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36

Considering just electrical properties, the most significant parameter for your selection of antenna conductor is resistance. You want to keep the resistance to a minimum, because when current flows through a resistance, the electrical power is converted to heat, according to Joule heating: $$ P = I^2 R $$ Any energy you use to make heat is energy you aren'...


18

The $492/f$ formula is for an ideal antenna in free space, the $468/f$ is an estimate for real antennas at a reasonable height over ground. The $492/f$ formula is a conversion from metric units to English units for the fundamental frequency and wavelength ($\lambda$) formula. $c = 3\times 10^8_{m/s}$ (the velocity of light) and $f =$ frequency -- \begin{...


14

Radio waves don't stop at a distance, they just get weaker; you've read this correctly. The reason that communications stop working at some distance is that the signals are too weak to be understood. Besides distance (and being absorbed or reflected by objects in the path) causing the signal to be weak in an absolute sense (how much power there is), there ...


9

This is called the "FM capture effect" and is mostly a characteristic of the demodulator design. Demodulators with a strong capture effect detect the zero crossings of the signal. The zero crossings are only slightly affected by mixing two signals until they are nearly equal amplitude. The steepness of the waveform at zero crossing means that the stronger ...


9

I understand why you think it should mix - that's what happens with SSB, for example. The difference with FM is that in FM, at any instant, each station's signal is only at one frequency. If you slowed the signals down and watched through a spectrum analyzer, you could watch the two signals, and you'd see two peaks at constant amplitude, moving independently....


8

Electrical wave propagation in wire is about 95% to 97% the speed of light. Since wavelength is most commonly used for building antennas, which involve conducting the wave from air into the wire and vice versa, the calculation is adjusted assuming the slower propagation in an unshielded conductor. However, this 3% to 5% discrepancy is small enough at ...


5

The field strength in volts/meter as a function of effective radiated power and distance is given by: $$E=\frac{7.01\sqrt{P_{ERP}}}{d} \tag 1$$ where d is the distance from the radiating antenna in meters and PERP is the effective radiated power in watts. So you can see your sample problem was chosen such that the 7 meter distance cancels out the upper 7....


5

If the birthrate in Canada is 400 thousand babies per year, and the birthrate is Russia is 1680 thousand babies per year, is there a population difference in the nations? Just as birthrate is a rate of stuff (births), power is a rate of energy. A watt is one joule per second, by definition. The frequency is irrelevant. To answer your question we need to ...


5

For someone who knows how to convert between inches, feet, and meters, it's really quite simple. You only really need to know one formula to do it all, and that formula is $300=f\times wavelength$. If you find the wavelength for the given frequency, then just find the type of antenna (quarter-wave), take the appropriate fraction of the wavelength, and ...


4

Legend $c$ = velocity of propogation = speed of light (299,792,458 meters/second) $f$ = frequency $\lambda $ = wavelength Formulas The basic formula for calculating wavelength is: \begin{equation} \lambda = \frac{c}{f} \end{equation} To make the math simpler, frequency ($f$) is expressed in megahertz (MHz) and the velocity of propogation in free space ($...


4

Telephones are designed as a system. So, yes, the plastic case, including a potential coating, have been incorporated in tests, and in consequent optimization (at least for large-scale mid- to high-end phones). Chances are you will not be improving anything. The metalized surface will be thinner than the skin depth at relevant frequencies, and quite ...


4

This is a magnetic flux transformer, not a transmission line transformer. It works by flux coupling just like a mains power transformer. So the bifilar part isn't too special, nothing like a transmission line transformer. In a regular flux transformer, the position of the winding doesn't really matter, only the number of turns. At RF though there is the ...


4

Skywave propagation is closely related to total internal reflection. This is easily demonstrated with visible light (which is also electromagnetic radiation) through a block of acrylic: Total internal reflection occurs when the angle of incidence is above a critical angle, which is a function of the refractive indexes of the two materials: $$ \theta_c = \...


4

Higher frequencies of electromagnetic radiation have higher energy per photon; “the wave” you transmit is made up of huge numbers of (coherent) photons — for example, at 5 GHz, 100 watts corresponds to $10^{25}$ photons per second. Therefore, the individual photons' energy is not all that relevant. For the same power (energy per time), count and frequency ...


3

The methodology needed to accurately calculate the radiated r-f power needed to generate a given field intensity at some distance along a terrestrial path is highly complex — probably beyond the practical possibility of posting in an answer here, even in outline form. However, the U.S. FCC has developed a set of propagation charts relating to this topic, ...


3

We have some clues from a historical perspective. In the January 1917 edition of The Radio Experimentor, author H. Winfield Secor noted that approximately 50 kW (kilo-watts) of transmitter power was required to reach a 3,000 to 4,000 mile range with CW. He then went on to note that 10 μA (micro-amperes) is considered a weak receive signal while 20 μA is a ...


3

Light waves and RF waves are both electromagnetic radiation. The only difference is that we can see the narrow band of frequencies that we call light. There is a common misconception that laser irradiance does not follow the inverse square law that describes the irradiance of an antenna. But in fact, they are exactly the same. The only difference is that ...


3

The curvature of the Earth also makes a difference between how a signal will propagate near the ground versus in outer space. And at a large enough distance, the random quantum behavior of the atoms and electrons in your receiver radio will drown out their reaction to any incoming radio waves.


3

There are two aspect you'll need to take into account. The first can be more easily dismissed as insignificant, and that is the velocity factor of the metal. For common conducting metals the difference between two metals is very small and is generally swamped by other things that affect velocity factor, such as insulation. The second, and more important ...


3

Thanks for the posting everyone! For licence test I took this approach after reading everyone's post & searching though all the technical details you posted. Convert Frequency to wavelength in meters. 300 ÷ frequency. Multiplying 39 × wavelength in meters = inches of full Antenna length. Divide inches by 2 for 1/2 wave. Divide 1/2 wave by 2 for 1/4 ...


2

That's a property of FM demodulation known as the Capture Effect: when multiple signals are present, only the strongest is demodulated. This does not occur with AM or SSB demodulation, where multiple signals can be simultaneously demodulated.


2

If you are interested in space communications (EME and satellites) even small losses are important since losses have room temperature. The sky as well as a good preamp should be well below 100K while losses are about 3 times warmer which makes them 3 times more destructive. When one designs a yagi one can trade between gain/pattern and element losses. The ...


2

Given the shape and desired band, I think what you want is a small loop antenna. This would simply be a coil of wire wound in the shape of your cylinder, preferably with a capacitor wired in parallel to make it resonant at the desired frequency and match impedance to the receiver. Small loops have the type of radiation pattern you are looking for (in the ...


2

I once removed the cover of an HTC One X for cleaning purposes. As curious as I am, I switched it on and noticed really poor reception. Upon further investigation, I concluded that some metal parts remaining on the cover are actually the antenna. Googling your phone, I found https://wiredornot.files.wordpress.com/2013/08/img_20130807_132347.jpg which shows ...


1

How do you calculate it? In the general case, you use a numerical EM field solver. Without a specification of the antenna geometry and exactly which point 7 meters away is under consideration, this is not an answerable question. Even with such specifications the answer is very complex to calculate, and not the sort of thing one would casually do with a ...


1

First of all, noisier or quieter, if you haven't made any contacts, it's probably time to change bands, regardless of anything else. As to the meat of your question, what does a quieter actually mean? Basically, it means one or more of the sources of RF noise has quieted down. What are these sources of noise? Well, here's a few: Aurora Borealis (Northern ...


1

Need to get a few things out of the way to understand why some things are. Velocity Factor - speed of a electromagnetic (EM) wave through ANY material. Frequency - Speed in Hz of a (EM) wave. 20Hz to 20kHz is Audio, 100kHz and above is considered Radio until nanometers then it's visible then on to other types. We are referring to RF. Antenna - item for ...


1

I always use the 468 formula and have for the past 30 years and ALWAYS had a perfect SWR and matched antenna. I have made over 200 dipoles and I have NEVER had to trim to get a perfect antenna. I also have NEVER made an antenna too short or too long!


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