Speaker Notes

The reference to Pages 23 through 64 is to the Gordon West General Class Study guide, 7th edition. This guide is available from HRO either in the store or order on-line, or through 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 reference long after you receive your license.

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.

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” 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 the question asks:

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:

.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? Answer: 468/7.1=65.9 feet.

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 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 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 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 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 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 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.

(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 radiated power (1 kW equivalent isotropically radiated power) without the authorization 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.

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

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 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 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 of your station records.

As a general rule, if you can hear them, they will be able to hear YOU. It’s best not to use 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 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 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 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.

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. 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 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

Read about space stations and Amateur Radio at www.amsat.org. AMSAT is a very active group of Amateur Radio operators who provision and operate repeater and telemetry 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 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 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.

This is a condition set forth by the FCC as part of the “punishment” of a licensee who’s 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 should print out this information and keep it with your operating aids in your station.

FYI:

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

The key to getting along in confined spaces in to act in a non-confrontation manner in all things. Don’t let road-rage type behavior get started on the ham bands.

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 (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 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 you MUST use upper sideband. All other bands, the sideband you use is recommended, not required by the rules.

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

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 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 slower?” Without the question mark it means “please send slower”.

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

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

Rule of thumb: In a genuine emergency, that is an immediate danger to human life or real 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 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 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 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).

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 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 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 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 surface. See the illustration on slide 59.

They are in alphabetical order but what happened to A, B, and C? They are atmospheric, 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.

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 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.

Since it combines signals, the balanced modulator is a type of mixer. It’s the “balancing” 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.

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 ruin it!

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

…and more bandwidth can cause interference to adjacent stations,

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 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 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 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 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 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 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 70.7 = 4998.5. Divide that by 50 and you get 100 Watts (rounded).

Why? Here’s the Ohm Law equivalent: P = E² / 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 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.

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.

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 power.

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 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 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 102 or 1 followed by 2 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, 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.

Refer to Slide 3-50 for a related question

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

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 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. 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, several portable transceivers are available in kit form that use direct conversion design.

See the notes on “neutralization”

FET stands for Field-Effect Transistor

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

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 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 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". 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.

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 only the 4 shown will be used in the Element-3 test. A description of these 4 formulas is shown on the next slide.

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.

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.

Referring to Slide 4 we see that P=E² / R will work for us. First we “square” E by 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 decimal point on the current (I).

This question uses the fourth formula; P=I² 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 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.

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 .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)

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

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 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 Guide.

High quality filter capacitors can stay charged for days, even weeks. If you are working 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 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 advantages of using an oscilloscope rather than a meter.

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 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 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 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 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 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 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 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 they were not of equal value? Use the reciprocal trick like in slide 113. You’ll have a chance to practice this very soon.

Remember, Inductors are treated numerically just like resistors. In series, simply add the values.

Duck soup! Convert 20 to it’s reciprocal by dividing it into 1. You’ll get 0.05. Do the same 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 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 decreases due to the mass of our hearing apparatus.

Now we’re talking about an inductor which behaves numerically just the opposite of a capacitor.

There is a picture of a wire wound resistor in the Gordon West Study Guide on page 136. 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 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 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 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, 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

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 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 poles), antenna mast guy wires not bonded.

Way back in Module-1 on slide 15 we introduced a bogus phone number. Have you 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.

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 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 difficult to adjust an HF antenna from a step ladder.

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 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 way up in the air. This view is referred to as the E-plane.

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 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 perpendicular to the wire.

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 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 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 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 length of a ½ wave antenna.

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

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 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. 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 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. 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 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. “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 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 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 the August 07 SARS newsletter

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

Slide Number	Footnote
1-2	The reference to Pages 23 through 64 is to the Gordon West General Class Study guide, 7th edition. This guide is available from HRO either in the store or order on-line, or through 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 reference long after you receive your license.
1-4	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. 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.
1-8	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” 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.
1-15	The formulas at the top are useful in converting frequency to wavelength. For example, if the question asks: 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: .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? Answer: 468/7.1=65.9 feet. 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.
1-16	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 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.
1-17	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 know them to operate properly. Get a copy of the band plan and keep it on your operating desk.
1-19	Thinking outside the box: The QPC seems to have overlooked the 60m band where all amateurs have equal privileges. No matter, the above information is what the examination requires.
1-23	It is probably easiest to remember that 1500 Watts PEP is the legal limit on all bands 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.
1-25,29	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 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.
1-30	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 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 radiated power (1 kW equivalent isotropically radiated power) without the authorization 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.
1-48	This band is so narrow and shared with other services that it is channelized. Very strict controls are in place restricting type of emissions.
1-54	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 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 of your station records.
1-60	As a general rule, if you can hear them, they will be able to hear YOU. It’s best not to use the frequency if you can copy other signals on it.
1-63,64	Notice the absence of decimal points in the frequency choices. For the purposes of this 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)
1-72	Don’t be fooled. You may be used to hearing CB operators on our 10 meter band. They aren’t “sharing” the band, they are operating illegally!
1-76	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 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.
1-81	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. 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.
1-97	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 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
1-99	Read about space stations and Amateur Radio at www.amsat.org. AMSAT is a very active group of Amateur Radio operators who provision and operate repeater and telemetry equipment on satellites. Control of this equipment is confined to “secret codes” for security reasons and is the only type of secret code permitted
1-101	CW operators use abbreviations extensively to speed up the conversation. You will also hear DX CW using something called “Cut numbers” to save time. All perfectly legal.
1-103	This is done quite often. There are “Swap nets” where amateur gear is listed for sale. It is 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.

Slide Number	Footnote
1-105	This is a condition set forth by the FCC as part of the “punishment” of a licensee who’s license has been revoked. He is not permitted to use Amateur Radio under any circumstances.
1-111	Third party agreements and reciprocal licensing agreements are published on the web. You should print out this information and keep it with your operating aids in your station.
1-133	FYI: On HF phone nets, “Break-Break” usually indicates priority traffic. A triple break as in C: above would indicate emergency traffic.
1-137,139	The key to getting along in confined spaces in to act in a non-confrontation manner in all 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.
1-147	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 (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.
1-165	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 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.
1-178	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 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.
1-186	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 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”.
1-192	In CW each Q-signal will have two meanings. If QRS? The meaning is “Shall I send 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

Slide Number	Footnote
2-3	Generally, the reference to “property” is meant to be homes and other buildings. How about a car? A bicycle? You decide.
2-6	Rule of thumb: In a genuine emergency, that is an immediate danger to human life or real property, pretty much anything goes.
2-27	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 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.
2-31	The illustration on slide 59 can be of help understanding this. There are only two questions on the test asking about maximum distances per region. The other question deals with the E-region.
2-33	This is because at noon during the summer, sunlight (and ultra violet radiation) will be at its maximum intensity. It is the effect of this radiation that changes the altitude of the F layers.
2-34	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). 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.
2-36	Notice the answer says “return a radio signal to earth”. Signals may “bounce” or “skip” but they are not “reflected”. The correct term is “refracted”.
2-44	On HF, beacon stations can be found on specific frequencies on 20 through 10 meters. See this URL for complete details: http://www.dxzone.com/cgi-bin/dir/jump2.cgi?ID=1112
2-46,48	Do you remember the conversion formula? mHz to meters: 300/f(mHz). Choose the answer that comes the closest. E.g. 300/22=13.6 (14) Choose 15.
2-54	This is the second of two questions referring to the maximum distance along the earth’s surface. See the illustration on slide 59.
2-58	They are in alphabetical order but what happened to A, B, and C? They are atmospheric, 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!
2-81	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 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

Slide Number	Footnote
3-105	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 deficiencies of early vacuum triodes used as high-frequency amplifiers in radio direction finding (RDF) equipment. Most modern receivers use these same principles.
3-118	Images can be difficult to handle. They can appear on both sum and difference signals. Read what Gordon West has to say about it on page 102 of the Study Guide.
3-130	Very early designs used direct conversion receivers for low cost and simplicity. Today, 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
3-154	Memory tip: Logic circuits use two states: on-off, high-low, one-zero, they all mean the 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.
3-181	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 makes it a NAND gate and reverses whatever the answer would have been. Contrary little devil!
3-183	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 thingy attached to it. Therefore, whatever the OR gate would have been, the NOR is just the opposite.
3-185	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". 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.
3-191,193	Abbreviation of complementary metal oxide semiconductor. Pronounced see-moss, CMOS is a widely used type of semiconductor. CMOS semiconductors use both NMOS (negative polarity) and PMOS (positive polarity) circuits. Since only one of the circuit types is on at 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

Slide Number	Footnote
4-3	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 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.
4-6	Referring back to slide 4 we see that the only formula solving for E will work here. Use E=SQR PR. 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.
4-8	Referring to Slide 4 we see that P=E² / R will work for us. First we “square” E by multiplying it by itself: 400 x 400 = 160,000. Now we divide that by R (800) to get 200 Watts.
4-10	This one is “Easy as PIE”. Multiply I times E. 0.2 x 12 = 2.4 Watts. Don’t forget the decimal point on the current (I).
4-12	This question uses the fourth formula; P=I² 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 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.
4-14	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 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.
4-18	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 .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.
4-24	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 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.
4-34	Read the description and see the graphics on pages 120 and 122 of the Gordon West Study Guide.
4-52	High quality filter capacitors can stay charged for days, even weeks. If you are working around a high voltage power supply don’t assume the bleeder resistors are doing their job. Check for voltage before you touch anything.
4-54	Well, at least the readout is more precise. With an analog meter you need a mirrored dial to 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!
4-56	With some things, just getting close enough to measure can change it. That’s one of the 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!
4-83	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 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?)
4-105	A special type of resistor that we want to change when the temperature changes. It is used primarily in thermostats, and weather stations.
4-107	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 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.
4-109	According to Kirchhoff's second law, everything in a circuit must be accounted for. You can’t end up with more or less than you started with.
4-110	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 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.
4-113	Now they’re making us do a little work. Convert each resistor to it’s reciprocal by dividing 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
4-117	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 picofarads…this would be the same if it was gigamegafarads. 5000 + 5000 + 750 = 10750 whatevers.
4-119	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 they were not of equal value? Use the reciprocal trick like in slide 113. You’ll have a chance to practice this very soon.

Slide Number	Footnote
4-121	Remember, Inductors are treated numerically just like resistors. In series, simply add the values.
4-123	Duck soup! Convert 20 to it’s reciprocal by dividing it into 1. You’ll get 0.05. Do the same with 50, get 0.02. Now add the two reciprocals and reconvert. 1 / 0.07 = 14.285 (round to 14.3)
4.130	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 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.
4-138	You might think of it as an audio tone. As the frequency increases, our ability to hear it decreases due to the mass of our hearing apparatus.
4-140	Now we’re talking about an inductor which behaves numerically just the opposite of a capacitor.
4-148	There is a picture of a wire wound resistor in the Gordon West Study Guide on page 136. Notice that it’s nothing but a coil of wire. Pure inductance!
4-154	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 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.
4-158	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 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.
4-166	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 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).
4-168	This is also a ratio problem but because we are invoking Ohms Law from the Power side, 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