If I were designing an RF circuit how do I determine what band the circuit should operate on? What are the trade offs? and why is the higher the frequency the lower range? It is quite un-intuitive for me since energy increases with frequency which should give a greater coverage or is it all dependent on the power of the transmitter?
Electronic – What are the trade-offs of transmission bands
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Best Answer
Energy increases as frequency increases, yes.
So of course, when you are generating a frequency you have to put that energy into it in the first place. The higher your frequency the more energy needed to make it go the same distance.
If you think about a wave being a simple two-state being, with one state HIGH and the other LOW - which represent a peak and a trough of a wave. It takes a finite amount of energy to set the wave to HIGH and a finite amount to LOW.
Say you generate 10 HIGH and 10 LOW. That's a fixed amount of energy that you put in.
With a long wavelength (low frequency) those HIGH and LOW points are spread over a long distance. With a short wavelength (high frequency) those HIGH and LOW points are all bunched up in a small space.
So to make the high frequency wave fill the same space as the low frequency you have to generate more HIGH and LOW points to fill the gap. So of course that takes more energy.
Now, which band should you pick? Well, there are a number of factors to consider, including: