# Tag Info

## Hot answers tagged path-loss

33

A link budget is a summary of a communications link that tries to take into account all factors which have an impact on the received signal strength. It is often used to determine the minimum amount of output power required at the transmitter for a given signal strength at the receiver, and takes into consideration power output, antenna gains, propagation ...

13

That's a pretty good answer but I can add some more details. The exact formula for path loss is $$20 \times \log_{10}\left( 4 \times \pi \times \frac{d}{\lambda{}} \right)$$ $20 \times \log_{10}\left(4\times\pi{}\right) \approx 21.98$, and that's where the '22' comes from. If your receiver specs give a minimum signal strength, then the analysis above is ...

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 ...

5

It's close enough to zero at 2m and 70cm that I wouldn't bother computing it. Judging by ITU-R P.676 the attenuation due to the atmosphere is less than 0.01 dB/km for all frequencies below 1 GHz, and the contribution of water vapor is negligible below 3 GHz. Figure 6 shows that the attenuation at the zenith is 0.03 dB at 1 GHz, and equation 28 says that for ...

4

UNDER CONSTRUCTION!!! A this answer is in need of a considerable edit, so please do not take it into account until it has been fixed. It seems to me as if you're referring to the two-ray propagation model, sometimes called signal ground bounce problem. The concept is relatively simple: You have one ray going from the antenna to the receiver and you have ...

4

How can I estimate the path loss The answer seems to be to pick one from a catalog of available path-loss models and apply the formula. Here, I'd try using the Hata model for urban areas. The model was originally made for portable cellular systems, so it uses a bit different terminology. The formula is not so complicated: $$L_{50} =69.55 +26.16 \cdot \... 4 At 8W, the power from the radio is 9.03dB. With the cable losses of the receiving station being 1.5dB (1.41W), and the antenna receiving 1.733e-8W, it appears the received signal will never make it to the receiving station's radio. This seems to be the crux of your misunderstanding. You can't convert watts into decibels, or decibels into watts. A ... 4 The gain becomes an issue only if its ridiculously low. The main purpose of an LNA is to lift the signal well above any noise of the following stages, and often that can be done with a gain of several dB. However, the fact that they didn't advertise it, makes it suspect. 4 First a bit of context regarding terminology. In antenna engineering, we use the term "loss" to refer to RF energy that is lost as heat. This is typically an undesirable loss that creates inefficiency in the antenna system. This is the loss that Phil addressed in his (now deleted) answer. When we speak about the directionality of an antenna, we use the ... 3 From the symptoms you state of not being able to hear transmissions from identical setups only 4 miles from each other, and then being able to have one person receive with the rubber duck antenna, it tells me your J-Pole antennas are not behaving ideally on the frequency you choose. Consider that an antenna like a J-Pole has about a 10MHz bandwidth in the ... 3 First consider what the Free Space Path Loss (FSPL) equation is modeling. As power propagates from the transmitting antenna in free space, it continues to spread out, lowering the power density (watts per square meter) at the receive antenna locus. On the receive side, the effective aperture (square meters) of the receive antenna is fixed for a given ... 2 A four element yagi shouldn't overload your receiver unless you're very close to the transmitter. If you were close enough for that to be an issue, your rubber duck antenna would work just fine. When you added the tee to the cable, you created a quarter wave stub trap that would filter out certain frequencies. It would probably work even better if you ... 2 My understanding of an RSSI reading is that it should equal Tx power - Tx feed loss + Tx antenna gain - pathloss + Rx antenna gain - Rx feed loss. If this is the case, if the radios at either end are identical and have the same Tx power, shouldn't the RSSI experienced at both sides be the same? Your intuition about reciprocity is correct but your ... 2 None of your formulas will mean anything because you are not considering the path loose. What is the SWR on the J-poles at the frequency you are using? How high are they? If they are in your basement instead of on the roof, that's a problem. J-poles that have good SWR, at 30 ft, with flat ground between them should easily work 10 miles apart with 8 watts. 2 Below is the output result of a NEC4.2 analysis of an electrically short monopole, showing its near-field boundary radius to be located about 30 meters from the base of that monopole. The groundwave field beyond about 30 meters decays at nearly a 1/r rate, but not exactly 1/r because of the surface wave propagation losses present for these conditions â€” even ... 1 I would suspect that the primary reason for the seemingly odd results is that you are likely operating the experiment in the near field of the antennas. As your distance changes, the mutual coupling of the antennas changes. As mutual coupling changes, antenna efficiencies, impedances, S11 and S22 parameters will change as well. You can execute experiments ... 1 Define the Antenna structure into a free available NEC Software like EZnec, MMANA-GAL or others and you are able to calculate everything of the radiation patterns, gain, losses, currents, voltages, resistance, the complex and real resistance of the entire antenna Building and the rf-field (E/H) shape. Also, you can see shape of the E and H field and the ... 1 The general formula for reactive field does not incorporate height, but rather maximum dimension of the antenna:$$R\approx 0.62\sqrt{\frac{D^3}{\lambda}} \tag 1 where D is the maximum dimension in meters and $\lambda$ is the wavelength in meters. Note that this results in an approximation in meters, not an exact answer. The radiative near field (...

1

I ended up switching from a 25ft RG-58A/U cable with F connectors on each end to an 18ft RG-58A/U cable with PL-259 connectors on each end. I can now receive, using my J-pole, my buddy's signal when he is transmitting with his J-pole. However, there is a lot of noise. This fixed the reception issue, but I'm still experimenting to figure out what the minimum ...

1

The reason loss generally goes down when the height of a receiver is increased is because channel fading is usually reduced because the increase in reciever height enables the propagating LOS signal to avoid more obstacles and travel further, hence reaching the receiver with higher power.This then increases the ratio of the power received in the direct line-...

1

Let me take the later question out of the title. how do I make one? By using a tool, such as Radio Mobile Online Help is here, you don't need a free account to read it This is a free tool, and will work for Amateur Radio frequencies. Once you get started, in order to display a "Link", you do the following Use "New Site" from the main menu to setup ...

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