Amplitude Modulation (AM)
by Erol Seke
For the course “Communications”
ESKİŞEHİR OSMANGAZİ UNIVERSITY
Initial Problem : Carry voice signal over distances without using cable
The Solution : Radiate its electromagnetic wave through air
and pick this up at the receiver locations
Resulting Problems : 1. There can only be one EM-wave in the air since the receiver picks up all.
2. Transmitter antenna must be very long.
Hints : 1. Human ear can only hear 20Hz-20kHz†.
2. EM-spectrum is very large compared to this.
Second Approach : Load voice onto HF EM-wave.
So that : HF EM needs shorter antenna (~inverse of wavelength).
Each voice channel use different EM-band.
The Question : How?
The Answer : Modulate EM-wave with voice signal(s).
That is : Change one or more properties of EM-wave by the voice/message signal
𝜓(𝑟, 𝑡) = 𝐴𝑐𝑜𝑠(𝑘𝑟 − 𝜔𝑡 + 𝜑)
Polarity Amplitude Frequency Phase
† : 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 Low bandwidth signal
Low bandwidth signal
Radio Frequency Radiating Antenna
Modulation property of Fourier Transform
x(t ) X ( )
1 is presumed to be cutoff freq. of x(t )
x(t ) cos(ot ) 12 X ( o ) 12 X ( o )
o 1
cos(o t ) is called the carrier signal or carrier. o is called carrier frequency
In time domain
x(t ) cos(c t )
envelope
x(t )
t t
Voice signal Multiplied signal
(modulated signal)
180º phase shift at zero crossings
cos(c t )
Carrier signal
t
Example : Find/Draw F x(t ) cos(ct ) for x (t ) sin( m t ) where c m
Solution X ( ) F sin(mt ) j ( ( m ) ( m ))
Y ( ) F x(t ) cos(ct )
1 1
X ( c ) X ( c )
2 2
j j j j
( c m ) ( c m ) ( c m ) ( c m )
2 2 2 2
I II III IV
j
I IV ( (c m )) ( (c m )) 1 sin((c m )t )
2 2
lower frequency components
j
II III ( (c m )) ( (c m )) 1 sin((c m )t )
2 2
upper frequency components
1 1
Entire signal y (t ) sin((c m )t ) sin((c m )t )
2 2
−𝜔𝑐 𝜔𝑐
III IV I II
Lower Side Band
Upper Side Band
Example : Find the modulated signal m(t) and its Fourier spectrum for x(t ) A1 cos(1t ) A2 cos(2t )
and c(t ) Ac cos(c t )
Let us apply the same multiplication operation on the modulated signal m(t)
m(t ) y (t )
x(t )
c (t ) c(t )
m(t ) x(t )c(t ) y(t ) m(t )c(t )
if c(t ) Ac cos(c t ) and c(t ) Ac cos(c t )
then y(t ) x(t ) Ac Ac cos2 (ct )
use cos 2 ( x) (1 cos(2 x)) / 2
Ac Ac x(t ) x(t ) cos(2c t )
1
2
Basic AM Modulator, Transmitter and Synchronous Receiver
Transmitter Receiver
x(t ) LPF xˆ (t )
baseband signal
cos(c t ) cos(c t )
This is called synchronous demodulation
typical Low Pass Filter freq. response
Xˆ ()
removed by LPF removed by LPF
Multiple Transmitters and Receivers
T1
R1
𝑥1 (𝑡)
air LPF 𝑥1 (𝑡)
cos(𝜔1 𝑡)
air cos(𝜔1 𝑡)
air
T2 air
𝑥2 (𝑡) …
air
cos(𝜔2 𝑡) air Rk
air LPF 𝑥𝑘 (𝑡)
air
… cos(𝜔𝑘 𝑡)
space
TN
𝑥𝑁 (𝑡) Sum of everything is in the air
cos(𝜔𝑁 𝑡)
𝑋(𝜔)
…
𝜔
𝜔2 𝑡 𝜔𝑖 𝑡 𝜔1 𝑡 𝜔𝑗 𝑡
Each receiver can adjust its own center frequency to pick up anyone of the signals
Problem is : how to create cos(c t ) at the receiver in phase with the transmitter oscillator
Let us use ( x(t ) mc ) cos(c t ) instead of x(t ) cos(c t ) at the transmitter where mc min{x(t )}
x(t ) mc
( x(t ) mc ) cos(c t )
no zero crossing
cos(c t )
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
Result of added DC
Notch filter to extract carrier
LPF
Carrier
Lower Side Band = LSB Upper Side Band = USB
Double Side Band, Suppressed Carrier = DSB-SC AM
Can we have single?
mc 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
In general y (t ) A(1 am xn (t )) cos(c t )
modulation index normalized signal
x(t )
xn (t ) so that 1 xn (t ) 1
max x(t )
or
y (t ) K (mc x(t )) cos(c t )
in order for x(t ) mc 0 mc minx(t )
minx(t )
am larger am smaller carrier power per signal power
mc
mc2
Carrier Power Pc mean square of mc cos(c t )
2
1 2
Sideband Power Ps mean square of x (t ) cos(c t ) ½ mean square of x(t ) x (t )
2
1 2
Power of a single sideband Pu PL x (t )
4
Total Power PT Ps Pc
1 2
2
mc x 2 (t )
Ps x 2 (t )
define (100) (100) as efficiency
Ps Pc mc x (t )
2 2
( am mc ) 2
for pure sinusoidal message signal x(t ) am mc cos( m t ) x (t )
2
2
am2
and (100) , am 1
2 am2
at am 1 (best case) m ax 33 % For conventional AM two thirds of power is wasted at
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 simpler 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?
Let us assume that x(t ) x (t ) x (t )
so that x (t ) x (t )
It can be written that x (t ) 1
2 x(t ) jx h (t )
and x (t ) 1
x(t ) jx h (t ) X ( ) X ( )U ( )
2
X ( ) 12 X ( ) 12 X ( ) sgn( )
jx h (t ) X ( ) sgn( ) or X h ( ) jX ( ) sgn( )
j
We know that (from tables) sgn( ) and F x(t ) y(t ) X ( )Y ( ) (convolution)
t
x( )
xh (t ) F -1 jX ( ) sgn( )
1
Hilbert Transform
t
d
yUSB (t ) Ac x(t ) cos(c t ) Ac xˆ (t ) sin(c t )
ideal 2 phase shift
yLSB (t ) Ac x(t ) cos(c t ) Ac xˆ (t ) sin(c t )
Example : Find yUS B (t ) for x (t ) cos( x t ) where c x
Solution xˆ (t ) is 90 degrees phase shifted version of x(t ) So xˆ (t ) sin( x t )
yUSB (t ) Ac cos( x t ) cos(c t ) Ac sin( x t ) sin(c t )
y USB (t ) Ac cos((c x )t ) and yLSB (t ) Ac cos((c x )t )
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
Ac cos(c t )
VSB – AM is used in modulation of analog monochrome television picture signals
modulation of analog color television picture/color signals is another story
Use of Nonlinear Circuits to Realize Multiplication
(Since Multiplication cannot be realized by Linear Circuits)
x1 (t ) log(x)
exp(x) x1 (t ) x2 (t )
x 2 (t ) log(x)
Generation of SSB
x(t ) cos(c t )
x(t )
cos(c t )
90 90 SSB
phase delay phase delay
sin(c t )
xh (t ) xh (t ) sin(c t )
Example : Draw y (t ) ( x(t ) am ) cos(c t ) am 0
for am 0 , am 2 and am 1
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.
am 2
am 1
In general
y (t ) A cos(c t )
Vary this with the message, you get Amplitude Vary this with the message signal, you get
Modulation (AM) Phase Modulation (PM)
If there are finite number of amplitude values, If there are finite number of phase values, it
it is called Amplitude Shift Keying (ASK) is called Phase Shift Keying (PSK)
Vary this with the message signal, you get
If both amplitude and phase modulation are Frequency Modulation (FM)
used at the same time, it is called
Quadrature Amplitude Modulation (QAM) If there are finite number of frequency
Digital version is also called QAM. values, it is called Frequency Shift Keying
(FSK)
In AM, amount of carrier and sidebands in the frequency spectrum determines the
modulation type : SSB, SSB-SC, DSB, DSB-SC, Conventional AM, VSB and their
sub-types.
END
