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Subelement E8

SIGNALS AND EMISSIONS

Section E8C

Digital signals: digital communication modes; information rate vs. bandwidth; error correction; constellation diagrams

What is Quadrature Amplitude Modulation or QAM?

  • A technique for digital data compression used in digital television which removes redundancy in the data by comparing bit amplitudes
  • Correct Answer
    Transmission of data by modulating the amplitude of two carriers of the same frequency but 90 degrees out of phase
  • A method of performing single sideband modulation by shifting the phase of the carrier and modulation components of the signal
  • A technique for analog modulation of television video signals using phase modulation and compression

"Quadrature" = 1/4, and a quarter of a circle is 90 degrees.

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What is the definition of symbol rate in a digital transmission?

  • The number of control characters in a message packet
  • The maximum rate at which the forward error correction code can make corrections
  • Correct Answer
    The rate at which the waveform changes to convey information
  • The number of characters carried per second by the station-to-station link

In digital communications, symbol rate, also known as baud or modulation rate, is the number of symbol changes, waveform changes, or signaling events, across the transmission medium per time unit using a digitally modulated signal or a line code. The symbol rate is measured in baud (Bd) or symbols per second.

SEE: https://en.wikipedia.org/wiki/Symbol_rate

Note that baud is a rate, so 'baud rate' is redundant, but it is colloquially used.

Hint: The only and correct answer has the word waveform in it.

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Why should the phase of a PSK signal be changed at the zero crossing of the RF signal?

  • Correct Answer
    To minimize bandwidth
  • To simplify modulation
  • To improve carrier suppression
  • All these choices are correct

Triggering a signal at a non-zero crossing is an "instantaneous" shift, locally similar to a square wave. From Fourier analysis we know that square waves require an infinite sum of frequencies... Long story short, larger jumps require more bandwidth.

Conversely, take the opposite view. If we want to use the minimum bandwidth, how do we achieve this? With a single sine wave. How does a sine wave begin? At the zero crossing.

Memory hint: Zero = minimize

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What technique minimizes the bandwidth of a PSK31 signal?

  • Zero-sum character encoding
  • Reed-Solomon character encoding
  • Correct Answer
    Use of sinusoidal data pulses
  • Use of linear data pulses

The PSK31 bandwidth is minimized by the special sinusoidal shaping of the transmitted data symbols in the form of pulses.

As seen in the image below, PSK pulses are a fixed length and may contain a phase reversal.

BPSK31 modulation

If these reversals were instantaneous, high-frequency square-wave-type components would appear in the signal, broadening the sidebands.

However, because the cosine shaping function is set so that its half period exactly matches the pulse length, the phase transitions are as slow (low-frequency) as possible. This keeps the sidebands as close to the carrier as possible, allowing the signal to occupy the narrowest possible bandwidth.

For more, see:

Image CC BY-SA 3.0 Albany45

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What is the approximate bandwidth of a 13-WPM International Morse Code transmission?

  • 13 Hz
  • 26 Hz
  • Correct Answer
    52 Hz
  • 104 Hz

Given:
CW Words Per Minute (WPMCW) = 13

What is the necessary bandwidth (BW) for this transmission?

For a CW transmission, remember:
Bandwidth (BWCW) ≈ 4 Hz * WPMCW

So in this case:
BWCW ≈ 4 Hz * 13 WPM
BWCW52 Hz

Test tip: To remind yourself that you must multiply by FOUR to calculate bandwidth in HERTZ from WORDS Per Minute of CW...just think,
"Four letter words hertz."

It may also help to remember that in a deck of cards there are 13 cards in each suit and 4 suits for 52 total cards.

Yet another way to remember it is that morse code is just 2 symbols, dot and dash, and a good rule of thumb from the General exam is to stay at least an extra bandwidth away from the edge of the band. Two doubled is 4!

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What is the bandwidth of an FT8 signal?

  • 10 Hz
  • Correct Answer
    50 Hz
  • 600 Hz
  • 2.4 kHz

FT8 ("Franke-Taylor design, 8-FSK modulation") is an extremely-weak-signal, digital, narrow bandwidth (50 Hz), QSO-only communication protocol used by amateur radio operators.

Sig ID Wiki - FT8

Memory Aid: If you pronounce “F T,” it sounds like “fifty,” which is the correct answer.

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What is the bandwidth of a 4,800-Hz frequency shift, 9,600-baud ASCII FM transmission?

  • Correct Answer
    15.36 kHz
  • 9.6 kHz
  • 4.8 kHz
  • 5.76 kHz

Given:
Frequency Shift = 4800 Hz
Transmission Rate = 9600 baud

What is the necessary bandwidth (BW)?

Remember:
Keying (\(K\)) should be 1.2 for most amateur radio purposes

\begin{align} \text{BW} &= (K \cdot \text{shift}) + \text{baud rate}\\ &=( 1.2 \cdot 4800\text{ Hz} ) + 9600\\ &= 15,360\text{ Hz}\\ &= 15.36\text{ kHz} \end{align}

** Test Tip - '4800' and '9600' consist of 4 digits. The answer is the only choice containing 4 digits '15.36'.

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How does ARQ accomplish error correction?

  • Special binary codes provide automatic correction
  • Special polynomial codes provide automatic correction
  • If errors are detected, redundant data is substituted
  • Correct Answer
    If errors are detected, a retransmission is requested

In automatic repeat-request (ARQ) systems the transmitter sends the data and also an error checking code. The receiver checks for errors and request retransmission of erroneous data.

The key thing to understand is just the name – Automatic Repeat Request, sometimes called Automatic Repeat Query. It automatically requests that the packet be retransmitted when it detects that the previously received packet was corrupted.

Hint: When you see ARQ, think "reQuesT".

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Which digital code allows only one bit to change between sequential code values?

  • Binary Coded Decimal Code
  • Extended Binary Coded Decimal Interchange Code
  • Extended ASCII
  • Correct Answer
    Gray code

For the values

0,1,2,3,4,5,6,7

Normal binary encoding looks like this:

000, 001, 010, 011, 100, 101, 110, 111

In some cases, all 3 values change between adjacent values. For example, from 3 to 4 the sequence goes from 011 to 100.

An example gray code is:

000, 001, 011, 010, 110, 111, 101, 100

In this case, there is still a unique encoding for each possible value, but only one bit changes between adjacent values. Gray codes are useful components in implementing hardware and in error correcting codes. This is because some technologies, like magnetic disks, can't reliably detect more than one change every so many bits.

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How can data rate be increased without increasing bandwidth?

  • It is impossible
  • Increasing analog-to-digital conversion resolution
  • Correct Answer
    Using a more efficient digital code
  • Using forward error correction

The best answer here given the restrictions (symbol rate must be increased, not bitrate, and bandwidth must not be increased) is using a more efficient digital code. A more efficient digital code would typically make the symbols shorter in time so that more can be sent in a given period of time without increasing the bandwidth.

"It is impossible" may become true at some point for bitrate once Shanon information theory limits for power and bandwidth have been exceeded, but generally speaking it is not "just impossible" to increase the symbol rate especially when the symbols are inefficient.

You might be tempted to say "Increasing analog-to-digital conversion resolution" but this would be wrong because increase ADC resolution would only help you add more symbols, not transmit and decode symbols faster or shorten the symbols. In other words, this answer might be true if we were talking about bitrate rather than symbol rate, but for this question we must increase the symbol rate not just the bitrate, and we are not allowed to use more bandwidth.

"Using forward error correction" is a particularly bad answer since this just introduces additional overhead, has no necessary effect on the symbol rate, and only lowers the net bitrate due to the transmission of redundant information.

Just remember that symbol rate is about efficiency to help you remember this answer.

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What is the relationship between symbol rate and baud?

  • Correct Answer
    They are the same
  • Baud is twice the symbol rate
  • Baud rate is half the symbol rate
  • The relationship depends on the specific code used

Baud is another name for symbol rate.

Symbol rate is the number of different transmission units sent per second over a link. If each symbol can contain two different values, it is equivalent to bits per second.

It is possible for a symbol to contain more different values. For example, if it contains four different possible values, each symbol contains 2 bits of information, and the number of bits per second is double the baud rate.

Silly hint: people with great bauds are often symbols of health and fitness that others admire. Therefore, baud and symbol [rate] are synonymous

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What factors affect the bandwidth of a transmitted CW signal?

  • IF bandwidth and Q
  • Modulation index and output power
  • Correct Answer
    Keying speed and shape factor (rise and fall time)
  • All these choices are correct

The bandwidth of a signal is directly related to the amount of information per second being transmitted. Obviously, if you increase the speed of the CW signal, the bandwidth will increase. Not as obvious is that if you make the rise and fall time of the CW elements faster, you increase the potential for harmonics (key clicks!) as well as increase the bandwidth.

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What is described by the constellation diagram of a QAM or QPSK signal?

  • How many carriers may be present at the same time
  • Correct Answer
    The possible phase and amplitude states for each symbol
  • Frequency response of the signal stream
  • The number of bits used for error correction in the protocol

A constellation diagram is a 2D plot where each point on the plot represents a unique combination of phase and amplitude of a signal.

The phase is measured as the angle counterclockwise from the horizontal axis to the point, while the amplitude is measured as the distance of a point from the origin.

Since digital modulation schemes most commonly distinguish symbols, or transmitted pieces of data, with unique combinations of amplitude and phase, constellation diagrams are a useful way to visualize how information is conveyed in a given modulation scheme.


Hint: The two example modulation schemes used in the question are QAM: Quadrature Amplitude Modulation and QPSK: Quadrature Phase Shift Keying, which together contain both keywords in the answer.

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What type of addresses do nodes have in a mesh network?

  • Email
  • Trust server
  • Correct Answer
    Internet Protocol (IP)
  • Talk group

A mesh network is made up of nodes which route traffic between each other in the same way that the backbone of the Internet works; Node A can talk to Node F as long as there are connected nodes between them that can route the traffic. They use the same protocol as the Internet itself -- IP, or Internet Protocol. That means they require IP Addresses.

Of the other options, the only one that is even an address type is Email, and while you could plausibly do something with email it wouldn't be inherent to the Mesh Network.

Note for the especially pedantic: yes, there are many kinds of mesh networking that do not use IP addressing. For this question, you'll just have to accept that there's a common colloquial meaning of "mesh network" specifically for this test.

See more here: Mesh Network

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What technique do individual nodes use to form a mesh network?

  • Forward error correction and Viterbi codes
  • Acting as store-and-forward digipeaters
  • Correct Answer
    Discovery and link establishment protocols
  • Custom code plugs for the local trunking systems

Mesh networks are usually either zero configuration or very little configuration; therefore the nodes (routers or hosts) need to discover each other and establish links between each other, since they have not been configured with this information ahead of time.

Network Discovery typically works by nodes either periodically broadcasting a message such as, "I'm here and my name is X" or "I'm here, my name is X, is anyone else there?" This allows nodes to learn what other nodes exist and how to send them messages. (Network discovery is also the means by which your computer learns of nearby Wi-Fi routers.)

But once nodes know of each other's existence, they need to establish links, which can be thought of like imaginary virtual cables over radio. This typically involves a message exchange between nodes, where capabilities (such as protocols and speeds) are negotiated between nodes, such that the best common capabilities are used and incompatible capabilities are rejected. (When you "join a Wi-Fi network" with your computer, you are establishing a link to one or more access points on that Wi-Fi network.)

The other answers don't even make sense for forming a mesh network. The only reasonable answer here is Discovery and link establishment.

Silly hint: Mesh is like linking

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