Frequency Modulation Term 3

Frequency Modulation

Relationship between FM and PMRelationship between FM and PM

• We can generate FM signal by using PM modulator and vice versa.

• From the above block diagrams, it can be shown that the generation of FM and PM signals are mutually related.

DifferentiatorFM

Modulatorvm(t) vPM(t)

dt

d

PMModulator

vm(t) vFM(t)Integrator

dtGeneration of FM

Generation of PM

)](cos[)( tvktEtv mpccPM

t

mfccFM dttvktEtv0

cos

• Demodulation process is used to get back the information signal.

• For FM demodulator in order to get back information signal from FM signal : PM modulator is used and the signal is pass through differentiator.

• In contrast for PM demodulator : FM demodulator is used and the signal is pass through the integrator.

• This shows the close relationship between FM and PM.

• Hence we can discuss only either one technique in angle modulation.

Differentiator

dt

dPM

Demodulatorvm(t)vFM(t)

FM Demodulator

FMDemodulator

vm(t)vPM(t) Integrator

dt

PM Demodulator

t

mfccFM dttvktEtv0

cos

)](cos[)( tvktEtv mpccPM

FM: Modulation Index

• Any modulation process produces sidebands.

• Side frequencies are the sum and difference of the carrier and modulating frequency.

• The bandwidth of an FM signal is usually much wider than that of an AM signal with the same modulating signal. (infinite number of pairs of upper and lower sidebands generate)

Modulation Index

– The ratio of the frequency deviation to the modulating frequency is known as the modulation index (m).

– In most communication systems using FM, maximum limits are put on both the frequency deviation and the modulating frequency.

)(

)(

Hzf

Hzfm

m

mm f

fm

mcmcmc nffffff ,2,

e.g. for m = 5,16 sidebands (8 pairs).

Determining the Number of Significant Sidebands

Bessel Functions – A complex mathematical process to

determining the number of significant sidebands

– The individual frequency components that make-up the modulated wave are not obvious. However, Bessel Function identities can be applied for this.

Bessel Function• Modulating wave is given by:

• By using Bessel Function, the angle-modulated wave can be written as:

• Jn(m) is the Bessel function of the first kind of nth order with argument m.

cos cos cos2n

n

nm J m n

)]cos(cos[)( tmtVtm mcc

• Thus, m(t) can be rewritten as

• Expanding the equation, becomes

• Where m = modulation index

Vc = peak amplitude of the unmodulated carrier

J0(m)= carrier component

J1(m)=first set of side frequencies displaced from the carrier by ωm

J2(m)=second set of side frequencies displaced from the carrier by 2ωm

Jn(m)=nth set of side frequencies displaced from the carrier by nωm

nmcmc

ntntmJVtm

2cos)()(

)...(.......)cos()(

)2cos()(]2

)cos[()(

]2

)cos[()(cos)()(

2

21

10

mJntmJ

tmJtmJ

tmJtmJVtm

mc

mcmc

mccc

Magnitudes of sidebands includes upper and lower sidebands are defined by J1(m), J2(m), and so on

• J can be defined as:

120) 5 4 3 2 1 Means5! (e.g.

index modulation m

frequency side theofnumber or J n

etc.)(1x2x3x4, factorial! where

...)!1(!3

62/

)!2(!2

42/

)!1(!1

2/1

2)(

2

n

m

n

m

n

m

n

mmJ

Figure 5-8: Carrier and sideband amplitudes for different modulation indexes of FM signals based on the Bessel functions.

Sidebands (Pairs)

Bessel Function PlotBessel Function Plot

Figure 5-9: Plot of the Bessel function data from Fig. 5-8.

Bessel Function PlotBessel Function Plot

Bessel Function PlotBessel Function Plot

Bandwidth

FM Signal Bandwidth– The higher the modulation index in FM, the greater the

number of significant sidebands and the wider the bandwidth of the signal.

– When spectrum conservation is necessary, the bandwidth of an FM signal can be restricted by putting an upper limit on the modulation index.

– E.g: In standard FM broadcasting, the maximum permitted frequency deviation is 75 kHz and the maximum permitted modulating frequency is 15 kHz (modulation index : 5)

Determining the Number of Significant Sidebands

mc mc c

m = 0.25

)( 1rads

BW

mc 4 mc 4c

m = 2

)( 1rads

BW=2nfm=8fm

mc 8 mc 8c

m = 5

)( 1rads

BW=2nfm=16fm

The number of sidebands depend on m value:

BW = 2fmNWhere N is the number of

significant* sidebands

*Significant sidebands are those that have an amplitude of greater than 1% (.01) in the Bessel table.

Bandwidth

Example:If the highest modulating frequency is 3 kHz and

the maximum deviation is 6 kHz, what is the modulation index?

m = 6 kHz/3 kHz = 2

What is the bandwidth?

BW = 2fmN

Where N is the number of significant* sidebands

BW = 2(3 kHz)(4) = 24 kHz

Solution:

Carlson’s RuleCarlson’s Rule

• Take into consideration only the power in the most significant sidebands whose amplitude are greater than 2 percent of the carrier (sidebands whose values are 0.02 or more)

• Therefore the BW needed for FM was :

maxmax2 mffBW

For FM Modulator with frequency deviation of 10 kHz, modulating frequency of 10 kHz, Carrier amplitude voltage of 10V and Carrier frequency of 500 kHz, determine the following:

(a) Minimum Bandwidth using Bessel table

(b) Minimum Bandwidth using Carson’s rule

(c) Amplitudes of the side frequencies and plot the output frequency spectrum

EXAMPLE 2 :

Solution:

a)

From Bessel function table, m=1 yields three sets of significant sidebands. Thus bandwidth is

b) Approx. minimum bandwidth is given by Carson’s rule. So

1kHz10

kHz10

mf

fm

Hz60)103(2 kkHzB

Hz40)1010(2 kkHzkHzB

c)

VJ

VJ

VJ

VJ

2.0)10(02.0

1.1)10(11.0

44)10(44.0

7.7)10(77.0

3

2

1

0

7.7V

44V

1.1V

0.2V

500 510 520 530490480470

0.2V

44V

1.1V

• Frequency spectrum consists of carrier component at fc and also sideband at fc±nfm where n is an integer (n = 1,2,3,…)

• The number of sideband depends on index modulation value, m.

• Magnitude of carrier signal decreases as m increases. • Amplitude of the frequency spectrum depends on

value of Jn(m).

• The bandwidth of modulated signal increases when index modulation, m increases. BW > 2∆fm is expected.

Summary of FM spectrum:

Power in FM signalPower in FM signal• Power signal depends on the amplitudes and not on the

frequencies. • The amplitude of the FM signal is constant and therefore the

power transmitted depends only on the amplitudes of the signal. It does not depends on the modulation index.

• For AM signal the power transmitted depends on the modulation index.

• It can be seen from the Bessel equation:

• In other word the total power of FM signal consists of the power in carrier component and all the power in the sidebands.

1

2

...2

0

3210

nJJ

JJJJJT

n

n

PP

PPPPPP

1

220

223

22

21

20 122...222

nnn JJJJJJJ

• FM equation is given by:

• And therefore the total power transmitted :

1

220

2

2223

222

221

220

2

2)(

2)(

2)(

2)(

2)(

)(

22

2...

2222

2

...2

...2

3210

3210

nn

c

nccccc

rmsJrmsJrmsJrmsJrmsJ

JJJJJFMT

JJR

E

R

JE

R

JE

R

JE

R

JE

R

JE

R

V

R

V

R

V

R

V

R

V

PPPPPP

n

n

])cos())[cos((...

])3cos()3)[cos((

])2cos()2)[cos((

])cos())[cos(()cos()()(

3

2

10

tntnJ

ttJ

ttJ

ttJtJEtv

mcmcn

mcmc

mcmc

mcmcccFM

Power Calculation

• The average power of modulated wave is defined as :

Pc = Vc2 / 2R

Vc=peak unmodulated carrier voltage (V)R = load resistance

The instantaneous power for angle modulation is defined as:Pt = m(t)2 / R W

or Pt = Vc

2 / R [1/2 +½ cos(2ωct + 2(t)]

• Average power of the second term is zero. Thus

Pc = Vc2 / 2R

Modulated carrier power = power of carrier + power of side frequency components

average power of modulated carrier = average power of unmodulated carrier.

Consists of an infinite number of sinusoidal

side frequency components

Power Calculation

• The total power for modulated wave is defined as:

• P0= modulated carrier power

• P1= power in the first set of sidebands

• P2= power in the second set of sidebands

• Pn= power in the nth set of sidebands

R

V

R

V

R

V

R

VP nct 2

)(2...

2

)(2

2

)(2

2

222

21

2

nt PPPP ...21

Narrowband and Wideband FM

From the graph/table of Bessel functions it may be seen that for small , ( 0.3) there is only the carrier and 2 significant sidebands, i.e. BW = 2fm.

FM with 0.3 is referred to as narrowband FM (NBFM) (Note, the bandwidth is the same as DSBAM).

For > 0.3 there are more than 2 significant sidebands. As increases the number of sidebands increases. This is referred to as wideband FM (WBFM).

Narrowband FM NBFM

Wideband FM WBFM

• Large modulation index m

c

f

f means that a large bandwidth is required – called

WBFM.

Narrow Band FM (NBFM)Narrow Band FM (NBFM)

• For FM signal with the small index modulation i.e m < 0.2, is called Narrow Band FM (FM jalur sempit)

• For FM signal that we have studied previously also known as WBFM and the equation is given by :

• Let :

• Hence, the equation yields:

• NBFM with m = small , therefore;

)sin(m)( tt m

)](sinm[sin)sin()]sin(mcos[)cos()( ttEttEtv mccmccFM

)]([sin)sin()](cos[)cos()( ttEttEtv ccccFM

1)sin(m)( tt m

])cos[(2

m])cos[(

2

m)cos(

)sin()sin(m)cos(

)sin()()cos()(

tE

tE

tE

ttEtE

ttEtEtv

mcc

mcc

cc

cmccc

ccccFM

1)](cos[ t )()](sin[ tt and

tmEt

mEtEtam mc

cmc

cccFCDSB cos

2cos

2cos)(

• Therefore :

• Hence NBFM equation yields :

• Compared with amDSB-FC signal:

• It is shown from both equations for NBFM and amDSB-FC consist of one carrier component and two sidebands components. But LSB component for NBFM the phase shift is varies for 90° (quadrature).

1)sin(m)( tt m

Differences between FM and AMDifferences between FM and AM

• Frequency spectrum)(VAmplitud

)( 1radsc mc mc 0

cA

2cmA

2cmA

22mc AmA

Di mana

)(VAmplitud

)( 1radsc mc mc 0

cA

2

cA

2

cA

AM FM

Figure: Carrier and sideband amplitudes for different modulation indexes of FM signals based on the Bessel functions.

Sidebands (Pairs)

Bessel Function PlotBessel Function Plot

Ex. 1 :

A carrier with a peak value of 2000 V is frequency modulated with a message signal of 5 kHz. The modulation index obtained is 2. Calculate the average power in:

(i) Highest sideband (ii) Lowest sideband . Given R = 50 Ω .

Solution : (β =m)

For β = 2 from Bessel table :

The highest sideband is : 58.01 JThe lowest sideband is : 01.05 J

03.02

58.02

4

1

J

J

=>

R

EP C 1

2

58.02

1

(i)

kW 5.1350

1

2

200058.02

50

1

2

200003.02

5

P

W4

(ii)

Ex. 2 :

(a) Determine the BW required to transmit FM signal when the modulating frequency, fm = 10 kHz and maximum frequency deviation is 20 kHz.

From Bessel table the components obtained is J0 , J1 , J2 , J3 , J4 and J5 That means J1 will be at 10 kHz, J2 at 20 kHz, J3 at 30 kHz etc.

Therefore BW = BFM = 2nfm = 2 x 5 x 10 = 100 kHz

210

20

mf

f

Amplitud

fc fc+fm fc+2fm

J0

J1

J5

fc-fm

f (kHz)

m

m

fffBW

212

Carson Rule

Solution :

(b) Repeat (a) with fm = 5 kHz

From Bessel table the highest component is J7

Therefore BW = 2 x 7 x 5 = 70 kHz

45

20

mf

f

m

m

fffBW

212

Carson Rule

Solution :

Equation for NBFM:

t

mfccFM dttvktEtv0

cos

Given, , tfEtv mmm 2sin kHz 10fkV 100cE

kHz 5dan V 1 , MHz 2.106 mmc fEf

(i) frequecy deviation

(ii) BW by Carson rule

(iii)Total Power

(iv)Equation for NBFM

Penyelesaian :

tfdttvkEtfEv

R

EP

ffBWf

f

Ekf

cmfcccNBFM

cFM

mm

mf

2sin22cos (iv)

1R gConsiderin ;kW 512

100

2 (iii)

kHz 3051022 25

10 (ii)

kHz 10110 (i)

22

)sin()(sin)cos()( ttEtEtv cmcccNBFM

Ex. 3 :

A FM signal, 2000 cos (2π x 108 t + 2 Sin π x 104t) is transmitted using an antenna with the resistance of 50 Ω. Determine

(i) Carrier frequency (ii) Modulation index (iii) Information signal

(iv) Power transmitted (v) Bandwidth (vi) Power in highest and lowest sidebandsPenyelesaian :

]sin[cos)( ttEtv mccFM

(i) fc = 108 Hz = 100 MHz

(ii) β = 2

(iii)fm = 104 / 2 = 5 kHz

i) (v) β = 2 => sideband 4

BW = BFM = 2nfm = 2 x 4 x 5 = 40 kHz

Carson - BW = 2(β + 1)fm = 2(2 + 1)5 = 30 kHz

i) (iv) Ec = 2000 V => Ec(rms) = 2000 / 2

PT = V2 (rms)

/ R

= (2000 / 2)2 / 50

= 40 kW

(vi) lowest side band J1 amplitude

for 1st side band = 0.58 x 2000

P = (0.58 x 2000/2)2 / 50 Ω

= 13.27 kW lowest sideband

For double side band

= 2 x 13.27 kW = 26.54 kW

Band limit : fc fm = 100 MHz 5 kHz

Highest sideband J4

P = (0.03 x 2000/2)2 / 50 Ω = 36 W