Sooo I wasn't sure if I should put this here or in the astrophysics section, but I figured electrical engineering was more relevant for my question.(adsbygoogle = window.adsbygoogle || []).push({});

I'm involved in a project at my university to build a (very) simple radio telescope, but I'm having a bit of trouble with the radiometer equation:

[itex]\sigma_T \approx \dfrac{T_{sys}}{\sqrt{\Delta \nu_{RF} \cdot \tau}}[/itex]

Where [itex]\sigma_T[/itex] is the "sensitivity" or the minimum detectable temperature, [itex]T_{sys}[/itex] is the system temperature, [itex]\Delta \nu_{RF}[/itex] is the bandwidth, and [itex]\tau[/itex] is the integration time.

I'm mostly having trouble with "integration time". This seems to imply that one has to use an integrating analog to digital converter (something like a dual-slope ADC I guess...), but this seems to prevent the application of Fourier analysis (am I wrong in thinking this?). However, I have seen some examples of radiometers that appear to take discrete samples, and apply some sort of "integration time".

Since the product [itex]\nu \cdot \tau[/itex] is effectively the number of samples that would have been taken over the integration time, I was wondering if this product could equated to some function of sample rate of a non-integrating ADC?

Rereading my above questions, they aren't the most coherent sentences in the world... But I'm not quite sure how else to pose them. I have a feeling I'm overthinking this, or missing something simple. If anyone needs anything clarified, please let me know!

Thanks so much for the help!

**Physics Forums - The Fusion of Science and Community**

# Radio astronomy - integration time vs. sample rate

Have something to add?

- Similar discussions for: Radio astronomy - integration time vs. sample rate

Loading...

**Physics Forums - The Fusion of Science and Community**