# Which of the following HF digital modes uses variable-length coding for bandwidth efficiency?

A. RTTY
B. PACTOR
C. MT63
D. PSK31

Now, I guess this answers my question about last weeks Extra question. Mainly why is it important.  It’s important because it does matter in transmission efficiency.  So which mode would be a good and efficient method of sending digital transmissions.  Well, once more lets examine the possibilities.

RTTY?  RTTY or Radio Teletype is more or less and ancient method of sending text via radio.  It actually dates back into the 1930’s.  RTTY uses a fixed-bit encoding.  It’s NOT variable.  So that basically rules it out as an answer.  (See: https://en.wikipedia.org/wiki/Radioteletype)

PACTOR?  PACTOR is a much more recent development, but its primary use is to send large messages, essentially being packet over HF.  It has built-in error correction, and other features.  Unfortunately, PACTOR I uses anywhere from 400 to 2600 Hz bandwidth, so not all that efficient.  (see: https://en.wikipedia.org/wiki/PACTOR)

MT63?  Now we’re getting somewhere in terms of bandwidth efficiency.  But again, has a bandwidth of anywhere from 500 to 2000 Hz. (https://en.wikipedia.org/wiki/MT63)

PSK31?  Hello!  Here’s what you need to know.  A proper PSK31 signal has a bandwidth of just 100 Hz.   That’s all.  Take a look at the image above, there are several PSK signals being seen here.  Yes, its a variable length coding.  See https://en.wikipedia.org/wiki/PSK31

[E2E09]

# What are the basic components of virtually all sine wave oscillators?

A. An amplifier and a divider
B. A frequency multiplier and a mixer
C. A circulator and a filter operating in a feed-forward loop
D. A filter and an amplifier operating in a feedback loop

To answer this question, lets look at what makes up a sine wave oscillator circuit.  For radio purposes, we will normally see what’s referred to as an “RC” oscillator, made of a network of resistors (R) and capacitors (C).

Essentially how it works, is that the output of a circuit, which is tuned using that RC network, or filter, is fed back into itself, so we’ve met the condition of feedback.  There is an amplifier in place to continuously compensate for the losses incurred while doing this, due to heat, radiation, etc.    That’s it.  That’s basically the textbook definition of an electronic oscillator.

There are several different types of oscillator circuits, but the image above is an example of a “phase shift” oscillator.  (https://en.wikipedia.org/wiki/Phase-shift_oscillator)

Therefore, the answer is D. A filter and an amplifier operating in a feedback loop

# What type of electrical component consists of two or more conductive surfaces separated by an insulator?*

A. Resistor
B. Potentiometer
C. Oscillator
D. Capacitor

Well, I guess I gave it away in the post title.  Yes, the answer is D. Capacitor.  But what exactly is going on here?

A capacitor is a device that has the “capacity” (get it?) to store energy in an electric field.  It does this by collecting charge on one of two plates, separated by an insulator called a dielectric.  The charge couples with the conductor on the other side and creates an electric field inside the dielectric. (See the image above.)  The most obvious type of this is the typical ceramic disc style capacitor.  It’s easy to visualize the construction that way.  A “rolled” foil regular or electrolytic capacitor is built the same way, except the whole 3-layer construct is rolled into a nice tight package.

Variable Capacitor (http://commons.wikimedia.org/wiki/File:Kombidrehko.jpg)

An “air” variable capacitor (see left) like ones that were used in early radios for tuning, simply use the air in between the fins as the dielectric.

Looking at the other possible answers, though:  A potentiometer is simply a type of variable resistor.  Neither is constructed in this manner.  An oscillator is a type of circuit, not a component therein.

*(This question is taken from the upcoming 2018-2022 Technician pool.  It may not be the same as the 2014-2018 pool.) [T6A05]

# What is one advantage of using ASCII code for data communications?

A. It includes built in error correction features
B. It contains fewer information bits per character than any other code
C. It is possible to transmit both upper and lower case text
D. It uses one character as a shift code to send numeric and special characters

This is yet another one of those questions that I’m not sure why it’s on an amateur radio exam, really.  But, there it is.  Let’s examine the options and figure out which is the “one” advantage.

A. It includes built in error correction features? FALSE.  ASCII is just a numeric representation of letters/numbers/symbols and doesn’t have any such thing.

B. It contains fewer information bits per character than any other code.  FALSE again.  There are much more efficient methods for this.  ASCII has its roots in computers, not radio.  It’s designed for programming, not efficiency in transmission.

C.  It is possible to transmit both upper and lower case text.  TRUE.  Different case of letters have different ASCII codes.  i.e. “A” is 65 where “a” is 97.

D.  It uses one character as a shift code to send numeric and special characters.  FALSE.  Look up any ASCII table and its plain to see that every symbol, whether its punctuation, numbers, letters, etc. has its own distinct code.  There is no “shift” modifier.

The answer then, would be C.  It’s possible to transmit both upper and lower case text.

I know, I know.  Why is this really important to amateur radio?  I guess its important working digital modes and needing to know the details behind how a data stream is coded.  Beyond that, its superficial knowledge.

But its on the test, and you should know the answer!

[E8D11]

(Image: https://commons.wikimedia.org/wiki/File:ASCII_full.svg)

# What is a use for an antenna analyzer other than measuring the SWR of an antenna system?

A. Measuring the front to back ratio of an antenna
B. Measuring the turns ratio of a power transformer
C. Determining the impedance of an unknown or unmarked coaxial cable
D. Determining the gain of a directional antenna

Well, you’re not going to use an antenna analyzer to try and figure out the windings on a power transformer.  At least I hope not.  Totally the wrong tool for the job.

Actually measuring the front to back ratio and the gain of a directional antenna requires specialized equipment, since it needs to measure the power output by a radio.  An antenna analyzer generally doesn’t do that.

So, we’re left with C. Determining the impedance of an unknown or unmarked coaxial cable.

But why are we able to do this with an antenna analyzer?

Simply put, to the point of view of the analyzer, the coax is just another antenna.  It has two components, the center conductor and the shield, and therefore a labeled 50Ω length of cable should have a characteristic impedance of 50Ω.  If you measure this differently, then it means there is a problem with the cable.  Either a section is open, shorted, or damaged in such a way that it allows the two conductors to interact.   In fact, many antenna systems rely on the antenna-like properties of coaxial cable to function, requiring specific feed-line lengths, etc.

[G4B13]

## What is the current flowing through a 24-ohm resistor connected across 240 volts? [T5D09]

A. 24,000 amperes
B. 0.1 amperes
C. 10 amperes
D. 216 amperes

Get out your calculators, because this is a cut-and-dried case of needing to use Ohm’s law.

Ohm’s law defines the relationship beteween voltage, current, and resistance in a circuit.  It looks like this:

$V = IR$

Where V is voltage, I is current, and R is resistance.  You might see it with and E for Voltage instead of V.  (Old habits die hard!)

All we need to do is substitute in our known values and then solve this simple equation for the current, like this:

$240 = (I)(24)$

re-arranging, we get:

$I = \frac{240}{24}$

which reduces to:

$I = 10$

and we get our answer, C. 10 amperes.

## What is the peak-inverse-voltage across the rectifier in a half-wave power supply? [G7A04]

A. One-half the normal peak output voltage of the power supply
B. One-half the normal output voltage of the power supply
C. Equal to the normal output voltage of the power supply
D. Two times the normal peak output voltage of the power supply

To answer this we must know what is going on in a half wave power supply.

A half-wave rectifier takes, as you might suspect, only half of a given alternating current (AC) input voltage, and turns that into (alright, well, sort of) direct current (DC.)  The above image shows what that basically looks like.  In this instance, it takes the positive voltage and discards the negative voltage.

The peak inverse voltage is the maximum input AC voltage that the device can withstand before it is operating outside of specifications, possibly even to the point of damage.  In a half-wave power supply, we are discarding one half of the wave, so it must be able to take a peak-to-peak input voltage of at least twice its output.

The answer then, is D. Two times the normal peak output voltage of the power supply.

## What information is traditionally contained in a station log? [G2D09]

A. Date and time of contact
B. Band and/or frequency of the contact
C. Call sign of station contacted and the signal report given
D. All of these choices are correct

When keeping logs for your station, you need certain pieces of information for the contact, or QSO to be considered valid.  If you don’t have the correct information in your log, when you go to confirm this contact with the other party, the information may not match, and the QSO might not be confirmed.  In a contest this can cost you points, or you might miss out on an award, or even something as simple as a QSL card.

At a bare minimum, you need the date and time of the contact, in UTC or “zulu” time.  This is also known as Greenwich Mean Time (GMT) and is basically the “world reference” for time.  You need the band, preferably the exact frequency (from which you can infer what band it was.)  You also need the call sign of the operator you contacted, and an additional piece of information that you (and they) can confirm as having been received (and sent.)  In contest settings this is usually a signal report.  Often times it will include other information relevant to the contest, like a location or sequence number.

Therefore, the answer is D. All of these choices are correct.  Now obviously you can record as much information about a QSO as you’d like, including recording the whole thing(!) But at a minimum you need the information outlined above.

## Which of the following electronic components can amplify signals? [T6B05]

A. Transistor
B. Variable resistor
C. Electrolytic capacitor
D. Multi-cell battery

The answer is, as luck would have it, A. Transistor.  The transistor is probably the miracle of 20th century technology.  It’s an amplifier, a switch, and so much more.

None of the other components listed have the ability to amplify a signal.

## Which of the following is an advantage of using a Schottky diode in an RF switching circuit rather than a standard silicon diode? [G6A06]

A. Lower capacitance
B. Lower inductance
C. Longer switching times
D. Higher breakdown voltage

I know what you’re thinking.  Where did this bit of word soup come from, right?

First off, we need to know the difference between a Schottky diode and a standard silicon diode.  To start, they have a much lower breakdown voltage than a regular diode, which makes them generally better in high speed applications.

Since they aren’t capable of storing as much electric potential, and therefore electric charge, (which is the definition of a capacitor,) we can deduce that it would also have a A. Lower capacitance than a standard silicon diode.  This could make it quite useful in high frequency RF applications.