Amplitude Modulation (AM) by Erol Seke For the course Communications - PowerPoint PPT Presentation

Amplitude Modulation (AM) by Erol Seke For the course “ Communications ” OSMANGAZI UNIVERSITY

Initial Problem : Carry voice signal over distances without using cable The Solution : Radiate its electromagnetic wave through air. The Problem : Human ear can only hear sounds in the frequency range of 20Hz - 20kHz approximately † . We would need a veerrrrry looonnnng antenna to efficiently radiate such low frequency signals into air The Solution : Use RF which radiates in air and carry voice with it. The Question : How? The Answer : Modulate RF with voice signal. † : Various sources give different ranges depending on some parameters (age, health etc) but we are not interested in an exact number here anyway.

Summary of Radio Transmission Electromagnetic Wave in media (air) RF signal RF Receiving Antenna air air RF RF Message signal Message signal Converter ? Receiver Transmitter air Converter ? RF signal LF signal LF signal Radio Frequency Radiating Antenna

Modulation property of Fourier Transform ⇔ ω x ( t ) X ( ) ω ⇔ ω − ω + ω + ω 1 1 x ( t ) cos( t ) X ( ) X ( ) o o o 2 2 ω is presumed to be cutoff freq. of x ( t ) 1 ω > ω o 1 ω ω cos( t ) is called the carrier signal or carrier. is called carrier frequency o o

In time domain ω envelope x ( t ) cos( t ) x ( t ) c t t Voice signal Multiplied signal (modulated signal) 180º phase shift at zero crossings ω cos( t ) c Carrier signal t

{ } ω = ω ω >> ω F x ( t ) cos( t ) x ( t ) sin( t ) Example : Find/Draw for where c m c m { } ω = F ω = π δ ω + ω − δ ω − ω X ( ) sin( t ) j ( ( ) ( )) Solution m m m { } 1 1 ω = F ω = ω − ω + ω + ω Y ( ) x ( t ) cos( t ) X ( ) X ( ) c c c 2 2 π π π π j j j j = δ ω − ω + ω − δ ω − ω − ω + δ ω + ω + ω − δ ω + ω − ω ( ) ( ) ( ) ( ) c m c m c m c m 2 2 2 2 I II III IV π ( ) j 1 + = δ ω − ω − ω − δ ω + ω − ω ⇔ − ω − ω I IV ( ( )) ( ( )) sin(( ) t ) c m c m c m 2 2 lower frequency components π ( ) j 1 + = δ ω + ω + ω − δ ω − ω + ω ⇔ ω + ω II III ( ( )) ( ( )) sin(( ) t ) c m c m c m 2 2 upper frequency components

1 1 = ω + ω − ω − ω y ( t ) sin(( ) t ) sin(( ) t ) Entire signal c m c m 2 2 III IV I II Lower Side Band Upper Side Band

= ω + ω + φ Homework : Find the modulated signal m ( t ) and its Fourier spectrum for x ( t ) A cos( t ) A cos( t ) 1 1 2 2 = ω c ( t ) A cos( t ) and c c

Let us apply the same multiplication operation on the modulated signal m ( t ) m ( t ) y ( t ) x ( t ) c ′ ( t ) c ( t ) ′ = = y ( t ) m ( t ) c ( t ) m ( t ) x ( t ) c ( t ) ′ ′ = ω = ω c ( t ) A cos( t ) c ( t ) A cos( t ) and if c c c c ′ = ω 2 y ( t ) x ( t ) A A cos ( t ) then c c c = + cos 2 ( x ) ( 1 cos( 2 x )) / 2 use 1 ( ) ′ = + ω A A x ( t ) x ( t ) cos( 2 t ) c c c 2

Basic AM Modulator, Transmitter and Synchronous Receiver Receiver Transmitter ˆ t x ( ) LPF x ( t ) baseband signal ω ω cos( t ) cos( t ) c c typical Low Pass Filter freq. response ˆ ω X ( ) removed by LPF removed by LPF ω cos( t ) Problem is : how to create at the receiver in phase with the transmitter oscillator c

+ ω ω m c > ( x ( t ) m ) cos( t ) x ( t ) cos( t ) min{ x ( t )} instead of at the transmitter where Let us use c c c + x ( t ) m + ω c ( x ( t ) m ) cos( t ) c c no zero crossing ω cos( t ) c no phase inversion Synchronous demodulation is easier now since we have a carrier signal to extract from input and use

Synchronous demodulation is easier now since we have a carrier signal to extract and use Notch filter to extract carrier LPF

Carrier Upper Side Band = USB Lower Side Band = LSB Double Side Band, Suppressed Carrier = DSB-SC AM Can we have single? m c < min{ x ( t )} Conventional Amplitude Modulation

Another Way to Demodulate Conventional AM Signal Half-Wave rectifier RC-discharge circuit (a LPF) better LPF DC blocking capacitor Note : Better LPF may not be enough. Much higher carrier frequency than illustrated would clearly improve the performance

= + ω + φ y ( t ) A ( 1 a x ( t )) cos( t ) In general m n c normalized signal modulation index x ( t ) = − < < x n ( t ) 1 x n ( t ) 1 so that max x ( t ) or = + ω + φ y ( t ) K ( m x ( t )) cos( t ) c c { } m c > + > min x ( t ) x ( t ) m 0 in order for c { } min x ( t ) = a a larger smaller carrier power per signal power m m m c

2 m ω = = c m cos( t ) P Carrier Power mean square of c c c 2 1 2 t ω = = = x ( t ) cos( t ) ½ mean square of x ( t ) x ( ) P mean square of Sideband Power c s 2 1 2 t = = P P x ( ) Power of a single sideband u L 4 ( ) 1 = + = + 2 2 P P P m x ( t ) Total Power T s c c 2 2 P x ( t ) η = × = × s ( 100 ) ( 100 ) as efficiency define + + P P 2 2 m x ( t ) s c c 2 ( a m m ) = = ω 2 c x ( t ) x ( t ) a m cos( t ) for pure sinusoidal message signal m c m 2 2 a η = × ≤ m ( 100 ) , a 1 and + m 2 2 a m = η = η max = a 1 33 % (best case) at For conventional AM two thirds of power is wasted at m best. That is, if we want to send 1 then we have to spend additional 2. So, why do we use conventional AM instead of other versions of AM (DSB-SC for example)?

Question : Why do we use conventional AM instead of other versions of AM (DSB-SC for example) even though we know the power disadvantage ? Simple Answer : Receiver is easier and cheaper (explain) Notice that information within USB and LSB are identical for real signals LSB USB Then, is it enough to transmit only one side to save power (and have the info transmitted of course)?

Single Side Band Suppressed Carrier AM LSB USB The power advantages are obvious. The question is; how do we generate these SSB-SC AM signals?

= + x ( t ) x ( t ) x ( t ) Let us assume that − + = ∗ so that x ( t ) x ( t ) − + [ ] = − 1 x ( t ) x ( t ) jx ( t ) It can be written that − h 2 [ ] ω = ω ω = + X ( ) X ( ) U ( ) 1 x ( t ) x ( t ) jx ( t ) and + + h 2 ω = ω + ω ω 1 1 X ( ) X ( ) X ( ) sgn( ) + 2 2 ⇔ ω ω ω = − ω ω jx h ( t ) X ( ) sgn( ) X h ( ) jX ( ) sgn( ) or j { } π ⇔ sgn( ω F ∗ = ω Y ω ) x ( t ) y ( t ) X ( ) ( ) We know that (from tables) and (convolution) t ∞ α { } 1 x ( ) - 1 ∫ F = − ω ω = α x h ( t ) jX ( ) sgn( ) d Hilbert Transform π − α t − ∞ = ω − ω y ( t ) A x ( t ) cos( t ) A x ˆ ( t ) sin( t ) c c c c USB π ideal phase shift 2 = ω + ω ˆ y ( t ) A x ( t ) cos( t ) A x ( t ) sin( t ) c c c c LSB

ω >> ω = ω x ( t ) cos( t ) y USB ( t ) where Example : Find for c x x = ω Solution ˆ t x ( ) x ( t ) ˆ x ( t ) sin( t ) is 90 degrees phase shifted version of So x = ω ω − ω ω y ( t ) A cos( t ) cos( t ) A sin( t ) sin( t ) c x c c x c USB = ω + ω = ω − ω y ( t ) A cos(( ) t ) y ( t ) A cos(( ) t ) and c c x c c x USB LSB Homework How can we demodulate SSB-AM signals? y USB t ( ) x ( t ) ? ? ?

Another way to generate SSB Very sharp filter USB We need to have very sharp filters to achieve this.!

VSB : Vestigial Side Band Instead we can allow a little bit of other sideband to pass; which means a relaxed version of the filter (cheaper) BPF x ( t ) VSB DSB ω A cos( t ) c c VSB – AM is used in modulation of television picture signals

Use of Nonlinear Circuits to Realize Multiplication x 1 t ( ) log( x ) exp( x ) x ( t ) x ( t ) × 1 2 log( x ) x 2 t ( ) Generation of SSB ω x ( t ) cos( t ) x ( t ) c ω cos( t ) c o o 90 90 SSB phase delay phase delay ω sin( t ) c ω x h ( t ) x ( t ) sin( t ) h c

= + ω = y ( t ) ( x ( t ) a ) cos( t ) a 0 Example : Draw m c m = = = a 1 a 0 a 2 and for , m m m if the binary message signal is given as shown below. Assume that carrier has high enough frequency that at least 3 cycles fit into a binary period. = a 2 m = a 1 m