AM Frequency Domain
Transcript of AM Frequency Domain
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Full AM:Frequency-Domain
Note that although both the carrier and the modulating signal may be
sign waves, the modulated AM waveform is nota sine wave.
v(t) =Ec( 1+ msinmt) sinct [from 2.2]
Expanding it and using a trigonometric identity will prove useful.
Expanding gives:
v(t) =Ecsinct + mEcsin mt sin ct [2.7]
The first term is just the carrier. The second can be expanded using two
trigonometric identities:
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sinAsinB= [cos(A-B) cos(A+B) ]and
cosA = cos (-A)to give
v(t) =Ecsinct + mEc[cos(c-m)t - cos(c+m)t ]
2which can be separated into three distinct terms:
v(t) =Ecsinct + mEccos(c-m)t - mEccos(c+m)t
[2.8]
2 2
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Cont.
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We now have, besides the original carrier, two additional sinusoidal
waves, one above the carrier frequency and one below.
When the complete signal is sketched in the frequency domain, we can
see the carrier and two additional frequencies (one to each side).
Ec
mEc2
fc - fm fc + fmfc
F ig: 7 AM in fr equency domain
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Mathematically, we have:
fusb= fc+ fm [2.9]
flsb= fc
fm [2.10]
Elsb= Eusb= mEc [2.11]
2where
fusb= frequency of the upper sidebandflsb = frequency of the lower sideband
Eusb= peak voltage of the upper-sideband component
Elsb = peak voltage of the lower-sideband component
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(a) A 1MHz carrier with an amplitude of 1V peak is modulated by a 1kHz
signal with m= 0.5. Sketch the voltage spectrum
(b) An additional 2 kHz signal modulates the carrier with m= 0.2. Sketch the
voltage spectrum
Example 2.4
1
0.25
0.999 1.000 1.001
Solution:
1
0.25
0.999 1.000 1.0010.998 1.002
0.1
f(MHz) f(MHz)
Lower
Sideband
Upper
Sideband
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Example 2.5
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Signal bandwidth one of important characteristics of any modulation
scheme.
How much bandwidth needed depends on the baseband frequency range.
In order to reduce interference from distant stations, many AM receivers dohave narrow bandwidth and limited audio frequency response.
An AM signal requires twice the bandwidth of the original signal.
For a video signal with a 4MHz maximum baseband frequency would need
8MHz of bandwidth. Mathematically, the relationship is:
B= 2 Fm [2.12]where
B = bandwidth in hertz
Fm= highest modulating frequency in hertz
Bandwidth
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Crucial SNR at the receiver depends as much on the signal power being
large and the noise power being small.
SinceEcis the peak carrier voltage, the powerPcdeveloped when this
signal appears across a resistanceRis simply;
Power Relationships
R
EP
c
c
2)2/(
WR
Ec
2
2
2
c
usblsb
mEEE
To find power in each sideband, referring to equation [2.11];
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Since the carrier and both sidebands are part of the same signal, the
sidebands will appear across the same resistanceRas the carrier. The two
sidebands will have equal power. Looking at the lower sideband,
R
EP
lsb
lsb
2
2
R
mEc
2
)2/( 2
R
Emc
2x4
22
R
Emc
2x4
22
usbc
PPm
4
2
[2.13]
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Since the two sidebands have equal power, the total sideband power is
given by;
csb Pm
P2
2
[2.14]
The total power,Ptin the whole signal is just the sum of the power in the
carrier and the sidebands, so it is;
2
1
2
2
2
mPP
or
P
m
PP
ct
cct
[2.15]
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Cont. At maximum modulation, the sideband
power is at most 33% of the total
transmitted power.
Percentage modulation (% m)
Percentageoftotalpower(%PT
)
Power in sidebands (PSB
)
Power in carrier (Pc)
100 90 80 70 60 50 40 30 20 10 00
20
40
60
80
100 2
4 c
mP
2
4 c
mP
c
P
2
12
T c
mP P
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An AM broadcast transmitter has a carrier power output of 50kW. What
total power would be proceed with 80% modulation?
Example 2.6
Solution:
21
2m
PPct
kW
kW
66
2
8.0150
2
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A broadcast radio transmitter radiates 10kW when the modulation
percentage is 60. How much of this is carrier power?
Example 2.7
Solution:2/1
2m
PP t
c
kW47.8
2/6.01
102
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Current Calculations
21
2m
P
P
c
t
21
22
2
2m
I
I
RI
RI
P
P
c
t
c
t
c
t
21
2m
I
I
c
t
21
2m
IIct [2.16]
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The antenna current of an AM transmitter is 8 amperes when only the
carrier is sent, but it increases to 8.93amperes when the carrier is
modulated by a single sine wave. Find the percentage modulation.
Determine the antenna current when the percent of modulation changes to0.8.
Example 2.8
Solution:
12
21
22
22
c
t
c
t
I
Im
m
I
I
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12
2
c
t
I
Im
Here;
93.82
2
m
For the second part we have;
21
2mII
ct
AIt
19.9
2
8.018
2
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Measuring Modulation Index inFrequency Domain
c
lsb
c
lsb
clsb
P
Pm
P
Pm
P
m
P
2
4
4
2
2
From eq [2.13];
[2.17]
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Frequency Domain
Unmodulated
frequency
SignalCarrier
Baseband
Modulated
frequency
Signal
Carrier
BasebandBaseband
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Improving on AM
Besides the 67% power loss due to thecarrier, the sidebands contain
redundant information. To maximize the efficiency of AM we
need to
Suppress the carrier Eliminate one of the sidebands
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Cont.
Upper and lowersidebands contain the
sameinformation.
AM modulated speech signal
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Suppressing the carrier
Eliminating the carrier results in adouble-sideband suppressed carrier
(DSSC or DSB) signal shown below.
-2
-1.5
-1
-0.5
0
0.5
1
1.5
Time (sec)
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Time (sec)
Voltage(V)
Full carrier AM signal Suppressed carrier AM signal (DSB)
Note the phase transitions
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Double-sideband suppressed carrier
Despite power savings, DSB AM is not widely usedbecause the signal is difficult to demodulate (recover)at the receiver.
One important application of DSB is the transmissionof color information in a TV signal.
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Why is AM still widely used?
AM is still widely used because it is simple andeffective.
AM broadcast radio
CB radio
TV broadcasting
Air traffic control radios
Garage door opens, keyless remotes