1
Chapter 4
Bandpass Signaling
Principles and Circuits
2
Introduction
(chapter objectives)
? Complex envelopes and modulated
signals
? Spectra of bandpass signals
? Nonlinear distortion
? Communication circuits(mixers,phase-
locked loops,frequency synthesizers,
and detectors)
? Transmitters and receivers
? Software radios
3
Introduction(main goals)
? Mastering how to represent complex envelope
of bandpass waveform,
? The principle of bandpass filtering and linear
distortion
? Bandpass sampling theorem
? Nolinear distortion
? Detector circuits
? Knowing PLL,DDC,general principle of
Transceiver,concept of software radios
4
Introduction
? The bandpass communication signal is
obtained by modulating a baseband
analog or digital signal onto a carrier,
? Complex envelope can represent any
type of bandpass signal,
5
4.1 Complex Envelope Representation of
Bandpass Waveforms
? general representation for bandpass
digital and analog signals
? Representation for modulated
signal
? Representation for bandpass noise
6
4.1 Complex Envelope Representation of
Bandpass Waveforms
? Definition1,baseband waveform
having a spectral magnitude that is
nonzero for frequencies in the vicinity of
the origin (i,e,..f=0) and negligible
elsewhere,
? For example,the information source
signal; audio signal in communication
system,
7
4.1 Complex Envelope Representation of
Bandpass Waveforms
? Definition2,bandpass waveform
having a spectral magnitude that is
nonzero for frequencies in some band
concentrated about at frequency ?= ±
?c,where ?c >>0,The spectral magnitude
is negligible eleswhere, ?c is called the
carrier frequency,
8
4.1 Complex Envelope Representation of
Bandpass Waveforms
? Definition3,Modulation
the process of imparting the source
information onto a bandpass signal with a
carrier frequency by the introduction of
amplitude or phase perturbations or both,
? This bandpass signal is called the
modulated signal s(t),and the baseband
source signal is called the modulating
signal m(t),
9
4.1 Complex Envelope Representation of
Bandpass Waveforms
? The definition of modulation indicates
that modulation may be visualized as a
mapping operation that maps the source
information m(t) onto the bandpass
signal,s(t),the bandpass signal will be
transmitted over the channel,
10
4.1 Complex Envelope Representation of
Bandpass Waveforms
? Fig.4-1 communication system
I n f o r m a t i o
n i n p u t
m ( t )
S i g n a l
p r o c e s s i n g
g ( t ) C a r r i e r
c i r c u i t s
s ( t ) T r a n s m i s s i
o n m e d i u m
( c h a n n e l )
r ( t ) )(
~ tg
C a r r i e r
c i r c u i t s
S i g n a l
p r o c e s s i n g
)((~ tm
T r a n s m i t t e r R e c e i v e r
? As the modulated signal passes through the channel,
noise corrupts it;
? the result is a bandpass signal-plus-noise waveform
that is available at the receiver input,r(t),
? The receiver has the job of trying to recover the
information that was sent from the source,
11
4.1 Complex Envelope Representation of
Bandpass Waveforms
(Complex Envelope Represention)
? All bandpass waveforms,whether they arise
from a modulated signal,interfering signal,or
noise,may be represented in a convenient form,
? v(t) will be used to denote the bandpass
waveform such as the signal when s(t)=v(t),
? the noise when n(t)=v(t),
? and the filtered signal plus noise at the channel
output when r(t)=v(t),or any other type of
bandpass waveform,
12
4.1 Complex Envelope Representation of
Bandpass Waveforms
(Complex Envelope Represention)
? Theorem,Any physical bandpass
waveform can be represented by,
? ? 1)-(4 )(Re)( tj cetgtv ??
1b)-(4 )]([)()( ttC o stRtv c ?? ??
1 c )-(4 s i n)(c o s)()( ttyttxtv cc ?? ??
13
4.1 Complex Envelope Representation of
Bandpass Waveforms
(Complex Envelope Represention)p235
? Where the complex envelope of v(t) is g(t),
)()( )(|)(|)()()( tjtgj etRetgtjytxtg ????? ?
? ? )(c o s)()(Re)( ttRtgtx ???
? ?
)()(|)(|)(
)(s i n)()(Im)(
22 tytxtgtR
ttRtgty
???
?? ?
???
?
???
?
??? ?
)(
)(
t a n)()( 1
tx
ty
tgt?
? ? cis associated carrier frequency(in hertz),ωc=2π ? c
14
4.1 Complex Envelope Representation of
Bandpass Waveforms
? Proof,Any physical waveform (it does not have to be
periodic) may be represented over all time,,by the
complex Fourier series,
??0T
00 /2,)( 0 Tectv
n
n
tjn
n ??
? ?? ????
???Because the physical waveform is real,c
-n=cn,and using
Re{.}=(1/2){.}+ (1/2){.}*,we get,
?
?
?
?
?
?
?? ?
?
?
tj
n
n ecctv
0
1
0 2Re)(
?
Because v(t) is a bandpass waveform,the cn have negligible
magnitudes for n in the vicinity of 0 and,in particular,
c0 =0,Thus,with the introduction of an arbitrary
parameter fc,Eqation above becomes,
15
4.1 Complex Envelope Representation of
Bandpass Waveforms
Because v(t) is a bandpass waveform with nonzero
spectrum concentrated near ? = ? c,the Fourier
coefficients cn are nonzero only for values of n in the
range ± n?0 ≈ ?c, Therefore,from Eq(4-8),g(t) has a
spectrum that is concentrated near ? =0, That is,g(t)
is a baseband waveform,
?
?
?
?
?
?? ???
?
? tjtnj
n
n
n
cc eectv ??? )2(Re)( )(
1
0
So that (1) follows,where
tnjn
n
n cectg
)(
1
02)( ?? ?
??
?
??
The waveform g(t),x(t),y(t),R(t),and θ(t) are all
baseband waveforms,and except for g(t),they are all
real waveform,R(t) is a nonnegative real waveform,
16
4.1 Complex Envelope Representation of
Bandpass Waveforms
? (1)(2)(3) are low-pass-to-band-pass transformation,the
factor ejωct in Eq.(1)-(3) translate the spectrum of the
baseband signal g(t) up to the carrier frequency ?c,in
communication terminology,the frequencies in the
baseband signal g(t) are said to be heterodyned up to ?c,
? In Cartesian coordinates,the complex envelope can be
represented as,g(t)=x(t)+jy(t),where x(t) is the in-phase
modulation associated with v(t),and y(t) is the
quadrature modulation associated with v(t),
? Alternatively,the polar form of g(t),represented by R(t)
and θ(t),R(t) is said to be the amplitude modulation
(AM) on v(t),θ(t) is said to be the phase modulation (PM)
on v(t),
? g(t) is the baseband equivalent of the bandpass signal,
17
4.2 Representation of Modulated Signals
? Modulation is the process of encoding the source
information m(t) (modulating signal) into a band-pass
signal s(t) (modulated signal),Consequently,the modulated
signal is just a special application of the bandpass
representation,The modulated signalis given by,
? ?tj cetgts ?)(Re)( ?
? Where ωc=2π?c,?c is carrier frequency,The complex
envelope g(t) is a function of the modulating signal m(t),
That is
)]([)( tmgtg ?
? g[.] performs a mapping operation on m(t),
18
4.3 Spectrum of Bandpass Signals
? The spectrum of a bandpass signal is directly related
to the spectrum of its complex envelope,
Theorem,If a bandpass waveform is represented by
? ? 1 1 )-(4 )(Re)( tj cetgtv ??then the spectrum of the bandpass waveform is
? ? 12)-(4 )()(21)( cc ffGffGfV ????? ?
and the PSD of the waveform is
? ? 13)-(4 )()(41)( cgcgv fffff ????? PPP
Where G(?)=F[g(t)] and Pg(?) is the PSD of g(t),
19
4.3 Spectrum of Bandpass Signals
? Proof,because
? ? tjtjtj ccc etgetgetgtv ??? ????? )(21)(21)(Re)(? Then,
])([21])([21)]([)( tjtj cc etgetgtvfV ?? ????? FFF
If we use F[g*(t)]=G*(- ?),and the frequency translation
property of Fourier transforms,we get,
? ?)]([)(21)( cc ffGffGfV ????? ?
Which is Equation(4-12); The PSD for v(t) is obtained by
first evaluating the autocorrelation for v(t),
? ? ? ?)()(Re)(Re)()()( ??? ??? ????? tjtjv cc etgetgtvtvR
20
4.3 Spectrum of Bandpass Signals
? Using the identity,
)R e(21)R e(21)R e()R e( 121212 CCCCCC ?? ?? and
tj cetgC ?)(2 ? )(1 )( ??? ??? tj cetgC
? we can get,
? ?
? ?)(
)(
)()(Re
2
1
)()(Re
2
1
)(
???
???
?
??
?
???
??
??
tjtj
tjtj
v
cc
cc
eetgtg
eetgtgR
? Realizing that both < > and Re{ } are linear operators,
we may exchange the order of the operators without
offecting the result,and the autocorrelation becomes,
21
4.3 Spectrum of Bandpass Signals
? or
? ?
? ????
?
?
??
cc
c
jtj
tj
v
eetgtg
etgtgR
2)()(Re
2
1
)()(Re
2
1
)(
??
?? ?
? ?
? ?
? ?
? ? 1 6 )-(4 ])(Re
2
1
])()(Re
2
1
)()(Re
2
1
])()(Re
2
1
)(
2
??
??
???
??
?
?
?
??
c
c
cc
c
j
g
j
jtj
j
v
eR
etgtg
eetgtg
etgtgR
?
??
??
??
?
?
? ?c is much larger
than the frequencies
in g(t),so the second
integral is negligible
according to
Riemann-Lebesque
Lemma
22
4.3 Spectrum of Bandpass Signals
The PSD is obtained by taking the Fourier transform of
Eq(4-16),
? ? ? ?)()(41)()( cgcgvv ffffRf ?????? ?PPFP ?
Since the PSD is a real function,Pg*(?)=Pg(?),Hence,the
PSD is given by Eq(4-13),
Notice,Eq.(4-16) has the same mathematical form as
Eq.(4-11) when t is replaced by τ,
23
4.4 Evaluation of Power
? Theorem,The total average normalized power of a
bandpass waveform v(t) is
17)-(4 |)(|21)0()()( 22 tgRdfftvP vvv ???? ? ??? P
where, normalized, implies that the load is
quivalent to 1 ohm
Proof,Substituting v(t) into Eq(2-67),we get
?????? dfftvP vv )()(2 Pbut
? ? ? ???? ?? dfeffR fjvvv ??? 21 )()()( PPF
so
????? dffR vv )()0( P
24
4.4 Evaluation of Power
? Also,from Eq(4-16)
? ? ? ?)0()(Re21)0(Re21)0( ??? ? tgtgRR gv
? ????? cjgv eRR )(Re21)( ?
? or
? ?2|)(|Re21)0( tgR v ?
Another type of power rating,called the peak envelope
power (PEP) is useful for transmitter specifications,
? But |g(t)| is always real,so
2|)(|
2
1)0( tgR
v ?
25
4.4 Evaluation of Power
? Definition,The peak envelope power (PEP) is the
average power that would be obtained if |g(t)| were to
be held constant at its peak value,
? The (PEP) is equivalent to evaluating the average
power in an unmodulated RF sinusoid that has a peak
value of Ap=max(v(t)),
? Theorem,The normalized PEP is given by
? ?2|)(|m a x21 tgP P E P＝
26
4.4 Evaluation of Power
? Example 4-1 Evaluate the magnitude spectrum for
an amplitude-modulated (AM) signal
Solution,the complex envelope of an AM signal is,
? ?)(1)( tmAtg c ??
so that the spectrum of the complex envelope is
)()()( fMAfAfG cc ?? ?According to representation of modulation signal,we
obtain the AM signal waveform
ttmAts cc ?c o s)](1[)( ??
and,using Eq (4-12),we get the AM spectrum
? ?)()()()(21)( ccccc ffMffffMffAfS ???????? ??
27
4.4 Evaluation of Power
the magnitude spectrum is
?
?
?
??
?
?
????
????
?
0 )(
2
1
)(
2
1
0 )(
2
1
)(
2
1
)(
fffMAffA
fffMAffA
fS
cccc
cccc
?
?
Suppose that the magnitude spectrum of the modulation
happens to be a triangular function,as shown in
Fig 4-2a, The resulting AM spectrum is shown in Fig
4-2b, The g(t)=Ac[1+m(t)] cause delta functions to
occur in the spectrum at ? =± ?c,where ? is the
assigned carrier frequency,Using Eq (4-17),we obtain
the total average signal power
?????? ??? )()(212
1 22 tmtmA
c
)()(2121)(121 222 tmtmAtmAP ccs ?????
28
4.4 Evaluation of Power
If we assume that the dc value of the modulation is zero,as
shown in Fig 4-2a,the average signal power becomes
]1[21 mcs PAP ??
where Pm=〈 m2(t) is
the power in the
modulation m(t),
(1/2)Ac2 is the
carrier power,
and (1/2)Ac2Pm is
the power in the
sidebands of s(t)
- 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
29
4.5 Bandpass Filtering and Linear
Distortion （ Equivalent Low-pass Filter)
? Fig 4-3 bandpass filtering
])(Re [)(
11
tj
c
etgtv
?
?
B a n d p a s s f i l t e r
])(R e [)(
11
tj
c
etkth
?
?
)(2/1)(2/1)(
*
cc
ffKffKfH ?????
])(Re [)(
22
tj
c
etgtv
?
?
a ) b a n d p a s s f i l t e r
- f c fc
1 / 2 K ( f - f c ) 1 / 2 K
*
( - f - f c )
| H ( f )
|f
|f
1 / 2 g 1 ( t )
1 / 2 G 1 ( f )
E q u i v a l e n t l o w - p a s s f i l t e r
1 / 2 k ( t )
1 / 2 K ( f )
1 / 2 g 2 ( t )
1 / 2 G 2 ( f )
b ) t y p i c a l b a n d p a s s f i l t e r f r e q u e n c y r e s p o n s e
c ) e q u i v a l e n t ( c o m p l e x i m p u l s e r e s p o n s e ) l o w - p a s s f i t l e r
1 / 2 | K ( f ) |
d ) t y p i c a l e q u i v a l e n t l o w - p a s s f i l t e r f r e q u e n c y r e s p o n s e
30
4.5 Bandpass Filtering and Linear
Distortion （ Equivalent Low-pass Filter)
Theorem,The complex envelopes for the input,output,
and impulse response of a bandpass filter are
related by,
2 2 )-(4 )(21*)(21)(21 12 tktgtg ?Where g
1(t) is the complex envelope of the input and k(t)
is the complex envelope of the impulse response,It also
follows that,
23)-(4 )(21)(21)(21 12 fKfGfG ?
Proof,We know that the spectrum of the output is
2 4 )-(4 )()()( 12 fHfVfV ?Because,v
1(t),v2(t) and h(t) are all bandpass waveforms,the spectra of these waveforms are related to the
spectra of their complex envelopes by Eq.(4-15),thus,
31
4.5 Bandpass Filtering and Linear
Distortion （ Equivalent Low-pass Filter)
? Thus,
? ?
? ? ? ?
)]()2/1)(()2/1()()2/1)(()2/1[(
)]()()()(
)()()()()[4/1(
)()()2/1()()(( 1 / 2 )
)()(
2
1
)(
**
11
**
1
*
1
*
11
**
11
*
222
cccc
cccc
cccc
cccc
cc
ffKffGffKffG
ffKffGffKffG
ffKffGffKffG
ffKffKffGffG
ffGffGfV
????????
?????????
???????
?????????
?????
)(21)(21)(21 12 fKfGfG ?
? Taking the inverse Fourier transform of both sides of
Equation above,Eq.(4-22) is obtained,
? Proof is over
32
4.5 Bandpass Filtering and Linear
Distortion （ Equivalent Low-pass Filter)
? This theorem indicates that any bandpass filter system
may be described and analyzed by using an equivalent
low-pass filter;
? Equation for equivalent low-pass filter are usually
much less complicated than those for bandpass filter,
so the equivalent low-pass filter system model is very
useful,it is the basic for computer programs that
using sampling to simulated bandpass communication
system,
? The equivalent low-pass filter with complex inpulse
response may be realized by using four low-pass filter
with real impulse response,
33
4.5 Bandpass Filtering and Linear
Distortion (Linear Distortion)
? For distortionless transmission of bandpass signals,
the channel transfer function,H(f)=|H(f)|ejθ(f) needs to
satisfy the following requirements,
(1) The amplitude response is constant,that is,
|H(f)|=A (4-27a)
(2) The derivative of the phase response is a constant,
that is,
2 7 b )-(4 T)(2 1 g?? df fd ??? Where T
g is constant called the complex envelope
delay or,more concisely,the group delay and
f H f? ?（ ）= （ ）
34
4.5 Bandpass Filtering and Linear
Distortion （ Linear Distortion)
Note,Eq(4-27a) is
identical to |H(f)|=A (2-
150a),but Eq.(4-27b)
is less restrictive than
θ(f)=-2πfTd (2-150b),
where Td=Tg,because
the integral of Eq.(4-27)
is,
θ(f)=-2πfTg+ θ0 (4-28)
(1)(2) are sufficient
requirements for
distortionless
transmission of
bandpass signal
? c
| H ( f ) |
A
θ ( f )
θ 0
? c
a ) M a g ni tude re s po ns e
b) ph a s e re s po ns e
S i gna l
ba ndw i dt h
?
?
? Fig.4-4 Tansfer characteristics of a
distortionless bandpass channel
35
4.5 Bandpass Filtering and Linear
Distortion （ Linear Distortion)
? If the input to the bandpass channel is represented by
ttyttxtv cc ?? s i n)(c o s)()(1 ??
? We find the output of the channel is
))(s i n ()())(c o s ()()( 002 ???? ???????? gcggcg TtTtAyTtTtAxtv
? Let
dcgcc TfTf ???? 2)( 0 ?????
? Then,
3 0 )-(4 )(s i n)()(c o s)(
))(s i n ()())(c o s ()()(2
dcgdcg
ccgccg
TtTtAyTtTtAx
ftTtAyftTtAxtv
??????
??????
??
????
36
4.5 Bandpass Filtering and Linear
Distortion ( Linear Distortion)
? where the modulation on the carrier (i.e,the x and y
components) has been delayed by the group time delay,
Tg,and the carrier has been delayed by the carrier time
delay,Td;
? This is distortionless transmission,which is obtained
when Eqs.(4-27a) and (4-27b) are satisfied
? Note that Tg will differ from Td,unless θ0 happens to be
zero,
? In summary,the general requirements for distortionless
transmission of either baseband or bandpass signals are
given by Eqs.(2-150a) and (2-150b),However for the
bandpasscase,Eq.(2-150b) is overly restrictive and may
be replaced by Eq.(4-27b),
37
4.6 Bandpass Sampling Theorem
? Sampling is used in software radios and for simulation of
communication,systems,If the sampling is carried out at
the Nyquist rate or larger(?c≥2B,where B is the highest
in the spectrum of the RF signal),the sampling rate can
be ridiculous,For example,consider a satellite
communication system with a carrier frequency of
?c=6GHz,The sampling raterequired can be at least
12GHZ
? For signals of this type (bandpass signal),it can be shown
that the sampling rate depends only on the bandwidth of
the signal,not on the absolute frequencies involved,that
is,we can reproduce the signal from samples of the
complex envelope,
38
4.6 Bandpass Sampling Theorem
Bandpass Sampling Theorem,If a (real) bandpass
waveform has a nonzero spectrum only over the
frequency interval ?1<| ?|< ?2,where the transmission
bandwith BT is taken to be the absolute bandwidth
BT= ?2- ?1,then the waveform may be reproduced form
sample values if the sampling rate is,
3 1 )-(4 2 Ts Bf ?
example,If the 6GHz bandpass signal previously discussed
had a bandwidth of 10MHz,a sampling frequency of
only 20MHz would be required instead of 12GHz,This is
a savings of three order magnitude,
39
4.6 Bandpass Sampling Theorem
? Proof,We know that the quadrature bandpass
representation is
32)-(4 s i n)(c o s)()( ttyttxtv cc ?? ??? let ?
c be the center of the bandpass,so that ?c = (?2- ?1)/2,Both x(t) and y(t) are baseband signals and are
absolutely bandlimited to B=BT/2,According to
sampling theorem for limited signal,an=ω(n/ ?s)—(2-
160),the sampling rate required to represent the
baseband signal is ?b≥ 2B=BT,then eq.(4-32) becomes,
?
?
??? ?
?
?
?
???
?
?
?
???
?
???
? ??
n bb
bb
c
b
c
b fntf
fntft
f
nyt
f
nxtv
))/((
) ) }/((s i n {s i n)(c o s)()(
?
???
? For the general case,where the x (n/ ?b ) and y (n/ ?b )
samples are independent,two real samples are
obtained for each,value of n,so that the overall
sampling rate for v(t) is
3 1 )-(4 2 Ts Bf ?
40
4.6 Bandpass Sampling Theorem
? The x and y samples can be obtained by sampling v(t)
at t= n/ ?b,but adjusting t slightly,so that cosωct=1
and sin ωct=-1,at exact sampling time for x and y,
respectively,that is,for t= n/ ?b,v(n/ ?b )=x(n/ ?b )
when cosωct=1 and v(n/ ?b )=y(n/ ?b ) when sin ωct=-1,
? Alternatively,x(t) and y(t) can first be obtained by the
use of two quadrature product detectors,The x(t) and
y(t) baseband signal can then be individually sampled
at a rate of ?b,and the overall equivalent sampling
rate is still ?s=2 ?b ≥2BT,
41
4.6 Bandpass Sampling Theorem
? Bandpass Dimensionality Theorem,Assume that a
bandpass waveform has a nonzero spectrum only over
the frequency Interval ?1<| ?|< ?2,where the transmission
bandwith BT is taken to be the absolute bandwidth BT=
?2- ?1,and BT<< ?1,The waveform may be completely
specified over a T0-second interval by,
3 4 )-(4 2 0TBN T?? independent pieces of information.N is said to be the
number of dimensions required to specify the waveform ? Computer simulation is often used to analyze
communication systems.The bandpass dimensionality
theorem tells us that a bandpass signal BT Hz wide can
be represented over a T0-s interval,provided that at
least N=2BTT0 samples are used,
42
4.7 Receiced Signal Plus Noise
Referring to above Fig,the signal out of the transmitter is
I n f o r m a t i o
n i n p u t
m ( t )
S i g n a l
p r o c e s s i n g
g ( t ) C a r r i e r
c i r c u i t s
s ( t ) T r a n s m i s s io n m e d i u m
( c h a n n e l )
r ( t ) )(~ tg C a r r i e r
c i r c u i t s
S i g n a l
p r o c e s s i n g
)((~ tm
T r a n s m i t t e r R e c e i v e r
)()(*)()( tnthtstr ??
where g(t) is the complet envelope for the particular type of
modulation used,if the channel is linear and time
invariant,the received signal plus noise is,
? ? )(Re)( tj cetgts ??
where h(t) is the impulse response of the channel and n(t) is
the noise at the receiver input
43
4.7 Receiced Signal Plus Noise
? Furthermore,if the channel is distortionless,its transfer
function is,
? and conesqudntly,the signal plus noise at the receiver
input is,
2 9 )-(4 )()( 2)2( 00 gg fTjjfTj eAeAefH ???? ??? ??
)]()(R e [)( ))(( tneTtAgtr cc ftjg ??? ? ??
where A is the gain of the channel,Tg is the channel group
delay,and θ(?c) is the carrier phase shift caused
by the channel,
44
4.8 Classification of Filters and Amplifiers
(Filters)
? Filters are devices that take an input waveshape and
modify the frequency spectrum to produce the output
waveshape,
? Filters may be classified in several ways,one is by the
type of construction used,Another is by the type of
transfer function that is realized,
? Filters use energy storage elements to obtain
frequency discrimination,In any physical filter,the
energy storage elements are imperfect,
? Two different measures of filter quality are used in the
technical literature,
45
4.8 Classification of Filters and Amplifiers
? The first definition is concerned with the efficiency of
energy storage in a circuit and is,
2 ( m a x im u m e n e r g y s to r e d d u r in g o n e c y c l e )Q
e n e r g y d is s ip a te d p e r c y c l e
??
? A large value for Q corresponds to a more perfect
storage element,that is,a perfect L,or C element
would have infinite Q,The second definition is
concerned with the frequency selectivity of a circuit
and is,
0fQ
B?
? Where f0 is the resonant frequency and B is the 3 dB
bandwidth,the larger the value of Q,the better is the
frequency selectivity,
46
4.8 Classification of Filters and Amplifiers
? Filters are also characterized by the type of transfer
function that is realized,
nn
kk
jajajaa
jbjbjbbfH
)(.,,)()(
)(.,,)()()(
2210
2210
???
???
????
?????
? Amplifier can be classified into two main categories,
Nonlinear and linear,And can be further classified into
the subcategories of circuit with memory and circuits
with no memory,circuits with memory contain inductive
and capacitive effects that cause the present output value
to be a function of previous input values as well as the
present input value,
? The transfer function of a distortionless amplifier is given
by,where K is voltage gain of the amplifier and
Td is the delay between the output and input waveform.,
dcTjKe ?
47
4.9 Nonlinear Distortion
? In addition to linear distortion,practical amplifiers
produce nonlinear distortion,To examine the effects of
nonlinearity and yet keep a mathematical model that is
tractable,In the following analysis we assume no memory,
If the amplifier is linear,we get,
4 1 )-(4 )()(0 tKvtv i?
? Where K is the voltage
gain of the amplifier,In
practical,the output of the
amplifier becomes (soft)
saturated at some value as
the amplitude of the input
signal is increased,
v 0 out put
S a t ur a t i on
l e ve l
v i i nput
F i g,4 - 5 N onl i ne a r a m pl i f i e r out put - t o i nput c ha r a c t e r i s t i c s
48
4.9 Nonlinear Distortion
? The output-to-input characteristic may be modeled by a
Taylor’s expansion about vi=0,that is,
42)-(4
0
2
2100 ?
?
?
?????
n
n
inii vKvKvKKv ?? where
4 3 )-(4
!
1
0
0
?
?
?
?
?
?
?
?
?
?
iv
n
i
n
n
dv
vd
n
K
? There will be nonlinear distortion on the output signal if
K2,K3,… are not zero,K 0 is the output dc offset level,K1v1
is the first-order(linear) term,K2vi2 is the second-order
term,and so on,Of course,K1 will be larger than K2,
K3,… if the amplifier is anywhere near to be linear,
49
4.9 Nonlinear Distortion
? The harmonic distortion associated with the amplifier
output is determined by applying a single sinusoidal test
tone to the amplifier input
4 4 )-(4 t in)( 00 ?sAtv i ?? The second-order output term is,
45)-(4 tc o s 2
2
K-
2
Kt)in(
t e r md i s t o r t i o n h a r m o n i c s e c o n d
0
2
02
t e r ml e v e l
2
022
002
?? ??? ?????
?? AAsAK
dc
?
? For a single-tone input,the output is,
46)-(4,,,)t3c o s (
)t2c o s ()tc o s ()(
3203
2021010
???
?????
??
????
V
VVVtv o u t
? Where Vn is the peak value of the output at the frequency
nf0 Hz,
50
4.9 Nonlinear Distortion
? The percentage of total harmonic distortion (THD) is
defined by,
4 7 )-(4 1 0 0( % )
1
2
2
?? ?
?
?
V
V
T H D n n? THD can be measured 1) by using a distortion analyzer;2)
evaluating by (4-47),with the vn’s obtained from a
spectrum analyzer,
? The intermodulation distortion(IMD) of the amplifier is
determined by applying two-tone test,
48)-(4 t int in)( 2211 ?? sAsAtv i ??? The second-order output term is,
t)intintin2t in(
t)int in(
2
22
221211
22
12
2
22112
????
??
sAssAAsAK
sAsAK
???
?
51
4.9 Nonlinear Distortion
? The first and last term produce harmonic distortion at the
frequency 2f1 and 2f2,the cross-product term produce
IMD,the second-order IMD is,
) t )c o s (-)t-( c o s (2tintin2 212121221212 ?????? ?? AAKssAAK? Which generates sum and difference frequencies,
? The third-order output term is,
t)intintin3
4 9 )-(4t intin3t in(
t)int in(
2
33
22
2
1
2
21
21
2
2
2
11
33
13
3
22113
3
3
???
???
??
sAssAA
ssAAsAK
sAsAKvK i
??
??
??
? The first and last term produce harmonic distortion at the
frequency 3f1 and 3f2,the cross-product term,the second-
order becomes,
52
4.9 Nonlinear Distortion
? Similarly,the third term is,) t ] )s i n ( 2-)t2( 1 / 2 ) [ s i n (tin(( 3 / 2 )
50)-(4 t ) )2 c o s ( 2-t ( 1in( 3 / 2 )
tintin3
212122
2
13
122
2
13
21
2
2
2
13
?????
??
??
????
?
sAAK
sAAK
ssAAK
) t ] )s i n ( 2-)t2( 1 / 2 ) [ s i n (tin(( 3 / 2 )
51)-(4 t intin3
12121
2
213
2
2
1
2
213
?????
??
???? sAAK
ssAAK
? The last two terms in Eqs.(4-50) and (4-51) are
intermodulation terms at nonharmonic frequencies,
? Example,For the case of bandpass amplifiers where f1 and f2
are within the bandpass with f1 close to f2,this distortion
product at 2f1+ f2 and 2f2+ f1 will usually fall outside the
passband,but the term 2f1- f2 and 2f2- f1 will fall inside the
passband and will be close to the desired frequencies this
will be the main distortion products for bandpass amplifier,
53
4.9 Nonlinear Distortion
? If A1 or A2 is increased sufficiently,the IMD will become
significant,since the desired output varies linearly with
A1 or A2 and the IMD output varies as A12A2 or A22A1,
? Of course the exact input level required for the
intermodulation products depends on the relative values
of K3 and K1,the level may be specified by the amplifier
third-order intercept point,If the input test tone’s
amplitude is equal,A,the desired linearly amplified
outputs have amplitudes of K1A,and each of the third-
order intermodulation products have amplitudes of
3K3A3/4,the ratio of the desired output to the IMD
output is,
5 2 )-(4
3
4
2
3
1
?
?
?
?
?
?
?
?
?
AK
KR
IM D
54
4.9 Nonlinear Distortion
55
4.10 Limiters
? A limiter is a nonlinear circuit with an output saturation
characteristic,
? A ideal limiter transfer function is essentially identical to
the output-to-input characteristic of an ideal comparator
with a zero reference level,
? A bandpass limiter is a nonlinear circuit with a saturating
characteristic followed by a bandpass filter,If a input of
bandpass signal is,
))(c o s ()()( tttRtv cin ?? ??? Where R(t) is the equivalent real envelope and θ(t) is the
equivalent phase function,The output of an ideal
bandpass limiter becomes,
55)-(4 ))(c o s ()( ttKVtv cLout ?? ??
56
4.10 Limiters
? Limiters are often
used in receiving
systems designed for
angle-modulated
signaling ---such as
PSK,FSK,and
analog FM--- to
eliminate any
variations in the real
envelope of the
receiver input signal
that are caused by
channel noise or
signal fading,
v out I d e a l l i m i t e r
c h a r a c t e r i s t i c
v in
v lim ( t )
v L
- v L
v L
- v L
v in ( t )
F i g, 4 - 7 i d e a l l i m i t e r c h a r a c t e r i s t i c w i t h
i l l u s t r a t i v e i n p u t a n d u n f i l t e r e d o u t p u t w a v e f o r m
57
4.11 Mixers,Up Converters,and Down
Converters
? an ideal mixer is an electronic circuit that functions as a
mathematical multiplier of two input signals,Mixers are
used to obtain frequency translation of the input signal,If
the input signal is,
L o c a l
o s c i l l a t o r
m i x e r
f i l t e r
v in ( t ) v 1 ( t )
v 2 ( t )
v Lo ( t ) = A 0 c o s ω 0 t
F i g, 4 - 8 M i x e r f o l l o w e d b y a f i l t e r
f o r e i t h e r u p a n d d o w n c o n v e r s i o n
5 6 )-(4 })(R e {)( tjinin cetgtv ??
? where gin is the
complex
envelope of the
signal.the
output of the
mixer is then,
)]([
4
A
t c o s})(R e {)(
00*0
001
tjtjtj
in
tj
in
tj
in
eeegeg
etgAtv
cc
c
????
? ?
?????
?
58
4.11 Mixers,Up Converters,and Down
Converters
? This illustrates that the input bandpass signal with a
spectrum near f=fc has been converted into two output
bandpass signal,One at the up-conversion frequency
band,where fu=fc+f0,and one at the down-conversion
band where fd=fc-f0,,a filter may be used to select
either the up-conversion component or the down-
conversion component,
57)-(4 })(R e {
2
A
})(R e {
2
A
)()()()([
4
A
)]()()([
4
A
t c o s})(R e {)(
)(0)(0
)(*)()(*)(0
*0
001
00
0000
00
tj
in
tj
in
tj
in
tj
in
tj
in
tj
in
tjtjtj
in
tj
in
tj
in
cc
cccc
cc
c
etgetg
etgetgetgetg
eeetgetg
etgAtv
????
????????
????
?
?
??
??????
??
??
????
???
?
59
4.11 Mixers,Up Converters,and Down
Converters
? A bandpass filter is used to selected the-conversion
component,but the down-conversion component is
selected by either a baseband filter or a bandpass filter,
For example,
58)-(4 })(R e {2A)( )(02 0 tjin cetgtv ?? ??? If f
c<f0 (4-57) can be represented by,
5 9 )-(4 })(R e {2A})(R e {2A)( )(*0)(01 00 tjintjin cc etgetgtv ???? ?? ??
? That is to saying that the sidebands have been exchanged;
the upper sideband of the input signal spectrum becomes
the lower sideband of the down-converted output signal,
and so on,The reason is,
)()]([ ** fGtg inin ??F
60
4.11 Mixers,Up Converters,and Down
Converters
? Summary,
? The complex envelope for the signal out of an up converter
is,
6 1 a )-(4 )()2/()( 02 tgAtg in?
? For the case of down conversion,there are two
possibilities,
61b)-(4 )()2/( )()2/()(
0
0
*
0
0
2 ff
ff
tgA
tgAtg
c
c
in
in
?
?
??
??
? We should notice that mixers used in communication
circuits are essentially mathematical multipliers,
? We should notice that mixers used in communication
circuits are essentially mathematical multipliers,They
should be not confused with the audio mixers(音频混音器 )
61
4.12 Frequency Multipliers
? Frequency multipliers is
essentially a nonlinear
amplifier followed by a
bandpass filter that is
designed to pass the nth
harmonic,they consist of
a nonlinear circuit
followed by a tuned
circuit,If a bandpass
signal is fed into a
frequency multiplier,the
output will appear in a
frequency band at the
nth harmonic of the
input carrier frequency,
62
4.12 Frequency Multipliers
? If the bandpass input signal is represented by,
7 2 )-(4 ))(c o s ()()( tttRtv cin ?? ??? The output signal of the nonlinear device is,
so t h e r t e r m ))(( t ) c o s ( n
))((c o s)()()(1
???
???
tntCR
tttRKtvKtv
c
n
c
nn
n
n
inn
??
??
? The output signal of the bandpass filter is,
7 3 )-(4 ))(( t ) c o s ( n)( tntCRtv cno u t ?? ??? This shows that the input amplitude variation R(t) appears
distorted on the output signal,but waveform of the angle
variation is not distorted by frequency multiplier,Thus
they are not used in signals if AM is to preserved; but are
very useful in PM and FM problem,n=2 doubler stage,
n=3 triple stage,
? Frequency multiplier is nonlinear,mixer is linear,
63
4.13 Detector Circuit
? The receiver contains carrier circuits that convert the
input bandpass waveform into an output baseband
waveform,These carrier circuits are called detector
circuits,This sections that follow will show how detector
circuits can be designed to produce R(t),θ(t),x(t),or y(t),
at their output for the corresponding bandpass signal that
is fed into the detector input,
I n f o r m a t i o
n i n p u t
m ( t )
S i g n a l
p r o c e s s i n g
g ( t ) C a r r i e r
c i r c u i t s
s ( t ) T r a n s m i s s i
o n m e d i u m
( c h a n n e l )
r ( t ) )(
~ tg
C a r r i e r
c i r c u i t s
S i g n a l
p r o c e s s i n g
)((~ tm
T r a n s m i t t e r R e c e i v e r
? Envelope detector; product detector; Frequency
modulation detector
64
4.13 Detector Circuit
(Envelope Detector)
? An ideal envelope detector is a circuit that produce a
waveform at its output that is proportional to the real
envelope R(t) of its input,If the bandpass input may be
represented by,
))(c o s ()()( tttRtv cin ?? ??
? Then the output of the ideal envelope detector is,
74)-(4 )()( tKRtv out ?
? Then the output of the ideal envelope detector is,
65
4.13 Detector Circuit (Envelope Detector)
66
4.13 Detector Circuit
(Envelope Detector)
? A simple diode detector circuit may approximates an ideal
envelope detector,but its RC time constant is chosen so
that the output signal will follow the real envelope R(t) of
the input signal,Consequently,the cutoff frequency of the
low-pass filter needs to be much smaller than the carrier
frequency fc,and much larger than the bandwidth of the
(detected) modulation waveform B,that is,
? B<<1/(2πRC)<<fc (4-75)
? Where RC is the time constant of the filter,The envelope
detector is typically used to detect the modulation on AM
signals,In this case,vin(t) has the complex envelope
g(t)=Ac(1+m(t)),If|m(t)|<1,then,
? vout(t)=KR(t)=K[g(t)]=KAc[1+m(t)]=KAc+KAcm(t) (4-75)
? Note,KAc is a DC voltage that is used to provide automatic
gain control (AGC),KAcm(t) is the detected modulation,
67
4.13 Detector Circuit
(Product Detector)
? A product detector is a mixer circuit that down-converts
the input (bandpass signal plus noise) to baseband,The
output of the multiplier is,
L o c a l
o s c i l l a t o r
L o w p a s s
f i l t e r
v in ( t ) = R ( t ) c o s [ ω c t+ θ ( t ) ] v 1 ( t ) v out ( t )
v Lo ( t ) = A 0 c o s ( ω 0 t+ θ 0 )
F i g, 4 - 1 4 p r o d u c t d e c t e c t o r
O r v in ( t ) = R e { g ( t ) e x p ( - j ω c t ) } ]
W h e r e g ( t ) = R ( t ) e x p ( j θ ( t ) )
v1(t)=R(t)cos[ωct+θ(t)] A0cos(ω0t+θ0)
=(1/2) A0 R(t) cos(θ(t)- θ0)
+(1/2) A0 R(t) cos(2ωct θ(t)+ θ0)
68
4.13 Detector Circuit
(Product Detector)
? After the low-pass filter,the output is,
vout(t) =(1/2) A0 R(t) cos(θ(t)- θ0)
=(1/2) A0Re{g(t)exp(-j θ0)} (4-76)
? Where g(t)=R(t)exp(j θ(t))=x(t)+jy(t),Note,if θ0 =0,the
oscillator is said to be phase synchronized with the in-
phase component,And the output becomes,
vout(t) =(1/2) A0 x(t) (4-77a)
If θ0 =900,
vout(t) =(1/2) A0 y(t) (4-77b)
This indicate that a product detector is senstive to AM
and PM,Example-1,if the input contains no angle
modulation,so that θ(t)=0,and if θ0 =0,then,
69
4.13 Detector Circuit
(Product Detector)
? Which implies that x(t) ≥0,and the real envelope is
obtained on the product detector output,Example-2 if an
angle-modulated signal Accos [ωct+θ(t)] is present at the
input and θ0 =900,the product detector output is,
vout(t) =(1/2) A0 R(t) (4-78a)
7 8 b )-(4 ))(s i n ()2/1(
}R e {)2/1()(
0
]90)([
0
0
tAA
eAAtv
c
tj
cout
?
?
?
? ?
? In this case,the product detector acts like a phase detector
with a sinusoidal characteristic,the reason is that the
output voltage is proportional to the sine of the phase
difference between the input signal and the oscillator
signal,A special case is that the phase difference is small,
70
4.13 Detector Circuit
(Product Detector)
? That is | θ(t) |<<900,then sin θ(t)= θ(t) and,
79)-(4 )()2/1( ))(s i n ()2/1()( 00 tAAtAAtv ccout ?? ??
Then the output of this phase detector is directly proportion
to the phase difference,
Summary,1) Unlike the envelope detector that is a
nonlinear device,the product detector acts as a linear
time-varying device with respect to the input vin(t),
2) Detector may also classified as being either coherent or
noncoherent,The former has two input- one for a
reference signal and one for the modulated signal that is to
be demodulated,The product detector is a coherent one,
The later has only one input,namely,the modulated singal
port,the envelope detector is a noncoherent one,
71
4.13 Detector Circuit
(Frequency Modulation Detector)
? An ideal frequency modulation (FM) detector is a
device that produces an output that is proportional to
the instantaneous frequency of the input,that is,if the
bandpass input is represented by R(t)cos(ωct+θ(t)),the
output of the ideal FM detector is,
8 0 )-(4 ))(θ+ω(=))(θ+ω(=)( dt tdKdt ttdKtv ccout
Usually,the FM detector is balanced,
81)-(4 )(θ=)( dt tdKtv o u tAlmost all of Frequency Modulation detector are based on
one of three principle,
1)FM-to-AM conversion
2)Phase-shift or quadrature detection
3) zero-crossing detection
72
4.13 Detector Circuit
(Frequency Modulation Detector)
? A slope detector
is one example of
the FM-to-AM
conversion principle,
it want a bandpass
limiter to suppress any
amplitude variations on
the input signal,the
differentiation operation
can be otained by any
circuit that acts like a
frequency-to-amplitude
converter,
73
4.13 Detector Circuit
(Frequency Modulation Detector)
? The quadrature detector is described as follows,a
Quadrature signal is first obtained form the FM signal;
then though the use of a product detector,the
quadrature signal is multiplied with the FM signal to
produce the demodulated signal,
)(
2
1
=
)(θ
2
1
)
)(θ
s i n (
2
1
=)(
8 8 )-(4 )
)(θ
+)(θ+ωs i n (=)(
))(θ+ωco s (=)(
2
21
2
212
2
1
21
tmKVKK
dt
td
VKK
dt
td
KVKtv
dt
td
KttVKtv
ttVtv
fLLLout
cLquad
cLin
＝
? Zero-crossing detector because the output of an ideal
FM detector is directly proportional to the
instantaneous frequency of the input,The linear
frequency-to-voltage characteristic may be obtained
directly by counting the zero-crossings of the input
waveform,
74
4.13 Detector Circuit
(Frequency Modulation Detector)
? Because,
dt
tdftf
ci
)(θ
π2
1+=)(
And nonostable
multivibrator is
triggered on positive
slope zero-crossings,
Vout(t) =0 f=fi
Vout(t) >0 fi>f
Vout(t) <0 fi<f
75
4.14 Phase-Locked Loops and Frequency
Synthesizers
? A Phase-locked loop consists of three basic components,
? A phase detector,
? a low-pass filter,
? a voltage-controlled oscillator
76
4.14 Phase-Locked Loops and Frequency
Synthesizers
? The first operating mode,PLL acts as a narrowband
tracking filter when the low-pass filter is a narrowband
filter,The frequency of the VCO will become that of noe
of the line components of the input signal spectrum,The
frequency of the VCO will track the input signal
component if it changes slightly in frequency,
? The second operating mode,the bandwidth of the LPF is
wider so that the VCO can track the instantaneous
frequency of the whole input signal,
? When PLL tracks the input signal in either of these ways,
the PLL is said to be,locked”,
? If the PLL is built using analog circuits,it is said to be an
analog PLL,conversely digital circuits and signals are
used,the PLL is said to be a digital PLL
77
4.14 Phase-Locked Loops and Frequency
Synthesizers
? If the input signal is,))(sin()(
0 ttAtv iiin ?? ??
? The VCO output signal is,))(sin()(
0 ttAtv ooo ?? ??
? where
94)-(4 )()( 2 dxxvKt tvo ?? ???? Then,
))()(2s i n (
2
))()(s i n (
2
))(s i n ())(s i n ()(
0
001
ttt
AAK
tt
AAK
ttttAAKtv
oi
oim
oi
oim
iooim
?????
????
?????
???
78
4.14 Phase-Locked Loops and Frequency
Synthesizers
? Then,
))()(2s i n (
2
))()(s i n (
2
))(s i n ())(s i n ()(
0
001
ttt
AAK
tt
AAK
ttttAAKtv
oi
oim
oi
oim
iooim
?????
????
?????
???
96)-(4 )(*)]([s i n
)(*))()(s i n (
2
)(
)(
2
tftK
tftt
AAK
tv
ed
t
oi
K
oim
e
d
?
??
?
?
??
? ?? ??
??? ??
? Where θe(t) is the phase error,Kd is the equivalent PD
constant,The overall equation describing the operation
of the PLL may be obtained by taking the derivative of
Eqs.(4-94) and θe(t),and combining the result by the
using of Eq.(4-96),the resulting nonlinear equation that
describes the PLL becomes,
79
4.14 Phase-Locked Loops and Frequency
Synthesizers
? In general this equation is difficult to solve,
99)-(4 )()(s i n)()( 0 ?????? dtfKKdt tddt td t evdie ? ???
? Application in communication system
? 1) FM detection
? 2) The generation of highly stable FM signals
? 3) Coherent AM detection
? 4) Frequency multiplication
? 5) Frequency synthesis
? 6) Providing bit synchronization and data detection
80
4.15 Direct Digital Synthesis
? Direct digital synthesis (DDS) is a method for generating a
desired waveform by using the computer technique,
? Application,
? Frequency synthesizer to generate local oscillator signals;
? Generating dial tones and busy signal and so on
81
4.16 Transmitter and Receiver
? Transmitter generate the modulated signal at the carrier
frequency fc from the modulating signal m(t),Any type of
modulated signal could be represented by,
1 1 4 )-(4 )]([)()( ttC o stRtv c ?? ??
1 1 5 )-(4 s i n)(c o s)()( ttyttxtv cc ?? ??
1 1 6 )-(4 )()()( tjytxtg ??
? ? 113)-(4 )(Re)( tj cetgtv ??
? There are two canonical forms for the generalized
transmitter
? Using the AM-PM generation technique
? Using the PM-AM generation technique
82
4.16 Transmitter and Receiver
83
4.16 Transmitter and Receiver
84
4.16 Transmitter and Receiver
? The receiver has the job of extracting the source
information from the received modulated signal that may
be corrupted by noise,
? There are two main classes of receivers,
? The tuned radio-frequency (TRF) receiver
? the superheterodyne receiver,
the TRF receiver consists of a number of cascaded high-gain
RF bandpass stages that are tuned to the carrier
frequency fc followed by and appropriate detector circuit,
85
4.16 Transmitter and Receiver
86
4.16 Transmitter and Receiver
? Notes,
? the IF filter is a bandpass filter that selects either the up-
conversion or down-conversion component,
? When up conversion is selected,the complex envelope of
IF filter output is the same as the complex envelope for
the RF input,
? When down conversion is selected with fLO>fc,the
complex envelope at the IF output will become the
conjugate of that for the RF input,if fLO<fc the sideband
are not inverted,
? Zero-IF receivers
? fLO=fc
87
4.16 Transmitter and Receiver
? The image response is the reception of an unwanted signal
located at the image frequency due to insufficient
attenuation of the image signal by the RF amplifier filter,
88
4.17 Software Radios
? Software radios use DSP hardware,microprocessors,
specialized digital ICs,and software to produce modulated
signals for transmission and to demodulate signals at the
receiver,
? The same hardware may be used for many different tpyes
of radios,since the software distinguishes one type form
another;
? If software radios are sold,the can be updated in the field
to include the latest protocols and features by downloading
revised software,
? The software radios concept is becoming more economical
and practical each day,it is the,way of the future”
89
Homework
? d
Chapter 4
Bandpass Signaling
Principles and Circuits
2
Introduction
(chapter objectives)
? Complex envelopes and modulated
signals
? Spectra of bandpass signals
? Nonlinear distortion
? Communication circuits(mixers,phase-
locked loops,frequency synthesizers,
and detectors)
? Transmitters and receivers
? Software radios
3
Introduction(main goals)
? Mastering how to represent complex envelope
of bandpass waveform,
? The principle of bandpass filtering and linear
distortion
? Bandpass sampling theorem
? Nolinear distortion
? Detector circuits
? Knowing PLL,DDC,general principle of
Transceiver,concept of software radios
4
Introduction
? The bandpass communication signal is
obtained by modulating a baseband
analog or digital signal onto a carrier,
? Complex envelope can represent any
type of bandpass signal,
5
4.1 Complex Envelope Representation of
Bandpass Waveforms
? general representation for bandpass
digital and analog signals
? Representation for modulated
signal
? Representation for bandpass noise
6
4.1 Complex Envelope Representation of
Bandpass Waveforms
? Definition1,baseband waveform
having a spectral magnitude that is
nonzero for frequencies in the vicinity of
the origin (i,e,..f=0) and negligible
elsewhere,
? For example,the information source
signal; audio signal in communication
system,
7
4.1 Complex Envelope Representation of
Bandpass Waveforms
? Definition2,bandpass waveform
having a spectral magnitude that is
nonzero for frequencies in some band
concentrated about at frequency ?= ±
?c,where ?c >>0,The spectral magnitude
is negligible eleswhere, ?c is called the
carrier frequency,
8
4.1 Complex Envelope Representation of
Bandpass Waveforms
? Definition3,Modulation
the process of imparting the source
information onto a bandpass signal with a
carrier frequency by the introduction of
amplitude or phase perturbations or both,
? This bandpass signal is called the
modulated signal s(t),and the baseband
source signal is called the modulating
signal m(t),
9
4.1 Complex Envelope Representation of
Bandpass Waveforms
? The definition of modulation indicates
that modulation may be visualized as a
mapping operation that maps the source
information m(t) onto the bandpass
signal,s(t),the bandpass signal will be
transmitted over the channel,
10
4.1 Complex Envelope Representation of
Bandpass Waveforms
? Fig.4-1 communication system
I n f o r m a t i o
n i n p u t
m ( t )
S i g n a l
p r o c e s s i n g
g ( t ) C a r r i e r
c i r c u i t s
s ( t ) T r a n s m i s s i
o n m e d i u m
( c h a n n e l )
r ( t ) )(
~ tg
C a r r i e r
c i r c u i t s
S i g n a l
p r o c e s s i n g
)((~ tm
T r a n s m i t t e r R e c e i v e r
? As the modulated signal passes through the channel,
noise corrupts it;
? the result is a bandpass signal-plus-noise waveform
that is available at the receiver input,r(t),
? The receiver has the job of trying to recover the
information that was sent from the source,
11
4.1 Complex Envelope Representation of
Bandpass Waveforms
(Complex Envelope Represention)
? All bandpass waveforms,whether they arise
from a modulated signal,interfering signal,or
noise,may be represented in a convenient form,
? v(t) will be used to denote the bandpass
waveform such as the signal when s(t)=v(t),
? the noise when n(t)=v(t),
? and the filtered signal plus noise at the channel
output when r(t)=v(t),or any other type of
bandpass waveform,
12
4.1 Complex Envelope Representation of
Bandpass Waveforms
(Complex Envelope Represention)
? Theorem,Any physical bandpass
waveform can be represented by,
? ? 1)-(4 )(Re)( tj cetgtv ??
1b)-(4 )]([)()( ttC o stRtv c ?? ??
1 c )-(4 s i n)(c o s)()( ttyttxtv cc ?? ??
13
4.1 Complex Envelope Representation of
Bandpass Waveforms
(Complex Envelope Represention)p235
? Where the complex envelope of v(t) is g(t),
)()( )(|)(|)()()( tjtgj etRetgtjytxtg ????? ?
? ? )(c o s)()(Re)( ttRtgtx ???
? ?
)()(|)(|)(
)(s i n)()(Im)(
22 tytxtgtR
ttRtgty
???
?? ?
???
?
???
?
??? ?
)(
)(
t a n)()( 1
tx
ty
tgt?
? ? cis associated carrier frequency(in hertz),ωc=2π ? c
14
4.1 Complex Envelope Representation of
Bandpass Waveforms
? Proof,Any physical waveform (it does not have to be
periodic) may be represented over all time,,by the
complex Fourier series,
??0T
00 /2,)( 0 Tectv
n
n
tjn
n ??
? ?? ????
???Because the physical waveform is real,c
-n=cn,and using
Re{.}=(1/2){.}+ (1/2){.}*,we get,
?
?
?
?
?
?
?? ?
?
?
tj
n
n ecctv
0
1
0 2Re)(
?
Because v(t) is a bandpass waveform,the cn have negligible
magnitudes for n in the vicinity of 0 and,in particular,
c0 =0,Thus,with the introduction of an arbitrary
parameter fc,Eqation above becomes,
15
4.1 Complex Envelope Representation of
Bandpass Waveforms
Because v(t) is a bandpass waveform with nonzero
spectrum concentrated near ? = ? c,the Fourier
coefficients cn are nonzero only for values of n in the
range ± n?0 ≈ ?c, Therefore,from Eq(4-8),g(t) has a
spectrum that is concentrated near ? =0, That is,g(t)
is a baseband waveform,
?
?
?
?
?
?? ???
?
? tjtnj
n
n
n
cc eectv ??? )2(Re)( )(
1
0
So that (1) follows,where
tnjn
n
n cectg
)(
1
02)( ?? ?
??
?
??
The waveform g(t),x(t),y(t),R(t),and θ(t) are all
baseband waveforms,and except for g(t),they are all
real waveform,R(t) is a nonnegative real waveform,
16
4.1 Complex Envelope Representation of
Bandpass Waveforms
? (1)(2)(3) are low-pass-to-band-pass transformation,the
factor ejωct in Eq.(1)-(3) translate the spectrum of the
baseband signal g(t) up to the carrier frequency ?c,in
communication terminology,the frequencies in the
baseband signal g(t) are said to be heterodyned up to ?c,
? In Cartesian coordinates,the complex envelope can be
represented as,g(t)=x(t)+jy(t),where x(t) is the in-phase
modulation associated with v(t),and y(t) is the
quadrature modulation associated with v(t),
? Alternatively,the polar form of g(t),represented by R(t)
and θ(t),R(t) is said to be the amplitude modulation
(AM) on v(t),θ(t) is said to be the phase modulation (PM)
on v(t),
? g(t) is the baseband equivalent of the bandpass signal,
17
4.2 Representation of Modulated Signals
? Modulation is the process of encoding the source
information m(t) (modulating signal) into a band-pass
signal s(t) (modulated signal),Consequently,the modulated
signal is just a special application of the bandpass
representation,The modulated signalis given by,
? ?tj cetgts ?)(Re)( ?
? Where ωc=2π?c,?c is carrier frequency,The complex
envelope g(t) is a function of the modulating signal m(t),
That is
)]([)( tmgtg ?
? g[.] performs a mapping operation on m(t),
18
4.3 Spectrum of Bandpass Signals
? The spectrum of a bandpass signal is directly related
to the spectrum of its complex envelope,
Theorem,If a bandpass waveform is represented by
? ? 1 1 )-(4 )(Re)( tj cetgtv ??then the spectrum of the bandpass waveform is
? ? 12)-(4 )()(21)( cc ffGffGfV ????? ?
and the PSD of the waveform is
? ? 13)-(4 )()(41)( cgcgv fffff ????? PPP
Where G(?)=F[g(t)] and Pg(?) is the PSD of g(t),
19
4.3 Spectrum of Bandpass Signals
? Proof,because
? ? tjtjtj ccc etgetgetgtv ??? ????? )(21)(21)(Re)(? Then,
])([21])([21)]([)( tjtj cc etgetgtvfV ?? ????? FFF
If we use F[g*(t)]=G*(- ?),and the frequency translation
property of Fourier transforms,we get,
? ?)]([)(21)( cc ffGffGfV ????? ?
Which is Equation(4-12); The PSD for v(t) is obtained by
first evaluating the autocorrelation for v(t),
? ? ? ?)()(Re)(Re)()()( ??? ??? ????? tjtjv cc etgetgtvtvR
20
4.3 Spectrum of Bandpass Signals
? Using the identity,
)R e(21)R e(21)R e()R e( 121212 CCCCCC ?? ?? and
tj cetgC ?)(2 ? )(1 )( ??? ??? tj cetgC
? we can get,
? ?
? ?)(
)(
)()(Re
2
1
)()(Re
2
1
)(
???
???
?
??
?
???
??
??
tjtj
tjtj
v
cc
cc
eetgtg
eetgtgR
? Realizing that both < > and Re{ } are linear operators,
we may exchange the order of the operators without
offecting the result,and the autocorrelation becomes,
21
4.3 Spectrum of Bandpass Signals
? or
? ?
? ????
?
?
??
cc
c
jtj
tj
v
eetgtg
etgtgR
2)()(Re
2
1
)()(Re
2
1
)(
??
?? ?
? ?
? ?
? ?
? ? 1 6 )-(4 ])(Re
2
1
])()(Re
2
1
)()(Re
2
1
])()(Re
2
1
)(
2
??
??
???
??
?
?
?
??
c
c
cc
c
j
g
j
jtj
j
v
eR
etgtg
eetgtg
etgtgR
?
??
??
??
?
?
? ?c is much larger
than the frequencies
in g(t),so the second
integral is negligible
according to
Riemann-Lebesque
Lemma
22
4.3 Spectrum of Bandpass Signals
The PSD is obtained by taking the Fourier transform of
Eq(4-16),
? ? ? ?)()(41)()( cgcgvv ffffRf ?????? ?PPFP ?
Since the PSD is a real function,Pg*(?)=Pg(?),Hence,the
PSD is given by Eq(4-13),
Notice,Eq.(4-16) has the same mathematical form as
Eq.(4-11) when t is replaced by τ,
23
4.4 Evaluation of Power
? Theorem,The total average normalized power of a
bandpass waveform v(t) is
17)-(4 |)(|21)0()()( 22 tgRdfftvP vvv ???? ? ??? P
where, normalized, implies that the load is
quivalent to 1 ohm
Proof,Substituting v(t) into Eq(2-67),we get
?????? dfftvP vv )()(2 Pbut
? ? ? ???? ?? dfeffR fjvvv ??? 21 )()()( PPF
so
????? dffR vv )()0( P
24
4.4 Evaluation of Power
? Also,from Eq(4-16)
? ? ? ?)0()(Re21)0(Re21)0( ??? ? tgtgRR gv
? ????? cjgv eRR )(Re21)( ?
? or
? ?2|)(|Re21)0( tgR v ?
Another type of power rating,called the peak envelope
power (PEP) is useful for transmitter specifications,
? But |g(t)| is always real,so
2|)(|
2
1)0( tgR
v ?
25
4.4 Evaluation of Power
? Definition,The peak envelope power (PEP) is the
average power that would be obtained if |g(t)| were to
be held constant at its peak value,
? The (PEP) is equivalent to evaluating the average
power in an unmodulated RF sinusoid that has a peak
value of Ap=max(v(t)),
? Theorem,The normalized PEP is given by
? ?2|)(|m a x21 tgP P E P＝
26
4.4 Evaluation of Power
? Example 4-1 Evaluate the magnitude spectrum for
an amplitude-modulated (AM) signal
Solution,the complex envelope of an AM signal is,
? ?)(1)( tmAtg c ??
so that the spectrum of the complex envelope is
)()()( fMAfAfG cc ?? ?According to representation of modulation signal,we
obtain the AM signal waveform
ttmAts cc ?c o s)](1[)( ??
and,using Eq (4-12),we get the AM spectrum
? ?)()()()(21)( ccccc ffMffffMffAfS ???????? ??
27
4.4 Evaluation of Power
the magnitude spectrum is
?
?
?
??
?
?
????
????
?
0 )(
2
1
)(
2
1
0 )(
2
1
)(
2
1
)(
fffMAffA
fffMAffA
fS
cccc
cccc
?
?
Suppose that the magnitude spectrum of the modulation
happens to be a triangular function,as shown in
Fig 4-2a, The resulting AM spectrum is shown in Fig
4-2b, The g(t)=Ac[1+m(t)] cause delta functions to
occur in the spectrum at ? =± ?c,where ? is the
assigned carrier frequency,Using Eq (4-17),we obtain
the total average signal power
?????? ??? )()(212
1 22 tmtmA
c
)()(2121)(121 222 tmtmAtmAP ccs ?????
28
4.4 Evaluation of Power
If we assume that the dc value of the modulation is zero,as
shown in Fig 4-2a,the average signal power becomes
]1[21 mcs PAP ??
where Pm=〈 m2(t) is
the power in the
modulation m(t),
(1/2)Ac2 is the
carrier power,
and (1/2)Ac2Pm is
the power in the
sidebands of s(t)
- 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
29
4.5 Bandpass Filtering and Linear
Distortion （ Equivalent Low-pass Filter)
? Fig 4-3 bandpass filtering
])(Re [)(
11
tj
c
etgtv
?
?
B a n d p a s s f i l t e r
])(R e [)(
11
tj
c
etkth
?
?
)(2/1)(2/1)(
*
cc
ffKffKfH ?????
])(Re [)(
22
tj
c
etgtv
?
?
a ) b a n d p a s s f i l t e r
- f c fc
1 / 2 K ( f - f c ) 1 / 2 K
*
( - f - f c )
| H ( f )
|f
|f
1 / 2 g 1 ( t )
1 / 2 G 1 ( f )
E q u i v a l e n t l o w - p a s s f i l t e r
1 / 2 k ( t )
1 / 2 K ( f )
1 / 2 g 2 ( t )
1 / 2 G 2 ( f )
b ) t y p i c a l b a n d p a s s f i l t e r f r e q u e n c y r e s p o n s e
c ) e q u i v a l e n t ( c o m p l e x i m p u l s e r e s p o n s e ) l o w - p a s s f i t l e r
1 / 2 | K ( f ) |
d ) t y p i c a l e q u i v a l e n t l o w - p a s s f i l t e r f r e q u e n c y r e s p o n s e
30
4.5 Bandpass Filtering and Linear
Distortion （ Equivalent Low-pass Filter)
Theorem,The complex envelopes for the input,output,
and impulse response of a bandpass filter are
related by,
2 2 )-(4 )(21*)(21)(21 12 tktgtg ?Where g
1(t) is the complex envelope of the input and k(t)
is the complex envelope of the impulse response,It also
follows that,
23)-(4 )(21)(21)(21 12 fKfGfG ?
Proof,We know that the spectrum of the output is
2 4 )-(4 )()()( 12 fHfVfV ?Because,v
1(t),v2(t) and h(t) are all bandpass waveforms,the spectra of these waveforms are related to the
spectra of their complex envelopes by Eq.(4-15),thus,
31
4.5 Bandpass Filtering and Linear
Distortion （ Equivalent Low-pass Filter)
? Thus,
? ?
? ? ? ?
)]()2/1)(()2/1()()2/1)(()2/1[(
)]()()()(
)()()()()[4/1(
)()()2/1()()(( 1 / 2 )
)()(
2
1
)(
**
11
**
1
*
1
*
11
**
11
*
222
cccc
cccc
cccc
cccc
cc
ffKffGffKffG
ffKffGffKffG
ffKffGffKffG
ffKffKffGffG
ffGffGfV
????????
?????????
???????
?????????
?????
)(21)(21)(21 12 fKfGfG ?
? Taking the inverse Fourier transform of both sides of
Equation above,Eq.(4-22) is obtained,
? Proof is over
32
4.5 Bandpass Filtering and Linear
Distortion （ Equivalent Low-pass Filter)
? This theorem indicates that any bandpass filter system
may be described and analyzed by using an equivalent
low-pass filter;
? Equation for equivalent low-pass filter are usually
much less complicated than those for bandpass filter,
so the equivalent low-pass filter system model is very
useful,it is the basic for computer programs that
using sampling to simulated bandpass communication
system,
? The equivalent low-pass filter with complex inpulse
response may be realized by using four low-pass filter
with real impulse response,
33
4.5 Bandpass Filtering and Linear
Distortion (Linear Distortion)
? For distortionless transmission of bandpass signals,
the channel transfer function,H(f)=|H(f)|ejθ(f) needs to
satisfy the following requirements,
(1) The amplitude response is constant,that is,
|H(f)|=A (4-27a)
(2) The derivative of the phase response is a constant,
that is,
2 7 b )-(4 T)(2 1 g?? df fd ??? Where T
g is constant called the complex envelope
delay or,more concisely,the group delay and
f H f? ?（ ）= （ ）
34
4.5 Bandpass Filtering and Linear
Distortion （ Linear Distortion)
Note,Eq(4-27a) is
identical to |H(f)|=A (2-
150a),but Eq.(4-27b)
is less restrictive than
θ(f)=-2πfTd (2-150b),
where Td=Tg,because
the integral of Eq.(4-27)
is,
θ(f)=-2πfTg+ θ0 (4-28)
(1)(2) are sufficient
requirements for
distortionless
transmission of
bandpass signal
? c
| H ( f ) |
A
θ ( f )
θ 0
? c
a ) M a g ni tude re s po ns e
b) ph a s e re s po ns e
S i gna l
ba ndw i dt h
?
?
? Fig.4-4 Tansfer characteristics of a
distortionless bandpass channel
35
4.5 Bandpass Filtering and Linear
Distortion （ Linear Distortion)
? If the input to the bandpass channel is represented by
ttyttxtv cc ?? s i n)(c o s)()(1 ??
? We find the output of the channel is
))(s i n ()())(c o s ()()( 002 ???? ???????? gcggcg TtTtAyTtTtAxtv
? Let
dcgcc TfTf ???? 2)( 0 ?????
? Then,
3 0 )-(4 )(s i n)()(c o s)(
))(s i n ()())(c o s ()()(2
dcgdcg
ccgccg
TtTtAyTtTtAx
ftTtAyftTtAxtv
??????
??????
??
????
36
4.5 Bandpass Filtering and Linear
Distortion ( Linear Distortion)
? where the modulation on the carrier (i.e,the x and y
components) has been delayed by the group time delay,
Tg,and the carrier has been delayed by the carrier time
delay,Td;
? This is distortionless transmission,which is obtained
when Eqs.(4-27a) and (4-27b) are satisfied
? Note that Tg will differ from Td,unless θ0 happens to be
zero,
? In summary,the general requirements for distortionless
transmission of either baseband or bandpass signals are
given by Eqs.(2-150a) and (2-150b),However for the
bandpasscase,Eq.(2-150b) is overly restrictive and may
be replaced by Eq.(4-27b),
37
4.6 Bandpass Sampling Theorem
? Sampling is used in software radios and for simulation of
communication,systems,If the sampling is carried out at
the Nyquist rate or larger(?c≥2B,where B is the highest
in the spectrum of the RF signal),the sampling rate can
be ridiculous,For example,consider a satellite
communication system with a carrier frequency of
?c=6GHz,The sampling raterequired can be at least
12GHZ
? For signals of this type (bandpass signal),it can be shown
that the sampling rate depends only on the bandwidth of
the signal,not on the absolute frequencies involved,that
is,we can reproduce the signal from samples of the
complex envelope,
38
4.6 Bandpass Sampling Theorem
Bandpass Sampling Theorem,If a (real) bandpass
waveform has a nonzero spectrum only over the
frequency interval ?1<| ?|< ?2,where the transmission
bandwith BT is taken to be the absolute bandwidth
BT= ?2- ?1,then the waveform may be reproduced form
sample values if the sampling rate is,
3 1 )-(4 2 Ts Bf ?
example,If the 6GHz bandpass signal previously discussed
had a bandwidth of 10MHz,a sampling frequency of
only 20MHz would be required instead of 12GHz,This is
a savings of three order magnitude,
39
4.6 Bandpass Sampling Theorem
? Proof,We know that the quadrature bandpass
representation is
32)-(4 s i n)(c o s)()( ttyttxtv cc ?? ??? let ?
c be the center of the bandpass,so that ?c = (?2- ?1)/2,Both x(t) and y(t) are baseband signals and are
absolutely bandlimited to B=BT/2,According to
sampling theorem for limited signal,an=ω(n/ ?s)—(2-
160),the sampling rate required to represent the
baseband signal is ?b≥ 2B=BT,then eq.(4-32) becomes,
?
?
??? ?
?
?
?
???
?
?
?
???
?
???
? ??
n bb
bb
c
b
c
b fntf
fntft
f
nyt
f
nxtv
))/((
) ) }/((s i n {s i n)(c o s)()(
?
???
? For the general case,where the x (n/ ?b ) and y (n/ ?b )
samples are independent,two real samples are
obtained for each,value of n,so that the overall
sampling rate for v(t) is
3 1 )-(4 2 Ts Bf ?
40
4.6 Bandpass Sampling Theorem
? The x and y samples can be obtained by sampling v(t)
at t= n/ ?b,but adjusting t slightly,so that cosωct=1
and sin ωct=-1,at exact sampling time for x and y,
respectively,that is,for t= n/ ?b,v(n/ ?b )=x(n/ ?b )
when cosωct=1 and v(n/ ?b )=y(n/ ?b ) when sin ωct=-1,
? Alternatively,x(t) and y(t) can first be obtained by the
use of two quadrature product detectors,The x(t) and
y(t) baseband signal can then be individually sampled
at a rate of ?b,and the overall equivalent sampling
rate is still ?s=2 ?b ≥2BT,
41
4.6 Bandpass Sampling Theorem
? Bandpass Dimensionality Theorem,Assume that a
bandpass waveform has a nonzero spectrum only over
the frequency Interval ?1<| ?|< ?2,where the transmission
bandwith BT is taken to be the absolute bandwidth BT=
?2- ?1,and BT<< ?1,The waveform may be completely
specified over a T0-second interval by,
3 4 )-(4 2 0TBN T?? independent pieces of information.N is said to be the
number of dimensions required to specify the waveform ? Computer simulation is often used to analyze
communication systems.The bandpass dimensionality
theorem tells us that a bandpass signal BT Hz wide can
be represented over a T0-s interval,provided that at
least N=2BTT0 samples are used,
42
4.7 Receiced Signal Plus Noise
Referring to above Fig,the signal out of the transmitter is
I n f o r m a t i o
n i n p u t
m ( t )
S i g n a l
p r o c e s s i n g
g ( t ) C a r r i e r
c i r c u i t s
s ( t ) T r a n s m i s s io n m e d i u m
( c h a n n e l )
r ( t ) )(~ tg C a r r i e r
c i r c u i t s
S i g n a l
p r o c e s s i n g
)((~ tm
T r a n s m i t t e r R e c e i v e r
)()(*)()( tnthtstr ??
where g(t) is the complet envelope for the particular type of
modulation used,if the channel is linear and time
invariant,the received signal plus noise is,
? ? )(Re)( tj cetgts ??
where h(t) is the impulse response of the channel and n(t) is
the noise at the receiver input
43
4.7 Receiced Signal Plus Noise
? Furthermore,if the channel is distortionless,its transfer
function is,
? and conesqudntly,the signal plus noise at the receiver
input is,
2 9 )-(4 )()( 2)2( 00 gg fTjjfTj eAeAefH ???? ??? ??
)]()(R e [)( ))(( tneTtAgtr cc ftjg ??? ? ??
where A is the gain of the channel,Tg is the channel group
delay,and θ(?c) is the carrier phase shift caused
by the channel,
44
4.8 Classification of Filters and Amplifiers
(Filters)
? Filters are devices that take an input waveshape and
modify the frequency spectrum to produce the output
waveshape,
? Filters may be classified in several ways,one is by the
type of construction used,Another is by the type of
transfer function that is realized,
? Filters use energy storage elements to obtain
frequency discrimination,In any physical filter,the
energy storage elements are imperfect,
? Two different measures of filter quality are used in the
technical literature,
45
4.8 Classification of Filters and Amplifiers
? The first definition is concerned with the efficiency of
energy storage in a circuit and is,
2 ( m a x im u m e n e r g y s to r e d d u r in g o n e c y c l e )Q
e n e r g y d is s ip a te d p e r c y c l e
??
? A large value for Q corresponds to a more perfect
storage element,that is,a perfect L,or C element
would have infinite Q,The second definition is
concerned with the frequency selectivity of a circuit
and is,
0fQ
B?
? Where f0 is the resonant frequency and B is the 3 dB
bandwidth,the larger the value of Q,the better is the
frequency selectivity,
46
4.8 Classification of Filters and Amplifiers
? Filters are also characterized by the type of transfer
function that is realized,
nn
kk
jajajaa
jbjbjbbfH
)(.,,)()(
)(.,,)()()(
2210
2210
???
???
????
?????
? Amplifier can be classified into two main categories,
Nonlinear and linear,And can be further classified into
the subcategories of circuit with memory and circuits
with no memory,circuits with memory contain inductive
and capacitive effects that cause the present output value
to be a function of previous input values as well as the
present input value,
? The transfer function of a distortionless amplifier is given
by,where K is voltage gain of the amplifier and
Td is the delay between the output and input waveform.,
dcTjKe ?
47
4.9 Nonlinear Distortion
? In addition to linear distortion,practical amplifiers
produce nonlinear distortion,To examine the effects of
nonlinearity and yet keep a mathematical model that is
tractable,In the following analysis we assume no memory,
If the amplifier is linear,we get,
4 1 )-(4 )()(0 tKvtv i?
? Where K is the voltage
gain of the amplifier,In
practical,the output of the
amplifier becomes (soft)
saturated at some value as
the amplitude of the input
signal is increased,
v 0 out put
S a t ur a t i on
l e ve l
v i i nput
F i g,4 - 5 N onl i ne a r a m pl i f i e r out put - t o i nput c ha r a c t e r i s t i c s
48
4.9 Nonlinear Distortion
? The output-to-input characteristic may be modeled by a
Taylor’s expansion about vi=0,that is,
42)-(4
0
2
2100 ?
?
?
?????
n
n
inii vKvKvKKv ?? where
4 3 )-(4
!
1
0
0
?
?
?
?
?
?
?
?
?
?
iv
n
i
n
n
dv
vd
n
K
? There will be nonlinear distortion on the output signal if
K2,K3,… are not zero,K 0 is the output dc offset level,K1v1
is the first-order(linear) term,K2vi2 is the second-order
term,and so on,Of course,K1 will be larger than K2,
K3,… if the amplifier is anywhere near to be linear,
49
4.9 Nonlinear Distortion
? The harmonic distortion associated with the amplifier
output is determined by applying a single sinusoidal test
tone to the amplifier input
4 4 )-(4 t in)( 00 ?sAtv i ?? The second-order output term is,
45)-(4 tc o s 2
2
K-
2
Kt)in(
t e r md i s t o r t i o n h a r m o n i c s e c o n d
0
2
02
t e r ml e v e l
2
022
002
?? ??? ?????
?? AAsAK
dc
?
? For a single-tone input,the output is,
46)-(4,,,)t3c o s (
)t2c o s ()tc o s ()(
3203
2021010
???
?????
??
????
V
VVVtv o u t
? Where Vn is the peak value of the output at the frequency
nf0 Hz,
50
4.9 Nonlinear Distortion
? The percentage of total harmonic distortion (THD) is
defined by,
4 7 )-(4 1 0 0( % )
1
2
2
?? ?
?
?
V
V
T H D n n? THD can be measured 1) by using a distortion analyzer;2)
evaluating by (4-47),with the vn’s obtained from a
spectrum analyzer,
? The intermodulation distortion(IMD) of the amplifier is
determined by applying two-tone test,
48)-(4 t int in)( 2211 ?? sAsAtv i ??? The second-order output term is,
t)intintin2t in(
t)int in(
2
22
221211
22
12
2
22112
????
??
sAssAAsAK
sAsAK
???
?
51
4.9 Nonlinear Distortion
? The first and last term produce harmonic distortion at the
frequency 2f1 and 2f2,the cross-product term produce
IMD,the second-order IMD is,
) t )c o s (-)t-( c o s (2tintin2 212121221212 ?????? ?? AAKssAAK? Which generates sum and difference frequencies,
? The third-order output term is,
t)intintin3
4 9 )-(4t intin3t in(
t)int in(
2
33
22
2
1
2
21
21
2
2
2
11
33
13
3
22113
3
3
???
???
??
sAssAA
ssAAsAK
sAsAKvK i
??
??
??
? The first and last term produce harmonic distortion at the
frequency 3f1 and 3f2,the cross-product term,the second-
order becomes,
52
4.9 Nonlinear Distortion
? Similarly,the third term is,) t ] )s i n ( 2-)t2( 1 / 2 ) [ s i n (tin(( 3 / 2 )
50)-(4 t ) )2 c o s ( 2-t ( 1in( 3 / 2 )
tintin3
212122
2
13
122
2
13
21
2
2
2
13
?????
??
??
????
?
sAAK
sAAK
ssAAK
) t ] )s i n ( 2-)t2( 1 / 2 ) [ s i n (tin(( 3 / 2 )
51)-(4 t intin3
12121
2
213
2
2
1
2
213
?????
??
???? sAAK
ssAAK
? The last two terms in Eqs.(4-50) and (4-51) are
intermodulation terms at nonharmonic frequencies,
? Example,For the case of bandpass amplifiers where f1 and f2
are within the bandpass with f1 close to f2,this distortion
product at 2f1+ f2 and 2f2+ f1 will usually fall outside the
passband,but the term 2f1- f2 and 2f2- f1 will fall inside the
passband and will be close to the desired frequencies this
will be the main distortion products for bandpass amplifier,
53
4.9 Nonlinear Distortion
? If A1 or A2 is increased sufficiently,the IMD will become
significant,since the desired output varies linearly with
A1 or A2 and the IMD output varies as A12A2 or A22A1,
? Of course the exact input level required for the
intermodulation products depends on the relative values
of K3 and K1,the level may be specified by the amplifier
third-order intercept point,If the input test tone’s
amplitude is equal,A,the desired linearly amplified
outputs have amplitudes of K1A,and each of the third-
order intermodulation products have amplitudes of
3K3A3/4,the ratio of the desired output to the IMD
output is,
5 2 )-(4
3
4
2
3
1
?
?
?
?
?
?
?
?
?
AK
KR
IM D
54
4.9 Nonlinear Distortion
55
4.10 Limiters
? A limiter is a nonlinear circuit with an output saturation
characteristic,
? A ideal limiter transfer function is essentially identical to
the output-to-input characteristic of an ideal comparator
with a zero reference level,
? A bandpass limiter is a nonlinear circuit with a saturating
characteristic followed by a bandpass filter,If a input of
bandpass signal is,
))(c o s ()()( tttRtv cin ?? ??? Where R(t) is the equivalent real envelope and θ(t) is the
equivalent phase function,The output of an ideal
bandpass limiter becomes,
55)-(4 ))(c o s ()( ttKVtv cLout ?? ??
56
4.10 Limiters
? Limiters are often
used in receiving
systems designed for
angle-modulated
signaling ---such as
PSK,FSK,and
analog FM--- to
eliminate any
variations in the real
envelope of the
receiver input signal
that are caused by
channel noise or
signal fading,
v out I d e a l l i m i t e r
c h a r a c t e r i s t i c
v in
v lim ( t )
v L
- v L
v L
- v L
v in ( t )
F i g, 4 - 7 i d e a l l i m i t e r c h a r a c t e r i s t i c w i t h
i l l u s t r a t i v e i n p u t a n d u n f i l t e r e d o u t p u t w a v e f o r m
57
4.11 Mixers,Up Converters,and Down
Converters
? an ideal mixer is an electronic circuit that functions as a
mathematical multiplier of two input signals,Mixers are
used to obtain frequency translation of the input signal,If
the input signal is,
L o c a l
o s c i l l a t o r
m i x e r
f i l t e r
v in ( t ) v 1 ( t )
v 2 ( t )
v Lo ( t ) = A 0 c o s ω 0 t
F i g, 4 - 8 M i x e r f o l l o w e d b y a f i l t e r
f o r e i t h e r u p a n d d o w n c o n v e r s i o n
5 6 )-(4 })(R e {)( tjinin cetgtv ??
? where gin is the
complex
envelope of the
signal.the
output of the
mixer is then,
)]([
4
A
t c o s})(R e {)(
00*0
001
tjtjtj
in
tj
in
tj
in
eeegeg
etgAtv
cc
c
????
? ?
?????
?
58
4.11 Mixers,Up Converters,and Down
Converters
? This illustrates that the input bandpass signal with a
spectrum near f=fc has been converted into two output
bandpass signal,One at the up-conversion frequency
band,where fu=fc+f0,and one at the down-conversion
band where fd=fc-f0,,a filter may be used to select
either the up-conversion component or the down-
conversion component,
57)-(4 })(R e {
2
A
})(R e {
2
A
)()()()([
4
A
)]()()([
4
A
t c o s})(R e {)(
)(0)(0
)(*)()(*)(0
*0
001
00
0000
00
tj
in
tj
in
tj
in
tj
in
tj
in
tj
in
tjtjtj
in
tj
in
tj
in
cc
cccc
cc
c
etgetg
etgetgetgetg
eeetgetg
etgAtv
????
????????
????
?
?
??
??????
??
??
????
???
?
59
4.11 Mixers,Up Converters,and Down
Converters
? A bandpass filter is used to selected the-conversion
component,but the down-conversion component is
selected by either a baseband filter or a bandpass filter,
For example,
58)-(4 })(R e {2A)( )(02 0 tjin cetgtv ?? ??? If f
c<f0 (4-57) can be represented by,
5 9 )-(4 })(R e {2A})(R e {2A)( )(*0)(01 00 tjintjin cc etgetgtv ???? ?? ??
? That is to saying that the sidebands have been exchanged;
the upper sideband of the input signal spectrum becomes
the lower sideband of the down-converted output signal,
and so on,The reason is,
)()]([ ** fGtg inin ??F
60
4.11 Mixers,Up Converters,and Down
Converters
? Summary,
? The complex envelope for the signal out of an up converter
is,
6 1 a )-(4 )()2/()( 02 tgAtg in?
? For the case of down conversion,there are two
possibilities,
61b)-(4 )()2/( )()2/()(
0
0
*
0
0
2 ff
ff
tgA
tgAtg
c
c
in
in
?
?
??
??
? We should notice that mixers used in communication
circuits are essentially mathematical multipliers,
? We should notice that mixers used in communication
circuits are essentially mathematical multipliers,They
should be not confused with the audio mixers(音频混音器 )
61
4.12 Frequency Multipliers
? Frequency multipliers is
essentially a nonlinear
amplifier followed by a
bandpass filter that is
designed to pass the nth
harmonic,they consist of
a nonlinear circuit
followed by a tuned
circuit,If a bandpass
signal is fed into a
frequency multiplier,the
output will appear in a
frequency band at the
nth harmonic of the
input carrier frequency,
62
4.12 Frequency Multipliers
? If the bandpass input signal is represented by,
7 2 )-(4 ))(c o s ()()( tttRtv cin ?? ??? The output signal of the nonlinear device is,
so t h e r t e r m ))(( t ) c o s ( n
))((c o s)()()(1
???
???
tntCR
tttRKtvKtv
c
n
c
nn
n
n
inn
??
??
? The output signal of the bandpass filter is,
7 3 )-(4 ))(( t ) c o s ( n)( tntCRtv cno u t ?? ??? This shows that the input amplitude variation R(t) appears
distorted on the output signal,but waveform of the angle
variation is not distorted by frequency multiplier,Thus
they are not used in signals if AM is to preserved; but are
very useful in PM and FM problem,n=2 doubler stage,
n=3 triple stage,
? Frequency multiplier is nonlinear,mixer is linear,
63
4.13 Detector Circuit
? The receiver contains carrier circuits that convert the
input bandpass waveform into an output baseband
waveform,These carrier circuits are called detector
circuits,This sections that follow will show how detector
circuits can be designed to produce R(t),θ(t),x(t),or y(t),
at their output for the corresponding bandpass signal that
is fed into the detector input,
I n f o r m a t i o
n i n p u t
m ( t )
S i g n a l
p r o c e s s i n g
g ( t ) C a r r i e r
c i r c u i t s
s ( t ) T r a n s m i s s i
o n m e d i u m
( c h a n n e l )
r ( t ) )(
~ tg
C a r r i e r
c i r c u i t s
S i g n a l
p r o c e s s i n g
)((~ tm
T r a n s m i t t e r R e c e i v e r
? Envelope detector; product detector; Frequency
modulation detector
64
4.13 Detector Circuit
(Envelope Detector)
? An ideal envelope detector is a circuit that produce a
waveform at its output that is proportional to the real
envelope R(t) of its input,If the bandpass input may be
represented by,
))(c o s ()()( tttRtv cin ?? ??
? Then the output of the ideal envelope detector is,
74)-(4 )()( tKRtv out ?
? Then the output of the ideal envelope detector is,
65
4.13 Detector Circuit (Envelope Detector)
66
4.13 Detector Circuit
(Envelope Detector)
? A simple diode detector circuit may approximates an ideal
envelope detector,but its RC time constant is chosen so
that the output signal will follow the real envelope R(t) of
the input signal,Consequently,the cutoff frequency of the
low-pass filter needs to be much smaller than the carrier
frequency fc,and much larger than the bandwidth of the
(detected) modulation waveform B,that is,
? B<<1/(2πRC)<<fc (4-75)
? Where RC is the time constant of the filter,The envelope
detector is typically used to detect the modulation on AM
signals,In this case,vin(t) has the complex envelope
g(t)=Ac(1+m(t)),If|m(t)|<1,then,
? vout(t)=KR(t)=K[g(t)]=KAc[1+m(t)]=KAc+KAcm(t) (4-75)
? Note,KAc is a DC voltage that is used to provide automatic
gain control (AGC),KAcm(t) is the detected modulation,
67
4.13 Detector Circuit
(Product Detector)
? A product detector is a mixer circuit that down-converts
the input (bandpass signal plus noise) to baseband,The
output of the multiplier is,
L o c a l
o s c i l l a t o r
L o w p a s s
f i l t e r
v in ( t ) = R ( t ) c o s [ ω c t+ θ ( t ) ] v 1 ( t ) v out ( t )
v Lo ( t ) = A 0 c o s ( ω 0 t+ θ 0 )
F i g, 4 - 1 4 p r o d u c t d e c t e c t o r
O r v in ( t ) = R e { g ( t ) e x p ( - j ω c t ) } ]
W h e r e g ( t ) = R ( t ) e x p ( j θ ( t ) )
v1(t)=R(t)cos[ωct+θ(t)] A0cos(ω0t+θ0)
=(1/2) A0 R(t) cos(θ(t)- θ0)
+(1/2) A0 R(t) cos(2ωct θ(t)+ θ0)
68
4.13 Detector Circuit
(Product Detector)
? After the low-pass filter,the output is,
vout(t) =(1/2) A0 R(t) cos(θ(t)- θ0)
=(1/2) A0Re{g(t)exp(-j θ0)} (4-76)
? Where g(t)=R(t)exp(j θ(t))=x(t)+jy(t),Note,if θ0 =0,the
oscillator is said to be phase synchronized with the in-
phase component,And the output becomes,
vout(t) =(1/2) A0 x(t) (4-77a)
If θ0 =900,
vout(t) =(1/2) A0 y(t) (4-77b)
This indicate that a product detector is senstive to AM
and PM,Example-1,if the input contains no angle
modulation,so that θ(t)=0,and if θ0 =0,then,
69
4.13 Detector Circuit
(Product Detector)
? Which implies that x(t) ≥0,and the real envelope is
obtained on the product detector output,Example-2 if an
angle-modulated signal Accos [ωct+θ(t)] is present at the
input and θ0 =900,the product detector output is,
vout(t) =(1/2) A0 R(t) (4-78a)
7 8 b )-(4 ))(s i n ()2/1(
}R e {)2/1()(
0
]90)([
0
0
tAA
eAAtv
c
tj
cout
?
?
?
? ?
? In this case,the product detector acts like a phase detector
with a sinusoidal characteristic,the reason is that the
output voltage is proportional to the sine of the phase
difference between the input signal and the oscillator
signal,A special case is that the phase difference is small,
70
4.13 Detector Circuit
(Product Detector)
? That is | θ(t) |<<900,then sin θ(t)= θ(t) and,
79)-(4 )()2/1( ))(s i n ()2/1()( 00 tAAtAAtv ccout ?? ??
Then the output of this phase detector is directly proportion
to the phase difference,
Summary,1) Unlike the envelope detector that is a
nonlinear device,the product detector acts as a linear
time-varying device with respect to the input vin(t),
2) Detector may also classified as being either coherent or
noncoherent,The former has two input- one for a
reference signal and one for the modulated signal that is to
be demodulated,The product detector is a coherent one,
The later has only one input,namely,the modulated singal
port,the envelope detector is a noncoherent one,
71
4.13 Detector Circuit
(Frequency Modulation Detector)
? An ideal frequency modulation (FM) detector is a
device that produces an output that is proportional to
the instantaneous frequency of the input,that is,if the
bandpass input is represented by R(t)cos(ωct+θ(t)),the
output of the ideal FM detector is,
8 0 )-(4 ))(θ+ω(=))(θ+ω(=)( dt tdKdt ttdKtv ccout
Usually,the FM detector is balanced,
81)-(4 )(θ=)( dt tdKtv o u tAlmost all of Frequency Modulation detector are based on
one of three principle,
1)FM-to-AM conversion
2)Phase-shift or quadrature detection
3) zero-crossing detection
72
4.13 Detector Circuit
(Frequency Modulation Detector)
? A slope detector
is one example of
the FM-to-AM
conversion principle,
it want a bandpass
limiter to suppress any
amplitude variations on
the input signal,the
differentiation operation
can be otained by any
circuit that acts like a
frequency-to-amplitude
converter,
73
4.13 Detector Circuit
(Frequency Modulation Detector)
? The quadrature detector is described as follows,a
Quadrature signal is first obtained form the FM signal;
then though the use of a product detector,the
quadrature signal is multiplied with the FM signal to
produce the demodulated signal,
)(
2
1
=
)(θ
2
1
)
)(θ
s i n (
2
1
=)(
8 8 )-(4 )
)(θ
+)(θ+ωs i n (=)(
))(θ+ωco s (=)(
2
21
2
212
2
1
21
tmKVKK
dt
td
VKK
dt
td
KVKtv
dt
td
KttVKtv
ttVtv
fLLLout
cLquad
cLin
＝
? Zero-crossing detector because the output of an ideal
FM detector is directly proportional to the
instantaneous frequency of the input,The linear
frequency-to-voltage characteristic may be obtained
directly by counting the zero-crossings of the input
waveform,
74
4.13 Detector Circuit
(Frequency Modulation Detector)
? Because,
dt
tdftf
ci
)(θ
π2
1+=)(
And nonostable
multivibrator is
triggered on positive
slope zero-crossings,
Vout(t) =0 f=fi
Vout(t) >0 fi>f
Vout(t) <0 fi<f
75
4.14 Phase-Locked Loops and Frequency
Synthesizers
? A Phase-locked loop consists of three basic components,
? A phase detector,
? a low-pass filter,
? a voltage-controlled oscillator
76
4.14 Phase-Locked Loops and Frequency
Synthesizers
? The first operating mode,PLL acts as a narrowband
tracking filter when the low-pass filter is a narrowband
filter,The frequency of the VCO will become that of noe
of the line components of the input signal spectrum,The
frequency of the VCO will track the input signal
component if it changes slightly in frequency,
? The second operating mode,the bandwidth of the LPF is
wider so that the VCO can track the instantaneous
frequency of the whole input signal,
? When PLL tracks the input signal in either of these ways,
the PLL is said to be,locked”,
? If the PLL is built using analog circuits,it is said to be an
analog PLL,conversely digital circuits and signals are
used,the PLL is said to be a digital PLL
77
4.14 Phase-Locked Loops and Frequency
Synthesizers
? If the input signal is,))(sin()(
0 ttAtv iiin ?? ??
? The VCO output signal is,))(sin()(
0 ttAtv ooo ?? ??
? where
94)-(4 )()( 2 dxxvKt tvo ?? ???? Then,
))()(2s i n (
2
))()(s i n (
2
))(s i n ())(s i n ()(
0
001
ttt
AAK
tt
AAK
ttttAAKtv
oi
oim
oi
oim
iooim
?????
????
?????
???
78
4.14 Phase-Locked Loops and Frequency
Synthesizers
? Then,
))()(2s i n (
2
))()(s i n (
2
))(s i n ())(s i n ()(
0
001
ttt
AAK
tt
AAK
ttttAAKtv
oi
oim
oi
oim
iooim
?????
????
?????
???
96)-(4 )(*)]([s i n
)(*))()(s i n (
2
)(
)(
2
tftK
tftt
AAK
tv
ed
t
oi
K
oim
e
d
?
??
?
?
??
? ?? ??
??? ??
? Where θe(t) is the phase error,Kd is the equivalent PD
constant,The overall equation describing the operation
of the PLL may be obtained by taking the derivative of
Eqs.(4-94) and θe(t),and combining the result by the
using of Eq.(4-96),the resulting nonlinear equation that
describes the PLL becomes,
79
4.14 Phase-Locked Loops and Frequency
Synthesizers
? In general this equation is difficult to solve,
99)-(4 )()(s i n)()( 0 ?????? dtfKKdt tddt td t evdie ? ???
? Application in communication system
? 1) FM detection
? 2) The generation of highly stable FM signals
? 3) Coherent AM detection
? 4) Frequency multiplication
? 5) Frequency synthesis
? 6) Providing bit synchronization and data detection
80
4.15 Direct Digital Synthesis
? Direct digital synthesis (DDS) is a method for generating a
desired waveform by using the computer technique,
? Application,
? Frequency synthesizer to generate local oscillator signals;
? Generating dial tones and busy signal and so on
81
4.16 Transmitter and Receiver
? Transmitter generate the modulated signal at the carrier
frequency fc from the modulating signal m(t),Any type of
modulated signal could be represented by,
1 1 4 )-(4 )]([)()( ttC o stRtv c ?? ??
1 1 5 )-(4 s i n)(c o s)()( ttyttxtv cc ?? ??
1 1 6 )-(4 )()()( tjytxtg ??
? ? 113)-(4 )(Re)( tj cetgtv ??
? There are two canonical forms for the generalized
transmitter
? Using the AM-PM generation technique
? Using the PM-AM generation technique
82
4.16 Transmitter and Receiver
83
4.16 Transmitter and Receiver
84
4.16 Transmitter and Receiver
? The receiver has the job of extracting the source
information from the received modulated signal that may
be corrupted by noise,
? There are two main classes of receivers,
? The tuned radio-frequency (TRF) receiver
? the superheterodyne receiver,
the TRF receiver consists of a number of cascaded high-gain
RF bandpass stages that are tuned to the carrier
frequency fc followed by and appropriate detector circuit,
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4.16 Transmitter and Receiver
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4.16 Transmitter and Receiver
? Notes,
? the IF filter is a bandpass filter that selects either the up-
conversion or down-conversion component,
? When up conversion is selected,the complex envelope of
IF filter output is the same as the complex envelope for
the RF input,
? When down conversion is selected with fLO>fc,the
complex envelope at the IF output will become the
conjugate of that for the RF input,if fLO<fc the sideband
are not inverted,
? Zero-IF receivers
? fLO=fc
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4.16 Transmitter and Receiver
? The image response is the reception of an unwanted signal
located at the image frequency due to insufficient
attenuation of the image signal by the RF amplifier filter,
88
4.17 Software Radios
? Software radios use DSP hardware,microprocessors,
specialized digital ICs,and software to produce modulated
signals for transmission and to demodulate signals at the
receiver,
? The same hardware may be used for many different tpyes
of radios,since the software distinguishes one type form
another;
? If software radios are sold,the can be updated in the field
to include the latest protocols and features by downloading
revised software,
? The software radios concept is becoming more economical
and practical each day,it is the,way of the future”
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Homework
? d
