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Lecture 1B (01/07)Signal Modulation
• Introduction • Signal Analysis• Advantages of Signal Modulation • Amplitude Modulation• Amplitude Modulation of Digital Signals • Reducing Power and Bandwidth of AM Signals • Frequency Modulation • Power and Bandwidth of FM Signals • Conclusion
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Introduction
• The tasks of communication is to encode information as a signal level, transmit this signal, then decode the signal at the receiver.
• An analog signal varies continuously with time, and has an infinite number of possible signal levels.
• A discrete signal changes only once during a certain time interval. The signal value during this time interval (sampling period) is one sample. Each sample can occupy a infinite number of possible levels.
• A digital signal is discrete, but each sample has a finite number of possible signal levels. The limited number of levels means that each sample transmits a single information. It also means that each sample can be represented as digital data, a string of ones and zeroes.
• A digital signal is preferred in computer communications because computers already store and process information digitally.
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S(t)
t
S(t)
tSampling Period
S(t)
tSampling Period
00000001001000110100010101100111100010011010101111001101
• In an analog signal, noise are added to the signal during transmission. The received signal will never be identical to the original signal. • A digital signal will still be subject to noise, but the difference between signal levels (an “0101” and an “0110”) will be sufficiently large so that the receiver can always determine the original signal level in each sampling period. The regenerated, so that the received digital data is an exact replica of the original digital data. • Analog to digital conversion reduces the amount of information of the signal by approximating the analog signal with a digital signal. The sampling period and number of levels of the digital signal should be selected in order to capture as much information of the original signal as possible
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S(t)
t
Sampling Interval
• A periodic signal satisfies the condition:
• “Period” of the signal.
• “Aperiodic.” signals
• An even function S (t) = S (-t)
• “Symmetric”.
• The phase of the signal.
tTtStS );()(
Signal Properties
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S(t)
A
T
t
A · cos( 2 · f · t )
A · cos( 2 · f · t - )
f
A
1/T
A
f
A
1/T
=0
1/T f
A
=/4
tftftftS mmmm
10cos
5
16cos
3
12cos
4V5)(
A
T
t
T
20
Square Wave
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Digital signals in the time and the frequency domain.
• The time domain is the signal level expressed as a function of time. • The frequency domain is comprised of amplitude and a phase for an
infinite number of sinusoidal functions. These correspond to the superposition of an infinite number of sinusoidal waveforms in the time domain.
• To convert from the time domain to the frequency domain, we take the Fourier transform of the time domain representation.
• The frequency domain representation of a signal does not change with time, but the Fourier transform for the signal over an infinitely long time period would be impossible. Instead, we assume that the signal has some finite duration, over a sampling interval:
1. We assume that outside of the sampling interval, the signal repeats itself, so that the signal is periodic.
2. An alternative assumption is that the time-domain signal has a zero value outside of the sampling interval, so that the signal is aperiodic.
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Fourier Series Expansion of Periodic Signals
a. There are a finite number of discontinuities in the period T.
b. It has a finite average value for the period T.
c. It has a finite number of posit. and negat. Max. in the period T.
1
00
0 )2(sin)2(cos)(n
nn
n tfnbtfnatS
f0 : Fundamental
frequency, n · f0 : Frequency of each
terman , bn : Fourier series
coefficients :
tdtST
aT
0
000 )(
1
tdtnftST
aT
n 0
00
0
)2(cos)(2
dttnftST
bT
n 0
0 00
)2(sin)(2
1
00 2cos)(n
nn tnfcatS
8
A
T
t
T
20
Square Wave
tftftftS mmmm
10cos
5
16cos
3
12cos
4V5)(
c(f)
ffrequency
c1
c2
c3
c4
c5 c
6c
7c
8c
9c
10c
11c
12
f0 2f
0 3f0 4f
0 5f0 6f
0 7f0 8f
0 9f0 10f
0 11f0 12f
0
tdtST
VT
RMS 0
2)(1
9
c(f) V
f frequency
5th harmonic missing
10th harmonic missing
f0 2f0 3f0 4f0 5f0 6f0 7f0 8f0 9f0 10f0 11f0 12f0
c(f) rms
f frequency
5th harmonic missing 10th harmonic missing
f0 2f0 3f0 4f0 5f0 6f0 7f0 8f0 9f0 10f0 11f0 12f0
tdtST
VT
RMS 0
2)(1
10
n = 1n = 3
n = 1 n = 5
n = 4 1
S ( t )
t
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According to Fourier analysis, any composite signal can be represented as
a combination of simple sine waves with different frequencies, phases, and
amplitudes.
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Introduction
A network and transport layers basic services comprise the end-to-end transport of the bit streams over a set of routers (switches).
They are produced using six basic mechanisms: a. Multiplexing b. Switching c. Error control d. flow control e. congestion control f. and resource allocation.
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Multiplexing
Combines data streams of many such users into one large bandwidth stream for long duration. Users can share communication medium.
a. Switch b. Multiplexer/Demultiplexer
a. Fully connected network b. network with shared links.
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Signal Modulation is: • The process of encoding a baseband (data), source
signal Sm (t) onto a carrier signal.
• The carrier waveform is varied in a manner related to the baseband signal.
• The carrier can be a sinusoidal signal or a pulse signal. The result of modulating the carrier signal is called the modulated signal.
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Signal Modulation Sc ( t ) = A · cos ( 2 f t+ )
A. Amplitude Modulation (AM), or Amplitude Shift Keying (ASK): The amplitude A of the carrier signal changes in proportion to the baseband signal.
B. Phase Modulation (PM), or Phase Shift Keying (PSK): The phase of the carrier signal changes in proportion to the baseband signal.
C. Frequency Modulation (FM), or Frequency Shift Keying (FSK): The frequency f of the carrier changes in proportion to the baseband signal.
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Amplitude, Phase, and Frequency Modulation of
a digital baseband signal
AM
1 0 0 0 1 0 1 1 0 0
PM
1 0 0 0 1 0 1 1 0 0
FM
1 0 0 0 1 0 1 1 0 0
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Advantages of Signal Modulation
• Radio transmission of the signal.
m103
s
11000
s
m103
;s
11000 5
8
f
cf
Sc ( t ) = A · cos ( 2 fc ) range (fc + f) and (fc - f) .
m30
s
11010
s
m103
;s
11010
6
8
6
f
cf
5GHz 6 cm
FDMWDM
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FDM Original Signals
M
Sc1 (t)
Sm1 (t)
M
Sc2 (t)
Sm2 (t)
M
Sc3 (t)
Sm3 (t)
DM
Sc1 (t)
DM
Sc2 (t)
DM
Sc3 (t)
Received Signals
Sm1 (t)
Sm2 (t)
Sm3 (t)
Transmission Line
Guard band
f
fc1-fm1 fc1 fc1+fm1 fc2-fm2 fc2 fc2+fm2 .......................
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Amplitude Modulation
• AM is simply the multiplication of the baseband signal with the carrier signal.
)2(cos)( tfAtS ccc
)2(cos)(1
)( tftSAk
tS cmc
)2(cos)()2(cosoffsetDC
)2(cos)(offsetDC)(11
1
tftSAtfA
tftSAtS
cmckcmck
cmmck
•Modulation coefficient k describeshow efficiently the modulator device multiplies the two signals.
Sm ( t ) X
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DSBTC AM• DSBTC AM. DC = Ac .
• Modulation index i of a DSBTC AM signal
• The simplest case of DSBTC AM, where the baseband signal is a sinusoidal signal with:
DC = Ac
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)2(cos)2(cos1)( 1 tftfA
AAAtS cm
c
mcck
c
m
A
Ai
)2(cos)2(cos1)(2
tftfik
AtS cm
c
)(cos)(cos2
1)(cos)(cos yxyxyx
componentfrequencycarrierNoncomponentfrequencyCarrier
tfftffik
Atf
k
A
tftfik
Atf
k
A
tftfik
AtS
mcmcc
cc
cmc
cc
cmc
))(2cos())(2cos(2
1)2cos(
)2cos()2cos()2cos(
)2(cos)2(cos1)(
22
22
2
c
m
A
tSi
)(max (without DC offset)
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Power spectrum of DSBTC AM signal with
sinusoidal baseband signal, i = 100%, k = 1.
Vrms
ffrequencyfc + fmfc - fm
2
fc
Ac2
2· 2
Ac2
Ac2
2· 2
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Sm (t)
t
t
Sc (t)
t
S(t)
Ac
24
)2(cos)2(cos)( tftfAAtS cmcc
))(2cos())(2cos(
2
1)2(cos)2(cos)(
2
2
tfftffA
tftfAtS
mcmcc
cmc
Vrms
ffrequencyfc + fmfc - fm fc
Ac2
2· 2
Ac2
2· 2
Sm ( t ) = Ac · cos ( 2 fm t)
DSBSC AM
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Sm (t)
t
t
Sc (t)
t
S(t)
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Amplitude Modulation of Digital Signals
Baseband Signal
Baseband Signal with DC shift
tftftftS mmmm
10cos
5
16cos
3
12cos
4V5)(
tftS cc 2cosV5)(
Carrier Signal
tftftftS mmmm
10cos
5
16cos
3
12cos
4V5V5)(
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Transmitted Signal
sidebandupper
tfftfftffk
V
sidebandlower
tfftfftffk
V
frequencycarrier
tfk
VtS
mcmcmc
mcmcmc
c
...52cos5
132cos
3
12cos
92.15
...52cos5
132cos
3
12cos
92.15
)2cos(25
)(
2
2
2
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DSBTC AM signal with square wave baseband signal, i = 10
0%, k = 10.
Vrms
ffrequencyfc - fm
2
fc
2.50V
fc - 3fm
2
1.59V
2
1.59V
fc + 3fmfc + fm
2
0.53V
2
0.53V
2
0.32V
2
0.32V
fc - 5fm fc + 5fm
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Sm (t)
t
t
Sc (t)
t
S(t)
5V
5V
10V
50V 1/2
1/2
625504
10
22
10
2
1
21
21
22
A
PPP
PowersignalModulated
Full
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DSBSC AM
tftftftS mmmm
10cos
5
16cos
3
12cos
4V5)(
sidebandupper
tfftfftffk
V
sidebandlower
tfftfftffk
VtS
mcmcmc
mcmcmc
...52cos5
132cos
3
12cos
92.15
...52cos5
132cos
3
12cos
92.15)(
2
2
Vrms
ffrequencyfc - fm fcfc - 3fm
2
1.59V
2
1.59V
fc + 3fmfc + fm
2
0.53V
2
0.53V
2
0.32V
2
0.32V
fc - 5fm fc + 5fm
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t
Sm (t)
t
Sc (t)
t
S(t) 180 degree phase shifts
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Binary Phase Shift Keying and Quadrature Phase Shift Keying
V5)(if180
V5)(if0)(
)(2cosV25
)(
tS
tSt
ttfk
tS
m
m
c
Since the phase of the transmitted signal can take one of two values (phase jumping), this type of modulation is called binary phase-shift keying (BPSK). It is a constant-amplitude method of modulation.
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If we add together two BPSK signals that are offset by a 90 degree phase shift, there are four possible phase jumping during every sampling period. Instead of one possible shift of 180º, there are three possible transitions: +90º, 180º, and ‑90º (+270º). The signal can be represented by the following formula:
V5)(V,5)(if315
V5)(V,5)(if225
V5)(V,5)(if135
V5)(V,5)(if45
)(
)(2cosV25
)(
21
21
21
21
tStS
tStS
tStS
tStS
t
ttfk
tS
mm
mm
mm
mm
c
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t
S(t) phase shifts
t
S1 (t)
t
S2 (t)
0 1 0 1 0
0 1 1 0 0
00 11 01 10 00
180º 90º 180º -90º
Time domain representation of QPSK signal, created: by combining BPSK signals S1 and S2 .
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Reducing Power and Bandwidth of AM Signals
• In DSBTC AM, the carrier frequency component carries no information.
• In all forms of amplitude modulation, the two sidebands contain identical information.
• Eliminating one of the two sidebands. • There are four possibilities for transmitting an amplitude
modulated signal • A. Dual sideband, transmitted carrier (DSBTC AM)• B. Vestigial sideband• C. Dual sideband, suppressed carrier (DSBSC AM)• D. Single sideband• Selecting the type of transmission can depend on the following
factors:• Power limitations• Bandwidth limitations• Simplicity of Demodulation Equipment
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• The power required for transmitting a sinusoidal baseband signal with DSBTC AM is:
Pt : Transmitted power, Ac :Carrier Signal Amplitude
• i : Modulation index
• i=100%, the Pt =1.5 times the power squared of the carrier frequency. Each sideband contributes 0.25 times the power of the carrier frequency.
• Vestigial sideband uses only one of the two sidebands of the signal.
21
222
12
22
1
22
222
iP
AAAP
c
cmct
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Four possible ways to transmit an AM signal.
fc
Vrms
(a) DSBTC AM
ffc
Vrms
(c) DSBSC AM
f
fc
Gain
ffc
Vrms
f
Bandpass Filter
fc
Vrms
(b) Vestigial sideband AM
ffc
Vrms
(d) Single sideband AM
f
Bandpass Filter
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Example Example We have an audio signal with a bandwidth of 4 KHz. What is the bandwidth needed if we modulate the signal using AM? Ignore FCC regulations.
SolutionSolution An AM signal requires twice the bandwidth of the original signal:
BW = 2 x 4 KHz = 8 KHz
Concept 1:
The total bandwidth required for AM can be determined from the bandwidth
of the audio signal: BWt = 2 x BWm.
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Frequency Modulation • Additive noise• Frequency modulation is a special case of phase
modulation, where the phase of the transmitted signal accumulates according to the amplitude of the baseband signal:
t
mcc dSktfAtS )(2cos)(
k : Constructional coefficient of the modulator
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• To find the maximum possible frequency deviation of an FM signal, we can assume the baseband signal is a constant voltage of +Am , then the integral will reduce to a linear function of t :
tAk
fAtAktfAtS mccmcc
2
2cos2cos)(
2max
mc
Akf
Sm ( t ) = Am · sin (2 fm t )
)2(cos2
)2(sin)()(0
tff
AkdfAkdSkt m
m
mt
mm
t
m
m
mf f
Akk
2
)2(cos2cos)( tfktfAtS mfcc
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• Equation can be expanded using Bessel's trigonometric identities into a series of phase-shifted sinusoidal terms.
• The amplitude of each term is determined by the Bessel function of the first kind Jn ( kf ), where kf is the modulation index.
n
mcfnmfcn
fnftkJtfktf2
2cos)()2(cos2cos
n
mcfncn
fnftkJAtS2
2cos)()(
)()1()( xJxJ nn
n
22cos
22cos)(
2cos)()(
1
0
ntfnf
ntfnfkJA
tfkJAtS
mc
n
mcfnc
cfc
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kf
1.0
0.6
0.4
0.8
0.2
0
-0.2
-0.4
J0 ( kf )
J1 ( kf ) J2 ( kf ) J3 ( kf ) J4 ( kf ) J5 ( kf )
10 6 8 4 2
First five Bessel coefficients for varying values
of kf .
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Power and Bandwidth of FM Signals
• Calculating the power of an FM signal is much simpler in the time domain.
1
220
22
)(2)(2
1
2 fnfcc
FM kjkjAA
P
12 fT kBB
In this equation B is effective bandwidth of baseband signal, but BT is absolute bandwidth of FM signal
)2(cos2cos)( tfktfAtS mfcc
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Ac · cos ( 2 fc + kf cos ( 2 fm ) ) at various values of kf . Vrms
f
kf = 0.5
Vrms
f
kf = 1
Vrms
f
kf = 2
Vrms
ffc +6fm
kf = 10
fc +9fm fc -9fm fc -6fm
fc fc + 3fm fc - 3fm
fc fc + 3fm fc - 3fm
fc fc + 3fm fc - 3fm
fc fc +3fm fc -3fm
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Example 1Example 1 We have an audio signal with a bandwidth of 4 MHz. What is the bandwidth needed if we modulate the signal using FM? Ignore FCC regulations.
SolutionSolution An FM signal requires 10 times the bandwidth of the original signal:
BW = 10 x 4 MHz = 40 MHz
Concept 2:
The total bandwidth required for FM can be determined from the bandwidth
of the audio signal: BWt = 10 x BWm.
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