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Subelement E8
SIGNALS AND EMISSIONS
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
• 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 a Forward Error Correction system, each character is sent with additional data. The receiver uses the additional data to verify the sent data, and potentially correct transmitted errors. The error correction is sent forward with the data, rather than depending 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 this.

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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
• 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 rate and 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.

akguido

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

-KE0IPR

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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?
• 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
• Use of sinusoidal data pulses
• Use of trapezoidal data pulses

Some PSK stuff:

PSK31’s name is derived from the modulation type – phase-shift keying – and data rate, which is actually 31.25 bauds.

PSK uses the 128-character code and the full 256 ANSI (American National Standards Institute).

PSK's emission designator for PSK31 is J2B

PSK63 is a faster variation.

PSK uses a variable-length code called Varicode.

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

-KE0IPR

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What is the necessary bandwidth of a 13-WPM international Morse code transmission?
• Approximately 13 Hz
• Approximately 26 Hz
• Approximately 52 Hz
• Approximately 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 * WPMCW

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

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What is the necessary bandwidth of a 170-hertz shift, 300-baud ASCII transmission?
• 0.1 Hz
• 0.3 kHz
• 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:

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

where:

• $K = 1.2$ (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}

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What is the necessary bandwidth of a 4800-Hz frequency shift, 9600-baud ASCII FM transmission?
• 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}

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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
• 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.

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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
• 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.

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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
• 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?
• 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 buad rate.

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