chapter 3 Angle Modulation updatefiruz - UniMAP...

CHAPTER 3

ANGLE MODULATION

● classified into two types such as◦ Frequency modulation (FM)◦ Phase modulation (PM)◦ Altering the frequency (FM) / phase (PM) of a

carrier signal to encode the massage signal.

●Used for :◦ Commercial radio broadcasting◦ Television sound transmission◦ Two way mobile radio◦ Cellular radio◦ Microwave and satellite communication system

2

ANGLE MODULATION

Advantages FM/PM over AM:➢Freedom from interference: all natural and

external noise consist of amplitude variations, thus receiver usually cannot distinguish between amplitude of noise or desired signal. AM is noisy than FM.

➢Operate in very high frequency band (VHF): 88MHz-108MHz

➢Can transmit musical programs with higher degree of fidelity.

3

Principles of FM● A sine wave carrier can be modified for

the purpose of transmitting information from one place to another by varying its frequency. This is known as frequency modulation (FM).

● In FM, the carrier amplitude remains constant and the carrier frequency is changed by the modulating signal.

4

● As the amplitude of the information signal varies, the carrier frequency shifts proportionately.

● As the modulating signal amplitude increases, the carrier frequency increases.

● With no modulation the carrier is at its normal center or resting frequency

5

Principles of FM

Principles of FM● Frequency deviation (∆f / fd) is the

amount of change in carrier frequency produced by the modulating signal.

● The frequency deviation rate is how many times per second the carrier frequency deviates above or below its center frequency.

● The frequency of the modulating signal determines the frequency deviation rate.

Principles of FM

Principles of FM

7

Carrier

Modulating Signal

FM signal

Principles of FM

8

Frequency-shift keying (FSK)● When the modulating signal has

only two amplitudes.● For example, when modulating

signal is a binary 0- the carrier frequency is the centre frequency level when modulating signal is a binary 1- the carrier frequency change to a higher frequency level.

● When the amount of phase shift of a constant-frequency carrier is varied in accordance with a modulating signal, the resulting output is a phase-modulation (PM) signal.

● Phase modulators produce a phase shift which is a time separation between two sine waves of the same frequency.

● The greater the amplitude of the modulating signal, the greater the phase shift.

Principles of PM

Principles of Phase Modulation● The maximum frequency deviation produced by a

phase modulator occurs during the time that the modulating signal is changing at its most rapid rate.

A frequency shift occurs in PM only when the modulating signal amplitude varies.

Principles of PM

Principles of Phase ModulationPhase-Shift Keying◦ The process of phase modulating a carrier

with binary data is called phase-shift keying (PSK) or binary phase-shift keying (BPSK).

◦ The PSK signal has a constant frequency, but the phase of the signal from some reference changes as the binary modulating signal occurs.

Principles of PM

Principles of Phase Modulation

Figure: Phase modulation of a carrier by binary data produces PSK.

Phase-shift keying (PSK)

Mathematical analysis of FM

● Mathematical analysis:● Let message signal:

● And carrier signal:

( ) tVt mmm ϖν cos=

( ) ]cos[ θϖν += tVt ccc

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Mathematical analysis of FM● During the process of frequency modulations the

frequency of carrier signal is changed in accordance with the instantaneous amplitude of message signal. Therefore the frequency of carrier after modulation is written as

● To find the instantaneous phase angle of modulated signal, integrate equation above w.r.t. t

( ) tcosVKtvK mm1Cm1ci ω+ω=+ω=ω

( ) tsinVK

tdttcosVKdt mm

m1Cmm1Cii ω

ω+ω=ω+ω=ω=φ ∫∫

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●Thus, we get the FM wave as:

●Where modulation index for FM is given by

)tsinVK

tcos(VcosVc)t(v mm

m1CC1FM ω

ω+ω=φ=

)sincos()( tmtVtv mfCCFM ωω +=

15m

m1f

VKmω

=

Mathematical analysis of FMMathematical analysis of FM

● K1 – deviation sensitivities Hz/V

mf

m

ffm

VKfΔ

=

=Δ ;1

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Mathematical analysis of FM

Example 1 (FM)● Determine the peak frequency

deviation (∆f) and modulation index (m) for an FM modulator with a deviation sensitivity K1 = 5 kHz/V and a modulating signal,

● Δf = 5k x 2 = 10 k Hz

●

)t20002cos(2)t(vm π=

17mf

m

ffm

VKfΔ

=

=Δ ;1

5210

==kkmf

Mathematical analysis of PM

● The process by which changing the phase of carrier signal in accordance with the instantaneous of message signal. The amplitude remains constant after the modulation process.

● Mathematical analysis: Let message signal:

And carrier signal:

( ) tVt mmm ϖν cos=

( ) ]cos[ θϖν += tVt ccc18

● Where = phase angle of carrier signal. It is changed in accordance with the amplitude of the message signal;

● i.e.

● After phase modulation the instantaneous voltage will be

● or

● Where mp = Modulation index of phase modulation● K is a constant and called deviation sensitivities of the phase

tKVtKV mmm ωθ cos)( ==

)coscos()(

)coscos()(

tmtVtvtKVtVtv

mpCCpm

mmCCpm

ωω

ωω

+=

+=

19

θ

Mathematical analysis of PM

EKT343 –Principle of Communication Engineering

Summary of Mathematical Equation for FM and PM

20

TomasiElectronic Communications Systems, 5e

Copyright ©2004 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

● Determine the peak phase deviation (m) for a PM modulator with a deviation sensitivity K = 2.5 rad/V and a modulating signal,

)20002cos(2)( ttvm π=

21

mp KVm =

radxmp 525.2 ==

Modulation Index and Sidebands

● Any modulation process produces sidebands.

● When a constant-frequency sine wave modulates a carrier, two side frequencies are produced.

● 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.

Modulation Index◦ The ratio of the frequency deviation to the

modulating frequency is known as the modulation index (mf).◦ In most communication systems using FM,

maximum limits are put on both the frequency deviation and the modulating frequency.◦ In standard FM broadcasting, the maximum

permitted frequency deviation is 75 kHz and the maximum permitted modulating frequency is 15 kHz.◦ The modulation index for standard FM

broadcasting is therefore 5.

Modulation Index and Sidebands

Modulation Index and Sidebands

Bessel Functions ◦ The equation that expresses the phase

angle in terms of the sine wave modulating signal is solved with a complex mathematical process known as Bessel functions.

◦ Bessel coefficients are widely available and it is not necessary to memorize or calculate them.

Modulation Index and Sidebands

FM&PM (Bessel function)

● Thus, for general equation:

)coscos()( tmtVtv mfCCFM ωω +=

⎟⎠

⎞⎜⎝

⎛ π+β+α=β+α ∑

∞

−∞= 2n

ncos)m(J)cosmcos(n

n

25

∑∞

−∞=

⎟⎠

⎞⎜⎝

⎛ π+ω+ω=

nmcnC 2

ntntcos)m(JV)t(m

B.F. (cont’d)

● 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

● n = number of the side frequency● mf = modulation index

( ) ( )( )

( )( ) ⎥

⎥⎦

⎤

⎢⎢⎣

⎡−

++

+−⎟⎟

⎠

⎞⎜⎜⎝

⎛= ....

!2!22/

!1!12/1

2

42

nm

nm

nm

mJ ffn

ffn

26

FM&PM (Bessel function)

Bessel Functions ◦ The symbol ! means factorial. This tells you to

multiply all integers from 1 through the number to which the symbol is attached. (e.g. 5! Means 1 × 2 × 3 × 4 × 5 = 120)◦ Narrowband FM (NBFM) is any FM system in

which the modulation index is less than π/2 = 1.57, or

mf < π /2.◦ NBFM is widely used in communication. It

conserves spectrum space at the expense of the signal-to-noise ratio.

FM&PM (Bessel function)

B.F. (cont’d)

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FM&PM (Bessel function)

Bessel Functions of the First Kind, Jn(m) for some value of modulation index

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Representation of frequency spectrum

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Example 3● F o r a n F M m o d u l a t o r w i t h a

modulation index m = 1, a modulating signal vm(t) = Vm sin(2π1000t), and an unmodulated carr ier vc( t ) = 10 sin(2π500kt). Determine the number of sets of significant side frequencies and their amplitudes. Then, draw the frequency spectrum showing their relative amplitudes.

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Solution for Example 3

● From the bessel table, m = 1, ◦ J0 = 0.77, J1= 0.44, J2= 0.11, J3 = 0.02◦ So, n = 3◦ Amplitudes:◦ V0 = J0Vc = 0.77 x 10 = 7.7 V◦ V1 = J1Vc = 0.44 x 10 = 4.4 V◦ V2 = J2Vc = 0.11 x 10 = 1.1 V◦ V3 = J3Vc = 0.02 x 10 = 0.2 V

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33

● Draw the frequency spectrum showing their relative amplitudes.

Comparison NBFM&WBFM

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WBFM NBFM

Modulation index greater than 10 less than π/2 / 1.57

Freq deviation 75 kHz 5 kHz

Modulation frequency

30 Hz- 15 kHZ 3 kHz

Spectrum Infinite no of sidebands and carrier

Two sidebands and carrier

Bandwidth 15 x NBFM2(δ*fm (max))

2 fm

Noise More suppressed Less suppressed

Application Entertainment & Broadcasting

Mobile communication

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:

● For n≥1 mfm nfB 2=

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FM Bandwidth

Another way to determine the BW of an FM signal is to use Carson’s rule:

Carson’s rule will always give a BW lower than that calculated with the previous formula.

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)1)........((2 mfm ffB +Δ=

FM Bandwidth

Example 4● For an FM modulator with a peak frequency

deviation, Δf = 10 kHz, a modulating-signal frequency fm = 10 kHz,

Vc = 10 V and a 500 kHz carrier, determine

◦ Actual minimum bandwidth from the Bessel function table.◦ Approximate minimum bandwidth using Carson’s

rule.◦ Plot the output frequency spectrum for the Bessel

approximation.

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Solution for Example 4

● ; n = 3●

BW = 2 n fm

= 2 x 3 x 10 k = 60 kHz

BW (carson rule) = = 2 (10k + 10 k) = 40 kHz

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mf f

fm Δ= 1

1010

==kkmf

)(2 mff +Δ

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Angle Modulation Part 2

■Power distribution of FM ■Generation & Demodulation of FM■Noise in FM■Application of FM

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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.

41

● 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.

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FM Power Distribution

Average Power● The average power in unmodulated carrier

● The total instantaneous power in the angle modulated carrier.

● The total modulated power

R2VP2c

c =

R2V

)]t(2t2cos[21

21

RV

P

)]t(t[cosRV

R)t(m

P

2c

c

2c

t

c2

2c

2

t

=⎭⎬⎫

⎩⎨⎧ θ+ω+=

θ+ω==

43

RV

RV

RV

RV

PPPPP nont 2

)(2..2)(2

2)(2

22..22

222

21

2

210 ++++=++++=

Example● For an FM modulator with a modulation

index m = 1, a modulating signal vm(t) = Vmsin(2π1000t)

and an unmodulated carrier vc(t) = 10sin(2π500kt)

Determine the unmodulated carrier power for the FM modulator given with a load resistance, RL = 50Ω. Determine also the total power in the angle-modulated wave.

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Solution for Example 6

45

0051.1)50(2)2.0(2

)50(2)1.1(2

)50(2)4.4(2

)50(27.7 2222

=+++=tP

WPc 1)50(2

102==

Exercise● For an FM modulator with modulation index,

m = 2, modulating signal, vm(t) = Vmcos(2π2000t)

and an unmodulated carrier, vc(t) = 10 cos(2π800kt)

Assume, RL=50Ω

a) Determine the number of sets of significant sidebands.b) Determine their amplitudes.c) Draw the frequency spectrum showing the relative amplitudes

of the side frequencies.d) Determine the bandwidth.e) Determine the total power of the modulated wave.

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Solution● Vm = 1 V, Vc = 10 V, fm = 2 kHz, fc = 800 kHz, RL =

50 Ω a) m = 2; n = 4 b) V0 = 10 (0.22) = 2.2 V V1 = 10 (0.58) = 5.8 V V2 = 10 (0.35) = 3.5 V V3 = 10 (0.13) = 1.3 V V4 = 10 (0.03) = 0.3 Vd) BW = 2n fm = 2 x 4 x 2k =16 kHz

e) PT = P0 + 2P1 + 2P2 + 2P3 +2P4

= 47

W0018.1)50(2

2)3.0(2)50(2

2)3.1(2)50(2

2)5.3(2)50(2

2)8.5(2)50(2

22.2=++++

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802k 804k 806k 808k800k798k796k794k792k

Generation of FM● Two major FM generation:i) Direct method:

i) straight forward, requires a Voltage-Controlled Oscillator (VCO) whose oscillation frequency has linear dependence on applied voltage.

ii) Advantage: large frequency deviationiii) Disadvantage: the carrier frequency tends to drift

and must be stabilized.iv) Common methods:

i) FM Reactance modulatorsii) Varactor diode modulators

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50

51

Indirect FM generation:● NBFM followed by frequency multiplier● Use nonlinear circuit to get multiplier● Can use mixer to change the carrier

frequency● Combination of mixer and multiplier

provides flexibilities

52

\

Generation of FM (cont’d)ii) Indirect method:

i. Frequency-up conversion.ii. Two ways:

a. Heterodyne methodb. Multiplication method

iii. One most popular indirect method is the Armstrong modulator

53

54

55

56

Note : LOW PASS FILTER (LPF) = SELECT LOWER BAND

84.00 MHz

57

A : f = 500 kHz X 162 = 81 MHz Δf = 15.432 Hz X 162 = 2499.984 Hz

B : f = 84 MHz - 81 MHz = 3.00 MHz Δf = 15.432 Hz X 162 = 2499.984 Hz

C: f = 3.00 MHz X 30 = 90.0 MHz Δf = 2499.984 X 30 Hz = 74.99952 kHz

Note : LOW PASS FILTER (LPF) = SELECT LOWER BAND

58Note : LOW PASS FILTER (LPF) = FILTER LOWER BAND

84.00 MHz

A complete Armstrong modulator is supposed to provide a 75kHz frequency deviation. It uses a balanced modulator and 90o phase shifter to phase- modulate a crystal oscillator. Required deviation is obtained by combination of multipliers and mixing, raise the signal from suitable for broadcasting.

Armstrong Modulator

59

kHz75MHz2.90toHz47.14kHz400 ±±

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Wideband Armstrong Modulator

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f= 400kHz X 81 = 32.4 MHzΔf =14.47 Hz X 81 = 1.172 kHz

f= 33.81 MHz - 32.4 MHz = 1.41 MHzΔf = 1.172 kHz

Note : LOW PASS FILTER (LPF) = SELECT LOWER BAND

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.

62

FM detection ● 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.

63

PLL Demodulator

64

Phase detector

VCO

Low pass filter Amplifier

FM input

Vc(t)

fvco

V0(t)

fi

● The phase detector produces an average output voltage that is linear function of the phase difference between the two input signals. Then low frequency component is pass through the LPF to get a small dc average voltage to the amplifier.

● 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.

65

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;

66

PLL Demodulator

● f0 + k2Vc(t)= fc +k1vm(t), so,

● If VCO can be tuned so that fc=f0, then

● Where Vc(t) is also taken as the output voltage, which therefore is the demodulated output

)()( 10 tvkfftV mcc +−∝

)()( 1 tvktV mc ∝

67

PLL Demodulator

Noise in FM● Noise is interference generated by

lightning, motors, automotive ignition systems, and power line switching that produces transient signals.

● Noise is typically narrow spikes of voltage with high frequencies.

● Noise (voltage spikes) add to a signal and interfere with it.

● Some noise completely obliterates signal information.

68

Noise in FM

● In AM systems, noise easily distorts the transmitted signal however, in FM systems any added noise must create a frequency deviation in order to be perceptible.

69

θ

Noise in FM(Cont’d)● The maximum frequency deviation due to random noise

occurs when the noise is at right angles to the resultant signal. In the worst case the signal frequency has been deviated by:

δ = θfm

● This shows that the deviation due to noise increases as the modulation frequency increases. Since noise power is the square of the noise voltage, the signal to noise ratio can significantly degrade.

● Noise occurs predominantly at the highest frequencies within the baseband

70

Noise-Suppression Effects of FM● FM signals have a constant modulated

carrier amplitude.● FM receivers contain limiter circuits that

deliberately restrict the amplitude of the received signal.

● Any amplitude variations occurring on the FM signal are effectively clipped by limiter circuits.

● This amplitude clipping does not affect the information content of the FM signal, since it is contained solely within the frequency variations of the carrier.

71

Noise-Suppression Effects of FM

72Figure 5-11: An FM signal with noise.

Noise-Suppression Effects of FMPreemphasis ◦ Noise can interfere with an FM signal and

particularly with the high-frequency components of the modulating signal.◦ Noise is primarily sharp spikes of energy

and contains a lot of harmonics and other high-frequency components.◦ To overcome high-frequency noise, a

technique known as preemphasis is used.◦ A simple high-pass filter can serve as a

transmitter’s pre-emphasis circuit.◦ Pre-emphasis provides more amplification

of only high-frequency components.73

Noise-Suppression Effects of FM

74Preemphasis circuit.

Noise-Suppression Effects of FMDe emphasis◦ A simple low-pass filter can operate as a

deemphasis circuit in a receiver.◦ A deemphasis circuit returns the frequency

response to its normal flat level.◦ The combined effect of preemphasis and

deemphasis is to increase the signal-to-noise ratio for the high-frequency components during transmission so that they will be stronger and not masked by noise.

75

Noise-Suppression Effects of FM

76Deemphasis circuit.

Application of FM

● FM is commonly used at VHF radio frequencies for high-fidelity broadcasts of music and speech (FM broadcasting). Normal (analog) TV sound is also broadcast using FM. The type of FM used in broadcast is generally called wide-FM, or W-FM

● A n a r r o w b a n d f o r m i s u s e d f o r v o i c e communications in commercial and amateur radio settings. In two-way radio, narrowband narrow-fm (N-FM) is used to conserve bandwidth. In addition, it is used to send signals into space.

77

Frequency Modulation Versus Amplitude Modulation

Major applications of AM and FM

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.

79

Disadvantages● Angle modulation requires a

transmission bandwidth much larger than the message signal bandwidth.

● Angle modulation requires more complex and expensive circuits than AM.

80

Summary of angle modulation-what you need to be familiar with

81

Summary (cont’d)

82

Summary (cont’d)

● Bandwidth:a) Actual minimum bandwidth from

Bessel table:

b) Approximate minimum bandwidth using Carson’s rule:

)(2 mfnB ×=

)(2 mffB +Δ=83

Exercise 1● Determine the deviation ratio and

worst-case bandwidth for an FM signal with a maximum frequency deviation 25 kHz and maximum modulating signal 12.5 kHz.

84

25.1225 == kkDR

BW = 2n fm = 2 x 4 x 12.5k

= 100 kHz

Exercise 2●For an FM modulator with 40-kHz

frequency deviation and a modulating-signal frequency 10 kHz, determine the bandwidth using both Carson’s rule and Bessel table.

● ; n = 7●Carson’s rule:

BW = 2 (40k + 10 k) = 100 kHz●Bessel table:

BW = 2 n fm = 2 x 7 x 10 k = 140 kHz

85

41040

==kkmf

Exercise 3● For an FM modulator with an

unmodulated carrier amplitude 20 V, a modulation index, m = 1, and a load resistance of 10-ohm, determine the power in the modulated carrier and each side frequency, and sketch the power spectrum for the modulated wave.

86

Solution for Exercise 3

87

Wx

P 858.11)10(2

2)77.020(0 ==

WxPP 872.3)10(2

2)44.020(11 ==−=

WxPP 242.0)10(2

2)11.020(22 ==−=

WxPP 008.0)10(2

2)02.020(33 ==−=

Exercise 4● A frequency modulated signal (FM)

has the following expression:

The frequency deviation allowed in this system is 75 kHz. Calculate the:◦ Modulation index◦ Bandwidth required, using Carson’s rule

)1010sin10400cos(38)( 36 tmttv ffm ×+×= ππ

88

15575

==kkmf

BW = 2 (75k + 5 k)

= 160 kHz

END OF ANGLE MODULATION

89