Amateur Extra: Reactive Power in an Ideal Circuit

What happens to reactive power in an AC circuit that has both ideal inductors and ideal capacitors? [E5D09]

A. It is dissipated as heat in the circuit
B. It is repeatedly exchanged between the associated magnetic and electric fields, but is not dissipated
C. It is dissipated as kinetic energy in the circuit
D. It is dissipated in the formation of inductive and capacitive fields

First, recall the definition of reactive power.  Reactive power is mathematically there, but doesn’t really “exist.”  It arises from the complex or “imaginary” components of reactance that arise in a circuit with inductors and/or capacitors exposed to alternating current.

In an ideal circuit, that is one where the capacitors and inductors don’t have any resistive components, the power B. is repeatedly exchanged between the associated magnetic and electric fields, but is not dissipated.  Of course there are always resistive physical elements at play, and this power gets bled off in the form of heat, since a portion manifests as real or non-reactive power.

Since it does exist in mathematical terms, reactive power from one circuit can affect and interact with the reactive power in another circuit, such as how power grids interact.

Here’s a more in depth explanation of True, Reactive, and apparent power: http://www.allaboutcircuits.com/textbook/alternating-current/chpt-11/true-reactive-and-apparent-power/

General: Voltmeter Impedance

Why is high input impedance desirable for a voltmeter? [G4B05]

A. It improves the frequency response
B. It decreases battery consumption in the meter
C. It improves the resolution of the readings
D. It decreases the loading on circuits being measured

Remember that impedance is the property of a circuit that prevents current from moving through a circuit.  When measuring a circuit, we do not want to alter the operation of the circuit.  In practice this is impossible, there will always be a certain amount of change when we measure something.

However, by using a voltmeter with a high input impedance, we prevent the current in the circuit from splitting off into the meter.  “Path of least resistance” and all that, or in our case, impedance.  If we were to attach a low-impedance meter to the circuit, it would “load” or pull power away from the circuit.

The answer then, is D. It decreases the loading on circuits being measured.

Technician: Correct Copy of Voice Traffic

What should be done to insure that voice message traffic containing proper names and unusual words are copied correctly by the receiving station? [T2C03]

A. The entire message should be repeated at least four times
B. Such messages must be limited to no more than 10 words
C. Such words and terms should be spelled out using a standard phonetic alphabet
D. All of these choices are correct

Let’s look at our possible answers and see if we can’t play a little elimination game.  A. The entire message should be repeated at least four times.  If you’re thinking that’s pretty inefficent and a general waste of time, you’re right.  B. Such messages must be limited to no more than 10 words.  That’s sort of the opposite end of the spectrum.  Limiting traffic to 10 word phrases is quite impractical.

So what do we do? If you find yourself needing to use words that may have unusual spelling, or names, or words that sound similar to other words that would confuse the meaning of your message, you need to fall back to spelling them out using a standard phonetic alphabet.

Now, there are several different types.  The “old school” one that you’ll hear used in, for example, old war movies.  The “police” one that is still used today by departments everywhere, and the one that is the nearly universally accepted standard in amateur radio, the NATO Phonetic alphabet.

That’s not to say you won’t hear different “soundings” of letters on the air, just realize that the correct way to do this is with the NATO alphabet.  Using “America” for “A” might sound nice, but could be confusing to somebody who doesn’t speak English and only uses the NATO alphabet.

The answer then, is C. Such words and terms should be spelled out using a standard phonetic alphabet.

Amateur Extra: Op Amp Output Impedance

What is the typical output impedance of an integrated circuit op-amp? [E7G15]

A. Very low
B. Very high
C. 100 ohms
D. 1000 ohms

A A. Very low output impedance is a property of an IC op-amp.   In an ideal case, it would be zero.  But of course that’s never possible.

In reality, the behavior of an op-amp depends on the circuit that its in, and it can be used in many different types of circuits (see image above.)

The reason for the low impedance, is so that the voltage gain that is output by the device acts on the rest of the circuit, and not the op amp itself.  If the voltage output had to contend with a high impedance across the output terminals, The power would go there, instead of actually being output.

More information: http://www.learningaboutelectronics.com/Articles/Why-does-an-op-amp-need-a-high-input-impedance-and-a-low-output-impedance

(image: https://commons.wikimedia.org/wiki/File:Op-amp_circuits.jpg)

General: Frequency Deviation

What is the frequency deviation for a 12.21 MHz reactance modulated oscillator in a 5 kHz deviation, 146.52MHz FM phone transmitter? [G8B07]

A. 101.75 Hz
B. 416.7 Hz
C. 5 kHz
D. 60 kHz

Whoa, what?  Guess what, dust off that calculator, because here’s where it starts to get mathy.  This question is also terribly worded.

What we’re looking for is the oscillator deviation.  The formula to calculate this is:

Oscillator Deviation = \frac{Transmitter Deviation}{(\frac{Output Frequency}{Oscillator Frequency})}

Simplified, it becomes this:

Oscillator Deviation = \frac{(Transmitter Deviation)(Oscillator Frequency)}{(Output Frequency)}

If we plug in our numbers, like so, don’t forget to use consistent units, so we must turn our 5kHz into .005MHz.

Oscillator Deviation = \frac{(.005)(12.21)}{(146.52)}

Solving…

Oscillator Deviation = .0004167 MHz, or 416.7 Hz, after we move the decimal six (mega!) places.

So our answer is B. 416.7 Hz

Now, mathematically you could think of this as a ratio between the oscillator and transmitter deviations and frequencies.  The equation for that looks like this, and winds up the same when you solve it….

\frac{Oscillator Deviation}{Oscillator Frequency} = \frac{Transmitter Deviation}{Transmitter Output Frequency}

Technician: Parallel Dipole

Which of the following describes a simple dipole mounted so the conductor is parallel to the Earth’s surface? [T9A03]

A. A ground wave antenna
B. A horizontally polarized antenna
C. A rhombic antenna
D. A vertically polarized antenna

Lets recall what we mean when we talk about an electromagnetic field being “polarized.”  The direction of polarization means the orientation of the electric field component of the wave.  We can also remember that the electric field component of a signal given off of an antenna such as a wire dipole is polarized in the direction of the wire.  A vertical antenna produces a vertically polarized signal, and a horizontal antenna produces a horizontally polarized signal.  If our dipole is parallel to the Earth’s surface, its horizontal.

Therefore, the answer is B. A horizontally polarized antenna.

It should be noted that the above logic is only a general approximation.  There are plenty of factors which can change a signal’s polarization, but as an estimation, it works.

Amateur Extra: Sine Wave Odd Harmonics

What type of wave is made up of a sine wave plus all of its odd harmonics? [E8A01]

A. A square wave
B. A sine wave
C. A cosine wave
D. A tangent wave

Well, the answer to this question is mathematically somewhat complex.  The answer is A. A square wave.

Rather than try and explain the formulation, which involves Fourier series, I think that this video I found on YouTube demonstrates the results perfectly.

So why would we want to do this?  Well, for “digital” anything, we need as close an approximation to a square wave as we can get, in order to represent those ones and zeros, or on and off states.  Since our oscillator circuits can’t on their own generate a square wave natively, only a sine wave, we need to take several different oscillator harmonics and combine them.  Just like in the above video.

If you’re really interested, here’s a great explanation, math and all: http://www.mathworks.com/help/matlab/examples/square-wave-from-sine-waves.html

General: Code in a PSK31 Signal

Which type of code is used for sending characters in a PSK31 signal? [G8C12]

A. Varicode
B. Viterbi
C. Volumetric
D. Binary

Alright, here we are again at a real head scratcher.  This is yet another example of something you just “need to know” for the test.  In real life you really wouldn’t use this information, although you’ll probably find yourself using PSK31 at some point.

PSK31 signals are encoded using A. Varicode.  It’s a fairly ingenious method where certain letters and characters use fewer bits than others, depending on their frequency of use.

For more information on Varicode, see the Wikipedia entry.

Technician: Mega vs. Giga

If a frequency readout shows a reading of 2425 MHz, what frequency is that in GHz? [T5B13]

A. 0.002425 GHZ
B. 24.25 GHz
C. 2.425 GHz
D. 2425 GHz

The best way to figure this out is to review our metric prefixes.  If you’re into computers at all, you already use “Mega” and “Giga” and probably even “Tera” more often than you think.

Break that down, and you get something like this:

Mega = million = 1,000,000
Giga = billion = 1,000,000,000
Tera = trillion = 1,000,000,000,000

you’ll notice that each one is 1,000 times larger than the one before. So…..

if we have 2425Mhz, or Megahertz we can take that number, divide it by 1,000, and we get our answer.

Wait, divide? I thought you said Giga was 1,000 times bigger?!  Correct!  If we want to convert Mega into Giga, we need to divide.  Think of it this way.  If we have 10 apples, how many hundreds of apples do we have?  .1! How did we get there? We divided by 10.  Same principle.  A single Gigahertz is 1,000 Megahertz.

Therefore, our answer is C. 2.425Ghz.

Amateur Extra: Satellite Mode Designators

What do the letters in a satellite’s mode designator specify? [E2A05]

A. Power limits for uplink and downlink transmissions
B. The location of the ground control station
C. The polarization of uplink and downlink signals
D. The uplink and downlink frequency ranges

Lets look at “mode designators” to understand what the question is about.  A mode designator is simply a letter that indicates what band to use when sending to (uplink) or listening to (downlink) a satellite.  For example, a U/U code means that both the uplink and downlink frequencies are in the 70cm band.  A V/U code means that the uplink is in the 2m VHF band, while the downlink is in the 70cm UHF band.

The answer then is D. The uplink and downlink frequency ranges.

Here is a list of mode designators:

Designator Band Frequency (General)
H 15 m 21 MHz
A 10 m 29 MHz
V 2 m 145 MHz
U 70 cm 435 MHz
L 23 cm 1.2 GHz
S 13 cm 2.4 GHz
S2 9 cm 3.4 GHz
C 5 cm 5 GHz
X 3 cm 10 GHz
K 1.2 cm 24 GHz
R 6 mm 47 GHz

Source: https://en.wikipedia.org/wiki/OSCAR#Mode_designators