Amateur Extra: JT-65 Encoding

Digital modes are the topic of this week’s Amateur Extra exam question….  [E8C13]

What is one advantage of using JT-65 coding?

A. Uses only a 65 Hz bandwidth
B. The ability to decode signals which have a very low signal to noise ratio
C. Easily copied by ear if necessary
D. Permits fast-scan TV transmissions over narrow bandwidth

Well, what the heck is JT-65 coding?  JT-65 is a digital mode of ham radio.  Initially developed for use in EME (Earth-Moon-Earth) VHF contacts, it was discovered that it could be used with great success on HF at low power.

I’ve personally used JT-65 at relatively low power to make QSO’s all over the world.

The great thing about JT-65, is the ability to pull a readable signal from a lot of noise.  You may not even be able to hear it with your ears, but the software is able to “see” the signal in the waterfall (the display of the audio frequencies at the top of the image.)  This is because of the relatively low amount of information (only 13 characters!) transmitted over a long amount of time (47 seconds!) with plenty of error correction and redundancy built in.

The answer, is, therefore, B. The ability to decode signals which have a very low signal to noise ratio

You should head on over to JT-65.com to get your JT-65 reference card, and stick around, eventually I’ll be releasing a JT-65 mini-course!

General: KN

What does it mean when a CW operator sends “KN” at the end of a transmission?

A. Listening for novice stations
B. Operating full break-in
C. Listening only for a specific station or stations
D. Closing station now

The answer is, C. Listening only for a specific station or stations.  You would use this “word,” which isn’t a word at all, if you are carrying on a QSO with another station or stations and signify that you’re done with your transmission and are awaiting a reply from them.

But why “KN?” Is it an abbreviation for something?  Well, not really.  Much like “CQ” (_._. __._) my theory is that its easily recognized by ear.  _._ _.    If you listen to CW long enough, you’ll start to recognize things like this quite easily… and when you hear it, its your turn!

[G2C03]

Technician: Connecting a Voltmeter

What is the correct way to connect a voltmeter to a circuit?

A. In series with the circuit
B. In parallel with the circuit
C. In quadrature with the circuit
D. In phase with the circuit

d’

Well, first off lets just get rid of C. and D. Neither “in quadrature” or “in phase” really make any sense at all.

So do we connect the probes of our voltmeter in parallel or in series to a circuit? We need to make sure we understand what each of those terms mean.

“In parallel” means we connect each probe to either end of the circuit in question.  You’ll hear the term “voltage across” a circuit, battery, or component quite a bit.  This is an easy way to remember parallel.  For example, you measure the voltage of a battery by connecting the voltmeter “across” the terminals, or simply one at each.  It would look something like the image below, where the voltmeter is “V.”

Series, on the other hand, involves inserting your meter into the circuit as part of the “chain” or “path” of the circuit, and looks something like the image below, where the ammeter is “A.”Voltmeter_AmmeterAttempting to measure voltage in this manner will  just not work.  Instead, you measure current, or amps this way.  By putting your ammeter “in the path” you can measure the amount of charge flowing through the circuit.

So, the correct answer is, B. In parallel with the circuit.

[T7D02]

Image Credits: https://commons.wikimedia.org/wiki/File:Voltmeter_Ammeter.jpg, http://commons.wikimedia.org/wiki/File:BASIC_stamp_and_analog_voltmeter.jpg

Amateur Extra: Vestigial Sideband Modulation

Ew, this sounds kind of gross, but this aspect of sideband modulation is on the test! [E2B06]

What is vestigial sideband modulation?

A. Amplitude modulation in which one complete sideband and a portion of the other are transmitted
B. A type of modulation in which one sideband is inverted
C. Narrow-band FM transmission achieved by filtering one sideband from the audio before frequency modulating the carrier
D. Spread spectrum modulation achieved by applying FM modulation following single sideband amplitude modulation

Vestigial sideband modulation? Sounds like something you’d pick up in the jungle, and have to take a lot of pills to get rid of.

It’s nothing that evil, and you’ll probably not run into it very much.  But hey, its on the test!

The definition of “vestigial” is: forming a very small remnant of something that was once much larger or more noticeable.

What this means for us, is that if you were to look at a sideband radio signal with a vestigial sideband on a scope, you would see something like the image above, where there is a full USB signal, and then part of the lower sideband.

That signal in the picture is what the old analog television signals looked like.  The audio was transmitted in a sideband, and the video in the other.  This is why it was possible, for example, to pick up the audio from old analog channel 6 on your car radio at 87.7 MHz.  Current ATSC digital television uses something called 8VSB.  The VSB stands for, you guessed it, vestigial sideband.

So, lets look at our answers.  D? Nope. No spread spectrum action going on here. C? Again, no. There is no requirement that the signal be narrow FM at all.  B? Getting warmer, but still wrong.

The correct answer is A. Amplitude modulation in which one complete sideband and a portion of the other are transmitted.

General: Data Packets

What part of a data packet contains the routing and handling information?

A. Directory
B. Preamble
C. Header
D. Footer

The answer is C. Header.  When dealing with data “packets”, no matter the type, there has to be a portion of data that contains information regarding what’s to be done with the data inside it.  This section is called the header, because it occupies the section at the front, or the head, of the packet.

TCP packets (used to deliver this web page to you!) are an example of a data packet that contains a header.  The above image shows how much information is packed into this section.  Data such as the source and destination address, TTL, protocol, error correction data, and more.

[G8C03]

Technician: VHF Exposure Limits

What is the maximum power level that an amateur radio station may use at VHF frequencies before an RF exposure evaluation is required?

A. 1500 watts PEP transmitter output
B. 1 watt forward power
C. 50 watts PEP at the antenna
D. 50 watts PEP reflected power

The answer to this question is C. 50 watts PEP at the antenna.  The answer lies in a FCC publication called OET Bulletin 65.  In this document is a lot of engineering and math, but most important is a table of frequencies and minimum power levels for which a station should be evaluated.

 

The reason for such a wide variance in the limit, is that the human body is affected by RF differently for different frequencies.  The same power level at one frequency may be completely harmless, while at a higher frequency, dangerous.

Think microwave ovens.  They simply use RF in a certain frequency range, and we all know what it does to stuff put in the oven, and not at a particularly high power.

Now, notice that this is not 50W PEP output from your radio.  This is at the antenna.  Depending on your station, you very likely see losses in your feedline and antenna which will make the actual output lower.   In practice, since most VHF radios are limited to 50W transmit power or lower anyway, you probably won’t have to worry about this.  If you run an amplifier or a radio capable of more than 50W output on VHF, then you should consider performing an RF evalution.

[T0C03]

Amateur Extra: Power Factor in an RL Circuit

Ah, another fine math problem from the Amateur Extra exam (don’t you just love these?) [E5D16]

What is the power factor of an RL circuit having a 30 degree phase angle between the voltage and the current?

A. 1.73
B. 0.5
C. 0.866
D. 0.577

awasdf

This question is a bit tricky, because we have to calculate the numeric value of the power factor due to the phase difference.  Luckily we can throw out A, because its greater than 1 (that can’t happen!)

But, we’re not actually given any values for the resistance or inductance in the circuit! How are we supposed to do this?

Simple.  All you need to remember, is how to plot the values for an RLC circuit on a set of coordinates.  We’re given the 30 degree phase difference.  So draw a line from the origin, one unit long, at an angle of 30 degrees from the X axis.  OK?

Now, this starts to look like a trig problem.  Let’s just remember that the power factor is defined as the actual resistance of the circuit (not the impedance!) divided by the total impedance.  It then represents the actual power dissipated by an equivalent resistance.  So, we need to find the value of the triangle leg along the X axis.  (Yes, I realize I need to update this with an actual image of a graph. Remind me to do this because I will forget!)

Therefore, we use the definition of a cosine.  (Remember your trig?)

\cos \theta = O / H

 

where O is the “opposite” side of the triangle vs. the angle, and H is the hypotenuse.  In our case, this looks like this:

\cos 30^{\circ} = O / 1

 

We can get rid of the 1, leaving us with only this, and since the cosine of 30 degrees is 0.866:

O = \cos 30^{\circ} = .866

 

Therefore C. 0.866 is our answer!

 

General: USB Frequency Range

What frequency range is occupied by a 3 kHz USB signal with the displayed carrier frequency set to 14.347 MHz?

A. 14.347 to 14.647 MHz
B. 14.347 to 14.350 MHz
C. 14.344 to 14.347 MHz
D. 14.3455 to 14.3485 MHz

This is really just a question of simple addition, so don’t let all the numbers there throw you for a loop.

Remember that USB, or Upper Side Band, puts the actual signal on the” upper” portion of the signal.  This means that the 3kHz signal is above the “displayed carrier frequency” (which really makes this question a bit misleading, since single sideband really doesn’t have a “carrier” signal….)

So, do the math.  Add the 3kHz, or .003 MHz (remember your metric prefixes!), to 14.347, and you get B. 14.347 to 14.350 MHz.

[G4D09]

Technician: Repeater Modulation

Which type of modulation is most commonly used for VHF and UHF voice repeaters?

A. AM
B. SSB
C. PSK
D. FM

The answer is D. FM, or Frequency Modulation.  I’m not sure there is a technical reason for this, as any modulation would work just as well.  Except PSK which really isn’t a modulation, but a digital mode, and you can use it in FM or SSB just as easily.

The reason probably lies more in a bit of circular effect.  VHF/UHF radios get built with FM capabilities, and therefore the repeaters get built with the same, and repeat.  FM is of course used in commercial applications in this range also, so adapting manufacturing for amateur radio products was probably a short move for producers.

FM also uses more bandwidth than SSB, 25kHz vs. 3kHz.  In upper 10m and beyond, this bandwidth becomes more and more usable.

SSB can still be found in the VHF/UHF range, though!  AM, not so much, but aircraft and airports do use AM in the 108-137 MHz range, and military aircraft are known to use AM from 225-400 MHz.

[T8A04]

Amateur Extra: HF Space Station Allocation

The Amateur Extra question of the week comes from the section on frequency allocations and amateur services… [E1D07]

Which amateur service HF bands have frequencies authorized to space stations?

A. Only 40m, 20m, 17m, 15m, 12m and 10m
B. Only 40m, 20m, 17m, 15m and 10m bands
C. 40m, 30m, 20m, 15m, 12m and 10m bands
D. All HF bands

Ah, man.  There is no good way to put this, but this is yet another question that you just have to memorize.

The correct answer is A. Only 40m, 20m, 17m, 15m, 12m and 10m.  Why? Because that’s how its defined in Part 97, section 207 of the FCC rules.

(c) The following frequency bands and segments are authorized to space stations:

(1) The 17 m, 15 m, 12 m, and 10 m bands, 6 mm, 4 mm, 2 mm and 1 mm bands; and

(2) The 7.0–7.1 MHz, 14.00–14.25 MHz, 144–146 MHz, 435–438 MHz, 2400–2450 MHz, 3.40–3.41 GHz, 5.83–5.85 GHz, 10.45–10.50 GHz, and 24.00–24.05 GHz segments.

Recall that HF only includes bands from 10m on up, and there’s your answer.

The truth is, if you’re even concerning yourself with working space stations on HF, you will know this.  The only other reason to know this, is so you don’t miss this question on the test!

Image Credit: http://www.scienceimage.csiro.au/image/11143