FM Transmission - MavDISK

5.0 Frequency Modulation : Transmission
5.1 Angle Modulation
Phase modulation (PM)
Frequency Modulation (FM)
1936 Major E. H. Armstrong demonstrated.
1939 first broadcast
5.2 A Simple FM Generator
Microphone capacitor used to vary the frequency.
Amplitude of voice input sound waves moves the microphone plates and changes the capacitance.
This changes the frequency.
The frequency of the sound waves determine the rate of change of frequency.
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fout
= fc +k*ei
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Example 5.1
25mV, 400Hz
deviation constant 750Hz/10mV
a) Frequency deviation generation by an input level of 25mV.
750Hz*25mV/10mV = 1.875 kHz
750Hz*-25mV/10mV = -1.875 kHz
Total frequency deviation = +- 1.875 kHz
Frequency deviation is 1.875 kHz
b) Rate of deviation = +- 2.25kHz at a rate of 400Hz.
An FM system is transmitting the signal, v = 1000sin( 556x106t + 5sin(94x103t) )
into a 50 ohm antenna. Determine,
a) Carrier frequency 88.5 MHz
b) Intelligence frequency 14.96 kHz
c) Transmitted power 10002/2*50 = 10 kW
e) Modulation index 5
d) Frequency deviation 5*14.96 = 74.8 kHz
In an FM transmitter, the output is changing between 89.003 and 88.997 MHz 1500 times a second. The
intelligence signal is 3 volts peak. What is the deviation constant?
Deviation constant = 3kHz/3V = 1kHz/V
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An FM system is transmitting the signal, v = 1000sin( 556x106t + 5sin(94x103t) ) into a 50 ohm antenna.
Determine the frequency deviation in kHz.
5*14.96 = 74.8 kHz

5.3 FM Analysis
Phase Modulation
e = Asin(ωct + mpsinωit)
modulation index for PM = mp
maximum phase shift caused by the intelligence signal
Frequency Modulation
e = Asin(ωct + mfsinωit)
modulation index for FM = mf = δ/fi
δ = maximum frequency shift caused by the intelligence signal
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FM mathematical solution - infinite number of frequency components space at multiples of the intelligence
frequency above & below fc. Bessel functions
e(t) = A*fc(t) fc(t) are the frequency components
fc(t) = J0(mf)*cosωct
– J1(mf)[cos(wc – wi)t – cos(wc + wi)t]
+ J2(mf)[cos(wc – 2wi)t – cos(wc + 2wi)t]
– J3(mf)[cos(wc – 3wi)t – cos(wc + 3wi)t]
+ J4(mf)[cos(wc – 4wi)t – cos(wc + 4wi)t]
– J5(mf)[cos(wc – 5wi)t – cos(wc + 5wi)t]………………..

Bessel Functions
EE3158 – Lecture 10 Fundamentals of Communications Slide 21
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An FM system has a maximum deviation of 12 kHz. The maximum signal frequency is 3 kHz. Calculate
the bandwidth required to transmit the FM signal by usng the Bessel function table below.
21*2 = 42 kHz
Carson’s rule
BW is approx = 2(δmax + fimax)
2% of the power is outside of the BW
Using Carson’s rule?
2(12 + 3) = 30 kHz

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An FM system has a maximum deviation of 60 kHz. The maximum signal frequency is 12 kHz. Calculate
the bandwidth required to transmit the FM signal by using the Bessel function table.
mod index = 5
BW = 2*8*12 = 192kHz
A 10kW FM transmitter has a modulation index of 5. Using the Bessel function table determine the power
in the J3 sidebands.
0.362*10kW = 1296W
An FM signal has the following characteristics,
modulation index 2.5
signal frequency 1200Hz
frequency deviation 3000Hz
Determine,
a) The bandwidth using the table.
BW = 2 x 6 x 1.2kHz = 14.4kHz
b) The bandwidth using Carson’s rule.
BW = 2(3000 + 1200) = 8.4kHz

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A 10 kW FM transmitter has a modulation index of 2.5. Using the Bessel function table determine the
power in,
a) Carrier J0
(-0.05)2(10kW) = 25 W
b) Sidebands J5
(0.02)2(10kW) = 4 W
f the modulation index is 5 for a 10 KW transmitter determine the power in the,
a) Carrier
Pc = 0.18*0.18*10kW = 324W
b) Highest significant sideband frequency
P8 = 0.02*0.02*10kW = 4W

PSPICE Simulation
V1 = 0
V2 = 6.28
TD = 0
TR = 0.99m
TF = .01m
PW = 1m
PER = 1m
600mV
400mV
200mV
V3
0
R1
1k
VOFF = 0
VAMPL = 2.0
FREQ = 100
V4
0V
0Hz 0.2KHz
V(SIN1:OUT)
0.4KHz 0.6KHz 0.8KHz
Frequency
1.0KHz 1.2KHz 1.4KHz
R2
1k
SIN
R3
1k
V
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LabVIEW Simulation
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V1 = 0
V2 = 6.28
TD = 0
TR = 0.99m
TF = .01m
PW = 1m
PER = 1m
600mV
400mV
200mV
V3
0
R1
1k
VOFF = 0
VAMPL = 2
FREQ = 100
0V
0Hz 0.2KHz
V(SIN1:OUT)
0.4KHz 0.6KHz 0.8KHz
Frequency
1.0KHz 1.2KHz 1.4KHz
V4
R2
1k
SIN
R3
1k
V
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Example 5.3
Bandwidth required?? Using just the Table of significant components.
FM signal fi = 10 kHz & δ = 20 kHz
mf = δ/fi
= 20 kHz/10 kHz
= 2
Looking at Table significant components are,
Jo, J1, J2, J3, J4
BW = 2( 4*fi)
= 2(40 kHz
= 80 kHz
Compared with Carson’s rule (98% of the power) which is
BW = 2(δ + fi)
= 2(20 + 10)
= 60 kHz
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An FM signal voltage applied to a 50-ohm load is, v(t) = 3000sin(2π*91.1(10 6 ) + 4sin(2000π))
What is the,
a) Signal frequency
1000 Hz
b) Frequency deviation
4000 Hz
c) Bandwidth using Carson’s rule.
BW = 2(4000 + 1000) = 10,000 Hz
Example 5.4
Repeat with fi changed to 5 kHz
mf = 20/5 = 4
Significant J’s are throught 7
BW = 2(7*5)
= 2*35
= 70 kHz
vs. Carson’s rule
BW = 2(20 + 5)
= 50 kHz
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An FM signal has the following characteristics,
modulation index 2.4
signal frequency 1000Hz
frequency deviation 4000Hz
Determine,
a) The bandwidth using the table below.
BW = 2*5*1000 = 10,000Hz
b) The bandwidth using Carson’s rule.
BW = 2(4000 + 1000) = 10,000Hz
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Example 5.5
FM sinal 2000sin(2π*10 8 t + 2sinπ*10 4 )t) applied to a 50ohm antenna
a) Carrier frequency, fc
fc = 10 8 Hz
b) Transmitted power
P = (2000*.707) 2
50
c) mf
mf = 2
d) Intelligence frequency
fi = π*10 4/2 2π = 5kHz
e) BW
mf = δ/fi
2 = δ/5kHz
δ = 10kHz
BW approx = 2(10kHz + 5kHz) = 30kHz
f) Power in the largest and smallest sidebands from Table 5.2.
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The modulation index is 0.5 for a 10 KW transmitter.
a) Determine the power in the carrier and all significant sideband frequencies
Pc = 0.942(10,000) = 8.836W
P1 = 2*0.242(10,000) = 1152W
P2 = 2*0.032(10,000) = 18W
b) What percentage of the 10KW is transmitted on frequencies where the power output level is not
significant?
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Example 5.6
a) Range of maximum modulation index for commercial FM where range of frequencies is
30 Hz to 15 kHz
mf = 75 kHz/100 Hz = 2500
mf = 75 kHz/15 kHz = 5
b) Repeat for a narrowband system where max deviation is 1 kHz.where range of requencies is
100 Hz to 2 kHz
mf = 1 kHz/100 Hz = 10
mf = 1 kHz/2 kHz = 0.5
c) Determine the DR for system in part b.
DR = max deviation/max signal freq
= 1 kHz/2 kHz = 0.5
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Example 5.7
Determine the relative total power of the carrier and side frequencies when
mf = 0.25 for a 10 kW FM transmitter
for mf = 0.25
J0 = 0.98
J1 = 0.12
and none other’s are significant
power is proportional to voltage squared so
0.98 2 * 10 kW = 9.604 kW
0.12 2 * 10 kW = 144 W
Total = 9.892 kW
Total power must be constant
Side frequency power is obtained from the carrier.
No additional energy is added.
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Zero-Carrier Amplitude
Broadcast FM
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Deviation Ratio (DR)
DR = maximum possible frequency deviation
maximum input frequency
DR (bradcast FM radio) = 75 kHz/15 kHz = 5
DR (TV NTSC) = 25 kHz/15 kHz = 1.67
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5.4 Noise Suppression
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Phase shift due to noise, φ
Frequency deviation
δ = φ * fi intelligence signal frequency
Looking at the INPUT signals and assume worst case, Right Angles to each other,
sin φ = N / S where N = noise and S is desired signal
φ = sin -1 N/S
If the S/N is 2:1 and Fi maximum of 15 kHz, then
φ = sin -1 ½ = 30 degrees = 0.52 radian
δ = φ * fi = 0.52*15kHz = 7.5kHz
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The output deviation at maximum signal frequency and max volume is 75kHz for broadcast FM (FCC reg).
Thus the output deviation is 7.5kHz / 75kHz = 0.1 and the output Signal / Noise ratio is 10:1.
This is compared to AM where a 2:1 ratio at the input is about that at the output.

FM Noise Analysis
Capture Effect
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Preemphasis
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Emphasis Circuits
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Dolby Dynamic Preemphsis
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5.5 Direct FM Generation
Varactor Diode
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LIC VCO FM Generation
Philips Semiconductors NE/SE566 VCO
Crosby Modulator (direct generation with AFC)
Direct FM modulation
Frequency multipliers
Feedback to keep the carrier at center frequency
Reactance modulator
Restricted to less than +/- 1/1000 change in center frequency for linear operation
5 MHz center frequency, +/- 4.167 kHz deviation
Multiply by 18 to get,
90 MHz center frequency, +/- 75 kHz deviation
Discriminator feedback dc level to the reactance modulator (maintains 2 MHz difference between
transmitted carrier and a crystal oscillator frequency)
Crystal osc set at 88 MHz
Difference from 2 MHz produces a dc voltage fed back to reactance modulator.
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5.6 Indirect FM Generation
Armstrong type (limited frequency deviation)
Modulation of a stable crystal oscillator
JFET biased in ohmic region
Voltage-controlled resistance
Reactance modulator. Simple circuit below, but not enough deviation.
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Wideband deviation using Armstrong FM system
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In an wideband Armstrong FM transmitter,
a) A carrier at 400 kHz with a deviation of 14.47 Hz is input to a x81 frequency multiplier. What is the
new carrier frequency and deviation at the output of the multiplier?
32400 kHZ and 1172 Hz
b) The output of the multiplier is fed to a mixer. The other input to the mixer comes from a 33.81 MHz
oscillator. What is the new carrier frequency and deviation at the output of the mixer?
1410 kHz and 1172 Hz

5.7 Phase-Locked-Loop FM Transmitter (Chapter 16.6 in Boylestad)
A narrow band system (200 Hz),
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5.8 Stereo FM
Input to modulator
Two separate 30 Hz to 15 kHz signals
L + R 30 Hz-15 kHz
L – R modulated with 38 kHz to 23-53 kHz
pilot carrier at 19 kHz
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Stereo is more suseptable to noise than mono
L-R is weaker
L-R is at higher frequency
S/N is 20 dB less than mono
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Stereo received on mono receiver is only 1 dB worse S/N than mono received. This is due to the 19 kHz
pilot.

5.9 FM Transmissions
Table of frequencies
Noncommercial broadcast 88 to 90 MHz
Commercial broadcast 200 kHz 90 to 108 MHz
Television audio 50 kHz 54 to 88 MHz, 174 to 216 MHz,
470 to 806 MHz
Narrowband public service 108 to 174 MHz, >806 MHz
Narrowband amateur many
Describe range of frequencies of operation (min/max) and bandwidth requirements for commercial
broadcast of,
a) AM 530-1700 kHz, 10 kHz
b) FM 90-108 MHz, 200 kHz
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5.10 Troubleshooting
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