A few weeks ago we discussed the notion of a Fresnel Radius which defines the “stay clear” area as the sort of official definition of “Line of Site.” In this case one of the things that I pointed out was that this stay clear zone got bigger with respect to wavelength and distance. In doing so I showed a picture similar to the one below.
The Fresnel Radius R above at the midpoint between transmitter A and receiver B was √(λ D) / 2, where λ was the wavelength and D and R are all measured in meters. Since that time I have received a number of queries about how this radius changes as one gets closer to the transmitter A and receiver B. All I noted was that it was a “football” or eliptical shape. So, here is the rest of the story - mind you I had no idea that the Fresnel Radius would be so exciting to so many people. And in providing the simple explanation, I must admit that I cheated you all with the simple calculation.
To understand the complexities better, note that the radius shrinks as you get closer to either the transmitter or receiver. This is in fact because the wave has yet to spread as far as you get closer to the transmitter and the receive aperture gets smaller as you get closer to the receiver. As such, tapping the knowledge from the free space loss calculations of many blogs ago, we note that a more general version of the Fresnel Radius can be thought of as:
R = √(λ dA dB / D)
where λ, dA , dB, and D are as defined above. Note that when dA = dB = D/2, then we get the simplified √(λ D) / 2.
The following graph illustrates the Fresnel Radius for different values of dA ranging from 0 to 2000 meters for both VHF and WiFi. As you can see, we get the familiar “football” or elliptical shape.
Note that when you are close to the receiver or transmitter, a small obstruction can do a lot of damage. Finally, the other thing look at is how the radius changes when you move the receiver closer. For example, look at two values for D, one at 2000 meters, the other at 1000 meters. How do the Fresnel Radius’ compare at 500 meters from the transmitter (dA = 500) for both distances? From the graph above we see that R is about 27 meters when D is 2000 meters. And when D is reduced to 1000, R is about 22 meters. The fact that it is smaller when the receiver is closer is intuitively obvious, but I’m always amazed that it really isn’t that much smaller - given that the total distance was reduced by half.