Transcript Wireless Communications and Networks

Frequency modulation and
circuits
Lecture 4
Review of Modulation Circuits



AM Generation
DSBSC Generation
FM Generation
AM Generation

The function of an AM modulator is to
modulate a carrier wave using an
intelligence signal, which results in sum
and difference frequencies, together
with the carrier.
A possible way to achieve this is to use an operational
amplifier to electrically sum the two signals as shown
in the figure below. In such an arrangement the two
signals will remain independent of each other.
The production of a typical AM wave requires the use of a
non-linear device, such as a transistor or diode. A non-linear
device is one that produces an output, which is not proportional
to the input. An example of a circuit, which produces an AM
wave, is shown.
The result of using the diode and resistor is to clip the
negative half of the composite signal. The result is the
output signal shown. It will be noticed that this signal is
only the top half of the AM wave form.
To produce the full AM wave will require the use
of a tank circuit. A diagram of the tank circuit is
shown together with its output
If this circuit is incorporated into the original nonlinear modulator circuit, the full AM wave can be
produced
A diode ha s no gain. To produce gain, a transistor could be
used in place of a diode. The circuit is as shown below.
DSBSC Generation
Circuits used to obtain DSBSC are called balanced
modulators. The figure below shows an example of a
balanced modulator. It uses diodes as the non-linear devices.
Circuits used to obtain DSBSC are called balanced modulators.
The figure below shows an example of a balanced modulator.
It uses diodes as the non-linear devices.
The diodes are paired and are turned on and off during the
positive and negative half of the carrier frequency cycle.
During the positive half, D1 and D4 are on and the other two
are off. The circuit will therefore look as follows
During the negative half, D2 and D3 are on and the other two
are off. The resulting circuit is shown.
The modulating signal undergoes a 180o phase shift.
The current produced
by the carrier signal
is split at the centre
taps of the
transformers and
flow in opposite
directions. This leads
to their cancellation
as they produce
magnetic fields,
which are equal in
magnitude but
opposite in phase.
Frequency Modulation

In this the instantaneous frequency of
the carrier is caused to vary by an
amount proportional to the amplitude of
the modulating signal. The amplitude is
kept constant.



More complex than AM this is because it
involves minute changes in frequency
FM is more immune to effects of noise
FM and PM are similar
Assume we have a carrier at a frequency of 100MHz,
called the resting frequency. If a modulating signal is
then applied, this will cause the carrier to shift
(deviate) from its resting frequency by a certain
amount. If the amplitude is increased then the amount
of deviation also increases. The rate is proportional to
frequency of the intelligence signal. If the signal is
removed then the carrier frequency shifts back to its
resting frequency.
This shift in frequency compared with the amplitude of
the modulating voltage is called the deviation ratio.
The deviation ratio is also called the deviation constant and
it defines how much the carrier frequency will change for a
given input voltage level.
The units are kHz/V
Example
Given that the deviation constant is 1kHz/10mV, what is the
shift in frequency for a voltage level of 50 mV?
Frequency deviation =
50 
1
 5kHz
10
Mathematical representation of FM
The following equation can be used to represent FM
f  f c (1  kVm cos t )
where fc is the unmodulated carrier frequency
k is a proportionality constant
and the last term is the modulating voltage
It will be seen that the maximum deviation will occur
when the cosine term is unity and hence we obtain
f  f c (1  kVm )
maximum deviation
  kVm f c
It can be shown that the instantaneous value of the FM
voltage is given by



v  A sin   ct  sin  mt 
fm


The modulation index for FM is defined as
max frequency deviation 
mf 

modulating frequency
fm
To solve for the frequency components of the FM
requires the use of Bessel functions.
This solution may be shown to be given by
v  A{J o m f sin  ct
 J1 m f [sin  c   m t  sin  c   m t ]
 J 2 m f [sin  c  2 m t  sin  c  2 m t ]
 J 3 m f [sin  c  3 m t  sin  c  3 m t ]
 J 4 m f [sin  c  4 m t  sin  c  4 m t ]...}
To evaluate the individual values of J is quite tedious and
so tables are used.
Observations
•Unlike AM where there are only three frequencies, FM has an infinite
number of sidebands
•The J coefficients decrease with n but not in any simple form and
represent the amplitude of a particular sideband. The modulation index
determines how many sideband components have significant
amplitudes
•The sidebands at equal distances from fc have equal amplitudes
•In AM increase depth of modulation increases sideband power and
hence total transmitted power. In FM total transmitted power remains
constant, increase depth of modulation increases bandwidth
•The theoretical bandwidth required for FM transmission is infinite.
Examples
In an FM system when the audio frequency is 300 Hz and the
audio voltage is 2.0V, the deviation is 5kHz. If the audio
voltage is now increased to 6V what is the new deviation? If
the voltage is now increased to 9V and the frequency
dropped to 100Hz what is the deviation? Find the modulation
index in each case.
  Vm
when V  9v
when V  6v
 5
  2.5kHz / V
  2.5  6  15kHz   2.5  9  22.5kHz
Vm
mf 
2

fm

5
 16.67
0.3
mf 

fm

15
 50
0.3

22.5
mf 

 225
f m 0.1
Find the carrier and modulating frequencies, the
modulating index, and the max. deviation of an FM wave
below. What power will the wave dissipate in a 10 ohm
resistor?
v  12 sin 6  10 t  5 sin 1250 t 
8
Compare this with:



v  A sin   ct  sin  mt 
fm


 m 1250
fm 

 199Hz
2
2
 c 6  108
fc 

 95.5MHz
2
2
Modulating index=5 as given.
2
Power,
12 
 72
Vrms 
2
 
P

 7.2W
R
10
10
What bandwidth is required to transmit an FM signal
with intelligence at 12KHz and max deviation 24 kHz
mf 

fm

24
2
12
Consult Bessel function table to note that for modulating
index of 2, components which exist are J1,J2,J3,J4 apart
from J0.
This means that apart from the carrier you get J1 at +/10kHz, J2 at +/- 20kHz, J3 at +/- 30kHz and J4 at +/40 kHz.
Total bandwidth is therefore 2x40=80kHz.
Carson’s Rule
This is an approximate method used to predict the
required bandwidth necessary for FM transmission
BW  2 max  f s max 
About 98% of the total power is included in the
approximation.
For an FM signal given by
v  60 sin 4  10 t  2 sin 2  10 t 
8
3
If this signal is input into a 30 ohm antenna, find
the carrier frequency
•the transmitted power
•the modulating index
•the intelligence frequency
•the required bandwidth using Carson's rule and tables
•the power in the largest and smallest sidebands
AM Vs FM systems
In both systems a carrier wave is modulated by an audio
signal to produce a carrier and sidebands. The
technique can be applied to various communication
systems eg telephony and telegraphy
Special techniques applied to AM can also be applied to
FM
Both systems use receivers based on the
superheterodyne principle
•In AM, the carrier amplitude is varied whereas in FM the
carrier frequency is varied
•AM produces two sets of sidebands and is said to be a
narrowband system. FM produces a large set of
sidebands and is a broad band system
•FM gives a better signal to noise ratio than AM under
similar operating conditions
•FM systems are more sophisticated and expensive than
AM systems
Transmitters
In an AM transmitter, provision must be made for varying the
carrier amplitude whilst for FM the carrier frequency is
varied.
AM and FM modulators are therefore essentially different in
design. FM can be produced by direct frequency modulation
or by indirectly phase modulation.
The FM carrier must be high usually in the VHF band as it
requires large bandwidth which is not available in the lower
bands.
Receivers
The FM and AM receivers are basically the same, however the FM
receiver uses a limiter and a discriminator to remove AM
variations and to convert frequency changes to amplitude
variations respectively. As a result they (FM) have higher gain than
AM.
FM receivers give high fidelity reproduction due to their large
audio bandwidth up to 15 kHz compared with about 8 kHz for AM
receivers.