Download Lecture 10

10. FM Generation and Transmission
Angle modulation
FM analysis, Bessel function
Bandwidth of FM waves
Noise suppression

Direct & Indirect FM generation
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FM Wave
Peak of intelligence wave
Peak amplitude of AM wave
Modulating intelligence wave
Difference between
AM Wave
and
FM Wave
Peak frequency of FM wave
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FM analysis, Bessel function
FM Analysis

PM e  A sin wct  mp sin wi t


FM e  A sin wct  mf sin wit

where :
e= instantaneous voltage
A= peak value of the carrier wave
wc= carrier angular velocity (2pfc )
mp= maximum phase shift caused by intelligence signal amplitude
wi= modulating(intelligence signal angular velocity (2pfi )
d = maximum frequency deviation (shift ) caused by intelligence signal
amplitude
mf = FM modulation index = ratio of maximum freq. deviation of the
carrier to the intelligence frequency
mf 
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d
fi
3
Simple FM Generator
capacitor
microphone
FM radio
waves
sound
waves
Oscillator
C
C + DC
C - DC
1
L
fO 
L
fL 
L
fH 
2 p LC
1
2 p L( C  DC )
1
2 p L( C  DC )
Sound wave will cause capacitor microphone to change it’s capacitance by DC
FM wave varies from fL to fH with fO at the center (when there is no sound waves)
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FM Equation
FM e  A sin wct  mf sin wit 
d
where mf  f
i
sine of sine
Bassel function solution of the FM equation is
fc ( t )  J0 ( mf ) cos wct  J1 ( mf ) cos wc  wi t  cos wc  wi t 
 J2 ( mf ) cos wc  2 wi t  cos wc  2 wi t 
 J3 ( mf ) cos wc  3 wi t  cos wc  3 wi t 

fc ( t )
 FM wave component
J0 ( mf ) cos wct
 carrier component
J1 ( mf ) cos wc  wi t  cos wc  wi t 
 first sideband  fi component
J2 ( mf ) cos wc  2 wi t  cos wc  2 wi t   sec ond sideband  2 fi component
2
4
6


 mf 
 mf 
 mf 


 2
 2
 2
mf
1







where JN ( mf ) 
n 



 ...............
n ! 1 ! n  1 ! 2 ! n  2 ! 3 ! n  3 !

2




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Solution of Bassel function (chart)
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Solution of Bassel function (graph)
2fi
fC
fi
2fi
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Bandwidth of FM waves
mf = 0.5
sin ce mf 
mf = 2.5
BW
4fi1
d
fi
 then mf  fi  d
BW
10fi2
If d is fixed the mf x fi is constant . For small fi , mf will be large, and more side frequencies
Example
If maximum deviation of the FM wave is fixed at d  20kHz. Determine the BW
required for an intelligence of fi = 10kHz.
mf = d / fi  20kHz/10kHz = 2 …..from table pair of important sideband is 4pairs (J0 to
J4) or 8 side frequencies. BW required for an intelligence of fi = 10kHz is 8x10kHz =
80kHz
Example
If maximum deviation of the FM wave is fixed at d  20kHz. Determine the BW
required if the intelligence is changed to fi = 5kHz.
mf = d / fi  20kHz/5kHz = 4 …..from table pair of important sideband is 7pairs (J0 to J7)
or 14side frequencies. BW required for an intelligence of fi = 5kHz is 14x5kHz = 70kHz
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Carson’s rule
Approximate BW prediction of a FM wave can be made by
Carson’s rule…….
BW = 2 (dmax + fi max )
Example
If maximum deviation of the FM wave is fixed at d  20kHz. Determine the BW
required for an intelligence signal having a higher cutoff frequency at fi max = 10kHz.
BW = 2 (dmax + fi max ) = 2 (20+ 10 )kHz = 60kHz
Note from important side frequency of Bessel function is BW= 80kHz.
( If neglecting sidebands having an amplitude less than 10% of maximum value of
fundamental amplitude component (unmodulated carrier J0 = 1) then there will
be 6 side frequencies (J0 to J3 ) as J4 = 0.03 can be neglected. Then BW =
6x10kHz = 60kHz)
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Example
Instantaneous voltage of the FM wave is e (t) = 2000 sin ( 2p x108 t+ 2 sin 2p x104t )
It is applied to an antenna having an impedance of 50W. Determine:
(a) Carrier frequency fC
(b) Transmitted power PT
(c) mf , (d) fi
(e) BW (by two methods)
(f) Power in the largest and the smallest sidebands predicted by the table of Bessel
function .
(a)
Carrier frequency fC = 108Hz =100MHz
(b) Transmitted power PT = (E
(c)
rms
)2/ R = (2000/1.414 )2/ 50W = 40kW
mf = 2 (from 2 sin p x104t )
(d) Information frequency (from 2 sin p x104t ) fi = (104/2)kHz = 5kHz
(e) (Carson’s rule) d = mf x fi =2x5k=10kHz
then BW = 2 (dmax + fi max ) = 2 (10+ 10 )kHz = 40kHz
Bessel function) table, 4 pair (or 8 side frequencies, J0 to J4 ) then BW =
8x5kHz = 40kHz
(f) (Bessel function) table largest SB2 is 58% of unmodulated carrier J0 = 1,
then PSB2 = (0.58E rms )2/ R = (0.58x2000/1.414 )2/ 50W = 13.5kW
smallest SB4 is 3% of unmodulated carrier J0 = 1,then PSB4 = (0.03E
(0.03x2000/1.414 )2/ 50W = 0.036kW=36W
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rms
)2/ R =
10
Broadcast FM (entertainment)
Channel width
200 kHz
Channel width
200 kHz
Channel width
200 kHz
Carrier1
Carrier2
100MHz
+75kHz
Maximum deviation
d = 75kHz
-75kHz
2x25kHz
guard band
Carrier accuracy 2kHz
Channel width
200 kHz
0.1MHz 0.1MHz
100.1MHz
100.2MHz
Channel width
200 kHz
100.4MHz
d= 75kHz = mf x fi = fixed to a maximum value.
Therefore mf decreases if fi increases
(at high intelligent frequency no. of sideband is less because of small mf )
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Narrowband FM (communications)
Narrowband FM is widely used in communications such as police, aircraft, taxicabs,
whether service, and private industry networks. Often voice transmission whose
highest frequency allowed is fi = 3kHz.
Narrowband FM has a (FCC) BW allocation of 10kHz to 30kHz.
Narrowband FM has a modulation index mf (or a deviation ratio) of (10kHz/3kHz)=3.3
to (30kHz/3kHz)= 10 compared to (75kHz/15kHz) = 5 in Broadcast FM
In FM, the waveform amplitude never varies just frequency. Therefore the
transmitted power must remain constant regardless of level of modulation. It is thus
seen that whatever the energy is contained in the side frequencies has been obtained
from the carrier. No additional energy is added during the modulation process. In
addition amplitude of the carrier in FM depends upon intelligence signal
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Example
(a)
Determine the permissible range in maximum modulation index for commercial
FM that has 30Hz to 15kHz modulating frequencies.
(b) Repeat for a narrowband system that allows a maximum deviation of 10kHz and
100Hz to 3kHz modulating frequencies.
d 75 kHz
d 75 kHz
( a ) mf (max)  
 2500 to mf (min)  
5
fi
30 Hz
fi
15 kHz
d 10 kHz
d 10 kHz
( b ) mf (max)  
 100 to mf (min)  
 3.3
fi 100 Hz
fi
3 kHz
Example
Determine the relative total power of the carrier and side frequencies when mf = 0.25
for a 10kW FM transmission
at mf  0.25 , J0  0.98 , J1  0.12 and for 10 kW transmitting power ,
Pcarrier  ( 0.98 ) 2  10 kW  9.604 kW and Psideband  ( 0.12 ) 2  10 kW  0.144 kW
Ptotal  Pcarrier  Psideand  9.604  0.144  9.892 kW  10 kW
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Noise suppression
FM limiter
and detector
Noise absent at AM detector output
External
noise
AM limiter
and detector
Noise at AM detector output
The most important advantage of FM over AM is the superior noise characteristics.
Static noise is rarely heard on FM although it is quite common in AM.
In FM intelligence is not carried by amplitude changes but instead by frequency
changes. The spikes of external noise picked up during transmission are clipped off by a
limiter circuit and through detector circuits insensitive to amplitude changes.
Unfortunately the noise spike still causes an undesired phase shift of the FM signal,
and this frequency shift cannot be removed.
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FM noise analysis
If FM receiver is tuned to 96MHz, the receiver provide gain only for frequencies near
96MHz. The noise also should be around this frequency since all other frequencies are
greatly attenuated.
Adding the noise to the desired signals will give a resultant signal with a different
phase angle then the desired FM signal alone. Since the noise can cause phase
modulation, it indirectly causes an undesired FM. Value of frequency deviation caused
by phase modulation is : d = f x fi
Where d = frequency deviation, f = phase shift (radians), fi = intelligence frequency
Consider the noise signal to be one-half of the amplitude of the desired signal. S/N = 2
intolerable in AM. But in FM it is not so bad as shown by the calculations below.
Since the noise “N” and signal “S” are at different frequencies (but in the same
range, as dictated by receiver tuned circuits) the noise is a rotating vector with “S” as
reference. The resultant phase “ f ” is maximum when “N” is perpendicular to
resultant vector “R”. Then f = sin-1 (N/S) = sin-1 (1/2) = 30deg = 0.52 rad. and maximum
possible frequency deviation is d = f x fi = 0.52 x 15kHz = 7.5kHzR
N
f
S
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FM S/N is better than AM S/N
In standard broadcast,maximum deviation is 75kHz at full intelligence signal amplitude.
Now the deviation from noise is 7.5kHz then S/N of the FM detector will be
(75kHz/7.5kHz) = 10 = S/N which is better than AM whose S/N = 2
Example
Determine the worst case output S/N for a broadcast FM program that has a
maximum signal frequency of 5kHz. The input is S/N = 2
f  sin -1 (N/S)  sin -1 (1/2)  30deg  0.52 rad  d  f x fi  0.52 x 5kHz  2.5kHz
S
75 kHz
S
( FM out ) 
 30  ( FM in )  2
N
2.5 kHz
N
Example
Determine the worst case output S/N for a narrowband FM communication that has a
maximum deviation of d = 10kHz and maximum signal frequency of 3kHz. The input is
S/N = 3
f  sin -1 (N/S)  sin -1 (1/3)  19.5deg  0.34 rad  d  f x fi  0.34 x 3kHz  1 kHz
S
10 kHz
S
( FM out ) 
 10  ( FM in )  3
N
1 kHz
N
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Direct FM generation
Reactance
Modulator
Varicap Modulator
• Reversed bias
capacitance of the
Varactor Diode D2 is
in parallel with tuned
capacitor C1 . C of D2
change due to change
of reverse bias by
the modulating
signal. Therefore
output freq. will
change due to change
of C of D2
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VCO
Modulator
FM out
Crosby
Modulator
FM
modulating
signal
C1
L1
Modulating
intelligence signal
Varying reverse
bias of Varactor
changes the
Varactor
capacitance
Carrier fC
oscillator
Varactor diode D2
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Reactance Modulator (equivalent capacitance)
C
R
e
e g  i1 R but i1 
R  jXc
iD
i1
eg
e




e R
R  jXc
g e R
id  gme g  m
R  jXc
Ceq= CRgm
Z 
 V
gm  gm0 1  GS
VP

 eg 
e  R  jXc 
jXc
e
1



id
gm  e  R
gm gm  R
If R is chosen to be R  Xc
jXc
1
1
Then Z  
 j

Then Ceq  CRgm
gm  R
wCRgm jwCeq
• When modulating signal is connected to Gate of FET, VGS will change. This will make
gm to change. Then Ceq = CRgm will change.
• If terminal “e” is connected to the LC tuned circuit of the carrier frequency
oscillator, then carrier frequency will change producing FM wave.
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Reactance Modulator (equivalent inductance)
Example
R
Show that the FET reactance
circuit shown will produce an
equivalent reactance of
Leq =(CR/gm)
e g  i1 (  jXC )
C
i1 
If
e
eg
e
R  jXC
id  gm e g 
z 
iD
i1
eg 
Leq=(CR/gm)
e  (  jXC )
R  jXC
 jXC ( gm e )
R  jXC
R  jXC
jwCR
e
e
1




 jXC ( gme )
id
 jXC  gm
gm
gm
R  jXC
jwCR
jwCR
1
CR

then z 
 jwLeq Where Leq 
gm
gm
gm
gm
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FET Reactance Modulator
Reactance
FM
modulator
VDD
RFC
FM
C
modulating
signal
C0
i1
iD
L0
R
DC0
modulating
signal
FET reactance circuit will provide changing equivalent inductance
proportional to modulating signal amplitude. As it is in parallel with
L0C0 tuned oscillator (for carrier frequency), the carrier frequency
will deviate according to dL produced by the FET reactance circuit,
creating an FM wave.
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FM
output
BJT Reactance Modulator
R6
Master
oscillator
R2
FM
C
C0
modulating
signal
• When modulating signal is connected to
Base of BJT, IB will change. This will make
bib to change. If current generator of
BJT having an equivalent in such a way
that bib = gmeg where gm=(bib )/eg = b/ bre
=1/re , then Ceq = CRgm = CR/re will change.
Note that re (=26mV/IC) will change with
IC (= bIB)
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R5
BJT
Ractance
modulator
L0
R
R3
R4
DC0
VCC
C3
modulating
signal
• When terminal “DC0=Ceq”
is connected to the L0C0
tuned circuit of the
carrier frequency
oscillator, then carrier
frequency will change
resulting an FM wave.
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Linear IC ,Voltage Controlled Oscillator (VCO) FM generation)
A voltage controlled oscillator (VCO) produces an output frequency
that is directly proportional to a control voltage level. The circuitry
necessary to produce such an oscillation with a high degree of
linearity between control voltage and frequency was formerly
productive on a discrete component bias. But now low cost
monolithic LIC VCO are available, they make FM generation
extremely simple.
Voltage
Controlled
Oscillator
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
V+
R1
modulating
input
6
Schmitt
trigger
566 VCO
C1
3 
V  VC  V
4



8
Current
source
VC 5

2 V  VC 
f0 
Note
 

R1 C1V
7
Buffer
amplifier
3
Buffer
amplifier
4
1
The figure shown is a 566VCO IC. It provide a high-quality FM generator with the
modulating voltage applied at VC terminal. The FM output can be taken as a square or a
triangular waveform from the IC. Feeding of any of these two outputs into an LC tank
circuit resonant at the center frequency of the VCO ( = carrier frequency of FM)
subsequently provide standard FM signal by flywheel effect.
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CROSBY Modulator
Audio
input
Reactance
modulator
fC =5 MHz
fC =30 MHz
fC =90 MHz
Df = 4.167
Df = 25 kHz
Df = 75 kHz
kHz (max)
(max)
(max)
Primary
Frequency
Frequency
oscillator
Tripler and
Tripler
5MHz
Doublers
Discrimina 2 MHz
tor
Frequency
stabilization
circuit
Crystal
oscillator
14.67 MHz
Frequency
Converter
(Mixer)
88 MHz
Frequency
Tripler and
Doublers
Power
Amplifier
90 MHz
To
Antenna
The mixer shown has the 90MHz carrier and 88MHz crystal oscillator signal as inputs.
The mixer output only accepts the difference component of 2MHz. Which is fed to the
discriminator. A Discriminator is the opposite of a VCO, in that it provides a dc level
output based upon the frequency input. Discriminator output is zero dc if it has an input
of exactly 2MHz which occurs when the transmitter is at precisely 90MHz. Any carrier
drift up or down causes the discriminator output to go positive or negative, resulting in
the appropriate primary oscillator readjustment.
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Frequency Doublers
fC =5 MHz
Df = 4.167
kHz (max)
High-distortion
amplifier (class C)
fC =10 MHz
Df = 8.334
kHz (max)
1
Tuned to f 
C
10MHz
2 p LC
Frequency Multipliers (doublers,triplers, etc.) are high distortion amplifier
which will produce high harmonic content at the output. Then the output is
tuned to the desired frequency multiple of the input frequency.
In the above example shown above, a 5MHz carrier having a frequency
deviation of Df=4.167kHz enters the class C high distortion amplifier. Tank LC
circuit of the amplifier output is tuned to 10MHz.
Note that Df=4.167kHz is also doubled to 8.334kHz. The reason is as follows:
Suppose a 1 MHz deviating +0.1MHz enters the amplifier. The output will double
of 10.1 x2 = 20.2 = a carrier of 2MHz and a double deviation of 0.2MHz.
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Frequency Converters
fC =5 MHz
Df = 0.1MHz
frequency
converter
Mixer
fC =10 MHz
Df = 0.1MHz
1
Tuned to f 
C
10MHz
2 p LC
Oscillator
fo =5 MHz
It is very important to note that in the case of a frequency converter only
the center frequency is changed, not the side frequency.
When 5.1MHz is mixed with 5MHz oscillator, the output will be 5.1+5=10.1MHz
meaning that the deviated frequency remains the same after the frequency
conversion.
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Indirect FM generation
Wideband Armstrong FM
FM
output
Audio
input
400kHz
Crystal
Oscillator
fC =400kHz
RF
Power
amplifier
Preemphasis
Balanced
modulator
90deg
phase
shifter
fC =90.1MHz
fd =75kHz
+
RF
amplifier
x81 first
Freq.
multipliers
fC =400kHz
fd =14.47Hz
fC =32.4MHz
fd =1.172Hz
x64 final
Freq.
multipliers
fC =1.41MHz
fd =1.172Hz
Mixer
f0=33.81MHz
Oscillator
Wideband Armstrong FM system
• Mixer is responsible for achieving the final carrier from the basic crystal oscillator
frequency
• Frequency multipliers are responsible for achieving the final deviation frequency from the
small deviation at the balanced modulator
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TE4201-Communication Electronics
Design of mixer freq. and multiplying factor of freq. multiplier
1. Given: final= 88MHz , 75kHz, final multiplier= x64, before first xlier is carrier
fC = 400k and fd=14.47Hz
2. then before multiplying fC =(88MHz/64) =1.375MHz, fd = (75kHz/64)
=1.172kHz
3. before first xlier is fd=14.47Hz
then first xlier is (1.172k/14.47)=81
4. before first xlier carrier fC =400k
then after first xlier 400kx81=32.4MHz, then f0 =32.4+1.375=33.81MHz
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HW:
A Wideband Armstrong FM system is to be transmitted at 88MHz. with maximum deviation
of 75kHz. A 400kHz crystal oscillator is used with a small deviation of 15Hz.at balanced
modulator output. The final multiplier has a multiplication factor of 64.
(a) Design the freq. of Oscillator and the freq. multiplying factor of the multiplier before the
mixer.
Note that a mixer will not change the deviation while the carrier is changed and that a
multiplier will change both carrier and deviation freq.
(b) If the crystal available is 410kHz, what changes will you make to the above FM system?
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