entc 4310 - electronics-communications lab report course projects/0000entc4310 lab... · optical...

1

ENTC 4310 - Electronics-Communications

Lab Report

Xiqiao Wang

East Tennessee State University

December 10, 2012

Instructor: Prof. Paul J. Sims

2

Contents

Lab 1 FM, AM and FSK……………………………….………………………………………3

Lab 2 Antenna Characteristics………………………………………………….…………….28

Lab 3 Phase Lock Loops Basics ...………………………………………………….…..........44

Lab 4 FM Generation…………………………………………………………………………60

Lab 5 FM Demodulation Using Phase-Locked Loop……………………………….………..68

Lab 6 AM Broadcast Transmitter and High performance point Boardcase amplifier……..…78

Optical Fiber Lab 1 Setting Up A Fiber Optics Analog Link………………………….…….101

Optical Fiber Lab 2 Setting Up a Fiber Optic Digital Link………………………….……….123

Fiber Optics Lab 5 Time division multiplexing of signals……………………….……….....146

Fiber Optics Lab 6 Framing in time division multiplexing……………………………….….163

Assignment 1………………………………………………………………………………....193

Assignment 2………………………………………………………………………….………198

3

ENTC4310 Lab 1 FM, AM and FSK

Xiqiao Wang

Lab Assignments

1. Use your Ramsey RF amplifier to amplify 1, 5, 10, 15, 20, 30, 40, 60, 100 MHz

2. Record input and output waveforms and put in lab book. (You will have to use the DSO

units for this.)

3. Make a spectrum of input and output waveforms at 1, 10, 20, MHz with FM, AM and

FSK.

4. Record spectrum and insert in lab book.

Equipment

Function signal generator, digital oscilloscope, AM-FM antenna amplifier.

Instructions

1. AM

In radio communication, a continuous wave radio-frequency signal (a sinusoidal carrier

wave) has its amplitude modulated by an audio waveform before transmission. The audio

waveform modifies the amplitude of the carrier wave and determines the envelope of the

waveform. In the frequency domain, amplitude modulation produces a signal with power

concentrated at the carrier frequency and two adjacent sidebands. Each sideband is equal

in bandwidth to that of the modulating signal, and is a mirror image of the other.

Amplitude modulation resulting in two sidebands and a carrier is called "double-sideband

amplitude modulation" (DSB-AM). Amplitude modulation is inefficient in power usage;

at least two-thirds of the power is concentrated in the carrier signal, which carries no

useful information (beyond the fact that a signal is present).

To increase transmitter efficiency, the carrier may be suppressed. This produces a

reduced-carrier transmission, or DSB "double-sideband suppressed-carrier" (DSB-SC)

signal. A suppressed-carrier AM signal is three times more power-efficient than AM. If

the carrier is only partially suppressed, a double-sideband reduced-carrier (DSBRC)

4

signal results. For reception, a local oscillator will typically restore the suppressed carrier

so the signal can be demodulated with a product detector.

Improved bandwidth efficiency is achieved at the expense of increased transmitter and

receiver complexity by completely suppressing both the carrier and one of the sidebands.

This is single-sideband modulation, widely used in amateur radio and other

communications applications. A simple form of AM, often used for digital

communications, is on-off keying: a type of amplitude-shift keying in which binary data

is represented by the presence or absence of a carrier. This is used by radio amateurs to

transmit Morse code and is known as continuous wave (CW) operation.

2. FM

Frequency modulation (FM) conveys information over a carrier wave by varying its

instantaneous frequency. This contrasts with amplitude modulation, in which the

amplitude of the carrier is varied while its frequency remains constant. In analog

applications, the difference between the instantaneous and the base frequency of the

carrier is directly proportional to the instantaneous value of the input-signal amplitude.

Digital data can be sent by shifting the carrier's frequency among a range of settings, a

technique known as frequency-shift keying (FSK).

3. FSK (Nicolls, 2010)

Frequency-shift keying (FSK) is a frequency modulation scheme in which digital

information is transmitted through discrete frequency changes of a carrier wave. The

simplest FSK is binary FSK (BFSK). BFSK uses a pair of discrete frequencies to transmit

binary (0s and 1s) information. With this scheme, the "1" is called the mark frequency

and the "0" is called the space frequency. The time domain of an FSK modulated carrier

is illustrated in the figures below.

5

The signal s transmitted for marks (binary ones) and spaces (binary zeros) are

This is called a discontinuous phase FSK system, because the phase of the signal s is

discontinuous at the switching times. A signal of this form can be generated by the

following system,

If the bit intervals and the phases of the signals can be determined (usually by the use of

phase-lock-loop), then the signal can be decoded by two separated filters as shown below,

6

The first filter is matched to the signal s1(t) and the second to s2(t). Under the assumption

that the signals are mutually orthogonal, the output of one of the matched filters will be

amplitude of the signal and the other will be zero. Decoding of the bandpass signal can be

therefore achieved by subtracting the outputs of the two filters and comparing the results

to a threshold voltage, which is set above the maximum allowable noise level for a

specific false rate.

4. Notice that always assign the input signal as the digital oscilloscope’s triggering signal.

Procesures

1. Built AM-FM antenna amplifier

In electronics, an antenna amplifier, also called antenna preamplifier, antenna

preamp or antenna booster, is a device that amplifies an antenna signal, usually into an

output with the same impedance as the input impedance. Typically 75 Ohm for coaxial

cable and 300 Ohm for twin lead cable.

We built valleman-kit K2622 AM-FM antenna amplifier in this lab. The amplifier

use supply voltage of 12-15 VDC, which do not need to be stabilized. From its official

manual, It has an gain of about 22 dB within 10 MHz to 150 MHz bandwidth, the input

impedance and output impedance are both 50-75 Ohm, which indicates that to achieve

the best power transmit property, we need to connect the amplifier though coaxial-cable.

Since this amplifier has to deal with high frequency signals, we should pay attention

during construction to keep the connection of the components as short as possible and

avoid using too much solder. The circuit diagram is shown as below,

7

Figure. The circuit diagram of the AM-FM antenna amplifier.

In the amplifier circuit, input and output stages are coupled with capacitors to block the

DC components of the input and output signal. R1 and R2 are tuned to set the proper bias

voltage for the base of the transistor. The impedance calculated from C4, R4 and R3’s

decides the gain of the transistor. Capacitor C3 is used to eliminate the noise component

from the amplified signal. Capacitor C5 is used to eliminate the noise components from

the DC supply voltage.

2. Plain Wave Amplification

In this step, we used plain sine waves at the listed frequencies to test the Ramsey RF

amplifier’s frequency response characteristics. If the frequency response is linear within

our test frequency range, the gain at each tested frequency should be equal. Otherwise,

the gains will differ from one another. As shown below are the input and output

waveforms. Channel 2 of the blue waves are input signals, and Channel 1 of the yellow

signals are output signals.

8

1 MHz 5MHz

10 MHz 15 MHz

20 MHz 30 MHz

9

40 MHz 60 MHz

100 MHz

The magnitudes of input and output signals and the corresponding numerical and decibel

gains are listed in the table below. The distortion of output waveforms compared with

standard sine wave should be attributed to the overloading effect at the RF amplifier’s

output stage, which is due to the big gain between 5 MHz and 40 MHz. The gain in DOS

unit are calculated as

And the log frequency is calculated as

10

3. AM spectrum measurement

freq (MHz) freq (log) Input (Vp, mV) output (Vp, mV) numeric gain gain in dB

1 0 100 50 0.5 -6.0205999

5 0.69897 100 736 7.36 17.337556

10 1 100 1500 15 23.521825

15 1.176091 100 1500 15 23.521825

20 1.30103 100 1500 15 23.521825

30 1.477121 100 1250 12.5 21.9382

40 1.60206 100 950 9.5 19.554472

60 1.778151 100 700 7 16.901961

100 2 56 100 1.785714286 5.0362395

-10

-5

0

5

10

15

20

25

30

0 20 40 60 80 100 120

Gai

n (

dB

)

Freq (MHz)

Gain vs Linear Freq

gain in dB

-10

-5

0

5

10

15

20

25

30

0 0.5 1 1.5 2 2.5

Gai

n (

dB

)

Log Freq (MHz)

Gain vs Log Freq

11

For 1 MHz, 10 MHz and 20 MHz carrier frequencies, we use Sine wave as the carrier

waveform. AM depths are all set to be 100%, and the AM frequencies for these three

cases are set to be 1 KHz. In order to avoid overloading, at 10 MHz and 20 MHz, we use

smaller input carrier amplitudes then the one used at 1 MHz. For 1 MHz, the signal

generator’s carrier amplitude is set as 100 mV. For 10 MHz and 20 MHz, it is set at 30

mV. All FFT analyses use Hanning Window.

3.1.AM-1 MHz

The following figures show the same AM waveforms with two different time

intervals. The first figure sets 100µs each interval, thus 10 intervals represent one AM

modulation cycle. The second figure sets 500µs each interval, thus two intervals

represent one AM modulation cycle. Two horizontal cursers indicate amplitudes of

input and output signals at modulation peaks, which equals to the AM carrier

magnitude. The blue input has a magnitude of 110 mV, and the output 54 mV. Gain

should be 0.5 in numerical scale, and -6 dB in decibel scale. This matches the

frequency response fairly well.

The following four figures show FFT analysis about the AM carrier frequency, 1

MHz. The peak is centered at 1 MHz. However, the primary frequency, 1 KHz, at

both sides stays too close to the carrier frequency to be distinguished by our current

oscilloscope. Also, since the amplitude is modulated, the central peak rises and falls.

So we have to use persist display function to identify the highest peak amplitude.

12

AM 1MHz//1KHz input FFT

AM 1MHz//1KHz output FFT

3.2.AM-10 MHz

At 10 MHz, cursors indicate the input amplitude to be 26.4 mV, and the output

amplitude to be 480 mV. The numerical gain is about 18, i.e. 25.19 dB. This matches

our frequency response predictions. The FFT analysis shows input and output

frequency components around 10 MHz. The magnitudes are obtained from 5 sec

persistent display function. For example, the output peak heights -11.7 dB, which is

about 260 mV. However, this is pretty far from the magnitude of 480 mV measured

from time-domain.

13



3.3.AM-20 MHz

Below is the screen shot when the AM carrier frequency is 20 MHz. With time

interval of 500 µs, we can see that two time interval represent one modulation cycle.

From time domain magnitude cursors, it is shown that the input carrier is 13.6 mV,

and the output carrier is 664 mV. Gain equals 7.94, i.e. about 33.8 dB. From the

persist display, it is indicated that the input signal has a magnitude of -39.3 dB at 20

14

MHz, or equivalently, 108 mV. Still, this is quit far from the time domain measured

value, 66.5 mV.



15

4. FM spectrum measurement

Here we use 20 mV as the carrier’s amplitude for all 1MHz, 10MHz and 20MHz cases.

The FM frequencies are set to be 1 KHz, and the FM deviation is set to be 10 KHz. The

relation between FM frequency and FM deviation frequency is as follows: The FM

frequency characterizes the FM modulation signal; when modulation signal is 0 Hz, the

modulated frequency will deviate 5 KHz negatively from the central frequency; when the

modulation signal is 500 Hz, the modulated signal will stay right at the carrier’s

frequency; while the modulation signal reaches 1 KHz, the modulated frequency will

deviate 5 KHz positively from the carrier’s frequency.

4.1.FM-1MHz

In time domain, the time interval is set to be 1 µs, which is the same with the carrier’s

period. The input magnitude is 33.6 mV, and the output 13.6 mV. Numerical gain is

0.405, i.e. -7.86 dB. This is close to the frequency response predicted value of -6 dB.

It is worth noting that, from FFT analysis, we can clearly observe the peak component

swings back and forth between the 990 KHz and 1.01 MHz, as it is shown below for

both input and output signals. However, as the peak swings, its amplitude stays

almost constant. For input signal, the peak amplitude is -36.1 dB, i.e. 15.6 mV. And

the output amplitude is -43.7 dB, this is about -7.6 dB gain. Here, the frequency

domain and time domain give the same result about magnitudes and gain.

16

FM 1MHz//10KHz input FFT

The peak component swings back and forth between the 990 KHz and 1.01 MHz.

17

FM 1MHz//10KHz output FFT

The peak component swings back and forth between the 990 KHz and 1.01 MHz.

4.2.FM-10MHz

The following time-domain waveform demonstrates clearly the varying of frequency

as the carrier is modulated. The FFT peak still swings between 10MHz ± 10KHz. But

our oscilloscope’s resolution is not high enough to show this motion in detail. The

input peak has an amplitude of -37.3 dB, and the output -14.2 dB. Thus the gain is

22.9 dB. This matches our frequency response prediction in the first step.

18



4.3.FM-20MHz

19

The frequency modulation is clearly shown in time-domain waveforms. The input

signal has a peak of amplitude -45.3 dB, and output -11.3 dB. This indicates a 34 dB

gain, which is not quit close to the 23.5 dB gain in first step.



5. FSK spectrum measurement

First we need to be clear about some of the terminology used in the signal generator in its

FSK function. The carrier frequency stands for the frequency representing the ones in the

20

a binary digital signal. The Hop Freq represent the frequency at zeros in the binary digital

signal. FSK rate is the date rate in bit/sec of the binary digital signal. For all the three

cases with different carrier frequencies, namely 1MHz, 10MHz and 20MHz, we adopt the

same hop frequency and FSK rate. That is

Hop Freq = 1 KHz

FSK rate = 500 Hz

5.1.FSK carrier frequency – 1MHz

In this case, the carrier frequency is set to be 1MHz and the generated FSK signal’s

amplitude is set to be 500 mVpp, which is relatively large signal amplitude in order to

compensate for the antenna amplifier’s low gain level at 1 MHz.

Since the carrier frequency is modulated by the FSK rate at 500 Hz rate, in the FFT

analysis of the input signal, the amplitude at the carrier frequency vibrate between

500 mVpp and 0 mVpp. Below is the FFT analysis using Hanning window on the

input FSk signal. We can see that there is a peak at 1 MHz, and it is the carrier

frequency, and there are two frequency components at 500 Hz and 1 kHz

respectively .

Figure. FFT analysis of the input FSK signal

As expected, the 1 MHz and 1 KHz frequency components vary between 0 mVpp and

500 mVpp at a rate of 500 Hz. The Figure below shown the amplitude interval as

between -15.4 dB and -24.2 dB, equivalently, 178 mV and 63 mV. The delta value is

8.8 dB and it is approximately 500 mV.

21

Figure. The amplitude interval of the input FSK signal at the carrier frequency.

The figure below shown the output signal component at 1 MHz. As expected, the

amplitude varies within interval -22.6 dB to -32.6 dB. The corresponding values are

79mV and 31 mV. The frequency response of the antenna amplifier at 1 MHz is -6

dB, which is about half the amplitude of the input signal. The data obtained here

confirm this prediction. It is worth notice that the output signal contains harmonic

components at each integer MHz frequency.

Figure. The output FSK signal at 1 MHz and its harmonic components.

22

The output signal components at 500 Hz and 1000 Hz are shown below, however, the

amplitudes for these two low frequencies are extremely small due to the limited

bandwidth of the antenna amplifier.

The input and output FSK signals in time domain is shown below, with the blue line

representing the input signal and the yellow line the output signal. The display is

presented with a persistent of 1 second. In the input signal, both the hop frequency

and carrier frequency are clearly captured. However, in the output signal, only the

carrier frequency is present. The hop frequency decayed to negligible level since it is

far off the pass band of our antenna amplifier. Besides, the carrier component shows a

saturation property at the upper level of the amplitude, this should be due to that the

equilibrium point of the transistor is not well balanced at 1 MHz. If the digital

oscilloscope is coupled by HF Reflection approach, then the carrier components in

both input and output signals disappear.

Figure. The input and output FSK signals in time domain

23

Figure. The input and output signals in time domain coupled by HF Reflection.

5.2.FSK carrier frequency – 10 MHz

Since the antenna amplifier has a big gain between 10 MHz to 150 MHz, the

amplitude of the input signal is chosen to be 30 mVpp in order to avoid overloading

the amplifier’s output stage.

Below is the FFT analysis of the input signal at 500 Hz, 1kHz and 10 MHz, among

which the 10 MHz FTT analysis is displayed with an infinite persistence.

Figure. The input signal in frequency domain.

The two figures below demonstrate the output signal in frequency domain. The 1000

Hz component in the output signal is clearly eliminated due to the amplifier’s decay.

The 500 Hz component is still present due to the bit rate. The amplitude of the output

signal is about 500 mVp. Thus the voltage gain at 10 MHz in this case is 500/30 = 18,

equivalently, 20*log(18)=25.1dB. This agrees with the bode-plot in section 1.

24

Figure. The output FSK signal in frequency domain.

Figure. The output FSK signal in time domain.

5.3.FSK carrier frequency – 20 MHz.

Here we also choose 30 mVpp as the amplitude of the input FSk signal. The input

signal in frequency domain is shown below. The peak amplitudes of the 20 MHz

carrier component and 1 kHz hop freq component are both about -33 dB, which is

22.3 mV.

25

Figure. The FFT analysis of input FSK signal with carrier at 20MHz.

Below is the output FSK signal in frequency domain. As expected, the 1 KHz

component is now eliminated. The amplitude at 20 MHz component is about -10 dB,

or equivalently, 316 mV. Again, this gain matches the prediction in the bode-plot.

Figure. The output FSK signal in frequency domain with carrier 20 MHz.

26

Figure. The output and input FSK signal in time domain with carrier 20 MHz.

27

Bibliography Nicolls, F. (2010, 5). EEE482F: Telecommunications (part 2). Retrieved 10 15, 2012, from Fred Nicolls:

http://www.dip.ee.uct.ac.za/~nicolls/lectures/eee482f/13_fsk_2up.pdf

28

ENTC4310 Lab 2 Antenna Characteristics

Xiqiao Wang

Objectives

Upon completion of this experiment, we are supposed to be able to

1. Construct a simple RF voltmeter or ammeter

2. Identify loops and nodes of a standing wave

3. Determine the resonant frequency of a half-wave dipole.

Equipment and components

RF signal generator, NTE109 Germanium Diode, 22-pF capacitor, 5-ft copper wire, wood board,

2 nails, 1-in plastic rod, 1-ft dry string, pin, 26 in of 300-ohm twine-lead transmission line

Introduction

Here I demonstrate some of the important concepts of antenna. Much of the following

information is quoted from the website http://www.antenna-theory.com/ by (Bevelacqua, 2005) .

1. Antenna impedance

Impedance relates the voltage and current at the input to the antenna. The real part

of the antenna impedance represents power that is either radiated away or absorbed

within the antenna. The imaginary part of the impedance represents power that is stored

in the near field of the antenna. This is non-radiated power. An antenna with a real input

impedance (zero imaginary part) is said to be resonant. Note that the impedance of an

antenna will vary with frequency.

1.1. Low frequency

In low frequency, the wavelength is very long. Usually the transmission line that

connects the transmitter or receiver to the antenna is short relative to a wavelength.

The impedance of the transmission line can be negligible. In other words, the power

loss due to the emission of the transmission line is negligible. Then the model of an

antenna connected to a voltage source can be simplified as follows,

29

Where ZS and ZA are impedances of the source and antenna. From circuit theory, we

know that P=I*V. The power that is delivered to the antenna is:

For maximum power to be transferred from the generator to the antenna, the ideal

value for the antenna impedance is given by:

1.2. High Frequency

In low-frequency circuit theory, the wires that connect things don't matter. Once

the wires become a significant fraction of a wavelength, they make things very

different. In this situation, the impedance of the transmission line cannot be neglected.

In fact, the impedance of a transmission line of length L and characteristic impedance

Z0 which is hooked to an antenna with impedance ZA can be written in forms of the

following formula,

30

Transmission line theory proves that if the antenna is matched to the transmission

line (ZA=ZO), then the input impedance does not depend on the length of the

transmission line. This theory greatly simplifies the problem. As we can see from the

formula above, when Z0 equals ZA, then the input impedance of the transmission line

with antenna is just the characteristic impedance of the transmission line.

If the antenna is not matched, the input impedance will vary widely with the

length of the transmission line. And if the input impedance isn't well matched to the

source impedance, not very much power will be delivered to the antenna. This power

ends up being reflected back to the generator, which can be a problem in itself

(especially if high power is transmitted). This loss of power is known as impedance

mismatch.

So, a good antenna design for high frequency application should match the

impedance of the antenna and the transmission line and the receiver or transmitter. In

general, impedance matching is very important in RF/microwave circuit design. It is

relatively simple at a single frequency, but becomes very difficult if wideband

impedance matching is desired.

2. The characteristic impedance of a transmission line

The voltage or current in a transmission line can be written in terms of the sum of a

forward travelling wave and a backward travelling wave as follows,

Where the and are the amplitudes of the forward travelling voltage or current wave,

and and are the amplitudes of the backward travelling voltage or current wave.

31

Then the characteristic impedance of the transmission line is defined as the ratio

of the magnitude of the forward traveling voltage wave to the magnitude of the forward

traveling current wave

3. Voltage reflection coefficient

Now consider a transmission line is hooked up with a load with impedance ZL, then the

voltage reflection ratio is defined as the ratio of the forward travelling voltage magnitude

versus to the backward voltage travelling magnitude. Calculations also indicate the

relation of the reflection coefficient with the load impedance ZL.

From this equation, we can see that if the load matches the transmission line, then the

reflection ratio will be zero.

4. Voltage Standing Wave Ratio (VSWR)

A common measure of how well matched the antenna is to the transmission line

or receiver is known as the Voltage Standing Wave Ratio (VSWR). VSWR is a real

number that is always greater than or equal to 1. A VSWR of 1 indicates no mismatch

loss (the antenna is perfectly matched to the transmission line). Higher values of VSWR

indicate more mismatch loss.

VSWR is a function of the reflection coefficient, which describes the power reflected

from the antenna. If the reflection coefficient is given by , then the VSWR is defined as:

In our experiment, VSWR is determined from the voltage measured along a transmission

line leading to an antenna. VSWR is the ratio of the peak amplitude of a standing wave to

the minimum amplitude of a standing wave, as seen in the following Figure:

32

When an antenna is not matched to the receiver, power is reflected (so that the

reflection coefficient, , is not zero). This causes a "reflected voltage wave", which

creates standing waves along the transmission line. The result is the peaks and valleys as

seen in Figure above. If the VSWR = 1.0, there would be no reflected power and the

voltage would have a constant magnitude along the transmission line.

One thing that becomes obvious is that the ratio of Vmax to Vmin becomes larger

as the reflection coefficient increases. That is, if the ratio of Vmax to Vmin is one, then

there are no standing waves, and the impedance of the line is perfectly matched to the

load. If the ratio of Vmax to Vmin is infinite, then the magnitude of the reflection

coefficient is 1, so that all power is reflected.

In general, if the VSWR is under 2 the antenna match is considered very good and

little would be gained by impedance matching. As the VSWR increases, there are 2 main

negatives. The first is obvious: more power is reflected from the antenna and therefore

not transmitted. However, another problem arises. As VSWR increases, more power is

reflected to the radio, which is transmitting. Large amounts of reflected power can

damage the radio. In addition, radios have trouble transmitting the correct information

bits when the antenna is poorly matched.

33

VSWR (s11)

Reflected

Power

(%)

Reflected

Power

(dB)

1.0 0.000 0.00 -Infinity

1.5 0.200 4.0 -14.0

2.0 0.333 11.1 -9.55

2.5 0.429 18.4 -7.36

3.0 0.500 25.0 -6.00

3.5 0.556 30.9 -5.10

4.0 0.600 36.0 -4.44

5.0 0.667 44.0 -6.02

6.0 0.714 51.0 -2.92

7.0 0.750 56.3 -2.50

8.0 0.778 60.5 -2.18

9.0 0.800 64.0 -1.94

10.0 0.818 66.9 -1.74

15.0 0.875 76.6 -1.16

20.0 0.905 81.9 -0.87

50.0 0.961 92.3 -0.35

Table. VSWR, Voltage reflection coefficient, and reflected power.

5. Bandwidth of antenna

Bandwidth is another fundamental antenna parameter. Bandwidth describes the

range of frequenciesover which the antenna can properly radiate or receive energy. Often,

the desired bandwidth is one of the determining parameters used to decide upon an

antenna. For instance, many antenna types have very narrow bandwidths and cannot be

used for wideband operation.

34

Bandwidth is typically quoted in terms of VSWR. For instance, an antenna may

be described as operating at 100-400 MHz with a VSWR<1.5. This statement implies that

the reflection coefficient is less than 0.2 across the quoted frequency range. Hence, of the

power delivered to the antenna, only 4% of the power is reflected back to the transmitter.

Alternatively, the return lossS11=20*log10(0.2)=-13.98 dB.

6. Polarization of antenna

The polarization of an antenna is the polarization of the radiated fields produced

by an antenna, evaluated in the far field. This simple concept is important for antenna to

antenna communication. First, a horizontally polarized antenna will not communicate

with a vertically polarized antenna. Due to the reciprocity theorem, antennas transmit and

receive in exactly the same manner. Hence, a vertically polarized antenna transmits and

receives vertically polarized fields. Consequently, if a horizontally polarized antenna is

trying to communicate with a vertically polarized antenna, there will be no reception.

In general, for two linearly polarized antennas that are rotated from each other by an

angle Φ, the power loss due to this polarization mismatch will be described by the

Polarization Loss Factor(PLF):

Hence, if both antennas have the same polarization, the angle between their radiated E-

fields is zero and there is no power loss due to polarization mismatch. If one antenna is

vertically polarized and the other is horizontally polarized, the angle is 90 degrees and no

power will be transferred.

Procedure

1. Construct a 50-uA RF ammeter by connecting the 1N34 diode across the 50-uA meter

movement. As shown below

35

The diode used here is NTE109 Ge diode. The NTE109 is a high conductance

device with good switching characteristics for low impedance circuits, high resistance–

high conductance for efficient coupling, clamping and matrix service, and forward and

inverse pulse recovery for critical pulse applications.

2. Construct the experimental diode antenna as shown by the diagram below. The length of

the antenna is 4 ft 6 in. Place the insulator in the exact center. Stretch the antenna

between the nail, set about 5 ft apart on the borad.

36

Figure. The center node of the xperimental dipole antenna.

3. According to theory, the resonant frequency of the antenna is given by the equation

f(MHz)=468/length (ft)

Uding the equation to calculate the resonant frequency of the 4-ft 8-in half0wave dipole.

Since 1 ft = 12 in, thus

f(MHz) = 468/(4+8/12)ft= 100.286 MHz

4. Couple the 26-in transmission line to the signal generator. Connect the RF ammeter

between the ungrounded lead and one leg of the diploe. As shown in the diagram below.

37

Figure. Connecting the ammeter and signal generator to the dipole.

5. Set the frequency of the signal generator to about 80 MHz. Adjust the amplitude of the

signal generator for about one quarter of full scale deflection on the RF ammeter.

6. Slowly increase the frequency of this signal generator until the ammeter reading peaks.

The antenna is now in resonance with Vin. The resonance frequency of our antenna

diploe is 102 MHz as shown below.

38

Figure. The resonance frequency of the experimental antenna dipole.

7. The measured resonance frequency (102MHz) is about 1.7 % higher than the theoretical

predicted resonance frequency (100.286 MHz). This deviation may be caused from the

introduced impedance at the joint points between the transmission line and antenna dipole.

The error on the dipole’s length may also cause this deviation.

8. Remove the RF ammeter from the transmission line and connect the line to the dipole, as

shown in the diagram below.

Figure. Signal generator connected directly to dipole.

39

9. Convert the RF ammeter to a relative indicating RF voltmeter by adding the 22-pF

capacitor in series with one terminal of the meter, as shown below. Ground the other

terminal by touching it the any instruments power supply line.

Figure. Relative indicating RF voltmeter circuit.

10. With the dipole excited by the signal generator, slide the voltmeter along the antenna

wire from left to right and observe the relative voltage reading. Note: It may be necessary

to increase the amplitude of the signal generator in order to get more meter deflection.

Record the relative voltage at the following points

Left end: 2.5 V

13 in from left: 4 V

2 ft 3 in from left: 7 V

Right end: 2 V

13 in from the right: 4.5 V

2 ft 3 in from the right: 5 V

11. Graph a curve from the voltage determined along the antenna from left to right.

40

12. The reason why the graph is not a nac wave form with positive and negative half-cycle is

that we convert a DC voltmeter to a AC voltmeter, which means the detected voltage

value is the RMS value instead. The diode is used to eliminate the negative component of

the ac voltage, and the capacitor is used for getting the mean value.

13. Push a pin through the ungrounded wire of the transmission line close to the signal

generator. Measure and record the voltage at this point and also on the same wire when it

connects to the dipole.

V at the signal generator = 6 V

V at the dipole = 4.2 V

14. The ratio of these two voltages a quarter-wave from the feedpoint is known as the

voltage-standing-wave ratio, or VSWR. The impedance of the center of a resonant dipole

should be about 70 ohm. The impedance of the transmission line is 300 ohm. The

impedance mismatch ration is 4.29 : 1.

15. Because the length of the transmission line is a quarter of the transmitted wavelength, and

the wavelength of the standing wave is half of the transmitted wavelength, thus the

relative voltage measured at each end of the transmission line stands for the Vmax an

Vmin of the standing wave. Thus their ration should be the VSWR. From our

measurement,

VSWR = 6/4.2 = 1.43

0

13

27

28 41

54, 2

0

1

2

3

4

5

6

7

8

0 20 40 60

Re

lati

ve v

olt

age

(V

)

Distance from the left (in.)

Relative voltage

Relative voltage

41

16. We can see that by -66.7 % does the Z mismatch at the dipole feed point differ from the

measured SWR. The possible explanation maybe that the transmission line we use does

not have the characteristic impedance of 300 ohm.

17. When we lay the antenna on top of the table, the polarization of the antenna is horizontal.

When it is rotated so that it points to the ceiling, it has a vertical polarization.

18. Disconnect the transmission line and connect the RF ammeter across the center insulator

of the dipole. Experiment with another lab group, here we use our antenna as an excited

dipole, and their antenna as a receiving one. The distance at which the signal can be

detected using horizontal-to-horizontal polarization is 18 inch at the resonance frequency

of 102 MHz.

19. Now, if we switch to the mode of horizontal-to-vertical polarization. Then detection

distance is 0 inch. In other words, the signal cannot be detected no matter how close these

two antennas stay.

Questions

1. Determine the wavelength of each of the following frequencies.

f = 25 kHz. Wavelength = 12000 meter = 39370 feet

f = 300 MHz. Wavelength = 1 meter = 3.28 feet

f = 5.8 GHz. Wavelength = 5..17 center meter = 0.1696 feet.

2. A half-wave dipole has an effective length of 25 in. At what frequency is the antenna

designed to operate. 25 in = 2.18 ft.

The resonance frequency of this antenna is

f (MHz) = 468 / (2.08 ft) = 225 MHz

3. The measured voltage at the loops and nodes of a certain transmission line are 5 and 1.25

V. Then the VSWR of the line is

VSWR = 5/1..25 = 4.

4. Determine the reflection coefficient for VSWR = 4.

Since we have the equation

42

The reflection coefficient can be calculated to be 0.6. This number can also be obtained

from the VSWR- table.

5. A certain transmission line has a reflection coefficient of 0.2. Then the VSWR of the line

is

VSWR = (1+0.2)/(1-0.2) = 1.5

43

Bibliography Bevelacqua, P. J. (2005). Introduction to Antennas. Retrieved 10 29, 2012, from Antenna Theory:

http://www.antenna-theory.com/

44

ENTC4310 Lab 3 Phase Lock Loops Basics

Xiqiao Wang

Introduction

Phase locked loops are used ain a large variety of applications within radio frequency

technology. PLLs can be used as FM demodulators and they also form the basis of indirect

frequency synthesizers. In addition to this they can be used for a number of applications

including the regeneration of chopped signals such as the colour burst signal on an analogue

colour television signal, for types of variable frequency filter and a host of other specialist

applications.

The operation of a phase locked loop, PLL, is based around the idea of comparing the

phase of two signals. This information about the error in phase or the phase difference between

the two signals is then used to control the frequency of the loop. When there two signals have

different frequencies it is found that the phase difference between the two signals is always

varying. If the phase difference is fixed it means that one is lagging behind or leading the other

signal by the same amount, i.e. they are on the same frequency. Locked means that the

oscillator’s phase maintains a constant relationship of that of the input signal.

Below are some concepts of phase lock loop,

Phase comparator:

As the name implies, this circuit block within the PLL compares the phase of two signals

and generates a voltage according to the phase difference between the two signals. The

phase comparator is a type of signal mixer.

Loop filter:

This filter is used to filter the output from the phase comparator in the PLL. It is used to

remove any components of the signals of which the phase is being compared from the

VCO line. It also governs many of the characteristics of the loop and its stability.

Voltage controlled oscillator (VCO):

The voltage controlled oscillator is the circuit block that generates the output radio

frequency signal. Its frequency can be controlled and swung over the operational

45

frequency band for the loop. The frequency of the VCO without any control signal

applied is called the free-running frequency, f0.

Taking phase into account, the mixing product at the difference of the two input signal

frequencies, fA and fB, is cos(2π[fA – fB]t + θ), with θ representing the difference in phase

between the signals. If the two signals have the same frequency and the phase difference is

constant, then fA – fB = 0, leaving cos (θ), a dc voltage that makes a fine VCO control signal.

The high frequency of the sum product at fA + fB is not suitable as a VCO control voltage

and so must be removed. That is the job of the low-pass loop filter — to remove everything but

the phase detector’s fA – fB product, along with the phase information. Depending

on the design of the phase detector and the nature of the signals (sine, square, pulse), the loop

filter may also need to convert short bursts of current into a smoothly varying voltage.

Figure 1. The basic structure of a phase locked loop. (Phase locked loop, PLL, tutorial, 2012)

A Phase-Locked Loop has basically three states:

1. Free-running.

After the PLL is turned on with no input signal, the VCO will oscillate at the free-running

frequency, f0, until an input signal is applied.

2. Capture.

This process of adjust and hold is called capture. The minimum and maximum input

frequencies to which the loop can move the VCO as it captures an input signal is called the

capture range.

3. Phase-lock.

When the frequency is locked, it means that the oscillator’s phase maintains a constant

relationship of that of the input signal. This also means the frequencies of the two signals

are the same, otherwise the phase difference would change.

46

Figure. The four frequency ranges that defines a PLL’s behavior (Phase locked loop, PLL, tutorial,

2012).

The range over which the loop system will follow changes in the input frequency is

called the lock range. If the control signal is proportional to the cosine of the phase difference, it

will be zero when the phase difference is 90° (cos 90° = 0). It will be a maximum when the two

signals are in phase (cos 0° = 1) or out of phase (cos 180° = –1). This defines the range over

which the PLL can keep the input and VCO frequencies locked together. As the input frequency

moves farther and farther from f0, the VCO’s free-running frequency, the loop’s control action

will keep the VCO frequency the same as the input frequency, but with a phase difference that

gets closer to 0 or 180°, depending on which direction the input frequency changes.

If the input frequency has moved so far that the phase difference between it and the VCO

frequency is either 0 or 180°, any further change will cause the control signal to move back

toward its 90° value and the VCO frequency away from the input signal. The loop is no longer

locked and the input and VCO frequencies are no longer the same. The range of input frequencies

between the value at which the loop is locked with a phase difference of 0° and 180° is called the

loop’s lock range. The lock range above and below f0 are called the loop’s hold ranges. The lock

range is not always centered on f0.

The LM565 is a general purpose Phase-Locked Loop IC containing a stable, highly linear voltage

controlled oscillator (VCO) for low distortion FM demodulation, and a double balanced phase

47

detector with good carrier suppression. The VCO frequency is set with an external resistor and

capacitor, and a tuning range of 10:1 can be obtained with the same capacitor. The characteristics

of the closed loop system--bandwidth, response speed, capture and pull in range--may be adjusted

over a wide range with an external resistor and capacitor. The loop may be broken between the

VCO and the phase detector for insertion of a digital frequency divider to obtain frequency

multiplication.

Figure. The internal and pin structure of LM565 (Roon, 2010).

Parts list

Capacitors: three 0.1uF capacitors, one 0.022uF ceramic or film capacitor, one 10uF 25 V

electrolytic capacitor.

Phase locked loop IC: NE565

Potentiometer: 10 kilo ohms

Resistor: three 4.7 kilo ohm ¼ W resistors.

Procedures

1. Build the PLL circuit as shown in the diagram below.

48

Figure. The 565 integrated circuit PLL contains almost all of the circuitry necessary to

build a PLL. Only a few discrete components are needed to set the VCO free-running

frequency and loop filter time constant.

2. Without applying any input signal, the PLL will operate in free-running state and the

free-running frequency can be calculated as

f0 = 1.2/(4*Rt*Ct) = 1360 Hz.

The figure below shows the free-running frequency matches the theoretical prediction.

Figure. The free-running frequency of the PLL, where the blue signal is the output of the

VCO.

49

3. Then apply a sine wave with a frequency around 1360 Hz from the function generator to

the PLL. We can observe the output square wave signal of VCO is locked to the input

frequency.

4. Slowly reduce the generator output frequency until the PLL loses lock — seen as one

trace suddenly becoming unstable. That frequency is the lower limit of the PLL’s lock

range. Return the generator frequency to f0 and then increase it until the PLL loses lock

again at the upper limit of the lock range. Total lock range is the difference between these

two frequencies. From our experiment, we can see that the lower limit of lock frequency

range is 883.4 Hz. And the lower limit of lock frequency range is 1779 Hz.

Figure. The lower limit of the lock range.

50

Figure. The upper limit of the lock range.

5. Change the generator’s frequency above the upper limit of lock range and slowly

decrease the frequency until the PLL suddenly capture the input frequency and locked on

it. This is the upper limit of the PLL’s capture range. Similarly, if we decrease the

generator’s frequency below the lower limit of lock range and slowly increase the input

signal frequency until the PLL captures the input frequency. This is the lower limit of the

capture range. From the measurement, the lower limit of the capture range is 1333 Hz

and the upper limit of the capture range is 1391 Hz.

Figure. The upper limit of capture range.

51

Figure. The lower limit of capture range.

6. Capture range depends on the time constant of the loop filter, determined by Cf and a 3.6

kΩ resistor connected inside the IC. Replace the capacitor Cf with a capacitor of 1 nF.

And measure the free-running frequency, lock range and capture range again.

From the measurements, we can see that the free-running frequency of the PLL does not

change since it is only decided by Rt and Ct. The upper limit of lock range is 2000 Hz

and the lower limit of the lock range is 800 Hz. The measure displayed by the

oscilloscope is not accurate due to we set the VCO signal as the trigger instead of the

generator. Thus when the PLL lose the lock, it is the display of the function generator’s

that masses up. Thus the oscilloscope cannot capture a stable frequency value of the

generator signal.

The lower limit of the capture range is 866 Hz, and the upper limit of the capture range is

1794 Hz. Compare with the capture range of 1333 Hz – 1391 Hz with Cf = 10 uF, we can

see that as the time constant of the loop filter decreases, the capture range is extended.

52

Figure. PLL’s free-running frequency with Cf=1 nF.

Figure. The upper limit of lock range with Cf = 1 nF.

53

Figure. The lower limit of lock range with Cf = 1 nF.

Figure. The upper limit of capture range with Cf = 1 nF.

54

Figure. The lower limit of capture range with Cf = 1 nF.

55

Bibliography Phase locked loop, PLL, tutorial. (2012). Retrieved 11 25, 2012, from radio-electronics:

http://www.radio-electronics.com/info/rf-technology-design/pll-synthesizers/phase-locked-

loop-tutorial.php

Roon, T. v. (2010, 11 7). Phase-Locked Loop. Retrieved 10 29, 2012, from

http://www.sentex.ca/~mec1995/gadgets/pll/pll.html

60

ENTC4310 Lab4 FM generation

Xiqiao Wang

Objective


1. Define the term carrier frequency when applied to FM

2. Explain the term deviation as applied to FM

Introduction

Frequency modulation (FM) may be accomplished using several different techniques. In

many commercial FM transmitters, frequency modulation is produced by varying the reactance

of a capacitor or varactor diode in a oscillator circuit. The oscillator may be a Colpitts, Hartley,

or any of a number of other oscillator circuits. If a voltage-controlled capacitor is placed in the

feedback loop of an oscillator and the modulation signal is applied across this capacitor, then the

oscillator will change in frequency and the resulting output signal is an FM signal. A varactor

diode may be used as a voltage-variable capacitance in a resonant circuit. Applying a varying

reverse voltage to varactor changes the junction capacitance, and hence the citcuit oscillation

frequency.

The most common FM generator experimental application uses the LM 566C to generate

the FM signal. The LM 566 is a linear voltage-to-frequency converter which can generate an FM

signal up to 1 MHz and for a +/- 10% deviation from the center frequency, it has an FM

distortion of less than 0.2%. The center frequency (fo) is set by a resistor (Ro) and a capacitor

(Co). The LM 566 delivers either an FM square wave or an FM triangular wave. However, this

is not important, since the square/triangular wave can be converted into a “good” sinewave with

the use of a simple RC filter. The input impedance for the modulation signal is 1 MΩ and the

output impedance of the square/triangular wave is 50 Ω. In both cases above, if the modulating

signal is a digital waveform, then the resulting FM signal is a frequency-shift-keying signal

(FSK).

61

The circuit used in this experiment is based on the 555 timer IC. The output waveshape

of the 555 timer is rectangular rather than sinusoidal and is not typical of most RF FM circuits;

however, it is useful in demonstrating the basic characteristic on FM generation.

Several definitions need to be cleared before we start the experiment.

The amplitude of the modulation signal determines the amount of the frequency change

from the center frequency.

The frequency of the modulation signal determines the rate of the frequency change from

the center frequency.

The amplitude of the FM signal is constant at all times and is independent of the

modulation signal.

Mathematically, an FM signal is written as

Where A is the amplitude of the signal, Wc is the center frequency for non-modulated signal

(carrier frequency); Wm is the modulation frequency; Mf is the FM modulation index

is the maximum frequency shift caused by the modulation signal, and fm is the frequency of

the modulation signal.

The bandwidth of an FM signal depends on the modulation index (Mf), and is approximated by

the well-known Carson’ s Rule:

where fm(max) is the maximum frequency of the modulating signal. The factor (2) in the

equation is to account for both the upper and lower sidebands (left and right of the carrier). This

equation gives the bandwidth which contains 98% of the signal power.


Dual-trace digital oscilloscope, 5-V DC power supply, audio signal generator, 555 timer IC, 6.8

kilo ohm resistor, 3.3 kilo ohm resistor, 0.01 uF capacitor, 47 uF 10 V electrolytic capactor, IC

prototype breadboard.

Procedure

62

1. Build the FM generation circuit as shown in the diagram below.

Figure. Experimental FM generation circuit.

When the modulation signal is inactive, the free-running frequency of the circuit (carrier

frequency) can be calculationed through the equation:

Substitute the resistor and capacitor values into this equation, we obtain the carrier

frequency of this circuit is 8520.71 Hz.

2. Construct the circuit and leave the modulation source disconnected.

3. Apply power to the circuit and using the oscilloscope to determine the peak amplitude

and frequency of ,

4. Connect the audio generator ( ) to the circuit. Using the scope, adjust the Vm for a sine

eave of 2 Vpp at 1 Hz.

5. Using channel 1 of the scope set for DC coupling and a sweep speed of 0.1 ms/div.

Observe the output of the 555 timer. Here we observed that the output square wave’s

63

frequency changes with the audio input sine wave’s amplitude. The timer 555’s output

signal’s frequency decreases as the audio signal’s amplitude increases, and the timer’s

output frequency increases as the amplitude of the audio signal decreases. The screen

shots of the scope are shown below. When the audio signal’s amplitude is about 0.6 V,

the timer frequency is about 5.94 kHz. When the audio signal’s amplitude drops to about

0.2 V, the timer frequency increases to 10 kHz.

Figure. The frequency of the 555 timer signal changes with the amplitude of the 1 Hz

audio signal.

64

6. Using channel 1 of the scope, adjust the Vm for a sine wave of 200 Hz, 2 Vpp. Adjust the

sweep speed of the scope so that approximately one cycle of the Vm is displayed. Leave

channel 1 connected to the output of the audio signal generator.

7. Observe the output of the 555 timer using channel 2 of the scope set for DC coupling.

The relationship between the oscillation frequency of the 555 and the amplitude of Vm is

shown in the figures below. As we can see, the frequency of timer peaks when the input

sine wave reaches the minimum voltage value, and the timer frequency decreases to the

slowest point as the sine wave reaches its maximum positive voltage value.

Figure. The relationship between the oscillation frequency of the 555 and the amplitude

of Vm.

8. Increase the amplitude if Vm to 4 Vpp. What effect does this have on the variation

(deviation) of the output frequency of the 555 timer.

From our observation, the qualitative relation between the timer frequency and the audio

signal amplitude obey the same rule as observed in step 7. The only difference is that the

amplitude of the frequency deviation of the timer output increases since the amplitude of

the audio signal increases. The FM frequency decreases more as the audio signal peaks

and the FM frequency increases more as the audio signal reaches its low ebb.

65

Figure. The relation between the FM signal and the audio input as the audio sine wave’s

amplitude increases to 4 Vpp.

Question

1. An FM transmitter has an unmodulated carrier frequency of 102.5 MHz. the transmitter

carrier frequency deviates at a rate of 5 kHz/V with an applied modulation voltage. What

is the oscillation frequency of the transmitter if Vm = -10 V, assuming a linear

modulation.

f = 102.5MHz + (-10 V)*(0.005 MHz/V) = 102.45 MHz.

2. Using the information of question 1, what value of Vm would result in a carrier deviation

of +25 kHz.

Vm = 5 V.

3. A sinusoidal voltage of 1 kHz, 2 Vpp is applied as Vm to the transmitter of question 1.

Determine the modulation index of the transmitter.

The maximum frequency shift caused by the modulation signal, , is 1V * 5 kHz/V = 5

kHz. The frequency of the modulation signal, fm, is 1 kHz. From the equation of the

modulation index

We can know that the modulation index in this case is 5.

66

4. Based on the information of question 3, how many significant sidebands would be

produced.

The spectrum of an FM signal is quite complicated and is dependent on mf. Actually, it

follows a Bessel Function

where J0(mf), J1(mf) etc. are in volts, and are the levels of the frequency components of

the FM signal for A = 1 V. The spectrum is dependent on mf, the modulation index, and

the table below gives the values of the Bessel-functions J0, J1, J2, etc.... for mf = 0 to 15.

Notice from the table that for mf = 2.4, there is no power in the center frequency

component (J0(2.4) =0). This also occurs at mf = 5.5, 8.6, ... . This does not mean that

there is no power transmitted in the signal. All that it means is that for m = 2.4, 5.5, ...,

there is no power at the center frequency and all of the power is in the sidebands.

Table. FM spectrum levels for mf = 0 to 15 (not in dB). A negative sign means a phase of

180° with respect to other components.

Thus we can see that for mf = 5, the number of significant sidebands is 8 on one sides and

16 on both sides.

5. What is the frequency of the highest significant sideband?

67

The frequency of the Nth sideband is denoted as

Thus for a carrier frequency of 102.5 MHz, and a modulation frequency of 1 kHz, and N

= 8, the highest significant sideband on either side is:

68

ENTC4310 Lab 5 FM demodulation – the phase-locked loop

Xiqiao Wang

Objective


1. Draw the block diagram of a phase-locked loop

2. Explain the function of a loop filter.

Introduction

The demodulation of an FM signal is much different from the demodulation of an AM

signal. Among the many FM demodulation techniques in use today, the phase-locked loop (PLL)

demodulators is quite popular. One of the major advantages of PLL demodulation is the

elimination of many complex and touchy tuned circuits used in the Froster-seely and ratio

detectors. This experiment will familiarize us with the basic operation of an IC phase-locked

loop. Since a brief review of PLL has been given in the introduction session in Lab 3, here I will

present a brief introduction to the basic idea of FM demodulation.

1. FM demodulation (Poole, 2010)

As the name suggests frequency modulation, FM uses changes in frequency to carry the

sound or other information that is required to be placed onto the carrier. As the

modulating voltage varies, so the frequency of the signal changes in line with it. This

type of modulation brings several advantages with it:

Interference reduction: When compared to AM, FM offers a marked

improvement in interference. In view of the fact that most received noise is

amplitude noise, an FM receiver can remove any amplitude sensitivity by driving

the IF into limiting.

Removal of many effects of signal strength variations: FM is widely used for

mobile applications because the amplitude variations do not cause a change in

audio level. As the audio is carried by frequency variations rather than amplitude

ones, under good signal strength conditions, this does not manifest itself as a

change in audio level.

Transmitter amplifier efficiency: As the modulation is carried by frequency

variations, this means that the transmitter power amplifiers can be made non-

69

linear. These amplifiers can be made to be far more efficient than linear ones,

thereby saving valuable battery power - a valuable commodity for mobile or

portable equipment.

In order to be able to convert the frequency variations into voltage variations, the

demodulator must be frequency dependent. The ideal response is a perfectly linear

voltage to frequency characteristic. Here it can be seen that the center frequency is in the

middle of the response curve and this is where the un-modulated carrier would be located

when the receiver is correctly tuned into the signal. In other words there would be no

offset DC voltage present.

Figure. Characteristic "S" curve of an FM demodulator

The ideal response is not achievable because all systems have a finite bandwidth and as a

result a response curve known as an "S" curve is obtained. Outside the bandwidth of the

system, the response falls, as would be expected. It can be seen that the frequency

variations of the signal are converted into voltage variations. To enable the best detection

to take place the signal should be centred about the middle of the curve. If it moves off

too far then the characteristic becomes less linear and higher levels of distortion result.

Often the linear region is designed to extend well beyond the bandwidth of a signal so

that this does not occur. In this way the optimum linearity is achieved. Typically the

bandwidth of a circuit for receiving VHF FM broadcasts may be about 1 MHz whereas

the signal is only 200 kHz wide (Poole, Phase locked loop, PLL, tutoria, 2010).

Below is a list of some of the main types of FM demodulator or FM detector.Each

of these different types of FM detector or demodulator has its own advantages and

disadvantages. In recent years, the Foster Seeley discriminator and the Ratio detector

70

have been less widely used. The main reason for this is that they require the use of wound

inductors and these are expensive to manufacture. Other types of FM demodulator have

overtaken them, mainly as a result of the fact that the other FM demodulator

configurations lend themselves more easily to being incorporated into integrated circuits.

Slope FM detector

Foster-Seeley FM detector

Ratio detector

PLL, Phase locked loop FM demodulator

Quadrature FM demodulator

Coincidence FM demodulator


Dual-trace oscilloscope, 10-V DC power supply, 565 PLL IC, 10 kilo ohm resistor, 4.7 kilo ohm

resistor, 0.47 uF capacitor, 0.0047 uF capacitor, 0.001 uF capacitor, IC prototype breadboard.

Procedure

1. The circuit is shown as in the figure below, is an FM demodulator, based on the 565 IC

PLL. The pin diagram and internal block diagram is also shown below. Construct the

circuit in this step.

71

Figure. Phase-locked loop FM demodulator

Figure. Pin diagram of 565 PLL IC.

Figure. Internal block diagram of the 565 PLL IC.

2. The FM signal to be demodulated is taken from the circuit used in Lab 4 the FM

generation. The R1 and R2 in the FM demodulation circuit diagram form a voltage

divider that attenuates the Fm signal.

3. Connect the output of the circuit used in Lab 4 to the input of the demodulator circuit (the

open end of R1). Notice that the FM generator constructed based on 555 timer IC has a

72

FM output signal of square wave. It is not the usual FM waveform in RF application.

Besides, the LM565 PLL IC is designed to receive an sine wave as the input signal. For

this reason, the AC component output component of the PLL demodulator circuit will be

extremely weak. The output signal measured at pin 7 is a oscillating DC signal averaged

by the capacitor C1, and it is used to be feed back to the VCO. If the input signal at pin 2

is only the unmodified carrier frequency, then the DC component is the voltage that

enable the VCO to generate the carrier frequency square wave signal. When FM

modulated signal is input from pin 2, then there is weak AC components rise on the DC

voltage at pin 7, which is used to adjust the VCO to keep in phase with the input FM

signal. Since the maximum frequency shift caused by the modulation signal is usually

very small compared with the carrier frequency, the AC components at pin 7 therefore

should be very weak compared with the DC component, which is used to set the central

carrier frequency.

Adjust Vm of the modulator circuit for a sine wave of 200 Hz at 2 Vpp using channel 2

of the oscilloscope. All the observed signals from pin 7 on the scope use AC coupling.

4. Observe the output of the demodulator circuit using the channel 2 of the oscilloscope.

Form our observation, the output of the demodulator is a sinusoidal wave with the same

frequency as the input sine wave signal of the FM generator. There is a constant phase

difference between these two signals, which shows that the PLL has been locked on the

generated FM signal.

5. Measure and record the peak-to-peak amplitude of the output signal.

Vout = 100 mVpp

Notice that we could have use the averaging math function to demonstrate a clearer

waveform of the demodulated signal.

73

Figure. Blue line: the input signal from the audio signal generator at 2 Vpp, input to the

FM generator. Yellow line: the demodulated signal from the PLL FM demodulator.

6. Increase Vm to 4 Vpp. Measure and record the peak-to-peak amplitude of Vout.

Vout = 200 mVpp.

Figure. Blue line: the input signal from the audio signal generator at 4 Vpp, input to the

FM generator. Yellow line: the demodulated signal from the PLL FM demodulator.

7. From Step 5 and 6, we can see that the amplitudes of the audio input signal and the FM

demodulated signal obey a linear relationship.

74

8. Decrease the frequency to 100 Hz. We then observed that the frequency of the

demodulated frequency dropped to 100 Hz.

Figure. As Vm’s frequency drops to 100 Hz, so does the FM demodulated signal’s

frequency.

9. Remove the capacitor C2 of the demodulator circuit. We did not observe any effect on

the demodulated signal.

Figure. The observed audio source and demodulated wave as the C2 capacitor is removed

from the demodulator.

75

Summary

The operation of the 565 PLL can be nest understood by referring to the block diagram

below. When no input signal is present, the voltage controlled oscillator (VCO) runs ata

frequency determined by the value of C0 and R0. If an input signal is applied to pin 2, the

phase detector produces a pulse train at pin 7 with a duty cycle proportional to the phase

difference between Vosc of the VCO and the input signal Vref. The pulse train error

signal is filtered by C1 (the loop filter) and applied to the control input of the VCO.

If the input to the PLL is frequency modulated, the control voltage applied to the VCO

varies in proportion to the deviation or variation of the incoming FM signal. This varying

control voltage forces the frequency of the VCO to track the frequency of the FM signal.

The varying VCO control voltage is a replica of the original modulating signal and is

used as the output of the demodulator.

Figure. Internal block diagram of the 565 PLL IC.

Question

1. The average output voltage Vo(DC) produced by a phase detector is given by the

equation Vo(DC) = , where is the conversion gain of the phase detector

76

in V/rad, and is the phase difference in radians between the input signals. One

radian equals 57.3 degrees. Using the information given, determine Vo(DC) for a

phase detector with the following specification: =1.5 V/rad, Vref= 0 rad, Vosc=π

rad.

Vo(DC) = π rad * 1.5V/rad = 4.71 V

2. Convert the phase shift obtained in question 1 from the radians to degrees.

3. A PLL FM demodulator has an input signal that deviations +/- 2kHz from an

unmodulated carrier frequency of 1.5 MHz. what voltage must the phase detector

produce in order to make the VCo track the deviation of the carrier, if the VCO has a

free-running frequency of 1.5 MHz and a sensitivity (on the control voltage input) of

500 Hz/V

Vo = [2000-(-2000)] Hz / (500 Hz/V) = 8 Vpp

77

Bibliography Poole, I. (2010, 5 6). FM demodulation. Retrieved 11 25, 2012, from radio-electronics: Ian Poole

Poole, I. (2010, 6 4). Phase locked loop, PLL, tutoria. Retrieved 11 25, 2012, from radio-electronics:

http://www.radio-electronics.com/info/rf-technology-design/pll-synthesizers/phase-locked-

loop-tutorial.php

78

ENTC 4310 Lab 6 AM Broadcast Transmitter and High performance point Boardcase amplifier

Part I High performance point Boardcase amplifier Technical Mannual

Xiqiao Wang

1. Introduction

An (A)mplitude (M)odulated signal is actually a combination of two signals. The high

frequency carrier is the frequency that one will tune on the radio receiver’s dial, from

530 to 1750 KHz. The modulation is the audio information that rides “on top” of the high

frequency carrier, resulting in a changing of the level, or amplitude, of the output

waveform.

2. Circuit description of Ramsey high intercept point FM broadcast amplifier

The components on the larger P14FMBA board are used to supply the voltage to

run the smaller amplifier/preamp board. The DC supplied to the control box is sent up the

coax to the amplifier board. This DC is supplied to VR1 which then provides a regulated

12 VDC for the rest of the circuit. The main components of the circuit are the MAR3 RF

amplifier and Q2, an MRF581 power transistor. The MAR3 is a state of the art amplifier

specially designed to work in a wide range of radio frequencies. The MAR3 amplifies the

signal, but does not have the power output capabilities to push out the kind of high power

we desire. For this reason, Q2 is placed in the circuit. Q2 amplifies the signal a bit more,

and is capable of generating a large amount of power out of the circuit. MAR3 is a

monolithic amplifier with wideband from DC to 2 GHz. It has a typical gain of 12.5 dB at

100 MHz as specified in its datasheet. MRF581 is a RF low power transistor. From its

datasheet, it is specified to have a power gain of 15.5dB at 500MHz.

D1 and D2 limit the input to U1 to 0.7 volts, to keep the amplifier from being

overdriven and damaged. L1, L2, C3, 4 and 5 form a lowpass filter that reduces unwanted

signals entering U1. U1 then amplifies the signal and couples it through C6 to Q2. Q1

and its associated components provide the proper bias for Q2. After being amplified by

Q2, the signal passes through another stage of filtering, to eliminate harmonics from the

79

output signal. This filtering allows the FMBA1 to produce clean, undistorted RF when

used with a transmitter or to “clean up” the input to your receiver when used as a preamp.

Figure 2. The functional blocks of Ramsey high intercept point FM broadcast

amplifier

3. Assembly of broadcast amplifier

There are several points we need to take serious when we solder the circuits and

connections, and the most important of all, always wear your safety glasses when you are

soldering and wash your hands each time after soldering, or Dr. Sims will take points

away from your final grades. Besides, it is recommended to use a 25-watt soldering

pencil with a clean, sharp tip. Use only rosin-core solder intended for electronics use.

Carefully brush away wire cuttings so they don't lodge between solder connections. Insert

the part, oriented correctly, into its correct holes in the PC board. If helpful, gently bend

the part's wire leads or tabs to hold it into place, with the body of the part snugly against

the top side ("component side") of the PC-board. Orient it correctly, follow the PC board

drawing and the written directions for all parts - especially when there's a right way and a

wrong way to solder it in. (Diode bands, electrolytic capacitor polarity, transistor shapes,

dotted or notched ends of IC's, and so forth.) And then solder all connections unless

directed otherwise. Use enough heat and solder flow for clean, shiny, completed

connections. If the temperature of your soldering pencil, usually you will have a hard

time to heat the solder spot on the board enough to melt with the soldering tin. Trim or

80

nip all excess wires extending beyond each solder connection, taking care that wire

trimmings do not become lodged in PC-board solder connections.

Since the FMBA1 runs at very high frequencies, it is extremely important that you

follow the instructions provided. Incorrectly installed components, excessively long

component leads, and bad solder joints may mean that your kit won’t work. Special

attentions should be paid when installing the MRF581 power amplifier. The long lead on

the part does not solder to the longer pad. The transistor should be placed so that the

longer lead is facing J2, the output “F” jack. We need to trim the long lead of the

transistor so that it will fit on the solder pad without hanging over.

The last 1.2 μH inductor will supply power to the smaller amplifier board we just

finished building. Here we choose the amplifier to be used as in preamp mode, which

means it is used to amplify the received signal from the antenna before the receiver. In

preamp mode we need to connect the antenna to J1 and the receiver to J2. In this mode

we must install the inductor in the place marked “preamp mode” on the PC board

silkscreen.

81

Figure 3. The broadcast amplifier assembled as working in preamp mode.

82

Figure 4. The power supply board.

4. Preamp Hookup

The use of coaxial cables for connections is to ensure the resistance matching at the input

and output of the amplifier in order to get maximum performance out of the amplifier.

We use two lengths of coaxial cables with “F” connectors to connect the J1 end of the

amplifier board to the antenna. The J2 end of the amplifier board is connected to the

receiver though the power supply board. In other words, the J2 connector of the amplifier

board is connected to the J2 connector of the power supply board through a coaxial cable.

The receiver is connected to the J1 connector of the power supply board. A 12V DC

voltage source is connected to the J3 connector on the power supply board, with the red

line being the negative and blue line being the positive end. The cable connecting J2 on

the power control boxand the J2 on the amplifier board will supply the 12 V DC voltage

to the amplifier board as well as carrying the RF down from the amp to the power control

83

board. The C1 capacitor on the power control board is used to eliminate DC voltage

component and only transmit the amplified RF signal to the receiver.

In this lab, the weak RF signal from the antenna is simulated by a weak signal

generated by a RF sine wave generator. As shown in the figure below, the weak RF

signal is input from the red and black clamps of the RF signal generator’s coaxial cable.

And the amplified signal is tested at the J1 connector of the power control box.

Figure 5. the weak RF signal is input to the amplifier’s J1 connector from the red and

black clamps of the RF signal generator’s coaxial cable.

84

Figure 6. The amplified signal is input to the control box at J2 connector and tested at the

J1 connector, where the DC voltage component is eliminated by the capacitor C1.

5. Measure the Bode-diagram of the RF amplifier

This FM broadcast amplifier is designed to amplify FM signal whose frequency ranges

from 80 to 120 MHz. This frequency band cannot be fully covered by the band capacity

of the digital oscilloscope in our lab. Thus we use the AC-to-DC RF probe and a DC

voltmeter to detect the gain of the amplifier at different frequencies. The AC-to-DC probe

used here is the FLUKE 85RF probe. The schematic diagram is shown below.

Figure 7. The schematic diagram of the AC-to-DC probe.

85

The 85RF high frequency probe is designed to convert a DC voltmeter into a high

frequency (100 kHz to 500 MHz ) AC voltmeter. Conversion from ac to dc is

accomplished on a one-to-one basis and includes a range of 0.25 to 30V rms. The probe’s

dc output is calibrated to be equivalent to the rms value of a sine wave input. The DC

voltmeter connection to the probe is shown in the figure below. The capacitor C1 in the

85RF probe diagram is used to block the DC component of the input voltage.

Figure 8. The 85RF high frequency probe connection with the DC voltmeter.

Frequency

(MHz)

input

RMS

(mV)

output

RMS

(mV)

Gain

(numeric)

Gain

(dB)

70 0.93 302.00 324.73 50.23

75 1.20 330.00 275.00 48.79

80 1.60 315.00 196.88 45.88

86

85 2.09 260.00 124.40 41.90

87.5 3.80 151.00 39.74 31.98

90 3.60 250.00 69.44 36.83

92.5 4.00 276.00 69.00 36.78

95 2.60 267.00 102.69 40.23

100 2.30 149.00 64.78 36.23

105 6.30 139.00 22.06 26.87

110 6.50 56.00 8.62 18.71

115 6.20 23.00 3.71 11.39

120 4.36 22.00 5.05 14.06

125 4.70 25.00 5.32 14.52

130 7.10 30.00 4.23 12.52

Table. The tested gain of the FM broadcast amplifier at various FM frequencies.

Figure 9. The Bode-plot of the FM broadcast amplifier.

From bode-plot above, we can see that the gain at 70 MHz is about 50dB. As the

frequency increases, the gain drops to 31.98 dB at 87.5 MHz. Then the gain increases to

about 36.23 dB at 100 MHz. This is very close to the specified gain of 38.5dB at

100MHz as indicated in the official manual. For the Frequency above 100 MHz, the

amplifier’s gain goes down till reaching a minimum of about 10 dB at around 115 MHz.

6. Testing Points and Trouble shooting.

0.00

10.00

20.00

30.00

40.00

50.00

60.00

65 75 85 95 105 115 125 135

Gai

n (

dB

)

Frequency (MHz)

Gain (dB)

87

Figure 10. Testing Points on the power control board.

We set two DC voltage test points on the power control board. The first testing

point is set at the connector J2. If the power supply is 12 Volt, then the DC voltage at

testing point 1 should be 12 volt. The second testing point is set at the connector J1.

Because the capacitor C1 blocks the voltage DC component, the DC voltage at testing

point 2 should be 0 volt.

88

Figure 11. Testing Points on the RF broadcast amplifier board.

We set 6 DC testing point on the amplifier board, namely testing point 3 through

testing point 8. Testing point 3 is set at the output of the VR7812, which is the common

point of the positive end of capacitor C18 and resistor R9. VR7812 is a voltage regulator

that regulate the input DC voltage to 12 Volts, which is supplied for the use on the

amplifier board. So if the power supply circuit on the amplifier board functions well, the

DC voltage value at testing point 3 should be 12 Volts. Testing point 4, 5, 6, 7 are used

for trouble shooting the functioning of the two RF amplifier MAR3 and MRF581.

Testing point 4 is set at the input of MAR1, which is the common point of capacitor C5

and MAR3. The DC voltage at this point should be 1.6 Volts. Testing point 5 is set at the

output of MAR3, which connecting both the capacitor C6 and inductor L3. The DC

voltage at this point should be 11 Volts. Testing point 6 is set at the input of MRF581,

which is the connecting point among capacitor C6, resistor R4 and MRF581. Transistor

Q2 provide the proper bias DC voltage for the base of MRF581. Thus the DC value at

testing point 6 should be about 0.6 volt. Testing point 7 is at the output of MRF581,

which joint the 6-hole bead inductor L4, capacitor C11 and MRF581. The DC voltage at

this point should be 5 Volts.

89

Part II Ramsey AM broadcast transmitter kit Circuit Test

1. Circuit Description of Ramsey AM broadcast transmitter kit

The schematic diagram is shown below.

Figure 12. The schematic diagram of Ramsey AM broadcast transmitter circuit.

90

Figure. The assembled AM transmitter board.

The RF oscillator consists of Q6 and associated components. The

frequency of operation is determined by selecting the proper values for C9

and C10, and adjusting the inductance of coil L2. The “buffer” amplifier (Q5)

is connected to the base of Q6 in order to use the undistorted oscillator output for the RF

carrier frequency.

The audio input path is routed from J1, the audio input source, to transistor Q2

to amplify the incoming signal. Notice that the transistor is biased to be linear

using resistors R3, R5, and R6. The incoming audio signal is therefore

amplified undistorted (for great sounding audio). Optional capacitor C4 is used

only when a microphone input is used to provide additional gain from transistor

Q2. The audio input level to the amplifier can also be adjusted using R12, the

input level adjustment.

The resulting audio output is fed to transistor Q1, which does not provide any

gain, but supplies enough current to modulate the RF carrier. Inductor L1

allows the low frequency audio to pass through but “chokes” the RF signal and

91

does not allow it to get “back into” the audio circuitry.

Transistors Q3 and Q4 comprise the “power amplifier” section of the circuit.

Their collector supply voltage is furnished by Q1, thus producing an AM output

waveform. This signal is then low pass filtered using C13, C14, and L3.

Notice also that the audio information is applied at the power amplifier stage.

This is referred to as “high level” modulation, and is commonly used for high

power AM broadcast stations. The distinct advantage to this is that the RF

amplifiers need not be biased for linear operation. It is much cheaper to

manufacture a linear amplifier for the relatively low frequency audio, than to

produce the AM waveform at a low level and amplify it to a higher power level

without distortion. The main disadvantage of high level modulation is that the

audio modulator’s power must be half that of the final transmitter, this is not too tough for

our low power kit.

It should also be stated that, due to the linear operation of the amplifiers in this

circuit (transistors Q1 and Q2 biased “on”), this circuit will consume some

power. It is not recommended that a common rectangular 9V battery be used

to power this kit. Instead, a battery “pack” consisting of eight 1.5 volt cells, a

12V sealed battery, or other external 12V DC supply may be used.

2. Circuit Testing According to the Technical Manual by Kolton Power.

Testing of the AM transmitter begins by connecting J3 to the 12V DC power supply, though

not switching the circuit on just yet. An AM radio is placed nearby to the circuit and is

tuned to a quiet spot within the chosen frequency range. An antenna wire is connected to

the RF OUT jack, which for these purposes is done by simply connecting an alligator clip to

the jack and wrapping the lead around the antenna of the radio loosely. The circuit is now

energized by switching the power button “on”. Using the plastic tuning tool, the L2 coil is

adjusted until the AM1C’s carrier signal is heard. The LEVEL ADJ is then rotated to its full

clockwise position in order to achieve the maximum level input position. An audio source

92

is then connected to the AUDIO IN jack via a function generator, set to 300 Hz, at 500

mVp-p with a 75Ω output load. R12 is then adjusted undistorted audio.

Figure 13 shows the AM1C schematic diagram with the locations of 10 various test points.

Figure 6-12 is a table that explains each point along with a screen shot taken from the

oscilloscope of what the output wave should appear like at each point.

Figure 13. Testing points of Ramsey AM broadcast transmitter circuit.

93

Figure 14. Testing points 1 at Negative lead of C2, the observed signal is of 300 Hz, at

about 500 mVp-p.

Figure 15. Testing points 2 at Base of Q2, the observed signal is of 300 Hz at 440 mVp-p;

+1.1VDC.

94

Figure 16. Testing points 3 at Emitter of Q2, the observed signal is of 300 Hz at 400 mVp-p;

+.4VDC.

Figure 17. Testing points 4 at Collector of Q2, the observed signal is of 3606 Hz at 2 Vp-p;

+8.5VDC.

Figure 18. Testing points 5 at Base of Q1, the observed signal is of 300 Hz at 1.6 Vp-p;

+5.6VDC.

95

Figure 19. Testing points 6 at Emitter of Q1, the observed signal is of 291 Hz at 1.52 Vp-p;

+3.8VDC.

Figure 20. Testing points 7 at Base of Q3 & Q4, the observed signal is of 1320 kHz at 1.04

Vp-p (non-sinusoidal); 0VDC.

96

Figure 21. Testing points 8 at Collector of Q3 & Q4, the observed signal is of 1320 kHz at

3.76 Vp-p (non-sinusoidal); +3.9VDC.

Figure 22. Testing points 9 at Base of Q5, the observed signal is of 1280 kHz at 11 Vp-p;

+1.2VDC.

Figure 23. Testing points 10 at RF OUT, the observed signal is of 1320 kHz at 4.2 Vp-p;

0VDC.

3. Testing conclusion and scoring.

The circuit is tested step by step according to the tech manual. The testing results matches

the description in the manual. The indicated waveforms in the manual are observed almost

the same at all the ten testing points. The manual gives an accurate and clear description of

the testing procedure and the expected testing results. So, I will rate this tech manual as

being at level A.

101

ENTC4310 Optical Fiber Lab 1 Setting Up A Fiber Optics Analog Link

Xiqiao Wang

Introduction

This experiment is designed to familiarize us with the optical fiber trainer (OFT). An analog

fiber optic link is to be set up in this experiment. The preparation of the optical finer for coupling

light into it and the coupling of the fiber to the LED and detector are described here. The LED

used is an 850 nm LED. The fiber is a multimode fiber with a core diameter of 1000 um. The

detector is a simple PIN detector.

The LED optical power is directly proportional to the current driving the LED. Similarly, for the

PIN diode, the current is proportional to the amount of light falling on the detector. Thus even

though the LED and diode are nonlinear devices, the current in the PIN diode is directly

proportional to the driving current of the LED. This makes the optical communication system a

linear system.

Analog link is used to transmit analog data in this type of communication link first bias point of

source is set approximately at the midpoint of the linear output region. The analog signal can be

transmitted by one of the several modulating techniques. The simplest form for optical fiber link

is intensity modulation, wherein the optical output from the source is modulated simply by

varying the current around the bias point in proportion to message signal. Thus signal is

transmitted in baseband. Generally amplitude-modulation (AM), frequency modulation (FM), or

phase-modulation (PM) are used for analog transmission for these type of modulation one should

pay careful attention to signal impairments in optical source which include harmonic distortions,

intermodulation products ,relative intensity noise(RIN) in the laser, and laser clipping. A block

diagram of analog link is shown below.

102

Figure. Optical fiber Analog communication link.

Using analog link one must take into account the frequency dependence of amplitude, phase and

group delay in the fiber. Thus the fiber should have a flat amplitude and group-delay response

within the passband required to send the signal free of linear distortion. Below shows some

analog drive circuits.

Figure. Analog drive circuits for simple application, linear low frequency and linear high

frequency applications.

103

Objective

The objective of this lab is to setup an 850nm fiber optics analog link. The linear relationship

between the input and received signals are observed. The effect of gain control on the received

signal is also observed, finally, the bandwidth of the link is measured.

Equipment

OFT, two channels, 20 MHz digital oscilloscope, function generator

Procedure

1. Identify the interfaces on the OFT with the help of layout diagram.

Figure. The setup for this experiment.

2. Set the switch SW8 to the analog position. Switch the power on.

3. Feed a 1Vpp sinusoidal signal at 1 KHz from a function generator, to the analog in post

P11 using the following procedure: i) connect a BNC-BNC cable from the function

generator to the BNC socket I/O3. ii) connect the signal post I/O3 to the analog in post

104

P11 using a patch cord. Then connect the 1 m fiber to the LED source LED 1 in the

optical Tx1 block. The light of wavelength of 850 nm is within the infra-red spectrum

and it cannot be seen by our eyes.

Figure. The observed light output (red tinge at a wavelength of 650 nm) at the end of the

fiber.

To observe a fed-in signal on an oscilloscope, use a 3-plug patch cord to connect the

signal post I/O3 to the required input post. Connect a BNC-BNC cable between the BNC

socket I/O2 and the oscilloscope.

Increase and decrease the amplitude level of the sinusoidal signal from 0Vpp to max 2

Vpp. We observe the input signal from the generator and the received signal at port P31

after the amplifier. In the scope display, the blue line is the signal after the receiving

amplifier, and the yellow line is the signal from the function generator. A cutoff

happened when the input signal has amplitude of 2 Vpp.

105

Figure. The received signal (blue) when the input signal has amplitude of 200 mVpp.

Figure. The received signal (blue) when the input signal has amplitude of 1 Vpp.

106

Figure. The received signal (blue) when the input signal has amplitude of 2 Vpp.

4. Feed a 5 Vpp rectangular signal at 0.5 Hz at P11. Observe the signal on the oscilloscope.

Then feed a 5 Vpp sine wave to the P11 and observe the output signal at port P31.

Figure. The amplified signal (blue) at port P31 when feed a 5 Vpp rectangular signal

(yellow) at 0.5 Hz at P11.

107

Figure. The amplified signal (blue) at port P31 when feed a 5 Vpp sinusoidal signal

(yellow) at 0.5 Hz at P11.

5. Connect the other end of the fiber to the detector PD1 in the optical Rx1 block.

6. Feed a sinusoidal wave of 1 KHz, 1 Vpp from the function generator to P11. The PIN

detector output signal is available at P32 in the optical Rx1 block before the signal is

amplified. Vary the input signal level driving the LED and observe the received signal at

the PIN detector. Plot the received signal peak-to-peak amplitude with respect to the

input peak-to-peak amplitudes.

Figure. The received signal at P32 (blue) versus the input signal from the generator.

108

7. Repeat step 6 using the 1m fiber. Plot the received signal peak-to-peak amplitude with

respect to the input peak-to-peak amplitudes. The LED output optical power is directly

proportional to the current driving it. The PIN diode current is also directly proportional

to the optical power incident on it. Therefore, the relationship between the input electrical

signal and the output electrical signal is linear. As shown in the plot below.

Figure. The received signal at P32 (blue) versus the input signal from the generator

though the 1m fiber with the input amplitude of 200 mVpp.


though the 1m fiber with the input amplitude of 1 Vpp.

109


though the 1m fiber with the input amplitude of 2 Vpp.

Figure. The linear relation between the received signal peak-to-peak amplitude with

respect to the input peak-to-peak amplitudes.

8. The PIN detector signal at P32 is amplified, with amplifier gain controlled by the GAIN

potentiometer. With a 3 Vpp input signal at P11, observe P31 as the gain potentiometer is

varied. As shown below, the signal at P31 gets clipped below 0 V and above 3.5 V.

0.2, 0.15

1, 0.6

2, 1.1

0

0.2

0.4

0.6

0.8

1

1.2

0 0.5 1 1.5 2 2.5

ou

tpu

t V

pp

Input Vpp

Output Vpp V.S. Input Vpp

output Vpp

110

Figure. The signal at P31 gets clipped below 0 V and above 3.5 V.

9. Measure the bandwidth of the link. Apply a 2 Vpp sinusoidal signal at P11 and observe

the output at P31. Adjust the gain such that no clipping takes place. Vary the frequency of

the input signal from 100 Hz to 5 MHz and measure the amplitude of the received signal.

Plot the received signal amplitude as a function of frequency (using a logarithimic scale

for frequency). Note that the frequency range for which the response is flat.

Figure. The output at P31 (blue) with the input signal of 2 Vpp, 6 Hz.

111



112


Figure. The output at P31 (blue) with the input signal of 2 Vpp, 5 kHz.

113



114



115

Figure. The output at P31 (blue) with the input signal of 2 Vpp, 1 MHz.


116



117



118

Figure. The plot of the received signal amplitude as a function of frequency (using a

logarithimic scale for frequency)

From the plot above, we can see that the bandwidth of the link is from 1000Hz to 3 MHz.

In this bandwidth, the output amplitude is between 0.8 Vpp and 1.2 Vpp, which has a

relatively flat pattern. The negative peak at 1 MHz is expected to be a measurement error.

10. Apply a square wave or a triangular wave with 1 Vpp and zero DC at the input of the

transmitter at P11. Note the frequency at which the received signal starts getting distorted.

freq (Hz) log freq P31 Amplitude (Vpp)

6 0.78 0.2

50 1.70 0.28

100 2.00 0.44

1000 3.00 0.84

5000 3.70 0.86

50000 4.70 0.9

100000 5.00 0.4

250000 5.40 1

500000 5.70 1.04

1000000 6.00 1.2

3000000 6.48 0.8

5000000 6.70 0.44

6000000 6.78 0.2

7000000 6.85 0.16

8000000 6.90 0.16

6.90, 0.16

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0.00 2.00 4.00 6.00 8.00

amp

litu

de

at

P3

1 (

Vp

p)

log (frequency)

P31 Amplitude (Vpp)

P31 Amplitude (Vpp)

119

Figure. The observed signal (blue) at P11 when the square wave input is at 6 Hz.

Figure. The observed signal (blue) at P11 when the square wave input is at 1 kHz.

120



121


Figure. The observed signal (blue) at P11 when the square wave input is at 5 MHz.

From the observations, the received signal starts to get distorted at 500kHz. In fact, this

frequency is within the predicted bandwidth in the previous step. But we need to notice

that the bandwidth is referred as to pure sine waves. The square wave can be expressed as

a sum of infinite number of harmonic sine waves. The lower the frequency of the square

wave, the greater the amplitude of the low frequency component of the sine wave. Thus

the distortion is milter for lower frequency square wave input signal. For higher

frequency square wave signal, the high frequency sine wave components outside the

122

bandwidth has more significant amplitude. As the pass band of the link cutoffs these

significant high frequency components, the waveform of the received square wave get

distorted.

123

ENTC4310 Optical Fiber Lab 2 Setting Up a Fiber Optic Digital Link

Xiqiao Wang

Introduction

The optical fiber trainer (ODT) can be used to set up two fiber optic digital links, one at a

wavelength of 650 nm and the other at 850 nm. LED1, in the Optical Tx1 block, is an 850 nm

LED, and LED2, in the Optical Tx2 block, is a 650 nm LED.

PD1, in the Optical Rx1 block, is a PIN detector which gives a current proportional to the optical

power falling on the detector. The received signal is amplified and converted to a TTL signal

using a comparator. The GAIN control plays a crucial role in this conversion. PD2, in the Optical

Rx block, is another receiver which directly gives out a TTL signal. Both the PIN detectors can

receive 650 nm as well as 850 nm signals, though their sensitivity is lower at 650 nm.

A digital communication link implies digital data is transmitted over optical fiber. The simplest

transmission link is point-point line that has a transmitter at one end and receiver on the other.

The general block diagram of a digital link is shown below.

Figure. block diagram of a digital link.

There are three types of drive circuits to drive digital data. For digital transmission LED will be

either ON or OFF state. There are special problems that need to be addressed when designing an

LED driver. The key concern is driving the LED so that the maximum speed is achieved. In

series driver circuit it is very simple to construct and when transistor is on then LED glows and

when transistor is off LED is off. But it takes large time for LED to reach the active low of

optical signal when 0 is transmitted. Shunt driver circuit offers much high speed compatibility

and uses transistor to quickly discharge the LED to turn it off. This circuit is several times faster

124

than series circuit. Here the power dissipation is very high than that of series circuit. Faster drive

circuit is used to increase the operating speed than the first two driver circuits. In this circuit two

RC elements are added to shunt drive circuit to improve the performance of the circuit.

Figure. three types of drive circuits to drive digital data.

Objective

The objective of this experiment is to learn to setup 850 nm and 650 nm digital links, and to

measure the maximum bit rates supportable on these links.

Equipment

OFT, two channel 20 MHz oscilloscope, function generator, 1 Hz – 10 MHz.

Procedure

1. Setup the circuit according to the block diagram.

125

Figure. The set up for the optic fiber digital communication link.

126

Figure. The block diagram of the experiment setup.

2. Set the switch SW8 tot the digital position.

3. Connect a 1 m optical fiber between LED 1 and the PIN diode PD1. Remove the shorting

plugs of the coded data shorting links S6 in the Manchester coder block and S26 in the

Decoder & clock recovery block. Ensure that the shorting plug of the jumper JP2 is

127

across the posts B and A1. By doing this, we connect the TTL output of the comparator

to the Manchester decoder.

4. Feed TTL signal of about 20 kHz from the function generator to post B of S6. Use the

BNC I/Os for feeding the observing signal. Observing the received signal at P31 (after

the amplifier) in channel 2 of the scope (triggering the oscilloscope with th e channel 1

signal). Increase and decrease the GAIN and observe the signal cutting off above 3.5 V.

Figure. The observed signal at P31 with the maximum GAIN.

Figure. The observed signal at P31 with the minimum GAIN.

128

Note that to generate a TTL signal using the function generator, we need to set a bias DC

voltage of 2.5 V for a square wave of 5 Vpp and 50% duty cycle.

5. Observe the received signal at post A of S26 on the channel 2 of the oscilloscope while

still observing the signal at P31 on channel 1. Note that the signal at S26 is inverted

version of the signal at P31. This is because the comparator is an inverse comparator.

Vary the gain potentiometer setting. Note that even though the received signal at P31

changes with gain, the output at S26 does not because the simple level comparator only

output the either its supply voltage or zero. Reduce the gain till the signal at P31 is less

than 0.5 V. Here we encountered the situation where the signal does not drop under 0.5 V,

so we pull the fiber out slightly at the receiver to reduce the level below 0.5 V. Note that

the signal at S26 now becomes all high. The comparator reference voltage is 0.5 V, and

unless the signal amplitude is greater than 0.55V, the comparator output is high.

Figure. When setting the potentiometer of GAIN as minimum, the signal at P31 is still

above 0.5 V.

129

Figure. Pull the fiber slightly out of the receiver port, and the amplitude at P31 drops

below 0.5 V. The signal at S26 (blue) becomes all high.

6. Set the gain such that the signal at P31 is about 2 V. Observe the input signal from the

function generator on channel 1 and the received TTL signal at post A of S26 on channel

2. Vary the frequency of the input signal and observe the output response. The screen

shoots of the scope are shown below, where the yellow lines are the TTL signal from

function generator and blue lines are from S26 (received signal after amplification and

comparison stages).

Figure. Set the gain such that the signal at P31 is about 2 V.

130

Figure. the received TTL signal at post A of S26 (blue line) with input signal at 20 kHz.

Figure. the received TTL signal at post A of S26 (blue line) with input signal at 400 kHz.

131

Figure. the received TTL signal at post A of S26 (blue line) with input signal at 1.5 MHz.


132



From the observations above, we can see that at 20 KHz, the received signal at S26 is in

phase with the TTL signal from the generator after the inverse amplifier and inverse

comparator. However, as the frequency increases, phase shift happens on the received

signal. At 1.5 MHz, the signal at S26 is 180 degrees out of phase with the input signal,

but the duty cycle still remains about 50 %. As the frequency increases to 2.2 MHz,

besides of continuing phase shift, the received signal’s duty cycle reduces. At 2.9 MHz,

133

the signal at S26 becomes all low, the transmitted digital signal cannot be recognized any

more.

7. Repeat steps 4, 5, 6 with the 3 m fiber.

7.1.Increase and decrease the GAIN and observe the signal cutting off above 3.5 V.

Figure. The observed signal at P31 with the maximum GAIN, with 3 m Fiber.

Figure. The observed signal at P31 with the minimum GAIN, with 3m Fiber.

7.2. The signal at S26 becomes all high as the signal at P11 drops below 0.5 V

134

Figure. With 3m Fiber, the signal at S26 (blue) becomes all high as the signal at P11

drops below 0.5 V

7.3.Set the gain such that the signal at P31 is about 2 V. Observe the input signal from the

function generator on channel 1 and the received TTL signal at post A of S26 on

channel 2. Vary the frequency of the input signal and observe the output response.

Figure. With 3m Fiber, the received TTL signal at post A of S26 (blue line) with

input signal at 2.7 MHz

135





From the observations above, we can find that the maximum bit rate that can be

transmitted though 3m fiber is about 6.3 MHz. This is much higher than the

maximum bit rate with 1m fiber, which is only 2.9 MHz.

8. Use the 1m fiber and insert it into LED2. Observe the light output at the other end of the

fiber (keep it away from the eyes). The output is a bright red signal. This is because the

136

light output at around 650 nm is in the visible range. The other end now should be

inserted into PD1.

9. Repeat steps 4, 5, 6 with this new link.

Figure. 1m fiber on the LED2-PD1 link, with the maximum GAIN.

Figure. 1m fiber on the LED2-PD1 link, as the signal at P31 drops below 0.5 V, the

signal at S26 becomes all high.

137

Figure. 1m fiber on the LED2-PD1 link, the maximum bit rate is about 7.4 MHz.

10. Use the 3m fiber ans setup the 650 nm digital link between LED2 and PD1. Repeat steps

4, 5, and 6.

Figure. 3m fiber on the LED2-PD1 link, with the maximum GAIN.

138

Figure. 3m fiber on the LED2-PD1 link, as the signal at P31 drops below 0.5 V, the

signal at S26 becomes all high.

Figure. 3m fiber on the LED2-PD1 link, the input TTL has a frequency of 3.2 MHz.

139



140

Figure. 3m fiber on the LED2-PD1 link, the input TTL has a frequency of 8 MHz.

From the observations on the maximum bit rate at 650 nm wavelength, we can see

that, with 1m fiber, the maximum bit rate is about 7.4 MHz. With 3m fiber, two

maximum bit rates are observed, at around 3.5 MHz, the signal at S26 drops to all

low, then as the frequency increases, the transmission resumes. As of the TTL

signal’s frequency increases to 8 MHz, the signal at S26 becomes all high, and it

reaches a second maximum bit rate.

In sum, using 650 nm wavelength will give a higher maximum bit rate than using

850 nm wavelength. And using 3m fiber will give a higher maximum bit rate than

using 1m fiber.

11. Change the shorting plug in jumper JP2 across the posts B and A2. Use the 1m fiber to

connect LED2 and optical receiver PD2. This time, the received signal does not go

through the amplifier.

12. Feed a TTL signal of 20 kHz at post B of S6 and observe the received signal at post A of

S26. Display both the signals on the oscilloscope on channels 1 and 2 (triggering with

channel 1). The receiver at PD2 is an integrated PIN diode and comparator that directly

gives out a TTL signal. Vary the frequency and find the maximum bit rate that can be

transmitted on this link.

141

Figure. With 1m fiber and LED2-PD2 link, the frequency increases to 3.3 MHz, the

received signal at S26 becomes 180 degrees out of phase with the input signal.


link reaches its maximum bit rate.

13. Repeat steps 11 and 12 using 3m fiber.

142


number of peaks of the received signal at S26 begins to mismatch the input signal.


number of peaks of the received signal at S26 further disappears.

143

Figure. With 3m fiber and LED2-PD2 link, the frequency increases to 13 MHz, the link

reaches its maximum bit rate.

In summary, without going through the amplifier stage, the maximum bit rate increases

for both 3m fiber and 1m fiber. But still, 3m fiber has a higher maximum bit rate (13

MHz) than that of 1m fiber (10.3 MHz).

14. Use the 1m fiber to connect LED1 and PD2. Feed a TTL signal of 20kHz at post B of S6

and observe the received signal at post A of S26. Display both the signal on the scope.

An 850 nm TTL to direct digital link is obtained. Vary the frequency and fine the

maximum bit rate.

144


link reaches its maximum bit rate.

15. Repeat step 14 with 3m fiber.

Figure. With 3m fiber and LED1-PD2 link, the frequency increases to 15.6 MHz, the link

reaches its maximum bit rate.

16. Change the shorting plug in JP2 to connect A1 and B (for PD1 receiver selection). Using

the 1m fiber connect LED1 (850 nm) and PD1. Let the GAIN control be at the minimum

level. Feed a 20 kHz TTL signal at post B od S6. Measure the peak-to-peak voltage at

P31 (after the amplifier), and designate it as V1.

145

Figure. The peak-to-peak voltage at P31 with a 850 nm link is V1=1.9Vpp.

17. Now connect the fiber between LED2 (650 nm) and PD1 without changing any other

setting. Measure the peak to peak voltage at P31, and designate it as V2.

Figure. The peak-to-peak voltage at P31 with a 650 nm link is V2=1.2Vpp.

18. The factory setting for the light output at the end of 1m fiber for LED1 is 3dB higher

(twice the power) than that for LED2. The PIN diode current “i” can be written as

i = ρ*P

where P is the optical intensity of the light falling on the detector and ρ is the

respinsitivity. The voltage at P31 is directly proportional to the PIN diode current “i”.

Using the results of step 16 and 17, compare the respinsitivity of the diode at 650 nm and

850 nm using the expression:

Where P1 is twice P2, and ρ1 and ρ2 are responsitivity of the diode at 850 nm and 650

nm. Thus

So, the respinsitivity of the PIN diode at 850 nm is 0.792 times its responsitivity at 650

nm.

146

ENTC4310 Fiber Optics Lab 5 Time division multiplexing of signals

Xiqiao Wang

Introduction

OFT is as much a synchronous time division multiplexing unit as it is a fiber optic

communication unit. The basic multiplexer has twelve 64 kbpsec channels which are time

multiplexed. The multiplexed data stream is Manchester coded and the resulting TTL signal is

fed to a Manchester decoder which recovers the clock and the data.

Time division multiplexing is also the basis of time-switching used today in Telecom switches.

While multiplexing, say the voice signal from port 1, V1 is transmitted before V2, the voice

signal from port 2. But at the receiver, the first received signal can be fed to port 2, and the later

signal to port 1, resulting in switch between the two ports.

If an asynchronous low bit rate signal is to be inserted in a synchronous Mux, the simplest

technique is to sample the input signal using a submultiple of the Mux output clock. However,

this gives rise to jitter in the received signal. This phenomenon is studied in this experiment.

Time-division multiplexing (TDM) is a type of digital (or rarely analog) multiplexing in which

two or more bit streams or signals are transferred apparently simultaneously as sub-channels in

one communication channel, but are physically taking turns on the channel. The time domain is

divided into several recurrent time slots of fixed length, one for each sub-channel. A sample byte

or data block of sub-channel 1 is transmitted during time slot 1, sub-channel 2 during time slot 2,

etc. One TDM frame consists of one time slot per sub-channel plus a synchronization channel

and sometimes error correction channel before the synchronization. After the last sub-channel,

error correction, and synchronization, the cycle starts all over again with a new frame, starting

with the second sample, byte or data block from sub-channel 1, etc.

Consider, for instance, a channel capable of transmitting 192 kbit/sec from Chicago to New York.

Suppose that three sources, all located in Chicago, each have 64 kbit/sec of data that they want to

transmit to individual users in New York. As shown in the figure below, the high-bit-rate

channel can be divided into a series of time slots, and the time slots can be alternately used by

the three sources. The three sources are thus capable of transmitting all of their data across the

147

single, shared channel. Clearly, at the other end of the channel (in this case, in New York), the

process must be reversed (i.e., the system must divide the 192 kbit/sec multiplexed data stream

back into the original three 64 kbit/sec data streams, which are then provided to three different

users). This reverse process is called demultiplexing.

Figure. Block diagram of time division multiplexing

Choosing the proper size for the time slots involves a trade-off between efficiency and delay. If

the time slots are too small (say, one bit long) then the multiplexer must be fast enough and

powerful enough to be constantly switching between sources (and the demultiplexer must be fast

enough and powerful enough to be constantly switching between users). If the time slots are

larger than one bit, data from each source must be stored (buffered) while other sources are using

the channel. This storage will produce delay. If the time slots are too large, then a significant

delay will be introduced between each source and its user. Some applications, such as

teleconferencing and videoconferencing, cannot tolerate long delays.

Objective

The objective of this experiment is to learn to setup the multiplexer and demultiplexer, and to

observe the simultaneous transmission of several channels (two voice and eight data channels)

using time division multiplexing. At the same time, some basic principles of time switching and

asynchronous data interfacing using oversampling are studied.

Equipment

OFT, two channel 20MHz oscilloscope, function generator.

Procedures

148

1. The interfaces used in the experiment are summarized in the lab manual. Set the jumpers

and switches, and short the shorting links as shown in the manual. This experiment

requires the setting up of an 850 nm or 650 nm digital link.

Figure 1. Experimental setup for this lab.

2. Turn on at least one of the switches SW0-SW1 in the 8-bit-data transmit block. This

ensures that the multiplexer is correctly aligned and should be the normal practice

whenever the mux-demux are being used.

3. Connect LED1 in the optical Tx1 block and PD1 in the optical Rx1 block. Use the 1m

optical fiber to set up the 850 nm digital link. Adjust the GAIN control until the LEDs

L0-L7 in the 8-bit-data receive block light up corresponding to the ON positions of SW0-

SW7. When the TDM link is working, the LEDs L8 and L9 in the marker detection block

will be off when out any flicker. The digital link and the TDM MUX-DEMUX are now

set up. Connect the telephone handsets at PHONE 1 and PHONE 2.

149

Figure. The transmitter block diagram for this experiment,

Figure. The receiver block diagram for this experiment.

150

4. OFT is now being used in the loop-back mode. The data and voice channels multiplexed

on the transmit side are demultiplexed on the receive side of the trainer. The voice input

at the mouse piece is now being looped back thought the fiber to the ear piece. We

checked this by disturbing the fiber link by removing the fiber from PD1, while speaking

into the mouse piece of one of the handsets. Now we can no longer hear ourself in the

earpiece.

5. Establish the fiber link again. Remove the shorting plugs of the voice enable shorting

links S7 and S8 in the timing and control block on the transmitter side. Using the patch

cords, interchange the voice slots by interconnecting the slot Select 1 signal [post A of S7]

to the voice Enable 2 [post B of S8] and the slot select 2 signal [post A of S8] to the voice

enable 1 [post B of S7] at the Tx side. Voice 1 and Voice 2 are now cross-connected and

the conversation can be carried out between two people using the two phones. The two

slots carrying voice data are now time-switched to provide the necessary connection.

Carrying on the conversation while at the same time turning the data switches on and off.

We can observe the simultaneous transmission of 8-bit data in one channel with two

voice channels on the link.

Reconnect the shorting links S7 and S8 to restore the original connection. However, now

remove the shorting plugs of the voice enable shorting links S27 and S28 in the timing

and control block on the receiver side, and cross connect them. Now, once again the

voice 1 and voice 2 are cross connected. This cross connection is now on the Rx side.

Note that voice 1 Tx signal is now connected back to Voice 1 Rx. Switching at both

transmitter and receiver cancel out each other.

6. Reconnecting shorting links S7, S8, S27, and S28. Remove the shorting plug of voice 1

shorting link in the voice coder block. Feed a sinusoidal signal of 1 kHz and 1 Vpp with

zero-DC at post B of S1 and display it on channel 1 of the oscilloscope. Trigger the scope

on channel 1. Observe the received signal at voice 1 signal post P23 on channel 2 of the

scope. Vary the frequency of the input signal and observe the received signal. Note the

lower frequency cutoff and the higher frequency cutoff when the output voltage falls to

0.7 Vpp (3 dB below 1 Vpp). From the observations below, the higher frequency cutoff

151

of the CODEC response is 3.57 kHz and the lower frequency cutoff of the CODEC

response is about 140 Hz.

Figure. The higher frequency cutoff of the CODEC response in the mux-demux link.

Channel one is the signal from function generator, channel two is the received signal.

(850 nm link)

Figure. The lower frequency cutoff of the CODEC response in the mux-demux link.

Channel one is the signal from function generator, channel two is the received signal.

(850 nm link)

152

The signal is being digitalized by a CODEC at 64 kbits/sec, multiplexed and transmitted

on the fiber link. The received optical signal is converted to a TTL signal and

demultiplexed to obtain the transmitted signal back. The signal at P23 is the reconstructed

version of the transmitted signal. The frequency response obtained is that of the CODEC

used to digitize and reconstruct the voice signal. If we observe the received signal as

shown in the figure below, we can find the step approximation of the original signal. This

is a typical output of a DAC device.

Figure. The received optical signal is converted to a TTL signal and demultiplexed to

obtain the transmitted signal back. (850 nm link)

7. The multiplexer also multiplexes the TTL signals controlled by switches SW0 – SW7. At

the receiver, the received signal is demultiplexed and the switch inputs are displayed at

the LEDs L0-L7 respectively. According the product manual, OFT also provides for

directly feeding in two low-frequency TTL signals instead of the static switch settings at

SW6 and SW7. If SW7 and SW 6 are kept in the ON position, then asynchronous TTL

signals from a function generator, which means the signals from the function generator

are not synchronized with the clock of the multiplexer, can be inserted at P1 and P2, the

received signal can then be observed at P21 and P22 respectively.

153

Insert a 100 Hz TTL signal at P1. Display the transmitted at channel 1 and the received

signal at P21 on channel 2. Vary the frequency of the TTL signal and observe the two

signals. Sketch the received signal at 500 Hz, 1000 Hz, 2000 Hz and 4000 Hz.

The input and output signal at 2000 Hz will appear as shown in the diagram below. The

output signal transitions between 0 and 1 levels will have an overlap of 125 usec.

Remember that the TTL signal at P1 is sampled at 125 usec (8 kbits/sec) and transmitted

on the fiber link. Increase and decrease the frequency of the transmitted signal in the

range of 100 Hz to 4000 Hz. Note that the overlap remains unchanged due to that they

used the same sample rate for all signals of all the frequencies. The reason for such delay

is because the function generator and the clock of the multiplexer is not synchronized,

this will results in a maximum phase difference of 1 sample period, etc. 125 usec.

Figure. The delay of the received TTL signal (channel 2) when compared with the input

TTL signal at 500 Hz. (850 nm link)

154

Figure. The overlap of the received TTL signal (channel 2) when compared with the

input TTL signal at 500 Hz. (850 nm link)



155





156





Here we use the display function of infinite persistence to demonstrate the overlap of the

received signal at 0 and 1 levels. From the observations above, we can see that for all the

tested frequencies within 100 Hz – 4000 Hz, the overlap remains the same as 125 usec,

and the delay between the received signal at P21 remains 12 usec unchanged. At 4000 Hz,

157

the overlap is the same with half period. For frequencies over 4000 Hz, the overlap

cannot be detected.

8. Now insert the TTL signal at both P1 and P2. Observe the outputs at P21 and P22 on

channel 1 and channel 2 of the oscilloscope. Note that simultaneous transmissions on

both the channels are observed since the two received signal are synchronized by the

same multiplexer clock.

Figure. The received signal at P21 and P22 at 500 Hz. (850 nm link)

The received signal at P21 and P22 at 2000 Hz. (850 nm link)

158

9. Repeat steps 2 to 7 using the 650 nm link, here we use LED2 and PD2. Note that the

MUX-DEMUX work equally well at both 650 nm and 850 nm wavelengths.

Figure. The upper frequency cutoff of the CODEC response in the mux-demux link.

Channel one is the signal from function generator, channel 2 is the received signal. (650

nm link)

Figure. The lower frequency cutoff of the CODEC response in the mux-demux link.

Channel one is the signal from function generator, channel 2 is the received signal. (650

nm link)

159


TTL signal at 100 Hz.(650 nm link)


input TTL signal at 100 Hz.(650 nm link)

160





161





From the observations above, we can see that the delay remains12 usec unchanged for

650 nm link, and the overlap remains 125 usec unchanged.

Summary

162

In this experiment, we observed time multiplexing of two digitized voice signals and

eight bit data signals. If the data signals to be multiplexed is not synchronized to the

multiplexer clock, jitter is observed in the received data. We also carried out time

switching of two voice channels in the experiment.

163

ENTC4310 Fiber Optics Lab 6 Framing in time division multiplexing

Xiqiao Wang

1. Introduction

This is an advanced experiment on time division multiplexing. The previous experiment

demonstrated the simultaneous transmission of several digital signals on a single fiber.

This experiment examines the methods of synchronous multiplexing.

The key to synchronous time division multiplexing is a frame which repeats itself every T

seconds. The frame has n bits, and since the frame rate is 1/T frames/second, the total

data rate is n/T bits/second. A signal occupying m bits per frame will have a data rate of

m/T per second.

2. Objective

The objective of this experiment is to examine the technique of time division

multiplexing. It explains the generation of frame clock, slot clock and bits clock, and the

method of insertion and removal of data from each slot. It also explains the concept of

time switching.

3. Equipment required

OFT, two channel 20MHz Oscilloscope, function generator

Figure . The transmitter block diagram for this experiment

164

Figure . The receiver block diagram for this experiment

4. Procedure

Figure. Experimental setup.

165

4.1.The experiment requires the fiber optics digital link to be set up at 650 nm or 850 nm

as they were set in experiment 2. The interfaces used are summarized in the figure

below.

4.2.Identification of the frames.

The Tx bit clock TxBCLK is available at S5. Observe the signal on the oscilloscope

on channel 1. Trigger the scope with channel 1. The frequency is measured as 768.8

kHz. This is the bit rate of the transmission.

Figure 1. The bit rate of the transmission at Tx bit clock TxBCLK, (port S5).

Now we can observe the signal Tx frame clock (TxFCo) which is available at post P9

in the timing and control block on channel 2 of the oscilloscope. This time, since the

signal we are going to observe is at channel 2, and the bit rate at channel 1 is far

higher than frame rate, we should trigger the scope with channel 2 at negative slope.

Synchronous time division multiplexing uses a repetitive frame, with each signal to

be multiplexed occupying a curtained fixed bit position in the frame. The HIGH as

well as the LOW of TxFCO each corresponds to one frame. The frame rate is two

times the frame clock frequency. Thus the frame rate is 8 KHz as shown below in

figure. 2. The bit rate is about 800 KHz. Thus there are about 100 bits per frame.

Within a second, there are 8000 frames.

166

Figure. 2 The bit rate is at channel 1, and the frame rate is at channel 2.

4.3.Bring the signal from P101 to the breadboard using the 40 core flat cable provided.

The details of the signal on this expansion port are given in Appendix F in the lab

manual.

Signal Name PIN

numbers

Description

TxBC2-

TxBC0

12-14 Gives the binary bit count (on the Tx side) within a

slot, taking binary equivalent values of 0 to 7 with

each successive Tx bit clock

TxSC3-

TxSC0

8-11 Gives the slot count (on the Tx side) within a frame,

taking binary equivalent of 1-11. Slot 0 is reserved for

the marker.

TxBCLK 15 Transmission bit clock. This pin corresponds to the

post B of S5 (Tx clock). This clock can be used

though this pin while S5 is shorted. When the shorting

plug is removed from S5, an external clock can be fed-

in to the Manchester Coder. The data in the Tx data

stream is expected to chnge at the rising edge of this

clock.

TxFCO 7 Tx Frame Clock ot Tx Frame Count 0. Each HIGH or

LOW period corresponds to slots 0 to 11 of a Tx

Frame. Slot 0 carries the marker pattern and when

TxFCO is LOW, even marker is sent out, whereas

when it is HIGH, odd marker is sent out.

Table 1. The PIN number and their functionality descriptions of P101.

167

Observe TxSC0 on channel 2 of the scope. This is the time clock of the slot 0 in each

frame. Trigger the scope on channel 1 with negative slope. Observe the signal

TXBCLK on channel 2 of the scope. Figure 3 below shows the screen shot. Note that

TxSC0 is HIGH for eight bits [eight bit transmission clock cycles] and LOW for eight

bits, whereas, one clock cycle of TxBC2 corresponds to eight data bits. Thus the

TxBC2 may be called slot clock with each slot containing eight bits of data. A source

which needs to transmit eight bits of data can place them in any one of the slot

positions in the frame. It can do so in every frame, using the same slot, and transmit at

a bit rate of eight times the frame rate. The frame and slot timings are shown after the

screen shots for reference.

Figure 3. Channel one is from the clock signal of slot 0 at post TxSC0, and channel 2

is the bit clock signal TxBCLK.

168


is the binary bit count with in two data bits per count at TxBC0.


is the binary bit count with in four data bits per count at TxBC1.

169


is the binary bit count with in eight data bits per count at TxBC2.

170

Figure 7. Frame and slot timing diagrams.

4.4.Now we connect Tx frame clock (TxFCO) at P9 on channel 1 of the oscilloscope

(triggering the scope with channel 1 with negative slope). Observe TxSC0, TxSC1,

TxSC2, and TxSC3 at the appropriate pins of P101, one after the other on channel 2

of the scope. Print these screen shots. Be careful to maintain their respective phase

shifts with reference to TxFC0. Note that there are 8000 frames per second and 12

slots within each frame and 8 bits within each slot.

171

Figure 8. Channel 1 is from the frame clock count at post TxFC0, and channel 2 is the

slot count within a frame at TxSC0.

Figure 9. Channel 1 is from the frame clock count at post TxFC0, and channel 2 is the

slot count within a frame at TxSC1.

172

Figure 10. Channel 1 is from the frame clock count at post TxFC0, and channel 2 is

the slot count within a frame at TxSC2.

Figure 11. Channel 1 is from the frame clock count at post TxFC0, and channel 2 is

the slot count within a frame at TxSC3.

The TxBC0, TXBC1, TxBC2 can be used to insert any bit in a slot, and TxSC0,

TxSC1, TxSC2 and TxSC3 can be used to enable any particular slot in the frame.

Together these lines enable one to select any specific bit within the 96 bit frame.

Time division multiplexing requires each analog signal to be digitalized first and then

placed in one or more bits in a frame. The above timing signals can be used to enable

a device to place data on a common Tx line as shown in the Figure 12. The full frame,

along with the marker marking the beginning of the frame, is shown in Figure 13.

173

Figure 12. Multiplexing several devices.

Figure 13. Frame with Marker and slots.

Data Insertion in Slot

4.5.With TxSC3 at P8 in the timing and control block on channel 1 and triggering on

channel 1 with negative slope, we observed the enable signals at S7, S8, S9 and S10

of the timing and control block on the oscilloscope and sketch them. Note that these

are eight bits active-low slot enable signals which enables slot 1, 2, 3, and 11

respectively in each frame. When slot select 1 at S7 is given to Voice CODEC1, the

eight bits generated by Voice CODEC 1 every 125 μs are placed in Slot 1 of the

outgoing frame. Similarly when Slot Select 3 at S9 is given to the 8-bit-data MUX,

eight bits of data corresponding to SW0 to SW7 are placed in Slot 3 of every frame.

174

Figure 14. Channel 1 is the TxSC3 at P8 as the slot count signal; channel 2 is the

enable signal at S7.



175





Now we connect S9, while it is shorted, to channel 1 of the oscilloscope, triggering

the scope on channel 1 with negative slope, and observe the data transmit signal (Tx

data) at S4 on channel 2. We can see that the eight bits of data in Slot 3 correspond to

the switch positin settings of SW0 to SW7. Toggle any of the switches and we can

see the changes in data on channel 2. Note that the bit corresponding to SW7 is

transmitted out first followed by SW6 to SW0.

176

Figure 18. Channel 1 is the connected with S9 (Tx Slot Select 3) when none of the

eight digital switches are on; Channel 2 is connected with S4 (Tx Data from MUX)

Figure 19. Channel 1 is the connected with S9 (Tx Slot Select 3) when switches SW3,

SW4 and SW5 are on; Channel 2 is connected with S4 (Tx Data from MUX)

Now we connect TxSC3 at P8 to channel 1 of the oscilloscope and trigger the scope

on channel 1 with negative slope; its transition starts slot 0, followed by slot 1 eight

bits later, and slot 2 eight bits later after that. The voice data in in slots 1 and 2 can be

observed on the scope, followed by the data bits corresponding to SW0 to SW7 in slot

3. The data line corresponding to slot 4 to slot 11 will be HIGH since no device on

the OFT places data on the Tx data line during these slots. These are expansion slots

and will be used in advanced experiments. However, we can notice that the data bits

177

in slot 1 and slot 2 are random. When speaking into phone 1, we can notice the flurry

of activity on the data line corresponding to slot 1. Similarly, speak into phone 2 and

we can notice the flurry of activity in the data corresponding to slot 2.

Figure 20. Channel 1 is connected to P8 (TxSC3); channel 2 is connected to S4

(TxData from MUX) with no voice and digital inputs.


(TxData from MUX) when there is no voice input, and the SW3 and SW5 are on.

178


(TxData from MUX) when there is no digital inputs, and voice input from phone 1

give flurries in during slot 1.


(TxData from MUX) with no voice and digital inputs. The data line corresponding to

slot 4 to slot 11 will be HIGH since no device on the OFT places data on the Tx data

line during these slots.

Slot Interchange

4.6.Remove the shorting plugs from S8 and S9. Connect post A of S8 to post B of S9 and

vice versa. Now Slot Select 2 is being given to the data MUX taking input from SW0

to SW7. Observe the eight bits of SW0 to SW7 now occupying the slot 2 position, on

179

channel 2 of the oscilloscope at S4 (Tx Data from MUX). Digitalized voice from

phone 2 now occupying the slot 2 position. Speak into phone 2 and notice that the

voice data causes flurries in slot 3.

Figure 24. In situation where shorting plugs from S8 and S9 are interchanged.

Channel 1 is connected to P8 (TxSC3); channel 2 is connected to S4 (TxData from

MUX) with no voice and digital inputs.



MUX) when there is no voice input, and the SW3 and SW5 are on.

180



MUX) when there is no digital inputs, and voice input from phone 1 give flurries in

during slot 1.

Remove the patch cords from S8 and replacing the shorting plug. Removing the

shorting plug from S7 and connect post A of S7 to post B of S9 and vice versa. Once

again, observe the data corresponding to SW0 to SW7 on channel 2 of the scope. The

Data now occupies Slot 1 as Slot Select 1 is now connected to the data mux. This

interchanging of slots is called Time Switching.

181



MUX) with no voice and digital inputs.



MUX) when there is no voice input, and the SW3 and SW5 are on.



MUX) when there is no digital inputs, and voice input from phone 1 give flurries in

during slot 1.

182

4.7.Turn SW6 on. Insert TTL data from the function generator at a very low rate. Here

we choose 1 Hz. Observe the data bit 2 in Slot 1 corresponding to SW6. Note the data

bit toggling as the TTL input at P2 changes. Below, Figure 30 and Figure 31 are

screen shot at the moments when the TTL input signal is HIGH and LOW.



MUX) when there is no voice input, and the TTL input signal at P2 is LOW.



MUX) when there is no voice input, and the TTL input signal at P2 is HIGH.

Observation of Receive Clocks

183

4.8.Replace the shoring plugs at S7 and S9. Ensure that the fiber optics link is set up and

that the multiplexer and demultiplexer are working. With Tx Clock (S5-TxBCLK) on

the channel 1, observe the signal at S25 on channel 2. The signal here is the Rx Clock

(RxBCLK). Note that the Rx Clock is equal to the Tx Clock, but is a delayed version.

The time delay is measured as 400 ns.

Figure 32. The time delay of the Rx Clock at channel 2 when compared with the

transmission bit data clock Tx Clock at channel 1. The delay is about 400 ns.

4.9.With Tx Frame clock (TxFCO) on channel 1 of the scope, and triggering on channel 1

with negative slope, observe the signal received at P29. This is the Rx frame clock

(RxFCO). Just like the TxFCO, its HIGH as well as its LOW correspond to one frame

each, equal to 96 bit clock.

184

Figure 33. Channel 1 is the Tx Frame clock (TxFCO), and channel 2 is the Rx frame

clock (RxFCO).

4.10. Bring the signals from P102 to the breadboard. With Rx frame clock (RxFCO) on

channel 1, and triggering the scope with channel 1 at negative slope, observe RxBC0,

RxBC1, RxBC2, RxSC0, RxSC1, RxSC2 and RxSC3 at pins 14, 13, 12, 11, 10, 9,

and 8 of P102 respectively. Sketch these signals and note that these received bit

clocks and slot clocks are exactly similar to the Tx bit clock and Tx slot clocks

observed at step 3 and 4. Note that each clock cycle of RxBC2 corresponds to eight

bits or an eight-bit-slot. There is, however, one important difference. Rx slot count

signals RxSC0 to RxSC3, counts slots 1 to 12 instead of slots 0 to 11. Slot 12 is

essentially the marker slot.

Figure 34. Channel 1 is connected to Rx frame clock (RxFCO), and channel 2 is

connected to Rx bit clock 0 (RxBC0).

185


connected to Rx bit clock 1 (RxBC1).


connected to Rx slot clock 0 (RxSC0).

186





187



Time Switching

4.11. Now observe the signals at S27, S28, S29 and S30 in the timing and control block

at the receiver side on the oscilloscope one after another, triggering the scope with

RxSC3 with negative slope. Note that they correspond to Slot 1, Slot 2, Slot 3 and

Slot 11 of the received frame respectively. Now observe the Rx data line at S24 in the

decoder and clock recovery block. Note that the data in Slot 3 still corresponds to

SW0 to SW7, since the Rx Slot Select 3 is given to the data demux; the data bits in

this slot then drive the LEDs L0 to L7.

188

Figure 40. Channel 1 is connected to RxSC3, and channel 1 is connected to post S27

(low-active enable signal of Slot 1).





189



Interchange receive Slot Selects 3 and 1 by removing the shorting plugs of S27 and

S29 and connecting post A of S27 to post B of S29 and vice versa. The data in Slot 3

is now driving the earpiece of Phone 1, while the output from Phone 1 mouthpiece

now drives the LEDs L0 to L7. We speak loudly into Phone 1 and we can observe the

LEDs L0 to L7 changing state vigorously. Turn ON SW7 and insert a TTL signal at

around 1 kHz into P1. The data from switches SW0 to SW6 and toggling data at P1

are now entering the decoder of CODEC 1 and the output analog signal is driving the

earpiece of phone 1. The toggling data at P1 (while the other bits remain constant)

causes a tone at the handset earpiece. Toggling the switches SW0 to SW6 and we can

note the increase and decrease of the tone level. Change the frequency of the signal

input at P1 and we can here the change in the tone.

190

Figure 44. Phone 1 mouthpiece now drives the LEDs L0 to L7.

4.12. Remove the shorting plugs of S7 and S9 and interchange transmit Slot Select 1

and 3 by connecting post A of S7 to post B of S9 and vice-versa. Observe the data at

S24 on the oscilloscope. The signals corresponding to SW0 to SW7 are now in Slot 1.

Since the Rx Slot Select 1 is also connected to the data demux, the signals from SW0

to SW7 now again drive LEDs L0 to L7. Also the voice from Phone 1 is digitalized

and placed in Slot 3 of the Tx Frame, and the data from Slot 3 of the receive frame

goes to the docoder of CODEC 1; gets converted to an analog signal and drives the

earpiece of Phone 1. The slot selects have been interchanged on both the transmit and

receive sides. The only effect of this is in the position that two signals occupy in the

data frame being transmitted. The external devices are not affected and the

multiplexer-demultiplexer now works as before, as if the interchange has not taken

place.

191

Figure 45. In situation where shorting plugs from S7 and S9 are interchanged, while

the shorting plugs from S27 and S29 are interchanged too. Channel 1 is connected to

RxSC3; channel 2 is connected to S24 (RxData from DeMUX) when there is no voice

and digital input signals.

Figure 46. In situation where shorting plugs from S7 and S9 are interchanged, while

the shorting plugs from S27 and S29 are interchanged too. Channel 1 is connected to

RxSC3; channel 2 is connected to S24 (RxData from DeMUX) when there is no voice

input, and SW4 and SW2 are ON.

4.13. Restore the removed shorting plugs on both the Tx and Rx sides. Now

interchange Voice 1 and Voice 2 on the Rx side by connecting post A of S27 to the

post B of S28 and vice versa. The voice data in Slot 1 will now drive the earpiece of

192

Phone 2 while the Voice data in slot 2 will drive the earpieve of Phone 1. Two people

can now talk to each other using the two handsets. Time switching is now taking

place for the two voice channels.


Channel 1 is connected to RxSC3; channel 2 is connected to S24 (RxData from

DeMUX) when there is no voice and digital input signals.

5. Summary

In this experiment, we examined in detail the framing aspects of time division

multiplexing. Frame clocks, slot clocks and bit clocks were observed at both the

multiplexer and the demultiplexer. The method to insert and remove synchronous data in

a frame were examines. The concept of time-switching was studies further.

193

Assignment 1

Xiqiao Wang

October 31, 2012

1.) Describe the different types of modulation used in U.S. Cellular Phone Networks. How

does this differ from EU Cellular Phone Networks.

Advanced Mobile Phone System (AMPS) is an analog mobile phone system standard

developed by Bell Labs. It was the primary analog mobile phone system in North

America through the 1980s and into the 2000s. The receiving frequency range is from

869 to 894 MHz, and its transmitting frequency ranges from 824 to 849 MHz. It uses

Frequency-division Multiplexing (FDMA), and uses Frequency Modulation (FM). Each

channel’s band width is 30 kHz, and it allows 832 channels for each user. In 2002, the

FCC decided to no longer require A and B carriers to support AMPS service as of

February 18, 2008. All AMPS carriers have converted to a digital standard such as

CDMA2000 or GSM.

Digital cellular telephone has been developed to greatly increase the number of

subscribers that can be served with the limited bandwidth available for cellular telephone

use. The main type of digital telephone systems in use in North America is the American

digital cellular (ADC) system, namely, IS-136, IS-95 and IS-54.

IS-136 uses TDMA modulation. It is the second generation of the TDMA digital cellular

system. TDMA operates in North America in the 800 MHz band and 1.9 GHz PCS band.

First introduced in 1994, IS-136 is also known as "Digital AMPS" and "D-AMPS." This

IS-136 digital cellular telephone system which replaced the IS-54 system uses the same

frequency bands as are used for AMPS, the same cell structure, and the same overall

control system design. One difference is the type of modulation used for voice channels.

The AMPS system uses analog FM for voice, while the IS-136 digital system uses

DQPSK modulation. The IS-136 provides three users per channel, while AMPS provides

only one. The AMPS systems have been gradually phased out because of the increased

capability of digital cellular systems.

The IS-95 digital cellular telephone system uses CDMA modulation. Like IS-136, it uses

the same receiving and transmission band range with AMPS. But it only assigns two

channels per user, which is due to that it adopt a wider band width for each channel,

namely, 1250 KHz per channel. IS-95 uses BPSK/DQPSK modulation for signals.

In Europe earlier cellular systems such as ETACS,NMT-450 and NMT-900 are now

being replaced by the pan-European digital cellular standard GSM (Global System for

194

Mobile) which was first deployed in 1990 in a new 900-MHz band which all of Europe

dedicated for cellular telephone service. The GSM standard has gained worldwide

acceptance as the first universal digital cellular system with modern network features

extended to each mobile user, and the leading digital air inter-face for PCS services

above 1,800 MHz throughout the world.

The major difference between the GSM system and the American digital cellular system

is that GSM uses GMSK modulation while IS-136 and IS-54 uses /4DQPSK

modulation and IS-95 uses BPSK/DQPSK modulation. GSM provides 8 users per

channel, while IS-136 and IS-54 only provide 3 users per channel, but IS-95 provides 118

users per channel. The most commonly used receiving range for GSM is 935 to 960 MHz

and transmission range from 890 to 915 MHz. But both the American digital and analog

cellular systems use a transmission frequency range of 824 to 846 MHz and a receiving

frequency range of 869 to 894 MHz.

2.) The transmitted power is 275 watts what is this in dBW?

3.) What organization is responsible for worldwide frequency allocation. How represents the

U.S.

The frequency allocations for the different communication services are provided on a

worldwide basis by the International Telecommunication Union (ITU). The allocations

for Region 2, which includes North and South America, have been adopted by the U.S.

Federal Communications Commission (FCC), which has responsibility for RF spectrum

management within the United States. Individual frequency assignments or authorizations

must be requested by the user and subsequently approved by the FCC before the user

may transmit at the requested frequencies. Other FCC regulations also must be followed,

such as the maximum transmitted power level, out-of-band harmonics, and spurious

signal levels.

4.) What was the name of the two navigation system that are no longer operational – Transit

is not one of them. Why were they allowed to end operation?

Omege system

OMEGA was the first truly global radio navigation system for aircraft, operated by the

United States in cooperation with six partner nations. It enabled ships and aircraft to

195

determine their position by receiving very low frequency (VLF) radio signals transmitted

by a network of fixed terrestrial radio beacons, using a receiver unit. It became

operational around 1971 and was shut down in 1997. The OMEGA system normally only

achieves an accuracy of 2 – 4 nautical miles within most of its coverage areas. Besides, it

suffers errors like propagation model errors, system errors due to the earth curvature,

ground conductivity errors, the diurnal effect and ambiguity. With the Global Positioning

System (GPS) being declared fully operational, and the numerous advantages of GPS

system over OMEGA, the use of OMEGA had dwindled to a point where continued

operation was not economically justified. Omega was permanently terminated on

September 30, 1997 and all stations ceased operation.

Loran-C

LORAN-C was originally developed to provide radionavigation service for U.S.

coastal waters and was later expanded. The system provided better than 0.25 nautical

mile absolute accuracy for suitably equipped users within the published areas.

Additionally, it provided navigation, location, and timing services for both civil and

military air, land, and marine users. It was approved as an en route supplemental air

navigation system for both Instrument Flight Rule (IFR) and Visual Flight Rule (VFR)

operations.

Because the high costs of maintaining the system can no longer be justified for a

small segment of the population using LORAN-C, and the signal is no longer required by

the armed forces, the transportation sector (air, land and maritime users), or the nation’s

security interests as the availability of GPS navigation has replaced the LORAN-C in

most applications, the Coast Guard published a Federal Register notice on Jan. 7, 2010,

regarding the termination of the transmission of the LORAN-C signal on Feb. 8, 2010. A

LORAN Programmatic Environmental Impact Statement Record of Decision states that

the environmentally preferred alternative is to decommission the LORAN-C Program and

terminate the North American LORAN-C signal. As noted on the US coast guard website,

the decision to cease transmission of the LORAN-C signal also reflects the president’s

pledge to eliminate unnecessary federal programs.

5.) What are the two primary GPS frequencies?

The GPS satellites transmit a microwave signal composed of two carrier frequencies (sine

waves) which are modulated by two digital codes and a navigation message. The two

carriers are called L1 carrier and L2 carrier. L1 carrier is generated at 1575.42 MHz, and

L2 carrier is generated at 1227.60 MHz. The two digital codes are called course

acquisition (C/A-code) for civilian use and precision code (P-code) for military use.

These two codes are generally called pseudorandom noise cod (PRN). Each satellite

196

transmits a unique PRN code, which does not correlate well with any other satellite's

PRN code. The C/A code is a 1,023 bit sequence which repeats every millisecond. The P-

code is a very long binary code sequence that repeats itself every 266 days and is

transmitted 10 times faster than the C/A-code. The L1 carrier is modulated by both the P-

code and C/A-code, while the L2 carrier is only modulated by P-code. The modulation

type for both L1 and L2 carrier is a bi-phase shift keying (BPSK) modulation (0 and 180

phase change as modulated by digital signal 0 and 1). A critical benefit of having two

frequencies transmitted from one satellite is the ability to measure directly, and therefore

remove, the ionospheric delay error for that satellite.

6.) Discuss the difference between Pseudorange measurements and Carrier Phase

measurement.

Both the pseudorange measurement and the carrier phase measurement are means for

GPS system to measure the range, or distance, between the GPS receiver’s antenna and

the GPS satellite’s antenna.

Pseudorange measurement uses the PRN code, either C/A-code or P-code to measure the

range. It assumes that the clocks, which control the signal generation on both the receiver

and satellite, are perfectly synchronized. Thus the two clocks will generate the same

sequence of signal simultaneously. Since the time it takes for the signal generated from

the satellite to travel thought the space and be picked up by the receiver can be measured

at the receiver by delaying the receiver’s locally generated signal until it matches the

received signal from the satellite, the distance then can be calculated by multiplying the

speed of light with the measured delay-time. Unfortunately, the clocks on the receiver

and satellite are generally not strictly synchronized, the real delay time is contaminated

by the synchronization error between the two clocks. Besides, as the transmitted signal

travels through the atmosphere, errors will be introduced by simply applying the speed of

light when travelling in vacuum along straight paths.

The carrier-phase measurement uses the carrier (pure sine wave) phase instead of the

digital code sequence to measure range. In principle, the range between the receiver’s

antenna and the satellite’s antenna will be the total number of the full carrier cycles plus

the fractional cycles, and then multiply it by the carrier wavelength. Since the

wavelengths of the carrier waves are very short, approximately 19cm for L1 and 24cm

for L2, compared to the C/A and P code lengths. Thus the precision of carrier phase

measurement can be 154 times (using L1 carrier wave) or 120 times (using L2 carrier)

higher than using P-code, and 1540 times or 1200 times higher than using C/A-code.

Unfortunately, tcarrier phase signals provides no time of transmission information. The

197

carrier signals, while modulated with time tagged binary codes, carry no time-tags that

distinguish one cycle from another. The measurements used in carrier phase tracking are

differences in carrier phase cycles and fractions of cycles over time. In other words, time-

of-transmission information for the L-band signal cannot be imprinted onto the carrier

wave as is done using PRN codes. At least two receivers track carrier signals at the same

time.

198

Assignment 2

Xiqiao Wang

October 31, 2012

1.) Given a transmitted frequency of 1.7 GHz and a radial velocity of 92 MPH what is the

Doppler frequency (watch units!)?

The radial velocity of the vehicle with respect to the radar v = 92 MPH = 41.128

meter/second, a the transmitted frequency f = Hz.

The Doppler frequency

2.) Calculate the effective capture area given a frequency of 645 MHz and a receive

directional antenna gain of 7 dBi (be sure to convert this to a numerical value).

The effective capture area of the receiver antenna is the antenna gain times the area of an

ideal isotropic antenna for the frequency of interest. An the gain of the antenna in a given

direction is a measure of how the power level in that direction compared with that which

would exist if the isotropic antenna had been present. The numerical receive antenna gain

is

Where g is the receiver’s gain in dBi. The wavelength of the frequency of interest is

Thus the effective capture area is

199

3.) Calculate the free space path loss for the following: Range 5.6 kilometers, and a

frequency of 267 MHz.

The free-space path loss (FSPL) is the loss in signal strength of an electromagnetic wave

that would result from a line-of-sight path through free space (usually air), with no

obstacles nearby to cause reflection or diffraction. It does not include factors such as the

gain of the antennas used at the transmitter and receiver, nor any loss associated with

hardware imperfections. FSPL is proportional to the square of the distance between the

transmitter and receiver. This is introduced by the spreading out of electromagnetic

energy in free space. FSPL is inversely proportional to the square of the frequency of the

radio signal. This relates to the antenna’s capture area.

Numerically, the FSPL is,

In decibel,

4.) Given:

Transmitted power 40 Watts

Transmitter and Receiver antenna gain of 8 dBi

Frequency 300 MHz

Range 27 Miles

Cable losses TX: 1.1dB RX:.8dB

What is the power received?

Assuming the excess path loss in this case is 0 dB.

The transmitter output power in decibel is

The wavelength is

200

The range is

The received power at the input to the receiver is

=-61.825 dBm

Where is the transmitter antenna gain, is the receiver antenna gain, is the excess

path loss, is the cable loss between transmitter an antenna, is the cable loss

between receiver and antenna.

5.) Find a commercial receiver that would work given the limits above. It may operate on a

frequency range of 110 MHz to 400MHz. (Hint – what is the sensitivity of the receiver?)

Receive sensitivity indicates how faint an RF signal can be successfully received by the

receiver. The lower the power level that the receiver can successfully process, the better

the receive sensitivity. Because receive sensitivity indicates how faint a signal can be

successfully received by the receiver, the lower power level, the better. In fact, the

receiver’s sensitivity is a function of overall gain. The more gain a receiver has, the

smaller the input signal necessary to produce a desired level of output. Another factor

that affects the sensitivity of a receiver is the signal-to-noise (S/N) ratio.

For the case here, the sensitivity requirement for the receiver is at least -91.825 dB, or -

61.825 dBm. Under a given accessible false alarm rate, this requirement may be tightened

by the noise temperature and signal-noise ratio of the given receiver. The specified 110

MHz to 400 MHz receiving band range is within Very High Frequency (VHF) range.

There are two possible commercial receivers that may be used for the application

mentioned above.

201

1. UAA3201T UHF/VHF remote control receiver (manufactured by PHILIPS

semiconductor)

The UAA3201T is a fully integrated single-chip receiver, primarily intended for use

in VHF and UHF systems employing direct AM Return-to-Zero (RZ) Amplitude

Shift Keying (ASK) modulation.

Wide frequency range from 150 to 450 MHz

input reference sensitivity is -105 dBm at the reference voltage of 433.92

MHz, at data rate = 250 bits/s, with the bit error rate (BER) less than

2. ADF7020-1 High Performance FSK/ASK Transceiver IC

Frequency range 80MHz ~ 650MHz

Typical receiver sensitivity -119dBm at reference voltage of 315 MHz and

BER=

Modulation: ASK, FSK

Typical application:

Low cost wireless data transfer, wireless medical applications, remote

control/security systems, wireless metering, keyless entry, home automation,

process and building control.

Here the bit error rate or bit error ratio (BER) is the number of bit errors divided by the

total number of transferred bits during a studied time interval. BER is a unitless

performance measure, often expressed as a percentage.

6.) Define Noise Power, Noise Temperature, Noise Figure

Noise Power:

At the receiver of a communication system, the additive noise can be of thermal origin

(thermal noise), or can be from other noise generating process. The sources of noise are

usually indistinguishable. We define the power the all the noise over bandwidth of

interest that is present at the detecting point as the noise power. The definition of noise

power in terms of the noise temperature (defined below) is

Where k is Boltzmann’s constant and T is noise temperature, B is bandwidth.

Noise temperature:

202

Most noise generating process will give noise which has a white spectrum, identical to

the thermal noise. Under the assumption that all the noises resemble the thermal noise

over the bandwidth of interest, the contribution of all noise sources can be lumped

together and regarded as a level of thermal noise. Thus the noise temperature can be

defined as the noise power density over Boltzmann’s constant.

The equivalent noise temperature of a receiving system represent an equivalent input

noise power. Here the equivalent input noise power will give the real output noise power

through a fictitious receiver whose components are noiseless. Since the noise temperature

is in linear dependence on the noise power, and the equivalent noise power can be

expressed as

Where is the noise power at the input of the antenna, and is the gain of antenna in

numerical form, is the noise power generate by the transmission line system, is the

gain of transmission line in numerical form. And is the noise power generated from

the low noise amplifer, and so on. Then the equivalent temperature of the receiving

system can be expressed as

Noise Figure:

The noise figure specifies the increase in equivalent noise power (referred to the input of

a receiving system) due to a components in the system when its input noise temperature

is The numerical form of noise figure is

Where is the equivalent temperature of the receiving system, and is call the

reference temperature, and it is usually chosen as 290 K. And the equivalent noise figure

for a system in cascade is given by

203

The noise figure can also be defined in terms of the signal-noise-ratio (SNR). It measures

the SNR degradation caused by the network. It can be defined as the ratio between the

SNR at the input port to the SNR at the output port, taking a two-port network for

example.

Where Si is the signal power at the amplifier input port, Ni is the noise power at the

amplifier input port, Nai is the amplifier noise referred to the input port, and G is the

amplifier gain.

7.) Given example shown in Section 6.8 if we need a S/N of 30 dB what physical parameters

could we change to make this happen. Show the new calculation for S/N (it may exceed

30 dB S/N).

There are mainly four parameters in the radar system that one can manage to change in

order to increase the signal-noise-ratio, namely, the transmitter output power, the

transmitter antenna gain, the receiver antenna gain, and the frequency of the radar signal.

The antenna gain of an antenna when it is used as a receiver antenna is the same as the

antenna gain when it is used as a transmitter antenna. Since the effective capture area of a

receiver antenna is proportional to the antenna gain as follows

So, if we will remain the signal frequency and the transmitter power unchanged, we can

increase the effective capture area of the antenna to

times, so that increases the

antenna gain by 5 dBi.

Or equivalently we can increase the directivity of the antenna, in other words, to

decrease the (-3) dB azimuth beam angle and (-3) dB elevation beam angle of the antenna

signal lobe. From the formula

By decreasing the product of the two angles by

times, the antenna gain will

increases by 5 dBi.

204

Based the two proposals which increase the antenna gain by 5 dBi, the new calculation

for the S/N value is

dB

8.) Describe one of the Colpitts oscillator, Hartley Oscillator, or Wein Bridge Oscillator.

Hartley Oscillator

A Basic Hartley Oscillator (Figure cited from http://www.electronics-

tutorials.ws/oscillator/hartley.html)

In the Hartley Oscillator the tuned LC circuit is connected between the collector and the

base of the transistor amplifier. The placement of the tap on the coil determines the

amount of energy fed back to sustain oscillations. The output signal can be taken directly

from the entire tuned circuit, from a tap on the coil, or from a secondary winding as

shown here. In each case, the output signal will be a good quality sine wave. The output

can also be taken from the drain (or collector) of the transistor, allowing the

semiconductor to amplify the signal. However, in that case the waveform is distorted,

containing significant harmonic energy.

http://www.electronics-tutorials.ws/oscillator/hartley.html

http://www.electronics-tutorials.ws/oscillator/hartley.html

205

When the circuit is oscillating, the voltage at point X (collector), relative to point Y

(emitter), is 180 degrees out-of-phase with the voltage at point Z (base) relative to point

Y. At the frequency of oscillation, the impedance of the Collector load is resistive and an

increase in Base voltage causes a decrease in the Collector voltage. Then there is a 180

degrees phase change in the voltage between the Base and Collector and this along with

the original 180 degrees phase shift in the feedback loop provides the correct phase

relationship of positive feedback for oscillations to be maintained.

The amount of feedback depends upon the position of the "tapping point" of the inductor.

If this is moved nearer to the collector the amount of feedback is increased, but the output

taken between the Collector and earth is reduced and vice versa. Resistors, R1 and R2

provide the usual stabilizing DC bias for the transistor in the normal manner while the

capacitors act as DC-blocking capacitors.

In this Hartley Oscillator circuit, the DC Collector current flows through part of the coil

and for this reason the circuit is said to be "Series-fed" with the frequency of oscillation

of the Hartley Oscillator being given as.

√

Where , M is the mutual inductance between the two separated coils

L1 and L2.

The frequency of oscillations can be adjusted by varying the "tuning" capacitor, C or by

varying the position of the iron-dust core inside the coil (inductive tuning) giving an

output over a wide range of frequencies making it very easy to tune. Also the Hartley

Oscillator produces an output amplitude which is constant over the entire frequency range.

206

ENTC4310 Assignment 3 and 4 (This will also count as 10 points extra Credit)

Xiqiao Wang

December 10, 2012

Note: This may sound easy but will take some time to finish.

1.) You have been requested by the local TV station to set up a Microwave relay system for a

remote news truck. The station microwave (Omni) antenna is atop Black Mountain at 3,000

feet the rest of the geography is flat. The news editor wants to also have a VHF voice link

with the the base antenna at Black Mountain.

His questions are:

a) How far out can the news truck go?

Since Transmitting and receiving antennas must be within line of sight, and due to the

refraction of atmosphere, an adjustment factor to account for refraction should be

included into our calculation of the maximum propagation range. Thus the effective

(radio) line of sight is

√ √

Where

d = distance between antenna and horizon (km)

h = antenna height (m)

K = adjustment factor to account for refraction, rule of thumb K = 4/3

b) What transmission power does he need for the Microwave and VHF link.

The transmission powers of both the Microwave signal and VHF signal are

determined by multitudes of factors. For a Line-of-Sight free space transmission model,

the required transmission power determined by free space loss (FSL) based upon the

distance between the transmitter and receiver and the transmitted wavelength.

Additionally, the sensitivity of the receiver and the antenna gain of both the transmitter

and receiver also determine the required transmission power in free space. The Friis Free-

207

Space Model equation is listed below, where the is the sensitivity of the receiver, and

L is the system loss (<=1).

However, the wireless transmission pass is not ideal practically, and there are

noises introduced externally from the environment and the internal noise generated by the

transmitter and receiver devices. So we have to leave a link margin, which requires extra

transmission power. The link margin equation is as follows,

Where EIRP is the effective isotropic radiation power, this value of EIRP depends on the

actual transmitter power and the directivity (azimuth and elevate beam angle) of the

transmission antenna. But it should be noted that, as being indicated in the question, the

base antenna is a microwave (omni) antinna. Eb/N0 is the bit energy per noise power

spectral density. The link margin relates the actual received Eb/N0 value with the

required Eb/N0 value to ensure a satisfactory error tolerance of the communication

requirement. T0 is the temperature at which the system is working. R is the digital data

transmission rate. L0 is the power loss due to other factors in the wireless pass, such as

atmospheric absorption, multipath, refraction and detraction, etc. For the TV signal

microwave transmission, the link margin should be higher than the VHF link margin.

In order to determine the transmission power, we can assume the worst cases for

each power loss and noise contribution factor.

c) If he buys a tower for the truck that will increase antenna elevation by 40 feet how

much will this help on distance from Black Mountain.

This will help on increasing the possible communication distance. The maximum

distance between two antennas for Ling-of-Sight propagation is

√ √ √

√

Where h1=3000 feet = 914.4 meters, and h2 = 40 feet = 12.192 meters.

The maximum distance will increase by (139.048-124.65) km = 14.4 km.

Ld

GGPdP rtt

r 22

2

)4()(

208

d) What type of transceiver should he buy for the VHF link? (Don’t worry about

Microwave for this one.)

I suggest them to buy Icom IC-2200H 65W vhf transceiver, which transceives at 144

MHz and has a stable 65 W output power.

Figure.1 Icom IC-2200H vhf transceiver.

Figure.2 Specification of Icom IC-2200H vhf transceiver.

209

From the specification above, we can see that the transmission range is from 144

to 148 MHz, and receiving range is from 136 to 174 MHz. The impedance of the antenna

connector is 50 ohms, thus we should use a coax transmission line with a characteristic

impedance of 50 ohms and matches the antenna in order to have the maximum power

transmission.

The receiver has a sensitivity of less than 0.18 uV at a SINAD of 12 dB. Here it

means that the receiver will produce intelligible speech with a signal at its input as low as

0.18μV, and the intelligible speech can be detected 12dB above the receiver's noise floor

(noise and distortion). Here SINAD stands for Signal-to-noise and distortion ratio, which

has the following expression,

.

The receiver’s sensitivity can be calculated as

e) What type of antennas should he buy and who makes it for the VHF link system –

this requires real data not made up data show manufacturer.

Figure.3 The PCTEL MWV1322S wide-band, field-tunable, VHF mobile antenna.

I suggest them to buy the PCTEL MWV1322S wide-band, field-tunable, VHF mobile

antenna. The antenna operates within the frequency rangef of 132-174 MHz, which

covers the transmission frequency of the Icom IC-2200H 65W vhf transceiver , 144

MHz. This vehicle antenna provides a gain of 2.4 dB with a ground plane or unity gain

without a ground plane. Maximum power input is 200 watts. As indicated in the data

210

sheet, for two-way communication, this PCTEL wideband VHF mobile antenna provides

26 MHz bandwidth without compromising performance.

f) Will weather conditions effect this link and how much.

The VHF (very high frequency) for voice communication ranges from 30 to 300 MHz.

Whereas the microwave ranges from 300 MHz to 300 GHz, among which the TV signal

bands ranges from 512 to 806 MHz, according to the spectrum chart of United States

Frequency Allocations.

Figure. 4 The frequency allocation of the broadcasting TV channels within microwave

band.

The two figure below show the atmosphere attenuation coefficient and rain attenuation

coefficient versus frequency. We can see that both rain and atmosphere absorption will

not have obvious attenuation effects on RF signal until the frequency increases beyond

5.0 GHz. Both the VHF frequency and the microwave TV signal frequency are below 1

GHz, and the attenuation coefficients due to weather effects are below 0.01 dB/km. So

we can conclude that the weather condition will only have very mild attenuation effects

on the transmitted signal and the link should not be affected.

211

Figure. 5 Horizontal attenuation due to oxygen and water vapor in atmosphere. (Chen,

1975)

212

Figure. 6 The logarithmic plot of the RF power attenuation by rain in dB/km. (Chen,

1975)

g) Show a link analysis for the VHF system with the knowledge that the Black

Mountain base station has a receiver sensitivity of -90dB and a transmit power of

100 Watts.

Since the effects from atmosphere attenuation and weather condition attenuation can be

negligible, we can adopt the free-space model to conduct the link analysis. So the one-

way excess propagation loss In the link analysis below, we assume the

transceiver’s antenna has a grounded plane, so it will have a gain of 2.4 dB. And we also

assume the antenna at the base station has a unit gain. We choose the distance to be the

maximum range D=124.65 km=124650 meters.

Where is the transmitted power,

is the transmitter antenna gain,

is the receiver antenna gain

is the wavelength of the transmitted carrier signal.

is the one-way excess propagation loss (>=1).

Down Link Analysis

For the link of transmitting signal from the base station and receiving by the

transceiver in the truck, we choose the frequency to be the middle value of the

transceiver’s receiving band range, f = (136+174)MHz / 2 = 155 MHz, so λ =

1.94 m.

This received power is much greater than the truck transceiver’s sensitivity, -

134.9 dB.

Up Link Analysis

For the link of transmitting signal from the truck transceiver and receiving by the

base station, we choose the frequency to be the middle value of the transceiver’s

213

transmission band range, f = (144+148)MHz / 2 = 146 MHz, so λ = 2.05 m. We

also set the transceiver to work at its maximum transmission power level, 65 Watt.

This value is less than the base station receiver sensitivity of -90 dB. Notice that

this is based on the assumption that the base station antenna has a unity gain, 0 dB.

So, we need to choose a base station antenna with a gain of at least 8 dB.

2.) What is the Q of an RF circuit?

In electronics communications, the Q (quality factor) is a dimensionless parameter that

indicates the energy losses relative to the amount of energy stored within an RF circuit. The

higher the Q the lower the rate of energy loss and hence oscillations will reduce more slowly.

In an RF circuits, energy losses are caused by resistance. Although this can occur anywhere

within the circuit, the main cause of resistance occurs within the inductor. The Q factor

definition equation is as follows,

In terms of the bandwidth in a RF circuit, the Q value relates directly in to the bandwidth of

the resonator with respect to its central frequency. As the Q increases, the bandwidth of the

tuned circuit will decrease, so that we can obtain a better selectivity of the circuit, which

could be RF filter or RF amplifier. Additionally, high-Q circuit can also help to eliminate the

image frequency in the IF amplifier stage in a superheterodyne receivers. For RF

applications where there is a requirement for wide bandwidth operation, a defined balance

between Q factor and the selectivity may be required. By controlling the Q of a resonant

circuit, you can set the desired selectivity. The optimum bandwidth is one that is wide

enough to pass the signal and its sidebands but narrow enough to eliminate signals on

adjacent frequencies.

214

Figure. 7 Q of a RF circuit with respect to its bandwidth (Poole, 2010).

3.) The input frequency is 101.5 MHz and the LO is 10.7 MHz what is the IF?

When the RF signal received by a superheterodyne receiver is amplified by the RF amplifier,

it is input to a mixer. The mixer also receives an input from a local oscillator or a frequency

synthesizer. The mixer output signals with frequencies of both the sum and difference of

these two signals. The IF amplifier selectsthe signal of the difference frequency, called the

intermediate frequency. Note that the local oscillator’s frequency may be above or below the

desired signal frequency based on different circuit designs.

Figure 8. The relation between the desired signal frequency, intermediate frequency, and

image frequency.

4.) What is the simplest mixer circuit

Mixers accept two inputs: The signal to be translated to another frequency is applied to one

input, and the sine wave from a local oscillator is applied to the other input. Like an

215

amplitude modulator, a mixer essentially performs a mathematical multiplication of its two

input signals. Any device or circuit whose output does not vary linearly with the input can be

used as a mixer. One of the simplest and most widely used types of mixer is the simple diode

modulator as shown below.

Figure. 9 circuit diagram for a single-ended diode mixer.

The input signal is applied to the primary winding of the transformer and coupled to the

secondary winding where it is linearly added with a local oscillator signal coupled through a

capacitor. the diode produces the sum and difference frequencies. The bandpass filter is

tuned to the desired intermediate frequency (IF), either the sum or the difference frequency.

However, there are several disadvantages of such a single ended diode mixer compared with

other types of mixers at the lower frequencies. It has a relatively high noise figure, a high

conversion loss, high-order nonlineararities, no isolation between the LO and the RF inputs,

and large output current at the LO frequency. Its main application is at microwave

frequencies, where other types of mixer may not be practical.

5.) In a direct conversion (zero IF) receiver at what frequency is the Local Oscillator set at?

The local oscillator (LO) frequency is set to the frequency of the incoming signal.

The direct conversion (DC), also called zero IF (ZIF) receiver, converts the incoming signal

directly to baseband without converting to an IF. It performs demodulation as part of the

translation. The low-noise amplifier (LNA) boosts the signal before the mixer, and Baseband

output is passed via a low-pass filter (LPF). The block diagram of the circuit is shown below.

216

Figure.10 A direct-conversion (zero-IF) receiver

6.) What is circuit sharing?

Circuit sharing is a design concept that a set of circuits are used by two or more functional

circuits with in the design to realize multiple functions. A typical application in RF

communication is the transceiver which is a device comprising both a transmitter and a

receiver which are combined and share common circuitry, a single housing and power supply.

Transceivers can share circuits, thereby achieve cost savings, and in some cases are smaller

in size. Similar concepts are used in duplexer which is a device that allows bi-directional

(duplex) communication over a single path. It isolates the receiver from the transmitter while

permitting them to share a common antenna.

7.) What is the phase constant for 468 MHz?

Phase constant B, also called wavelength number or wavelength constant, is defined as the

number of wavelengths within a length of 2π.

8.) Calculate Skin Depth for the cable spec given in D2L.

Skin Effect states that as the transmitted signal’s frequency increases, depth of penetration

into adjacent conductive surfaces decreases for boundary currents associated with

electromagnetic waves. This results in the confinement of the voltage and current waves at

the boundary of the transmission line, thus making the transmission more lossy.

The skin depth is given by:

217

where f = frequency, Hz

μ = permeability, H/m

γ = conductivity, S/m

From the data sheet of RG-58A/U type coaxial, we know that the inner conductor wire is

made of 035" tinned copper.

Table 1. The description of the coax inner conductor (upper), and the conductivity and

permeability of copper (middle and below) (Education, 2001).

Pure copper’s conductivity is 5.8E+7 simens/m, and its magnetic permeability is 1.257E-6

H/meter.

Now we substitute these values into the equation above,

√

The following is a table of skin depth for some frequency values.

218

Table 2. Skin depth for sample frequencies.

9.) Pi is 50 watts and Pr is 3.4 watts what is the mismatch and how could you fix this?

When an electromagnetic wave travels down a transmission line and encounters a

mismatched load or a discontinuity in the line, part of the incident power is reflected back

down the line. The return loss is defined as

The term mismatch loss is used to describe the loss caused by the reflection due to a

mismatched line. It is defined as

In the case of current problem, the mismatch will be

We can add devices between the transmission line and the load to match the impedance

of the effective load to the characteristic impedance of the transmission line, .

Notice that if the characteristic impedance of the transmission line has a reactive component,

then the Z values here are both complex numbers. But usually the characteristic impedance

of a transmission line, like coax, is purely resistive.

There are a variety of devices used between a source, of energy and a load that perform

"impedance matching". To match electrical impedances, engineers use combinations of

transformers, resistors, inductors, capacitors and transmission lines. These passive (and

active) impedance-matching devices are optimized for different applications.

219

10.) In 1977 a light signal was sent on an early fiber optic cable the cable was 32 miles

long what was the loss in dB and in percentage of the original signal at the receiver?

Upon the optic cable set up in 1977, telephone signals used infrared light with a wavelength

of 850 nm to send data at 6.2 Mbps and 45 Mbps. The loss was 2 dB per km, and repeaters

were required every few kilometers.

Since 32 mile =51.5 km

The total Gain in dB is

(

)

11.) Using the formula provided in the selected reading and question 10 above what is

the power ratio?

The power ratio ( ) over an optical fibre of length x (km) is defined as

where P(x) is the power received, P(0) is power transmitted, and a is the fibre attenuation in

dB/km. So,

12.) What is a step indexed fiber and why is it used?

Step-index fibers have a uniform core with one index of refraction, and a uniform cladding

with a smaller index of refraction. When plotted on a graph as a function of distance from

the center of the fiber, the index of refraction resembles a step-function. The figure below

illustrates how the index of refraction varies with location in a cross-section of a step-index

fiber.

220

Figure.11 The index of refraction varies with location in a cross-section of a step-index fiber

(Neets, 2011).

Now let us assume a step-index fiber has a core of radius R1 and a constant refractive

index n1. A cladding of slightly lower refractive index n2 surrounds the core. The step decrease

occurs at a radius equal to distance R1. The difference in the core and cladding refractive index

is the parameter , which is the relative refractive index difference.

The ability of the fiber to accept optical energy from a light source is related to ; additionally,

also relates to the numerical aperture by

The bare glass fiber in air is a kind of step-index fiber with the environmental air being

the cladding material. The purpose of using step-index fiber is to increase the relative refractive

index difference between the core and the cladding in order to have a larger brewster angle for

full reflection of the light signal within the core fiber. In this case, the fiber will allow the light

reaching the boundary of the core fiber with a larger incidence angle still be full reflected back.

Additionally, the larger relative refractive index difference can increase the numerical aperture,

which means that light can have a larger incident angle into the core fiber and can still stay

within the core fiber during the transmission.

Some characteristics of step-index multimode fiber are

• Large core size, so source power can be efficiently coupled to the fiber

• High attenuation (4-6 dB / km)

• Low bandwidth (50 MHz-km)

• Used in short, low-speed datalinks

• Also useful in high-radiation environments, because it can be made with pure silica core

221

Bibliography Chen, C. (1975, April). The Attenuation of Electromagnetic Radiation by Haze, Fog, Clouds, and Rain.

Rand Corporation.

Education, N. (2001). Conductivity and Resistivity Values for Copper & Alloys. Collegepark, Maryland,

USA.

Neets. (2011, 6). MULTIMODE STEP-INDEX FIBERS. Retrieved 12 3, 2012, from The Integrated Publishing:

http://www.tpub.com/neets/tm/107-1.htm

Poole, I. (2010). Q Quality Factor Tutorial. Retrieved 12 9, 2012, from radio-electronics:

http://www.radio-electronics.com/info/formulae/q-quality-factor/basics-tutorial.php

entc 4310 - electronics-communications lab report course projects/0000entc4310 lab... · optical...

Documents