III. Amplitude Modulation
AM schemes
Double sideband suppressed carrier (DSB-SC)
Double sideband transmitted carrier (DSB-WC)
Modulation efficiency & index
Single sideband (SSB)
Vestigial sideband (VSB)
Amplitude shift keying (ASK)
Modulators
Gated modulator
Square law modulator
SSB modulator
Vestigial modulator
Demodulators
Coherent demodulation
gated demodulation
frequency mismatch effects
quadrature receiver
SSB demodulation
VSB demodulation
ASK demodulation
Incoherent demodulation
rectifier detector
envelope detector
SSB incoherent demodulation
ASK incoherent demodulation
AM Broadcast
10/25/03
EC2500.MPF/Fall-FY04
1
III. Amplitude Modulation
1) Double Sideband Suppressed Carrier
s(t)
Ä
sm(t)
cos (2pfct)
S(f)
– fm
Sm(f) =
fm
f
Sm(f)
f
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EC2500.MPF/Fall-FY04
2
•
Multiple Signal Transmission
s1(t)
Ä
(
cos 2p f c1 t
s1(t)
Ä
(
cos 2p f c2 t
S1(f)
sm1 ( t )
)
sm2 ( t )
Å
f
sm(t)
S2(f)
f
)
Sm(f)
f
•
10/25/03
Transmission constraints ?
EC2500.MPF/Fall-FY04
3
•
Receiver
Sm(f)
f
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•
Example: information signal
s (t ) =
s(t)
sin ( 2p t )
t
Ä
sm(t)
cos (2pfct)
fc = 10 Hz
1) Plot sm(t)
2) Compute and plot Sm(f)
3) Design the receiver needed to recover s(t)
from sm(t)
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6
2) Double Sideband Transmitted Carrier
Ä
s1(t)
Å
sm(t)
A
cos ( 2p f ct )
sm(t) =
Sm(f) =
s(t)
t
s(t) + A
t
sm(t)
t
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•
Modulation Efficiency
efficiency
•
signal power
h=
total power
=
Modulation Index
m=
max s ( t )
A
sm(t)
m=
t
s ( t ) = cos ( 2p f 0t )
A=?
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EC2500.MPF/Fall-FY04
8
3) Single Sideband
•
Recall double sideband AM
Sm(f)
f
–fc
fc
Are both sides really needed ?
SUSB(f)
–fc
fc
f
SLSB(f)
–fc
10/25/03
fc
EC2500.MPF/Fall-FY04
f
9
4) Vestigial Sideband (VSB)
10/25/03
•
•
SSB needs less frequency bandwidth than DSB
•
VSB is a trade-off between DSB and SSB
SSB transmitter and receivers are complicated
(expensive)
EC2500.MPF/Fall-FY04
10
5) Amplitude Shift Keying (ASK)
Ä
s(t)
sm(t)
cos (2pfct)
s(t)
…
T 2T 3T
t
sm(t)
t
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11
•
ASK Spectrum
Sm(f)
- Typical s(t) not periodic, due to random “0” &
“1” bits
1) Periodic s(t) case
2) Extension to non periodic s(t) case
1) Periodic s(t): 1 0 1 0 1
s(t)
Tb=T/2
Tb
S( f ) =
t
¥
å a d ( f - kf
k
0
)
k =-¥
Recall Delta’s are located at kf0=k/2Tb=kRb/2
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2) Extension to non periodic s(t) case
Recall the Power Spectrum of random NRZ was
defined as (section II.A):
Def: The power spectral density (PSD) for a non
periodic signal s(t) is defined as:
æ 1
G ( f ) = lim ç
DT ®¥ DT
è
10/25/03
ò
DT / 2
-DT / 2
EC2500.MPF/Fall-FY04
s (t )e
- j 2p ft
ö
dt ÷
ø
2
13
•
10/25/03
ASK Spectrum for random bits signal
EC2500.MPF/Fall-FY04
14
7) Modulators
a) Gated modulator
• Introduction
y1(t)
s(t)
BPF
y2(t)
p(t)
S(f)
A
–fm
fm
f
y1(t) =
Y1(f)=
f
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• Implementation
Note:
ì0
• we need to implement s(t).p(t) with p(t ) = í
î1
s(t)
~+
-
e2(t)
•Process corresponds to a switch operating at a rate
of fc times/sec
•Too fast for a mechanical switch, must be electric.
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16
C
s(t)
D1 & D2 are a matched pair
D3 & D4
+
D3
D1
~+ A
-
B
D2
D4
D
- ~+
e2(t)
-
Acos(2pfct)
1) Acos(2pfct)>0 =>
1) Acos(2pfct)<0 =>
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b) Square law modulator
+
y1(t)
Å
s(t)
(
)2
y2(t)
BPF
y3(t)
sc(t)
S(f)
A
–fm
fm
f
y2(t) =
Y2(f)=
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S(f)
A
–fm
fm
f
Y2(f)
f
Y3(f)
f
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b) SSB modulator
sm(t)
s(t)
yUSB(t)
H1(f)
cos (2pfct)
S(f)
A
–fc
f
fc
Sm(f)
H1(f)
f
0
s(t)
H1(f)
-
Ä+
sLSB(t)
H2(f)
cos (2pfct)
Sm(f)
0
10/25/03
f
0
H2(f)
f
EC2500.MPF/Fall-FY04
0
f
20
•
10/25/03
Potential problem with above SSB modulator
EC2500.MPF/Fall-FY04
21
•
Phase shift SSB modulator
y1(t)
Ä
~
cos (2pfct)
s(t)
–90°
Ä
–90°
y3(t)
+
Å+
y(t)
-
Hp(f)
y4(t)
y2(t)
S(f)
Hp(f)
A
j
f
–j
–fm
f
fm
Hp(f) = –j sgn (f)
Y1(f)
f
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Y3(f)
f
Y4(f)
f
Y2(f)
f
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23
Y1(f) + Y2(f)
f
Y1(f) – Y2(f)
f
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• SSB time domain expression
y1(t)
Ä
~
cos (2pfct)
s(t)
–90°
–90°
Ä
y3(t)
Hp(f)
y4(t)
+
Å+
y(t)
-
Hp(f)
y2(t)
j
f
–j
• SSB frequency domain expression
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•
Vestigial Sideband (VSB) modulator
s(t)
Å
H(f)
sm(t)
cos (2pfct)
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27
8) Demodulators
•
Two different types:
–
–
coherent: requires synchronization
incoherent: simple to implement
a) Coherent demodulation
– fm
sm(t)
S(f)
Sm(f)
A
A/2
fm
Ä
f
y2(t)
?
f
fc
– fm
y(t)
c(t) = cos (2pfct)
Y1(f)
f
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28
y2(t) =
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29
¬ Gated Demodulator
•
Do we have to use a cos (2pfct) to recover
s(t) ?
Assume c(t) defined as
c(t)
T
2T
t
T/2
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30
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31
¬ Square Law Demodulator
sm(t)
( . )2
y(t)
y (t ) = ( sm (t ) + A cos(2p f c t ) )
2
DBB-WC modulation case
sm(t)=
y(t)=
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32
¬ Frequency & Phase Mismatch Effects
Ä
sm(t)
y1(t)
LPF
y(t)
cos ( 2p ( f c + Df ) t + Dq )
y1 ( t ) = ( s ( t ) cos 2p f c t ) cos ( 2p ( f c + Df ) t + Dq )
=
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33
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34
•
Quadrature Receiver
Ä
sm(t)
sµ1 (t )
LPF
s1(t)
( · )2
cos ( 2p f c t + Dq )
Ä
sµ2 (t )
+
Å
+
LPF
s2(t)
so(t)
Ö
( · )2
cos ( 2p f c t + Dq )
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35
10/25/03
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36
¬ Single Sideband Demodulation
S(f)
f
SUSB(f)
sUSB(t)
f
fm
– fm
Ä
y1(t)
?
y(t)
cos (2pfct)
Y1(f)
f
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37
y1 ( t ) =
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38
¬ Vestigal Sideband Demodulation
SV ( f ) = S m ( f ) H ( f ) =
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39
l ASK Demodulation
sm(t)
Tb
A
…
1
•
0
1
0
1
t
1
We know which signal shape is sent originally
take advantage of it
•
Matched filter detector
Recall: binary matched filter detector
Ä
sm(t)
Tb
ò ( × ) dt
0
+
s1(t)
Tb
Å
-
Ä
compare to
threshold
Tb
ò ( × ) dt
0
s0(t)
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40
sm(t)
t
Tb
sm(t)
Ä
Tb
ò ( × ) dt
Y
compare to
threshold Th
0
A cos ( 2p f c t )
Y
f
•
f
How to compute the threshold Th ?
Tb
Y = ò Asm ( t ) cos ( 2p f c t ) dt
0
when sm ( t ) = A cos 2p f c t
Tb
= ò A2 cos 2 ( 2p f c t ) dt
0
=A
2
ò
Tb
0
é1 + cos ( 4p f c t ) ù
ê
ú dt
2
ë
û
ì A2Tb
ï
Y =í 2
ï 0
î
when sm ( t ) = A cos 2p f c t
when sm ( t ) = 0
Threshold =
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EC2500.MPF/Fall-FY04
41
•
Carrier frequency recovery in AMTC
when fc is transmitted
–
–
narrowband BP filter
phase lock loop
éë s ( t ) + Aùû cos ( 2p f ct )
Ä
BPF
10/25/03
LPF
y(t) =
~ cos ( 2p f ct )
EC2500.MPF/Fall-FY04
42
b) Incoherent (asynchronous) Demodulation
t
•
Square Law Detector
sm(t)
( · )2
y1(t)
LPF
y2(t)
Ö(· )
y1 ( t ) =
y2 ( t ) =
10/25/03
•
LPF cutoff frequency:
•
Constraint needed on signal amplitude:
EC2500.MPF/Fall-FY04
43
•
Rectifier Detector
sm(t)
Rect
y1(t)
H(f)
LPF
–fm
fm
f
sm(t)
sm ( t ) = ( A + s ( t ) ) cos ( 2p f c t )
t
0
y1(t)
y1 ( t ) =
t
0
•
Expand cos ( 2p f c t ) in a Fourier series expansion
Fundamental period for cos ( 2p f c t )
cos ( 2p f ct )
t
cos ( 2p f c t ) =
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EC2500.MPF/Fall-FY04
44
•
Envelope Detector
t
Need to follow envelope only !
+
ei(t)
-
A
B
+
e0(t)
-
(1) ei(t) < e0(t)
VA < VB
(2) ei(t) ³ e0(t)
VA ³ VB
diode shuts off
capacitor discharges in R
diode transmits
capacitor loads
sm(t)
e0(t)
t
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45
•
Envelope Detector Constraints
–
filter RC detector time constraint must be
small enough to be able to track changes in
sm(t) peak values
sm(t)
Problem when RC
time constraint is too
large! Demodulated
signal doesn’t follow
envelope.
–
t
filter RC detector time constraint must be large
enough to follow the envelope trend only.
sm(t)
High frequency
components are
generated when
RC time constraint
is too small
(capacitor discharges
too fast).
t
What is the maximum charge in prior values
in sm(t)?
10/25/03
EC2500.MPF/Fall-FY04
46
•
Single Sideband (non-coherent) Demodulator
(1) Add carrier to make demodulation easier.
S(f)
fm
– fm
f
Sm(f)
f
®
10/25/03
sLSB ( t ) + A cos ( 2p f c t )
s ( t ) cos 2p f c t + sˆ ( t ) sin 2p f c t
=
+ A cos ( 2p f c t )
2
é s (t )
ù
1
=ê
+ Aú cos 2p f c t + sˆ ( t ) sin ( 2p f c t )
2
ë 2
û
EC2500.MPF/Fall-FY04
47
(2) Use envelope detector to recover information
signal.
envelope of
é s (t )
ù
1
+
A
ê
ú cos 2p f c t + sˆ ( t ) sin ( 2p f c t )
2
ë 2
û
2
æ s (t )
ö æ1
ö
+ A ÷ + ç sˆ ( t ) ÷
E= ç
ø
è 2
ø è2
;
10/25/03
s (t )
+A
2
2
when A ? s ( t )
EC2500.MPF/Fall-FY04
48
•
ASK Incoherent Demodulator
sm(t)
sm(t)
A
Envelope
Detector
BPF
1
0
1
y(t)
T
1
compare to
threshold Tk
How to select Th ?
…
t
–A
y(t)
t
0
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49
9) Applications to the AM superheterodyne receiver
•
Application of modulation property
p ( t ) = s ( t ) × cos 2p f 0 t
[
P( f ) =
•
Applications: communication systems; Amplitude
modulation (AM) systems.
•
•
Speech exist in the range 300Hz~5KHz
Atmosphere attenuates signals rapidly in the range
10Hz-->20KHz, and propagates much better at high
frequencies
shift speech to higher frequency range
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50
• Recall what the AM signal looks like
Example:
0
0
10
x(t)
10
x ( t ) = cos ( 40p t ) × cos ( 400p t )
20
20
30
30
40
50
60
40 50 60
Time t (msec)
70
70
80
80
90
90
100
envelope
100
1
écos ( 440p t ) + cos ( 360p t ) ùû
2ë
1
= éëe j 440p t + e- j 440p t + e - j 360p t + e - j 360p t ùû
4
x (t ) =
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52
Spectrum of x(t) for generic frequencies:
y ( t ) = cos ( 2p f1t ) cos ( 2p f 2 t )
f 2 ? f1
1
écos 2p ( f1 + f 2 ) t + cos 2p ( f1 - f 2 ) t ùû
2ë
1
= exp éë j 2p ( f1 + f 2 ) t ùû + exp éë - j 2p ( f1 + f 2 ) t ùû
4
+ exp éë j 2p ( f1 - f 2 ) t ùû + exp éë - j 2p ( f1 - f 2 ) t ùû
=
{
Y(f)
f2
f2
-(f2+f1)
-(f2-f1)
f2-f1
f1+f2
f
called carrier frequency
Note:
Change f2 ® you change where the frequency’s components
are for a constant f1
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•
How to recover the original speech signal ?
Demodulate….
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54
Several basic operations are needed in a broadcast receiver:
1. Station separation: must be able to pick out a specific signal
and reject others
2. Amplification: needed when the signal picked up by the radio
antenna is too weak to drive the loudspeakers
3. Demodulation: The received signal is centered around the
carrier frequency and must be demodulated before it is fed into the
speakers
In standard AM:
- The maximum audio signal frequency is around 5kHz
- Each station is assigned 10kHz by the FCC (i.e., each
adjacent carriers are separated by 10kHz)
- The AM frequency band assignment is 540kHz -->
1600kHz
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55
• Filter constraints: We need tuneable filters with sharp
cutoff frequencies to select the station we want
impossible to realize!
• What is done instead: We build a fixed bandpass filter
and shift the input frequencies so that the frequencies of
interest falls within the fixed passband of the filter
• Such a shifting process is called heterodyning
• The receiver doing this operation is called a
superheterodyne receiver
• Basic AM superheterodyne receiver diagram
Intermediate
frequency
(IF) amplifier
Radio frequency
(RF) amplifier
Detector
Baseband
Amplifier
Speaker
LO
Receiver Components:
- RF amplifier: amplifies a portion of the spectrum (tuneable)
- Local oscillator (Mixer): shifts the signal to a specific
frequency range
- IF amplifier: filters and amplifies around a fixed freqency (for
AM systems around 455kHz)
- Detector: demodulates (i.e., extracts) the audio signal
- Baseband amplifier: amplifies the audio signal
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56
Example:
- Assume no RF amplifier (will be added and discussed later)
- Consider the case of a 1kHz AM wave modulated by a carrier
at 1MHz (i.e., the station center frequency is at 1MHz)
The generated AM signal has frequencies at:
Mixer frequencies at:
(IF) amplifier
AM
signal
LO
frequency
at:
Note: only
components around
455kHz are accepted
by the IF amplifier
What is the IF
amplifier required
bandwidth? _______
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57
• What happens if we want to accept a station located at
1600kHz ? Assume the IF amplifier is centered around
455KHz.
Mixer frequencies at:
(IF) amplifier
AM
signal
Note: only
components around
455kHz are accepted
by the IF amplifier
LO
frequency
at:
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58
Summary:
- The key is to make the LO track the incoming
signal so hat the difference between the incoming
signal and the LO frequency is a constant frequency
(called the IF frequency) equal to 455KHz.
- By convention the LO has to be at a frequency
455KHz above the incoming carrier frequency.
- Once we have the IF amplifier output, we can
demodulate.
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59
•The Potential Image Frequency problem: Sometimes
we can get a signal other than that desired at the IF
amplifier
• Example: Assume we have a desired signal of
frequency 1KHz modulated at 620KHz and an
undesired signal of frequency 1 KHz modulated at
1530KHz
AM modulated frequencies located at:
Mixer frequencies at:
(IF) amplifier
AM
signal
Note: only
components around
455kHz are accepted
by the IF amplifier
LO
frequency
at:
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Desired signal modulated @ 620K
Undesired signal modulated @ 1530K
Y(f)
2K
f
-1530K
-620K
620K
1530K
Shift to the right
Shift to the left
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• Notes: 1. Both desired and undesired modulated signals
have identical components after the mixer. These
components won’t be separated.
2. Such undesired modulated components are called
image frequencies (they are the frequencies which appear
in the correct range to the IF amplifier, while they are
undesirable to start with.
• Question: how to determine the image frequency which will
be a problem to a specific AM signal ?
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10/25/03
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63
• How to use the RF amplifier to remove the image
frequency problem
Mixer frequencies at:
(IF) amplifier
RF
AM
signal
Note: only components
around 455kHz are
accepted by the IF
amplifier
LO
frequency
at:
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10/25/03
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65
• Application to Frequency Division Multiplexing
cos(2pfat)
x1(t)
cos(2pfbt)
y(t)
x2(t)
cos(2pfct)
x3(t)
Y(f)
f
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• Demodulation
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