Post on 26-Mar-2020
ECE 461/561
COMMUNICATIONS I
Dr. Mario Edgardo Magaña
School of EECS
Oregon State University
2
This course deals with the fundamental aspects of analog electronic
telecommunication systems. The first part of the course focuses on learning how
to model, analyze, and design analog amplitude and angle modulated
communication systems (and their variants) and to review the principles of
random processes, in order to model a random noisy communication channel.
The last part of the course is devoted to evaluate analog communication system
performance when a modulated signal is transmitted through a random noisy
channel and to introduce the discretization of analog communication signals to
understand the transition to digital communication systems.
3
4
OBJECTIVE: The objective of a communication system is to efficiently transmit
an information-bearing signal (message) from one location to another through a
communication channel (transmission medium).
WISH LIST FOR EFFICIENT COMMUNICATION:
• High reliability
• Minimum transmitted signal power
• Minimum transmission bandwidth
• Low implementation complexity (low cost)
EFFICIENT TRANSMISSION: Can be achieved by processing the signal through
a technique called modulation.
MODULATION TYPES:
• Analog: Amplitude and angle modulation
• Digital: Amplitude, angle and hybrid (amplitude and angle together) modulation
5
MATHEMATICAL PRELIMINARIES
Let a complex signal in polar form be described by 0( )
j tx t Ae
. Then,
0 0( ) cos sin ( ) ( )r ix t A t jA t x t jx t .
Thus, the real and imaginary parts of ( )x t are
0( ) Re ( ) cosrx t x t A t and 0( ) Im ( ) sinix t x t A t .
Clearly, ( )( ) ( ) j tx t x t e , where
2 2( ) ( ) ( )r ix t x t x t and 1 ( )
( ) tan( )
i
r
x tt
x t
.
Even and Odd signals
( )x t is an even signal iff ( ) ( )x t x t and ( )x t is an odd signal iff ( ) ( )x t x t .
In general,
( ) ( ) ( )e ox t x t x t ,
6
where
( ) ( )( )
2e
x t x tx t
and
( ) ( )( )
2o
x t x tx t
.
Example: Let ( )x t be described by 0( ) cosx t A t . Then
0 0( ) cos cos( ) sin sin( )x t A t A t .
Since 0 0cos( ) cos( )t t and 0 0sin( ) sin( )t t , we get
0
( ) ( )( ) cos cos( )
2e
x t x tx t A t
and 0
( ) ( )( ) sin sin( )
2o
x t x tx t A t
.
Energy and Power Signals
A signal ( )x t is an energy signal iff its energy /2 2
/2lim ( )
T
xTT
E x t dt
.
A signal ( )x t is a power signal iff its power xP is such that /2 2
/2
10 lim ( )
T
xTT
P x t dtT
.
Example: If ( )x t is bounded over a finite time interval and zero outside that interval, then ( )x t
is an energy signal.
7
Example: Let 0( ) cosx t A t . Then we can show that ( )x t is a power signal, i.e.
2
/2 /2 22
0/2 /2
1 1lim ( ) lim cos
2
T T
xT TT T
AP x t dt A t dt
T T
.
Fourier Series and Fourier Transforms:
Let ( )x t be a periodic signal, i.e. ( ) ( )x t x t nT , where T is the period of repetition. Then if
( )x t satisfies the Dirichlet conditions, i.e.
a) ( )x t is absolutely integrable over one period T, i.e.
0
0
( )t T
tx t dt
, for arbitrary 0t .
b) The number of maxima and minima of ( )x t in each period is finite.
c) The number and size of the discontinuities of ( )x t in each period is finite.
the periodic signal ( )x t has a Fourier series described by
0
0
2( ) ,
jn t
n
n
x t x eT
,
where 0
0
0
1( )
t Tjn t
nt
x x t e dtT
.
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Let ( )x t be an aperiodic signal. Then if ( )x t satisfies the Dirichlet conditions, i.e.
a) ( )x t is absolutely integrable over the real line, i.e.
( )x t dt
.
b) The number of maxima and minima of ( )x t in any finite interval over the real line is
finite.
c) The number and size of the discontinuities of ( )x t in any finite interval over the real line
is finite.
the signal ( )x t has a Fourier transform in the Hz frequency domain described by
2( ) ( ) j f tX f x t e dt
and the original signal can be recovered from
2( ) ( ) j f tx t X f e df
Parseval’s Theorm:
Let ( )x t be a periodic signal, i.e. ( ) ( )x t x t nT , where n is an integer and T is the period of
repetition. Then the power content of ( )x t is given by
9
0
0
2 21( )
t T
x nt
n
P x t dt xT
.
Let ( )x t be an aperiodic signal which possesses a Fourier transform ( )X f . Then
2 2( ) ( )x t dt X f df
.
Lowpass and Bandpass Signals:
Let ( )x t be described by ( ) ( )cos 2 ( )cx t A t f t t , where ( )A t and ( )t are slowly
varying signals of time (they have small frequency content around zero frequency, i.e. they are
low pass signals). Then,
( ) ( )cos ( ) cos 2 ( )sin ( ) sin 2
( )cos 2 ( )sin 2
c c
I c Q c
x t A t t f t A t t f t
x t f t x t f t
,
where the in-phase and quadrature components ( )Ix t and ( )Qx t are given by
( ) ( )cos ( )Ix t A t t and ( ) ( )sin ( )Qx t A t t ,
respectively.
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Hence, ( )Ix t and ( )Qx t are lowpass signals and ( )x t is a bandpass signal with its spectrum
around the center frequency cf . The lowpass complex envelope of ( )x t is given by
( ) ( ) ( )l I Qx t x t jx t
and
2
2
( ) Re ( ) Re ( ) ( ) cos 2 sin 2
Re ( )cos 2 ( )sin 2 ( )cos 2 ( )sin 2
Re ( )cos 2 ( )sin 2 ( )sin 2 ( )cos 2
( )cos 2 ( )sin 2
cj f t
l I Q c c
I c I c Q c Q c
I c Q c I c Q c
I c Q c
x t x t e x t jx t f t j f t
x t f t jx t f t jx t f t j x t f t
x t f t x t f t j x t f t x t f t
x t f t x t f
( ) cos 2 ( )c
t
A t f t t
.
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Standard Amplitude Modulation (AM)
In this type of modulation the amplitude of a sinusoidal carrier is varied
according to the transmitted message signal. Let m(t) be the message signal
we would like to transmit, ka be the amplitude sensitivity (modulation index), and
c(t) = Accos(2πfct) be the sinusoidal carrier signal, where Ac is the amplitude of the
carrier and fc is the carrier frequency. Then the transmitted standard AM signal
waveform is described by
Requirements:
1. , to avoid overmodulation and phase reversals
2. , where Wm is the message bandwidth
( ) 1 ( ) cos(2 )AM c a cs t A k m t f t
ttmka ,1)(
mc Wf
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Taking the Fourier transform of the modulated waveform, we get
Let be described by
Magnitude spectrum of m(t)
( ) ( ) 1 ( ) cos(2 )
cos(2 ) ( )cos(2 )
( ) ( ) ( ) ( )2 2
AM AM c a c
c c c a c
c a cc c c c
S f s t A k m t f t
A f t A k m t f t
A k Af f f f M f f M f f
( )M f
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Then the magnitude spectrum of sAM(t) is
Magnitude spectrum of sAM(t)
Observations:
1. The transmission bandwidth is
2. Carrier signal is transmitted explicitly (delta functions are present in
frequency spectrum)
mAM WB 2
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Let the message signal be the frequency tone , then with )2.0cos()( ttm
( ) 1 cos(2 ) cos(2 ) 10 1 0.5cos 0.2 cos 2AM c a m cs t A k f t f t t t
15
16
To conserve transmitted power, let us suppress the carrier, i.e., let the
transmitted waveform be described by
This is called double side-band suppressed carrier (DSB-SC) modulation.
).2cos()()( tftmAts ccDSB
17
In the frequency domain,
The transmitted modulated waveform now has the following spectral
characteristics:
Magnitude spectrum of sDSB(t)
( ) ( )
( )cos 2
( )) ( ) ( )2 2
DSB
c c
c cc c c c
S f s t
A m t f t
A AM f f f f f M f f M f f
F
F
18
Observations:
1. Transmission bandwidth is (same as standard AM)
2. Transmitted power is less than that used by standard AM
Example: Let a sinc pulse be transmitted using DSB-SC modulation.
mDSB WB 2
19
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Receivers for AM and DSB
Receivers can be classified into coherent and non-coherent categories.
Definition: If a receiver requires knowledge of the carrier frequency and phase to
extract the message signal from the modulated waveform, then it is called coherent.
Definition: If a receiver does not require knowledge of the phase (only rough
knowledge of the carrier frequency) to extract the message signal from the
modulated waveform, then it is called non-coherent.
Non-coherent demodulator (receiver) for standard AM
Peak envelope detector (Standard AM demodulator)
)(ˆ tm)(tS AMR C
+
- -
+D
21
Observations:
• The net effect of the diode is the multiplication (mixing) of the signals applied
to its input. Therefore, its output will contain the original input frequencies, their
harmonics and their cross products.
• The network is a lowpass filter with a single pole (lossy integrator) that
removes most of the high frequencies. R and C have to be judiciously picked
so that the time constant = RC is neither too short (rectifier distortion ) nor
too long (diagonal clipping ).
• A rule of thumb for choosing is based on the highest modulating signal
(message) frequency that can be demodulated by a peak envelope detector
without attenuation, that is,
where ( ) is the maximum modulating signal (message) frequency in
Hz (rad/s).
2 2
(max) (max)
1/ 1 1/ 1, 1
2
a a
a
m m
k kk
f
(max)mf (max)m
22
Let the message signal be described by the sinusoidal tone .
Then, the following graphs show the types of distortion that can occur when is
properly and improperly chosen.
tftm m2cos)(
Properly chosen
23
Illustration of rectifier distortion and diagonal clipping (with
improperly chosen )
Rectifier distortion
Diagonal clipping
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Coherent demodulator for DSB-SC: Consider the following demodulator which
assumes fc has been estimated perfectly at the receiver, though is not known.
At the output of the mixer,
and
cos4cos)(2
ˆ
2cos)(2cosˆ
)(2cosˆ)(
tftmAA
tftmAtfA
tstfAtv
ccc
cccc
DSBcc
( )
ˆ ˆ2 2 cos
4 2
j jc c c cc c
V f F v t
A A A AM f f e M f f e M f
25
4
Let the message signal have the following magnitude spectrum
Then, if fc >>Wm,
-2 2
26
Suppose is such that
Then, if ,
fHlpf
mcm WfBW 2 cos2
ˆˆ fM
AAfM cc
27
Coherent Costas loop receiver for DSB-SC :
28
I-channel:
After downconverwsion,
At the output of the lowpass filter, with |H(0)| = 1,
Q-channel:
cos2cos)(2
coscos)()(
ttmA
tttmAtv
c
c
cccI
)(cos2
)( tmA
tm c
I
sin2sin)(2
)( ttmA
tv c
c
Q
)(sin2
)( tmA
tm c
Q
29
Feedback path:
At the output of the multiplier,
At the output of the feedback lowpass filter,
The purpose of hf (t) is to smooth out fast time variations of me(t)
The output of the VCO is described by
2sin)(8
cossin)(4
)(
22
2
2
tmA
tmA
tm
c
ce
dthmtm feef )()()(
( ) cos ( ) ,VCO cx t t t
30
where c is the VCO’s reference frequency and is the
residual phase angle due to the tracking error. The constant kv is the frequency
sensitivity of the VCO in rad/s/volt (it depends on the circuit implementation).
The instantaneous frequency in radians/sec of the VCO’s output is given by
Clearly, if (t) were small and slowly varying, then the instantaneous frequency
would be close to c and the output of the I-path would also be proportional to
the desired message signal m(t), since , for (t) small.
,)()(0t
efv dmkt
0( )( )
( ),
t
c v efc
c v ef
d t k m dd t tk m t
dt dt
cos 1
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Angle Modulation
Let i(t) be the instantaneous phase angle of a modulated sinusoidal carrier,
i.e.,
where Ac is the constant amplitude.
The instantaneous frequency of the sinusoidal carrier is
Observation: The signal s(t) can be thought of as a rotating phasor of length Ac
and angle i(t), i.e.
),(cos)( tAts ic
.)(
)(dt
tdt i
i
( ) ( )c is t A t
32
If s(t) were an unmodulated carrier signal, then the instantaneous angle would be
where c Constant angular velocity in rad/s
c Constant but arbitrary phase angle in radians
Let i(t) be varied linearly with the message signal m(t) and c = 0, then
where kp Phase sensitivity of the modulator in rad/volt (circuit dependent)
In this case we say that the carrier has been phase modulated.
The phase modulated waveform is given by
Let the instantaneous frequency i(t) be varied linearly with the message signal
m(t), i.e.,
cci tt )(
),()( tmktt pci
))(cos()( tmktAts pccPM
)()( tmkt ci
33
where k Frequency sensitivity of the modulator in rad/s/volt (circuit dependent)
In this case we say that the carrier has been frequency modulated and the
instantaneous angle is obtained by integrating the instantaneous frequency, i.e.,
The modulated waveform is therefore described by
Observation: Both phase and frequency modulation are related to each other and
one can be obtained from the other. Hence, we could deduce the properties of one
of the two modulation schemes once we know the properties of the other. This is
illustrated in the following 2 block diagrams:
t
c
t
c
t
ii dmktdmkdt000
)()()()(
t
ccFM dmktAts0
)(cos)(
34
FM modulation
0( )
t
m d
35
PM modulation
( )dm t
dt
36
The figure below shows a comparison between AM, FM and PM modulation of
the same message waveform:
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Frequency Modulation
Consider the frequency modulation of a message signal (frequency tone)
The instantaneous frequency (in Hz) of the FM signal is
Define the maximum frequency deviation as
Since
The instantaneous phase angle of the FM signal is
),2cos()( tfAtm mm
( ) cos(2 ), since 2 and 2 .i c f m m c c ff t f k A f t f k k
max ( ) ,i c f mf f t f k A
0 0
( ) 2 ( ) 2 cos 2
2 sin(2 )
2 sin(2 )
t t
i i c f m m
f m
c m
m
c m
t f d f k A f d
k Af t f t
f
f t f t
0
( ) cos cos 2
t
FM c c m ms t A t k A f d
max ( ) max cos(2 ) .i c f m m c f m cf t f k A f t f k A f f
38
where , is known as the FM modulation index (for a tone) or
the maximum phase deviation (in rad) produced by the tone in question
The FM modulated tone is therefore given by
This sFM(t) signal is nonperiodic unless fc = nfm, where n is a positive integer.
For the general case, fc is not necessarily a multiple of fm. Now,
where the complex envelope of the FM signal is described by
Observation: Unlike sFM(t), se(t) is periodic with period 1/fm.
/ /f m m mk A f f f
( ) cos 2 sin(2 )
cos(2 )cos( sin(2 )) sin(2 )sin( sin(2 )) .
FM c c m
c c m c m
s t A f t f t
A f t f t f t f t
2 sin(2 )
sin(2 ) 2 2
( )
( ) ,
c m
m c c
j f t f t
FM c
j f t j f t j f t
c e
s t e A e
e A e e e s t e
sin(2 )( ) mj f t
e cs t A e
39
Since se(t) meets the Dirichlet conditions, we can compute its Fourier series, i.e.,
where the Fourier series coefficients are given by
,)(2 tfnj
n
nemects
.dtefA
dteeAf
dte)t(sf
f/T,dte)t(sT
c
m
m
mm
m
m
mm
m
m
m
m
f/
f/
tfn)tfsin(j
mc
f/
f/
tfnj)tfsin(j
cm
f/
f/
tfnj
em
m
/T
/T
tfnj
en
21
21
22
21
21
22
21
21
2
2
2
21
1
40
Let
Then,
and
where is the Bessel function of the first kind of order n.
Alternatively,
where the Gamma function (t) is defined by
.tfx m2
dxf
dtm2
1
),(JA
dxeA
c
nc
nxxsinjcn
2
dxe)(J nxxsinj
n2
1
2
0
( 1)( ) ,
! ( 1) 2
m nm
n
m
Jm m n
1
0
( ) t xt x e dx
41
Plot of Jn(x), n = 0, 1, 2
Therefore,
and the FM tone waveform is described by
n
tnfj
ncemeJAts
2)()(
2 2( ) ( ) ( )cos 2 ( ) .m cj nf t j f t
FM c n c n c m
n n
s t e A J e e A J f nf t
42
Spectrum of the FM modulated frequency tone:
Average power of the FM waveform:
The average power delivered to a 1 ohm load resistor by the FM waveform is
Now,
is also the average power of the FM waveform delivered to a 1 ohm resistor.
( ) ( )
( ) cos 2 ( )
( ) cos 2 ( )
( )( ) ( )
2
FM FM
c n c m
n
c n c m
n
nc c m c m
n
S f s t
A J f nf t
A J f nf t
JA f f nf f f nf
2/2
cAP
/2 2
/2
1lim ( )
T
FMTT
P s t dtT
43
Examples:
44
It can be shown, in the limit, that
Let us now take a look at the properties of the Bessel function.
1.
2.
Hence, the average power of an FM tone is, as expected,
Suppose is small, i.e., 0 < ≤ 0.3, then
•
•
•
Under the assumption that is small, the Fourier series representation of the FM
waveform can be simplified to three terms.
)()1()( n
n
n JJ
1)(2
n
nJ
.2/2
cA
1)(0 J
2/)(1 J
2,0)( nJ n
2/2 2 2
/2
1lim ( ) ( )
2
Tc
FM nTT
n
As t dt J
T
45
Thus, for small, the FM tone may be described by
In the frequency domain,
)t)ff(cos(A)t)ff(cos(A)tfcos(A
)t)ff(cos()t)ff(cos()tfcos(A)t(s
mccmcccc
mcmcccFM
22
22
2
22
22
2
mcmc
c
mcmc
c
cc
c
FM
ffffffA
ffffffA
ffffA
fS
4
42)(
46
A plot of the magnitude spectrum of the FM tone with small is shown below
The time domain FM waveform can be represented in phasor form as follows:
For arbitrary t = t0, and small , we can illustrate graphically the phasor
representation and arrive at some conclusion.
tfAtfAAS mcmccFM 22
12
2
10
47
The following figure shows an example of the phasor representation
Observation:
The resultant phasor , has magnitude and is out of phase with
respect to the carrier phasor
Analytically,
FMS
,cFM AS
.0cA
)2sin()2cos()2sin()2cos(2
1 tfjtftfjtfAAS mmmmccFM
48
But,
and
Consequently, the resultant phasor (in rectangular form) is given by
The magnitude of the resultant may be approximated by
since , n = ½ and in our case.
)2cos(
)2sin(sincos)2cos()2cos(0
tf
tftftf
m
mmm
)2sin(
)2cos(sincos)2sin()2sin(0
tf
tftftf
m
mmm
)2sin( tfjAAS mccFM
1/22 2 2 2 2 2
2 2
sin (2 ) 1 sin (2 )
11 sin (2 ) ,
2
FM c c m c m
c m
S A A f t A f t
A f t
(1 ) 1 , 1nx nx nx 2 2sin 2 mx f t
49
Finally, the magnitude and phase of the resultant are found to be
Observation:
• For an FM tone, the spectral lines sufficiently away from the carrier may be
ignored because their contribution (amplitude) is very small when β is small.
FM Transmission Bandwidth:
For an FM tone, as becomes large Jn() has significant components only for
All significant lines are contained in the frequency range
where f is the peak frequency deviation.
)4cos(
441)(
22
tfAtS mcFM
1 1sin(2 )( ) tan tan sin(2 )c m
FM mS
c
A f tS t f t
A
.// mmmf fffAkn
,ffff cmc
50
Let be small, i.e., 0 < ≤ 0.3, then
and only the first pair of spectral lines are significant, i.e., the significant lines are
contained in the range fc± fm
Observation: The previous analysis of an FM tone suggests that
1. For large the FM bandwidth is
2. For small the FM bandwidth is
In general, the FM transmission bandwidth may be approximated by
This is known as the Carson’s rule.
Observation: Carson’s rule underestimates the transmission bandwidth by about
10%.
0),()(0 nJJ n
2FMB f
.2 mFM fB
)/11(2
)/1(2
22
f
fff
ffB
m
mT
51
Alternative definition of FM tone transmission bandwidth:
A band of frequencies that keeps all spectral lines whose magnitudes are greater
than 1% of the unmodulated carrier amplitude Ac, i.e.,
where
,2 max mT fnB
.01.0)(:maxmax nJnn
52
General Case:
Let an arbitrary message signal m(t) have bandwidth Wm.
Define the peak frequency deviation and the deviation ratio by
and
Then Carson’s rule can be used to define the transmission bandwidth of an
arbitrary FM signal, i.e., when m(t) is arbitrary.
Specifically, the FM transmission bandwidth can be defined by
)(maxˆ tmkft
f
./ˆ mWfD
2 2
12 1 2 1
2 1 2 1
m
m
m m
m
BT f W
Wf f
f D
fW W D
W
53
Example: In commercial FM in the US, f = 75 kHz, Wm = 15 kHz.
Therefore, the deviation ratio is D = 75 kHz/15 kHz = 5.
Using Carson’s rule, the transmission bandwidth is
Using the Universal curve, the transmission bandwidth is
In practice, FM radio in the US uses a transmission bandwidth of BT = 200 kHz.
Generation of FM
The frequency of the carrier can be varied by the modulating signal m(t) directly or
indirectly.
,180)/11(2 kHzDfBT
.kHzf.BT 24023
54
Direct generation of FM
If a very high degree of stability of the carrier frequency is not a concern, then we
can generate FM directly using circuits without external crystal oscillators. Examples
of this method are VCO’s, varactor diode modulators, reactance modulators, Crosby
modulators (modulators that use automatic frequency control), etc..
Reactance FM modulator
m(t)
+
-
R2
R1
R3
C1
RFC1
C2
R2 C3
R6
R5RFC2
R7 C4
C7
C6
C5
L1 L2
+
-
+VCC
)(tsFM
55
Indirect generation of FM
Commercial applications of FM (as established by the FCC and other spectrum
governing bodies) require a high degree of stability of the carrier frequency. Such
restrictions can be satisfied by using external crystal oscillators, a narrowband
phase modulator, several stages of frequency multiplication and mixers.
Let us begin with the synthesis of narrow-band FM.
Narrowband frequency modulator
The narrow band FM signal is given by
t
fccNB dmktfAts0
)(22cos)(
56
with kf (and thus fNB) small, since m(t) is in the mV range.
Let us now consider a nonlinear technique to increase the FM signal bandwidth.
Let sNB(t) be input to a nonlinear device with transfer characteristic y(t) = axn(t),
where x(t) is its input, namely,
Nonlinear device.
Let , then at the output of the nonlinear device, we
observe
Let us expand this last equation to infer the effect of this nonlinear device.
t
fci dmktft0
)(22)(
)(cos)( taAty i
nn
c
n
a( )NBs t ( )y t
57
cosni(t) can be expanded as follows:
Likewise,
Thus,
Expanding the last term of the last equation, we get
)(cos)(2cos2
1)(cos
2
1
)(cos)(2cos12
1
)(cos)(cos)(cos
22
2
22
ttt
tt
ttt
i
n
ii
n
i
n
i
i
n
ii
n
)(cos)(2cos2
1)(cos
2
1)(cos 442 tttt i
n
ii
n
i
n
)(cos)(2cos4
1)(cos)(2cos
4
1)(cos
2
1)(cos 4242 tttttt i
n
ii
n
ii
n
i
n
2 4 41cos 2 ( )cos ( ) 1 cos 4 ( ) cos ( ).
2
n n
i i i it t t t
58
Rewriting the equation before the last one, we get
The last term in the expansion of cosni(t) is given by
)(cos)(6cos32
1
)(cos)(2cos32
1)(cos)(4cos
16
1
)(cos)(2cos4
1)(cos
8
1)(cos
2
1
)(cos)(4cos8
1)(cos
8
1
)(cos)(2cos4
1)(cos
2
1)(cos
6
66
442
44
42
tt
tttt
tttt
ttt
tttt
i
n
i
i
n
ii
n
i
i
n
ii
n
i
n
i
n
ii
n
i
n
ii
n
i
n
).(cos)(cos2
11
ttk i
kn
ik
59
Let n be an even number, then, when k = n, the last term is
If, on the other hand, n is an odd number, then when k = n-1, the last term is
Therefore, the last term in the expansion of cosni(t) is always
So, y(t) can be expanded as
)(cos2
11
tn in
)(cos2
1)()2cos(
2
1)(cos)()1cos(
2
1112
tntnttn ininiin
)(cos2
11
tn in
)(cos2
)(2cos)(cos)(1210 tn
Aatctccty in
n
c
ii
60
Example: Consider the cases when n = 2 and n = 3.
Let n = 2, then
or
Let n = 3, then
)(cos)( 22 taAty ic
)(2cos222
)(2cos1)(
22
2 taAaAt
aAty i
cci
c
)(3cos4
)(cos4
3
)(cos2
1)(3cos
2
1
2
1)(cos
2
1
)(cos)(2cos2
1)(2cos
2
1)(
33
3
3
taA
taA
tttaA
tttaAty
i
c
i
c
iiic
iiic
61
Finally,
Let y(t) be input to an ideal bandpass filter with unity gain, bandwidth wide enough
to accommodate spectrum of a wide band signal with center frequency nfc, i.e.,
Ideal bandpass filter
Then,
t
fcn
n
c
t
fc
t
fc
dmkntfnA
a
dmktfcdmktfccty
0
1
0
2
0
10
)(22cos2
)(44cos)(22cos)(
t
fcn
n
cWB dmkntfn
aAts
0
1)(22cos
2)(
( )BP
H f ( )WB
s t( )y t
Ideal filter
62
The instantaneous frequency in Hz of sWB(t) is
Observations about sWB(t) :
1. The carrier frequency is nfc
2. The peak frequency deviation is nfNB
These are the desired properties of the WB FM signal.
The overall frequency multiplier device is shown below:
Complete frequency multiplier
( ) ( )i c ff t nf nk m t
( )BP
H f 1
cos2
n
cin
aAn t
Ideal filter
center cf nf
( )na cosc iA t
63
Example: Noncommercial FM broadcast in the US uses the 88-90 MHz band and
commercial FM broadcast uses the 90-108 MHz band (divided into 200 kHz
channels). In either case f = 75 kHz. Suppose we target a station with fc = 90.1
MHz. Then the indirect FM generation method suggested by Armstrong enables
us to achieve our goals.
Let us start with a 400 kHz crystal oscillator and a narrow band phase modulator
with f = 14.47 Hz.
64
Armstrong indirect method of FM generation
33.808
fc = 1.408
fc = 1.408
65
Demodulation of FM signals
Consider the following receiver architecture
Frequency discriminator implementation of an FM demodulator
The slope circuit is characterized by purely imaginary transfer functions
.2,1),( isH i
66
Let H1(f ) be described by
Graphically,
elsewhere,0
2/2/),2/(2
2/2/),2/(2
)(1 TcTcTc
TcTcTc
BffBfBffaj
BffBfBffaj
fH
2
Tc
Bf
2
Tc
Bf cf
2
Tc
Bf
cf2
Tc
Bf
1( )H f
2 Tj aB
2 Tj aB
f
67
where a > 0 is a constant that determines the slope of H1(s ).
Define G1(f ) H1(f )/j, then is the impulse response of a real
bandpass system described by G1(f ).
In the time domain,
where g1,I(t) and g1,Q(t) are the in-phase and quadrature components of g1(t).
Therefore, the complex envelope of g1(t) is described by
which implies that
Using this information, we get
1 1, 1,( ) ( )cos(2 ) ( )sin(2 )I c Q cg t g t f t g t f t
)()()(~,1,11 tjgtgtg QI
2
1 1( ) ( ) cj f tg t e g t e
)(2
)2sin()(2)2cos()(2
)()()()()(~)(~
1
,1,1
2
,1,1
2
,1,1
2
1
2
1
tg
tftgtftg
etjgtgetjgtgetgetg
cQcI
tfj
QI
tfj
QI
tfjtfj cccc
1
1 1( ) ( )g t G f
68
But,
or
since
which implies that has a lowpass frequency response limited to
and that
Observations: can be obtained by taking the part of G1( f ) that corresponds
to positive frequencies, shifting it to the origin and then scaling it by a factor of 2.
In the next figure is replaced by and replaced by
2 2
1 1 12 ( ) ( ) ( )c cj f t j f tg t F g t e g t e
))((~
)(~
)(2 111 cc ffGffGfG
)(~
1 fG 2/TBf
1 1( ) 2 ( ), 0.cG f f G f f
)(~
1 fG
)(1 fG jfH /)(1 )(~
1 fG
./)(~
1 jfH
* *
1 1( ) ( )g t G f
69
Frequency responses of H1(f ), and H2(f )),(~
1 fH
70
From the previous derivation,
But,
which implies that
If s1(t) is the output of the slope filter H1(f ) when the input is sFM(t) then the complex
envelope of the output is
1
4 ( / 2), / 2( )
0, elsewhere
T Tj a f B f BH f
cos ( )
( )
tf 0 c
c
tj2πk m(τ)dτ j2π f t
FM c c f c
0
j2π f t
FM
s (t)= A 2π f t+2πk m τ dτ = e A e ×e
= e s t ×e .
)(~)(~
2
1)(~
11 tsthts FM
( )t
f 0j2πk m(τ)dτ
FM cs t A e
71
In the frequency domain,
But,
which implies that in the time domain we get
1 1
1( ) ( ) ( )
2
2 ( / 2) ( ) 2 ( ) ( ), / 2
0,
FM
T FM FM T FM T
S f H f S f
j a f B S f a j f S f j aB S f f B
elsewhere
( )2 ( )
dx tj f X f
dt
F
0 0
0
1
2 ( ) 2 ( )
2 ( )
( )( ) ( )
2 ( )
21 ( )
t tf f
tf
FMT FM
j k m d j k m d
c f c T
j k m df
c T
T
ds ts t a ja B s t
dt
j aA k m t e j aA B e
kj aA B m t e
B
72
Therefore,
Observation: s1(t ) contains both AM and FM.
However, if
0
2
1 1
2 2 ( )
0
0
( ) ( )
21 ( )
21 ( ) sin 2 2 ( )
21 ( ) cos 2 2 ( )
2
c
tc f
j f t
j f t j k m df
c T
T
tf
c T c f
T
tf
c T c f
T
s t e s t e
ke j aA B m t e
B
kaA B m t f t k m d
B
kaA B m t f t k m d
B
ttmB
k
T
f ,1)(2
73
Then a distortionless envelope detector can extract m(t) plus a bias, i.e.,
Finally, if
then
Moreover,
The cascade of a slope circuit and an envelope detector is known as a frequency
discriminator.
.)(2
1)(1
tm
B
kBaAts
T
f
Tce
)(~
)(~
12 fHfH
.)(2
1)(2
tm
B
kBaAts
T
f
Tce
1 2ˆ ( ) ( ) ( )
4 ( )
e e
c f
m t s t s t
aA k m t
74
Frequency discriminator
FM demodulation via phase-locked loops
We consider the phase-locked loop (PLL) FM detector shown below
Phase-locked loop FM detector
75
From the previous block diagram, assuming an ideal propagation channel, and
The phase detector is modeled by
then,
kd depends on the multiplier implementation.
))(sin()(
))(cos()(
0 ttAte
ttAtx
cv
ccr
)()(sin)()(2sin2
))(sin())(cos()(1
tttttAAk
kttAttAte
c
vcd
dcvcc
76
with the proper choice of lowpass filter, the output of the phase detector is
A VCO is an FM modulator with peak frequency deviation
where implies that
An equivalent nonlinear model is now shown
Nonlinear model of PLL FM demodulator
)()(sin2
)( ttAAk
te vcd
d
( )max
t
d t
dt
)(ˆ)(
tmkdt
tdvco
.)(ˆ)(
0
t
vco dmkt
77
Assuming the PLL is operating in the near lock condition, i.e., , or that
is small. Then,
and the linear approximation of the PLL is given by
Linear model of the phase-locked loop
Let the loop filter have the transfer function HLF(s) = 1, then
)()( tt
)()( tt
)()()()(sin tttt
ˆ ( ) ( ) ( )2
( ) ( ) ,2
d c v
d c vt t
k A Am t t t
k A Ak t t k
78
Thus, the output of the VCO is given by
Let k0 = ktkvco, then
or
In the s-domain, assuming zero initial conditions,
The closed-loop transfer function from input (t) to output (t) is therefore given by
0 0
ˆ( ) ( ) ( ) ( )
t t
vco t vcot k m d k k d
)()()(
0 ttkdt
td
),()()(
00 tktkdt
td
0 0( ) ( ) ( ),s k s k s
0 0
0 0
( )( ) ( ) ( )
( )cl
k ksH s s s
s s k s k
79
The corresponding impulse response is
, where u(t) is the unit step function.
Let us now find out what happens when the loop gain k0 is increased, i.e.,
Clearly,
or , faster for large .
Example: Let the message signal be a step function, i.e., m(t) = Au(t), then
In this case,
In the s-domain, and
1lim)(lim0
0
00
ks
ksH
kcl
k
)()( ss
)()( tt
t
ccFM dAuktAtx0
)(cos)(
0
( ) ( )
t
t k A u d k At
2)(
s
Aks
)()(
0
2
0
kss
kAks
0 0k
01
0( ) ( ) ( )k t
cl clh t H s k e u t L
80
The Laplace transform of is then given by
Let k1 = kω/kvco, then in the time domain,
Clearly, as t , the estimate
Observation: This result is valid when the initial phase error is small.
Remark: A large loop gain k0 results in practical difficulties, hence, a different loop
filter has to be used.
Consider the loop filter described by
Then the output of the VCO is given by
)(ˆ tm
0 0 0 0
1 1 1 1 1ˆ ( ) ( ) ( ) .( )
tt t
vco
Ak k AkM s k s s Ak k
s s k k s s k k s s k
)()()()()(ˆ 00
1111 tueAktmktueAktuAktmtktk
).()(ˆ 1 tmktm
0,/)( asassH LF
0ˆ ( ) ( )( )
( ) ( ) ( ) ( ) ( )t LF LFvco vco
k H s k H sM ss k k s s s s
s s s
81
Let be small, then the closed-loop transfer function is
Define the phase angle error by then
where
Consider again the step function message m(t) = Au(t). Then
)()( tt
aksks
ask
sHks
sHk
s
ssH
LF
LF
cl
00
2
0
0
0
)(
)(
)(
)()(
),()(ˆ)( sss
)(2
)()()(
)()()(
22
2
00
2
2
0
0 sss
ss
aksks
ss
sHks
sHkss
nnLF
LF
akn 0
02 kn
a
k
ak
kk
n
0
0
00
2
1
22
tAkdAkdAukt
tt
00
)()(
82
In the complex frequency domain,
Let kA, then (s) = /s2
and
If 0 < < 1, then
and
Hence the steady-state phase difference error is zero and (t) (t) as t.
A typical value of is 0.707.
2)(
s
Aks
22 2)(
nnsss
tet n
t
n
n 2
21sin
1)(
.0)(lim
tt
83
Superheterodyne Receiver
Definition: To heterodyne means to combine a radio frequency wave with a locally
generated wave of different frequency, in order to produce a new frequency equal
to the sum or difference of the two.
Specifically, a superheterodyne receiver is one that performs the operations of
carrier frequency tuning of the desired signal, filtering it to separate it from
unwanted signals, in most instances, amplifying it to compensate for signal power
loss due to propagation medium.
Generic superheterodyne receiver
84
Example: Let m(t) modulate a sinusoidal carrier with frequency fc = 10 MHz. Let the
bandwidth of the modulated carrier be BT = 200 kHz and let fIF = 1 MHz. Let the
local oscillator frequency be fLO = 11 MHz and let an interferer have its spectrum
located at 11.95 ≤ f ≤ 12.05 MHz. Then, if no bandpass filter is used in the RF
section, at the output of the RF block and of the mixer we have (for f 0 ).
Spectra when no RF filter is used
output