ece eee f311 pm and fm sept 3 -11 2014

8/11/2019 Ece Eee f311 Pm and Fm Sept 3 -11 2014

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Topic # 5: Phase & Frequency Modulation

T1. B.P. Lathi, Modern Digital and Analog Communication Systems, 3rd

Edition, Oxford University Press, 1998: OR 4thEdition 2010 Chapter 5

T2. Simon Haykin & Michael Moher: Communication Systems; John Wiely, 4th

Edition OR 5thEdition, 2010, 5/e. : Chapter 4

Sept 311, 2014


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ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION

Introduction

Amplitude modulation Process of varying a carrier signals amplitude according to message

signal

Phase Modulation Can we vary of the phase of a carrier according to message signal ?

Frequency Modulation

Can we vary of the frequency of a carrier according to message signal ?

Is it for fancy or any benefits over AM exist ? Noise immunity over AM at the cost of increased BW

Constancy of the transmitted signal envelop.2


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Instantaneous Frequency

3

()

= w

c

Given instantaneous frequency, the

angle function is then

And the instantaneous Frequency is

(t) is the generalized angle

and is a function of t.

The generalized angle for a conventionalSinusoid A cos( wct + o) is

Instantaneous Frequency

(t) = wct + o

o

Dt

()

= wi(t)

ois the accumulated phase upto t = 0.


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Phase Modulation

4

Instantaneous frequency varies linearly with derivative of the message signal

The instantaneous frequency is

PM Signal

Phase varies linearly with message signal


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Example.

5

In an angle modulating system, for themodulating signal m(t) shown in figure,

the constant kp = 10p. Carrier frequency

fc= 100 MHz. Find the frequency

excursion when PM is used. Sketch PM

wave

PM Case:

Slope of m(t) = 2 / 10-4= (+/-)20,000


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Frequency Modulation

FM Signal

The phase varies with respect to the integral of the message signal.

Instantaneous Frequency varies linearly with message signal

The Phase is


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Example.

7

In an angle modulating system, for the

modulating signal m(t) shown in figure,

the constant kf = 2pX 105and Carrier

frequency fc= 100 MHz. Find the

frequency excursion when FM is used.

Sketch FM wave The instantaneous frequency increases

linearly from 99.9 to 100.1 MHz over ahalf cycle and decreases from 100.1 to

99.9 MHz, over the rest of the half cycle

of m(t).


8/36ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION

Characteristics of PM & FM signals

8

1. Constancy of Transmitter Power

The amplitude of the PM & FM

waves is constant, irrespective of

the deviation factors km& kf

Hence, the transmitted power is

constant.

2. Non linearity of the modulationprocess

If m(t) = m1(t) + m2(t);

s1(t)= ( + pm1(t))

s2(t)= ( + pm2(t))

s(t)= ( + p(m1(t) + m2(t)))

Clearly s(t) s1(t) + s2(t)

For Phase Modulation

Same applies to FM Also.

Compare this with DSB-SC



Characteristics of PM & FM signals

3. Immunity of angle modulation tononlinearities of the system.

9

Even with Higher order nonlinearities.

Check for DSB-SC Case

for non linearity of type



Bandwidth of Frequency Modulated Waves

10

For the FM Signal

Let

That is, the FM signal is like several

DSB-SC signals with modulating

signals a(t), a2(t), a3(t) an(t).

Hence, the spectrum consists of un

modulated carrier plus spectra of

{an(t) , n= 1,2,3}, centered at wc.

Expanding in power series



Bandwidth of Frequency Modulated Waves

11

If M(w) is band limited to B Hz, A(w)

is also band limited to B Hz

Spectrum of a2(t) is A(w)*A(w) and is

band limited to 2B. Similarly, an(t) will

be band limited to nB.

Clearly, the theoretical bandwidth of

Frequency Modulated signal is infinite

Since, from FT properties,

However, for practical signals withbounded | a(t)|, |kf a(t)| will remain finite.

For large n 0

Hence, many of the higher order

terms of FM(t)

Based on Bandwidth we have

1. Narrowband FM and

2. Wideband FM

vanish and power remains in finitebandwidth. BW is finite for practical

Signals



Narrowband Frequency Modulation

12

This has similarities toAM signal with

Carrier.

If |kfa(t) |



Narrowband Phase Modulation

13

If |kpm(t) |



Wideband Frequency Modulation

14

If |kfa(t) |



BW - Wideband Angle Modulation

15

Each cell has a constant amplitude m(tk) and duration 1/2B

The FM Spectrum of one burst is then

The FM BW

The FM Spectrum of () is then sum

of the spectra of short burst of sinusoids

of duration (1/2B) and freq.


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Example.

17

Step 1: Find the essential

bandwidth of the given waveform

Consider only upto third harmonic.

Estimate BFMfor the

modulating signal m(t) seefigure. Use kf = 2pX 10

5.

In this case the amplitude of harmonics decay rapidly.

Third harmonic is 11% and 5this only 4% of the

fundamental..

B = 3 X (104/ 2 ) = 15 KHz.

Step 2: Find the frequency deviation, based on kfand peak amplitude of the modulating waveform

The bandwidth of the FM signal is then



Example.

18

Estimate BPMfor the modulating

signal m(t) see figure. Use kp = 5p.

What if amplitude of m(t) is doubled ?

Doubling the amplitude of m(t) changes its peak

value and not the BW. Hence for FM signal

The bandwidth of the FM signal is then


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Single Tone Frequency Modulation

20

Bessel Function of the first kind and nth order.

Jn(b)



Single Tone Frequency Modulation

21

n = 0 : Themodulated

signal has a

carrier

component.

n 0 : It also has

infinite number of

sidebands

The strength of nthsideband at w= wc+ nwm is Jn(b)

Carsons formula


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Demodulation of FM Signals.

23

The instantaneous frequency

of the FM signal is

Can a simple network with

transfer function H() = j

produce an output proportionalto instantaneous frequency?

A circuit with transfer function

as H(w) = jw, is a

differentiator in time domain.

wc+ kf m(t) > 0 for all t

Dw= kf mp< wc,

What if A is not constant ? Use band pass Limiting


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Practical FM Demodulation Circuits

25

A tuned circuit (tuned to wc) , followed by

an envelope detector can then be used.

Frequency response of a tuned circuit is

almost linear around the tuned frequency.

Hence, it acts like a slope detector

Linearity zone is increased by a

balanced discriminator.

Zero Crossing Detectors: Rate of zero

crossings is the measure of instantaneous

frequency.


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Interference & Angle Modulated systems

27

I cos (wc

+w

)t

Assume an unmodulated Carrier at

frequency wc is being transmitted.

the received signal is

If the demodulator is a Phase Detector, the output i

Its like a single tone with

amplitude, scaled by A, the

carrier amplitude.

If the demodulator is an FM detector, It will produc

output from the Instantaneous Frequency:

the output is

Its again like a single tone with amplitude, scaled

by A, the carrier amplitude and the frequency of

the interferer

In both cases, for A >> I, effect of

interference is negligible.


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Interference due to Channel Noise

29

The channel noise acts as an interference

White noise has a constant power spectral

density: Implies Interference amplitude ,I,

is constant for all w

Voice signals have low PSD higher

frequencies.

Effect of Noise interference is high for

higher frequencies,.

Noise Amplitude in Phase Demodulated

output = (I/ 2 A) and is independent of

the interferer frequency

The Phase detector Output is

The FM Detector Output is

Noise Amplitude in Frequency

Demodulated output = (Iw

/2 A)

SNR of FM output suffers !!

Consequences ?


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Remedy: Pre Emphasis & De Emphasis

30

Pre Emphasis: : Amplify the higher

frequencies of m(t) before modulation

De Emphasis:After demodulation, apply

an inverse operation to the pre emphasis.


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SNR Performance of Angle Modulated Systems

Phase Modulation ( Small noise Case)Frequency Modulation ( Small noise Case)

31

is Baseband SNR


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Problem # 2

34


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Problem # 3

35

Composite N = (final deviation / initial deviation)

= 75000/10 = 7500 : N = N1 * N2

For carrier : (L1 - 100Khz * N1 ) * N2 = 98.1 MHz. :

The local Oscillator frequency L1 is constrained

by 10MHz L1 11 MHz

Case1:If N1 = 100 then N2 = 75,

( L110Mhz) * 75 = 98.1 MHz. :

L1= 11.308 MHz.

Does not work !!

X N1 BPF X N2

L1

Case2: If N1 = 125 then N2 = 60,

(12.5 MHz - L1 ) * 60 = 98.1 MHz. :

L1= 10.865 MHz.May be this will work

Case3: If N1 = 75 & N2 = 100,

( L1 - 7.5 MHz) * 100 = 98.1 MHz. :

L1= 8.48 Mhz MHz.

Does not work !!


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Problem # 3

X N1 BPF X N2L1

ece eee f311 pm and fm sept 3 -11 2014

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