# Tag Info

14

Short answer: Math says max link rate is 2Mb/s if you knew the perfect channel coding. Which is still an unsolved puzzle. Long answer: You're calculating a link rate. That is fine, and can be answered using Shannon's Channel Capacity, which gives us the upper limit for bits per second that we can get across a given channel: $$C= b\cdot \log_2\left(1+ \... 14 The ITU bands are actually delineated along plain powers of ten! They're just hiding a bit. From the description above a table of all the bands on Wikipedia: As a matter of convention, the ITU divides the radio spectrum into 12 bands, each beginning at a wavelength which is a power of ten (10^n) metres… So the HF band is from 100–10m, or the UHF band ... 12 It depends on what you mean by "stronger". Ignoring actual limitations of your hardware and just considering the theory of communications, if you have 25 W of transmit power, then you can spread that over as much bandwidth as you want, and you're always transmitting 25 W of power. The quantity which is 25 times higher in the 100 Hz case than the 2500 Hz ... 11 That's a good question---and one that is heavily under debate currently. The FCC originally limited symbol rates as a way of limiting bandwidth for data modes indirectly (it made sense at the time). But now that there are more advanced modulations (like the various forms of phase shift keying) that can exceed the symbol rate limitations in less bandwidth ... 11 The Current FCC Regulations (as of July 2018) Parts 97.305(c) and 97.307(f) regulate digital modes primarily by symbol rate. The table in 97.305(c) maps bands/frequencies to specific symbol rate limitations in 97.307(f) as follows: (2) No non-phone emission shall exceed the bandwidth of a communications quality phone emission of the same modulation type. ... 10 The usual figure used in this context is, apparently, fractional bandwidth, defined as the bandwidth divided by the center frequency, and therefore having a range of 0 to 2.$$ \text{fractional bandwidth} \,=\, \frac{f_{\text{max}} - f_{\text{min}}}{f_{\text{center}}} \,=\, 2\frac{f_{\text{max}} - f_{\text{min}}}{f_{\text{max}} + f_{\text{min}}} $$Given ... 9 The increased bandwidth of a folded dipole is almost entirely due to the extra thickness. Two parallel elements behave as a thicker single element. There is a small contribution too from the combination of the reactances of the transmission-line mode and radiator-mode acting in opposite directions. 9 Summary: In all cases, no more bandwidth than necessary. §97.307(a), (b) In data portions of LF, MF, and HF bands, 500 Hz. §97.3(c)(2) In the phone portions of LF, MF, and HF bands, no more than "the bandwidth of a communications quality phone emission of the same modulation type." 4 kHz is a good number to put on it, being a typical limit of an ... 9 The signal is not at exactly one frequency. The only signal that exists exactly at one frequency is an unmodulated carrier, and such a signal contains no information. As soon as the carrier is modulated, the signal's energy is spread over a wider bandwidth. So to receive any signal containing information, the receiver must listen to some range of ... 8 Loaded Q-factor Like any resonant circuit, the bandwidth of an antenna is determined by its loaded quality factor, defined by Q_{\ell}\overset{.}{=}\frac{X}{R}. The lower the loaded Q-factor, the broader the antenna's bandwidth will be: BW_{-3dB}=\frac{f_{res}}{Q_{\ell}}, with f_{res} the resonant frequency. Analysis of the loaded Q_{\ell} of a ... 8 There's more space at higher frequencies so the allocations can be bigger. Consider the difference between 15 MHz and 30 MHz, and the difference between 5,015 MHz and 5,030 MHz. It's a 15 MHz change either way, but in the first case involves a halving of wavelength, whereas in the later case the wavelength changes only by a factor of 0.997. Since the way ... 7 Someone was wrong. The 300 GHz band is the biggest, because it has no upper bound. You can fit an unlimited number of channels in it. See Which band will I be authorized the most bandwidth? There is no theoretical requirement that signals be more "spread out" at higher frequencies. However, as VHF and up tend to be larger bands, the band plans tend to ... 7 Summary: Theoretical maximum in the neighborhood of 10s of megabits per second. Less than that in practice, perhaps a lot less depending on budget. Let's start with the Friis transmission equation:$$ P_{r(\mathrm{dB})} = P_{t(\mathrm{dB})} + G_{t(\mathrm{dB})} + G_{r(\mathrm{dB})} + 147.6 - 20 \log_{10} (rf)  Insert the values you've given for ...

7

Bandwidth: Broadcast FM stations are spaced 200 KHz apart, with the intention of allowing this kind of signal structure without interference: The FM broadcast bandwidth of just more than 50 KHz is quite sufficient for near-CD quality stereo. Normal broadcast FM uses multiplexing so the main channel includes L+R - both channels - what we call Mono. Then ...

7

There are some straightforward reasons I can think of which apply to all spectrum allocations, not just amateur radio: In a lot of cases, the bandwidth of a RF device is relative; if you are using the same physical principles, but change the dimensions/values to double the operating frequencies, everything scales up equally so you have a device with twice ...

7

Before getting to the questions you asked, an observation: the design doesn't appear to include a balun. Adding one is probably a good idea. Is there a way to flatten the SWR in this range [80m] or is this graph typical? It's typical. Because 80 meters is such a low frequency, the fractional bandwidth of the entire band is quite large. For this reason ...

6

A little capacitive reactance is what gives you the greater bandwidth. In a regular ½λ dipole, the current that flows along the conductors are in phase. When we add the second conductor in a folded dipole, what we are really doing is extending the dipole. As a result the current in the new section flows in the same direction as those in the original dipole....

6

You raise an excellent question and your thought processes are indeed on the right track. First some background. An ideal, uninterrupted sinusoidal carrier has zero bandwidth. Real world factors such as phase noise, amplifier distortion, etc. produce a measurable bandwidth of the carrier. When the carrier is keyed on and off as it is with Morse code, this ...

6

It may be easier to think of the receiver as receiving all frequencies. The job of the receiver is not to tune to a single frequency. Its job is to filter out everything that's not at that frequency, or in a band around it. So when you tune your FM receiver to 144.2500 with a 25 kHz bandwidth, you're telling it to reject all frequencies that are more than ...

6

It means the range of frequencies in which the antenna is designed to operate. Operated outside that range, the antenna may not meet its specifications. This particular antenna is a telescopic whip, a kind of monopole antenna. Such antennas are typically 1/4 wavelength long. It is advertised for 300 to 1100 MHz, and it's length can be adjusted from 24.5 cm ...

6

The elements of a Yagi are dipoles. If you look at a dipole at any instant, the voltage at one tip will be some value, and the voltage at the other tip has the same magnitude, but opposite polarity. So let's say at some instant, the left side is at +50V and the right side is at -50V. Consider these relative to the soil at the base of the mast, if you like. ...

6

Binary PSK with instantaneous phase shifts would be equivalent to multiplying a sine wave (the carrier) with a square wave with values at 1 or -1. When two signals are multiplied, this forms a frequency mixer. A mixer with inputs at frequencies $f_1$ and $f_2$ creates outputs at $f_1 + f_2$ and $|f_1 - f_2|$. A sine wave is just one frequency, let's call ...

5

Easy enough to graph at wolframalpha.com: This shows two signals: one at some frequency (blue), and another at 15 times that frequency (magenta). The brown line is the two added together. If you took this brown signal and put it through a low-pass filter, you could recover the blue signal. If you put it through a high-pass filter, you could recover the ...

5

The receiver "listens" to any and all signals in its receive bandwidth (talking about AM, CW, or sideband here, FM handles everything differently). Generally, there will be only one strong signal that gets through the receiver's filter, and that (plus atmospheric or man-made noise) is all you'll hear. If there are two stations within the filter width, ...

5

The data port does not carry FM signals: it carries demodulated audio. The radio performs FM demodulation and de-emphasis, and the output is heard on the speaker and is also present on the 1200 baud data output. On most multi-mode radios this port caries the demodulated audio for other mode selections as well. The data input is similar, but in reverse. The ...

5

Marcus and Richard have answered the question that you asked, how to triple the bandwidth of a dipole, with the correct answer that it can't be done. I suggest that you edit your question to give us more information about the problem that you're trying to solve and its constraints, and then we can offer more targeted suggestions. In the meantime, here are ...

4

Without a doubt, the 300GHz band. All frequencies above 300GHz are authorized for amateur use. There is no bigger band, amateur or otherwise. Here's all the national band plan has to say about it: All modes and licensees (except Novices) are authorized Amateur Bands above 10.5 GHz. Also, it has the coolest band designation: Tremendously high frequency. ...

4

It may help to consider the two arms of the folded dipole as end-shorted (ROUGHLY quarter-wave) sections of 300 ohm transmission line. Any current which attempts to flow in opposite directions on the two conductors in each arm will not radiate (which is why the transmission line works - the two currents generate opposing fields). Only a "common-mode" ...

4

Assuming the antenna is in free space, you only need to know the length and diameter of the wire used to construct the dipole. The math is hairy but I wrote a program to do the calculations. Here is the SWR (assuming a 50 ohm source) and feedpoint impedance for a dipole 10 meters long, with a diameter of 2.053mm: 14.100 MHz: SWR 2.07, 61.3 + j-39.6 14.130 ...

4

There is no way to transmit information in a signal with zero bandwidth. Switching the carrier at the zero crossings would reduce bandwidth but not take it to zero. There's a mathematical explanation on DSP StackExchange: Does “keying on” a sine wave at a zero-crossing reduce its bandwidth? Summary: in the worst case of switching on/off at the peak ...

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