[SOLVED] I need help understanding this counterintuitive RF phenomena.

[unbuildpain]

Below is radio frequency chart from Wikipedia. Everywhere on the net, it is said that, as the frequency increases, they can't travel further, they face most obstacles, they get absorbed by oxygen, etc, this is the reason given for why 5G needs more towers, etc. If that is true, why are frequencies higher than 5GHz used in long distance communication, satellite communication, radar, etc.. This seems counter-intuitive to me. If higher frequencies can't travel further, then they shouldn't be used in long distance communication, radar, etc. So why are they being used in these things and why 5G frequencies are said to can't travel further?

 

[betwixt]

You answered your own question. How many obstacles are there between a ground based dish and a satellite?
Radar is designed to pick up reflected signals bouncing off obstacles.
And yes, 5G does need more cell stations because the frequencies it uses are more prone to suffering dead spots from obstacles and reflections. Note that 5G means 'Fifth Generation' not '5 GHz' though.

Brian.

 

[unbuildpain]

You answered your own question. How many obstacles are there between a ground based dish and a satellite?

Many kilometers of clouds, dust, air, oxygen, etc. I don't understand why it would work when 5G frequencies don't work.

Radar is designed to pick up reflected signals bouncing off obstacles.
And yes, 5G does need more cell stations because the frequencies it uses are more prone to suffering dead spots from obstacles and reflections. Note that 5G means 'Fifth Generation' not '5 GHz' though.
Brian.

My bad, I thought 5G and 5GHz are same. In the chart, F and D bands are used for amateur radio, doesn't that mean it has to travel far and through objects.

 

[BigBoss]

The answer is Bandwidth due to higher data rate.
They use higher frequencies to achieve more bandwidth and historically some bands are reserved by broadcasting,military,meteorology etc.

 

[unbuildpain]

The answer is Bandwidth due to higher data rate.
They use higher frequencies to achieve more bandwidth and historically some bands are reserved by broadcasting,military,meteorology etc.

Still doesn't explain why higher frequencies which are believed to not travel further are used in satellite communication.

 

[Georgy.Moshkin]

Antennas have smaller dimensions at higher frequency. Easier to make narrow beam, can use less power for transmitter.

 

[betwixt]

High and low frequencies travel the same distance for a given source power and receiver sensitivity.
The difference in range is not due to the frequency, it is due to the medium through which the signal passes.
For satellite communication where for practical purposes it is difficult to maintain high power while in orbit, VHF and above is used simply because the low signal levels arriving at ground level dictate that antennas with good signal gathering capabilities are used. The only practical antenna is one with a parabolic reflector to direct a wide aperture to a small pick-up probe. The size of the dish to achieve the needed 'gain' (gathering ability) is proportional to the wavelength of the signal so it is only feasible to use them at very short wavelengths (= high frequencies).
Different medium have different attenuation, dry air has quite low loss, humid air a little more, clouds a little more again and metals present almost total attenuation. At microwave frequencies there are particular molecular resonances that absorb specific wavelengths, for example water molecules in microwave ovens.

Brian.

 

[BigBoss]

High and low frequencies travel the same distance for a given source power and receiver sensitivity.
The difference in range is not due to the frequency, it is due to the medium through which the signal passes.
Brian.

It's wrong.. Free Path Space Loss is absolutely depended on the wavelength so frquency..

Free-space path loss - Wikipedia

en.wikipedia.org

 

[vfone]

The original Friis transmission formula is: (Pr/Pr)=(Ar*At)/[(d^2)(λ^2)]

Friis transmission equation - Wikipedia

en.wikipedia.org

But not many people follow this formula because it contains the Aperture of the RX/TX antennas, and not the Directivity (gain) of the antennas.
The reason is that, the Aperture (dimension) of the antenna is the same for all frequencies, when the Directivity of an antenna can be specified for any frequency.

The Free-space path loss formula (Pr/Pt)=(Dt*Dr)[λ/(4*Pi*distance)] is more used and meaningful.
From this, the free space path loss can be reduced to: [(4*Pi*distance)/λ]^2, which makes the path loss results to depend only by frequency and by distance.
Double the frequency or double the distance the path loss increase with 6dB.
The path loss will increase with the medium that passes, but this is not the main contributor to the free-space path loss.