am frequency domain

8/12/2019 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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7

(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;

%[email protected]

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

am frequency domain

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