Frequency Modulation

Frequency versus Amplitude Modulation

Frequency Modulation (FM)
 Most popular analog modulation technique
 Amplitude of the carrier signal is kept constant (constant
envelope signal), the frequency of carrier is changed
according to the amplitude of the modulating message signal;
Hence info is carried in the phase or frequency of the carrier.
 Has better noise immunity: In frequency modulation the
amplitude of the carrier signal does not change. There is a
certain advantage of this. Noises due to atmospheric or man
made disturbances usually appear as fluctuations in the
amplitude of the carrier signal. But in FM the receiver used is
insensitive to amplitude variations. So at the receiving end the
received signal is free from the effects of such disturbances,
which makes FM to have better noise immunity.
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 Performs better in multipath environment.
 Small-scale fading cause amplitude fluctuations as we have
seen earlier. But FM system is insensitive to amplitude
variations. Therefore, frequency modulation performs well
in multipath environment. (S.N.: Multipath fading occurs when
signals reach a receiver via many paths & their relative strengths & phases
change.)

 FM system can occupy large bandwidth. Increasing the

bandwidth occupied increases the SNR ratio and hence
improves noise performance.

 In FM the relationship between received signal power and

received signal quality is non-linear. Therefore there is a
rapid increase in quality for an increase in received power.

 Resistant to co-channel interference (capture effect).

 Amplitude Modulation (AM)

 Changes the amplitude of the carrier signal according to

the amplitude of the message signal
 All info is carried in the amplitude of the carrier

 There is a linear relationship between the received signal

quality and received signal power.
 AM systems usually occupy less bandwidth then FM
systems.
 AM carrier signal has time-varying envelope.

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 Angle Modulation x  A sint   
 Angle of the carrier is varied according to the
amplitude of the modulating baseband signal.
 Two classes of angle modulation techniques:
Frequency Modulation
 Instantaneous frequency of the carrier signal is
varied linearly with message signal m(t)

Phase Modulation
 Thephase q(t) of the carrier signal is varied linearly
with the message signal m(t).

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Mathematical Representation of FM (kenedy page 80):

The instantaneous frequency f of the frequency modulated wave is

given by
f  f c (1  kVm cos mt )
Where fc = unmodulated carrier frequency
k= proportionality constant

Vm cos mt = instantaneous modulating voltage

(cosine being preferred for simplicity in calculations)

The maximum deviation for this particular signal will occur
when the cosine term has its maximum value, that is  1 .
Under this condition the instantaneous frequency will be

f  f c (1  kVm )
So that the maximum deviation δ will be given by

δ=kVm fc.

The instantaneous amplitude of the FM signal will be given

by a formula of the form

v  A sin[ F (c , m )]  A sin 

In order to find θ, ω must be integrated with respect to time. And
we know that
  c (1  kVm cos mt )
Thus

    dt   c (1  kVm cos mt )dt c  (1  kVm cos mt )dt

kVm sin mt kVmc sin mt
 c (t  )  c t 
m m
kVm f c sin mt 
 c t   c t  sin mt
fm fm
The instantaneous value of FM voltage becomes:


v  A sin(c t  sin mt )
fm

The modulation index for FM, mf is defined as

δ
m maximum frequency deviation 
f
modulating frequency fm
Modulated voltage
v  A sin(c t  m f sin mt )
 4

FM Example:

- + - -+

-4 0.5 1 1.5 2
Message signal m(t )  4 cos(2t )
FM Signal s (t )  cos2 8t  4 sin( 2t )
Carrier Signal cos(2 8t )

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 FREQUENCY MODULATION

 t

sFM (t )  Ac cos(2f c t   (t )]  Ac cos 2f c t  2k f  m( x)dx 
  
kf is the frequency deviation constant (kHz/V)

If modulation signal is a sinusoid of amplitude Am, frequency fm:

k f Am
sFM (t )  Ac cos(2f c t  sin( 2f mt )]
fm
PHASE MODULATION
sPM (t )  Ac cos2f c t  k m(t )
kq is the phase deviation constant

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FM Index mf 
k f Vm


fm fm
fm: the maximum bandwidth of the modulating signal
δ : peak frequency deviation of the transmitter.
Vm: peak value of the modulating signal

Example: Given m(t) = 4cos(2p4x103t) as the message signal and

a frequency deviation constant gain (kf) of 10kHz/V;
Compute the peak frequency deviation and modulation index!
Answer: fm=4kHz

δ = 10kHz/V * 4V = 40kHz.
mf = 40kHz / 4kHz = 10

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It is important to note that as the modulating frequency
decreases and the modulating voltage amplitude remains
constant, the modulation index increases. This will be the basis
for distinguishing frequency modulation from phase modulation.

Example: In an FM system, where the audio frequency

is 500 Hz and the AF voltage is 2.4 V, the deviation is
4.8kHz. If the AF voltage is now increased to 7.2V,
what is the new deviation? If the AF voltage is raised to
10V while the AF is dropped to 200Hz, what is the
deviation? Find the modulation index in each case.
As we know, δ proportional to Vm

δ/V=4.8/2.4=2kHz/V
Thus when Vm is 7.2V then δ=2x7.2=14.4kHz.

Similarly , when Vm is 10V, then δ=2x10=20kHz

Note that the change in modulating frequency made no
difference to the deviation since it is independent of the
modulating frequency.

Modulation index mf1 = δ1/fm1 =4.8/0.5=9.6

Modulation index mf2 = δ2/fm2 =14.4/0.5=28.8
Modulation index mf3 = δ3/fm3 =20/0.2=100

Thus the modulating frequency change have to taken in

Example: The voltage equation of FM wave is

v  12 sin(6  10 t  5 sin 1250t )

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Find the carrier and modulating frequency, the modulation index

and maximum deviation. What power will this FM wave
dissipate in a 10-Ω resistor?

fc= 95.5 MHz and fm= 199Hz

mf=5 as given
δ=mfm = 5x199=995 Hz
P=Vrms 2 /R = 72/10=7.2 W.
 The solution of FM modulated wave for
frequency is not easy. It is function of sine of sine.
The only solution involves the use of Bessel
functions.

Unlike AM, where there are only three frequencies

(the carrier and the first two side bands). FM has an
infinite number of sidebands as well as the carrier .

Theoretically bandwidth required in FM is

infinite.
 Example: What is the bandwidth required for an FM signal in
which the modulating frequency is 2 kHz and the maximum
deviation is 10 kHz?

Ans: modulation index= δ/fm = 10/2=5.

From the Bessel function table, it is seen that the highest J

coefficient for this value of modulation index is J8 .

The required bandwidth = fm * highest needed sideband * 2

= 2kHz * 8 *2 = 32 kHz.

A rule of thumb (Carson’s rule) states that the bandwidth required to

pass an FM is twice the sum of the deviation and the highest
modulating frequency.
Spectra and Bandwidth of FM Signals
An FM Signal has 98% of the total transmitted power in a RF bandwidth BT

Carson’s Rule

BT  2(m f  1) f m Upper bound

BT  2 Lower bound

Example:
Analog AMPS FM system uses modulation index of mf = 3 and fm = 4kHz.
Using Carson’s Rule: AMPS has 32kHz upper bound and 24kHz lower
bound on required channnel bandwidth.
Example: A 25 MHz carrier is modulated by a 400 Hz audio
sine wave. If the carrier voltage is 4 V and the maximum
deviation is 10kHz, write the equation of this modulated wave
for (a) FM and (b) PM. If the modulating frequency is now
changed to 2 kHz, all else remaining constant , write a new
equation for (c) FM and (d) PM.

ωc =

ωm =

Modulation index =mf=mp=δ/fm=10,000/400=25.