There are disagreements as to the use of the term "digital" in this context, but don't be thrown off by thinking that the only time you can use "digital" is referring to computers.

Essentially the argument is that "digital" indicates two basic states; on and off. Whether you agree with this definition or not, the only one here that could be arguably digital but still has unequal lengths (longer or shorter tone for the "dit" or "dah") is morse code.

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Baudot uses five data bits per character, ASCII uses seven or eight; Baudot uses two characters as shift codes, ASCII has no shift code.

More information can be found below:

ASCII

Baudot

Hint: Baudot is a five-bit code. Only one answer has this.

One Word Key "5".

\(2^5 = 32\) is the smallest size to represent 26 letters.

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ASCII, or "American Standard Code for Information Interchange" uses numbers in place of all characters, both uppercase and lowercase.

65-90 are A-Z, 97-122 are a-z

ASCII is an 8-bit code providing a potential for 256 possible characters. As such, it does not require a shift code.

In contrast, BAUDOT code is a 5-bit code only allowing 32 possible combinations. BAUDOT therefore can only support upper case letters.

To see more info visit: http://en.wikipedia.org/wiki/ASCII http://en.wikipedia.org/wiki/Baudot_code

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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.

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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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AS PER wikipedia, spread-spectrum techniques are methods by which a signal (e.g. an electrical, electromagnetic, or acoustic signal) generated with a particular bandwidth is deliberately spread in the frequency domain, resulting in a signal with a wider bandwidth.

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In some respects, the best way to know the answer to this is to understand why each of the wrong answers can be eliminated.

In the case of any frequency hopping, there isn't a single RF carrier.

In binary phase-shift keying, there are only two (binary) phases used at all times, and the operation can only flip between one and the other as needed, so it can't support a high-speed bit stream.

In phase compandored spread spectrum, the process of phase companding takes place. That's a compression process, where the amount of phase variation of the original signal is compressed into a smaller phase range at the transmitter, then restored at the receiver. Naturally that means a lower bit rate is used to send a higher bit rate signal, but compression is not possible for all source signals without deliberately discarding data, so either phase compandored spread spectrum is lossy to achieve a given bit rate, or the bit rate must vary depending on the source data -- which doesn't directly eliminate phase compandored spread spectrum, but it does make it less probable than direct sequence spread spectrum.

In direct sequence spread spectrum, there is a single carrier, and more than two phase positions are used so that multiple bits can be represented at one time, and no compression is used, allowing for a high-speed bit stream at the full, raw rate required by the source signal.

Memory Trick: High-Speed to get there Directly

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A true parity bit is a Single Error Detecting Code. It can detect when one bit has been changed, but conveys nothing about which bit. Parity is also a very simple error detecting code that provides rather poor characteristics at dealing with double errors and beyond.

Simple parity rules have to be agreed-upon by the sender and receiver. Typically, parity is either ignored or set to either an "ODD" or "EVEN" state. The state of the parity chosen for the communication tells the sender the format for how the parity bit will be computed and appended.

If ODD parity is selected, the transmitter will append a parity bit set so that the quantity of "Ones" in the transmission is an odd number.

If EVEN parity is selected, the transmitter will append a parity bit set so that the quantity of "Ones" in the transmission is an even number.

The receiver then independently computes parity for the received value and compares it to what it expects.

Example: ODD PARITY

7-byte value to be transmitted: 100100

This contains an EVEN number of 1's (there are two).

Sender appends a "1" as the parity bit, forcing the count to three (ODD number).

Transmitted value becomes: 1001001 (the right-most bit is the parity bit).

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JT65 [...] is intended for extremely weak but slowly varying signals, such as those found on troposcatter or Earth-Moon-Earth (EME, or "moonbounce") paths. It can detect signals several dB underneath the noise floor [...] –Wikipedia

Note that the 65 doesn't have anything to do with the bandwidth, it's because the encoded messages are transmitted using MFSK with 65 tones.

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