What are the trade-offs of transmission bands?

Question

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?

Answer 1

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:

1. Energy available - if you are making a low powered system then using a lower frequency band will save power.
2. Environment - What obstacles are there in the way? Lower frequencies penetrate things than higher frequencies. They can even bounce off the ionosphere to bend around the globe.
3. Bandwidth - Do you need to send a lot of data quickly? If so then you need a higher frequency to fit all that data in.
4. Regulations - There are laws governing what you can transmit and on what frequencies. This severely limits your choice of frequencies depending on what it is you're transmitting.

Answer 2

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?

A perfect transmit antenna is sometimes referred to as an isotropic antenna - it converts the electrical power at its terminals to EM power equally in all directions. It's a bit like a lightbulb - the power fed to it results in light being produced at nearly all angles - as you walk further away the light decreases (because the incident light hitting your retina has spread itself a bit thinner). Even so, at a thousand miles distant, if a gazillion eyes were all set looking back at the light source and you could collect all that light from those gazillion eyes it would be the same total power as that collected at 1 metre distant.

OK the term "gazillion isn't very meaningful and there will be some attenuation of the light in the atmosphere but hopefully you see where I'm coming from.

It doesn't matter what frequency you transmit at, the power is still produced by a transmitting antenna but spreads thinner as distance increases. So, nothing hopefully unintuitive here.

The unintuitiveness arises when you try and build an antenna for detecting the received power. Any antenna has, what is called, an effective aperture. This is measured in square metres or sq inches - it's the area which an antenna collects power over and is like the size of your retina - a bigger retina collects more power.

The effective aperture of any antenna (including the theoretical isotropic antenna) is determined by the frequency it is "tuned" to. For an isotropic antenna this is: -

Effective aperture = \$\dfrac{\lambda^2}{4\pi}\$

For any other antenna type the same relationship holds true thus, the power received diminishes with falling wavelength. For a great read and a decent (and intuitive) explanation see this. It's a document called "Essentials of Radio Wave propagation" written by Christopher Haslett. It's a really good document to take a copy of too.