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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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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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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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.

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:

• https://en.wikipedia.org/wiki/PSK31
• http://aintel.bi.ehu.es/psk31theory.html

_{Image CC BY-SA 3.0 Albany45}

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Given:
CW Words Per Minute (WPM_CW) = 13

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

For a CW transmission, remember:
Bandwidth (BW_CW) ≈ 4 Hz * WPM_CW

So in this case:
BW_CW ≈ 4 Hz * 13 WPM
BW_CW ≈ 52 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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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}

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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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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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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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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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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Baud is another name for symbol rate. If you chose “it depends on the code” you were probably thinking of “bit rate” since that does differ from symbol rate/baud.

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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