1 CHAPTER 2 AMPLITUDE MODULATION (AM) 2-2 AM Single Side Band Communications.

1

CHAPTER 2

AMPLITUDE MODULATION(AM)

2-2 AM Single Side Band Communications

EKT 231 Communication System 2

Review: conventional AM(DSB-FC)

Frequency spectrum:

Bandwidth=2Xfmmax

Total Power=Pcarrier +Pusb +Plsb

fc

fc+fmfc-fm


Two major Disadvantages/Drawbacks of

DSBFC

Large power consumption, where carrier power constitutes >2/3 transmitted power.{remember:carrier does not contain any information}

Large bandwidth utilized.

Thus, DSBFC is both power and bandwidth inefficient


Double side band suppressed carrier(DSB-SC)

Frequency spectrum:

Bandwidth:2 x fmmax

Total Power= Pusb + Plsb

fc

fc+fmfc-fm


Single Side Band Full Carrier (SSB-FC)

Frequency spectrum:

Bandwidth=fmmax

Total Power=Pcarrier +Pusb

fc

fc+fmfc-fm


Single Side band Suppress Carrier (SSB-SC)

Frequency spectrum:

Bandwidth=fmmax

Total Power=+Pusb

fc

fc+fmfc-fm


Comparison of time domain representation of three common AM transmission systems:

TomasiElectronic Communications Systems, 5e

Copyright ©2004 by Pearson Education, Inc.

Upper Saddle River, New Jersey 07458All rights reserved.


Example 1

For an AM DSBFC wave with a peak unmodulated carrier voltage Vc = 10 Vp,frequency of 100kHz, a load resistor of RL = 10 , frequency of modulating signal of 10kHz and m = 1, determine the following

i) Powers of the carrier and the upper and lower sidebands.

ii) Total power of the modulated wave.

iii) Bandwidth of the transmitted wave.

iv) Draw the power and frequency spectrum.


Example 1..cont’d

For the same given values, determine questions (ii)-(iv) for a AM DSB-SC, AM SSB-FC and AM SSB-SC systems. Determine also the percentage of power saved in each of the system design.


Example 1..cont’d Solution for DSBFC;i)

ii)

iii) Bandwidth=2xfmmax=2(10kHz)=20kHz

W

Pm

Pm

PPccct

5.7)5(4

1)5(

4

15

4422

22

WPm

PP

WR

V

R

VP

c

lsbusb

cc

c

25.14

5102

)10(

2

)2/(

2

222


Example 1..cont’d Solution:For DSB-SC

ii)

iii)Bandwidth=2xfmmax=2(10kHz)=20kHz

iv)

W

Pm

Pm

Pcct

5.2)5(4

1)5(

4

14422

22

fc

110kHz90kHz

%67.66

%1005.7

5%

5

5.25.7

xW

WPower

W

WWPower

saved

saved


Example 1..cont’d Solution:For SSB-FC

ii)

iii)Bandwidth=fmmax=10kHz

iv)

W

Pm

PPcct

25.6)5(4

15

42

2

%67.16

%1005.7

25.1%

25.1

25.65.7

xW

WPower

W

WWPower

saved

saved

100kHz 110kHzfc-fm


Example 1..cont’d Solution:For SSB-SC

ii)

iii)Bandwidth=fmmax=10kHz

iv)

W

Pm

Pct

25.1)5(4

14

2

2

%33.83

%1005.7

25.6%

25.6

25.15.7

xW

WPower

W

WWPower

saved

saved

fc

110kHzfc-fm


Comparison

DSB-FC DSB-SC SSB-FC SSB-SCPower (W)

7.5 2.5 6.25 1.25

BW(kHz) 20 20 10 10

Power Saving(%)

66.67 16.67 83.33

Modulating signal = 10kHzUnmodulated carrier frequency=100kHzUnmodulated carrier voltage = 10Vp


Methods of Generating SSB

4 methods,i) Balanced Ring Modulator

ii) FET Push-Pull Balanced Modulator

iii) Balanced Bridge Modulator

iv) Linear IC Balanced Modulators


(a) schematic diagram; (b) D1 and D2 biased on; (c) D3 and D4 biased on




(i) Balanced ring modulator





Balanced Modulator Waveforms





(ii) FET Push-Pull Balanced Modulator





FET Push-Pull Balanced Modulator Phasor Diagrams





(iii) Balanced Bridge Modulator





(iv) Linear IC Balanced Modulators


SSB Transmitters 2 types,i) Filtering method A filter removes the undesired sideband producing

SSB. Quartz crystal filters are the most widely used

sideband filters since they are very selective and inexpensive.

ii) Phasing method

A balanced modulator eliminates the carrier and provides DSB.


Filtering method


Phasing methods-using two balance modulator Another way to produce SSB uses a

phase shift method to eliminate one sideband.

Two balanced modulators driven by carriers and modulating signals 90º out of phase produce DSB.

Adding the two DSB signals together results in one sideband being cancelled out.


Phasing method..cont’dBalanced

Modulator 1

Balanced Modulator 2

Phase shifter

Phase shifter

+

Information signal

Carrier signalOutput Signal, ao(t)

Am cos wmt

Am cos (wmt + 90)

Ac cos (wct + 90)

a2(t)

a1(t)

Blok digram 1


Phasing method..cont’d

)2()90cos()90cos(2

1

cos*)90cos()(

)1()()()(

00

0

1

210

twtwAAtwtwAA

twAtwAta

tatata

mcmcmcmc

mmcc

)3()90cos()90cos(2

1

)90cos(*)cos()(

00

0

2

twtwAAtwtwAA

twAtwAta

mcmcmcmc

mmcc

)90cos(

)3()2()(0

0

twtwAA

ta

mcmc


Phasing method..cont’d

Blok digram 2


AM VESTIGIAL SIDEBAND (VSB)

Also called asymmetric sideband system. Compromise between DSB & SSB. Easy to generate. Bandwidth is only ~ 25% greater than

SSB signals. Derived by filtering DSB, one pass band is

passed almost completely while just a trace or vestige of the other sideband is included.


Cont’d…vsb

AM wave is applied to a vestigial sideband filter, producing a modulation scheme – VSB + C

Mainly used for television video transmission.


Cont’d…vsb

VSB Frequency Spectrum

fcfc

LSB MSB

CarrierVSB


Advantages/Benefits of SSB

Power consumption Bandwidth conservation Selective fading Noise reduction


Disadvantages of SSB

Complex receivers Tuning difficulties

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1 CHAPTER 2 AMPLITUDE MODULATION (AM) 2-2 AM Single Side Band Communications.

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