Chapter Three(FM)Update

8/3/2019 Chapter Three(FM)Update

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CHAPTER 3

Frequency Modulation (fm)



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• Frequency is the measurement of the number oftimes that a repeated event occurs per unit oftime.

• It is also defined as the rate of change of phase ofa sinusoidal waveform.



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Phase

• The instantaneous phase at time t is , whichhas units of radians. t has units of seconds, so f has units of cycles [ = 2π radians = 360 degrees]per second, which represents the frequency of theoscillations.

• Frequency defines (or reflects) the rate at whichthe instantaneous phase changes. The duration ofone cycle of the wave, called its period , is givenby: T=1/f (seconds per cycle).

• When phase is referred to as the phase or phase shift , the implied reference is:



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• The illustration below shows phase shift. The horizontal axisrepresents an angle (phase) that is increasing with time.

•

A phase shift is also a difference or change in the initial phase . If is delayed (time-shifted) by of its cycle, it becomes:• ,whose initial phase is . Thus, a shift in time is equivalent to

a phase shift. Conversely, a change in the phase isequivalent to a shift in time.



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• In phase waves

• Out of phase waves



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• ANGLE MODULATION is often called as FM.Frequency & Phase Modulation (FM & PM) areboth forms of ANGLE MODULATION.

• Because of its superior performance than AM, it is

used extensively for commercial broadcastingradio broadcasting, television soundtransmission, 2-way mobile radio, cellular radio,microwave and satellite communications systems.

Angle Modulation



• Frequency Modulation is the process of changingcarrier frequency by the modulating signal whilethe carrier amplitude remains constant.

• As the modulating signal amplitude increases,the carrier frequency increases and vice versa.

• The amount of change in carrier frequencyproduced by the modulating signal is calledFrequency Deviation ( f). Meanwhile, the change inphase is called Phase Deviation ( )

• The deviation is proportional to the amplitude ofthe modulating signal.

Frequency Modulation (FM)



• In PM, the phase shift of the carrier is varied bythe amplitude of the modulating signal.

• FM produces pairs of sidebands spaced from thecarrier in multiples of the modulating frequency.

• The modulation index m of FM signal is the ratioof the frequency deviation f d to the modulatingfrequency, f m (m = f d / fm)

• The modulation index determines the number ofsignificant pairs of sidebands in FM signals.



FM Bandwidth

•

The Bandwidth of FM signal isproportional to the modulationindex and defined as below:

BW = 2(f d max + fm max )



A cell phone transmitter has a maximum

frequency deviation of 12 kHz. Calculate themodulation index if it operates at maximumdeviation with a voice frequency of :(a) 300 Hz

(b) 2.5 kHz

SOLUTION :

(a) m = f d / fm = 12 kHz/ 300 Hz = 40

(b) m = f d / fm = 12 kHz/ 2500 Hz = 4.8

EXAMPLE 1



Advantage of FM over AM

• The primary advantage of FM over AM is itsimmunity to noise.

• Noise is short-duration amplitude variationscaused by lightning, motors, auto ignitions,power transients and other sources.

• Another advantage of FM over AM is the captureeffect that allows the strongest signal onfrequency to dominate without interference fromthe other signal.

• FM has greater transmitter efficiency since classC Amplifier may be used



• FM has wide bandwidth compared to AM.

• The spectrum space taken up by FM signal maybe limited by carefully controlling the deviationratio.

• The circuits used in FM to produce anddemodulate FM is more complex and expensivethan AM circuits.

Disadvantage of FM over AM



Angle Modulation• The angle modulation can be expressed

mathematically as:

• The magnitude of the frequency ( f) and phasedeviation ( ) is proportional to the amplitude of themodulating signal, V m and the rate at which the changesare occurring is equal to the modulating signalfrequency, f m .

m(t) = angle modulated waveV c = peak carrier amplitude (Volt) c = carrier radian frequency (rad/sec) (t )= instantaneous phase deviation (radians)

m(t) = V c cos [ ω c t + (t)]



The instantaneous phase of a harmonic carriersignal is varied in such a way that the instantaneousphase deviation i.e. the difference between theinstantaneous phase and that of the carrier signal islinearly related to the size of the modulating signalat a given instant of time.

Phase modulation

rad)(t

Instantaneous phase rad)(t t c

Instantaneous phase deviation

)( t KV m

Instantaneous phase deviation, θ is given by



V V rad

K

)]cos(cos[)( t KV t V t m mmcc

Where K is deviation sensitivity for phase

modulation

)cos()(bygivenis(t)Vm t V t V mmm

Thus

)cos( t m m

θ can also be written as

Where m is phase modulation index or alsoknown as peak phase deviation.



Phase modulation

Where K = deviation sensitivity (rad per volt)

V m = peak modulating signal amplitude (volt)

m = KV m (radians)

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

)]cos(cos[)(

t mt V t t V

t KV t V t m

mcc

mcc

mmcc

Thus

Phase modulation index, m is defined by



Frequency ModulationThe frequency of a harmonic carrier signal is varied in such away that the instantaneous frequency deviation i.e. thedifference between the instantaneous frequency and thecarrier frequency is linearly related to the size of themodulating signal at a given instant of time.

zt

t

H)('

rad/s)('

Instantaneous frequency

rad)('

)]([

t t

t t dt d

c

c

Instantaneous frequency deviation

rad/s)(1t V K

m

Instantaneous phase deviation, θ is given by



V V srad K

/ 1

)]sin(cos[

])cos(cos[

])cos(cos[

)(cos[)(

1

1

1

t V K

t V

t V K t V

t V K t V

t t V t m

mm

mcc

mmcc

mmcc

cc

Where K 1 is deviation sensitivity for phase modulation

)cos()(bygivenis(t)Vm t V t V mmm

Thus



Frequency modulation index• Frequency modulation index is defined as

• Frequency deviation which is the change in carrierwhen acted on by a modulating signal frequency isgiven by:

– Peak frequency shift in hertz – Peak-to-peak frequency deviation of carrier swing

• Therefore m can be rewritten as:

(unitless)1

m

mV K m

f f 2

(unitless))()(

Hz f Hz f

mm



• Thus

)]sin(cos[

)]sin(cos[

)]sin(cos[)( 1

t mt V

t f

f t V

t V K

t V t m

mcc

mm

cc

mm

m

cc

Percent Modulation

100modulation%(max)

)(

f f actual



PM and FM of sine-wave signal

FM

PM

Carrier

Modulatingsignal



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EXERCISE 1

(a) Determine the frequency deviation ( f or

fd) and modulation index (m) for an FMModulator with Deviation Sensitivity,

K1 = 5 kHz/V and the modulating signal isvm(t) = 2cos(2 2000t)

(b) Determine the phase deviation (m) for aPM Modulator with deviation sensitivity K =2.5 rad/v and a modulating signal ofvm(t) = 2cos(2 2000t)



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FM & PM Modulator & Demodulator

• A Phase Modulator is a circuit in which the

carrier is varied in such a way that itsinstantaneous phase is proportional to themodulating signal.

•

A Frequency Modulator is a circuit in whichthe carrier is varied in such a way that itsinstantaneous phase is proportional to theintegral of the modulating signal.



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•

PM Modulator = Differentiator followed byan FM Modulator

• PM Demodulator = FM demodulator followedby an integrator

• FM modulator = integrator followed by a PMModulator

• FM Demodulator = PM demodulator followedby a differentiator



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• For a given input signal, an FM broadcast-band transmitter has frequency deviation, f= 20 kHz. Determine the frequency deviation,

f if the amplitude of the modulating signalincreases by a factor of 2.5

EXERCISE 2



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Bessel Function• Modulating wave is given by:

• The individual frequency components that make-upthe modulated wave are not obvious. However,Bessel Function identities can be applied for this.• By using Bessel Function, the angle-modulatedwave can be written as:

• J n (m) is the Bessel function of the first kind of n thorder with argument m .

cos cos cos2n

n

nm J m n

)]cos(cos[)( t mt V t m mcc



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• Thus, m(t) can be rewritten as

• Expanding the equation, becomes

Where m = modulation indexVc = peak amplitude of the unmodulated carrierJ

0(m)= carrier component

J 1(m)=first set of side frequencies displaced from thecarrier by ω m

J 2(m)=second set of side frequencies displaced from thecarrier by 2 ω m

J n(m)= n th set of side frequencies displaced from the

carrier by n ω

m

n

mcmc

nt nt m J V t m

2cos)()(

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

)2cos()(]2

)cos[()(

]2

)cos[()(cos)({)(

2

21

10

m Jnt m J

t m J t m J

t m J t m J V t m

mc

mcmc

mccc



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• A single-frequency modulating signal produces an infinitenumber of side frequencies each displaced from the carrierby an integral multiple of the modulating frequency. A set of

sidebands includes upper and lower sidebands:and so onSuccessive set of sidebands are called first order sidebands,second order sidebands and so on. Their magnitudes aredefined by J 1(m), J 2(m), and so on

• J can be defined as:

• Refer to handout given for the value of modulation index andits Bessel values or refer to table 7.3 in the textbook.

indexmodulationm frequencysidetheof numberorJn

etc.)(1x2x3x4,factorial!where

...)!1(!362 /

)!2(!242 /

)!1(!12 / 1

2)(

2

nm

nm

nm

nm

m J

mcmcmc nf f f f f f ,2,



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Bessel function of the first kind

d d h f l d l



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Bandwidth of angle modulation• The Bandwidth of angle modulation can be determined byusing Bessel Table and defined as:-

Where n = number of significant sidebandsfm = modulating signal frequency (Hz)

By using Carson’s rule , the bandwidth of angle modulation isdefined as:

Where ∆ f = peak frequency deviation (Hz)fm = modulating signal frequency (Hz)

Hz)(2 m f f B

Hz)(2 m f n B

EXAMPLE 2



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For FM Modulator with frequency deviation of 10kHz, modulating signal frequency of 10 kHz,Carrier amplitude voltage of 50V and Carrierfrequency 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 plotthe output frequency spectrum

EXAMPLE 2 :

S l i



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Solution:a)

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

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

rule . So

1kHz10kHz10

m f f

m

Hz60)103(2 k kHz B

Hz40)1010(2 k kHzkHz B



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c)

V J

V J

V J

V J

1)50(02.0

5.5)50(11.02

12)50(24.0

5.38)50(77.0

0

1

0

38.5V

12V12V

5.5V5.5V

1V1V

500 510 520 530490480470

EXAMPLE 3



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Determine the deviation & ratio andbandwidth for FM transmitter withfrequency deviation of 75 kHz andmaximum modulating signal of 15 kHz.

EXAMPLE 3

P C l l ti



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Power Calculation• In angle modulation, power of angle modulated wave = powerof unmodulated carrier. The power from unmodulated carriersignal is redistributed among carrier and sidebands•

The average power of angle modulation is defined as :

P c = Vc2 / 2R

Vc=peak unmodulated carrier voltage (V)

R = load resistance

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

or

P t = Vc2

/ R [1/2 +½ cos(2 ω ct + 2 (t)• Average power of the second term is zero. Thus

P c = Vc2 / 2R



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• The total power for modulated wave isdefined as:

• P 0= modulated carrier power•

P 1= power in the first set of sidebands• P 2= power in the second set of sidebands• P n= power in the n th set of sidebands

RV

RV

RV

RV

P nct 2)(2

...2)(2

2)(2

2

22

2

2

1

2

nt PPPP ...21

EXAMPLE 4



Determine the unmodulated carrier power for

FM Modulator with modulation index of 1,modulating signal of v m(t) = V msin(2 1000t)and carrier signal of v c(t) = V csin(2 500kt),assume load resistance, R L = 50 . Alsodetermine the total power for the anglemodulation.

EXAMPLE 4

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Chapter Three(FM)Update