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Subelement E8
Section E8C
Digital signals: digital communication modes; information rate vs bandwidth; error correction
How is Forward Error Correction implemented?
  • By the receiving station repeating each block of three data characters
  • By transmitting a special algorithm to the receiving station along with the data characters
  • Correct Answer
    By transmitting extra data that may be used to detect and correct transmission errors
  • By varying the frequency shift of the transmitted signal according to a predefined algorithm

In Forward Error Correction (FEC), an algorithm computes "parity" information from the user data that is "sent forward" to the receiver so it can verify and correct transmission errors without necessarily relying on retransmission. Note the answers: An algorithm is used to encode and decode the data, but the algorithm itself is not sent, as the receiver should already have it.

HINT: Both the question & and the correct answer only contains the word "error."

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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 duration of each bit in a message sent over the air
  • Correct Answer
    The rate at which the waveform of a transmitted signal 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 rate in it.

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When performing phase shift keying, why is it advantageous to shift phase precisely at the zero crossing of the RF carrier?
  • Correct Answer
    This results in the least possible transmitted bandwidth for the particular mode
  • It is easier to demodulate with a conventional, non-synchronous detector
  • It improves carrier suppression
  • All of these choices are correct

When the phase is changed precisely at 0, and the modulation scheme is BPSK, there is effectively 0 energy involved in the transition from one state to another when performed at a zero crossing. At any other time, the signal level would need to instantaneously jump to match the new phase, but at a zero crossing, the levels of two opposite phases are equal. When a sharp jump happens, this creates an edge that results in increased bandwidth.

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What technique is used to minimize the bandwidth requirements of a PSK31 signal?
  • Zero-sum character encoding
  • Reed-Solomon character encoding
  • Correct Answer
    Use of sinusoidal data pulses
  • Use of trapezoidal 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 necessary bandwidth of a 13-WPM international Morse code transmission?
  • Approximately 13 Hz
  • Approximately 26 Hz
  • Correct Answer
    Approximately 52 Hz
  • Approximately 104 Hz

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

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." -W9MPX

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 necessary bandwidth of a 170-hertz shift, 300-baud ASCII transmission?
  • 0.1 Hz
  • 0.3 kHz
  • Correct Answer
    0.5 kHz
  • 1.0 kHz

The necessary bandwidth of a 170-hertz shift, 300-baud ASCII transmission is 0.5 kHz.

The ARRL Extra Class License manual states: bandwidth (BW) is:

\[\text{BW}_{(\text{Hertz})} = (K \times \text{shift}) + B\]


  • \(K = 1.2\) (an estimated empirical factor)
  • \(B\) = baud, or symbol rate.

Therefore, \begin{align} \text{BW} &= (1.2 \times 170\text{ Hz}) + 300\text{ baud}\\ &= 504\text{ Hz}\\ &\approx 0.5\text{ kHz} \end{align}

Hint1: It's the only answer that has a number not in the question, "5".

Hint2: 'shift' to the number not in the question :)

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What is the necessary bandwidth of a 4800-Hz frequency shift, 9600-baud ASCII FM transmission?
  • Correct Answer
    15.36 kHz
  • 9.6 kHz
  • 4.8 kHz
  • 5.76 kHz

Frequency Shift = 4800 Hz
Transmission Rate = 9600 baud

What is the necessary bandwidth (BW)?

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.

This is also how TCP works in TCP/IP on the Internet. (TCP is used for your connection to this web site.)

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

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Which is the name of a digital code where each preceding or following character changes by only one bit?
  • Binary Coded Decimal Code
  • Extended Binary Coded Decimal Interchange Code
  • Excess 3 code
  • Correct Answer
    Gray code

For the values


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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What is an advantage of Gray code in digital communications where symbols are transmitted as multiple bits
  • It increases security
  • It has more possible states than simple binary
  • It has more resolution than simple binary
  • Correct Answer
    It facilitates error detection

If a symbol contains two bits, and normal binary values are used, such that for example 100Hz represents 00, 150Hz represents 01, 200Hz represents 10, and 250Hz represents 11... misdetecting 200Hz as the adjacent value 150Hz because of noise causes both output bits to flip (from 10 to 01). This makes things relatively hard on an error correction/error detection system.

If a gray code is used, where only one bit changes for adjacent values-- for example, 100Hz represents 00, 150Hz represents 01, 200Hz represents 11, and 250Hz represents 10-- then a similar error only flips one bit.

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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
  • Symbol rate is only used for packet-based modes
  • Baud is only used for RTTY

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.

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