CHAPTER 2 AMPLITUDE MODULATION AM 2 2 AM

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

Review: conventional AM(DSBFC) Frequency spectrum: fc-fm fc fc+fm Bandwidth=2 Xfmmax Total Power=Pcarrier +Pusb +Plsb

Two major Disadvantages/Drawbacks of DSBFC n n 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) n Frequency spectrum: fc-fm n n fc fc+fm Bandwidth: 2 x fmmax Total Power= Pusb + Plsb

Single Side Band Full Carrier (SSB-FC) Frequency spectrum: fc-fm fc fc+fm Bandwidth=fmmax Total Power=Pcarrier +Pusb

Single Side band Suppress Carrier (SSB-SC) Frequency spectrum: fc-fm Bandwidth=fmmax Total Power=+Pusb fc fc+fm

Comparison of time domain representation of three common AM transmission systems: Tomasi Electronic Communications Systems, 5 e Copyright © 2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Example 1 For an AM DSCFC wave with a peak unmodulated carrier voltage Vc = 10 Vp, frequency of 100 k. Hz, a load resistor of RL = 10 , frequency of modulating signal of 10 k. Hz 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 n 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 n Solution for DSBFC; i) iii) Bandwidth=2 xfmmax=2(10 k. Hz)=20 k. Hz

Example 1. . cont’d n Solution: For DSB-SC ii) iii)Bandwidth=2 xfmmax=2(10 k. Hz)= 20 k. Hz iv) 90 k. Hz 110 k. Hz

Example 1. . cont’d n Solution: For SSB-FC ii) iii)Bandwidth=fmmax=10 k. Hz iv) fc-fm 100 k. Hz 110 k. Hz

Example 1. . cont’d n Solution: For SSB-s. C ii) iii)Bandwidth=fmmax=10 k. Hz iv) fc-fm fc 110 k. Hz

Methods of Generating SSB 2 methods, i) Filtering method n n n 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 DSB signal Carrier oscillator Balanced modulator Antenna SSB signal Sideband filter Linear amplifier Microphone Audio amplifier Filter response curve Upper Lower sidebands

Phasing methods-using two balance modulator n n n 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’d Am cos wmt Balanced Modulator 1 A 1(t) Ac cos (wct + 90) Information signal Output Signal, aot Phase shifter + Carrier signal Phase shifter Am cos (wmt + 90) Balanced Modulator 2 A 2(t)

Phasing method. . cont’d

VESTIGIAL SIDEBAND (VSB) Ø Ø Ø 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 Ø Ø AM wave is applied to a vestigial sideband filter, producing a modulation scheme – VSB + C Mainly used for television video transmission.

Cont’d…vsb Ø VSB Frequency Spectrum VSB Carrier LSB MSB fc fc

Advantages/Benefits of SSB n n Power consumption Bandwidth conservation Selective fading Noise reduction

Disadvantages of SSB n n Complex receivers Tuning difficulties