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CHAPTER 2

AMPLITUDE MODULATION

(AM)

Part 3 - AM Single Side Band

Communications

OBJECTIVES

 To define and describe AM single sideband  (^) To compare single sideband transmission to conventional double sideband AM  The explain the methods of generating SSB  To define and describe AM vestigial sideband  To describe the advantages and disadvantages of SSB

REVIEW: CONVENTIONAL

AM(DSB-FC)

Frequency spectrum: Bandwidth=2 x fmmax Total Power=Pcarrier +Pusb +Plsb fc fc-fm fc+fm

TWO MAJOR

DISADVANTAGES/DRAWBACKS

OF DSBFC

 (^) Large power consumption, where carrier power constitutes 2/3 or more of the total transmitted power. Carrier does not contain any information, the sidebands contain information.  (^) Large bandwidth utilized. With double- sideband transmission, the information contained in the upper sideband is identical to the information contained in the lower sideband. Therefore, transmitting both sidebands is redundant. Thus, DSBFC is both power and bandwidth inefficient

CONT’D…

 A form of amplitude modulation in which the carrier is transmitted at full power but only one of the sidebands (either the upper or lower) is transmitted  Requires less bandwidth than DSBFC but also produces a demodulated signal with a lower amplitude

SINGLE SIDE BAND

SUPPRESS CARRIER (SSB-

SC)

Frequency spectrum: Bandwidth=fm(max) Total Power=+Pusb fc fc-fm fc+fm

COMPARISON OF TIME DOMAIN

REPRESENTATION OF THREE COMMON AM

TRANSMISSION SYSTEMS:

Tomasi

EXAMPLE 1

For an AM DSCFC 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

 (^) Solution for DSBFC; i) ii) iii) Bandwidth=2xfm(max)=2(10kHz)=20kHz W P m P m Pt Pc c c ( 5 ) 7. 5 4 1 ( 5 ) 4 1 5 4 4 2 2 2 2       

W

m P

P P

W

R

V

R

V

P

c usb lsb c c c

2 2 2 2

EXAMPLE 1..CONT’D

 (^) Solution:For DSB-SC ii) iii)Bandwidth=2xfm(max)=2(10kz)=20k Hz iv)

66. 67 % 100 %
67. 5 5 % 5
68. 5 2. 5      x W W Power W Power W W saved saved W P m P m P t c c ( 5 ) 2. 5 4 1 ( 5 ) 4 1 4 4 2 2 2 2      90kHz 110kHz

EXAMPLE 1..CONT’D

 (^) Solution:For SSB-SC ii) iii)Bandwidth=fm(max)=10kHz iv)

83. 33 % 100 %
84. 5
85. 25 %
86. 25
87. 5 1. 25      x W W Power W Power W W saved saved W P m P t c ( 5 ) 1. 25 4 1 4 2 2    fc fc-fm^ 110kHz

METHODS OF GENERATING

SSB

i) Filtering method



A filter removes the undesired

sideband producing SSB.



Balanced modulators is used to

suppress the unwanted carrier and

filters to suppress the unwanted

sidebands



Quartz crystal filters are the most

widely used sideband filters since

they are very selective and

inexpensive.

CONT’D…

ii) Phasing method 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.

BLOCK DIAGRAM OF PHASING

METHOD

Balanced Modulator 1 Balanced Modulator 2 Phase shifter Phase shifter

Information signal Carrier signal Output Signal, aot Am cos wmt Am cos (wmt + 90) Ac cos (wct + 90) A 2 (t) A 1 (t)