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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 rate in it.

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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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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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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.

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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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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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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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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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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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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