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Transcript angle modulation

FREQUENCY MODULATION(FM )
CHAPTER 3
FREQUENCY MODULATION
(FM)
Part 1
FM :Introduction


FM is the process of varying the
frequency of a carrier wave in
proportion to a modulating signal.
The amplitude of the carrier wave is
kept constant while its frequency and a
rate of change are varied by the
modulating signal.
FM :Introduction (cont…)

Fig 3.1 : Frequency
Modulated signal
FM :Introduction (cont…)

i.
ii.
iii.
iv.
The important features about FM
waveforms are :
The frequency varies.
The rate of change of carrier frequency changes is
the same as the frequency of the information
signal.
The amount of carrier frequency changes is
proportional to the amplitude of the information
signal.
The amplitude is constant.
FM :Introduction (cont…)




The FM modulator receives two signals,the information signal
from an external source and the carrier signal from a built in
oscillator.
The modulator circuit combines the two signals producing a FM
signal which passed on to the transmission medium.
The demodulator receives the FM signal and separates it,
passing the information signal on and eliminating the carrier
signal.
Federal Communication Coporation (FCC) allocation for a
standard broadcast FM station is as shown in Fig.3.2
Figure 3.2: FM frequency allocation by FCC
Analysis of FM





Mathematical analysis:
Let message signal:
 m t   Vm cos mt
(3.1)
And carrier signal:
 c t   Vc cos[ ct   ]
(3.2)
Where carrier frequency is very much higher
than message frequency.
Analysis of FM (cont’d)

In FM, frequency changes with the change of the
amplitude of the information signal. So the
instantenous frequency of the FM wave is;
(3.3)

K is constant of proportionality

i  c  Kvm t   C  KVm cos mt
Analysis of FM(cont’d)

Thus, we get the FM wave as:
vFM (t )  Vc cos 1  VC cos(C t 
KVm
m
sin mt )
vFM (t )  VC cos(C t  m f sin mt )

3.4
Where modulation index for FM is given by
mf 
KVm
m
Analysis of FM(cont’d)

Frequency deviation: ∆f is the relative
placement of carrier frequency (Hz) w.r.t its
unmodulated value. Given as:
max  C  KVm
min  C  KVm
d  max  C  C  min  KVm
d KVm
f 

2
2
FM(cont’d)

Therefore:
KVm
f 
;
2
f
mf 
fm
f  Vm ;
Example 3.1

FM broadcast station is allowed to have
a frequency deviation of 75 kHz. If a 4
kHz (highest voice frequency) audio
signal causes full deviation (i.e. at
maximun amplitude of information
signal) , calculate the modulation index.
Example 3.2


Determine the peak frequency
deviation,  f , and the modulation
index, mf, for an FM modulator with a
deviation Kf = 10 kHz/V. The
modulating signal to be transmitted is
Vm(t) = 5 cos ( cos 10kπt).
Equations for Phase- and Frequency-Modulated Carriers
Tomasi
Electronic Communications Systems, 5e
Copyright ©2004 by Pearson Education,
Inc.
Upper Saddle River, New Jersey 07458
FM&PM (Bessel function)

Thus, for general equation:
vFM (t )  VC cos(C t  m f cos mt )
vFM (t )  VC sin( C t  m f sin mt )
 VC sin C t cos( m f sin mt )  cos c t sin( m f sin mt )
Bessel function
vt FM  VC [ J 0 (m f ) sin C t  J1 (m f )sin( C  m )t  sin( C  m )t 
 VC [ J 2 (m f ) sin( C  2m )t  sin( C  2m )t
 VC [ J 3 (m f ) sin( C  3m )t  sin( C  3m )t ]  ..
B.F. (cont’d)

It is seen that each pair of side band is preceded by J
coefficients. The order of the coefficient is denoted by
subscript m. The Bessel function can be written as
 mf
J m m f   
 2





n
 1 m f / 22 m f / 24


 ....
 
 n! 1!n  1! 2!n  2!

N=number of the side frequency
M=modulation index
Bessel Functions of the First Kind, Jn(m)
for some value of modulation index
B.F. (cont’d)
Representation of frequency spectrum
Angle Modulation
Part 2
FM Bandwidth
Power distribution of FM
Generation & Detection of FM
Application of FM

FM Bandwidth



Theoretically, the generation and transmission of FM requires
infinite bandwidth. Practically, FM system have finite bandwidth
and they perform well.
The value of modulation index determine the number of sidebands
that have the significant relative amplitudes
If n is the number of sideband pairs, and line of frequency
spectrum are spaced by fm, thus,, the bandwidth is:
B fm  2nf m

For n=>1
FM Bandwidth (cont’d)



Estimation of transmission b/w;
Assume mf is large and n is approximate mf +2; thus
Bfm=2(mf +2)fm
=2(
f
 2) f m
fm
B fm  2(f  2 f m )........(1)
(1) is called Carson’s rule
Deviation Ratio (DR)

The worse case modulation index which produces the widest
output frequency spectrum.
DR 

f (max )
f m (max )
Where
 ∆f(max) = max. peak frequency deviation
 fm(max) = max. modulating signal frequency
Example 3.3




An FM modulator is operating with a peak
frequency deviation  f =20 kHz.The
modulating signal frequency, fm is 10 kHz,
and the 100 kHz carrier signal has an
amplitude of 10 V. Determine :
a) The minimum bandwidth using Bessel
Function table.
b)The minimum bandwidth using Carson’s
Rule.
Sketch the frequency spectrum for (a), with
actual amplitudes.
Example 3.4




For an FM modulator with a modulation index
m=1, a modulating signal Vm(t)=Vm
sin(2π1000t), and an unmodulated carrier
Vc(t) = 10sin(2π500kt), determine :
a) Number of sets of significant side
frequencies
b) Their amplitudes
c) Draw the frequency spectrum showing
their relative amplitudes.
Example 3.4 (solution)

a) From table of Bessel function, a modulation index
of 1 yields a reduced carrier component and three
sets of significant side frequencies.





b) The relative amplitude of the carrier and side
frequencies are
Jo = 0.77 (10) = 7.7 V
J1 = 0.44 (10) = 4.4 V
J2 = 0.11 (10) = 1.1 V
J3 = 0.02 (10) = 0.2 V
Example 3.4 (solution) :
FREQUENCY SPECTRUM
7.7 V
4.4V
4.4V
1.1V
1.1V
0..2


497
J3
0..2
498
J2
499
500
501
502
503
J1
JO
J1
J2
J3
FM Power Distribution


As seen in Bessel function table, it shows that as the
sideband relative amplitude increases, the carrier
amplitude,J0 decreases.
This is because, in FM, the total transmitted power is
always constant and the total average power is equal to
the unmodulated carrier power, that is the amplitude of
the FM remains constant whether or not it is
modulated.
FM Power Distribution
(cont’d)


In effect, in FM, the total power that is originally in the
carrier is redistributed between all components of the
spectrum, in an amount determined by the modulation
index, mf, and the corresponding Bessel functions.
At certain value of modulation index, the carrier
component goes to zero, where in this condition, the
power is carried by the sidebands only.
Average Power
Vc2
Pc 
R

The average power in unmodulated carrier

The total instantaneous power in the angle modulated carrier.
m(t ) 2 Vc2
Pt 

cos 2 [ct   (t )]
R
R
2
Vc2  1 1
 Vc
Pt 
  cos[ 2ct  2 (t )] 
R 2 2
 2R

The total modulated power
Vc2 2(V1 ) 2 2(V2 ) 2
2(Vn ) 2
Pt  P0  P1  P2  ..  Pn 


 .. 
2R
2R
2R
2R
Example 3.5


a) Determine the unmodulated carrier
power for the FM modulator and
condition given in example 3.4,
(assume a load resistance RL = 50 Ώ)
b) Determine the total power in the
angle modulated wave.
Example 3.5 : solution

a) Pc = (10)(10)/(2)(50) =1 W

b)


7.7 2
2(4.4) 2
2(1.1) 2
2(0.2) 2
Pt 



2(50)
2(50)
2(50)
2(50)
Pt  0.5929  0.3872  0.0242  0.0008  1.0051W
Generation of FM

Two major FM generation:
i)
Direct method:
straight forward, requires a VCO whose oscillation frequency
has linear dependence on applied voltage.
Advantage: large frequency deviation
Disadvantage: the carrier frequency tends to drift and must
be stabilized.
example circuit:
i)
ii)
iii)
iv)
i)
ii)
Reactance modulator
Varactor diode
Generation of FM (cont’d)
ii) Indirect method:
Frequency-up conversion.
Two ways:
i.
ii.
a.
b.
iii.
Heterodyne method
Multiplication method
One most popular indirect method is the Armstrong
modulator
Armstrong modulator
Vm(t)
fm
Integrator
Balanced
modulator
Phase
shifter
Vc(t)
fc
Frequency
multiplier
(x n)
Down
converter
Crystal oscillator
Armstrong modulator

For example:
Let fm =15Hz and fc= 200kHz
At frequency deviation= 75kHz,it need a frequency
multiplication by a factor, n,
n=75000/15=5000;
So it need a chain of four triplers (34) and six
ie:n= (34) x (26)=5184,
But,
n x fc=5000 x 200kHz=1000MHz
So, down converter with oscillating
needed to put fc in the
FM band of 88MHz-108MHz
doublers (26),
frequency=900MHz is
FM Detection/Demodulation

FM demodulation



is a process of getting back or regenerate the
original modulating signal from the modulated FM
signal.
It can be achieved by converting the frequency
deviation of FM signal to the variation of
equivalent voltage.
The demodulator will produce an output where its
instantaneous amplitude is proportional to the
instantaneous frequency of the input FM signal.
FM detection (cont’d)


To detect an FM signal, it is necessary to have
a circuit whose output voltage varies linearly
with the frequency of the input signal.
The most commonly used demodulator is the
PLL demodulator. Can be use to detect either
NBFM or WBFM.
PLL Demodulator
V0(t)
FM input
Phase
detector
Low pass
filter
Amplifier
fVc0
VCO
Vc(t)
PLL Demodulator


The phase detector produces an average output voltage that is
linear function of the phase difference between the two input
signals. This low frequency component is selected by LPF.
After amplification, part of the signal is fed back through VCO
where it results in frequency modulation of the VCO frequency.
When the loop is in lock, the VCO frequency follows or tracks
the incoming frequency.
PLL Demodulator


Let instantaneous freq of FM Input,
fi(t)=fc +k1vm(t),
and the VCO output frequency,
f VCO(t)=f0 + k2Vc(t);
f0 is the free running frequency.
For the VCO frequency to track the
instantaneous incoming frequency,
fvco = fi; or
PLL Demodulator

f0 + k2Vc(t)= fc +k1vm(t), so,
Vc (t )  f c  f 0  k1vm (t )

If VCO can be tuned so that fc=f0, then
Vc (t )  k1vm (t )

Where Vc(t) is also taken as the output voltage,
which therefore is the demodulated output
Comparison AM and FM


Its the SNR can be increased without increasing transmitted
power about 25dB higher than in AM
Certain forms of interference at the receiver are more easily to
suppressed, as FM receiver has a limiter which eliminates the
amplitude variations and fluctuations.

The modulation process can take place at a low level power
stage in the transmitter, thus a low modulating power is needed.

Power content is constant and fixed, and there is no waste of
power transmitted

There are guard bands in FM systems allocated by the
standardization body, which can reduce interference between
the adjacent channels.
Application of FM

used by most of the field VHF portable,
mobile and base radios in exploration
use today. It is preferred because of its
immunity to noise or interference and at
the frequencies used the antennas are
of a reasonable size.
Summary of angle modulation
-what you need to be familiar with
Summary (cont’d)
Summary (cont’d)

a)
Bandwidth:
Actual minimum bandwidth from Bessel
table:
B  2(n  f m )
b) Approximate minimum bandwidth using
Carson’s rule:
B  2(f  f m )
Summary (cont’d)

Multitone modulation (equation in general):
i  c  Kvm1  Kvm 2
i  c  2f1 cos 1t  2f 2 cos 2t....
f1
f 2
i  C t 
sin 1t 
sin 2t......
f1
f2
Summary (cont’d)
v fm t   VC sin i
f1
f 2
v fm t   VC sin[ C t 
sin 1t 
sin 2t ]
f1
f2
 VC sin[ C t  m f 1 sin 1t  m f 2 sin 2t ]...........
Summary (cont’d)Comparison NBFM&WBFM
ANGLE MODULATION
Part 3
Advantages
Disadvantages

Advantages





Wideband FM gives significant improvement in the SNR at the output
of the RX which proportional to the square of modulation index.
Angle modulation is resistant to propagation-induced selective fading
since amplitude variations are unimportant and are removed at the
receiver using a limiting circuit.
Angle modulation is very effective in rejecting interference. (minimizes
the effect of noise).
Angle modulation allows the use of more efficient transmitter power in
information.
Angle modulation is capable of handing a greater dynamic range of
modulating signal without distortion than AM.
Disadvantages



Angle modulation requires a transmission
bandwidth much larger than the message
signal bandwidth.
The capture effect where the wanted signal
may be captured by an unwanted signal or
noise voltage.
Angle modulation requires more complex and
inexpensive circuits than AM.
END OF ANGLE
MODULATION