ekt 231 : communication system chapter 3 : angle modulation

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Transcript ekt 231 : communication system chapter 3 : angle modulation

EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
ANGLE MODULATION
CHAPTER 3
Review on Part 1
Part 2
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EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Review on
Part 1
ANGLE MODULATION
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EKT 231 : COMMUNICATION SYSTEM
CHAPTER 3 : ANGLE MODULATION
Define the following…..
Modulation
Demodulation
Amplitude modulation (AM)
Angle modulation (FM,PM)
Modulating signal
Carrier signal
Bandwidth
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Introduction
• Angle modulation is the process by which
the angle (frequency or Phase) of the
carrier signal is changed in accordance
with the instantaneous amplitude of
modulating or message signal.
• also known as “Exponential modulation"
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Cont’d…
• Classified into two types ;
– Frequency modulation (FM)
– Phase modulation (PM)
• Used for :
–
–
–
–
–
Commercial radio broadcasting
Television sound transmission
Two way mobile radio
Cellular radio
Microwave and satellite communication system
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Cont’d…
Advantages 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):88M-108MHz
Can transmit musical programs with higher
degree of fidelity.
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FREQUENCY MODULATION
PRINCIPLES
• In FM the carrier amplitude remains constant, the
carrier frequency varies with the amplitude of
modulating signal.
• The amount of change (the relative displacement of
carrier frequency in hertz in respect to it un-modulated
value) in carrier frequency produced by the modulating
signal is known as frequency deviation.
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PHASE MODULATION (PM) PRINCIPLES
• The process by which changing the phase of carrier
signal in accordance with the instantaneous of
message signal. The amplitude and frequency
remains constant after the modulation process.
• Mathematical analysis:
Let message signal:
 m t   Vm cos mt
And carrier signal:
 c t   Vc sin[ ct   ]
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PM (cont’d)
• After phase modulation, with
  KVm (t )  KVm cosmt
the instantaneous voltage will be
vPM (t )  VC sin(Ct  KVm cosmt )
vPM (t )  VC sin(Ct  mp cosmt )
• where
mp = Modulation index of phase modulation
K = a constant and called deviation sensitivities of
the phase
 = phase angle of carrier signal .It is changed in
accordance with the amplitude of the message signal
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FREQUENCY MODULATION(FM)
• A process where the frequency of the carrier wave
varies with the magnitude variations of the
modulating or audio signal.
• The amplitude of the carrier wave is kept constant.
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FM(cont’d)
• Mathematical analysis:
• Let message signal:
 m t   Vm cos mt
• And carrier signal:
 c t   Vc cos[ ct   ]
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FM (cont’d)
• During the process of frequency modulations the
frequency of carrier signal is changed in
accordance with the instantaneous amplitude of
the message signal .Therefore the frequency of
carrier after modulation is written as
i  c  Kvm t   C  KVm cosmt
• To find the instantaneous phase angle of modulated
signal, integrate equation above w.r.t. t
i   i dt   C  KVm cosmt dt  C t 
KVm
m
sin mt
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FM(cont’d)
• Thus, we get the FM wave as:
vFM (t )  Vc cos1  VC cos(C t 
KVm
m
sin mt )
vFM (t )  VC cos(Ct  m f sin mt )
• Where
mf 
KVm
m
Known as the modulation index for the FM
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FM(cont’d)
• Frequency deviation: ∆f is the relative
displacement of carrier frequency (Hz) w.r.t its
unmodulated value. Given as:
max  C  KVm
min  C  KVm
d  max  C  C  min  KVm
d KVm
f 

2
2
• Where K is the deviation sensitivity in Hz/V
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FM(cont’d)
• The peak to peak frequency deviation (2Δf) is
called the carrier swing.
• Therefore:
KVm
f 
2
• The modulation index is
f
mf 
fm
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•
Example 3.1
(a) An FM modulator with
Deviation sensitivity, K1=5kHz/V
Modulating signal Vm(t)=2cos(2π2000t)
Determine
(i) The peak frequency deviation(Δf) : Ans:10kHz.
(ii) The modulation index (m) : 5 (unitless)
Hint:
KVm
f 
 K1Vm
2
f
mf 
fm
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Angle-modulated wave in the frequency domain.
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•
Example 3.2
(a) A PM modulator with
Deviation sensitivity, K=2.5rad/V
Modulating signal Vm(t)=2cos(2π2000t)
Determine
(i) The peak phase deviation(m) : Ans: 5 rad
Hint: Peak phase shift for modulated wave is the modulation index itself.
m  KV m
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Phase and frequency modulation of a sine-wave carrier by a sine-wave signal: (a) unmodulated
carrier; (b) modulating signal; (c) frequency-modulated wave; (d) phase-modulated wave
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FM&PM (Bessel function)
• Thus, for general equation:
vFM (t )  VC cos(Ct  m f cosmt )
vFM (t )  VC sin(Ct  m f sin mt )
 VC sin Ct cos(m f sin mt )  cosct sin(m f sin mt )
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Bessel function
vt FM  VC [ J 0 (m f ) sin Ct  J1 (m f )sin(C  m )t  sin(C  m )t 
 VC [ J 2 (m f ) sin(C  2m )t  sin(C  2m )t
 VC [ J 3 (m f ) sin(C  3m )t  sin(C  3m )t ]  ..
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Representation of frequency spectrum
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Bessel Functions of the First Kind, Jn(m)
for some value of modulation index
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•
Example 3.2
(a) An FM modulator with
m=1
Modulating signal Vm(t)=Vmsin(2π1000t)
Unmodulated carrier V c(t)=10sin(2π500kt)
Determine
(i) Number of sets of significant side frequencies
(Ans: 1 carrier + 3 sets sides freq.)
(ii) Their amplitudes
(iii) Draw the frequency spectrum
Hint: (i), (ii), (iii) use table of Bessel functions of the First Kind, Jnm
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Solution
(ii) Their amplitude
J0=0.77(10)
J1=0.44(10),
J2=0.11(10)
J3=0.02(10)
; Carrier
; 1st sideband pairs
; 2nd sideband pairs
; 3rd side band pairs
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Solution (cont’d)
(iii) Draw the frequency spectrum.
6kHz
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Angle Modulation
Part 2
•FM Bandwidth
•Power distribution of FM
•Generation & Detection of FM
•Application of FM
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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 actual minimum
bandwidth using Bessel table:
B fm  2nfm
• With n = number of significant sidebands
fm
= modulating signal frequency
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FM Bandwidth (cont’d)
Carson’s rule also can be used to estimate the
bandwidth regardless of modulation index
B fm  2(f  f m )
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•Example 3.3
An FM modulator has the following information;
É f = 10kHz;
Vc = 10V
f m = 10kHz
f c = 500kHz
Determine
(a)Actual minimum bandwidth from the B.F table : Ans=60kHz.
(b)Approximate minimum bandwidth using Carson’s rule.
(c)Plot the output frequency spectrum for the Bessel
approximation
Hint: m  f
B  2nf
f
fm
fm
m
B fm  2(f  f m )
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Solution
Given;
É f = 10kHz;
Vc = 10V f m = 10kHz f c = 500kHz
(a)Actual minimum bandwidth from the B.F table.
m=10kHz/10kHz=1, from B.F table, B=2(3x10kHz)=60kHz.
(b) Approximate minimum bandwidth using Carson’s rule.
B fm  2(f  f m )  2(10kHz 10kHz)  40kHz.
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(c) Plot the output frequency spectrum for the Bessel
approximation
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Deviation Ratio (DR)
• The worse case modulation index which produces the
widest output frequency spectrum.
DR 
f (max)
f m (max)
• Where
∆f(max) = max. peak frequency deviation
fm(max) = max. modulating signal frequency
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Example 3.4
An FM broadcast-band transmitter has a maximum
frequency deviation of 75kHz and a maximum
modulating signal frequency of 15kHz. Determine the
deviation ratio bandwidth.
Solution
The deviation ratio (i.e the worst case modulation index).
DR 
f (max)
f m (max)
75 kHz
 15
kHz  5
From the B.F table, a modulation index of 5 gives 8
significant sidebands.
Then B=2(8x15kHz)=240kHz.
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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 it is modulated or not.
• The total power in angle-modulated wave is equal to the
power of the un-modulated wave.
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FM Power Distribution (cont’d)
• 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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Average Power
•
The average power in unmodulated carrier
•
The total power in the angle modulated carrier.
m(t ) 2 Vc2
Pt 

cos2 [ct   (t )]
R
R
2
Vc2  1 1
 Vc
Pt 
  cos[2ct  2 (t )] 
R 2 2
 2R
•
Vc2
Pc 
2R
The modulated carrier power is the sum of the powers of the carrier and
the side frequency components as follow;
Vc2 2(V1 ) 2 2(V2 ) 2
2(Vn ) 2
Pt  P0  P1  P2  ..  Pn 


 .. 
2R
2R
2R
2R
Vc=peak unmodulated carrier voltage (volts), Vn= sidebands voltage
(volts), Pc=Carrier power, R=Resisitive load (ohms)
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Example 3.5
An FM modulator has the following information;
m = 1; vm( t ) = Vmsi n (2ù1000t )
vc( t ) = 10si n (2ù500kt )
Determine;
(i) The unmodulated carrier power for the FM modulator
(Assume RL=50Ω)
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Vc2
Pc 

 1W
2R 2(50)
(i) The total power in the angle modulated wave.
Vc2 2(V1 ) 2 2(V2 ) 2 2(V3 ) 2
Pt 



 1.0051W
2R
2R
2R
2R
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