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ENTC 4310 - Electronics-Communications
Lab Report
Xiqiao Wang
East Tennessee State University
December 10, 2012
Instructor: Prof. Paul J. Sims
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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
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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)
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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.
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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,
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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,
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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.
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1 MHz 5MHz
10 MHz 15 MHz
20 MHz 30 MHz
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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
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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
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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.
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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.
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AM 10MHz//1KHz input FFT
AM 10MHz//1KHz output FFT
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
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MHz, or equivalently, 108 mV. Still, this is quit far from the time domain measured
value, 66.5 mV.
AM 20MHz//1KHz input FFT
AM 20MHz//1KHz output FFT
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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.
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FM 1MHz//10KHz input FFT
The peak component swings back and forth between the 990 KHz and 1.01 MHz.
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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.
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FM 10MHz//10KHz input FFT
FM 10MHz//10KHz output FFT
4.3.FM-20MHz
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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.
FM 20MHz//10KHz input FFT
FM 20MHz//10KHz output FFT
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
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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.
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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.
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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
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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.
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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.
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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.
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Figure. The output and input FSK signal in time domain with carrier 20 MHz.
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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
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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,
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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,
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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.
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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:
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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.
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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.
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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
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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.
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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.
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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.
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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.
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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
56
57
58
59
60
ENTC4310 Lab4 FM generation
Xiqiao Wang
Objective
Upon completion of this experiment, we are supposed to be able to
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.
Equipment and components
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
Upon completion of this experiment, we are supposed to be able to
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
Equipment and components
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
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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
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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
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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
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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.
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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.
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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
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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)
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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.
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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.
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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.
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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
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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
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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.
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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.
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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.
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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.
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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.
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98
99
100
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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.
Figure. The received signal at P32 (blue) versus the input signal from the generator
though the 1m fiber with the input amplitude of 1 Vpp.
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Figure. The received signal at P32 (blue) versus the input signal from the generator
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
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 50 Hz.
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 100 Hz.
112
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 1000 Hz.
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 5 kHz.
113
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 50 kHz.
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 100 kHz.
114
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 250 kHz.
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 500 kHz.
115
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 1 MHz.
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 3 MHz.
116
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 5 MHz.
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 6 MHz.
117
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 7 MHz.
Figure. The output at P31 (blue) with the input signal of 2 Vpp, 8 MHz.
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
Figure. The observed signal (blue) at P11 when the square wave input is at 250 kHz.
Figure. The observed signal (blue) at P11 when the square wave input is at 500 kHz.
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Figure. The observed signal (blue) at P11 when the square wave input is at 850 kHz.
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
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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.
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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
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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.
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Figure. The set up for the optic fiber digital communication link.
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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
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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.
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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.
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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.
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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.
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Figure. the received TTL signal at post A of S26 (blue line) with input signal at 1.5 MHz.
Figure. the received TTL signal at post A of S26 (blue line) with input signal at 2.2 MHz.
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Figure. the received TTL signal at post A of S26 (blue line) with input signal at 2.7 MHz.
Figure. the received TTL signal at post A of S26 (blue line) with input signal at 2.9 MHz.
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,
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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
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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
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Figure. With 3m Fiber, the received TTL signal at post A of S26 (blue line) with
input signal at 5.6 MHz
Figure. With 3m Fiber, the received TTL signal at post A of S26 (blue line) with
input signal at 6.3 MHz
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
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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.
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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.
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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.
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Figure. 3m fiber on the LED2-PD1 link, the input TTL has a frequency of 3.5 MHz.
Figure. 3m fiber on the LED2-PD1 link, the input TTL has a frequency of 7.5 MHz.
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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.
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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.
Figure. With 1m fiber and LED2-PD2 link, the frequency increases to 10.13 MHz, the
link reaches its maximum bit rate.
13. Repeat steps 11 and 12 using 3m fiber.
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Figure. With 3m fiber and LED2-PD2 link, the frequency increases to 11.5 MHz, the
number of peaks of the received signal at S26 begins to mismatch the input signal.
Figure. With 3m fiber and LED2-PD2 link, the frequency increases to 12.3 MHz, the
number of peaks of the received signal at S26 further disappears.
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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.
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Figure. With 1m fiber and LED1-PD2 link, the frequency increases to 13.14 MHz, the
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.
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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.
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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
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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
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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.
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Figure. The transmitter block diagram for this experiment,
Figure. The receiver block diagram for this experiment.
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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
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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)
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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.
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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)
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Figure. The overlap of the received TTL signal (channel 2) when compared with the
input TTL signal at 500 Hz. (850 nm link)
Figure. The delay of the received TTL signal (channel 2) when compared with the input
TTL signal at 1000 Hz. (850 nm link)
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Figure. The overlap of the received TTL signal (channel 2) when compared with the
input TTL signal at 1000 Hz. (850 nm link)
Figure. The delay of the received TTL signal (channel 2) when compared with the input
TTL signal at 2000 Hz. (850 nm link)
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Figure. The overlap of the received TTL signal (channel 2) when compared with the
input TTL signal at 2000 Hz. (850 nm link)
Figure. The overlap of the received TTL signal (channel 2) when compared with the
input TTL signal at 4000 Hz. (850 nm link)
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,
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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)
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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)
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Figure. The delay of the received TTL signal (channel 2) when compared with the input
TTL signal at 100 Hz.(650 nm link)
Figure. The overlap of the received TTL signal (channel 2) when compared with the
input TTL signal at 100 Hz.(650 nm link)
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Figure. The delay of the received TTL signal (channel 2) when compared with the input
TTL signal at 1000 Hz.(650 nm link)
Figure. The overlap of the received TTL signal (channel 2) when compared with the
input TTL signal at 1000 Hz.(650 nm link)
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Figure. The delay of the received TTL signal (channel 2) when compared with the input
TTL signal at 2000 Hz.(650 nm link)
Figure. The overlap of the received TTL signal (channel 2) when compared with the
input TTL signal at 2000 Hz.(650 nm link)
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
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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.
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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
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Figure . The receiver block diagram for this experiment
4. Procedure
Figure. Experimental setup.
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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.
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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.
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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.
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Figure 4. Channel one is from the clock signal of slot 0 at post TxSC0, and channel 2
is the binary bit count with in two data bits per count at TxBC0.
Figure 5. Channel one is from the clock signal of slot 0 at post TxSC0, and channel 2
is the binary bit count with in four data bits per count at TxBC1.
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Figure 6. Channel one is from the clock signal of slot 0 at post TxSC0, and channel 2
is the binary bit count with in eight data bits per count at TxBC2.
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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.
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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.
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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.
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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.
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Figure 14. Channel 1 is the TxSC3 at P8 as the slot count signal; channel 2 is the
enable signal at S7.
Figure 15. Channel 1 is the TxSC3 at P8 as the slot count signal; channel 2 is the
enable signal at S8.
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Figure 16. Channel 1 is the TxSC3 at P8 as the slot count signal; channel 2 is the
enable signal at S9.
Figure 17. Channel 1 is the TxSC3 at P8 as the slot count signal; channel 2 is the
enable signal at S10.
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.
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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
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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.
Figure 21. Channel 1 is connected to P8 (TxSC3); channel 2 is connected to S4
(TxData from MUX) when there is no voice input, and the SW3 and SW5 are on.
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Figure 22. Channel 1 is connected to P8 (TxSC3); channel 2 is connected to S4
(TxData from MUX) when there is no digital inputs, and voice input from phone 1
give flurries in during slot 1.
Figure 23. Channel 1 is connected to P8 (TxSC3); channel 2 is connected to S4
(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
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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.
Figure 25. 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) when there is no voice input, and the SW3 and SW5 are on.
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Figure 26. 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) 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.
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Figure 27. In situation where shorting plugs from S7 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.
Figure 28. In situation where shorting plugs from S7 and S9 are interchanged.
Channel 1 is connected to P8 (TxSC3); channel 2 is connected to S4 (TxData from
MUX) when there is no voice input, and the SW3 and SW5 are on.
Figure 29. In situation where shorting plugs from S7 and S9 are interchanged.
Channel 1 is connected to P8 (TxSC3); channel 2 is connected to S4 (TxData from
MUX) when there is no digital inputs, and voice input from phone 1 give flurries in
during slot 1.
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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.
Figure 30. In situation where shorting plugs from S7 and S9 are interchanged.
Channel 1 is connected to P8 (TxSC3); channel 2 is connected to S4 (TxData from
MUX) when there is no voice input, and the TTL input signal at P2 is LOW.
Figure 31. In situation where shorting plugs from S7 and S9 are interchanged.
Channel 1 is connected to P8 (TxSC3); channel 2 is connected to S4 (TxData from
MUX) when there is no voice input, and the TTL input signal at P2 is HIGH.
Observation of Receive Clocks
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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.
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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).
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Figure 35. Channel 1 is connected to Rx frame clock (RxFCO), and channel 2 is
connected to Rx bit clock 1 (RxBC1).
Figure 36. Channel 1 is connected to Rx frame clock (RxFCO), and channel 2 is
connected to Rx slot clock 0 (RxSC0).
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Figure 37. Channel 1 is connected to Rx frame clock (RxFCO), and channel 2 is
connected to Rx slot clock 1 (RxSC1).
Figure 38. Channel 1 is connected to Rx frame clock (RxFCO), and channel 2 is
connected to Rx slot clock 2 (RxSC2).
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Figure 39. Channel 1 is connected to Rx frame clock (RxFCO), and channel 2 is
connected to Rx slot clock 3 (RxSC3).
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.
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Figure 40. Channel 1 is connected to RxSC3, and channel 1 is connected to post S27
(low-active enable signal of Slot 1).
Figure 41. Channel 1 is connected to RxSC3, and channel 1 is connected to post S28
(low-active enable signal of Slot 2).
Figure 42. Channel 1 is connected to RxSC3, and channel 1 is connected to post S29
(low-active enable signal of Slot 3).
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Figure 43. Channel 1 is connected to RxSC3, and channel 1 is connected to post S30
(low-active enable signal of Slot 11).
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.
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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.
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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
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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.
Figure 47. In situation where shorting plugs from S27 and S28 are interchanged.
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.
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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
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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
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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
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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
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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.
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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
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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
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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.
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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:
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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
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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.
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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.
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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.
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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-
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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()(
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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.
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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
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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.
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Figure. 5 Horizontal attenuation due to oxygen and water vapor in atmosphere. (Chen,
1975)
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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
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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.
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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
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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.
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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:
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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.
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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.
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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.
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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
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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