Looking at the Formula for FSPL we see that it increases with Frequency.
Why is X-Band used for Deep Space Communications instead of lower bands like S-Band? Is it a question of data-rates? Is the noise floor lower in this band? Is it just a question of infrastructure, as the DSN uses X-Band?
Lower bands would lower the FSPL quite a bit, so I want to know the reason for this decision.
The wavelength dependence of the definition of free space path loss (FSPL) is an artifact of the way the receiver's antenna gain is defined in the same link budget calculation. It's referenced to an ideal isotropic antenna with a receive area of roughly 1 square wavelength, which for high frequency gets very small. If you do them together (transmit gain, path loss, receive gain) you'll see that the higher frequency wins because the gain of the transmit antenna go up, and the combination of the FSPL and the receive antenna together remain the same.
So the reason higher frequencies are used is because the total link budget is better.
You can't just take the FSPL and ignore the other terms in the budget, it won't make sense because of the way things are defined.
You can see real-world examples of link budgets in this answer (short one) and in this answer (longer one) and in this question. You may find this one interesting as well.
From here"
From this answer which is from this answer:
$$G \sim 20 \times \log_{10}\left( \frac{\pi d}{\lambda} \right)$$
$$L_{FS} = 20 \times \log_{10}\left( 4 \pi \frac{R}{\lambda} \right).$$
$$ P_{RX} - P_{TX} = G_{TX} - L_{FS} + G_{RX} $$
Change from dB to linear scale:
$$ \frac{P_{RX}}{P_{TX}} = \frac{G_{TX}G_{RX}}{L_{FS}} = \frac{\pi^4 d_{RX}^2 d_{TX}^2}{\lambda^4}\frac{\lambda^2}{16 \pi^2 R^2} = \frac{\pi^2 d_{RX}^2 d_{TX}^2}{\lambda^2}\frac{1}{4^2 R^2} = \frac{\pi^2 d_{RX}^2 d_{TX}^2}{4 \lambda^2 R^2}$$
So the fraction of the transmitted power that is received depends on $\lambda^{-2}$ ; it improves as the frequency goes up and the wavelength goes down.
