# Why does NASA use higher frequencies even though they have worse Free Space Path Loss (FSPL)?

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.

• You should rephrase the title of your question (maybe "why using low frequencies for deep space communication?") so that it is easier to browse the website without opening each question about deep space communication – Manu H Jun 2 '19 at 8:28
• The wavelength dependence of the definition of free space path loss (FPSL) is an artifact of he 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. – uhoh Jun 2 '19 at 10:28
• @ManuH how does that look? – uhoh Jun 2 '19 at 15:38

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"

## $$P_{RX} = P_{TX} + G_{TX} - L_{FS} + G_{RX}$$

• $$P_{RX}$$: received power by spacecraft
• $$P_{TX}$$: transmitted power by wristwatch
• $$G_{TX}$$: Gain of wristwatch's transmitting antenna (compared to isotropic)
• $$L_{FS}$$: Free space Loss, what we usually call $$1/r^2$$
• $$G_{RX}$$: Gain of spacecraft's receiving antenna (compared to isotropic)

$$G \sim 20 \times \log_{10}\left( \frac{\pi d}{\lambda} \right)$$

$$L_{FS} = 20 \times \log_{10}\left( 4 \pi \frac{R}{\lambda} \right).$$

$$P_{RX} - P_{TX} = G_{TX} - L_{FS} + G_{RX}$$

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.

• thank you so much! I see that now, did not consider the gain formula, but you are right it is the key to the problem! – Clex Jun 2 '19 at 15:43
• @Clex ya it was confusing for me at first too. Running through a whole example ike I did here takes some time the first time (it took me hours!), but it helps a lot in the long run. – uhoh Jun 2 '19 at 15:48
• Yeah, I've always thought that the definition of FSPL (@uhoh, note the acronym isn't FPSL) was kludgy. But once a peer-reviewed paper using it was accepted, saying it should be redefined to be more physically intuitive has the telecom specialists tilting their heads back some while still looking at me, pointing, and uttering, "BLASPHEMY!!". – Tom Spilker Jun 3 '19 at 19:53
• @TomSpilker ha! I can picture it. Thanks for catching my misacronymization also. – uhoh Jun 3 '19 at 21:52