Angle Modulation Objectives To introduce phase modulation (PM) and frequency modulation (FM) To...

Angle ModulationAngle Modulation

Objectives

• To introduce phase modulation (PM) and frequency modulation (FM)

• To define frequency deviation• To describe the relationship between PM and FM• To determine the bandwidth of FM signal• To explain the methods for generating wideband FM signals• To explain wideband FM demodulation method• To illustrate frequency division multiplexing (FDM) method• To compare AM and FM


Angle modulation • Angle modulation is the process of varying the total phase angle

of a carrier wave in accordance with the instantaneous value of the message signal, while keeping the amplitude of the carrier constant.

Because the amplitude of the carrier does not change, angle modulation has the advantage of using the peak power of a transmitter (P = A2 /2).

However, angle modulation requires more complicated transmitters and receivers as compared to amplitude modulation.


A generalized sinusoidal signal of constant peak amplitude A can be expressed as

s(t) = A cos(t)where (t) is the total phase angle (instantaneous or generalized an

gle).

Types of angle modulationThere are two types of angle modulation, according to two ways of

varying the phase angle (t).


Phase modulation (PM)The phase angle is varied linearly with a modulating signal m(t)

PM(t) = ct + 0 + kPMm(t)where kPM is known as phase sensitivity of the modulator.

The PM waveform can then be expressed assPM(t) = A cos[ct + 0 + kPMm(t)]


Frequency modulation (FM)The instantaneous frequency (t) is varied linearly with a modulati

ng signal m(t) about an unmodulated frequency c, i.e.FM(t) = c + kFMm(t)

where kFM represents the frequency sensitivity of the modulator.

The total phase angle can be obtained by integrating the instantaneous frequency FM(t)

The corresponding FM wave is given by

0( ) ( ) [ ( )] ( )FM FM c FM c FMt t dt k m t dt t k m t dt

0( ) cos[ ( ) ]FM c FMs t A t k m t dt


Frequency deviationThe instantaneous frequency of FM signal, FM(t) = c + kFMm(t), v

aries with time. Frequency deviation is the maximum frequency change from the av

erage c, i.e. = kFMm(t)max

The frequency deviation is a useful parameter for determining the bandwidth of the FM signals.


Relationship between PM and FM• Both the net effect of PM and FM is variation in total phase angle.• In PM, phase angle varies linearly with m(t) while in FM, phase an

gle varies linearly with the integral of m(t). • FM signal can be obtained by using phase modulator and PM wa

ve can be obtained by using frequency modulator.


Example:Tone modulation

The message signalm(t) = Am cosmt

PM (Tone modulation)sPM(t) = A cos[ct + 0 + kPMm(t)]

= A cos[ct + 0 + kPM Amcosmt]= A cos[ct + 0 + mp cosmt]

where mp = kPMAm is called the phase modulation index, representing the maximum phase deviation .


FM (Tone modulation)

where mf = kFMAm / m = / m, i.e. the ratio of frequency deviation to the modulating frequency, is called the frequency modulation index.

mf = / m

The frequency deviation in FM represents the maximum phase deviation

= mf = / m

0

0

0

0

( ) cos[ ( ) ]

cos[ cos ]

cos[ sin ]

cos[ sin ]

FM c FM

c FM m m

FM mc m

m

c f m

s t A t k m t dt

A t k A tdt

k AA t t

A t m t


In the case of tone modulation, it is difficult to distinguish PM and FM waves.


We therefore focus our discussion on FM.

• The bandwidth of an FM signal depends on the frequency deviation.

• When the frequency deviation (and hence mf) is high, the bandwidth will be large.

Types of frequency modulationIf mf < 1, this is known as narrowband FM. If mf > 1 it is wideband FM ( also known as broadcast FM).

What is the significance of bandwidth for FM signal?


Since the amplitude of an FM signal is kept constant,the power of an FM signal is the same as that of anunmodulated carrier, i.e., PFM = A2/2.

In other words, the power is independent of thefrequency modulation process since the power is notdependent on the signal’s frequency.

Most of the FM signal power is contained in itsbandwidth.


It can be shown that 98% of the total power of an FM signal is contained in the bandwidth. ( Then, how to calculate the bandwidth?)

Bandwidth of a sinusoidally modulated FM signalWFM = 2(mf + 1)m = 2( + m)

or BFM = 2(mf + 1)fm = 2(f + fm)

The bandwidth is twice the sum of the maximum frequency deviation and the modulating frequency ---- Carson’s rule.


Bandwidth of FM signal with arbitrarymodulating signals:

WFM = 2( + m)or

BFM = 2(f + fm)


ExampleA 10 MHz carrier is frequency modulated by a sinusoidal signal

such that the peak frequency deviation is 50 kHz. Determine the bandwidth of the FM signal if the frequency of the modulating sinusoidal is (a) 500 Hz; (b) 10 kHz.

Solution(a) mf = f / fm = 50000/500 = 100

B = 2(mf + 1)fm = 2(101)500 = 101 kHz

(b) mf = f / fm = 50/10 = 5B = 2(mf + 1)fm = 2(6)10 = 120 kHz


Exercise

If f = 75 kHz and fm = 15 kHz, determine the FMbandwidth and compare this with the AM bandwidth.

What is your finding?


Wideband modulation methodsDirect method, voltage-controlled oscillator (VCO)The direct method depends on varying the frequency of an oscillat

or linearly with m(t) for FM.


In the VCO, the modulating signal varies the voltage across the capacitor, as a consequence, the capacitance changes and causes a corresponding change in the oscillator frequency, i.e.

C = C0 + C = C0 + k0m(t), C = C0 (1 + C/C0)where k0 is a constant. Assume that

0

0

1

LC

000

0

0 0 00 0 0

0 0

1 1 1(1 )

21

( )(1 ) ( ) ( )

2 2

C

CLC LC C

C

k m t km t km t

C C

where k is also a constant and (1 - x)-1/2 1 – x/2 is used.


Wideband FM demodulation methodApply sFM(t) to a differentiator, the output is

{ cos[ ( ) ]}( )

[ ( )]sin[ ( ) ]

c FMFM

c FM c FM

d A t k m t dtds t

dt dt

A k m t t k m t dt

The response of an envelope detector becomes

A[c + kFMm(t)]Blocking the dc term Ac, the output is so(t) = AkFMm(t)


Multiplexing Multiplexing is combining several messages into a compositesignal for transmission over a single communication channel.

When the multiplexing is carried out in the frequency domain, it is called Frequency division multiplexing

(FDM).

How to separate different messages at the receiver side?Provided that there is no spectral overlap of the sidebands of one signal with those of the adjacent signal, the messages canbe separated at the receiver by bandpass filtering.

In order to make efficient use of transmitter power and of bandwidth, SSB modulation would normally be used with FDM.


Frequency division multiplexing (FDM)


FDM can be used in long distance telephone, radio/TV broadcasts, etc. But adjacent FDM signals are vulnerable to crosstalk and the requirement for bandpass filters increase the cost and complexity of the FDM system.

Main advantages and disadvantages of FDM• Ability to convey many simultaneous messages• Expense and size of the band-pass filters• Vulnerability to crosstalk


Comparison between AM and FMNoise performanceWideband FM has better noise performance than AM. The greate

r the bandwidth, the better is the noise performance. Narrowband FM has a noise performance equivalent to AM.

Channel bandwidthThe wideband FM has a larger bandwidth as compared to AM.

ExampleIn a typical broadcast system, each channel bandwidth in AM is 1

5kHz, whereas, in FM, it is 150kHz.

Questions (Angle Modulation)Questions (Angle Modulation)

Questions

1. What is FM ? Write the waveform expression for FM.2. What is PM? Write the waveform expression for PM.3. How to generate FM signal by using phase modulator?4. How to generate PM signal by using frequency modulator?5. What is frequency modulation index?6. How to determine the bandwidth of FM signal?7. How to generate wideband FM signal in a direct method?8. How to demodulate wideband FM signal?9. What is FDM and what is its main advantage?

Exercise Problems (Angle Exercise Problems (Angle Modulation)Modulation)

1. An angle modulated signal is s(t) = Acos(ct + 100cosft).(a) For a PM, what is m(t) when kPM = 2?(b) What is the peak frequency change caused by m(t) in (a)?

2. Work problem 1 assuming modulation is FM and kFM = 2.

3. A 100 MHz carrier is frequency modulated by a sinusoidal signal of frequency 5kHz. If the peak frequency deviation is 50kHz, what is the approximate band of frequencies occupied by the FM waveform.


4. Twelve speech signals are to be combined using FDM for transmission over a wideband radio link. Each signal is concentrated in 300Hz < f < 3.4kHz and is allocated a nominal low pass channel bandwidth of 4kHz to allow for the use of practical filters with finite transition regions. Estimate the radio bandwidth required if the signals are combined using SSB-SC frequency division multiplexing and the radio receiver employs conventional amplitude modulation.


5. For a carrier wave with frequency of 1 MHz and amplitude of 3V, if the message signal is a cosine function, has unit amplitude and its frequency is 2 kHz, and the maximum frequency deviation is 4 kHz, determine the time domain expression of the FM wave.

6. A carrier wave has fc = 50 MHz, is frequency modulated by a sinusoidal wave of frequency fm = 20 kHz,

(a) if the frequency deviation is 20Hz, determine the minimum bandwidth;

(b) if the frequency deviation is increased to 1 MHz, what is the corresponding bandwidth?


7. The carrier wave of frequency 100MHz, is frequency modulated by a sinusoidal message signal of frequency 10kHz, the maximum frequency deviation is 1MHz,

(a) determine the approximate bandwidth of the FM wave;(b) if the amplitude of modulating signal is doubled, determine the

bandwidth of the FM wave;(c) if the frequency of the modulating wave is also doubled,

determine the bandwidth of the FM wave.

8. A sinusoidal signal of 10kHz is used to modulate a carrier wave of 100MHz, assuming that the maximum frequency deviation is 50kHz, determine the bandwidth for

(a) AM wave; (b) SSB wave; (c) FM wave.

Date post:	22-Dec-2015
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Angle Modulation Objectives To introduce phase modulation (PM) and frequency modulation (FM) To...

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