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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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ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
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).
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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
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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
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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
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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
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Narrowband Frequency Modulation
12
This has similarities toAM signal with
Carrier.
If |kfa(t) |
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Narrowband Phase Modulation
13
If |kpm(t) |
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Wideband Frequency Modulation
14
If |kfa(t) |
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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
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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)
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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
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