posted 05-02-2013 10:11 AM ET (US)        

http://continuouswave.com/whaler/reference/graphics/ FCC_PropagationChart.pdf

The chart is very interesting because it shows signal loss with distance with several curves. There is a curve labeled "Free Space", which represents the theoretical best-case loss curve for a radio wave traveling a certain distance. There are also curves for real-world situations, with various antenna heights.

The curves show that for very short distance paths and very tall antennas, the path loss in the real world is the same as in free space, but once the path distance becomes more than a few miles, the characteristics of the path loss change remarkably.

I investigated this chart to see if I could deduce the path loss factor for the distance that was being used. I will describe my method.

First, I looked closely at the free-space curve. I found the signal levels (dBu) for the distances (d) of 10 and 100 miles.

  d    dBu
 10    83
 100   63

When the distance changed by a factor of ten, the signal decreased by -20dB. This suggests the relationship is

dB = x log(d)
-20 = x log(10)
x = -20 / log(10) = -20

The curve on the chart seems to show that path loss as a function of distance behaves according to the relationship

dB = -20 log(d)

This is exactly what would be expected. The factor -20 log(d) is used in the free-space model of path loss.

Next, I investigated the other curves to see what sort of relationship between signal level and distance they represented. Using the curve for an antenna of 100-feet, I found the following values of signal and distance:

  d    dBu
  5    75
 50    19

When the distance changed by a factor of ten, the signal decreased by -56dB. This suggests the relationship is

dB = x log(d)
-56 = x log(10)
x = -56 / log(10) = -56

The curve on the chart seems to show that path loss as a function of distance behaves according to the relationship

dB = -56 log(d)

A relationship between signal and distance where the path loss increases at that rate is significantly more lossy than some of the other estimates I have seen. In another article,

Estimating Path Loss on Marine Non-Line-of-Sight Paths

the path loss factor was found to be often cited in the range of -40 log(d) to perhaps -46 log(d), but not nearly as steep as the -56 log(d) suggested by the FCC propagation chart.

posted 05-03-2013 11:44 AM ET (US)            

When the signal loss or path loss for a certain distance is described as being

20 log(d)

we are saying that the path loss increases with distance accord to that rate. Let us explore the increase in path loss when the path length doubles, as in a path of 10-miles increasing to a path of 20-miles. The path loss due to distance will be:

For 10-miles

loss = 20 log(10)
loss = 20dB

For 20-miles

loss = 20 log(20)
loss = 26dB

When the path distance doubled, the path loss changed 6dB. Because there was decrease in signal, we say the path loss changed -6dB

The notation -6dB represent a ratio. In terms of a ratio of power, we can find the power ratio from the definition of dB for power, which is

dB = 10 log (P1/P2) where P1, P2 are two power levels

We can also evaluate that relationship to find the power ratio represented by the dB expression. That becomes

(P1/P2) = 10^(dB/10)

Now we evaluate for -6dB to find the ratio of powers:

(P1/P2) = 10^(-6/10)
(P1/P2) = 10^-0.6
(P1/P2) = 0.25

In other words, when the path distance doubled, the signal power decreased to 0.25 of the original power, or one-fourth the original power. This is another way of describing the inverse square law relationship. I think many readers will be familiar with that behavior, but may have not realized that it was represented in the notation 20 log(d).

When models for propagation use a factor for distance of 20 log(d), they are assuming that the decrease in signals will just obey a simple inverse square law relationship. This is the behavior that would occur in free space, an ideal environment in which nothing else affects the radio wave as it propagates. In the real world there are other factors in the environment that affect the signal loss in propagation, causing the real-world signal decrease to be greater. This behavior is reflected in having a path loss model in which the distance factor is greater than 20 log(d).

posted 05-08-2013 07:55 AM ET (US)            

Pr = 0.66*Gb*Gm * [(Hb*Hm)^2/d^4] *[40/f]^2 *Pt

where

Pr = received power
Pt = transmit power
Gm= gain of mobile antenna (not in dB)
Gb = gain of base antenna (not in dB)
Hm =height of mobile antenna (meters)
Hb = height of base antenna (meters)
d = path distance (meters)
f = Mhz

This model contains several significant observations

--the path loss increases at a rate of 40 log(d) (from the 1/d^4 factor)

--there is decrease in path loss with antenna height at a rate of 6dB per octave, that is, doubling the height improves the received signal by a factor of four;

--and frequency influences these effects. There is an empirical factor for frequency which is applicable for frequencies in the range of 30-MHz to 1,000-Mhz, and for distances in the range of 1 to 50-km.

For more on Egli's model, and propagation path loss models in general, see

http://people.seas.harvard.edu/~jones/es151/prop_models/propagation. html#egli

posted 05-08-2013 09:18 AM ET (US)            

Antenna gain = 2 for both antennas, i.e. 3-dB
Antenna height = 3-meters, i.e. about 10-feet, for both antennas
Transmitter power = 25-watts
Frequency = 156-MHz

the model calculates the received signal power as follows

Miles__dBm
0.6 = -94.5
1.2 = -106.5
2.4 = -119.6
5.0 = -130.6

As you can see, when the distance doubles the path loss increases by 12dB, which is what is suggested by the 40 log(d) factor.

A really good receiver sensitivity is 0.25-microVolt, which is a signal level of -119dBm. Egli's model suggests that we will reach that signal level at a range of only 2.5-miles. The model does much better when antenna height is increased. For example, if one antenna is raised to 10-meters (33-feet), then the path distance for a -119dBm received signal doubles to 5-miles.

posted 05-08-2013 09:45 AM ET (US)            

http://www.netcontrol.com/eng/services/radio-networking-help-tools/ egli-calculator/

posted 05-12-2013 10:34 AM ET (US)            

Antenna Height-1 = 100-feet (e.g., a shore station)
Antenna Height-2 = 10-feet (e.g., a small boat)
Frequency = 157-MHz
Transmission distance = 30-miles

is calculated to be -160dB. We can now predict the signal level we might have for this path if we have the following characteristics in each station:

Transmitter power = 25-Watts
Antenna gain = 3dB

A transmitter power of 25-Watts is a signal of +44dBm. Both the transmit antenna and the receive antenna have 3-dB gain. We can now predict the received signal level as follows:

TxPower + TxAntGain + RxAntGain +PathLoss = RxPower

+44dBm + 3dB + 3dB + -160dB = -110dBm

A received signal level of -110dBm in a 50-Ohm system corresponds to a 0.7-microVolt of signal. Most modern VHF Marine Band radios have receivers with sensitivity of 0.5-microVolt for 10-dB SINAD, so the received signal on this path should be above the receiver's rated sensitivity.

posted 05-14-2013 01:46 PM ET (US)            

I inspected the FCC chart of path predicted coverage (linked above) and found that the same behavior was used in their predictive model: when the antenna height is doubled the predicted signal increases by 6-dB.

There seems to be some agreement in the two models on the influence of antenna height, and that influence seems really very generous. It is my understanding that both of these models are adaptations of real measurements of signal transmission over real terrain in the VHF and UHF spectrum, that is, they are not theoretical approaches but rather embodiments of actual observed phenomenon which have been then reduced to an empirical formula.

The influence of antenna height seems very significant in these two methods of predicting radio propagation, and this seems to suggest that antenna height really is very important, particularly when it is possible without too much mechanical difficulty to create a doubling of the antenna height, as would occur, say, if an antenna were raised to 10-feet from 5-feet in height. There is evidence in both the FCC data and Egli's data that such an antenna height increase will pay back a substantial dividend in increased signal (and thus increased range).