Timeline for Why doesn't thermal radio emission from a DSN "hot dish" completely swamp the benefits of a cold LNA?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 28, 2017 at 6:45 | comment | added | uhoh | @gosnold the "astrophysical sky" is presumably 2.73K, no? Oh, maybe not, there's atmospheric water. OK I don't care about simply, I just care about calculate :) so maybe I have to ask a separate question about this. Hope to hit the library this weekend first. | |
| Apr 28, 2017 at 6:42 | comment | added | gosnold | @uhoh That means emissivity * 300k> 1/10* whatever the sky temperature is at 1mm (you cannot calculate that one simply, but there are charts of sky temps vs frequency in many books) | |
| Apr 28, 2017 at 5:27 | comment | added | uhoh | @gosnold don't forget the "uhoh" after the "(at)" sign. This is a great reference, thanks! I'm stuck not knowing how to do the math even behind this sentence in the introduction: "The thermal emission from a simple 300 K mirror surface with an emissivity $\epsilon=2.5\times10^{-3}$ is, for example, 10 times brighter than the astrophysical sky at $\lambda=1mm$." | |
| Apr 27, 2017 at 18:55 | comment | added | gosnold | @ right, it could be IR. Found this paper (Emissivity measurements of reflective surfaces at near-millimeter wavelengths, pdfs.semanticscholar.org/edab/…) that says emisiivity of aluminium is 3.10-3 at 1.17mm, so it adds 1K of noise temperature. | |
| Apr 27, 2017 at 16:14 | comment | added | uhoh | @gosnold thanks! The book's title suggest microwaves, but I couldn't find from the table if those are microwave or infrared emissivities. Also notable that even a thin coating might be enough - it wouldn't have to be solid gold luckily. With a 300K reflector and a 4K front-end, even 0.04 might be rather high though. | |
| Apr 27, 2017 at 11:48 | comment | added | gosnold | This has a table of metal emissivities, alumium can go as low as 0.04: books.google.co.uk/… | |
| Apr 27, 2017 at 6:14 | comment | added | uhoh | This is more of a comment than a proper stackexchange answer. Handwaving arguments (as in the "skipped-step" definition used by people running out of chalk at a blackboard) are great among friends and to save time, but for a helpful stackexchange answer it's best to back up quantitative statements of something being large or small with either some math or a supporting link. Without that, future readers have no way to verify the validity of the answer, and stackexchange is all about good answers. | |
| Apr 27, 2017 at 6:02 | comment | added | uhoh | Invoking transparency does not help by itself. This hot glass is transparent to visible wavelengths but can simultaneously glow quite brightly at visible wavelengths! | |
| Apr 27, 2017 at 5:53 | review | Low quality posts | |||
| Apr 27, 2017 at 6:14 | |||||
| Apr 27, 2017 at 5:53 | comment | added | pericynthion | Please be less rude. You don't get to dictate how I answer. Write your own if you want. | |
| Apr 27, 2017 at 5:46 | comment | added | pericynthion | White paint is transparent to RF. Emissivity is low enough that the antenna noise contribution from its self temperature is negligible - it's dominated by the ground temperature from the portion of the sidelobes that see the earth. Typically single-digit Kelvins noise temp for a well-designed reflector at moderate to high elevation angles. | |
| Apr 27, 2017 at 5:43 | comment | added | uhoh | Numbers please, and a reference, so we can believe it. Also, I see white paint everywhere btw. Let's build a good answer for future readers. How low is "very low"? What would the "effective temperature" contribution be of a 300K dish with an emissivity of $\epsilon_{dish}$, where $\epsilon_{dish}$ is the "very low" emissivity that you supply in the answer? 150K? 50K? The front end could potentially be as cold as 4K. At least ballpark numbers would help demonstrate the principle and make it believable that a "hot dish" isn't really so hot after all! | |
| Apr 27, 2017 at 5:35 | history | answered | pericynthion | CC BY-SA 3.0 |