 &c1B = baud rate 2 &c1−−
&c0&c0
Baud rate = BBaud rate = B −− (&c1(&c1 −− &c0 ) = 6000&c0 ) = 6000 −− '000 = 4000'000 = 4000
But because the baud rate is the sa#e as the bit rate, the
bit rate is 4000 bps$
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Figure 5.8 SK
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Figure 5.9 SK constellation
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Figure 5.10 The .-SK method
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Figure 5.11 The .-SK characteristics
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Figure 5.12 The 8-PSK characteristics
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Figure 5.13 +elationship %et,een %aud rate and %and,idth in SK
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'ample 2 'ample 2
Find the bandwidth for a 4-PSK signal transmitting at
2000 bps. Transmission is in half-duplex mode.
Solution Solution
or 7!" the baud rate is the sa#e as the band+idth,
+hich #eans the baud rate is 5000$ But in 7!" the bit
rate is 3 ti#es the baud rate, so the bit rate is 15,000 bps$
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'ample 3 'ample 3
Given a bandwidth of 5000 Hz for an 8-PSK signal, what
are the baud rate and bit rate?
Solution Solutionor 7!" the baud rate is the sa#e as the band+idth,
+hich #eans the baud rate is 5000$ But in 7!" the bit
rate is 3 ti#es the baud rate, so the bit rate is 15,000 bps$
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
!uadrature amplitude modulation is a
com%ination of ASK and SK so that amaimum contrast %et,een each
signal unit (%it4 di%it4 tri%it4 and so on)
is achieved&
Note: Note:
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Figure 5.14 The .-!A" and 2-!A" constellations
Fi 5 15 Ti d i f 2 !A" i l
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
Figure 5.15 Time domain for an 2-!A" signal
Fi 5 16 0 !A" ll i
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Figure 5.16 0-!A" constellations
Fi 5 17 #it d % d
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Figure 5.17 #it and %aud
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Ta%le /& #it and %aud rate comparison
ModulationModulation UnitsUnits Bits/BaudBits/Baud Baud rateBaud rate Bit Rate
A!" F!" 2#$! A!" F!" 2#$! Bit 1 % %
4#$!" 4#&AM4#$!" 4#&AM 'i(it 2 % 2%
8#$!" 8#&AM8#$!" 8#&AM )ri(it 3 % 3%
16#&AM16#&AM &uad(it 4 % 4%
32#&AM32#&AM $enta(it 5 % 5%
64#&AM64#&AM *e+a(it 6 % 6%
128#&AM128#&AM e,ta(it 7 % 7%
256#&AM256#&AM -ta(it 8 % 8%
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
'ample 5 'ample 5
A constellation diagram consists of eight equally spaced
points on a circle. If the bit rate is 4800 bps, what is the
baud rate?
Solution Solution
%he consteation indicates 7!" +ith the points 45
derees apart$ !ince '3 = , 3 bits are trans#itted +ith
each sina unit$ %here&ore, the baud rate is 400 / 3 = 1600 baud
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
'ample 'ample
Compute the bit rate for a 1000-baud 16-QAM signal.
Solution Solution
1689 sina has 4 bits per sina unit sinceo'16 = 4$
%hus,
(1000)(4) = 4000 bps
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'ample * 'ample *
Compute the baud rate for a 72,000-bps 64-QAM signal.
Solution Solution
6489 sina has 6 bits per sina unit sinceo' 64 = 6$
%hus,
'000 / 6 = 1',000 baud
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5.2 Telephone Modems
"odem Standards
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A telephone line has a %and,idth of
almost *.55 67 for data transmission&
Note: Note:
Figure 5 18 Telephone line %and,idth
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Figure 5.18 Telephone line %and,idth
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"odem stands for
modulator$demodulator&
Note: Note:
Figure 5 19 "odulation$demodulation
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Figure 5.19 "odulation$demodulation
Figure 5 20 The 8 * constellation and %and,idth
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Figure 5.20 The 8&* constellation and %and,idth
Figure 5 21 The 8 *%is constellation and %and,idth
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Figure 5.21 The 8&*%is constellation and %and,idth
Figure 5.22 Traditional modems
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Figure 5.22 Traditional modems
Figure 5.23 /0K modems
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Figure 5.23 /0K modems
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5.3 Modulation of Analog Signals5.3 Modulation of Analog Signals
Amplitude "odulation (A")
Frequency "odulation (F")
hase "odulation (")
Figure 5.24 Analog-to-analog modulation
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Figure 5.24 Analog to analog modulation
Figure 5.25 Types of analog-to-analog modulation
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g yp f g g
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The total %and,idth required for A"
can %e determined from the %and,idthof the audio signal9
#:t ; * #:m&
Note: Note:
Figure 5.26 Amplitude modulation
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g p
Figure 5.27 A" %and,idth
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g
Figure 5.28 A" %and allocation
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g
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'ample 'ample
We have an audio signal with a bandwidth of 4 KHz.
What is the bandwidth needed if we modulate the signalusing AM? Ignore FCC regulations.
Solution Solution
n 9 sina re*uires t+ice the band+idth o& the
oriina sina:
B = ' x 4 "-. = "-.
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The total %and,idth required for F"
can %e determined from the %and,idthof the audio signal9
#:t ; 5 #:m&
Note: Note:
Figure 5.29 Frequency modulation
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Figure 5.30 F" %and,idth
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The %and,idth of a stereo audio signal
is usually / K67& Therefore4 an F"
station needs at least a %and,idth of/5 K67& The FCC requires the
minimum %and,idth to %e at least *55
K67 (5&* "67)&
Note: Note:
Figure 5.31 F" %and allocation
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McGraw-Hill © The McGraw-Hill Companies, Inc., 2004
' l .' l .
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'ample . 'ample .
We have an audio signal with a bandwidth of 4 MHz.
What is the bandwidth needed if we modulate the signalusing FM? Ignore FCC regulations.
Solution Solution
n 9 sina re*uires 10 ti#es the band+idth o& the
oriina sina:
B = 10 x 4 9-. = 40 9-.