I've been playing around with the capacity planner and have a question.
Keeping all things the same and just changing from FCC 5.8ghz to FCC 2.4ghz, I've noticed that 5.8ghz has about half the predicted range of the 2.4ghz. (MCS15 for 2.4 is 4.45 Miles, 5.8 is 2.07miles) Why?
This is common to all radio systems and is due to Free Space Path Loss (FSPL). The greater the frequency causes the path loss to increase and hence the coverage of the radio system to decrease.
If you look at the ePMP Capacity Planner Guide, a formula is given for FSPL. This formula is :
- FSPL [dB] = 36.6 + 20logf + 20logd
Where f is expressed in MHz and d is expressed in miles.
When you start factoring in real life scenarios things change as well. You'll find much less noise in 5 GHz than 2.4 GHz as well as more usable spectrum. In regions with trees you'll get more penetration with 2.4 than 5. Things go back and forth.
@Daniel Sullivan wrote:
The greater the frequency causes the path loss to increase
What exactly is the process that is contributing to the alleged increased loss at higher frequencies?
The short description in the link below may help explain the two components considered in the equation described by Dan. First component is the distance between the transmitter and the receiver while the other is how well the transmitted signal can be picked up at the receiver. It is the second component the one that is frequency dependent.
Hope that helps,