Amateur Extra: Digital Time Division Multiplexing

The Extra-Extra-Extra Read All About it question of the week is a real beauty. [E8B12]

What is digital time division multiplexing?

A. Two or more data streams are assigned to discrete sub-carriers on an FM transmitter
B. Two or more signals are arranged to share discrete time slots of a data transmission
C. Two or more data streams share the same channel by transmitting time of transmission as the sub-carrier
D. Two or more signals are quadrature modulated to increase bandwidth efficiency

Much like last time, whoa, come on!  Trust me, though, this one is pretty easy to remember.

Take a deep breath, close your eyes, and contemplate your navel, I mean the question.  We can automatically throw answers A and D out, because they make no mention of time.  Time is the key here.

Now take another breath, and think of the next word, division.  Discrete?  Bingo.  You just solved it.  The correct answer is B. Two or more signals are arranged to share discrete time slots of a data transmission.  C. Two or more data streams share the same channel by transmitting time of transmission as the sub-carrier is close, at least it sounds close, but is actually meaningless.  Transmitting the time of the transmission is rather pointless.

But what is “Time Division Multiplexing” anyway.  Short version:  If your mobile phone uses TDMA, then you’re using it on a daily basis.  It’s what allows all those conversations to co-exist on the same radio frequencies simultaneously.

Think of it like this:  Pretend there are 10 people sitting in a circle.  Each of them has one second to talk before the next one starts talking.  The cycle goes round-and-round, and if you were to then remove every other voice except for the one you wanted to listen to, you could understand what was being said.  Now imagine that happening at a much higher speed.  All you have to do is grab the “chunk” of the signal that happens every .002 seconds (as opposed to .001 or .005) and ignore the rest for that cycle.

 

General: Power Dissipation

Which value of an AC signal results in the same power dissipation as a DC voltage of the same value? [G5B07]

A. The peak-to-peak value
B. The peak value
C. The RMS value
D. The reciprocal of the RMS value

This is, unfortunately a “definition” question, so we’ll just come right out and tell you the answer, then explain a bit….

The correct answer is C. The RMS Value.  RMS stands for Root Mean Square.  Mathematically, when speaking about electrical power, it is the peak voltage of the sine wave, divided by the square root of 2.  So it looks like this, where V is the peak voltage.

RMS = V / √2

The exact derivation of this can be found here on Wikipedia.

Physically, you can think of this as similar to a saw being pulled back and forth through a tree or board.  The voltage increases sharply, levels off, decreases sharply again to negative, levels off, and repeats in the sine wave pattern.  Just as the saw blade stops at the end of each cutting motion, and cuts in both directions.

This is how an AC voltage dissipates the equivalent power of the same RMS DC voltage.

Technician: Electrical Current in the Body

How does current flowing through the body cause a health hazard [T0A02]

A. By heating tissue
B. It disrupts the electrical functions of cells
C. It causes involuntary muscle contractions
D. All of these choices are correct

Lets run down the possible answers, and see what we come up with….

A. By heating tissue.  Your body has a native resistance value to it.  Since you’re basically made of water and other stuff, its actually pretty low, but current flowing in your body will generate heat.  Higher current == higher temperatures.  Not cool.

B. It disrupts the electrical functions of cells.  True. Combined with the heating effect cells can be permanently damaged.  Nerve cells which rely on electrical impulses, are especially vulnerable.

C. It causes involuntary muscle contractions.  Also true.  See answer B.  Those nerve cells can interpret that electricity as instructions, and that can cause involuntary contractions.  This is why you NEVER try to grab somebody who is being electrocuted and cannot let go of what is causing it…  you will likely be unable to let go as well.

Since these are all correct, the correct answer is, of course, D!

Amateur Extra: Blocking Dynamic Range

Today’s question for you studying to be Extra’s is a real doozy, about receivers. [E4D01]

What is meant by the blocking dynamic range of a receiver?

A. The difference in dB between the noise floor and the level of an incoming signal which will cause 1 dB of gain compression
B. The minimum difference in dB between the levels of two FM signals which will cause one signal to block the other
C. The difference in dB between the noise floor and the third order intercept point
D. The minimum difference in dB between two signals which produce third order intermodulation products greater than the noise floor

Whoa, man. Lots of big words to digest in that one.  First off, just what in the heck do we mean by “Blocking Dynamic Range?”

Blocking Dynamic Range (BDR) indicates how well the receiver handles strong nearby
signals before desensitization occurs.  By definition (or, something you unfortunately have to memorize!) its the value of an input signal, relative to the MDS (or Minimum Discernable Signal, also called the noise floor) of the receiver, that will cause the gain to decrease by 1dB.

For example, if a – 25 dBm input signal causes 1 dB of gain compression for a receiver with a
MDS of -135 dBm, the blocking dynamic range is 110 dB.

Therefore, the correct answer is: A. The difference in dB between the noise floor and the level of an incoming signal which will cause 1 dB of gain compression.

Questions like these are what makes the Amateur Extra exam just that much more challenging, when you have to memorize definitions such as this!

General: Antenna Installation Precautions

What precaution should be taken when installing a ground-mounted antenna? [G0A13]

A. It should not be installed higher than you can reach
B. It should not be installed in a wet area
C. It should limited to 10 feet in height
D. It should be installed so no one can be exposed to RF radiation in excess of maximum permissible limits

To accurately answer this question, we should run down the list of possible answers one by one….

A. It should not be installed higher than you can reach.  This is obviously false, as you know by now the higher you can get an antenna, is usually better.  Granted you’ll need to reach it eventually, but if its on the roof or the top of a tower… good for you!

B. It should not be installed in a wet area.  Now, I’m sure this might be true, just from an engineering or convenience point of view. Who wants to slog around in the mud, right?  Putting an antenna in a wet area isn’t necessarily a problem.  Actually, depending on the water, it may actually be beneficial…  So not B.

C. It should limited to 10 feet in height.  Nope, nope, nope.  My 43′ ground-mounted vertical says nope.

So, that leaves us with D. It should be installed so no one can be exposed to RF radiation in excess of maximum permissible limits.

Now this is usually more of a concern if you’re running a high-power amplifier.  You don’t want anyone near your radiating elements with 1kW of RF coming out of them.  But even some antennas can pose a danger at lower power, say 100W.  You don’t want to put them where they could cause a problem.  It’s just the right thing to do.  D. is the correct answer.

Technician: Definition of Telecommand

What is the FCC Part 97 definition of telecommand? [T1A13]

A. An instruction bulletin issued by the FCC
B. A one-way radio transmission of measurements at a distance from the measuring instrument
C. A one-way transmission to initiate, modify or terminate functions of a device at a distance
D. An instruction from a VEC

The definition of a telecommand is set in FCC Part 97, specifically 97.3(a)(45)…

(45) Telecommand station. An amateur station that transmits communications to initiate, modify or terminate functions of a space station.

This is one of those “its the rules” questions, but the answer makes perfect sense if you think about it.  A and D are patently false.  B is the definition telemetry, not telecommand….  And if you think about C. if you are initiating, modifying, or terminating the functions of a device, you are giving it a command, or, a telecommand.  The rule specifies a “space station” but in general terms, the result is the same regardless of the end receiving the command.

The device pictured, just an average radio remote control for a car or airplane, would be a very basic example of a device that is issuing a “radio telecommand” to another device.

Amateur Extra: VHF/UHF Contest Activity

This week in the Amateur Extra spectrum, we’ll look at a question from the “band plan” area of the exam [E2C06]

During a VHF/UHF contest, in which band segment would you expect to find the highest level of activity?

A. At the top of each band, usually in a segment reserved for contests
B. In the middle of each band, usually on the national calling frequency
C. In the weak signal segment of the band, with most of the activity near the calling frequency
D. In the middle of the band, usually 25 kHz above the national calling frequency

We’ll look at the incorrect answers first, and explain why they’re wrong.  This is often the best approach to the exam questions, as the incorrect answers will be blatantly incorrect.

It’s not A. At the top of each band, usually in a segment reserved for contests.  This is because there is no section of the band (any band actually) reserved for contests.  It’s up to the operator to stick to the band plan for your area, but there isn’t a “contest only” section.

B. In the middle of each band, usually on the national calling frequency?  They’re trying to trick you now, you see.  The calling frequency shouldn’t be used to conduct QSO’s of any kind, really.  It’s called the “calling frequency” for a reason.  You make an initial contact, and then agree to move to a different frequency (or QSY) after that, freeing up the calling frequency for more “callers” as it were.

Is it D. In the middle of the band, usually 25 kHz above the national calling frequency?  Well, we’re getting warmer, but still no dice.  While we’re not sitting on the calling frequency, its still close enough that a strong FM signal could drown everyone out.  Plus, “the middle of the band” is pretty vague.

The remaining answer, C. In the weak signal segment of the band, with most of the activity near the calling frequency is correct.  But why? Didn’t we just say we didn’t want to be near the calling frequency? and that there is no “reserved” portion of the band?

It’s true there isn’t a “contest only” section of the bands.  However, there are plenty of conventions that are used.  One of them being that the weak signal part of VHF/UHF should be used for contests.  Primarily since (unless you’ve got an absolutely killer station, or the propagation gods are smiling) those VHF/UHF signals will be weak signals, and you’ll probably be working sideband (SSB) anyway.  Maybe not, depends on the particular contest and its rules I suppose.  Again, the conventional, agreed upon band plan for VHF, 2m in particular, puts the weak signal section (144.1 MHz – 144.2 MHz) directly below the SSB calling frequency (144.2 MHz), not above.

General: Using the “AG” Suffix

This question comes from the General exam, the “rules” section…. [G1D06]

When must you add the special identifier “AG” after your call sign if you are a Technician Class licensee and have a CSCE for General Class operator privileges, but the FCC has not yet posted your upgrade on its Web site?

A. Whenever you operate using General Class frequency privileges

B. Whenever you operate on any amateur frequency

C. Whenever you operate using Technician frequency privileges

D. A special identifier is not required as long as your General Class license application has been filed with the FCC

The answer is, A. Whenever you operate using General class frequency privileges.  While you may have passed your General exam, until that upgrade goes “live” in the FCC’s database, you’ll need to let everyone know.  Of course if you’re still operating in the bands and ranges available to Technician class operators, that isn’t strictly necessary.

The reason you’d want to do this is simple.  People will look you up in databases like the FCC’s, or on other displays of the same database, like QRZ.com.  If they spot you operating in the General class range, but your license still shows as Technician, then they could give you grief about it on the air, or worse, actually file a complaint.

The same situation applies if you upgrade from General to Amateur Extra, you’ll want to use an “/AE” suffix when operating in the Extra ranges.  As well as upgrading from the older, obsolete licensing classes (i.e. Novice, Advanced.)

The exact regulation [97.119(f)(2)] states.

(f) When the control operator is a person who is exercising the rights and privileges authorized by §97.9(b) of this part, an indicator must be included after the call sign as follows:.

(2) For a control operator who has requested a license modification from Novice or Technician to General Class: AG;

 

Technician: Satellite Tracking Software

Lets turn our gaze skyward for this question from the Technician pool… [T8B03]

Which of the following are provided by satellite tracking programs?

A. Maps showing the real-time position of the satellite track over the earth

B. The time, azimuth, and elevation of the start, maximum altitude, and end of a pass

C. The apparent frequency of the satellite transmission, including effects of Doppler shift

D. All of these answers are correct

Working or even just monitoring satellite radio transmissions can be fun and challenging. It’s also important that you’re armed with all the information you need to listen in and make contact.  There are numerous software programs available for you to use, and they all provide various information on the many satellites up there….  So lets look at the answers and see what we can come up with.

A. Maps showing the real-time position of the satellite track over the earth.  Your software should show you your location on the map, and the location of the satellite(s) overhead, so you can visually determine the passes, when they occur, etc.

B. The time, azimuth, and elevation of the start, maximum altitude, and end of a pass.  You will also need to know the time (of course) when the passes are.  You also need the starting and ending elevation and azimuth, so that you can aim your antenna accordingly, and if possible track the satellite.  The maximum altitude is also useful to know just how far away from you its going to be.

C. The apparent frequency of the satellite transmission, including effects of Doppler shift.  Satellites are moving incredibly fast relative to you.  This causes the radio frequency to “Doppler shift” depending on the speed of the satellite, and whether its moving towards you or away from you.  Think of the sound a train makes as it approaches you.  As it passes you, the sound changes frequency from high to low.  The same thing happens with the radio signals on satellites.  You’ll need to adjust high as it approaches you, and adjust lower as it moves away.

So, it looks like you really need to know all of these parameters to effectively utilize amateur radio satellites.  Your software should show you all of these things.  The correct answer, therefore is:

D. All of these answers are correct

 

Amateur Extra: Power Factor

Ah, another fine math problem from the Amateur Extra exam (don’t you just love these?) [E5D18]

How many watts are consumed in a circuit having a power factor of 0.71 if the apparent power is 500 VA?

A. 704 W

B. 355 W

C. 252 W

D. 1.42 mW

Remember that the “power factor” of a circuit is the percentage of power that is actually dissipated by the circuit.  Theoretically, if you apply 500 watts (or volt-amps, VA, its the same thing!) to a circuit, it should use all 500, right?  In reality, its not so.

So, to figure the actual power usage in this circuit, we just multiply the apparent power by the power factor, to get the correct answer, which is B. 355 W.  So what looks like another difficult math problem, is really quite simple.

(500)(0.71) = 355

 

So where did the rest go?  It basically disappeared, into the reactance of the circuit, and did no actual work. I know, it can be a bit strange to wrap your head around, but that’s how it works.