1
Chapter 5
AM,FM,and digital
modulation systems
2
5.0 Introduction
(Chapter objectives)
? Amplitude modulation and single sideband
? Frequency and phase modulation
? Digitally modulated signals(OOK,BPSK,
FSK,MSK MPSK,QAM,QPSK,
π/4QPSK,and OFDM)
? Spread sprectrum and CDMA system
3
5.0 Introduction
(the goal of this chapter)
? Study g(t) and s(t) for various types of
analog and digital modulations
? Evaluate the spectrum for various types of
analog and digital modulations
? Examine some transmitter and receiver
structures
? Learn about spread spectrum systems
4
5.0 Introduction
(the key of this chapter)
? Grasping phase modulation and frequency modulation ? Grasping Binary modulation and Bandpass signaling
? Grasping Multilevel Modulated bandpass signaling
? Grasping Minimum-shift keying (MSK)and GMSK
? Knowing the principle of FDM,OFDM and Spread
Spectrum Systems
5
5.0 Introduction
(band-pass signal and its spectra)
? Communication system constructure
input signal processing carrier circuits
Transmission
Medium
(channel)
transmitter
g(t) s(t)
carrier
circuits
signal
processing
r(t) g(t) output
? Modulation is the process of imparting the source
information onto a Band-pass signal with a carrier
frequency fc by the introduction of amplitude or
phase perturbations or both,
6
5.0 Introduction
(band-pass signal and its spectra)
Theorem:Any physical band-pass waveform can be representative by
} R e { g ( t ) eS ( t ) tj c??? g(t) is called the complex envelope of s(t),and f
c is
carrier frequency (in hertz),ωc=2πfc
? Notice,The desired type of modulated signal,s(t),is
obtained by selecting the appropriate modulation
mapping function g[m(t)],where m(t) is the analog or
digital base-band signal,
? The voltage (or current)spectrum of the band-pass
signal is
21 ) ] G * ( - f- f)[ G ( f - f/S ( f ) cc ??? and the PSD is
41 ) ] ( - f - f )( f - f[/( f ) c*gcgs PPP ??
7
5.1 Amplitude Modulation
? The complex envelope of an
AM signal is
given by
? Where Ac specifies
the power level and
m(t) is the
modulating signal,
The representation
for AM signal is
given by,
]1[ m ( t )Ag ( t ) c ??
tm ( t)As ( t) cc ?c o s]1[ ??
? Fig.5-1
8
5.1 Amplitude Modulation
(modulation percentage for AM signal )
? Definition %positive modulation= (A
max-Ac)/ Ac × 100=max[m(t)] × 100
%negative modulation= (Ac-Amin) / Ac× 100=-min[m(t)]
× 100
? %overall modulation= (Amax- Amin)/ (2Ac)× 100
={max[m(t)]-min[m(t)]}/2 × 100
? where Amax is the maximum value of Ac [1+m(t)],and Ac is
the level of the AM envelope in the absence of modulation,
? notice:The percentage of modulation can be over100
%(Amin will have a negative value).If the transmitter uses
a two-quadrant multiplier that produces a zero output
when Ac[1+m(t)] is negative,the output signal will be,
9
5.1 Amplitude Modulation
? Which is a distorted AM signal which is call as over
modulated signal, Its bandwidth is much wider than
that of the undistorted AM signal,
?
?
?
?
???
-1m ( t ) if 0
-1m ( t ) if c o s)](1[)( ttmAts cc ?
? if the percentage of negative modulation is less than
100%,an envelope detector may be used to recover the
modulation without distortion;if the percentage of
negative modulation is over 100%,undistorted
modulation can still be recovered provided the product
detector is used,
? A product detector is superior to an envelope detector
when the input signal-to-noise ratio is small,
10
5.1 Amplitude Modulation
? because the total average normalized power of a band-pass waveform v(t) is given by
22 )(
2
1)0()()( tgRdffPtvP vvv ???? ??
??? Where ―normalized‖ implies that the load is equivalent
to 1 ohm,The normalized average power of the AM
signal is,
)(2/1)(2/1
)()(212/1
)](1[2/1)(2/1)(
2222
22
222
tmAtmAA
tmtmA
tmAtgts
ccc
c
c
???
???
???
? If the modulation contains no dc level,then 〈 m(t) 〉
=0,and the normalized power of the AM signal is
9)-(5 2/12/1
)(2
p o w e rc a r r i e r
2)( 22
?? ??? ?????
p o w e rs i d e b a n d
tc
d i s c r e t e
ct mAAs ??
11
5.1 Amplitude Modulation
? modulation efficiency:the percent of the total power of the modulated signal that conveys information,
%1 0 0
)(1
)(
2
2
?
?
?
tm
tm
E
? The highest efficiency that can be attained for a 100%
AM signal would be 50%.(For the case when square-
wave modulation is used),The normalized peak
envelope power (PEP)of the AM signal,
2
2
)]}(m a x [1{2 tmAP cPEP ??
12
5.1 Amplitude Modulation
? The voltage spectrum of the AM signal is given by
)]()()()([2)( ccccc ffMffffMffAfS ???????? ??? The AM spectrum is just a translated version of the
modulation spectral component.The bandwidth is
twice that of the modulation
? Example 5-1 Suppose that a 5000w AM transmitter is
connected to a 50Ω load; If the transmitter is 100%
modulated by a 1000Hz test tone,Then compute
(1)the total actually average power,(2)actually PEP,(3)
The modulation efficiency,
? Solution,the constant Ac is given by (1/2)Ac2/50=5000,
thus peak voltage across the load will be Ac=707V
during the times when there is no modulation,
13
5.1 Amplitude Modulation
(2)then across the 50 Ω load,the PEP is,
50/]1[)2/1(
/)2/12/1(
)(2
)(22)(
2
22
tc
tcct
mA
RmAAs
??
??
WRA
Rtm
A
P
c
c
PEP
2 0 0 0 050/7 0 7*4*)2/1(/*4*)2/1(
/)]}(m a x [1{
2
22
2
2
???
??
(3)The modulation efficiency would be,
%33%100
)(1
)(
2
2
??
?
?
tm
tm
E
(1)the actual average power
14
5.1 Amplitude Modulation
15
5.1 Amplitude Modulation
- B 0 B
f
| M ( f ) |
- f c - B - f c - f c +B
f
A c /2
f c - B f c f c +B
W e i ght = A c /2 | S ( f ) | W e i ght = A c /2
a ) M a gni t ude s pe c t r um of m odul a t i on
b) M a gni t ude s pe c t r um of A M s i gna l
U ppe r
s i de ba nd
l ow
s i de ba nd
Fig,4-2 Spectrum of AM signal
16
5.3 Double-sideband Suppressed
Carrier
? A double-sideband suppressed carrier(DSB-SC) signal is an AM signal that has a suppressed discrete carrier,
tm ( t)As ( t) cc ?c o s?? The voltage spectrum of the DSB-SC signal is
21 ) ] fG * ( f)[ G ( f- f/S ( f) cc ???? The percent of modulation of a DSB-SC is,infinite
? The modulation efficiency of a DSB-SC is 100%
? A product detector (which is more expensive than an
? Envelope detector) is required for demodulation of the
DSB-SC signal
? The sideband power of a DSB-SC signal is four times
that of a comparable AM signal with the some peak
level,that is to say,the DSB-SC signal has a four_fold
power advantage over that of an AM signal
17
5.3 Double-sideband Suppressed
Carrier
- B 0 B
f
| M ( f ) |
- f c - B - f c - f c +B
f
A c /2
f c - B f c f c +B
W e i ght = A c /2 | S ( f ) | W e i ght = A c /2
a ) M a gni t ude s pe c t r um of m odul a t i on
b) M a gni t ude s pe c t r um of A M s i gna l
U ppe r
s i de ba nd
l ow
s i de ba nd
Spectrum of DSB-CS signal
18
5.5Asymmetric sideband signals
(Signal sideband)
? An upper single sideband (USSB) signal has a zero-valued spectrum for|f|<f
c,where fc is the carrier
frequency,
? A lower signal sideband (LSSB) signal has a zero-
valued spectrum for |f|>fc,where fc is the carrier
frequency,
? SSB-AM:The bandwidth is the same as that of the
modulating signal (which is half the bandwidth of an
AM or DSB-SC signal),
? The term SSB refers to the SSB-AM type of signal,
unless otherwise denoted,
19
5.5Asymmetric sideband signals
? Theorem:The complex envelope of an SSB signal is
given by
)](?)([)( tmjtmAtg c ??? Which results in the SSB signal waveform,
]s in)(?c o s)([)( ttmttmAts ccc ?? ??
? Where the upper(-) sign is used for USSB and the
lower (+) sign is used for LSSB,Denotes the
Hilbert transform of m(t),which is given by,)(? tm
)()()(? thtmtm ??? Where h(t)=1/(πt),and H(f)=F[h(t)] corresponds to a
? -900 phase-shift network,
?
?
?
?
???
0
0 )(
fj
fjfH
20
5.5Asymmetric sideband signals
Fig.5-4 Spectrum for a USSB signal
| M ( f) |
- B B
1
( a ) B a s e b a n d M a g n i t u d e S p e c t r u m
| G ( f) |
- B B
2A c
( b ) M a g n i t u d e o f c o r r e s p o n d i n g S p e c t r u m
o f t h e c o m p l e x E n v e l o p e fo r U S S B
| S ( f) |
- fc - B - fc fc fc + B
A c
( b ) M a g n i t u d e o f c o r r e s p o n d i n g S p e c t r u m
o f U S S B s i g n a l
? Proof,Taking the Fourier transform
of the complex envelope of an SSB
signal,we get,
) ]}(?[)({)( tmjfMAfG c F??
? Because of Hilbert transform,we can
find that the equation becomes,
)](1)[()( fjHfMAfG c ??
? For USSB case,choose the upper sign,
then,
??
?
?
????
0
0
0
)(2)](1)[()(
f
ffMAfjHfMAfG c
c
? Substituting it into eq.(4-15),we have,
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
?
?
??
?
c
c
cc
c
ccc
ff
ff
ffMA
ff
ffffMA
fS
)(
0
0
)(
)(
21
5.5Asymmetric sideband signals
? The normalized average power of the USSB signal is
)(
)(?)(
2
1
|)(?)(|
2
1
)(
2
1
)(
22
)()(?
222
2222
22
tmA
tmtmA
tmtmAtgts
c
tmtm
c
c
?
?
??
???
? The SSB signal power is the power of the modulating
signal〈 m2(t)〉 multiplied by the power gain factor Ac,
? The normalized peak envelope power (PEP) is,
})](?[)(m a x {)2/1(})(m a x {)2/1( 222 2 tmtmAtg c ??
22
5.5Asymmetric sideband signals
? Fig,5-5 Generation of SSB
m ( t )
M o d u l a t i o n
i n p u t
- 90
0
p h a s e
s h i f t a c r o s s
b a n d o f m ( t )
B a s e b a n d
p r o c e s s i n g
O s c i l l a t o r
f = f c
- 90
0
p h a s e
s h i f t a t f = f c
+
+
s ( t )
S S B s i g n a l
m ( t )
v 1 (t)
v 1 ( t ) = A c c o s ( ω c t)
v 2 ( t ) = A c s i n ( ω c t)
a) p h a s e m e t h o d
S i d e b a n d f i l t e r
( b a n d p a s s f i l t e r o n
e i t h e r u p p e r o r
l o w e r s i d e b a n d )
O s c i l l a t o r
f = f c
m ( t )
M o d u l a t i o n
i n p u t
s ( t )
S S B s i g n a l
b) f i l t e r m e t h o d
23
5.5Asymmetric sideband signals
? SSB signals have both AM and PM,The AM
component (real envelope) is,
22 )](?[)()()( tmtmAtgtR c ???? The PM component is
])( )(?[t a n)()( 1 tm tmtgt ???? ??? SSB signals may be received by using a super- heterodyne
receiver that incorporates a product detector with θ0=0,
thus the receiver output is,
)(})(R e { 0 tmKAetgKv cjout ?? ? ?? Where K depends on the gain of the receiver and the loss in
the channel,In detecting SSB signals with audio
modulation,the reference phase θ0 does not have to be zero,
because the same intelligence is heard regardless of the
value of the phase used,For digital modulation,the phase
has to be exactly correct so that the digital waveshape is
preserved
24
5.5Asymmetric sideband signals
(Vestigial sideband)
25
5.5Asymmetric sideband signals
(Vestigial sideband)
? In certain application,a DSB modulation technique takes
too much bandwidth for the channel,and an SSB
technique is too expensive to implement,although it takes
only half the bandwidth,In this case,a compromise
between DSB and SSB,called vestigial sideband(VSB),is
often chosen,
? VSB is obtained by partial suppression of one of the
sidebands of a DSB signal,
? sideband of the DSB signal is attenuated by using a band-
pass filter,called a vestigial sideband filter,The VSB signal
is given by,
)(*)()( thtsts vV S B ?? Where s(t) is a DSB signal,and h
v(t) is the impulse response
of the VSB filter.The spectrum of the VSB signal is,
)()()( fvHfSfV S BS ?
26
5.5Asymmetric sideband signals
(Vestigial sideband)
? The modulation on the VSB signal can be recovered by
a receiver that uses product detection or,if a large
carrier is present,by the use of envelope detection,For
recovery of undistorted modulation,the transfer
function for VSB filter must satify the constraint,
BfCffHffH cvcv ?????,)()(? Proof,
)]()()()()[2/()( fHffMfHffMAfS vcvccV S B ????
? Thus only when the constraint satisfy,Vout(f)=KM(f) hold
true,
BfffHffHfMAA
BfffHfMffHfMAAfV
fHfffffSAfV
tsAtV
cvcvc
cvcvco u t
ccV S Bo u t
V S Bo u t
o
?????
?????
????
?
) ],()()[()4/(
) ],()()()()[4/()(
)() ] }()2/1()()2/1[(*)({)(
h ( t )*t]c o s)([)(
0
0
0 c
??
?
? the product detector output is
27
Summary
? AM
DSB-SC
SSB-AM
VSB
]1[ m ( t )Ag ( t ) c ??
m ( t )Ag ( t ) c?
)](?)([)( tmjtmAtg c ??
)](?)([)( tmjtmAtg c ?? )(*)()( thtsts vAMS S BV S B ??
tm ( t)As ( t) cc ?c o s]1[ ??
tm ( t)As ( t) cc ?c o s?
]s i n)(?c o s)([)( ttmttmAts cccAMS S B ?? ???
)(*]s i n)(?c o s)([)( thttmttmAts vcccV S B ?? ??
21 ) ] fG * ( f)[ G ( f- f/S ( f) cc ???
)]()()()([2)( ccccc ffMffffMffAfS ???????? ??
??
?
??
?
??
??
????
?
??
?
?
???
c
c
ccc
ccc
ff
ff
ffMAff
ffffMAfS
)(
0
0
)()(
)]()()()()[2/()( fHffMfHffMAfS vcvccV S B ????
28
Summary
? Example 5-2 if linear modulated signal is as following,
(1) s1(t)=cos(ω0t)cos (ωct)
(2) s2(t)(1+0.5sin (ω0t)) cos (ωct)
Where ωc=6ω0,please giving their spectrum,
? Solution, (1) the spectrum is,
)]5()5()7()7([
2
)]}()([*)]()([{
2
1
)(
0000
001
????????????
?
??????????????
?
?
????????
??????? ccS
-7ω0 -5ω0 5ω0 7ω0
29
Summary
? (2) the spectrum is
)]7()7()5()5({
4
)]6()6([
)]}()([*)]()([{
2
5.0
)()([)(
0000
00
00
2
????????????
?
???????
?????????????
?
?
????????
????????
????
???????
????
j
j
S
cc
cc
-7ω0 -5ω0 5ω0 7ω0
j ω
ω
S2(f)
30
5.6 Phase Modulation and Frequency Modulation
(Representation of PM and FM signal)
? PM and FM are special cases of angle-modulated
signaling,In this kind signaling the complex envelope is,
)()( tjc eAtg ??? Here the real envelope R(t)=|g(t)|=A
c,is a constant,and the phase θ(t) is a linear function of the modulating
signal m(t),However,g(t) is a nonlinear function of the
modulation,The resulting angle-modulated signal is,
)](c o s [)( ttAts cc ?? ??? For PM,the phase is directly proportional to the modulating
signal
)()( tmDt p??? Where the proportionality constant D
p is the phase sensitivity of the phase modulator,its units is radians per
volt,For FM,the phase is proportional to the integral of
m(t),so that,
31
5.6 Phase Modulation and Frequency
Modulation (Representation of PM and FM
signal)
? Where the frequency deviation constant Df has units
of radians/volt-sencond,From the last two equations,
we can see that if we have a PM signal modulated by
mp(t),there is also FM on the signal,corresponding to
a different modulation waveshape that is given by,
?
??
? ??? dmDt tf )()(
]
)(
[
dt
tdm
D
D
m p
f
p
f ?? Where the subscripts f and p denote frequency and
phase,Similarity,if we have an FM signal modulated
by mf(t),the corresponding phase modulation on this
signal is,
? ??? t f
p
f
p dmD
D
m ?? )(
32
5.6 Phase Modulation and Frequency
Modulation (Representation of PM and FM
signal)
? Fig,5-7 Generation of FM from PM,and vice versa
m f ( t ) I n t e g r a t o r
g a i n = D f / D
p
m p ( t ) P h a s e m o d u l a t o r
( c a r r i e r f r e q u e n c y = f c )
s ( t ) F M
s i g n a l o u t
m p ( t ) d i f f e r e n t i a t o r
g a i n = D p /D f
m f ( t ) f r e q u e n c y m o d u l a t o r
( c a r r i e r f r e q u e n c y = f c )
s ( t ) P M
s i g n a l o u t
a ) G e n e r a t i o n o f F M u s i n g a p h a s e M o d u l a t o r
b ) G e n e r a t i o n o f P M u s i n g a f r e q u e n c y M o d u l a t o r
? Direct PM circuit are realized sinusoidal signal
through a time-varying circuit which introduces a
phase shift that varies with the applied modulation
voltage,Similarly,a direct FM circuit is obtained by
varying the tuning of an oscillator tank circuit
according to the Modulation voltage,
33
5.6 Phase Modulation and Frequency
Modulation (Representation of PM and FM
signal)
? Definition,If a band-pass signal is represented by,
)(c o s)()( ttRts ??Where Ψ(t)=ω
ct+θ(t),then the instantaneous frequency (hertz) of s(t) is,
])([2 1])([2 1)(2 1)( dt tdfdt tdttf cii ?????? ????
? For the case of FM,the instantaneous frequency is,
)(2 1])([2 1)( tmDfdt tdftf fcci ??? ?? ??
? The instantaneous frequency varies about the assigned
carrier frequency fc in a manner that is directly
proportional to the modulated signal m(t),
34
5.6 Phase Modulation and Frequency
Modulation (Representation of PM and FM
signal)
? The frequency deviation from the carrier frequency is,
])([2 1)( dt tdftff cid ?????
? And the peak frequency deviation is,
]})([21m a x { dt tdF ??? ?
? The peak-to-peak deviation is
]})([2 1m i n {]})([2 1m ax { dt tddt tdF pp ????? ??
? For FM signaling,the peak frequency deviation is
related to the peak modulation voltage by,
pf VDF ?? 2
1?
? Where Vp=max[m(t)]
35
5.6 Phase Modulation and Frequency Modulation
(Representation of PM and FM signal)
? An increase in the amplitude of
the modulation signal Vp will
increase ΔF and the bandwidth of
the FM signal,But will not affect
the average power level of the FM
signal which is Ac2/2
? As Vp is increased,spectral
components will appear farther
and farther away from the carrier
frequency,and the spectral
components near the carrier
frequency will decrease in
magnitude,since the total power in
the signal remains constant.This
situation is distinctly different
form AM signaling,where the level
of the modulation affects the
power in the AM signal,but does
not affect it bandwidth,
? Fig.5-9
36
5.6 Phase Modulation and Frequency
Modulation
(Representation of PM and FM signal)
? The peak phase deviation is defined by,
)](m a x [ t??? ?? the peak modulation voltage is
pp DD???? Where V
p=max[m(t)],? Definition,The phase modulation index is given by,
??? ?p? Where Δθ is the peak phase deviation
? Definition,The frequency modulation index is given by,
B
F
f
?? ?
? Where ΔF is the peak frequency deviation and B is the
bandwidth of the modulating signal,which,for the case
of sinusoidal modulation,is fm,the frequency of the
sinusoid,
37
5.6 Phase Modulation and Frequency
Modulation
(Spectra of Angle-Modulated Signals)
? As the spectrum of the band-pass waveform is
)]()([21)( * cc ffGffGfS ?????? Where G(f)=F[g(t)]=F[A
cejθ(t)] is the spectra of complex
envelope signal,For the type of AM modulation signaling,
we able to obtain relatively simple formulas relating S(f) to
M(f),For angle modulation signaling,this is not the case,
because g(t) is a nonlinear function of m(t),thus,a general
formula relating G(f) to M(f) cannot be obtained,because
G(f)=F[g(t)]=F[Acejθ(t)] must be evaluated on a case-by-case
basic for the particular modulating waveshape of interest,
? Since g(t) is nonlinear function of m(t),superposition does
not hold,and FM spectra for the sum of two modulating
waveshape is not the same as summing the FM spectra
that were obtained when the individual waveshape were
used
38
5.6 Phase Modulation and Frequency
Modulation
(Spectra of Angle-Modulated Signals)
? Example 5-2 spectrum of a PM or FM signal with
sinusoidal modulation
tAtm mmp ?s i n)( ?
tt m??? sin)( ?
? If the modulation on PM signal is,
? Then,
? Where the phase modulation index is βp =DpAm= β,
? When mf(t)=Amcosωmt in FM modulation,we can get the
same phase function,Where β = βf =DfAm/ωm,The peak
frequency deviation would be ΔF=DfAm/2π,
? The complex envelope is
tjctjc meAeAtg ??? s i n)()( ??
? Which is periodic with period Tm=1/fm,Its Fourier series
can be represented by,
5 5 )-(5 )( tjn
n
n
n nectg
?? ??
???
?
39
5.6 Phase Modulation and Frequency
Modulation
(Spectra of Angle-Modulated Signals)
? Where
dteeTAc m
m
mmT
T
tjntj
m
c
n ??
?? 2/
2/
s i n )( ???
)(]2 1[ )s i n( ??? ?? ??? ncnjcn JAdeAc ?? ?? ?
Here Jn(β) is the Bessel function of
the first kind of the nth,and J-n(β)
=(-1)n Jn(β) hold true,Taking the
Fourier transform of Eq.(5-55),
we get,
)()( m
n
n
n nffcfG ?? ?
??
???
?
)()()( m
n
n
nc nffJAfG ?? ?
??
???
?? Fig.5-10
40
5.6 Phase Modulation and Frequency Modulation
(Spectra of Angle-Modulated Signals)
? Fig,5.11
? Carson’s rule:the bandwidth
of the angle-modulated signal
depends on the phase
modulation index or the
frequency modulation index β,
and bandwidth of the
modulating signal,That is,
BB T )1(2 ?? ?? In fact,98%of the total power
is constrained in that
bandwidth
41
5.6 Phase Modulation and Frequency
Modulation
(Narrowband Angle Modulation)
? When θ(t) is restricted to a small value,say θ(t)<0.2rad,
the complex envelope may be approximated by,
)](1[)( tjAtg c ???? Then a narrowband angle-modulated signal is,
?? ??? ????? ??
t e r mrmc a r r i e r t e
s i n)(c o s)(
s i d e b a n d
cc
d i s c r e t e
cc ttAtAts ??? ??
? This signal is similar to an AM-type signal,except that
the sideband term is 900 out of phase with the discrete
carrier term,The spectrum of the narrowband angle-
modulation signal is,
) ] }()([)]()({[2)( ccccc ffffjffffAfS ???????? ????
s i g n a l i n g FMf o r
s i g n a l i n g PMf o r
2
( t ) ][( f )
??
?
?
?
?? M ( f )
πj
Df
M ( f ) D p
?? ?
? where
42
5.6 Phase Modulation and Frequency
Modulation
(Narrowband Angle Modulation)
? Generation of NBFM Fig.5-12
43
5.6 Phase Modulation and Frequency
Modulation
(Widthband Angle Modulation)
? Theorem:For WBFM signaling,where
? and B is the bandwidth of m(t),the normalized PSD of
the WBFM signal is approximated by
])(c o s [)( ??? ????? t dmDtAts fcc
12 )](m a x [ ?? ??? tmD ff
? where fm(.) is the PDF of the modulating signal,
) ) ](2()(2([2)(
2
c
f
mc
f
m
f
c ff
DfffDfD
Af ????? ????
44
5.6 Phase Modulation and Frequency
Modulation
(Widthband Angle Modulation)
? Generation of WBFM fig.5-13
45
5.6 Phase Modulation and Frequency
Modulation
(Widthband Angle Modulation)
? Example 5-3 spectrum for WBFM with Triangular
modulation,
? Fig.5-14
? The PDF for triangular modulation is,
??
?
?
?
?
?
o t h e r w i s e m,0
,
2
1
)( ppm
Vm
Vmf
46
5.6 Phase Modulation and Frequency
Modulation
(Widthband Angle Modulation
? Where Vpis the peak voltage of the triangular waveform
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
???
o t h e r w i s e f,0
)(
2
,
2
1
o t h e r w i s e f,0
)(
2
,
2
1
2
)(
2
pc
fp
pc
fp
f
c Vff
DV
Vff
DVD
A
f
???
Ρ
? The PSD of the WBFM signal becomes
??
???
??
??? ???????
??
???
??
??? ?????
o t h e r w i s e f,0
)()(,
8
o t h e r w i s e f,0
)()(,
8)(
22
FffFf
F
AFffFf
F
A
f cc
c
cc
c ??
?
??
?P
? Where the peak frequency deviation is,
?? 2
pVfDF ?
47
5.6 Phase Modulation and Frequency
Modulation
(Widthband Angle Modulation
? Other example Fig,5-15
48
5.6 Phase Modulation and Frequency
Modulation
? some important properties of angle-modulated signals
are,
? An angle-modulated is a nonlinear function of the
Modulation and the bandwidth of the signal increases
as the modulation index increases;
? The discrete carrier level changes depending on the
modulating signal;
? The bandwidth of a narrowband angle-modulated
signal is twice the modulating signal bandwidth
? The real envelope of an angle-modulated signal is
constant,and does not depend on the level of the
modulating signal
49
5.6 Phase Modulation and Frequency
Modulation (Preemphasis and Deemphasis in
Angle-Modulated system)
? preemphasis:If the level of the modulation is boosted at
the top end of the spectrum,the SNR at the output of
the receiver can be improved,
? Deemphasis:The level of the the modulated is
attenuated at high frequencies on the receiver output,
? Preemphasized FM is actually a combination of FM
and PM and combines the advantages of both with
respect to noise performance
50
5.6 Phase Modulation and Frequency Modulation
(Preemphasis and Deemphasis in Angle-Modulated system)
? Fig.5-16 Angle-modulated system with preemphasis and dedmphasis
51
5.7 Frequency-Division multiplexing and FM
stereo
? Frequency-division multiplexing (FDM)is a
technique for transmitting multiple message
simultaneously over a wideband channel by
first modulating the message signals onto
several sub-carriers and forming a composite
base-band signal that consists of the sum of
these modulated sub-carriers.This composite
signal may then be modulated onto the main
carrier,
52
5.7 Frequency-Division multiplexing and FM
stereo
? Fig.5-17 FDM system
53
5.7 Frequency-Division multiplexing and FM
stereo
? The composite signal spectrum must consist of
modulated signals that do not have overlapping
spectra,the composite base-band signal then modulates
a main transmitter to produce the PDM signal that is
transmitted over the wide-band channel,
? The received FDM signal is first demodulated to
reproduce the composite base-band signal that is passed
through filters to separate the individual modulated
sub-carriers.Then the sub-carriers are demodulated to
reproduce the message signals m1(t),m2(t),and so on,
54
5.7 Frequency-Division multiplexing and FM
stereo
? Fig,5-18
? To obtain the compatibility
feature,the left-and right -
channel audios are
combined to produce the
monaural signal,and the
difference audio is used to
modulate a 38-KHz DSB-
SC signal.A 19-KHz pilot
tone is added to the
composite base-band signal
mb(t) to provide a reference
signal for coherent sub-
carrier demodulation in the
receiver
55
5.9 Binary Modulated Bandpass
Signaling
? Digitally modulated bandpass signals
are generated by using the complex
envelopes for AM,PM,FM,or QM
signaling,But the modulating signal m(t)
is a digital signal given by the binary or
multilevel line codes
56
5.9 Binary Modulated Bandpass
Signaling
? binary bandpass
signaling
techniques,
? on-off keying
(OOK) or
Amplitude shift
keying(ASK),
? Binary phase-
shift keying
(BPSK) and
? Frequency-shift
keying(FSK),
m ( t )
m ( t )
s ( t )
s ( t )
s ( t )
s ( t )
a) U n i p ol ar
m od u l at i on
b ) p ol ar
m od u l at i on
c ) O O K
s i gn al
d ) B P S K
s i gn al
e ) F S K
s i gn al
f ) D S B - SC
s i gn al
? Fig.5-19 Bandpass digitally
modulated signals
57
5.9 Binary Modulated Bandpass
Signaling (OOK signal)
? OOK signal,
tftmAttmAts cccc ?? 2c o s)(c o s)()( ??where is the unipolar baseband data signal,It is shown
that OOK is identical to unipolar binary modulation
on a DSB-SC signal,
? complex envelope
)()( tmAtg c?
PSD is,
?
?
?
?
?
?
?
?
???
?
???
??? 22 s i n)(
2)( b
b
b
c
g fT
fTTfAfP
?
??
? PSD
comparison with(3.39b)
?????? ?????????? )(11s i n4)(
22
fTfT fTTAfP
bb
bbu n i p o l a r ?
?
?
58
5.9 Binary Modulated Bandpass
Signaling
? When m(t) has a peak value of A=2(1/2),s(t) has an
average normalized power of ; It shows that there
is a frequency line in the PSD of complex envelope for
OOK,and the OOK is AM-type signaling,
? The null-to-null bandwidth is2R,and the transmission
bandwidth is∞,while the absolute bandwidth is,
? Bit rate is R=1/Tb,
? If raised cosine-rolloff filtering is used(to conserve
bandwidth),the absolute baseband bandwidth is
B=(1/2)(1+r)R,where D=R=2B/(1+r) and r is the
rolloff factor of the filter,The absolute transmission
bandwidth is BT=(1+r)R
2/2cA
59
5.9 Binary Modulated Bandpass Signaling
(demodulation for OOK)
? Fig.5-21
60
5.9 Binary Modulated Bandpass
Signaling (demodulation for OOK)
? 1)Envelope detector(noncoherent detection)
? 2)product detector(coherent detection) with
Low-pass Filter Processing
? 3) product detector(coherent detection) with
Matched Filter Processing.This is an optimum
detection of OOK—that is,to obtain the
lowest BER when the input OOK signal is
corrupted by additive white Gaussian
noise(AWGN)
61
5.9 Binary Modulated Bandpass
Signaling
? Note,the optimum coherent OOK detector is
more costly to implement than the
noncoherent OOK detector,If the input noise
is small,the noncoherent receiver may be the
best solution,considering both cost and noise
performance,The trade-off in BER
performance between optimum coherent
detection and nonoptimum noncoherent
detection,
62
5.9 Binary Modulated Bandpass
Signaling （ BPSK）
? Bandpass signal for BPSK,
)](c o s [)( tmDtAts pcc ?? ?
? Where m(t) is a polar baseband data signal,
For convenience,let m(t) have peak values of
± 1 and a rectangular pulse shape
? complex envelope representation
Expanding equation above we get,
ttmDAtDA
ttmDAttmDA
tmDtAts
cPccPc
cPccPc
pcc
??
??
?
s i n)()s i n (c o s)c o s (
s i n))(s i n (c o s))(c o s (
)](c o s [)(
??
??
??
63
5.9 Binary Modulated Bandpass
Signaling
? it can be shown that BPSK is also a form of AM-type
signaling like OOK signaling,But when digital
modulation index,
122 ???? ?? ? pDh
Conclusion,For this optimum BPSK signal of h=1,
the complex envelope is
2/??? ?? pD
ttmA
ttmDAtDA
tmDtAts
cc
cPccPc
pcc
?
??
?
s i n)(
s i n)()s i n (c o s)c o s (
)](c o s [)(
??
??
??
)()( tmjAtg c?
and it equivalent to DSB-SC signaling with a polar
baseband data waveform
64
5.9 Binary Modulated Bandpass
Signaling
? PSD of complex envelope for optimum BPSK
Because m(t) is polar signaling,the PSD for the complex
envelope as
2
2 s i n)(
???
?
???
??
b
b
bcg fT
fTTAfP
?
?
where s(t) has an average normalized power of Ac2/2 2/2cA
the null-to-null bandwidth is 2R,like OOK
Detection for BPSK,
1) coherent detection
2) A optimum detection (the low-pass filter in fig.5.22a is
replaced by an integrate-and-dump matched filter
processing,)
65
5.9 Binary Modulated Bandpass Signaling
? Fig.5-21
66
5.9 Binary Modulated Bandpass
Signaling
? Notice:unlike OOK signaling,synchronous detection
must be used to detect BPSK,the main reason is that
there is no discrete carrier term in the BPSK signal;if
a low-level pilot carrier is transmitted together with
the BPSK signal,a PLL may be used to extract the
carrier reference,Otherwise,a Costas loop or
squaring loop may be used to synthesize the carrier
reference because BPSK is DSB-SC signal
? if Costas loop and square loop are used,the phase
ambiguity must be resolved,This can be accomplished
by using differential coding at the transmitter input
and differential decoding at the receiver output,
67
5.9 Binary Modulated Bandpass
Signaling(DPSK )
? Although phase-shift-keyed signals cannot be detected
incoherently,a partially coherent technique can be
used whereby the phase reference for the present
signaling interval is provided by a delayed version of
the signal that occurred during the previous signaling
interval
? if,the symbol phase
relationship between binary sequence and 2DPSK
signal is represented as,
??
??
0 i g i t a l0
1 i g i t a l
d
d???
00000
000000
1010011100
or
s i g n a l2 D P S K of p h a s e
s e q u e n c y d a t aB i n a r y
??????
?????
68
5.9 Binary Modulated Bandpass
Signaling
? detection
partially coherent detection
O n e - b i t
d e l a y,T b
L o w - p a s s
f i l t e r
D e c o d e d
b i n a r y
o u t p u t
D P S K
i n p u t
69
5.9 Binary Modulated Bandpass Signaling
(Frequency-shift keying (FSK))
? According to the method used to generate it,FSK
signal can be classified as two different types,
Discontinuous-phase FSK and Continuous-pahse FSK
? Discontinuous-phase FSK signal is represented by
??
?
?
????
0f o r )co s (
1f o r )co s ())(co s ()(
22
11
??
????
tA
tAttAts
c
c
cc
? when m(t) is polar binary pulse waveform,we can get,
)c o s ()()c o s ()())(c o s ()( 2211 ?????? ?????? ttmAttmAttAts cccc
? The discontinuous phase function is
?
?
?
??
???
0f o r
1f o r )(
22
11
tt
ttt
c
c
???
????
70
5.9 Binary Modulated Bandpass Signaling
(Frequency-shift keying (FSK))
? Continuous-phase FSK In order to generate the
continuous-phase FSK signal,the data signal are
feed into a frequency modulator,This FSK
signal is,
? ?tjtfcc cetgdmDtAts ???? )(Re))(c o s ()( ??? ? ??
? thus its complex envelope can be represented by,
)()( tjc eAtg ??? where
? ??? tf dmDt ??? )()(
71
5.9 Binary Modulated Bandpass Signaling
(Frequency-shift keying (FSK))
? It can be shown that although m(t) is
discontinuous at the switching time,the phase
function θ(t) is proportional to the integral of
m(t),If the serial data input waveform is
binary,such as a polar baseband signal,the
resulting FSK signal is called a binary FSK
signal,Of course,a multilevel input signal
would produce a multilevel FSK signal
72
5.9 Binary Modulated Bandpass Signaling
(Frequency-shift keying (FSK))
? Bandwidth of FSK signaling:For the FSK signal the
approximate transmission bandwidth is give by
Carson’s rule,
BFB T 22 ???? where B is the bandwidth of the digital modulation
waveform,When modulating waveform is square-
wave,the bandwidth is B=R,we can find that the this
FSK transmission bandwidth becomes,
RFB T 22 ?? ?? if a raised cosine-rolloff premodulation filter is used,
the transmission bandwidth of the FSK signal
becomes,
RrFB T )1(2 ??? ?
73
5.9 Binary Modulated Bandpass Signaling
(Frequency-shift keying (FSK))
? Generation of FSK(Fig 5-23)
F r e q u e n c y
m o d u l a t o r
( c a r r i e r
f r e q, = f c )
B i n a r y d a t a
i n p u t m ( t )
F S K o u t p u t
b ) c o n t i n u o u s - p h a s e F S K
O s c i l l a t o r
F r e q, = f 2
B i n a r y d a t a i n p u t m ( t )
C o n t r o l l i n e
E l e c t r o n i c
s w i t c h
a ) d i s c o n t i n u o u s - p h a s e F S K
F S K o u t p u t
O s c i l l a t o r
F r e q, = f 1
74
5.9 Binary Modulated Bandpass Signaling
(Detection of FSK Fig,5-28)
F r e q u e n c y
d e t e c t o r
F S K i n p u t
b i n a r y i n p u t
a ) n o n c o h e r e n t D e t e c t i o n
F S K i n p u t C o s ( ω 1 t) F S K
i n p u t
L o w - p a s s
f i l t e r
L o w - p a s s
f i l t e r
C o s ( ω 2 t)
+
-
b ) c o h e r e n t ( s y n c h r o n o u s ) D e t e c t i o n
75
5.9 Binary Modulated Bandpass Signaling
? Example 1
? Example 2
? Example 3
? Example 4
76
5.10 Multilevel Modulated Bandpass Signaling
? If a level stands for a k bits binary digits,the number
of levels in the multilevel signal is, M=2k,The
symbol rate(baud) of the multilevel signal is D=R/k，
for example,if k=2,M=4 and baud D=R/2,where the
bit rate is R=1/Tb。
? Multilevel Modulated bandpass signaling used often
have,
? QPSK; MPSK;QAM; OQPSK and π/4QPSK
M od ul a t e d
ou t pu t
D i gi t a l - to - a na l
og c on ve r t e r l
bi t
B i na r y i np ut
M = 2
l
- l e ve l
di gi t a l s i gn a l
D = s ym bo l s / s
= R / l
F i g,5 - 29 m od ul a t e d di gi t a l t r a ns m i s s i on s ys t e m
T r a ns m i t t e r
77
5.10 Multilevel Modulated Bandpass Signaling
(MPSK)
? In general,MPSK can also generated by using two
quadrature carriers modulated by in-phase and
quadrature components of the complex envelope
except using a phase modulator,in that case,
)()()( )( tjytxeAtg tjc ??? ?
? where the permitted values of x and y are
ici Ax ?c o s? ici Ay ?sin?
Mii,,2,1,???
? is the permitted phase angles of the
MPSK signal,
78
5.10 Multilevel Modulated Bandpass Signaling
(QPSK)
? Fig 5-31
79
5.10 Multilevel Modulated Bandpass Signaling
? M-ary phase-shift keying(MPSK) with M=4 is
called QPSK,its permitted values of complex
envelope contains four points
)()( tjc eAtg ??? A plot of two possible sets of g(t) is shown as
following
00 01
11 10
00
01
11
10
80
5.10 Multilevel Modulated Bandpass Signaling
(QAM)
? in general,the QAM signal constellations are not
restricted to have permitted signaling points only on a
circle,The general QAM signal is,
ttyttxts cc ?? s i n)(c o s)()( ??? its complex envelope is
)()()()()( tjetRtjytxtg ????
? where
2)()()( tjytxtR ??
))(/)((t a n)( 1 txtyt ???
81
5.10 Multilevel Modulated Bandpass Signaling
(QAM)
? Fig.5-32 16 symbol
QAM constellaltion
y i m a g i n a r y
a x i s
( q u a d r a t u r e )
x r e a l
a x i s
( i n p h a s e )
? The waveforms of I
and Q components are
represented by,
? ??
n
n D
nthxtx )()(
? ??
n
n D
nthyty )()(
? where h(t) is the pulse
shape that is used for
each symbol,and
baud rate D=R/k
82
5.10 Multilevel Modulated Bandpass Signaling
? In some applications,the timing between the x(t) and
y(t) components is offset by Ts/2=1/(2D),that is,
? ??
n
n D
nthxtx )()( ? ???
n
n DD
nthyty )
2
1()(
? this signal is called offset QAM(OQAM)
? Offset quadrature phase-shift keying(OQPSK)
One popular type of offset signaling is offset
QPSK(OQPSK) which is identical to offset QAM
when M=4,The AM is reduced because a maximum
phase transition of only π/2 occurs (as opposed to for
QPSK) since the I and Q data cannot change
simultaneously,because the data are offset,
?
83
5.10 Multilevel Modulated Bandpass Signaling
(π/4 QPSK modulation)
? A π/4 QPSK signal is generated
by alternating between two
QPSK constellations that are
rotated by π/4 with respect to
each other,the two signal
constellations are used
alternately as follows,Given a point on one of the signal
constellations that corresponds to two bits of input data,
two new bits are read to determine the next point that is
selected from the other constellation,For example an
input data is ―00,11,10,10,01‖,if ―00‖correspond to a
phase shift of Δθ=450,a ―11‖ to Δθ=1350,a ―10‖ to
Δθ=1350,a ―10‖ to Δθ=-450 and so on,
?
00 01
11 10
00
01
11
10
84
5.10 Multilevel Modulated Bandpass Signaling
(relationship between QPSK π/4 QPSKand
OQPSK )
? comparison between QPSK,QPSK and OQPSK for
non-rectangle pulse waveform
Q P S K
QPSK)4/(?O Q P S K
?)
43,41(43 ??? ????21
Type of
modulation
Maximum phase
shift
AM big small smallest
Q A M M P S K Q P S K
O Q P S K
( π / 4 ) Q P S K
I n a c i r c l e M = 4
D e l a y
A l t e r n a t i n g
85
5.10 Multilevel Modulated Bandpass Signaling
? PSD for MPSK,QAM,QPSK,OQPSK and π/4 QPSK
? the complex envelope for MPSK and QAM is
represented by,
??
??
?? )()( sn nTtfctg? where c
n is a complex-valued random variable representing the multilevel value during the nth
symbol pulse,And f(t)=Π(t/Ts) is the rectangular
symbol pulse with symbol duration Ts,and its Fourier
transform is
???
?
???
??
???
?
???
??
b
bb
s
ss
fTk
fTkkT
fT
fTTfF
?
?
?
? s i ns i n)(
? thus the PSD for the complex envelope of MPSK or
QAM signal with data modulation of rectangular bit
shape is
2s i n
)( ??
?
?
???
??
s
sbg
fT
fTC k TfP
?
?
86
5.11 Minimum-Shift Keying (MSK)
and GMSK
? background
? In practice,the communication system need to
conserve efficiently bandwidth and improve
the efficiency of power supply,for example
Class C amplifiers without distortion can be
used,
? Minimum-shift keying is another bandwidth
conservation technique that has been
developed,Its advantage of producing a
constant-amplitude signal and,consequently,
can be amplified with Class C amplifiers
without distortion
87
5.11 Minimum-Shift Keying (MSK)
and GMSK
? Definition
MSK is continuous 2FSK with minimum
modulation index h=0.5
? Properties of MSK
1)continuous binary FSK；
2)modulation index being minimum
3)two signals are orthogonal over the bit
interval；
4)being identical with OQPSK when the pulse
shape is sinusoidal type
88
5.11 Minimum-Shift Keying (MSK)
and GMSK
? Obtaining modulation index h
For FSK
)Tt(0 0f o r )c o s ()( 1f o r )c o s ()()( b
222
111 ??
??
?
??
???
??
??
tAts
tAtsts
c
c
According to property 1),we get,
0 t t i m es w i t ch i n g at t h e 21 ?? ??
And according to 3),we have,
? ?bT dttsts0 21 0)()(
89
5.11 Minimum-Shift Keying (MSK)
and GMSK
? That is,
dtttAdttsts bb T cT ?? ??? 0 221120 21 )c o s ()c o s ( )()( ????
? ?
? ?
?
?
?
?
?
?
?
?????
??
?
?
?
?
?
?
?????
?
21
212121
2
21
212121
2
)s in ()()(s in
2
)s in ()()(s in
2
??
??????
??
??????
bc
bc
TA
TA
? ? ?
?
??
?
?
?
??????
21
2121212 )sin()()(sin
2 ??
?????? bc TA
? ? ?
?
??
?
?
?
??????
F
FTA bc
22
)sin()(22sin
2
21212
?
?????
? ? 0
2
)s i n ()(2s i n
2
2121
2
??
?
?
??
?
?
?????
F
hA c
?
?????
90
5.11 Minimum-Shift Keying (MSK)
and GMSK
? And obtain,
? ? )s i n ()(2s i n 2121 ????? ????h
? when θ1≠ θ2
?1,2,k 2 ?? ?? kh
?1,2,k 22 ?? ?? kh
? when θ1= θ2
? Thus,the modulation index for MSK
bFTh ??? 22/1 4/)4/(1 RTF b ????
91
5.11 Minimum-Shift Keying (MSK)
and GMSK
? complex envelope for MSK
For the FSK signal over the signaling interval(0,Tb),the
complex envelope is,
)Tt(0 )( b2)(2)( 0 ?????? ?? tTjcdmFjctjc b
t
eAeAeAtg
????
?
or
)Tt(0 )()()( b???? tjytxtg
where
)Tt(0 ),2T t1c o s ()( b
b
????? ?cAtx
)Tt(0 ),2T t1s in ()( b
b
????? ?cAtyThe MSK signaling is,
ty ( t ) s i n-tc o s)()( cctxts ???
92
5.11 Minimum-Shift Keying (MSK)
and GMSK
? Because,
)Tt(0 ),2T t1c o s ()( b
b
????? ?cAtx
)2Ttc o s (
b
?
cA?and
)Tt(0 ),2T t1s in ()( b
b
????? ?cAty
)2T ts in (
b
?
cA??
93
5.11 Minimum-Shift Keying (MSK)
and GMSK
94
5.11 Minimum-Shift Keying (MSK)
and GMSK
? Type I MSK,Basic pulse waveform alternates
between a positive and a negative half-cosinusoid
? Type II MSK:Basic pulse waveform always a
positive half cosinusoid
? The instantaneous frequency is,
Ffdt tdff cci ????? )(2 1 ??
? where
???
?
???
?? ?
)(
)(t a n)( 1
tx
tyt?
95
5.11 Minimum-Shift Keying (MSK)
and GMSK
? PSD for MSK
? Because x(t) and y(t) have independent data and their
dc value is zero,and since g(t)=x(t)+jy(t),the PSD for
the complex envelope is,
)f()f()f()f( xyxg 2 PPPP ???
2
g 2
2 )f(F
T)f( b?P
? Since the pulse width is 2Tb,this PSD becomes,
??
??? ??
el s ew h er e
Tt
T
tA
tf bbc
0
)
2
co s ()( ?
? where,is the pulse shape,For the MSK
half-cosinusoidal pulse shape,we have,[f ( t )]fF F?)(
96
5.11 Minimum-Shift Keying (MSK)
and GMSK
? and its Fourier transform is,
? ?2)4(1 2c o s4)( fT fTTAfF b bbc ?? ? ?
? ?222
22
)4(1
2c o s16)(
fT
fTTAf
b
bbc
g
?
?
?
?P
? Thus,the PSD for the complex envelope for an
MSK signal is,
97
5.11 Minimum-Shift Keying (MSK)
and GMSK
? Generation of Fast MSK signals
FM t ra nsmit te r
△ F= (1 /4 )R
m (t)
in p u t
s(t)
FFSK s ig n al
? Parallel Generation of Type I MSK signals
S erial - to
- p aral lel
co n v ert e
r( 2 bit)
m ( t)
inpu t
s ( t)
M S K s ignal
O s cill ator
f 0 = △ F =R /4
co s ( π t/2T b )
- 90 0
p h as e sh ift
S y nc
s ignal
x data
y data
y ( t)
x( t)
C arr ier
o s cill ator,f c
p h as e sh ift
- 90 0
p h as e sh ift
A c cos ( ω c t)
A c s in( ω c t)
? Serial Generation of MSK
m ( t)
in p u t
s ( t)
M S K s i g n a l a t
c a r r i e r f r e q u e n c y f c C a r r i e r o s c i l l a r a t
f r e q u e n c y f 2 =f c - △ F
B P S K s i g n a l a t c a r r i e r
f r e q u e n c y f c M S K b a n d p a s s f i l t e r
a b o u t f 1 H ( f )
98
Minimum-Shift Keying (MSK) and
GMSK
? Gaussian-filtered MSK (GMSK)
Before the data are frequency modulated onto the
carrier,the data are filtered by a filter having a
Gaussian-shaped frequency response characteristic,
The transfer function of the Gaussian low-pass filter is
? ?)2/2( l n)/( 2)( BfefH ??
? Properties of GMSK
1) Low spectral sidelobes on the transmitted MSK signal
2) constant envelope
99
5.11 Minimum-Shift Keying (MSK)
and GMSK
? Comparison of the spectral efficiencies of
various types of digital signal
? Table 5-7 spectral efficiency of digital signals
100
5.11 Minimum-Shift Keying (MSK)
and GMSK
101
5.12 Orthogonal Frequency Division
Multiplexing (OFDM)
? comparion between FDM and OFDM
a
c
a,FDM b,OFDM
102
5.12 Orthogonal Frequency Division
Multiplexing (OFDM)
? principle of OFDM
I n p u t b i t s
C h a n n e l
c o d i n g
S y m b o l
m a p p i n g
S / p I F F T
R a i s i n g - c o s i n e
w i n d o w
cp
P / s
D / A
RF - TX
a ) t a ns m i t t e r
O u t p u t b i t s
C h a n n e l
d e c o d i n g
d e m a p pi n
g
P / s FFT
s yn t h e s i s
De - cp
P / s
A / D
RF - RX
b) re c e i ve r
F r e,c or re c
te
C S I e s t i m a t o r
E q u a l i z a t -
i o n
P i l o t
s ym b o l
103
5.12 Orthogonal Frequency Division
Multiplexing (OFDM)
d( n)
S
/P
c
on
ve
rt
e
r
mN
tfj
e
1
2
?
?
m
tfj
e
1
2 ?
)( ms
?
?
?
?
?
?
?
?
1
0
1
0
)2e x p ()(
)2e x p ()()(
N
n
mn
N
n
tfjnd
N
nmjndms
?
?
s
n NT
nf ?
sm mTt ?
)1,...,1,0( ?? Nm
OFDM modulator
104
5.12 Orthogonal Frequency Division
Multiplexing (OFDM)
? complex envelope of OFDM
)2 1(1)()( 1
0
21
0
????? ?? ?
?
?
?
Nn
TfeaAtaAtg n
N
n
tfj
nc
N
n
nnc n
??
? Where is a constant,
2na
0?na? PSD for OFDM
? Because the PSD of each carrier is of the form
|Sa(π(?-?n))|2,the overall PSD for the complex
envelope of the OFDM signal is
21
0 )(
))(s i n ()( ??
? ?
?? N
n n
n
g Tff
TffCf
?
?P
TaAC nc 2?
105
5.12 Orthogonal Frequency Division
Multiplexing (OFDM)
? Application of OFDM
? Wireless LAN
? Digital Audio broadcasting in the European
? Digital Video broadcasting in European
? Broad wireless communication
? DMT in ADSL
?
106
5.13 Spread Spectrum System
? In communication systems mentioned above,we have
concerned primarily with the performance of
communication systems in terms of bandwidth efficiency
and energy efficiency with respect to natural noise,
However,in some application,we also need to consider
multiple-access capability,antijam capability,
interference rejection and covert operation,or low-
probability of intercept(LPI) capability,These
performance objectives can be optimized by using spread
spectrum techniques,
? Advantage of spread spectrum techniques,
? 1 Low-probability of intercept;
? 2 Anti-jamming;
? 3 Multi-access
107
5.13 Spread Spectrum System
? An system which can be considered an SS system must
satisfy two criteria,
? 1 The bandwidth of the transmitted signal s(t) needs
to be much greater than that of the message m(t),
? 2 The relatively wide bandwidth of s(t) must be
caused by an independent modulating waveform c(t)
called the spreading signal,and this signal must be
known by the receiver in order for the message signal
m(t) to be detected,
? The SS signal is,
? ?tj cetgts ?)(Re)( ?
its complex envelope is a function of both m(t) and c(t),
In most cases,a product function is used,so that,
g(t)=gc(t)gm(t)
108
5.13 Spread Spectrum System
? SS signals can be classified by the type of mapping
functions that are used for gc(t),The mose common
types of SS signals have,
? 1) Direct Sequence(DS)
? A DSB-SC type of spreading modulation is used,and
gc(t)=c(t) is a polar NRZ waveform,
? 2) Frequency Hopping(FH)
? gc(t) is of FM type where there are M=2k hop
frequencies determined by the k-bit words obtained
from the spreading code waveform c(t),
? 3) Hybrid techniques
? A system include both DS and FH
109
5.13 Spread Spectrum System
(Direct Sequence-DC )
if information waveform m(t)=± 1,gm(t)=Acm(t),for DC g
c(t)=c(t),the complex envelope for the SS singal
becomes,
)()()( tctmAtg c?
the DS signal is
? ?tj cetgts ?)(Re)( ?
called a direct sequence spreading,spread spectrum
signal(DS-SS)Fig.5-39a,c(t) is a polar spreading signal,
which is generated by using a pseudonoise(PN) code
generator,as shown in Fig.5-39b
110
5.13 Spread Spectrum System
111
5.13 Spread Spectrum System
? maximum-length sequence,called m-sequence is a PN
code which is generated by r shift register stages and
has a maximum period of N=2r-1 chips,A m-sequence
has some properties as following,
? P1,In one period,the number of 1’s is always one
more than the number of 0’s;
? P2,the modulo-2 sum of any m-sequence,when
summed chip by chip with a shifted version of the
same sequence,produces another shifted version of
the same sequence,
? P3,if a window of width r is slid along the sequence
for N shifts,then all possible r-bit words will appear
exactly once,except for the all 0 r-bit word,
? P4,if the 0’s and 1’s are represented by –1 and +1V,
the autocorrelation of the sequence is,
112
5.13 Spread Spectrum System
? P4,if the 0’s and 1’s are
represented by –1 and +1V,
the autocorrelation of the
sequence is,
??
???
??
?
? Nk 1
Nk 1
)( ?
?
N
kR c
? ?? ?? 10)/1()( Nn knnc ccNkR
1??nc
? where
113
5.13 Spread Spectrum System
? The PSD for the complex envelope of BPSK-DS-SS
signal is,
(f )P*(f )P(f )P cmg 2cA?? Fig,5-41
? Without
spreading,the
level of the in-
band PSD would
be proportional to
Ac2/(2Rb),but
with spreading,
the in-band
spectral level
drops to Ac2/(2Rc)
114
5.13 Spread Spectrum System
(Receiver )
? If the input to the receiver consists of the SS
signal plus a narrowband(sine wave) jammer
signal,then,
)(c o s)()()()()( tnttctmAtntstr jcc ???? ?? where the jamming signal is The
output of despreader is
tAtn cjj ?c o s)( ?
ttcAttmAv cjcc ?? c o s)(c o s)(1 ??? After LPF,then
)()(2 tcAtmAv jc ??? the jammer power at the receiver output is
bc
jR
R cjn RR
Adf
RAP
b
b /2
1 22 ?? ?
?
115
5.13 Spread Spectrum System
(Receiver )
? The principle of LPI and anti-jamming
116
5.13 Spread Spectrum System
(Frequency hopping )
? In a frequency-hopping(FH) SS system,there are M=2k
hop frequencies controlled by the spreading code,in
which k chip words are taken to determine each hop
frequency,FH is accomplished by using a mixer circuit
wherein the LO signal is provided by the output of a
frequency synthesizer that is hopped by the PN
spreading code,The serial-to-parallel converter
programmable dividers in the frequency synthesizer,
The k-chip word speifies one of the possible M =2k hop
frequencies,
117
5.13 Spread Spectrum System
(Frequency hopping )
? The FH signal receiver has the knowledge of the
transmitter,c(t),so that the frequency synthesizer in
the receiver can be hopped in synchronism with that
at the transmitter,This despreads the FH signal,and
the source information is recovered from the
dehopped signal with the use of a conventional FSK or
BPSK demodulator,as appropriate,
118
5.13 Spread Spectrum System
(Frequency hopping )
? Fig.5-42
119
5.14 Summary
? AM,SSB,PM,and FM signaling techniques were
considered in detail
? Digital signaling techniques such as OOK,BPSK,FSK,
MSK and OFDM were developed,The spectra for
these digital signals were evaluated in terms of the bit
rate of the digital information source
? Multilevel digital signaling techniques such as QPSK,
MPSK and QAM,were also studied,and their spectra
were evaluated,
? The key parts is 5.6,5.9,5.10,5.11
120
5.14 Homework
Problem,
5-1 5-10 5-11 5-12
Chapter 5
AM,FM,and digital
modulation systems
2
5.0 Introduction
(Chapter objectives)
? Amplitude modulation and single sideband
? Frequency and phase modulation
? Digitally modulated signals(OOK,BPSK,
FSK,MSK MPSK,QAM,QPSK,
π/4QPSK,and OFDM)
? Spread sprectrum and CDMA system
3
5.0 Introduction
(the goal of this chapter)
? Study g(t) and s(t) for various types of
analog and digital modulations
? Evaluate the spectrum for various types of
analog and digital modulations
? Examine some transmitter and receiver
structures
? Learn about spread spectrum systems
4
5.0 Introduction
(the key of this chapter)
? Grasping phase modulation and frequency modulation ? Grasping Binary modulation and Bandpass signaling
? Grasping Multilevel Modulated bandpass signaling
? Grasping Minimum-shift keying (MSK)and GMSK
? Knowing the principle of FDM,OFDM and Spread
Spectrum Systems
5
5.0 Introduction
(band-pass signal and its spectra)
? Communication system constructure
input signal processing carrier circuits
Transmission
Medium
(channel)
transmitter
g(t) s(t)
carrier
circuits
signal
processing
r(t) g(t) output
? Modulation is the process of imparting the source
information onto a Band-pass signal with a carrier
frequency fc by the introduction of amplitude or
phase perturbations or both,
6
5.0 Introduction
(band-pass signal and its spectra)
Theorem:Any physical band-pass waveform can be representative by
} R e { g ( t ) eS ( t ) tj c??? g(t) is called the complex envelope of s(t),and f
c is
carrier frequency (in hertz),ωc=2πfc
? Notice,The desired type of modulated signal,s(t),is
obtained by selecting the appropriate modulation
mapping function g[m(t)],where m(t) is the analog or
digital base-band signal,
? The voltage (or current)spectrum of the band-pass
signal is
21 ) ] G * ( - f- f)[ G ( f - f/S ( f ) cc ??? and the PSD is
41 ) ] ( - f - f )( f - f[/( f ) c*gcgs PPP ??
7
5.1 Amplitude Modulation
? The complex envelope of an
AM signal is
given by
? Where Ac specifies
the power level and
m(t) is the
modulating signal,
The representation
for AM signal is
given by,
]1[ m ( t )Ag ( t ) c ??
tm ( t)As ( t) cc ?c o s]1[ ??
? Fig.5-1
8
5.1 Amplitude Modulation
(modulation percentage for AM signal )
? Definition %positive modulation= (A
max-Ac)/ Ac × 100=max[m(t)] × 100
%negative modulation= (Ac-Amin) / Ac× 100=-min[m(t)]
× 100
? %overall modulation= (Amax- Amin)/ (2Ac)× 100
={max[m(t)]-min[m(t)]}/2 × 100
? where Amax is the maximum value of Ac [1+m(t)],and Ac is
the level of the AM envelope in the absence of modulation,
? notice:The percentage of modulation can be over100
%(Amin will have a negative value).If the transmitter uses
a two-quadrant multiplier that produces a zero output
when Ac[1+m(t)] is negative,the output signal will be,
9
5.1 Amplitude Modulation
? Which is a distorted AM signal which is call as over
modulated signal, Its bandwidth is much wider than
that of the undistorted AM signal,
?
?
?
?
???
-1m ( t ) if 0
-1m ( t ) if c o s)](1[)( ttmAts cc ?
? if the percentage of negative modulation is less than
100%,an envelope detector may be used to recover the
modulation without distortion;if the percentage of
negative modulation is over 100%,undistorted
modulation can still be recovered provided the product
detector is used,
? A product detector is superior to an envelope detector
when the input signal-to-noise ratio is small,
10
5.1 Amplitude Modulation
? because the total average normalized power of a band-pass waveform v(t) is given by
22 )(
2
1)0()()( tgRdffPtvP vvv ???? ??
??? Where ―normalized‖ implies that the load is equivalent
to 1 ohm,The normalized average power of the AM
signal is,
)(2/1)(2/1
)()(212/1
)](1[2/1)(2/1)(
2222
22
222
tmAtmAA
tmtmA
tmAtgts
ccc
c
c
???
???
???
? If the modulation contains no dc level,then 〈 m(t) 〉
=0,and the normalized power of the AM signal is
9)-(5 2/12/1
)(2
p o w e rc a r r i e r
2)( 22
?? ??? ?????
p o w e rs i d e b a n d
tc
d i s c r e t e
ct mAAs ??
11
5.1 Amplitude Modulation
? modulation efficiency:the percent of the total power of the modulated signal that conveys information,
%1 0 0
)(1
)(
2
2
?
?
?
tm
tm
E
? The highest efficiency that can be attained for a 100%
AM signal would be 50%.(For the case when square-
wave modulation is used),The normalized peak
envelope power (PEP)of the AM signal,
2
2
)]}(m a x [1{2 tmAP cPEP ??
12
5.1 Amplitude Modulation
? The voltage spectrum of the AM signal is given by
)]()()()([2)( ccccc ffMffffMffAfS ???????? ??? The AM spectrum is just a translated version of the
modulation spectral component.The bandwidth is
twice that of the modulation
? Example 5-1 Suppose that a 5000w AM transmitter is
connected to a 50Ω load; If the transmitter is 100%
modulated by a 1000Hz test tone,Then compute
(1)the total actually average power,(2)actually PEP,(3)
The modulation efficiency,
? Solution,the constant Ac is given by (1/2)Ac2/50=5000,
thus peak voltage across the load will be Ac=707V
during the times when there is no modulation,
13
5.1 Amplitude Modulation
(2)then across the 50 Ω load,the PEP is,
50/]1[)2/1(
/)2/12/1(
)(2
)(22)(
2
22
tc
tcct
mA
RmAAs
??
??
WRA
Rtm
A
P
c
c
PEP
2 0 0 0 050/7 0 7*4*)2/1(/*4*)2/1(
/)]}(m a x [1{
2
22
2
2
???
??
(3)The modulation efficiency would be,
%33%100
)(1
)(
2
2
??
?
?
tm
tm
E
(1)the actual average power
14
5.1 Amplitude Modulation
15
5.1 Amplitude Modulation
- B 0 B
f
| M ( f ) |
- f c - B - f c - f c +B
f
A c /2
f c - B f c f c +B
W e i ght = A c /2 | S ( f ) | W e i ght = A c /2
a ) M a gni t ude s pe c t r um of m odul a t i on
b) M a gni t ude s pe c t r um of A M s i gna l
U ppe r
s i de ba nd
l ow
s i de ba nd
Fig,4-2 Spectrum of AM signal
16
5.3 Double-sideband Suppressed
Carrier
? A double-sideband suppressed carrier(DSB-SC) signal is an AM signal that has a suppressed discrete carrier,
tm ( t)As ( t) cc ?c o s?? The voltage spectrum of the DSB-SC signal is
21 ) ] fG * ( f)[ G ( f- f/S ( f) cc ???? The percent of modulation of a DSB-SC is,infinite
? The modulation efficiency of a DSB-SC is 100%
? A product detector (which is more expensive than an
? Envelope detector) is required for demodulation of the
DSB-SC signal
? The sideband power of a DSB-SC signal is four times
that of a comparable AM signal with the some peak
level,that is to say,the DSB-SC signal has a four_fold
power advantage over that of an AM signal
17
5.3 Double-sideband Suppressed
Carrier
- B 0 B
f
| M ( f ) |
- f c - B - f c - f c +B
f
A c /2
f c - B f c f c +B
W e i ght = A c /2 | S ( f ) | W e i ght = A c /2
a ) M a gni t ude s pe c t r um of m odul a t i on
b) M a gni t ude s pe c t r um of A M s i gna l
U ppe r
s i de ba nd
l ow
s i de ba nd
Spectrum of DSB-CS signal
18
5.5Asymmetric sideband signals
(Signal sideband)
? An upper single sideband (USSB) signal has a zero-valued spectrum for|f|<f
c,where fc is the carrier
frequency,
? A lower signal sideband (LSSB) signal has a zero-
valued spectrum for |f|>fc,where fc is the carrier
frequency,
? SSB-AM:The bandwidth is the same as that of the
modulating signal (which is half the bandwidth of an
AM or DSB-SC signal),
? The term SSB refers to the SSB-AM type of signal,
unless otherwise denoted,
19
5.5Asymmetric sideband signals
? Theorem:The complex envelope of an SSB signal is
given by
)](?)([)( tmjtmAtg c ??? Which results in the SSB signal waveform,
]s in)(?c o s)([)( ttmttmAts ccc ?? ??
? Where the upper(-) sign is used for USSB and the
lower (+) sign is used for LSSB,Denotes the
Hilbert transform of m(t),which is given by,)(? tm
)()()(? thtmtm ??? Where h(t)=1/(πt),and H(f)=F[h(t)] corresponds to a
? -900 phase-shift network,
?
?
?
?
???
0
0 )(
fj
fjfH
20
5.5Asymmetric sideband signals
Fig.5-4 Spectrum for a USSB signal
| M ( f) |
- B B
1
( a ) B a s e b a n d M a g n i t u d e S p e c t r u m
| G ( f) |
- B B
2A c
( b ) M a g n i t u d e o f c o r r e s p o n d i n g S p e c t r u m
o f t h e c o m p l e x E n v e l o p e fo r U S S B
| S ( f) |
- fc - B - fc fc fc + B
A c
( b ) M a g n i t u d e o f c o r r e s p o n d i n g S p e c t r u m
o f U S S B s i g n a l
? Proof,Taking the Fourier transform
of the complex envelope of an SSB
signal,we get,
) ]}(?[)({)( tmjfMAfG c F??
? Because of Hilbert transform,we can
find that the equation becomes,
)](1)[()( fjHfMAfG c ??
? For USSB case,choose the upper sign,
then,
??
?
?
????
0
0
0
)(2)](1)[()(
f
ffMAfjHfMAfG c
c
? Substituting it into eq.(4-15),we have,
?
?
?
?
?
?
??
??
?
?
?
?
?
?
?
?
?
??
?
c
c
cc
c
ccc
ff
ff
ffMA
ff
ffffMA
fS
)(
0
0
)(
)(
21
5.5Asymmetric sideband signals
? The normalized average power of the USSB signal is
)(
)(?)(
2
1
|)(?)(|
2
1
)(
2
1
)(
22
)()(?
222
2222
22
tmA
tmtmA
tmtmAtgts
c
tmtm
c
c
?
?
??
???
? The SSB signal power is the power of the modulating
signal〈 m2(t)〉 multiplied by the power gain factor Ac,
? The normalized peak envelope power (PEP) is,
})](?[)(m a x {)2/1(})(m a x {)2/1( 222 2 tmtmAtg c ??
22
5.5Asymmetric sideband signals
? Fig,5-5 Generation of SSB
m ( t )
M o d u l a t i o n
i n p u t
- 90
0
p h a s e
s h i f t a c r o s s
b a n d o f m ( t )
B a s e b a n d
p r o c e s s i n g
O s c i l l a t o r
f = f c
- 90
0
p h a s e
s h i f t a t f = f c
+
+
s ( t )
S S B s i g n a l
m ( t )
v 1 (t)
v 1 ( t ) = A c c o s ( ω c t)
v 2 ( t ) = A c s i n ( ω c t)
a) p h a s e m e t h o d
S i d e b a n d f i l t e r
( b a n d p a s s f i l t e r o n
e i t h e r u p p e r o r
l o w e r s i d e b a n d )
O s c i l l a t o r
f = f c
m ( t )
M o d u l a t i o n
i n p u t
s ( t )
S S B s i g n a l
b) f i l t e r m e t h o d
23
5.5Asymmetric sideband signals
? SSB signals have both AM and PM,The AM
component (real envelope) is,
22 )](?[)()()( tmtmAtgtR c ???? The PM component is
])( )(?[t a n)()( 1 tm tmtgt ???? ??? SSB signals may be received by using a super- heterodyne
receiver that incorporates a product detector with θ0=0,
thus the receiver output is,
)(})(R e { 0 tmKAetgKv cjout ?? ? ?? Where K depends on the gain of the receiver and the loss in
the channel,In detecting SSB signals with audio
modulation,the reference phase θ0 does not have to be zero,
because the same intelligence is heard regardless of the
value of the phase used,For digital modulation,the phase
has to be exactly correct so that the digital waveshape is
preserved
24
5.5Asymmetric sideband signals
(Vestigial sideband)
25
5.5Asymmetric sideband signals
(Vestigial sideband)
? In certain application,a DSB modulation technique takes
too much bandwidth for the channel,and an SSB
technique is too expensive to implement,although it takes
only half the bandwidth,In this case,a compromise
between DSB and SSB,called vestigial sideband(VSB),is
often chosen,
? VSB is obtained by partial suppression of one of the
sidebands of a DSB signal,
? sideband of the DSB signal is attenuated by using a band-
pass filter,called a vestigial sideband filter,The VSB signal
is given by,
)(*)()( thtsts vV S B ?? Where s(t) is a DSB signal,and h
v(t) is the impulse response
of the VSB filter.The spectrum of the VSB signal is,
)()()( fvHfSfV S BS ?
26
5.5Asymmetric sideband signals
(Vestigial sideband)
? The modulation on the VSB signal can be recovered by
a receiver that uses product detection or,if a large
carrier is present,by the use of envelope detection,For
recovery of undistorted modulation,the transfer
function for VSB filter must satify the constraint,
BfCffHffH cvcv ?????,)()(? Proof,
)]()()()()[2/()( fHffMfHffMAfS vcvccV S B ????
? Thus only when the constraint satisfy,Vout(f)=KM(f) hold
true,
BfffHffHfMAA
BfffHfMffHfMAAfV
fHfffffSAfV
tsAtV
cvcvc
cvcvco u t
ccV S Bo u t
V S Bo u t
o
?????
?????
????
?
) ],()()[()4/(
) ],()()()()[4/()(
)() ] }()2/1()()2/1[(*)({)(
h ( t )*t]c o s)([)(
0
0
0 c
??
?
? the product detector output is
27
Summary
? AM
DSB-SC
SSB-AM
VSB
]1[ m ( t )Ag ( t ) c ??
m ( t )Ag ( t ) c?
)](?)([)( tmjtmAtg c ??
)](?)([)( tmjtmAtg c ?? )(*)()( thtsts vAMS S BV S B ??
tm ( t)As ( t) cc ?c o s]1[ ??
tm ( t)As ( t) cc ?c o s?
]s i n)(?c o s)([)( ttmttmAts cccAMS S B ?? ???
)(*]s i n)(?c o s)([)( thttmttmAts vcccV S B ?? ??
21 ) ] fG * ( f)[ G ( f- f/S ( f) cc ???
)]()()()([2)( ccccc ffMffffMffAfS ???????? ??
??
?
??
?
??
??
????
?
??
?
?
???
c
c
ccc
ccc
ff
ff
ffMAff
ffffMAfS
)(
0
0
)()(
)]()()()()[2/()( fHffMfHffMAfS vcvccV S B ????
28
Summary
? Example 5-2 if linear modulated signal is as following,
(1) s1(t)=cos(ω0t)cos (ωct)
(2) s2(t)(1+0.5sin (ω0t)) cos (ωct)
Where ωc=6ω0,please giving their spectrum,
? Solution, (1) the spectrum is,
)]5()5()7()7([
2
)]}()([*)]()([{
2
1
)(
0000
001
????????????
?
??????????????
?
?
????????
??????? ccS
-7ω0 -5ω0 5ω0 7ω0
29
Summary
? (2) the spectrum is
)]7()7()5()5({
4
)]6()6([
)]}()([*)]()([{
2
5.0
)()([)(
0000
00
00
2
????????????
?
???????
?????????????
?
?
????????
????????
????
???????
????
j
j
S
cc
cc
-7ω0 -5ω0 5ω0 7ω0
j ω
ω
S2(f)
30
5.6 Phase Modulation and Frequency Modulation
(Representation of PM and FM signal)
? PM and FM are special cases of angle-modulated
signaling,In this kind signaling the complex envelope is,
)()( tjc eAtg ??? Here the real envelope R(t)=|g(t)|=A
c,is a constant,and the phase θ(t) is a linear function of the modulating
signal m(t),However,g(t) is a nonlinear function of the
modulation,The resulting angle-modulated signal is,
)](c o s [)( ttAts cc ?? ??? For PM,the phase is directly proportional to the modulating
signal
)()( tmDt p??? Where the proportionality constant D
p is the phase sensitivity of the phase modulator,its units is radians per
volt,For FM,the phase is proportional to the integral of
m(t),so that,
31
5.6 Phase Modulation and Frequency
Modulation (Representation of PM and FM
signal)
? Where the frequency deviation constant Df has units
of radians/volt-sencond,From the last two equations,
we can see that if we have a PM signal modulated by
mp(t),there is also FM on the signal,corresponding to
a different modulation waveshape that is given by,
?
??
? ??? dmDt tf )()(
]
)(
[
dt
tdm
D
D
m p
f
p
f ?? Where the subscripts f and p denote frequency and
phase,Similarity,if we have an FM signal modulated
by mf(t),the corresponding phase modulation on this
signal is,
? ??? t f
p
f
p dmD
D
m ?? )(
32
5.6 Phase Modulation and Frequency
Modulation (Representation of PM and FM
signal)
? Fig,5-7 Generation of FM from PM,and vice versa
m f ( t ) I n t e g r a t o r
g a i n = D f / D
p
m p ( t ) P h a s e m o d u l a t o r
( c a r r i e r f r e q u e n c y = f c )
s ( t ) F M
s i g n a l o u t
m p ( t ) d i f f e r e n t i a t o r
g a i n = D p /D f
m f ( t ) f r e q u e n c y m o d u l a t o r
( c a r r i e r f r e q u e n c y = f c )
s ( t ) P M
s i g n a l o u t
a ) G e n e r a t i o n o f F M u s i n g a p h a s e M o d u l a t o r
b ) G e n e r a t i o n o f P M u s i n g a f r e q u e n c y M o d u l a t o r
? Direct PM circuit are realized sinusoidal signal
through a time-varying circuit which introduces a
phase shift that varies with the applied modulation
voltage,Similarly,a direct FM circuit is obtained by
varying the tuning of an oscillator tank circuit
according to the Modulation voltage,
33
5.6 Phase Modulation and Frequency
Modulation (Representation of PM and FM
signal)
? Definition,If a band-pass signal is represented by,
)(c o s)()( ttRts ??Where Ψ(t)=ω
ct+θ(t),then the instantaneous frequency (hertz) of s(t) is,
])([2 1])([2 1)(2 1)( dt tdfdt tdttf cii ?????? ????
? For the case of FM,the instantaneous frequency is,
)(2 1])([2 1)( tmDfdt tdftf fcci ??? ?? ??
? The instantaneous frequency varies about the assigned
carrier frequency fc in a manner that is directly
proportional to the modulated signal m(t),
34
5.6 Phase Modulation and Frequency
Modulation (Representation of PM and FM
signal)
? The frequency deviation from the carrier frequency is,
])([2 1)( dt tdftff cid ?????
? And the peak frequency deviation is,
]})([21m a x { dt tdF ??? ?
? The peak-to-peak deviation is
]})([2 1m i n {]})([2 1m ax { dt tddt tdF pp ????? ??
? For FM signaling,the peak frequency deviation is
related to the peak modulation voltage by,
pf VDF ?? 2
1?
? Where Vp=max[m(t)]
35
5.6 Phase Modulation and Frequency Modulation
(Representation of PM and FM signal)
? An increase in the amplitude of
the modulation signal Vp will
increase ΔF and the bandwidth of
the FM signal,But will not affect
the average power level of the FM
signal which is Ac2/2
? As Vp is increased,spectral
components will appear farther
and farther away from the carrier
frequency,and the spectral
components near the carrier
frequency will decrease in
magnitude,since the total power in
the signal remains constant.This
situation is distinctly different
form AM signaling,where the level
of the modulation affects the
power in the AM signal,but does
not affect it bandwidth,
? Fig.5-9
36
5.6 Phase Modulation and Frequency
Modulation
(Representation of PM and FM signal)
? The peak phase deviation is defined by,
)](m a x [ t??? ?? the peak modulation voltage is
pp DD???? Where V
p=max[m(t)],? Definition,The phase modulation index is given by,
??? ?p? Where Δθ is the peak phase deviation
? Definition,The frequency modulation index is given by,
B
F
f
?? ?
? Where ΔF is the peak frequency deviation and B is the
bandwidth of the modulating signal,which,for the case
of sinusoidal modulation,is fm,the frequency of the
sinusoid,
37
5.6 Phase Modulation and Frequency
Modulation
(Spectra of Angle-Modulated Signals)
? As the spectrum of the band-pass waveform is
)]()([21)( * cc ffGffGfS ?????? Where G(f)=F[g(t)]=F[A
cejθ(t)] is the spectra of complex
envelope signal,For the type of AM modulation signaling,
we able to obtain relatively simple formulas relating S(f) to
M(f),For angle modulation signaling,this is not the case,
because g(t) is a nonlinear function of m(t),thus,a general
formula relating G(f) to M(f) cannot be obtained,because
G(f)=F[g(t)]=F[Acejθ(t)] must be evaluated on a case-by-case
basic for the particular modulating waveshape of interest,
? Since g(t) is nonlinear function of m(t),superposition does
not hold,and FM spectra for the sum of two modulating
waveshape is not the same as summing the FM spectra
that were obtained when the individual waveshape were
used
38
5.6 Phase Modulation and Frequency
Modulation
(Spectra of Angle-Modulated Signals)
? Example 5-2 spectrum of a PM or FM signal with
sinusoidal modulation
tAtm mmp ?s i n)( ?
tt m??? sin)( ?
? If the modulation on PM signal is,
? Then,
? Where the phase modulation index is βp =DpAm= β,
? When mf(t)=Amcosωmt in FM modulation,we can get the
same phase function,Where β = βf =DfAm/ωm,The peak
frequency deviation would be ΔF=DfAm/2π,
? The complex envelope is
tjctjc meAeAtg ??? s i n)()( ??
? Which is periodic with period Tm=1/fm,Its Fourier series
can be represented by,
5 5 )-(5 )( tjn
n
n
n nectg
?? ??
???
?
39
5.6 Phase Modulation and Frequency
Modulation
(Spectra of Angle-Modulated Signals)
? Where
dteeTAc m
m
mmT
T
tjntj
m
c
n ??
?? 2/
2/
s i n )( ???
)(]2 1[ )s i n( ??? ?? ??? ncnjcn JAdeAc ?? ?? ?
Here Jn(β) is the Bessel function of
the first kind of the nth,and J-n(β)
=(-1)n Jn(β) hold true,Taking the
Fourier transform of Eq.(5-55),
we get,
)()( m
n
n
n nffcfG ?? ?
??
???
?
)()()( m
n
n
nc nffJAfG ?? ?
??
???
?? Fig.5-10
40
5.6 Phase Modulation and Frequency Modulation
(Spectra of Angle-Modulated Signals)
? Fig,5.11
? Carson’s rule:the bandwidth
of the angle-modulated signal
depends on the phase
modulation index or the
frequency modulation index β,
and bandwidth of the
modulating signal,That is,
BB T )1(2 ?? ?? In fact,98%of the total power
is constrained in that
bandwidth
41
5.6 Phase Modulation and Frequency
Modulation
(Narrowband Angle Modulation)
? When θ(t) is restricted to a small value,say θ(t)<0.2rad,
the complex envelope may be approximated by,
)](1[)( tjAtg c ???? Then a narrowband angle-modulated signal is,
?? ??? ????? ??
t e r mrmc a r r i e r t e
s i n)(c o s)(
s i d e b a n d
cc
d i s c r e t e
cc ttAtAts ??? ??
? This signal is similar to an AM-type signal,except that
the sideband term is 900 out of phase with the discrete
carrier term,The spectrum of the narrowband angle-
modulation signal is,
) ] }()([)]()({[2)( ccccc ffffjffffAfS ???????? ????
s i g n a l i n g FMf o r
s i g n a l i n g PMf o r
2
( t ) ][( f )
??
?
?
?
?? M ( f )
πj
Df
M ( f ) D p
?? ?
? where
42
5.6 Phase Modulation and Frequency
Modulation
(Narrowband Angle Modulation)
? Generation of NBFM Fig.5-12
43
5.6 Phase Modulation and Frequency
Modulation
(Widthband Angle Modulation)
? Theorem:For WBFM signaling,where
? and B is the bandwidth of m(t),the normalized PSD of
the WBFM signal is approximated by
])(c o s [)( ??? ????? t dmDtAts fcc
12 )](m a x [ ?? ??? tmD ff
? where fm(.) is the PDF of the modulating signal,
) ) ](2()(2([2)(
2
c
f
mc
f
m
f
c ff
DfffDfD
Af ????? ????
44
5.6 Phase Modulation and Frequency
Modulation
(Widthband Angle Modulation)
? Generation of WBFM fig.5-13
45
5.6 Phase Modulation and Frequency
Modulation
(Widthband Angle Modulation)
? Example 5-3 spectrum for WBFM with Triangular
modulation,
? Fig.5-14
? The PDF for triangular modulation is,
??
?
?
?
?
?
o t h e r w i s e m,0
,
2
1
)( ppm
Vm
Vmf
46
5.6 Phase Modulation and Frequency
Modulation
(Widthband Angle Modulation
? Where Vpis the peak voltage of the triangular waveform
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
?
?
?
?
?
?
?
?
???
o t h e r w i s e f,0
)(
2
,
2
1
o t h e r w i s e f,0
)(
2
,
2
1
2
)(
2
pc
fp
pc
fp
f
c Vff
DV
Vff
DVD
A
f
???
Ρ
? The PSD of the WBFM signal becomes
??
???
??
??? ???????
??
???
??
??? ?????
o t h e r w i s e f,0
)()(,
8
o t h e r w i s e f,0
)()(,
8)(
22
FffFf
F
AFffFf
F
A
f cc
c
cc
c ??
?
??
?P
? Where the peak frequency deviation is,
?? 2
pVfDF ?
47
5.6 Phase Modulation and Frequency
Modulation
(Widthband Angle Modulation
? Other example Fig,5-15
48
5.6 Phase Modulation and Frequency
Modulation
? some important properties of angle-modulated signals
are,
? An angle-modulated is a nonlinear function of the
Modulation and the bandwidth of the signal increases
as the modulation index increases;
? The discrete carrier level changes depending on the
modulating signal;
? The bandwidth of a narrowband angle-modulated
signal is twice the modulating signal bandwidth
? The real envelope of an angle-modulated signal is
constant,and does not depend on the level of the
modulating signal
49
5.6 Phase Modulation and Frequency
Modulation (Preemphasis and Deemphasis in
Angle-Modulated system)
? preemphasis:If the level of the modulation is boosted at
the top end of the spectrum,the SNR at the output of
the receiver can be improved,
? Deemphasis:The level of the the modulated is
attenuated at high frequencies on the receiver output,
? Preemphasized FM is actually a combination of FM
and PM and combines the advantages of both with
respect to noise performance
50
5.6 Phase Modulation and Frequency Modulation
(Preemphasis and Deemphasis in Angle-Modulated system)
? Fig.5-16 Angle-modulated system with preemphasis and dedmphasis
51
5.7 Frequency-Division multiplexing and FM
stereo
? Frequency-division multiplexing (FDM)is a
technique for transmitting multiple message
simultaneously over a wideband channel by
first modulating the message signals onto
several sub-carriers and forming a composite
base-band signal that consists of the sum of
these modulated sub-carriers.This composite
signal may then be modulated onto the main
carrier,
52
5.7 Frequency-Division multiplexing and FM
stereo
? Fig.5-17 FDM system
53
5.7 Frequency-Division multiplexing and FM
stereo
? The composite signal spectrum must consist of
modulated signals that do not have overlapping
spectra,the composite base-band signal then modulates
a main transmitter to produce the PDM signal that is
transmitted over the wide-band channel,
? The received FDM signal is first demodulated to
reproduce the composite base-band signal that is passed
through filters to separate the individual modulated
sub-carriers.Then the sub-carriers are demodulated to
reproduce the message signals m1(t),m2(t),and so on,
54
5.7 Frequency-Division multiplexing and FM
stereo
? Fig,5-18
? To obtain the compatibility
feature,the left-and right -
channel audios are
combined to produce the
monaural signal,and the
difference audio is used to
modulate a 38-KHz DSB-
SC signal.A 19-KHz pilot
tone is added to the
composite base-band signal
mb(t) to provide a reference
signal for coherent sub-
carrier demodulation in the
receiver
55
5.9 Binary Modulated Bandpass
Signaling
? Digitally modulated bandpass signals
are generated by using the complex
envelopes for AM,PM,FM,or QM
signaling,But the modulating signal m(t)
is a digital signal given by the binary or
multilevel line codes
56
5.9 Binary Modulated Bandpass
Signaling
? binary bandpass
signaling
techniques,
? on-off keying
(OOK) or
Amplitude shift
keying(ASK),
? Binary phase-
shift keying
(BPSK) and
? Frequency-shift
keying(FSK),
m ( t )
m ( t )
s ( t )
s ( t )
s ( t )
s ( t )
a) U n i p ol ar
m od u l at i on
b ) p ol ar
m od u l at i on
c ) O O K
s i gn al
d ) B P S K
s i gn al
e ) F S K
s i gn al
f ) D S B - SC
s i gn al
? Fig.5-19 Bandpass digitally
modulated signals
57
5.9 Binary Modulated Bandpass
Signaling (OOK signal)
? OOK signal,
tftmAttmAts cccc ?? 2c o s)(c o s)()( ??where is the unipolar baseband data signal,It is shown
that OOK is identical to unipolar binary modulation
on a DSB-SC signal,
? complex envelope
)()( tmAtg c?
PSD is,
?
?
?
?
?
?
?
?
???
?
???
??? 22 s i n)(
2)( b
b
b
c
g fT
fTTfAfP
?
??
? PSD
comparison with(3.39b)
?????? ?????????? )(11s i n4)(
22
fTfT fTTAfP
bb
bbu n i p o l a r ?
?
?
58
5.9 Binary Modulated Bandpass
Signaling
? When m(t) has a peak value of A=2(1/2),s(t) has an
average normalized power of ; It shows that there
is a frequency line in the PSD of complex envelope for
OOK,and the OOK is AM-type signaling,
? The null-to-null bandwidth is2R,and the transmission
bandwidth is∞,while the absolute bandwidth is,
? Bit rate is R=1/Tb,
? If raised cosine-rolloff filtering is used(to conserve
bandwidth),the absolute baseband bandwidth is
B=(1/2)(1+r)R,where D=R=2B/(1+r) and r is the
rolloff factor of the filter,The absolute transmission
bandwidth is BT=(1+r)R
2/2cA
59
5.9 Binary Modulated Bandpass Signaling
(demodulation for OOK)
? Fig.5-21
60
5.9 Binary Modulated Bandpass
Signaling (demodulation for OOK)
? 1)Envelope detector(noncoherent detection)
? 2)product detector(coherent detection) with
Low-pass Filter Processing
? 3) product detector(coherent detection) with
Matched Filter Processing.This is an optimum
detection of OOK—that is,to obtain the
lowest BER when the input OOK signal is
corrupted by additive white Gaussian
noise(AWGN)
61
5.9 Binary Modulated Bandpass
Signaling
? Note,the optimum coherent OOK detector is
more costly to implement than the
noncoherent OOK detector,If the input noise
is small,the noncoherent receiver may be the
best solution,considering both cost and noise
performance,The trade-off in BER
performance between optimum coherent
detection and nonoptimum noncoherent
detection,
62
5.9 Binary Modulated Bandpass
Signaling （ BPSK）
? Bandpass signal for BPSK,
)](c o s [)( tmDtAts pcc ?? ?
? Where m(t) is a polar baseband data signal,
For convenience,let m(t) have peak values of
± 1 and a rectangular pulse shape
? complex envelope representation
Expanding equation above we get,
ttmDAtDA
ttmDAttmDA
tmDtAts
cPccPc
cPccPc
pcc
??
??
?
s i n)()s i n (c o s)c o s (
s i n))(s i n (c o s))(c o s (
)](c o s [)(
??
??
??
63
5.9 Binary Modulated Bandpass
Signaling
? it can be shown that BPSK is also a form of AM-type
signaling like OOK signaling,But when digital
modulation index,
122 ???? ?? ? pDh
Conclusion,For this optimum BPSK signal of h=1,
the complex envelope is
2/??? ?? pD
ttmA
ttmDAtDA
tmDtAts
cc
cPccPc
pcc
?
??
?
s i n)(
s i n)()s i n (c o s)c o s (
)](c o s [)(
??
??
??
)()( tmjAtg c?
and it equivalent to DSB-SC signaling with a polar
baseband data waveform
64
5.9 Binary Modulated Bandpass
Signaling
? PSD of complex envelope for optimum BPSK
Because m(t) is polar signaling,the PSD for the complex
envelope as
2
2 s i n)(
???
?
???
??
b
b
bcg fT
fTTAfP
?
?
where s(t) has an average normalized power of Ac2/2 2/2cA
the null-to-null bandwidth is 2R,like OOK
Detection for BPSK,
1) coherent detection
2) A optimum detection (the low-pass filter in fig.5.22a is
replaced by an integrate-and-dump matched filter
processing,)
65
5.9 Binary Modulated Bandpass Signaling
? Fig.5-21
66
5.9 Binary Modulated Bandpass
Signaling
? Notice:unlike OOK signaling,synchronous detection
must be used to detect BPSK,the main reason is that
there is no discrete carrier term in the BPSK signal;if
a low-level pilot carrier is transmitted together with
the BPSK signal,a PLL may be used to extract the
carrier reference,Otherwise,a Costas loop or
squaring loop may be used to synthesize the carrier
reference because BPSK is DSB-SC signal
? if Costas loop and square loop are used,the phase
ambiguity must be resolved,This can be accomplished
by using differential coding at the transmitter input
and differential decoding at the receiver output,
67
5.9 Binary Modulated Bandpass
Signaling(DPSK )
? Although phase-shift-keyed signals cannot be detected
incoherently,a partially coherent technique can be
used whereby the phase reference for the present
signaling interval is provided by a delayed version of
the signal that occurred during the previous signaling
interval
? if,the symbol phase
relationship between binary sequence and 2DPSK
signal is represented as,
??
??
0 i g i t a l0
1 i g i t a l
d
d???
00000
000000
1010011100
or
s i g n a l2 D P S K of p h a s e
s e q u e n c y d a t aB i n a r y
??????
?????
68
5.9 Binary Modulated Bandpass
Signaling
? detection
partially coherent detection
O n e - b i t
d e l a y,T b
L o w - p a s s
f i l t e r
D e c o d e d
b i n a r y
o u t p u t
D P S K
i n p u t
69
5.9 Binary Modulated Bandpass Signaling
(Frequency-shift keying (FSK))
? According to the method used to generate it,FSK
signal can be classified as two different types,
Discontinuous-phase FSK and Continuous-pahse FSK
? Discontinuous-phase FSK signal is represented by
??
?
?
????
0f o r )co s (
1f o r )co s ())(co s ()(
22
11
??
????
tA
tAttAts
c
c
cc
? when m(t) is polar binary pulse waveform,we can get,
)c o s ()()c o s ()())(c o s ()( 2211 ?????? ?????? ttmAttmAttAts cccc
? The discontinuous phase function is
?
?
?
??
???
0f o r
1f o r )(
22
11
tt
ttt
c
c
???
????
70
5.9 Binary Modulated Bandpass Signaling
(Frequency-shift keying (FSK))
? Continuous-phase FSK In order to generate the
continuous-phase FSK signal,the data signal are
feed into a frequency modulator,This FSK
signal is,
? ?tjtfcc cetgdmDtAts ???? )(Re))(c o s ()( ??? ? ??
? thus its complex envelope can be represented by,
)()( tjc eAtg ??? where
? ??? tf dmDt ??? )()(
71
5.9 Binary Modulated Bandpass Signaling
(Frequency-shift keying (FSK))
? It can be shown that although m(t) is
discontinuous at the switching time,the phase
function θ(t) is proportional to the integral of
m(t),If the serial data input waveform is
binary,such as a polar baseband signal,the
resulting FSK signal is called a binary FSK
signal,Of course,a multilevel input signal
would produce a multilevel FSK signal
72
5.9 Binary Modulated Bandpass Signaling
(Frequency-shift keying (FSK))
? Bandwidth of FSK signaling:For the FSK signal the
approximate transmission bandwidth is give by
Carson’s rule,
BFB T 22 ???? where B is the bandwidth of the digital modulation
waveform,When modulating waveform is square-
wave,the bandwidth is B=R,we can find that the this
FSK transmission bandwidth becomes,
RFB T 22 ?? ?? if a raised cosine-rolloff premodulation filter is used,
the transmission bandwidth of the FSK signal
becomes,
RrFB T )1(2 ??? ?
73
5.9 Binary Modulated Bandpass Signaling
(Frequency-shift keying (FSK))
? Generation of FSK(Fig 5-23)
F r e q u e n c y
m o d u l a t o r
( c a r r i e r
f r e q, = f c )
B i n a r y d a t a
i n p u t m ( t )
F S K o u t p u t
b ) c o n t i n u o u s - p h a s e F S K
O s c i l l a t o r
F r e q, = f 2
B i n a r y d a t a i n p u t m ( t )
C o n t r o l l i n e
E l e c t r o n i c
s w i t c h
a ) d i s c o n t i n u o u s - p h a s e F S K
F S K o u t p u t
O s c i l l a t o r
F r e q, = f 1
74
5.9 Binary Modulated Bandpass Signaling
(Detection of FSK Fig,5-28)
F r e q u e n c y
d e t e c t o r
F S K i n p u t
b i n a r y i n p u t
a ) n o n c o h e r e n t D e t e c t i o n
F S K i n p u t C o s ( ω 1 t) F S K
i n p u t
L o w - p a s s
f i l t e r
L o w - p a s s
f i l t e r
C o s ( ω 2 t)
+
-
b ) c o h e r e n t ( s y n c h r o n o u s ) D e t e c t i o n
75
5.9 Binary Modulated Bandpass Signaling
? Example 1
? Example 2
? Example 3
? Example 4
76
5.10 Multilevel Modulated Bandpass Signaling
? If a level stands for a k bits binary digits,the number
of levels in the multilevel signal is, M=2k,The
symbol rate(baud) of the multilevel signal is D=R/k，
for example,if k=2,M=4 and baud D=R/2,where the
bit rate is R=1/Tb。
? Multilevel Modulated bandpass signaling used often
have,
? QPSK; MPSK;QAM; OQPSK and π/4QPSK
M od ul a t e d
ou t pu t
D i gi t a l - to - a na l
og c on ve r t e r l
bi t
B i na r y i np ut
M = 2
l
- l e ve l
di gi t a l s i gn a l
D = s ym bo l s / s
= R / l
F i g,5 - 29 m od ul a t e d di gi t a l t r a ns m i s s i on s ys t e m
T r a ns m i t t e r
77
5.10 Multilevel Modulated Bandpass Signaling
(MPSK)
? In general,MPSK can also generated by using two
quadrature carriers modulated by in-phase and
quadrature components of the complex envelope
except using a phase modulator,in that case,
)()()( )( tjytxeAtg tjc ??? ?
? where the permitted values of x and y are
ici Ax ?c o s? ici Ay ?sin?
Mii,,2,1,???
? is the permitted phase angles of the
MPSK signal,
78
5.10 Multilevel Modulated Bandpass Signaling
(QPSK)
? Fig 5-31
79
5.10 Multilevel Modulated Bandpass Signaling
? M-ary phase-shift keying(MPSK) with M=4 is
called QPSK,its permitted values of complex
envelope contains four points
)()( tjc eAtg ??? A plot of two possible sets of g(t) is shown as
following
00 01
11 10
00
01
11
10
80
5.10 Multilevel Modulated Bandpass Signaling
(QAM)
? in general,the QAM signal constellations are not
restricted to have permitted signaling points only on a
circle,The general QAM signal is,
ttyttxts cc ?? s i n)(c o s)()( ??? its complex envelope is
)()()()()( tjetRtjytxtg ????
? where
2)()()( tjytxtR ??
))(/)((t a n)( 1 txtyt ???
81
5.10 Multilevel Modulated Bandpass Signaling
(QAM)
? Fig.5-32 16 symbol
QAM constellaltion
y i m a g i n a r y
a x i s
( q u a d r a t u r e )
x r e a l
a x i s
( i n p h a s e )
? The waveforms of I
and Q components are
represented by,
? ??
n
n D
nthxtx )()(
? ??
n
n D
nthyty )()(
? where h(t) is the pulse
shape that is used for
each symbol,and
baud rate D=R/k
82
5.10 Multilevel Modulated Bandpass Signaling
? In some applications,the timing between the x(t) and
y(t) components is offset by Ts/2=1/(2D),that is,
? ??
n
n D
nthxtx )()( ? ???
n
n DD
nthyty )
2
1()(
? this signal is called offset QAM(OQAM)
? Offset quadrature phase-shift keying(OQPSK)
One popular type of offset signaling is offset
QPSK(OQPSK) which is identical to offset QAM
when M=4,The AM is reduced because a maximum
phase transition of only π/2 occurs (as opposed to for
QPSK) since the I and Q data cannot change
simultaneously,because the data are offset,
?
83
5.10 Multilevel Modulated Bandpass Signaling
(π/4 QPSK modulation)
? A π/4 QPSK signal is generated
by alternating between two
QPSK constellations that are
rotated by π/4 with respect to
each other,the two signal
constellations are used
alternately as follows,Given a point on one of the signal
constellations that corresponds to two bits of input data,
two new bits are read to determine the next point that is
selected from the other constellation,For example an
input data is ―00,11,10,10,01‖,if ―00‖correspond to a
phase shift of Δθ=450,a ―11‖ to Δθ=1350,a ―10‖ to
Δθ=1350,a ―10‖ to Δθ=-450 and so on,
?
00 01
11 10
00
01
11
10
84
5.10 Multilevel Modulated Bandpass Signaling
(relationship between QPSK π/4 QPSKand
OQPSK )
? comparison between QPSK,QPSK and OQPSK for
non-rectangle pulse waveform
Q P S K
QPSK)4/(?O Q P S K
?)
43,41(43 ??? ????21
Type of
modulation
Maximum phase
shift
AM big small smallest
Q A M M P S K Q P S K
O Q P S K
( π / 4 ) Q P S K
I n a c i r c l e M = 4
D e l a y
A l t e r n a t i n g
85
5.10 Multilevel Modulated Bandpass Signaling
? PSD for MPSK,QAM,QPSK,OQPSK and π/4 QPSK
? the complex envelope for MPSK and QAM is
represented by,
??
??
?? )()( sn nTtfctg? where c
n is a complex-valued random variable representing the multilevel value during the nth
symbol pulse,And f(t)=Π(t/Ts) is the rectangular
symbol pulse with symbol duration Ts,and its Fourier
transform is
???
?
???
??
???
?
???
??
b
bb
s
ss
fTk
fTkkT
fT
fTTfF
?
?
?
? s i ns i n)(
? thus the PSD for the complex envelope of MPSK or
QAM signal with data modulation of rectangular bit
shape is
2s i n
)( ??
?
?
???
??
s
sbg
fT
fTC k TfP
?
?
86
5.11 Minimum-Shift Keying (MSK)
and GMSK
? background
? In practice,the communication system need to
conserve efficiently bandwidth and improve
the efficiency of power supply,for example
Class C amplifiers without distortion can be
used,
? Minimum-shift keying is another bandwidth
conservation technique that has been
developed,Its advantage of producing a
constant-amplitude signal and,consequently,
can be amplified with Class C amplifiers
without distortion
87
5.11 Minimum-Shift Keying (MSK)
and GMSK
? Definition
MSK is continuous 2FSK with minimum
modulation index h=0.5
? Properties of MSK
1)continuous binary FSK；
2)modulation index being minimum
3)two signals are orthogonal over the bit
interval；
4)being identical with OQPSK when the pulse
shape is sinusoidal type
88
5.11 Minimum-Shift Keying (MSK)
and GMSK
? Obtaining modulation index h
For FSK
)Tt(0 0f o r )c o s ()( 1f o r )c o s ()()( b
222
111 ??
??
?
??
???
??
??
tAts
tAtsts
c
c
According to property 1),we get,
0 t t i m es w i t ch i n g at t h e 21 ?? ??
And according to 3),we have,
? ?bT dttsts0 21 0)()(
89
5.11 Minimum-Shift Keying (MSK)
and GMSK
? That is,
dtttAdttsts bb T cT ?? ??? 0 221120 21 )c o s ()c o s ( )()( ????
? ?
? ?
?
?
?
?
?
?
?
?????
??
?
?
?
?
?
?
?????
?
21
212121
2
21
212121
2
)s in ()()(s in
2
)s in ()()(s in
2
??
??????
??
??????
bc
bc
TA
TA
? ? ?
?
??
?
?
?
??????
21
2121212 )sin()()(sin
2 ??
?????? bc TA
? ? ?
?
??
?
?
?
??????
F
FTA bc
22
)sin()(22sin
2
21212
?
?????
? ? 0
2
)s i n ()(2s i n
2
2121
2
??
?
?
??
?
?
?????
F
hA c
?
?????
90
5.11 Minimum-Shift Keying (MSK)
and GMSK
? And obtain,
? ? )s i n ()(2s i n 2121 ????? ????h
? when θ1≠ θ2
?1,2,k 2 ?? ?? kh
?1,2,k 22 ?? ?? kh
? when θ1= θ2
? Thus,the modulation index for MSK
bFTh ??? 22/1 4/)4/(1 RTF b ????
91
5.11 Minimum-Shift Keying (MSK)
and GMSK
? complex envelope for MSK
For the FSK signal over the signaling interval(0,Tb),the
complex envelope is,
)Tt(0 )( b2)(2)( 0 ?????? ?? tTjcdmFjctjc b
t
eAeAeAtg
????
?
or
)Tt(0 )()()( b???? tjytxtg
where
)Tt(0 ),2T t1c o s ()( b
b
????? ?cAtx
)Tt(0 ),2T t1s in ()( b
b
????? ?cAtyThe MSK signaling is,
ty ( t ) s i n-tc o s)()( cctxts ???
92
5.11 Minimum-Shift Keying (MSK)
and GMSK
? Because,
)Tt(0 ),2T t1c o s ()( b
b
????? ?cAtx
)2Ttc o s (
b
?
cA?and
)Tt(0 ),2T t1s in ()( b
b
????? ?cAty
)2T ts in (
b
?
cA??
93
5.11 Minimum-Shift Keying (MSK)
and GMSK
94
5.11 Minimum-Shift Keying (MSK)
and GMSK
? Type I MSK,Basic pulse waveform alternates
between a positive and a negative half-cosinusoid
? Type II MSK:Basic pulse waveform always a
positive half cosinusoid
? The instantaneous frequency is,
Ffdt tdff cci ????? )(2 1 ??
? where
???
?
???
?? ?
)(
)(t a n)( 1
tx
tyt?
95
5.11 Minimum-Shift Keying (MSK)
and GMSK
? PSD for MSK
? Because x(t) and y(t) have independent data and their
dc value is zero,and since g(t)=x(t)+jy(t),the PSD for
the complex envelope is,
)f()f()f()f( xyxg 2 PPPP ???
2
g 2
2 )f(F
T)f( b?P
? Since the pulse width is 2Tb,this PSD becomes,
??
??? ??
el s ew h er e
Tt
T
tA
tf bbc
0
)
2
co s ()( ?
? where,is the pulse shape,For the MSK
half-cosinusoidal pulse shape,we have,[f ( t )]fF F?)(
96
5.11 Minimum-Shift Keying (MSK)
and GMSK
? and its Fourier transform is,
? ?2)4(1 2c o s4)( fT fTTAfF b bbc ?? ? ?
? ?222
22
)4(1
2c o s16)(
fT
fTTAf
b
bbc
g
?
?
?
?P
? Thus,the PSD for the complex envelope for an
MSK signal is,
97
5.11 Minimum-Shift Keying (MSK)
and GMSK
? Generation of Fast MSK signals
FM t ra nsmit te r
△ F= (1 /4 )R
m (t)
in p u t
s(t)
FFSK s ig n al
? Parallel Generation of Type I MSK signals
S erial - to
- p aral lel
co n v ert e
r( 2 bit)
m ( t)
inpu t
s ( t)
M S K s ignal
O s cill ator
f 0 = △ F =R /4
co s ( π t/2T b )
- 90 0
p h as e sh ift
S y nc
s ignal
x data
y data
y ( t)
x( t)
C arr ier
o s cill ator,f c
p h as e sh ift
- 90 0
p h as e sh ift
A c cos ( ω c t)
A c s in( ω c t)
? Serial Generation of MSK
m ( t)
in p u t
s ( t)
M S K s i g n a l a t
c a r r i e r f r e q u e n c y f c C a r r i e r o s c i l l a r a t
f r e q u e n c y f 2 =f c - △ F
B P S K s i g n a l a t c a r r i e r
f r e q u e n c y f c M S K b a n d p a s s f i l t e r
a b o u t f 1 H ( f )
98
Minimum-Shift Keying (MSK) and
GMSK
? Gaussian-filtered MSK (GMSK)
Before the data are frequency modulated onto the
carrier,the data are filtered by a filter having a
Gaussian-shaped frequency response characteristic,
The transfer function of the Gaussian low-pass filter is
? ?)2/2( l n)/( 2)( BfefH ??
? Properties of GMSK
1) Low spectral sidelobes on the transmitted MSK signal
2) constant envelope
99
5.11 Minimum-Shift Keying (MSK)
and GMSK
? Comparison of the spectral efficiencies of
various types of digital signal
? Table 5-7 spectral efficiency of digital signals
100
5.11 Minimum-Shift Keying (MSK)
and GMSK
101
5.12 Orthogonal Frequency Division
Multiplexing (OFDM)
? comparion between FDM and OFDM
a
c
a,FDM b,OFDM
102
5.12 Orthogonal Frequency Division
Multiplexing (OFDM)
? principle of OFDM
I n p u t b i t s
C h a n n e l
c o d i n g
S y m b o l
m a p p i n g
S / p I F F T
R a i s i n g - c o s i n e
w i n d o w
cp
P / s
D / A
RF - TX
a ) t a ns m i t t e r
O u t p u t b i t s
C h a n n e l
d e c o d i n g
d e m a p pi n
g
P / s FFT
s yn t h e s i s
De - cp
P / s
A / D
RF - RX
b) re c e i ve r
F r e,c or re c
te
C S I e s t i m a t o r
E q u a l i z a t -
i o n
P i l o t
s ym b o l
103
5.12 Orthogonal Frequency Division
Multiplexing (OFDM)
d( n)
S
/P
c
on
ve
rt
e
r
mN
tfj
e
1
2
?
?
m
tfj
e
1
2 ?
)( ms
?
?
?
?
?
?
?
?
1
0
1
0
)2e x p ()(
)2e x p ()()(
N
n
mn
N
n
tfjnd
N
nmjndms
?
?
s
n NT
nf ?
sm mTt ?
)1,...,1,0( ?? Nm
OFDM modulator
104
5.12 Orthogonal Frequency Division
Multiplexing (OFDM)
? complex envelope of OFDM
)2 1(1)()( 1
0
21
0
????? ?? ?
?
?
?
Nn
TfeaAtaAtg n
N
n
tfj
nc
N
n
nnc n
??
? Where is a constant,
2na
0?na? PSD for OFDM
? Because the PSD of each carrier is of the form
|Sa(π(?-?n))|2,the overall PSD for the complex
envelope of the OFDM signal is
21
0 )(
))(s i n ()( ??
? ?
?? N
n n
n
g Tff
TffCf
?
?P
TaAC nc 2?
105
5.12 Orthogonal Frequency Division
Multiplexing (OFDM)
? Application of OFDM
? Wireless LAN
? Digital Audio broadcasting in the European
? Digital Video broadcasting in European
? Broad wireless communication
? DMT in ADSL
?
106
5.13 Spread Spectrum System
? In communication systems mentioned above,we have
concerned primarily with the performance of
communication systems in terms of bandwidth efficiency
and energy efficiency with respect to natural noise,
However,in some application,we also need to consider
multiple-access capability,antijam capability,
interference rejection and covert operation,or low-
probability of intercept(LPI) capability,These
performance objectives can be optimized by using spread
spectrum techniques,
? Advantage of spread spectrum techniques,
? 1 Low-probability of intercept;
? 2 Anti-jamming;
? 3 Multi-access
107
5.13 Spread Spectrum System
? An system which can be considered an SS system must
satisfy two criteria,
? 1 The bandwidth of the transmitted signal s(t) needs
to be much greater than that of the message m(t),
? 2 The relatively wide bandwidth of s(t) must be
caused by an independent modulating waveform c(t)
called the spreading signal,and this signal must be
known by the receiver in order for the message signal
m(t) to be detected,
? The SS signal is,
? ?tj cetgts ?)(Re)( ?
its complex envelope is a function of both m(t) and c(t),
In most cases,a product function is used,so that,
g(t)=gc(t)gm(t)
108
5.13 Spread Spectrum System
? SS signals can be classified by the type of mapping
functions that are used for gc(t),The mose common
types of SS signals have,
? 1) Direct Sequence(DS)
? A DSB-SC type of spreading modulation is used,and
gc(t)=c(t) is a polar NRZ waveform,
? 2) Frequency Hopping(FH)
? gc(t) is of FM type where there are M=2k hop
frequencies determined by the k-bit words obtained
from the spreading code waveform c(t),
? 3) Hybrid techniques
? A system include both DS and FH
109
5.13 Spread Spectrum System
(Direct Sequence-DC )
if information waveform m(t)=± 1,gm(t)=Acm(t),for DC g
c(t)=c(t),the complex envelope for the SS singal
becomes,
)()()( tctmAtg c?
the DS signal is
? ?tj cetgts ?)(Re)( ?
called a direct sequence spreading,spread spectrum
signal(DS-SS)Fig.5-39a,c(t) is a polar spreading signal,
which is generated by using a pseudonoise(PN) code
generator,as shown in Fig.5-39b
110
5.13 Spread Spectrum System
111
5.13 Spread Spectrum System
? maximum-length sequence,called m-sequence is a PN
code which is generated by r shift register stages and
has a maximum period of N=2r-1 chips,A m-sequence
has some properties as following,
? P1,In one period,the number of 1’s is always one
more than the number of 0’s;
? P2,the modulo-2 sum of any m-sequence,when
summed chip by chip with a shifted version of the
same sequence,produces another shifted version of
the same sequence,
? P3,if a window of width r is slid along the sequence
for N shifts,then all possible r-bit words will appear
exactly once,except for the all 0 r-bit word,
? P4,if the 0’s and 1’s are represented by –1 and +1V,
the autocorrelation of the sequence is,
112
5.13 Spread Spectrum System
? P4,if the 0’s and 1’s are
represented by –1 and +1V,
the autocorrelation of the
sequence is,
??
???
??
?
? Nk 1
Nk 1
)( ?
?
N
kR c
? ?? ?? 10)/1()( Nn knnc ccNkR
1??nc
? where
113
5.13 Spread Spectrum System
? The PSD for the complex envelope of BPSK-DS-SS
signal is,
(f )P*(f )P(f )P cmg 2cA?? Fig,5-41
? Without
spreading,the
level of the in-
band PSD would
be proportional to
Ac2/(2Rb),but
with spreading,
the in-band
spectral level
drops to Ac2/(2Rc)
114
5.13 Spread Spectrum System
(Receiver )
? If the input to the receiver consists of the SS
signal plus a narrowband(sine wave) jammer
signal,then,
)(c o s)()()()()( tnttctmAtntstr jcc ???? ?? where the jamming signal is The
output of despreader is
tAtn cjj ?c o s)( ?
ttcAttmAv cjcc ?? c o s)(c o s)(1 ??? After LPF,then
)()(2 tcAtmAv jc ??? the jammer power at the receiver output is
bc
jR
R cjn RR
Adf
RAP
b
b /2
1 22 ?? ?
?
115
5.13 Spread Spectrum System
(Receiver )
? The principle of LPI and anti-jamming
116
5.13 Spread Spectrum System
(Frequency hopping )
? In a frequency-hopping(FH) SS system,there are M=2k
hop frequencies controlled by the spreading code,in
which k chip words are taken to determine each hop
frequency,FH is accomplished by using a mixer circuit
wherein the LO signal is provided by the output of a
frequency synthesizer that is hopped by the PN
spreading code,The serial-to-parallel converter
programmable dividers in the frequency synthesizer,
The k-chip word speifies one of the possible M =2k hop
frequencies,
117
5.13 Spread Spectrum System
(Frequency hopping )
? The FH signal receiver has the knowledge of the
transmitter,c(t),so that the frequency synthesizer in
the receiver can be hopped in synchronism with that
at the transmitter,This despreads the FH signal,and
the source information is recovered from the
dehopped signal with the use of a conventional FSK or
BPSK demodulator,as appropriate,
118
5.13 Spread Spectrum System
(Frequency hopping )
? Fig.5-42
119
5.14 Summary
? AM,SSB,PM,and FM signaling techniques were
considered in detail
? Digital signaling techniques such as OOK,BPSK,FSK,
MSK and OFDM were developed,The spectra for
these digital signals were evaluated in terms of the bit
rate of the digital information source
? Multilevel digital signaling techniques such as QPSK,
MPSK and QAM,were also studied,and their spectra
were evaluated,
? The key parts is 5.6,5.9,5.10,5.11
120
5.14 Homework
Problem,
5-1 5-10 5-11 5-12
