Categories: quicknotes,satellite

Edit(2017/09): this article was written early in my PhD, with little knowledge of actual systems. It happens that the Uplink/Downlink frequency bands are not so linked together. I might update this article later, or publish a new one with more accurate assumptions.

Let \(\Delta F_{FL}\), \(\Delta F_{DL}\) and \(\Delta F_{UL}\) be the Frequency bands for, respectively, the whole forward link (i.e. Gateway → Satellite → Users), downlink (Satellite → Users) and uplink (Gateway → Satellite).

In the simple scheme where a satellite covers a full continent with a single beam (traditional TV Broadcast model) we have this relation: $$\Delta F_{DL} = \Delta F_{UL} = \frac{1}{2} \Delta F_{FL}$$

Multi-spot beam satellites allow to have different coverage beam, where the same frequencies can be used and not interfere with each other (the beams are sufficiently well focused and spaced).

In this case, a frequency can be used \(k\) times, where k is the frequency reuse factor.

For example, in a 4-color distribution scheme, with 120 spots, there is a theoretical reuse factor of 20.
In reality, the reuse factor is closer to 12 than 20, but that's another topic.

In the case where the reuse factor is \(k=2\) (meaning that a frequency is used twice over the whole coverage), taking the hypothesis of full use of the bandwidth over each spot beam, the necessary bandwidth in the uplink would be: $$\Delta F_{UL} = 2\Delta F_{DL}$$

Extending this to \(k\), we have: $$\Delta F_{UL} = k\Delta F_{DL}$$

Provided that the global bandwidth for this forward link is constant \(\Delta F_{FL}=cst\), we have the following relation of the uplink bandwidth according to the reuse factor: $$\Delta F_{UL} = k(\Delta F_{FL} - \Delta F_{UL})$$ $$\Delta F_{UL} = \frac{k}{k+1}\Delta F_{FL}$$

Sources:

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