III. Amplitude Modulation - Facultyfaculty.nps.edu/fargues/teaching/ec2500/EC2500-IIIFY04.pdf ·...
Transcript of III. Amplitude Modulation - Facultyfaculty.nps.edu/fargues/teaching/ec2500/EC2500-IIIFY04.pdf ·...
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III. Amplitude Modulation
AM schemesDouble sideband suppressed carrier (DSB-SC)Double sideband transmitted carrier (DSB-WC)Modulation efficiency & indexSingle sideband (SSB)Vestigial sideband (VSB)Amplitude shift keying (ASK)
ModulatorsGated modulatorSquare law modulatorSSB modulatorVestigial modulator
DemodulatorsCoherent demodulationgated demodulationfrequency mismatch effectsquadrature receiverSSB demodulationVSB demodulationASK demodulation
Incoherent demodulationrectifier detectorenvelope detectorSSB incoherent demodulationASK incoherent demodulationAM Broadcast
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III. Amplitude Modulation
1) Double Sideband Suppressed Carrier
�s(t) sm(t)
cos (2�fct)
S(f)
f– fm fm
Sm(f)
f
Sm(f) =
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• Multiple Signal Transmission
Sm(f)
f
sm(t)
�s1(t)� �
1ms t
�s1(t)� �
2ms t�� �1
cos 2 cf t�
� �2cos 2 cf t�
S2(f)
f
S1(f)
f
• Transmission constraints ?
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• Receiver
Sm(f)
f
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• Example: information signal
�s(t) sm(t)
cos (2�fct)fc = 10 Hz
� �� �sin 2 t
s tt�
�
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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2) Double Sideband Transmitted Carrier
sm(t) =
Sm(f) =
sm(t)�s1(t) �
� �cos 2 cf t�
A
s(t)
t
s(t) + A
t
sm(t)
t
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• Modulation Efficiency
efficiencysignal powertotal power
� �
�
• Modulation Index
� �max s tm
A�
sm(t)
m =
t
� � � �0cos 2?
s t f tA
��
�
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3) Single Sideband
• Recall double sideband AM
Sm(f)
f
–fc fc
Are both sides really needed ?
SUSB(f)
f–fc fc
SLSB(f)
f–fc fc
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4) Vestigial Sideband (VSB)
• SSB needs less frequency bandwidth than DSB
• SSB transmitter and receivers are complicated (expensive)
• VSB is a trade-off between DSB and SSB
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5) Amplitude Shift Keying (ASK)
�s(t) sm(t)
cos (2�fct)
s(t)
t…
3T2TT
sm(t)
t
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• ASK Spectrum Sm(f)
- Typical s(t) not periodic, due to random “0” &“1” bits
1) Periodic s(t) case2) Extension to non periodic s(t) case
1) Periodic s(t): 1 0 1 0 1
0( ) ( )kk
S f a f kf�
�
���
� ��
t
s(t)
Tb
Tb=T/2
Recall Delta’s are located at kf0=k/2Tb=kRb/2
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Recall the Power Spectrum of random NRZ wasdefined as (section II.A):
2/ 2 2
/ 2
1( ) lim ( )T j ft
TTG f s t e dt
T�
��
��� ��
� �� � ��� �
�
Def: The power spectral density (PSD) for a nonperiodic signal s(t) is defined as:
2) Extension to non periodic s(t) case
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• ASK Spectrum for random bits signal
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7) Modulators
a) Gated modulator
S(f)
f–fm fm
A
y1(t) =
Y1(f)=
p(t)
s(t) BPFy1(t) y2(t)
• Introduction
f
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• ImplementationNote:
• we need to implement s(t).p(t) with 0
( )1
p t �� ��
•Process corresponds to a switch operating at a rateof fc times/sec
•Too fast for a mechanical switch, must be electric.
s(t) e2(t)+-~
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s(t) e2(t)+-~
~
D1
D2
D3
D4
C
A
D
B
+-
+
-
Acos(2�fct)
D1 & D2 are a matched pairD3 & D4
1) Acos(2�fct)>0 =>
1) Acos(2�fct)<0 =>
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b) Square law modulator
S(f)
f–fm fm
A
y2(t) =
sc(t)
s(t) � BPFy1(t) ( )2
y3(t)y2(t)+
Y2(f)=
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S(f)
f–fm fm
A
Y2(f)
f
Y3(f)
f
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b) SSB modulator
cos (2�fct)
s(t)sm(t)
H1(f) yUSB(t)
S(f)
f–fc fc
A
Sm(f)
f0
H1(f)
f0
Sm(f)
f0
H2(f)
f0
cos (2�fct)
s(t)
H2(f)
H1(f) sLSB(t)�+
-
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• Potential problem with above SSB modulator
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• Phase shift SSB modulator
S(f)
f–fm fm
A
Y1(f)
f
Hp(f) = –j sgn (f)
Hp(f)
f–j
j
cos (2�fct)
s(t)–90° Hp(f)
�
�
�
�
~
–90°y4(t)
y2(t)
y(t)�
�
y3(t)
y1(t)
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Y3(f)
f
Y4(f)
f
Y2(f)
f
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Y1(f) + Y2(f)
f
f
Y1(f) – Y2(f)
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cos (2�fct)
s(t)–90° Hp(f)
�
�
�
�
~
–90°y4(t)
y2(t)
y(t)�
�
y3(t)
y1(t)
• SSB time domain expression
Hp(f)
f–j
j
• SSB frequency domain expression
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• Vestigial Sideband (VSB) modulator
cos (2�fct)
s(t) � H(f) sm(t)
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8) Demodulators
• Two different types:
– coherent: requires synchronization
– incoherent: simple to implement
a) Coherent demodulation
S(f)
f– fm fm
A
f– fm fc
A/2
Sm(f)
f
Y1(f)
c(t) = cos (2�fct)
sm(t) ? y(t)y2(t)
�
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y2(t) =
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� Gated Demodulator
• Do we have to use a cos (2�fct) to recover s(t) ?
Assume c(t) defined as
tT
c(t)
2TT/2
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� Square Law Demodulator
sm(t) ( . )2 y(t)
� �2( ) ( ) cos(2 )m cy t s t A f t�� �
DBB-WC modulation case
sm(t)=
y(t)=
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� Frequency & Phase Mismatch Effects
� � � �� � � �� �1 cos 2 cos 2c cy t s t f t f f t� � �� � � � �
�
sm(t) LPFy1(t) y(t)�
� �� �cos 2 cf f t� ��� ��
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• Quadrature Receiver
sm(t) ��
��
s1(t)�
� �cos 2 cf t� �� �
LPF ( · )2
s2(t)�
� �cos 2 cf t� �� �
LPF ( · )2
�1 ( )s t
�2 ( )s t
so(t)
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� Single Sideband Demodulation
S(f)
f
SUSB(f)
f– fm fm
cos (2�fct)
sUSB(t) ? y(t)y1(t)
�
f
Y1(f)
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� �1y t �
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� � � � � �V mS f S f H f� �
� Vestigal Sideband Demodulation
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� ASK Demodulation
• We know which signal shape is sent originally
take advantage of it
• Matched filter detector
Recall: binary matched filter detector
tA
Tb
1
sm(t)
11 10 0
…
s1(t)
s0(t)
sm(t)
Tb
compare tothreshold�
� �0
bTdt���
�
�
�
� �0
bTdt��
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• How to compute the threshold Th ?
� � � � � �
� �
� �
� �
� �
0
2 2
0
2
0
2
cos 2 when cos 2
cos 2
1 cos 42
when cos 220 when 0
b
b
b
T
m c m c
T
c
T c
bm c
m
Y As t f t dt s t A f t
A f t dt
f tA dt
A T s t A f tY
s t
� �
�
�
�
� �
�
�� �� � �
� �
�
� � ��
�
�
�
Threshold =
f
tTb
sm(t)
sm(t) Y compare tothreshold Th
� �0
bTdt���
� �cos 2 cA f t�
f
Y
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10/25/03 EC2500.MPF/Fall-FY04 42
• Carrier frequency recovery in AMTC
when fc is transmitted
– narrowband BP filter
– phase lock loop
LPF y(t) =�
BPF� �~ cos 2 cf t�
� � � �cos 2 cs t A f t��� �� �
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b) Incoherent (asynchronous) Demodulation
t
• Square Law Detector
� �
� �
1
2
y t
y t
�
�
• Constraint needed on signal amplitude:
• LPF cutoff frequency:
sm(t) �y1(t)LPF( · )2
y2(t)( · )
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• Rectifier Detector
sm(t)y1(t)
LPFRect
H(f)
f–fm fm
t
sm(t)
0
t
y1(t)
0
� � � �� � � �cos 2m cs t A s t f t�� �
� �1y t �
• Expand in a Fourier series expansion� �cos 2 cf t�
� �cos 2 cf t� �
Fundamental period for � �cos 2 cf t�
� �cos 2 cf t�
t
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t
• Envelope Detector
Need to follow envelope only !
�
e0(t)
BA
ei(t)�
�
�
(1) ei(t) < e0(t) VA < VB
diode shuts offcapacitor discharges in R
(2) ei(t) � e0(t) VA � VB
diode transmitscapacitor loads
sm(t) e0(t)
t
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• Envelope Detector Constraints
Problem when RCtime constraint is toolarge! Demodulatedsignal doesn’t followenvelope.
– filter RC detector time constraint must be small enough to be able to track changes in sm(t) peak values
– filter RC detector time constraint must be large enough to follow the envelope trend only.
High frequencycomponents aregenerated whenRC time constraintis too small(capacitor dischargestoo fast).
What is the maximum charge in prior valuesin sm(t)?
sm(t)
t
sm(t)
t
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• Single Sideband (non-coherent) Demodulator
(1) Add carrier to make demodulation easier.
f
Sm(f)
S(f)
f– fm fm
� � � �
� � � �� �
� �� � � �
cos 2ˆcos 2 sin 2
cos 22
1 ˆcos 2 sin 22 2
LSB c
c cc
c c
s t A f t
s t f t s t f tA f t
s tA f t s t f t
�
� �
�
� �
�
�� �
� �� � �� �� �
�
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(2) Use envelope detector to recover informationsignal.
� �� �
� �� �
2 21 ˆ2 2
when2
s tE A s t
s tA A s t
� � � �� � �� � � �
� �� �
�� �
envelope of
� �� � � �
1 ˆcos 2 sin 22 2c c
s tA f t s t f t� �
� �� �� �
� �
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• ASK Incoherent Demodulator
y(t)sm(t)
Tcompare tothreshold Tk
EnvelopeDetectorBPF
t
y(t)
0
t
A1sm(t) 1 10
…
–A
How to select Th ?
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• Application of modulation property
� � � �
� �
0cos 2p t s t f t
P f
�� �
�
�
• Applications: communication systems; Amplitudemodulation (AM) systems.
• Speech exist in the range 300Hz~5KHz
• Atmosphere attenuates signals rapidly in the range10Hz-->20KHz, and propagates much better at highfrequencies
shift speech to higher frequency range
9) Applications to the AM superheterodyne receiver
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� � � � � �cos 40 cos 400x t t t� �� �
� � � � � �
440 440 360 360
1 cos 440 cos 360214
j t j t j t j t
x t t t
e e e e� � � �
� �
� � �
� �� �� �
� �� � � �� �
60 70 80 1009010 20 30 40 500
60 70 80 1009010 20 30 40 500
x(t)
Time t (msec)
envelope
• Recall what the AM signal looks like
Example:
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Spectrum of x(t) for generic frequencies:
Note:
Change f2 � you change where the frequency’s componentsare for a constant f1
� � � � � �
� � � �
� � � ��
� � � �
1 2 2 1
1 2 1 2
1 2 1 2
1 2 1 2
cos 2 cos 21 cos 2 cos 221 exp 2 exp 24
exp 2 exp 2
y t f t f t f f
f f t f f t
j f f t j f f t
j f f t j f f t
� �
� �
� �
� �
�
� � � �� �� �
� � � � �� � � �� � � �
� � � � �� � � �� � � �
�
�(f2+f1) �(f2�f1) f2�f1 f1+f2
f2�
f2
Y(f)
called carrier frequency
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10/25/03 EC2500.MPF/Fall-FY04 54
• How to recover the original speech signal ?
Demodulate….
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Several basic operations are needed in a broadcast receiver:1. Station separation: must be able to pick out a specific signaland reject others2. Amplification: needed when the signal picked up by the radioantenna is too weak to drive the loudspeakers3. Demodulation: The received signal is centered around thecarrier frequency and must be demodulated before it is fed into thespeakers
In standard AM:- The maximum audio signal frequency is around 5kHz- Each station is assigned 10kHz by the FCC (i.e., eachadjacent carriers are separated by 10kHz)- The AM frequency band assignment is 540kHz -->1600kHz
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• Filter constraints: We need tuneable filters with sharpcutoff frequencies to select the station we want
impossible to realize!
• What is done instead: We build a fixed bandpass filterand shift the input frequencies so that the frequencies ofinterest falls within the fixed passband of the filter
• Such a shifting process is called heterodyning• The receiver doing this operation is called asuperheterodyne receiver
• Basic AM superheterodyne receiver diagram
Radio frequency (RF) amplifier
LO
Intermediate frequency
(IF) amplifierDetector
Baseband Amplifier Speaker
Receiver Components:- RF amplifier: amplifies a portion of the spectrum (tuneable)- Local oscillator (Mixer): shifts the signal to a specificfrequency range- IF amplifier: filters and amplifies around a fixed freqency (forAM systems around 455kHz)- Detector: demodulates (i.e., extracts) the audio signal- Baseband amplifier: amplifies the audio signal
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Example:- Assume no RF amplifier (will be added and discussed later)- Consider the case of a 1kHz AM wave modulated by a carrierat 1MHz (i.e., the station center frequency is at 1MHz)
The generated AM signal has frequencies at:
LO frequency
at:
AMsignal
Mixer frequencies at:
(IF) amplifier
Note: onlycomponents around
455kHz are acceptedby the IF amplifier
What is the IFamplifier requiredbandwidth? _______
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• What happens if we want to accept a station located at1600kHz ? Assume the IF amplifier is centered around455KHz.
LO frequency
at:
AMsignal
Mixer frequencies at:
(IF) amplifier
Note: onlycomponents around
455kHz are acceptedby the IF amplifier
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Summary:- The key is to make the LO track the incomingsignal so hat the difference between the incomingsignal and the LO frequency is a constant frequency(called the IF frequency) equal to 455KHz.
- By convention the LO has to be at a frequency455KHz above the incoming carrier frequency.
- Once we have the IF amplifier output, we candemodulate.
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•The Potential Image Frequency problem: Sometimeswe can get a signal other than that desired at the IFamplifier
• Example: Assume we have a desired signal offrequency 1KHz modulated at 620KHz and anundesired signal of frequency 1 KHz modulated at1530KHz
LO frequency
at:
AMsignal
Mixer frequencies at:
(IF) amplifier
Note: onlycomponents around
455kHz are acceptedby the IF amplifier
AM modulated frequencies located at:
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620K 1530K-620K-1530K
Y(f)
f
2K
Shift to the right
Shift to the left
Desired signal modulated @ 620KUndesired signal modulated @ 1530K
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• Notes: 1. Both desired and undesired modulated signalshave identical components after the mixer. Thesecomponents won’t be separated. 2. Such undesired modulated components are calledimage frequencies (they are the frequencies which appearin the correct range to the IF amplifier, while they areundesirable to start with.
• Question: how to determine the image frequency which willbe a problem to a specific AM signal ?
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• How to use the RF amplifier to remove the imagefrequency problem
LO frequency
at:
AMsignal
Mixer frequencies at:
(IF) amplifier
Note: only componentsaround 455kHz areaccepted by the IF
amplifier
RF
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• Application to Frequency Division Multiplexing
x1(t)
cos�2pfat)
x2(t)
cos�2�fbt)
x3(t)
cos�2�fct)
y(t)
Y(f)
f
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• Demodulation