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```									                                Speaker Notes

Element 3 Training

July 2007

Slide
Footnote
Number

The reference to Pages 23 through 64 is to the Gordon West General Class Study guide, 7 th
edition. This guide is available from HRO either in the store or order on-line, or through
1-2      the W5YI VEC http://www.w5yi.org
Cost is \$21 plus tax and shipping.
This guide is not only a great aid to passing your test but will remain a valuable desk

CSCE: Certificate of Successful Completion. This document is sort of like a “receipt” to
prove that you passed a particular element. It‟s main purpose is to serve as proof of
completion if a candidate wishes to upgrade before his permanent license is issued.
1-4      At the current time, VECs and FCC are processing license applications in less than two
weeks. Once you have the permanent “hard copy” of your license the CSCE has little
importance.

Once your license has been “Granted” (FCC speak for Issued) you don‟t need the AG. This
is not the same as being “Filed” as in answer D. In reality, the difference between “Filed”
1-8      and “Granted” is usually only a day or two but be careful; sometimes things will hold up an
application, such as a previously pending application. Your license is not “Granted” until
the ULS database says it is.

The formulas at the top are useful in converting frequency to wavelength. For example, if
Which of the following frequencies is in the 12 meter band?
A: 3.940 mHz
B: 12.940 mHz
C: 17.940 mHz
D: 24.940 mHz
You can divide each into 300 until you get a reasonable answer. For example, 300/24.940 =
12.03..Close enough?
The next line looks like a phone number. I did that to make it easy to memorize. Here‟s
what it does:
1-15     .707 (point seven oh seven) is the number you use to convert a peak AC voltage to RMS
(Root Mean Squared). This needs to be done since an AC voltage can not be used in a DC
calculation. Converting it to RMS first “makes the wave stand still” so it can be measured.
Read the questions carefully as they sometimes ask for Peak-to-peak voltage. RMS
calculations are made on Peak voltages only! Example: What is the RMS value of a peak-
to-peak AC voltage of 5 volts? Answer: divide 5 by 2 and then multiply 2.5 times .707.
1.77 volts RMS is the answer.
468 is a “constant” used to calculate the length of a half-wave dipole. Divide 468 by the
frequency in mHz. For example: What is the length of a half-wave dipole cut for 7.1 mHz?
1005 is used in the same way to calculate the length of the wire needed for a “quad”
antenna. The question might ask you “How long is one side of a quad loop cut for 7.1 mHz?
The answer is 1005/7.1=141.5 (feet) divided by 4=35.4 feet.
The bottom collection of formulae are all the ones you need for the test even though there
are actually 12 variations. Memorize these and you will have it made. More on Ohms law
later.

A chronological guide.
The original Ham bands were those shown on the left side of the screen in white. Notice
that they all were related harmonically. This was to simplify construction since the
operating frequencies were determined by crystals.
In 1953 15 meters was added. It was also harmonically related.
In 1967 there were five classes of license; Novice, Technician, Conditional, General and
Extra. There was very little incentive for a General to upgrade to Extra since the only thing
you got for it was the right to request a one by two callsign such as W6CO. The FCC in an
effort to get Generals to upgrade, thus improving their skills, decided to take away blocks
of frequencies for everyone except Extras. Once we got over the outrage, we bit the bullet
and upgraded getting our frequencies back. You can still see those bands that were affected
1-16      today: They are the ones stating with nn025. Notice that only 80, 40, 20 and 15 have the
missing blocks. The FCC left 160 and 10 alone since they were too lightly used to apply
“incentives” to.
The bands shown in “rust” color were added during a WARC (World Amateur Radio
Conference) in 1979. Incentive licensing had been in effect for more than ten years and
was not needed any more.

Notice that the “WARC” bands are quite narrow compared to the older bands. 30 meters is
so narrow that only CW is permitted. No phone and no image. It is also a shared band so
power is limited to 200 Watts max.
Finally in 2003 60 meters was opened up for shared use under very strict operating
standards.

These are the bands and the corresponding frequencies. The question pool asks about those
shown in green. For some reason they skip over the ones in white although you need to
1-17      know them to operate properly. Get a copy of the band plan and keep it on your operating
desk.

Thinking outside the box: The QPC seems to have overlooked the 60m band where all
1-19      amateurs have equal privileges. No matter, the above information is what the examination
requires.

It is probably easiest to remember that 1500 Watts PEP is the legal limit on all bands
1-23      except: 30 meters (200 Watts) and 60 meters (50 Watts). In all cases however, the operator
has the responsibility to use no more power than necessary to maintain the contact.

10 meter repeaters can be found between 29.5 and 29.7 mHz. This sub-band is open only to
General class and above operators but Technician class operators can use the facilities in a
cross-band configuration when the repeater has a General class or higher operator at the
1-25,29   control point. The Technician class operator can either transmit on 2-meters, where he does
have privileges or on Echolink. In either case it is the responsibility of the control operator
to be present when ever the repeater is in operation.

Amateurs are allowed 1500 watts PEP with a few exceptions. There are only a few bands
where power restrictions are universal. Only the 30m question (200 watts) will be on the
1-30      test. Other power restrictions are based on operating class (Novice/Technician) and
location (near the White Sands Missile Test Range or the Canadian border).
§97.313 Transmitter power standards.
(a) An amateur station must use the minimum transmitter power necessary to carry out the
desired communications.
(b) No station may transmit with a transmitter power exceeding 1.5 kW PEP.
(c) No station may transmit with a transmitter power exceeding 200 W PEP:
(1) On the 10.10-10.15 MHz segment; (2) When the control operator is a Novice Class
operator or a Technician Class operator who has received credit for proficiency in
telegraphy in accordance with the international requirements; or
(3) The 7.050-7.075 MHz segment when the station is within ITU Regions 1 or 3.
(d) No station may transmit with a transmitter power exceeding 25 W PEP on the VHF 1.25
m band when the control operator is a Novice operator.
(e) No station may transmit with a transmitter power exceeding 5 W PEP on the UHF 23
cm band when the control operator is a Novice operator.
(f) No station may transmit with a transmitter power exceeding 50 W PEP on the UHF 70
cm band from an area specified in footnote US7 to §2.106 of the FCC Rules, unless
expressly authorized by the FCC after mutual agreement, on a case-by-case basis, between
the District Director of the applicable field facility and the military area frequency
coordinator at the applicable military base. An Earth station or telecommand station,
however, may transmit on the 435-438 MHz segment with a maximum of 611 W effective
otherwise required. The transmitting antenna elevation angle between the lower half-power
(-3 dB relative to the peak or antenna bore sight) point and the horizon must always be
greater than 10°.
(g) No station may transmit with a transmitter power exceeding 50 W PEP on the 33 cm
band from within 241 km of the boundaries of the White Sands Missile Range. Its
boundaries are those portions of Texas and New Mexico bounded on the south by latitude
31° 41' North, on the east by longitude 104° 11' West, on the north by latitude 34° 30'
North, and on the west by longitude 107° 30' West.
(h) No station may transmit with a transmitter power exceeding 50 W PEP on the 219-220
MHz segment of the 1.25 m band.

1-36      This is one of the basic rules for operating on any band.

1-38,40   This band is only 50 KHz wide so Phone and broad data is not permitted.

This band is so narrow and shared with other services that it is channelized. Very
1-48      strict controls are in place restricting type of emissions.

The reason for this is because we are restricted to only 50 Watts ERP (Effective Radiated
Power). If you use a dipole, which is considered a “Unity Gain” antenna, your PEP and
your ERP will be the same. If you use a gain antenna however, such as a Yagi, your ERP
1-54      will be your PEP multiplied by the gain. You must then reduce your PEP so that your ERP
is back within limits. You must note this in your logbook, which becomes a permanent part

As a general rule, if you can hear them, they will be able to hear YOU. It‟s best not to use
1-60      the frequency if you can copy other signals on it.

Notice the absence of decimal points in the frequency choices. For the purposes of this
1-63,64   course, both forms are the same but if you want to convert to wavelength, by dividing into
300, put the decimal where it belongs first. (after the first digit)

Don‟t be fooled. You may be used to hearing CB operators on our 10 meter band. They
1-72      aren‟t “sharing” the band, they are operating illegally!
It may be difficult to find printed guidelines indicating the “DX window”. As a rule of
thumb, these exist primarily in the bottom 25 kHz of the HF bands with the exception of
1-76       3790 to 3800 in the 80 meter band. If you confine ragchewing to the General class portion
of any band you won‟t be in the DX window.

1-79       This is a very unlikely circumstance, but it‟s the answer they want.

There are much better reasons for keeping a log than in the previous question, and they are
all personal reasons. If you participate in exchanging QSL cards, a log is vital since it
contains vital information you will need to verify a contact sometimes years after the fact.
1-81       Logs also serve as “journals” and can be fun to review years after they are closed.

Ignore the green “happy face”. Its used in the “live” course to signal a change in topic.

The retransmission of a space shuttle or ISS traffic would be done by a bulletin station.
This would be one example of an allowed “broadcast” or one-way transmission which
1-97       would otherwise be illegal. Remember, one-way transmissions are allowed when they are
“of general interest to the amateur radio community.” 97.3(a)10

group of Amateur Radio operators who provision and operate repeater and telemetry
1-99       equipment on satellites. Control of this equipment is confined to “secret codes” for security
reasons and is the only type of secret code permitted

CW operators use abbreviations extensively to speed up the conversation. You will also
1-101      hear DX CW using something called “Cut numbers” to save time. All perfectly legal.

This is done quite often. There are “Swap nets” where amateur gear is listed for sale. It is
1-103      considered poor practice to conduct the actual sale on the air, rather get the other party‟s
email address or telephone number and finalize the deal off the air.

Module-1 (continued)

Slide
Footnote
Number
This is a condition set forth by the FCC as part of the “punishment” of a licensee who‟s
1-105     license has been revoked. He is not permitted to use Amateur Radio under any
circumstances.

Third party agreements and reciprocal licensing agreements are published on the web. You
1-111     should print out this information and keep it with your operating aids in your station.

FYI:
1-133     On HF phone nets, “Break-Break” usually indicates priority traffic. A triple break as in C:
above would indicate emergency traffic.

1-        The key to getting along in confined spaces in to act in a non-confrontation manner in all
137,139   things. Don‟t let road-rage type behavior get started on the ham bands.

1-145     Helpful hint: Q-signals often includes abbreviations. QRP stands for “Reduce Power”
There are many hams who enjoy QRP operation and they even have contests to see how far
they can make a contact (miles per milliwatt). Some consider 5 Watts to be the upper limit
of QRP operation. Give it a try, you may be amazed at what you can accomplish with just a
fraction of a Watt.

Voice Operated (VOX) headsets and speaker/mikes are available. These permit hands-free
operation by detecting the voice of the user and automatically activating the push-to-talk
(PTT). Most of these unit offer a way to switch to manual operation to eliminate falsing
1-147      (inadvertent transmissions) in a noisy environment. They also generally offer some means
to control the sensitivity (gain) of the VOX circuitry. Better units include VOX delay to
eliminate falsing on pulse noises and anti-VOX to prevent received audio from activating
the transmitter.

This is the appearance of an AM signal on a spectrum scope. There is a carrier in the center
that uses power but contains no information (it can be used for tuning). Each of the two
sidebands (upper and lower) are identical but inverted. HF amateur communications are
generally on a single sideband (SSB) by suppressing the carrier and opposite sideband. This
1-165      is more efficient by using less bandwidth and power is not wasted transmitting a dead
carrier and duplicate sideband.

Because of radio design simplicity, the convention has been to use lower sideband (LSB)
below 14 MHz and upper sideband (USB) above 14 MHz.

Make a mental note: These 3 bands, (the lower 3) use Lower Sideband, all others use Upper
Sideband. A recent exception is 60 meters which also uses Upper Sideband. On 60 meters
1-178      you MUST use upper sideband. All other bands, the sideband you use is recommended, not
required by the rules.

1-180      Once again, this is a STRONG recommendation, not the rule.

1-182      On CW, “de” is the abbreviation for “from”. CW ops use many other abbreviations as well.

In matching one tone to another (yours versus his) as you get close to being matched you
will notice a harmonic caused by the two audio tones mixing. As you get closer and closer
1-186      to matching the harmonic will get lower and lower in frequency until it gets so low you
can‟t hear it. This is “zero beat”.

In CW each Q-signal will have two meanings. If QRS? The meaning is “Shall I send
1-192      slower?” Without the question mark it means “please send slower”.

1-238      In fact, the code is referred to as “Varicode”.

End Notes for Module-1

Module-2

Slide
Footnote
Number
Generally, the reference to “property” is meant to be homes and other buildings. How about
2-3       a car? A bicycle? You decide.

Rule of thumb: In a genuine emergency, that is an immediate danger to human life or real
2-6       property, pretty much anything goes.

The key points are the D, E and F layers. The F layer splits and becomes F1 and F2 with the
additional energy from the daytime sun. The D and E layers disappear at night without the
sun‟s energy. The D layer absorbs most RF power, particularly at low frequencies. The
other layers will reflect RF energy. The higher the layer, the further the contact possible.
Propagation conditions are a balance between the D layer absorption of lower frequencies
and the reflection of the higher frequencies by the upper layers. As more of the sun‟s
energy (primarily electromagnetic and particle emissions) strikes the ionosphere, all the
layers grown in size and density. The D layer will absorb more low frequency RF and the
2-27      upper layers will reflect more high frequency RF. This creates a window of frequencies
above the Lowest Usable Frequency (LUF) and Maximum Usable Frequency (MUF) that
will work for long distance communications.
Sunspots provide huge amounts of energy to reflect radio waves in the ionosphere and are
especially important in the upper bands. Sunspots occur in 11-year cycles of high and low
activity.

For additional information refer to Page 70 in the Gordon West Study Guide.

The illustration on slide 59 can be of help understanding this. There are only two questions
2-31      on the test asking about maximum distances per region. The other question deals with the
E-region.

This is because at noon during the summer, sunlight (and ultra violet radiation) will be at its
2-33      maximum intensity. It is the effect of this radiation that changes the altitude of the F layers.

Refer to the illustration on Slide 34: Note that the altitude and angle of refraction varies with
frequency. Generally the higher frequencies will be refracted at higher altitudes and at
shallower angles. If the frequency is high enough, it won‟t be refracted enough to arrive
back on the surface. The highest frequency that can be refracted is the “critical frequency”
and generally the Maximum Usable Frequency (MUF).
2-34      If the angle if the wave is too steep, it will continue into space. The steepest angle that will
be refracted back to the surface is the “critical angle”.
Think of skipping a stone on a pond. If you hit the water at just the right angle, the stone
will skip multiple times. If the angle of your toss is too oblique, the stone will penetrate the
surface of the water and sink without skipping.

Notice the answer says “return a radio signal to earth”. Signals may “bounce” or “skip” but
2-36      they are not “reflected”. The correct term is “refracted”.

On HF, beacon stations can be found on specific frequencies on 20 through 10 meters. See
2-44      this URL for complete details: http://www.dxzone.com/cgi-bin/dir/jump2.cgi?ID=1112

Do you remember the conversion formula? mHz to meters: 300/f(mHz). Choose the
2-46,48   answer that comes the closest. E.g. 300/22=13.6 (14) Choose 15.

This is the second of two questions referring to the maximum distance along the earth‟s
2-54      surface. See the illustration on slide 59.

They are in alphabetical order but what happened to A, B, and C? They are atmospheric,
2-58      not Ionospheric. In Ham radio we only care about D, E and the F layer which splits in two
during the day. See slide 27 for the illustration.
2-61,63    The Darn D layer is Disruptive During Daylight!

An Azmuthal map is a map projection centered on a particular location, used to determine
the shortest path between points on the surface of the earth. From this map, centered on
Seattle, it can be seen that San Diego is 165 degrees, Hawaii about 230 degrees and
2-81       Anchorage, AK will be about 325 degrees for the shortest path. These bearings will likely
be different than those from a standard Mercator projection because they account for the
earth‟s curvature.

End Notes for Module-2

Module-3

Slide
Footnote
Number
This is the appearance of an AM signal on a spectrum scope. There is a carrier in the center
that uses power but contains no information (it can be used for tuning). Each of the two
sidebands (upper and lower) are identical but inverted. HF amateur communications are
3-5        generally on a single sideband (SSB) by suppressing the carrier and opposite sideband. This
is more efficient by using less bandwidth and power is not wasted transmitting a dead
carrier and duplicate sideband.

Since it combines signals, the balanced modulator is a type of mixer. It‟s the “balancing”
3-10       that removes the carrier. Some radios allow the carrier to pass through (for CW or AM
phone) by “un-balancing” the modulator.

In this illustration you can see why SSB is the narrowest of the choices. A phase-modulated
or FM signal would be at least as wide as the carrier and both sidebands shown here.
3-18       A double sideband signal, (they do exist) would use both sidebands shown above but the
carrier would have been removed.

A speech processor does not necessarily improve your signal. Improperly adjusted, it can
3-22       ruin it!

ALC stands for Automatic Level Control. Properly adjusted, your ALC indication on your
3-26       meter should show little or no motion.

3-28       …and more bandwidth can cause interference to adjacent stations,

Illustration on slide 33:
Two tone audio tests are used on an oscilloscope to test proper linearity. The pure tones fed
in will give you a stable picture on the scope if the amplifier is properly adjusted. The lower
image shows a signal with some noise that it also overdriven (over modulated) and is
3-33       clipping or flat topping. Any two audio tones may be used, but they must be within the
transmitter audio passband, and should not be harmonically related.
Speech Processors are used to keep input audio close to a perfect level for 100%
modulation. If they are not adjusted properly, they can cause the distortion seen here.
Overdriving causes very distorted audio and no additional power output.

This is a current model Tektronix oscilloscope. It has four signal inputs along the lower
right edge and can display the waveforms from those inputs. There are three main sections
3-38       to the right of the screen; vertical, horizontal and trigger. The horizontal and vertical
controls adjust how the waveform appears on the screen by adjusting the horizontal and
vertical amplifiers. The trigger controls allow the oscilloscope to synchronize to external
events.

The term “monitoring oscilloscope” simply refers to the way the oscilloscope is used. It is
3-42   connected into the transmitter‟s output during operation in order to evaluated the transmitted
signal. It‟s just an oscilloscope.

Self oscillations can occur within a vacuum tube and are caused by the internal components
3-50   of the tube being in proximity to one another. If allowed to continue, self oscillations will
not only distort the signal but can actually damage the tube.

A simple formula for calculating the total bandwidth of an FM signal: The sum of the
3-58   deviation (in kHz) and the highest modulating frequency, times 2. In this example, 5+3 x
2= 16

Looks tough right? But it‟s easy. Find the ratio of the modulated oscillator to the
3-61   transmitters frequency: 146.52/12.21=12. Now apply that ration to the 5,000 Hz deviation:
5,000/12 = 416.7 Hz

Peak Voltage = ½ peak-to-peak or 100 Volts. Now multiply 100 x .707 to convert it to
RMS. 100 x .707 = 70.7 Volts. Now square that by multiplying 70.7 times itself. 70.7 x
3-71   70.7 = 4998.5. Divide that by 50 and you get 100 Watts (rounded).
2
Why? Here‟s the Ohm Law equivalent: P = E / R Review Slide 15 in Module-1, and
Slide 3 in Module-4. We'll see this same formula used again.

There are three key voltage measurements, RMS (Vrms), Peak (Vpk) and Peak to Peak
(Vpp). The Peak voltage is relative to ground. RMS or Root Mean Square is more of an
“average” voltage over the duration of the AC cycle. RMS voltage is calculated by
3-72   multiplying 0.707 times the peak voltage. The Peak to Peak voltage measures from the peak
of the positive swing to the peak of the negative swing.
The 120 volt AC line voltage in the wall is a peak value. That means the RMS voltage is 85
volts and Peak to Peak is 240 volts.

3-74   Is this a trick? Without modulation there is no envelope!

Here‟s another one just like on Slides 70-71. Same Ohms Law formula: P = E2 / R.
3-76   Start with 500/2 so we get Vp. Remember, we can‟t use Ohms Law on Vpp. Multiply 250
x .707 and get 176.75. Square t hat by multiplying it by itself. You should have 31,240.56
right? Divide that by 50 and you should get 624.81, which rounds nicely to 625.

Just like in slides 73 and 74, the PEP of an unmodulated carrier is the same as the average
3-78   power.

3-82   Refer to the next slide for a 1 minute course in dB.

Don‟t be overwhelmed by all those numbers. You only need to remember a few of them.
We‟ll start by getting rid of the ones in parenthesis. They are there to show you the actual
value of 3 and 6 dB when rounded.
Forget about the numbers from 0 down to the bottom of the chart; they are the same as the
ones above the 0 except they are reciprocals. What‟s a reciprocal? Simply what you get
3-83   when you divide any number into 1. Try it: 1 divided by 1.995 (3 dB) is .5012 (-3 dB)
Now for the purpose of this course, forget the far right column; we‟re only going to be using
the P2/P1 column. After you pass your test, come back here and memorize the V2/V1
column.
So what are we left with? –1 db equals .7943 or nearly .8. -3 dB equals .5012 or just about
.5. -6 equals .2512 or right at .25 and finally –10 equals .1 or 1/10th. So what? Apply these
losses to 100 Watts of your precious transmitter output: one dB loss equates to 20 Watts
lost. 3 dB loss equates to 50 Watts lost or one half! 6 dB loss and you‟ve given up three
quarters of your power. Finally, 10 dB loss and you have only 10 Watts left! We'll explore
this subject in more detail in the Feedlines section; where dB really counts!

More on the subject of dB. By definition, if you apply 50 microvolts to the antenna terminal
of a modern receiver, and it‟s input impedance is 50 Ohms (most are) then your S-meter
3-85      should read S-9. Also by definition, one S-unit is 6 dB, so to increase the S-meter reading
from S-8 to S-9 you must increase the power by 4.

You could go back to the chart and you would see that 20 dB is 10 2 or 1 followed by 2
3-87      zeros. You could also say that it‟s 10 times stronger (10dB) twice. Each time you do the 10
dB calculation you add a zero.

What does linear mean? If you plot the changes, and the result forms a straight line, the plot
is said to be linear. If the line curves then the plot is said to be non-linear. In amplification,
3-95      we don‟t want anything to change except the power. We want a linear output so we use a
linear amplifier. If the amplifier produces a non-linear output, our signal will be distorted.

3-101     Refer to Slide 3-50 for a related question

By adding Negative Feedback to the same amount of Positive Feedback, we cancel the
3-103     effects. Net sum zero

Module-3 continues

Slide
Footnote
Number
Superheterodyne or simply Superhet: The superheterodyne principle was originally
conceived in 1918 by Edwin Armstrong during World War I as a means of overcoming the
3-105     deficiencies of early vacuum triodes used as high-frequency amplifiers in radio direction
finding (RDF) equipment. Most modern receivers use these same principles.

Images can be difficult to handle. They can appear on both sum and difference signals.
3-118     Read what Gordon West has to say about it on page 102 of the Study Guide.

Very early designs used direct conversion receivers for low cost and simplicity. Today,
3-130     several portable transceivers are available in kit form that use direct conversion design.

3-148     See the notes on “neutralization”

3-150     FET stands for Field-Effect Transistor

Memory tip: Logic circuits use two states: on-off, high-low, one-zero, they all mean the
3-154     same thing. In bipolar transistors, these state correspond to saturation and cut-off.

3-177     From the SARS Newsletter A short article describing digital logic circuits.

From the handout: If input A is “one” AND input B is “one” then output “Q” should be
“one”, but this is a NAND gate. The little “NOT” thingy that attaches to the AND gate
3-181     makes it a NAND gate and reverses whatever the answer would have been. Contrary little
devil!
From the handout: Just like the previous example, if input A is “zero” OR input B is “zero”
then output “Q” should be “zero” but this is a NOR gate which is an OR gate with a NOT
3-183      thingy attached to it. Therefore, whatever the OR gate would have been, the NOR is just
the opposite.

From the handout. Basic digital circuits are binary because they deal with only two
(bi) states: ON/OFF or HIGH/LOW or TRUE/FALSE. The terms all mean the same
thing and are used interchangeably. Early computers actually used mechanical
switches that were switched on or off, one at a time, to "program" the computer. It
was found that a combination of 8 single pole switches could provide 256 different
combinations which was sufficient to describe all the letters in the alphabet both
upper and lower case, all numbers from 0 to 9, and several "special characters".
3-185
We can answer all the Logic Circuit questions in the new pool with only 3 switches.
Here's why: One switch, by itself yields only two states, TRUE or FALSE. It's either
on or off. When you combine that with a second switch you now have 4 possible
conditions: All FALSE or OFF, All TRUE or ON, Switch A TRUE, Switch B FALSE,
and Switch A FALSE, Switch B TRUE. Combine these two switches with a third
and you double that number of possibilities. Each new switch added doubles the
number of possible states.

Abbreviation of complementary metal oxide semiconductor. Pronounced see-moss, CMOS
is a widely used type of semiconductor. CMOS semiconductors use both NMOS (negative
3-         polarity) and PMOS (positive polarity) circuits. Since only one of the circuit types is on at
191,193    any given time, CMOS chips require less power than chips using just one type of transistor.
This makes them particularly attractive for use in battery-powered devices, such as portable
computers.

End Notes for Module-3

Module-4

Slide
Footnote
Number
This illustration is similar to one that can be found on page 116 of the Gordon West Study
Guide. There are 12 different formulas to help with Ohms Law and Power calculations but
4-3        only the 4 shown will be used in the Element-3 test. A description of these 4 formulas is
shown on the next slide.

4-4        Here are the 4 formulas that are found in the Element-3 test.

Referring back to slide 4 we see that the only formula solving for E will work here. Use
E=SQR PR.
4-6        We first multiply P times R or 1200 x 50 to get 60,000. Now simply press your Square
Root (SQR) key to get 244.94. Round up to 245.
2
Referring to Slide 4 we see that P=E / R will work for us. First we “square” E by
4-8    multiplying it by itself: 400 x 400 = 160,000. Now we divide that by R (800) to get 200
Watts.

This one is “Easy as PIE”. Multiply I times E. 0.2 x 12 = 2.4 Watts. Don‟t forget the
4-10   decimal point on the current (I).
2
This question uses the fourth formula; P=I R but be careful. These formulas only work
with whole units, that is Ohms, Volts, Amps, Watts. When working with fractional units
such as milliamps and kilohms we must use decimal points. Refer to the next slide to see
4-12   where the decimal points go.
7.0 milliamps is really .007 Amps. Multiply .007 times itself and you get 0.000049 Amps.
Now multiply that by 1250 Ohms (1.2 kilohms) and you
get 0.06125 Watts. Can you convert that to Milliwatts? The choices of the answers will do
it for you: 0.06125 Watts is approximately 61 milliwatts.

There are three key voltage measurements, RMS (Vrms), Peak (Vpk) and Peak to Peak
(Vpp). The Peak voltage is relative to ground. RMS or Root Mean Square is more of an
“average” voltage over the duration of the AC cycle. RMS voltage is calculated by
4-14   multiplying 0.707 times the peak voltage. The Peak to Peak voltage measures from the peak
of the positive swing to the peak of the negative swing.
The 120 volt AC line voltage in the wall is a peak value. That means the RMS voltage is 85
volts and Peak to Peak is 240 volts.

We talked about reciprocals in Module-1. A reciprocal is developed when you divide a
number into 1. We know that to find the RMS value of Vp you multiply it by .707. But
what if you already have the RMS and you want to put it back? You use the reciprocal of
4-18   .707 (1 / .707) which is 1.414. Try multiplying 120 Vrms by 1.414. You should get 169.68
BUT WAIT! Those tricky question pool guys. The answer is not B.169.7. Look at the
question again: They want Vpp. You have to double your answer: 169.68 x 2 = 339.36 (
round to 339.4)

4-20   What could be easier: 17 x .707 = 12.019. Did you do this one in your head?

4-22   When they say “Linear” power supply they are referring to the old fashion transformer type.

Most components have an equivalent series resistance (ESR). This is composed of the
various parts of the component such as the leads and internal connections. In some devices
4-24   the ESR is not an issue but in switching power supplies we want it to be as low as possible
because it affects the operation of the system.

Read the description and see the graphics on pages 120 and 122 of the Gordon West Study
4-34   Guide.

High quality filter capacitors can stay charged for days, even weeks. If you are working
4-52   around a high voltage power supply don‟t assume the bleeder resistors are doing their job.
Check for voltage before you touch anything.

Well, at least the readout is more precise. With an analog meter you need a mirrored dial to
4-54   eliminate parallax. That‟s not a requirement with a digital meter. But they can be wrong
also. Buy a cheap-0 and it might be miles off!

With some things, just getting close enough to measure can change it. That‟s one of the
4-56   advantages of using an oscilloscope rather than a meter.

4-82   Keyed connectors: Another name for polarized connectors. If you rely only on color, you
will eventually hook it up backwards. Trust me!

This is the schematic that is handed out at the test session. You will be asked to identify
various components, for example “G7A22 which symbol in figure G7-1 represents a
4-83      variable capacitor? Answer: symbol 5. There are 6 questions pertaining to this handout, the
most you can expect to see on any test will be one of them. (but which one?)

A special type of resistor that we want to change when the temperature changes. It is used
4-105     primarily in thermostats, and weather stations.

Refer to slide 97: In calculating the value of components in series or parallel, resistors and
Inductors are treated alike, capacitors just the opposite. Since we are dealing with 3
4-107     identical components we can simply divide 3 into 10 to get the answer. Bu what if they
were not identical? Convert each to it‟s reciprocal, add them up, reconvert.

According to Kirchhoff's second law, everything in a circuit must be accounted for. You
4-109     can‟t end up with more or less than you started with.

This looks intimidating! But they chose a value that makes it easy. Go back to slide 105
and review. If three equal value resistors in parallel produce 50 Ohms of resistance then the
4-110     resistors must be 150 Ohms each. You need go no further, but test it if you wish: three
resistors of any value add up to the total of their values, in this case 450 Ohms.

Now they‟re making us do a little work. Convert each resistor to it‟s reciprocal by dividing
4-113     it into 1. Now add up the three values and reconvert. You should be adding: 0.1, 0.05, and
0.02 for a total of 0.17. Now divide 0.17 into 1 which will give you 5.88. Round to 5.9

You can do this one in your head. Referring to slide 97, capacitors in parallel are treated
numerically like resistors in series. Just add „em up. Don‟t be intimidated by
4-117     picofarads…this would be the same if it was gigamegafarads. 5000 + 5000 + 750 = 10750
whatevers.

This one is very much like the resistor question in slide 110. Capacitors in series are treated
numerically like resistors in parallel. 100 divided by 3 gives you 33.3 microfarads. What if
4-119     they were not of equal value? Use the reciprocal trick like in slide 113. You‟ll have a
chance to practice this very soon.

Module-4 continues

Slide
Footnote
Number
Remember, Inductors are treated numerically just like resistors. In series, simply add the
4-121     values.

Duck soup! Convert 20 to it‟s reciprocal by dividing it into 1. You‟ll get 0.05. Do the same
4-123     with 50, get 0.02. Now add the two reciprocals and reconvert. 1 / 0.07 = 14.285 (round to
14.3)

Illustration of a resonant circuit. Inductors and Capacitors both exhibit Reactance or a
resistance to AC measured in ohms. Inductive reactance (Xl) increases with frequency and
4.130     capacitive reactance (Xc) decreases with frequency. When inductors and capacitors are
used in the same circuit, the circuit is said to be resonant when inductive and capacitive
reactance are equal.
Matching load impedance is important to insure maximum power delivery. Nice chart, but
relax. No questions pertain to this material.

You might think of it as an audio tone. As the frequency increases, our ability to hear it
4-138      decreases due to the mass of our hearing apparatus.

Now we‟re talking about an inductor which behaves numerically just the opposite of a
4-140      capacitor.

There is a picture of a wire wound resistor in the Gordon West Study Guide on page 136.
4-148      Notice that it‟s nothing but a coil of wire. Pure inductance!

The term “solenoid inductor” is used to differentiate it from a toroidal inductor. It is simply
an inductor wound around a straight form. Some contain a core material some don‟t. See a
4-154      picture of a piece of “airdux” (pretty well beat up) in the Gordon West Study Guide on page
135. This is an example of a solenoid inductor.

Any time you have two or more wires side by side you will have interstitial capacitance.
Coaxial cable for example has a specific capacitance between the center conductor and the
4-158      shield. Telephone cable has a specific capacitance between the conductors as well as
between the conductors and the shield. The capacitance is reduced greatly by separating the
conductors.

Most transformer questions are simple ratio problems. For example, if the transformer had
the same number of turns in the primary as the secondary, the ratio would be 1:1. This
would be the design of an isolating transformer. What ever voltage is applied to the primary
4-166      will appear at the secondary. In this example, the ratio is 4.5:1 (2250 / 500) meaning that
what ever voltage is applied to the primary, 4.5 times less will appear on the secondary
winding. Try it: 120 / 4.5 = 26.66 (round up).

This is also a ratio problem but because we are invoking Ohms Law from the Power side,
4-168      square root gets involved. First find the ratio by dividing the secondary impedance into the
primary impedance: 600 / 4 = 150. Now find the square root of 150. 12.247 or 12.2:1

End Module-4

Module-5

Slide
Footnote
Number
For a proper RF ground, it‟s important to use a heavy gauge copper (not aluminum) braid or
foil and keep the run length to a minimum. Standard wire, even heavy gauge, or long runs
can cause your ground lead to act as an antenna and resonate. A good ground will reduce
5-5        RFI to other equipment, minimize the risk of shock and reduce electrical interference. At a
minimum a ground rod should be 8 feet long. This can be difficult to achieve in some
locations and multiple shorter rods tied together can be used.

Some examples: Corroded gutters and downspouts, Loose pole-line hardware (PG&E
5-25       poles), antenna mast guy wires not bonded.

Way back in Module-1 on slide 15 we introduced a bogus phone number. Have you
5-33       memorized it? .707-468-1005. 468 is the number we use to calculate the length of a half-
wave antenna element. 468 / f(mHz). In this example, 468 / 14.250 = 32.842 feet.
5-35   Another example using the constant: 468. 468 / 3.55 = 131.83 feet.

You may already be aware that many Hams use 50 Ohm coaxial cable to feed their
antennas. Most Hams think that a typical HF antenna has a characteristic impedance of 50
Ohms. Actually, a half-wave dipole‟s characteristic impedance is closer to 70 Ohms if you
5-39   get it way up in the air. If you install this antenna at the height most Hams do, around 30 to
50 feet, it‟s impedance will have dropped to around 50 Ohms. For this reason, it is very

You can match a high impedance feedline such as twin lead or Ladderline to a dipole by
deliberately moving the feed point to one side. If you get your 300 Ohm feedline attached at
just the right place you will have a nearly perfect match.
When using coaxial cable (50 Ohms) you must be careful to get it exactly at the center of
5-41   the half-wave point. Many Hams adjust the SWR of a new dipole by trimming the ends of
the wire. Shortening it up raises the resonant frequency, but you must be careful to keep
both sides exactly the same length. If one is longer than the other, you have, in effect
attached the feed point off-center and you‟ll never get your match.

See the illustration on the next slide. Imagine you are looking down on the antenna from
5-43   way up in the air. This view is referred to as the E-plane.

Illustration of the radiation pattern of a horizontal dipole.
If the dipole antenna was the red line, the lobes above and below show the radiation pattern
at ½ wavelength above the ground. This pattern actually rotates around the wire like a
5-44   donut. As the antenna is lowered below ½ wavelength, the lobes flatten until it becomes
nearly omni directional.

Almost, but not quite. Radiation off the ends of the wire will still be less than that
5-46   perpendicular to the wire.

Illustration of how the radiation angle changes with height above ground.
Illustration of the H-plane. Now we are looking at the antenna off the end of the wire. At ¼
wave, the radiation is as strong straight up as it is in a horizontal direction. As you raise the
antenna even further, the upward radiation diminishes and the power is added to the front to
back lobes as seen in the center diagram. Notice that the maximum radiation in this diagram
5-47   appears to be about 45 degrees above the horizon. Not bad for DX work. The diagram on
the right shows what happens when you raise the antenna even further. Additional lobes
appear which will give you a DX advantage, sort of like having a “broader paint brush”.
The lower lobe will maximize your distance. Go back to Page 70 in the Gordon West Study
Guide and you‟ll have a better understanding of “radiation angles”.

Sometimes this is exactly what we want. If our goal is to work stations just beyond our
5-49   ground wave capability, we may do well with an NVIS antenna.

At heights below ¼ wave you may have trouble matching it to your feedline. If this is the
5-53   case, use low loss coax and feed it with a tuner.

Use the 468 constant again, but don‟t forget to divide the answer by 2. 468 gives you the
5-57   length of a ½ wave antenna.

A typical vertical antenna cut for the Ham bands will have a feed point impedance of 35
5-59   Ohms more or less. By sloping the radials downward we can raise the impedance to match
the feedline.

5-64   This would seem to be at odds with the previous question. How are we to raise the feed
point impedance to 50 Ohms if our buried radials extend straight out from the base of the
antenna? A matching network does the trick.

Makes sense. An omni directional antenna receives signals and noise from every point of
5-66   the compass. A directional antenna receives signals and noise from only a few points on the
compass. (Of course you have to be able to rotate it.)

Here‟s why: A resonant antenna, properly matched to the feedline and transmitter radiates
nearly everything that is fed to it. A random antenna, is not resonant anywhere (probably)
and if you feed it directly, much of the energy that would normally be radiated is bounced
back to the transmitter, often on the outside of the feedline. This can be a bad situation.
5-68   The reflected energy can flow all over your station equipment including your microphone!
Pass the chap stick!
This is not to imply that a random antenna is a bad thing. It can be quite useful when
properly installed and fed.

The driven element is nothing more than a dipole. The only difference, usually, is a dipole
5-74   would be made of wire and the driven element of a Yagi antenna is usually made from
tubing or rod material. The illustration on the next slide shows this graphically.

Note the driven element with the balun attached, the smaller directors, larger reflector and
the pattern with a large front lobe. The driven element is approximately a half wave dipole.
As parasitic elements (directors and reflectors) are added and the boom length is increased,
the more gain is provided by the antenna. Reflectors are placed behind the driven element
to reflect the power forward and the directors act as conductors in front of the wave.
Reflectors are generally ~3% longer than the driven element and directors ~3% shorter.
5-75   Five element Yagis like this one yield 10-12dB gain over a dipole, based on the design of
the antenna.
Antenna designers have to balance forward gain, front-to-back ratio, bandwidth and feed
point impedance when designing an antenna. Some antennas will be optimized for
maximum forward gain to improve their output and others may have better front-to-back
ratios to improve their noise rejection. Antenna bandwidth can be increased by using larger
diameter elements.

Take another look at the radiation pattern on slide 75 to see a (distorted) picture of a main
lobe radiated from a directive antenna. Regarding directivity, notice that the pattern is not
5-83   pointed the same direction as the antenna. This is a drawing oversight. In reality, the main
lobe points the same direction as the antenna‟s boom.

Gain is like gas mileage. It all depends on how far the manufacturer can stretch the truth.
5-87   “Theoretical” is a great weasel-word.

A typical SWR Bandwidth chart for 20 meters. The arbitrary limit of 2.0 was set by the
station operator based on data provided by the equipment manufacturer. SWR plots are
5-91   made at nine points between 14.000 and 14.100. An acceptable bandwidth is shown
between 14.025 and 14.060

In Gamma matching, the shield of the driven element is attached to the driven element in the
normal manner but the center conductor, instead of being attached to the element on the
opposite side of the boom, is attached some distance out from the boom to match the
5-94   impedance. A capacitor is inserted in the center conductor to tune out the inductance of the
extended center conductor. For a photo of a gamma match on a UHF Yagi see the article in

True enough, and that‟s the answer they‟re looking for, but in reality insulation is sometimes
5-96   used to prevent bi-metallic contact of the various antenna components. If your antenna has
insulation, that‟s the main purpose.
Module-5 continues

Slide
Footnote
Number
See the illustration on the next slide. This has been modified from the dipole graphic to
5-98      show the effects on a Yagi.

A Ham transmitter, operating at 7.000 mHz could also be heard on 14.000, 21.000, 28.000
and even higher. Hopefully the harmonic signals would be extremely weak and not heard
5-105     far from your station. A monoband antenna would not radiate the harmonic signals very
well, but a multiband antenna might.

Illustration of a cubical quad antenna:
The driven element of a quad is a full wave loop, so each side is quarter wave. In this
picture the feed is along the bottom, horizontal wire. This means the polarization of this
5-112     antenna is horizontal. If the antenna were rotated 90 degrees so the feed were on a vertical
side, the polarization would be vertical. Like the Yagi, reflector elements are ~5% larger
and director elements are ~5% smaller than driven elements.

Disclaimer: The previous slide says 5% more (longer). This is a change from the previous
syllabus which used 3% as the difference between the driven element and the reflector or
director. If you read the text in the Gordon West Study Guide you will find that he still
5-116     uses 1030 as the constant used to calculate the length of the reflector which is about 3%.
Has the question pool committee changed the laws of physics?
No matter, you no longer need to calculate the lengths of quad or delta elements. Just
remember “slightly more” for reflectors and “slightly less” for directors.

Delta loop antennas also have full wave elements and offer performance similar to quad
antennas. Because of the large elements, 2 element quads and deltas offer similar
5-121     performance to a 3 element Yagi. In reference to the test questions, quads and deltas have
nearly identical characteristics to Yagis, just a bit more gain.

You beer drinkers can quit your snickering. The antenna gets its name from Harold H.
5-131     Beverage who invented it in 1919.

The Beverage antenna is particularly useful for 160 meters. Not widely used by urban
5-133     dwellers as the require a lot of room!

As you might imagine, a quarter wave vertical antenna for 75 meters would normally be 59
feet tall. It‟s hard enough to install an antenna of that height in your back yard much less on
the bumper of your car. Mobile antennas for 75 meters do exist however, and to make them
5-137     short enough to clear overhead objects like wires and freeway overpasses, they are heavily
loaded with inductors. This tends to make them relatively inefficient although they do
work.

“L” being the electrical symbol for inductance (a coil) and “C” the symbol for a capacitor,
5-141     an LC network is made up of at least one inductor and one capacitor.

Refer to the drawing on page 160 of the Gordon West Study Guide. The matching device
shown does two things: It matches the impedance of the 50 Ohm coax to the 600 Ohm
Ladderline. But it also provides the transition between the unbalanced coaxial cable and the
5-151     balanced Ladderline. There are devices available that only do the balancing, not the
impedance transformation. These devices are called baluns. A common use would be
between the 50 Ohm coax cable and the feed point of a dipole antenna. No matching
required, but balancing can help.
Here are two digital Field Strength Meter models. Some SWR meters and frequency
counters can also be used as field strength meters. Field Strength Meters are used to test the
5-165      density of an RF field. This allows testing of antenna patterns and ambient RF levels at
repeater and home sites to insure safe levels.

Chart: Notice how losses go up as frequency increases for each coax type. Coax losses
shown above are for 100 feet lengths. Loss is a length multiplier, so a 200 ft length would
5-173      have twice the loss shown above and a 50 ft length would have half the loss. This multiplier
factor is why you should keep cable installation lengths between radios and antennas as
short as practical!

Impedance of feedlines is generally determined by the distance between conductors and
their size. Amateur antennas are generally fed with 50 ohm coax or 450 ohm
ladder/window line. Occasionally TV-style 300 ohm twin lead or 75 ohm coax is used for
5-180      certain antenna types. Ultimately, these feedlines must all be matched to the typical 50 ohm
input on the radio to insure proper power transfer to the antenna. Many dipole style
antennas use a balun (BALanced to UNbalanced) to match the balanced antenna feedpoint
to an unbalanced 50 ohm coaxial feed.

A clarification: Sometimes the antenna impedance is less than that of the feedline,
5-192      sometimes it is more. In any case, the resulting SWR is always displayed with the larger of
the two numbers in the ratio first as shown above. 1:2.5 would be incorrect.

5-194      The ratio is always determined by dividing the smaller number into the larger one.

Important detail. The matching device (antenna tuner) does not change anything. It simply
5-198      “tunes” the feedline to match the radio so the transmitter can pass the power to the feedline.

Note the PL-259 is a Plug and is therefore male. The SO-238 is the corresponding chassis
mount Socket. The PL-259 is by far the most common connector in amateur use. It‟s
simple to install and cheap. Unfortunately, its impedance isn‟t stable – especially at high
frequencies. For lower power work, BNCs and SMAs are more common because of their
5-203      compact size.
N connectors are used by hams at higher (UHF) frequencies because they handle high power
levels and match coax perfectly into the GHz range. N Connectors are also used by the
military because they are gasketed making them water and dust-proof.

End Module-1

Module-6

Slide
Footnote
Number
OET bulletin 65 requires us to PERFORM A ROUTINE RF EXPOSURE EVALUATION