Amateur Extra: Estimating RF Fields

Which of the following would be a practical way to estimate whether the RF fields produced by an amateur radio station are within permissible MPE limits? [E0A03]

A. Use a calibrated antenna analyzer
B. Use a hand calculator plus Smith-chart equations to calculate the fields
C. Use an antenna modeling program to calculate field strength at accessible locations
D. All of the choices are correct

Lets examine all the possible answers and see if we can get the correct one by elimination….

A. Use a calibrated antenna analyzer.  This might tell you if you’ll be able to transmit through your antenna with a low SWR on a given frequency, but it won’t tell you anything about the radiation pattern and field strength.

(We just eliminated D. All of the choices are correct!)

B. Use a hand calculator plus Smith-chart equations to calculate the fields.  Much like A. Using a Smith chart won’t help you determine the RF fields coming off of the antenna.

The only way to estimate the RF pattern and field strength coming from an antenna or antenna system is to use software (or do it the very hard way!) to model the system and provide theoretical field values at given locations.

C. Use an antenna modeling program to calculate field strength at accessible locations is the correct answer.


General: NVIS Height

At what height above ground is an NVIS antenna typically installed? [G9D03]

A. As close to 1/2 wavelength as possible
B. As close to one wavelength as possible
C. Height is not critical as long as it is significantly more than 1/2 wavelength
D. Between 1/10 and 1/4 wavelength

First off, what the heck is an NVIS antenna?

NVIS is an acronym for Near Vertical Incident Skywave.  The military uses HF antennas of this design in order to facilitate short-range communications.  As the name suggests, the idea is to send the signal almost nearly straight up to the sky, where it will be reflected down again.

The answer is D. Between 1/10 and 1/4 wavelength. At HF frequencies, this is pretty low. Low enough to make it easily deployable.  There’s a lot of math and physics involved, and rather than really try and dive into it without a lifejacket, lets just say that the other answers are incorrect.  The reason is actually pretty obvious when you think about the wavelengths involved.  To mount a horizontal dipole, which is how this technique is normally used, at such distances above ground is often impractical, if not impossible.

Technician: Monitoring SWR

Where should an in-line SWR meter be connected to monitor the standing wave ratio of the station antenna system? [T4A05]

A. In series with the feed line, between the transmitter and antenna
B. In series with the station’s ground
C. In parallel with the push-to-talk line and the antenna
D. In series with the power supply cable, as close as possible to the radio

OK, lets put on our “common sense” caps and come up with the answer.  A lot of times the exam questions are like this, painfully obvious.  If you’re not ready for them, they can throw you off.

In series with the station’s ground? Why would you do that? It doesn’t even make sense. Neither does in parallel with the PTT line and the antenna. How would you even wire that up?  In series with the power supply cable? The only meter you would want there, is a DC ammeter, to measure the current going into your radio.

Therefore, the answer is A. In series with the feed line, between the transmitter and the antenna.  The SWR or Standing Wave Ratio meter shows how much of your signal is being reflected back from the antenna.  The only way you will know this, is by placing the meter in the signal path from your radio to your antenna.

Amateur Extra: ASCII Parity Bit

What is the advantage of including a parity bit with an ASCII character stream? [E8C12]

A. Faster transmission rate
B. The signal can overpower interfering signals
C. Foreign language characters can be sent
D. Some types of errors can be detected

The definition of a “parity bit” according to Wikipedia is: a bit added to the end of a string of binary code that indicates whether the number of bits in the string with the value one is even or odd. Parity bits are used as the simplest form of error detecting code.

Given the other answers don’t deal with error detection at all, its a safe bet that the answer is D. Some types of errors can be detected.

General: Overmodulation

Which of the following is an effect of overmodulation? [G8A08]

A. Insufficient audio
B. Insufficient bandwidth
C. Frequency drift
D. Excessive bandwidth

The answer is D. Excessive bandwidth.  Now, for the purposes of getting this question correct on the exam, I’ll just say this: Just remember the connection between overmodulation and excessive bandwidth.

The real reason this occurs, is technically fairly complex, and rather than try and rewrite and research it on my own, I’ll just point you at this informative piece at stackexchange:

They have some animations that clearly show what is going on there, showing the development of the spurious emissions that develop in an AM radio signal when overmodulated.

The above image shows a normally modulated FM signal.  If I locate a decent example or demo of how this also applies in the FM realm, I’ll add to this entry.

Technician: Switch Components

Which of these components can be used as an electronic switch or amplifier? [T6B03]

A. Oscillator
B. Potentiometer
C. Transistor
D. Voltmeter

Piece of cake this one is….  Lets go through the false answers first, just to make sure.

A. Oscillator.  By definition, an oscillator simply takes a signal and, well, oscillates it.  on/off, high/low, whatever.  While definitely important in radio (you can’t get an alternating waveform without one) you can’t really make a switch out of one.  It is possible to use an oscillator as an amplifier.  It’s how vacuum tube radios get the work done (and that’s our first clue!)

B. Potentiometer.  A potentiometer is nothing more than an adjustable value resistor.  While it may have a switch built into it, such as an on/off/volume switch combination, a potentiometer by itself doesn’t make a particularly good switch.  I suppose you could use it as one, by increasing a signal via the “pot” up to or below a given “cutoff” value…. but that’s a bit of a stretch.

D. Voltmeter.  A voltmeter simply measures the voltage across a given circuit or component.  Nothing more.  It’s quite impossible to use one as a switch.  (I think!)

So all that remains is C. Transistor.  A transistor is a solid state device that toggles between states depending on an external input.  By placing (for example) a 5V load on a transistor you can essentially “flip the switch,” and removing the voltage will turn it off again.


Amateur Extra: EME Communications

Which of the following digital modes is especially useful for EME communications? [E2D03]

A. FSK441
C. Olivia
D. JT65

Due to the time it takes for signals to get to the moon and back, (EME = Earth-Moon-Earth) we need a mode that is highly dependent on time synchronization.  This way everyone knows when to transmit, and when to listen. D. JT65 is just that mode.

JT65 is a highly structured digital mode that requires the operator’s station computer to be synchronized to the world time servers.  Communications are then alternated at 60 second intervals, 47 seconds transmit, with 13 seconds for decoding.  Transmissions are limited to 12 characters, and the QSO follows a strict format of exchanges.

All of the other modes might be capable of making the EME trip, but all are independent of any external reference that makes them a reliable option.


General: 80m Privileges

Which of the following frequencies is within the General Class portion of the 80-meter band? [G1A08, 97.301(d)]

A. 1855 kHz
B. 2560 kHz
C. 3560 kHz
D. 3650 kHz

Well, we can eliminate answers A. 1855 kHz and B. 2560 kHz as neither one of those is in the 80-meter band, which is at 3500 kHz to 4000 kHz.  The answer is C. 3560 kHz.  This lies in the CW/data range of the 80m band.  The “phone” or SSB (single sideband) voice portion of the 80m General class portion doesn’t start until 3800 kHz.

This is defined in FCC Part 97.301(d)

Unfortunately, there isn’t a good way to remember the band privileges, except to memorize them.  Sorry!  (If you have a good way to do this, please let me know!)


Technician: Spin Fading

What causes spin fading when referring to satellite signals? [T8B09]

A. Circular polarized noise interference radiated from the sun
B. Rotation of the satellite and its antennas
C. Doppler shift of the received signal
D. Interfering signals within the satellite uplink band

Well, to start with, I suppose we should define “spin fading.”  Spin fading refers to the (sometimes rapid) change in received signal strength you will encounter when listening to amateur radio satellite signals.

It occurs because the satellite is not “fixed” in position during its orbit.  It’s tumbling end-over-end as it makes its pass.  Therefore, the antenna is always moving.

I suppose you could say we’ve just provided our answer, also.  B. Rotation of the satellite and its antennas.  Interfering signals won’t exhibit “fading” characteristics.  Doppler shift will be observed, but it manifests as a slight change in the signal frequency, depending on if the satellite is moving towards you or away from you.  And answer A. Circular(ly) polarized noise interference radiated from the sun just doesn’t make much sense at all.

Amateur Extra: Rectangular Coordinates

What do the two numbers represent that are used to define a point on a graph using rectangular coordinates? [E5C11]

A. The magnitude and phase of the point
B. The sine and cosine values
C. The coordinate values along the horizontal and vertical axes
D. The tangent and cotangent values

This might seem like a strange and really lame question, but I think the reason for it is to make sure you understand the difference between rectangular and polar coordinates.

In a rectangular coordinate system, the two numbers represent C. The coordinate values along the horizontal and vertical axes.  That might seem rather obvious.  For example, if you think of a checkerboard, for example, and measure over three squares, and up four, then those coordinates would be 3,4.

Polar coordinates on the other hand, are represented by a radius and an angle.  For example if you were to take a string and pin it to the center of the checkerboard, then rotate that string around the center, at a given length.  So if your string was 6″ long and it was at 90º from the horizontal, your polar coordinates would be 6∠90º.

See the image above for what this looks like.