5_frequency modulation and transmission

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5.0 Frequency Modulation : Transmission

5.1 Angle ModulationPhase modulation (PM)

Frequency Modulation (FM)

1936 Major E. H. Armstrong demonstrated.

1939 first broadcast

5.2 A Simple FM GeneratorMicrophone 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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out = fc +k*eif


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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


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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)*cosct

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]..


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EE3158 Lecture 10 Fundamentals of Communications Slide 21

Bessel Func t ions


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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

Carsons rule

BW is approx = 2(max + fimax)

2% of the power is outside of the BW

Using Carsons 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 = 5BW = 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 Carsons 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 frequencyP8 = 0.02*0.02*10kW = 4W


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LabVIEW Simulation


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R2

1k

V4

FREQ = 100

VAMPL = 2

VOFF = 0R1

1k

V

V3

TD = 0

TF = .01m

PW = 1m

PER = 1m

V1 = 0

TR = 0.99m

V2 = 6.28

0

SIN

R3

1k

Frequency

0Hz 0.2KHz 0.4KHz 0.6KHz 0.8KHz 1.0KHz 1.2KHz 1.4KHz

V(SIN1:OUT)

0V

200mV

400mV

600mV


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Example 5.3

Bandwidth required?? Using just the Table of significant components.

FM signal fi = 10 kHz & = 20 kHz

mf = /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 Carsons rule (98% of the power) which is

BW = 2( + fi)

= 2(20 + 10)

= 60 kHz


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An FM signal voltage applied to a 50-ohm load is, v(t) = 3000sin(2*91.1(106) + 4sin(2000))

What is the,

a) Signal frequency

1000 Hz

b) Frequency deviation

4000 Hz

c) Bandwidth using Carsons rule.

BW = 2(4000 + 1000) = 10,000 Hz

Example 5.4

Repeat with fi changed to 5 kHz

mf = 20/5 = 4

Significant Js are throught 7

BW = 2(7*5)

= 2*35

= 70 kHz

vs. Carsons ruleBW = 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 Carsons rule.

BW = 2(4000 + 1000) = 10,000Hz


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Example 5.5

FM sinal 2000sin(2*108t + 2sin*10

4)t) applied to a 50ohm antenna

a) Carrier frequency, fc

fc = 108 Hz

b) Transmitted power

P = (2000*.707)2

50

c) mf

mf = 2

d) Intelligence frequency

fi = *104/2

2 = 5kHz

e) BW

mf = /fi

2 = /5kHz

= 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 others are significant

power is proportional to voltage squared so

0.98

2

* 10 kW = 9.604 kW

0.122

* 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,

Frequency deviation

= * fi intelligence signal frequency

Looking at the INPUT signals and assume worst case, Right Angles to each other,

sin = N / S where N = noise and S is desired signal

= sin-1

N/S

If the S/N is 2:1 and Fi maximum of 15 kHz, then

= sin-1

= 30 degrees = 0.52 radian

= * fi = 0.52*15kHz = 7.5kHz

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.


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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 oscillatorJFET 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

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5.7 Phase-Locked-Loop FM Transmitter (Chapter 16.6 in Boylestad)

A narrow band system (200 Hz),

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5 8 St FM


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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 s septable to noise than mono


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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

Stereo received on mono receiver is only 1 dB worse S/N than mono received. This is due to the 19 kHz

pilot.

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5 9 FM Transmissions


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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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5.10 Troubleshooting


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5_frequency modulation and transmission

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