MODULATION
1. What is modulation?
1. What is modulation?
• Modulation is the process of putting information
onto a high frequency carrier for transmission
(frequency translation).
• Once this information is received, the low frequency
information must be removed from the high frequency
carrier. This process is known as “ Demodulation”.
2. What are the reasons for
modulation?
2. What are the reasons for
modulation?
1. Frequency division multiplexing (To support multiple
transmissions via a single channel)
To avoid interference
M1(f)
Multiplexed
signal
M(f)
f
0
+
M2(f)
0
f
0
f1
f2
f
2. Practicality of Antennas
Transmitting very low frequencies require antennas with
miles in wavelength
3.What are the Different of
Modulation Methods?
3. What are the Different of
Modulation Methods?
1. Analogue modulation- The modulating signal and
carrier both are analogue signals
Examples: Amplitude Modulation (AM) , Frequency
Modulation (FM) , Phase Modulation (PM)
2. Pulse modulation- The modulating signal is an
analogue signal but Carrier is a train of pulses
Examples : Pulse amplitude modulation (PAM), Pulse
width modulation (PWM), Pulse position modulation
(PPM)
3.What are the Different of
Modulation Methods?
3. Digital to Analogue modulation- The modulating
signal is a digital signal , but the carrier is an
analogue signal.
Examples: Amplitude Shift Keying (ASK), FSK, Phase
Shift Keying (PSK)
4. Digital modulation Examples: Pulse Code Modulation, Delta
Modulation,Adaptive Delta Modulation
ANALOG AND DIGITAL
Analog-to-analog conversion is the representation of
analog information by an analog signal. One may ask
why we need to modulate an analog signal; it is
already analog. Modulation is needed if the medium is
bandpass in nature or if only a bandpass channel is
available to us.
Topics discussed in this section:
Amplitude Modulation
Frequency Modulation
Phase Modulation
Figure Types of analog-to-analog modulation
Figure Amplitude modulation
Note
The total bandwidth required for AM
can be determined
from the bandwidth of the audio
signal: BAM = 2B.
Figure AM band allocation
Note
The total bandwidth required for FM can
be determined from the bandwidth
of the audio signal: BFM = 2(1 + β)B.
Figure Frequency modulation
Figure FM band allocation
Figure Phase modulation
Note
The total bandwidth required for PM can
be determined from the bandwidth
and maximum amplitude of the
modulating signal:
BPM = 2(1 + β)B.
4. What are the Basic Types of
Analogue Modulation Methods ?
4. What are the Basic Types of
Analogue Modulation Methods ?
Consider the carrier signal below:
sc(t ) = Ac(t) cos( 2fc t +  )
1.
Changing of the carrier amplitude Ac(t) produces
Amplitude Modulation signal (AM)
2.
Changing of the carrier frequency fc produces
Frequency Modulation signal (FM)
3.
Changing of the carrier phase  produces
Phase Modulation signal (PM)
Analogue Modulation Methods
5. What are the different Forms
of Amplitude Modulation ?
5. What are the different Forms
of Amplitude Modulation ?
1.
Conventional Amplitude Modulation (DSB-LC)
(Alternatively known as Full AM or Double Sideband
with Large carrier (DSB-LC) modulation
2.
Double Side Band Suppressed Carrier (DSB-SC)
modulation
3.
Single Sideband (SSB) modulation
4.
Vestigial Sideband (VSB) modulation
Conventional Amplitude Modulation
(Full AM)
6. Derive the Frequency Spectrum for Full-AM
Modulation (DSB-LC)
6. Derive the Frequency Spectrum for Full-AM
Modulation (DSB-LC)
1 The carrier signal is
s c ( t )  Ac cos(  c t ) where  c  2 f c
2 In the same way, a modulating signal (information
signal) can also be expressed as
s m ( t )  Am cos  m t
3 The amplitude-modulated wave can be expressed as
s ( t )   Ac  s m ( t )  cos(  c t )
4 By substitution
s ( t )   Ac  Am cos(  m t )  cos(  c t )
5 The modulation index.
m 
Am
Ac
6
Therefore The full AM signal may be
written as
s ( t )  Ac (1  m cos(  m t )) cos(  c t )
cos A cos B  1 / 2[cos( A  B )  cos( A  B )]
s ( t )  Ac (cos  c t ) 
mA c
2
cos(  c   m ) t 
mA c
2
cos(  c   m ) t
7. Draw the Frequency Spectrum of the above AM
signal and calculate the Bandwidth
7. Draw the Frequency Spectrum of the above AM
signal and calculate the Bandwidth
fc-fm
fC
2fm
fc+fm
8. Draw Frequency Spectrum for a complex input
signal with AM
8. Draw Frequency Spectrum for a complex input
signal with AM
fc-fm
fc
fc+fm
Frequency Spectrum of an AM signal
The frequency spectrum of AM waveform contains
three parts:
1. A component at the carrier frequency fc
2. An upper side band (USB), whose highest frequency
component is at fc+fm
3. A lower side band (LSB), whose highest frequency
component is at fc-fm
The bandwidth of the modulated waveform is twice the
information signal bandwidth.
• Because of the two side bands in the frequency spectrum its
often called Double Sideband with Large Carrier.(DSB-
LC)
• The information in the base band (information) signal is
duplicated in the LSB and USB and the carrier conveys no
information.
Example
We have an audio signal with a bandwidth of 5 KHz.
What is the bandwidth needed if we modulate the signal
using AM?
Example
We have an audio signal with a bandwidth of 5 KHz.
What is the bandwidth needed if we modulate the signal
using AM?
Solution
An AM signal requires twice the bandwidth of the
original signal:
BW = 2 x 5 KHz = 10 KHz
AM Radio Band
(m)
9.Modulation
What is theIndex
significance
of modulation index ?
• m is merely defined as a parameter, which determines the
amount of modulation.
• What is the degree of modulation required to establish a
desirable AM communication link?
Answer is to maintain m<1.0 (m<100%).
• This is important for successful retrieval of the original
transmitted information at the receiver end.
(m)
9.Modulation
What is theIndex
significance
of modulation index ?
• If the amplitude of the modulating signal is higher than the
carrier amplitude, which in turn implies the modulation
index
. This will cause severe distortion to the
m  1 . 0 (100 %)
modulated signal.
Power
distribution
in full
AM
10.
Calculate
the power
efficiency
of AM signals
10. Calculate the power efficiency of AM signals
• The ratio of useful power, power efficiency :
sidebands
power
total power
2

m /2
1 m / 2
2

m
2
2m
2
• In terms of power efficiency, for m=1 modulation, only
33% power efficiency is achieved which tells us that only
one-third of the transmitted power carries the useful
information.
Double Side Band Suppressed Carrier
(DSB-SC) Modulation
•
The carrier component in full AM or DSB-LC does not convey any
information. Hence it may be removed or suppressed during the
modulation process to attain higher power efficiency.
•
The trade off of achieving a higher power efficiency using DSB-SC
is at the expense of requiring a complex and expensive receiver due
to the absence of carrier in order to maintain transmitter/receiver
synchronization.
11. Derive the Frequency Spectrum for Double Sideband
Suppressed Carrier Modulation (DSB-SC)
1 Consider the carrier
s c ( t )  Ac cos(  c t ) where  c  2 f c
2 modulated by a single sinusoidal signal
s m ( t )  Am cos  m t
where  m  2 f m
3 The modulated signal is simply the product of these two
s ( t )  A c cos(  c t ) A m cos( 
 A c A m cos(  c t ) cos( 
since
cos A cos B 
1
2

A m Ac
 cos(
m
m
t)
t)
A  B )  cos( A  B ) 
A m Ac
cos(  c   m ) t 
cos(  c   m ) t
2
    

2
    

USB
LSB
s c ( t )  Ac cos  c t
s m ( t )  Am cos  m t
s ( t )  Ac cos(  c t ) Am cos(  m t )
X
Frequency Spectrum of a DSB-SC AM Signal
fc-fm
fc
fc+fm
• All the transmitted power is contained in the two sidebands
(no carrier present).
• The bandwidth is twice the modulating signal bandwidth.
• USB displays the positive components of sm(t) and LSB
displays the negative components of sm(t).
Generation and Detection of DSB-SC
• The simplest method of generating a DSB-SC signal is
merely to filter out the carrier portion of a full AM (or
DSB-LC) waveform.
• Given carrier reference, modulation and demodulation
(detection) can be implemented using product devices or
balanced modulators.
BALANCED MODULATOR
Sm(t)
S1(t)
AM Modulator 1
Accos(ct)
Sm(t)
S(t)
Carrier
DSB-SC
Accos(ct)
AM Modulator 2
-Sm(t)
S2(t)
• The two modulators are identical except for the sign
reversal of the input to one of them. Thus,
s1 ( t )  Ac (1  m cos(  m t )) cos(  c t )
s 2 ( t )  Ac (1  m cos(  m t )) cos(  c t )
s ( t )  s1 ( t )  s 2 ( t )
 2 mA c cos(  m t ) cos(  c t )
COHERENT (SYNCHRONOUS) DETECTOR OR
DSB-SC (PRODUCT DETECTOR)
v(t)
DSB-SC Signal s(t)
X
vo(t)
LPF
Cosct
Local Oscillator
• Since the carrier is suppressed the envelope no longer
represents the modulating signal and hence envelope
detector which is of the non-coherent type cannot be used.
v ( t )  s ( t ) cos(  c t )  2 mA c cos(  m t ) cos(  c t )  cos(  c t )
2
Am
Ac
Ac cos(  m t ) cos ( c t )
2
 1  cos 2 c t 
 2 A m cos(  m t ) 

2


 A m cos(  m t )  A m cos(  m t ) cos( 2 c t )
since s m ( t )  A m cos(  m t )
 s m (t) 
s m (t ) cos ( 2 c t)
   
Unwanted t erm(remove d by LPF)
• It is necessary to have synchronization in both frequency
and phase between the transmitter (modulator) & receiver
(demodulator), when DSB-SC modulation ,which is of the
coherent type, is used.
Both phase and frequency must be known to demodulate
DSB-SC waveforms.
LACK OF PHASE SYNCHRONISATION
Let the received DSB-SC signal be
s DSB  SC ( t )  s m ( t ) cos  c t    Ac
if  is unknown,
v ( t )  s DSB  SC ( t ) cos  c t
 Ac s m ( t ) cos  c t    cos  c t

Ac
2
s m ( t ) cos   cos  2 c t  
Output of LPF
v o (t ) 
Ac
2
s m ( t ) cos 

But we want just
v o (t ) 
Ac
2
s m (t )
Due to lack of phase synchronization, we will see that the
wanted signal at the output of LPF will be attenuated by an
amount of cos.
In other words, phase error causes an attenuation of the
output signal proportional to the cosine of the phase error.
The worst scenario is when =/2, which will give rise to
zero or no output at the output of the LPF.
LACK OF FREQUENCY SYNCHRONISATION
Suppose that the local oscillator is not stable at fc but at
fc+D f, then
v ( t )  s DSB  SC ( t ) cos  c  D  t
 Ac s m ( t ) cos  c t cos  c  D  t

Ac
2
s m ( t ) cos D  t  cos  2 c t  D  
Output of LPF
v o (t ) 
Ac
2
s m ( t ) cos D  t
Thus, the recovered baseband information signal will vary
sinusoidal according to cos D t
This problem can be overcome by adding an extra
synchronization circuitry which is required to detect  and
D t and by providing the carrier signal to the receiver.
A synchronizer is introduced to curb the synchronization
problem exhibited in a coherent system.
Let the baseband signal be
s m ( t )  Am cos  m t
Received DSB-SC signal
s ( t )  Ac s m ( t ) cos  c t
SYNCHRONISER
( )2
PLL
BPF
2
Mathematical analysis of the synchronizer is shown below:
s ( t )  Ac A m cos  m t cos  c t
2

2
2
2
2
Ac A m
4
2

2
Ac A m
4
2
2
2
1  cos
2 m t 1  cos 2 c t 
1  cos
2 m t  cos 2 c t  cos 2 m t cos 2 c t 
2
A A 
1
1

 c m 1  cos 2 m t  cos 2 c t  cos 2  c   m t  cos 2  c   m t 
4 
2
2

Output of BPF
2
2
Ac A m
4
cos 2 c t
Output of frequency divider
k cos  c t
where k is a constant of proportionality.
DISADVANTAGE OF USING COHERENT SYSTEMS
• The frequency and phase of the local oscillator signal must
be very precise which is very difficult to achieve.
It requires additional circuitry such as synchronizer circuit
and hence the cost is higher.
Single-Sideband
Modulation
Single Side Band
Modulation (SSB)
How to generate SSB signal?
• Generate DSB-SC signal
• Band-pass filter to pass only one of the sideband
and suppress the other.
For the generation of an SSB modulated signal
to be possible, the message spectrum must have
an energy gap centered at the origin.
• Example of signal with -300 Hz ~ 300 Hz energy gap
Voice : A band of 300 to 3100 Hz gives goodarticulation
• Also required for SSB modulation is a highly selective filter
• Vestigial
Sideband
Modulation
Vestigial
Side Band
Modulation (VSB)
Instead of transmitting only one sideband as SSB, VSB
modulation transmits a partially suppressed sideband and a
vestige of the other sideband.
Comparison of Amplitude Modulation methods
Comparison of Amplitude Modulation methods
-
Full AM (or DSB-LC)
Sidebands are transmitted in full with the carrier.
Simple to demodulate / detect
Poor power efficiency
Wide bandwidth ( twice the bandwidth of the information
signal)
Used in commercial AM radio broadcasting, one
transmitter and many receivers.
Comparison of Amplitude Modulation methods
DSB-SC
- Less transmitted power than full AM and all the transmitted
power is useful.
- Requires a coherent carrier at the receiver; This results in
increased complexity in the detector(i.e. synchroniser)
- Suited for point to point communication involving one
transmitter and one receiver which would justify the use of
increased receiver complexity.
Comparison of Amplitude Modulation methods
-
-
SSB
Good bandwidth utilization (message signal bandwidth =
modulated signal bandwidth)
Good power efficiency
Demodulation is harder as compares to full AM; Exact
filter design and coherent demodulation are required
Preferred in long distance transmission of voice signals
Comparison of Amplitude Modulation methods
-
•
VSB
Offers a compromise between SSB and DSB-SC
VSB is standard for transmission of TV and similar signals
Bandwidth saving can be significant if modulating signals
are of large bandwidth as in TV and wide band data
signals.
For example with TV the bandwidth of the modulating
signal can extend up to 5.5MHz; with full AM the
bandwidth required is 11MHz