Hot answers tagged

19

Mathematically yes, the value of that equation increases with frequency. However, that's not to say there's some physical mechanism for frequency-dependent attenuation in free space. Rather, the frequency term is in the equation due to the assumption of unity gain antennas at each end. A larger antenna is required to get the same gain at a lower frequency. ...


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{...


13

You didn't include antenna gain at all. That number for the path loss is a starting point assuming zero gain (isotropic antennas) on both ends. This isn't a realistic situation, because of course the vast majority of the signal from an isotropic antenna doesn't reach the moon at all, but it's conventional for calculating link budgets. A 9-element Yagi has a ...


12

Baud is another name for symbols per second (a unit of symbol rate), so you can't convert between it and the others (which are units of bit rate) without knowing how many symbols are in use. A symbol is the smallest unit of a digital modulation. For example, suppose we're doing on-off keying (also known as CW): the signal can be either present or absent, so ...


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 ...


8

The diagram is a polar plot where 0 degrees (the 0 on the right-hand edge) corresponds to the front of the antenna and 180 degrees corresponds to the back. Within the diagram, a logarithmic scale is utilized to represent radiation strength. This is in addition to the fact that the Decibel is a logarithmic unit. (The diagram really should explicitly list its ...


7

A neper, just like a decibel, is a logarithmic expression of ratios. The decibel uses the base-10, or decadic, logarithm while the neper uses the natural, or Euler constant, logarithm. The decibel is strictly defined as the ratio of two powers. $$dB=10\log_{10}\left(\frac{P_1}{P_2}\right) \tag 1$$ While it is common to see a decibel formula based on ...


7

The short answer: \begin{equation} \frac{V_{p-p}}{V_{rms}} = 2\sqrt{2} \end{equation} The long answer, or how to derive the above: As noted on the Wikipedia page for root mean square, the RMS of a sine wave is equal to its amplitude divided by the square root of two. (You can also derive this by doing the integral over a sine wave yourself.) \begin{...


7

Try the Friis noise formula: $$ F_{eq} = F_1 + {F_2-1 \over G_1} + {F_3-1 \over G_1 G_2} + \cdots \tag 1 $$ $F_n$ is the noise factor of the n-th component, and likewise $G_n$ is the gain. The noise factor $F$ is the linear ratio form of the noise figure which is given in decibels. For example, the first component may be an LNA, the second component a ...


7

Decibels are all "ratios" at their core. A unitless dB is a simply a ratio of one number to another, perhaps input power relative to output power. We can also use decibels for absolute values, by fixing the denominator to a standard reference — e.g. one milliwatt in dBm. But the most convenient thing about decibels is that, although they are ratios, because ...


6

There are a lot of ways to approach this problem, but here's one: we can calculate the power density of that field, and determine the area from which the antenna captures power, and multiply them together. Calculating the power density If we know the electric field to be 1 mV/m, and the transmitting antenna is distant, then the magnetic field must be ...


6

Let's imagine that there are two antennas, entirely in free space. There is no Earth or any other object near enough to affect propagation. Let's also assume that these antennas are far enough apart that the antennas interact only by their far fields. To start, let's assume that both antennas are isotropic, meaning that they radiate equally in each ...


6

TL;DR: $\frac{V}{m}$ and $\text{dB}(\mu V/m)$ are units for the field strength of an electric field. For a practical application skip to the end! Derivation of the field strength A point charge $q_1$ generates a field strength* of $E = \frac{1}{4\pi\varepsilon_0}{q_1\over r^2}$ at a distance of $r$. This is derived from Coulomb's law that is $F=\frac{1}{...


6

Whatever modulation we use, there's a baseband signal we wish to transmit (music, voice recording, whatever), which somehow modulates a carrier to produce the output signal. Your question suggests you are primarily concerned about how the frequency domain representation of the baseband translates to the frequency domain of the output. This is a valid thing ...


6

I convert the 27dBm and 3dBm -> 10^2.7-10^0.3. But how can we simply subtract the two dBm If you convert to exponential form then you must simultaneously replace subtraction with division (or addition with multiplication), so you have $10^{2.7} / 10^{0.3}$ instead of $10^{2.7} - 10^{0.3}$. Then you will see that you get the same result: $$2.7 - 0.3 = 2.4$$ ...


6

The SWR is related to the reflection coefficient $\Gamma$: $$ \Gamma = {Z_L - Z_0 \over Z_L + Z_0 } $$ $$ \text{VSWR} = {1+|\Gamma| \over 1 - |\Gamma|} $$ where: $Z_0$ is the feedline impedance, usually 50 ohms, and $Z_L$ is the load (nominally, the antenna) impedance. The reflection coefficient is a complex number and thus takes into account the phase ...


6

Think about it this way. 27 dBm means 27dB above a milliwatt. Take 6dB away. Now you have something that's 21 dB above a milliwatt. Or, 21 dBm.


5

Given the matched loss of the feedline and the SWR at the transmitter, we can calculate the SWR at the antenna in three simple steps. First convert the SWR at the transmitter to the corresponding magnitude of the reflection coefficient (Gamma), or MRC for short within the context of this answer. The MRC is the magnitude of the complex ratio of the reflected ...


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 ...


5

The classic dipole is a half-wave antenna. This means that the total length of the antenna is lambda/2. So writing it as 1/2-lambda is OK from an English language point of view, but not IMO as a rigorous mathematical formula. For a half-wave dipole, each side of the feedpoint is one-half of that or a quarter-wave.


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

I generally understand that AM side bands occupy sufficiently wide band width to recreate the spectrum of the modulating audio signal (while varying amplitude at each frequency in the spectrum of the sound wave)--so the spectrum of a side band should look basically like the spectrum of the underlying modulating audio wave. Is this understanding wrong? This ...


4

There are a few possibilities. First of all, in larger assemblies (mostly important in repeater design) each component will have a parameter called insertion loss. This is the loss, in dB, through that componenet - the loss caused by inserting that component in your feed line. Filters have insertion loss, as do duplexers, and even inline wattmeters and ...


4

An FSK signal which is the same symbol repeated is an unmodulated carrier, and like an unmodulated carrier, it contains no information. Making some assumptions about the bit shaping filter it might be possible to make a reasonable estimate of the symbol rate judging by the growth and decay of the envelope at the start and end of the transmission, but it's an ...


4

If you make a dipole exactly a half-wavelength long, then it will be too long and out of resonance. The formula for determining the length of a half-wavelength dipole in feet is 468÷frequency in MHz. That's shorter than the actual wavelength in free space which is 492÷MHz.


3

Introduction In amateur radio, we are used to giving signal reports in the form of "S-units". The S-unit is based on the amount of power at the receiver's input terminals. Thus we can see that the gain of the receive antenna plays a role in the level of the S-unit. In professional circles, it is more common to be concerned with the "field strength" of the ...


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 ...


3

This would suggest that longer wavelength antennas are more efficient collectors of electromagnetic energy. Why is this? I think of it like in optics. A larger lens will collect more light because more light is incident on the lens. More photons hit the lens. A radio antenna is a lens for a different wavelength of light, i.e. radio waves, not visible ...


3

I'd like to offer a parallel and simplified explanation to W8II's correct answer above, for the mathematically-challenged VHF+ enthusiasts among us. :-) As was mentioned in recent threads here, effective aperture is a measure of the power captured by an antenna. Effective aperture can be expressed as a function of the antenna gain and the wavelength. Now, ...


3

There are already a couple of nice answers to your question. I thought I would add a little more context - no need to vote for this as it is tangential to the question. A 1/2 wavelength (1/2 $\lambda$) long dipole is one of the most basic and easy to deploy antennas. It is popular because when it is cut and fed in the center, it very closely matches the ...


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