5 ms·

"Makes it so clear what it is." Well.., I've been more busy with writing code lately so that the first question coming to mind was, how many bytes is an array

"Makes it so clear what it is." Well.., I've been more busy with writing code lately so that the first question coming to mind was, how many bytes is an array of one square kilometer? And I assume it's a two-dimensional array.

Etheryte 9mo ago
It's a radio telescope, how would you imagine translating that to bytes?
- stargazer-3 9mo ago
  Here's an article mentioning the data transmission rates in SKA, up to 20 terabits per second: https://www.skao.int/en/explore/big-data https://www.skao.int/en/explore/big-data
- touisteur 9mo ago
  Every sensor in the array is sampling at frequency, so - first order - you can use that sampling frequency and the sample size, you get an idea of the input bandwidth in bytes/second. There are of course bandwidth reduction steps (filtering, downsampling, beamforming)...
  - jacquesm 9mo ago
    Aren't they sampling broadband for later processing?
    - touisteur 9mo ago
      On SKA from what I understand they're sampling broadband but quickly beamform and downsample as the datarates would be unsustainable to store over the whole array.
      - jacquesm 9mo ago
        Right, that makes sense, you'd be looking at an insane amount of data across the ranges that these sensors can look at. But they would still need to preserve phase information if they want to use the array for what it is best at and that alone is a massive amount of data.
        touisteur 9mo ago
        I think they preserve timestamped I,Q data. Know some people looking at down-sampling, preselecting those signals for longer term storage and deeper reprocessing and they seem to have a 24h window to 'analyze and keep what you need'. We're still in technological phase where ADCs are far more advanced than storage and online processing systems, which means throwing away a lot. But I have high hopes for a system where you upgrade computing, network, storage (and maybe ADCs...) and you get an improved sensor. Throw man-hours at some GPU kernel developers and you get new science. The limit seems more now about enough people and compute to fully exploit the data than technological...
        jacquesm 9mo ago
        Fascinating stuff, thank you for the details and the view of a possible path forward.
        touisteur 8mo ago
        BTW if you're interested in the concept of upgrading a sensor without retooling the RF part, and the impact of 'just' putting new COTS racked server hardware and engineering man-hours to get a 'new' sensor with new capabilities, have a look at Julien Plante's work on NenuFAR (which isn't like the SKA at all :-) : https://cnrs.hal.science/USN/obspm-04273804v1 https://cnrs.hal.science/USN/obspm-04273804v1 . Damien Gratadour, his PhD supervisor is an amazing technologist, dedicated to improving astronomy instruments, and I was very lucky to work with him and his team... the things the French can string together with small teams and thin budgets...
        jacquesm 9mo ago
  - Etheryte 9mo ago
    This makes no sense though? Given the Nyquist theorem, simply increasing sampling frequency past a certain step doesn't change the outcome.
    - labcomputer 9mo ago
      Actually, it does. You can decimate the higher sample rate to increase dynamic range and S/N ratio. Also, for direct down conversion, you can get better mirror frequency rejection by oversampling and filtering in software.
      - Etheryte 9mo ago
        None of this changes the actual real amount of data you have at the end of the day though after all is said and done, that's what I mean, so long as you don't botch it and capture too little. In computing terms, the amount of real data in a compressed archive and the uncompressed original is the same, even if the file size is larger for the latter.
    - touisteur 8mo ago
      Sorry, not sure I follow from what I said (explaining how much data sensors produce) to 'increasing the sampling frequency' ? You're usually sampling at larger width to then put specifically taylored pass-band filter and removing aliasing effects and then downsampling. This is a classic signal acquisition pattern : https://dsp.stackexchange.com/questions/63359/obtain-i-q-components-from-a-real-signal-on-the-fly-hilbert-transform-or-digit https://dsp.stackexchange.com/questions/63359/obtain-i-q-com...
- dylan604 9mo ago
  Are you deliberately obtuse to the play on words of an array being used from a programmer's use of the word in contrast to an array of antennas?
  - JamesTRexx 8mo ago
    If you have to ask, you know the answer. :-p
dylan604 9mo ago
I'm thinking that you'd need to print out the array to be able to properly measure this. So we'd need to decide on the print itself. Do we use A4 paper, old school green bar? What size font is used? We'd also need to decide on the contents of each element in the array. Let's say they are floats with a set limit of precision. So an A4 is 210mm x 297mm. 1km / 210mm = 4761.9 1km/297mm = 3367.0. If we go with 4bytes per float, that's 4761.9 * 3367.0 * 4 = 64,133,269.2 bytes => 61.16 MB It's also before coffee, so my logic might not be right for basic math yet

Etheryte 9mo ago

It's a radio telescope, how would you imagine translating that to bytes?

stargazer-3 9mo ago
Here's an article mentioning the data transmission rates in SKA, up to 20 terabits per second: https://www.skao.int/en/explore/big-data https://www.skao.int/en/explore/big-data
touisteur 9mo ago
Every sensor in the array is sampling at frequency, so - first order - you can use that sampling frequency and the sample size, you get an idea of the input bandwidth in bytes/second. There are of course bandwidth reduction steps (filtering, downsampling, beamforming)...
- jacquesm 9mo ago
  Aren't they sampling broadband for later processing?
  - touisteur 9mo ago
    On SKA from what I understand they're sampling broadband but quickly beamform and downsample as the datarates would be unsustainable to store over the whole array.
    - jacquesm 9mo ago
      Right, that makes sense, you'd be looking at an insane amount of data across the ranges that these sensors can look at. But they would still need to preserve phase information if they want to use the array for what it is best at and that alone is a massive amount of data.
      - touisteur 9mo ago
        I think they preserve timestamped I,Q data. Know some people looking at down-sampling, preselecting those signals for longer term storage and deeper reprocessing and they seem to have a 24h window to 'analyze and keep what you need'. We're still in technological phase where ADCs are far more advanced than storage and online processing systems, which means throwing away a lot. But I have high hopes for a system where you upgrade computing, network, storage (and maybe ADCs...) and you get an improved sensor. Throw man-hours at some GPU kernel developers and you get new science. The limit seems more now about enough people and compute to fully exploit the data than technological...
        jacquesm 9mo ago
        Fascinating stuff, thank you for the details and the view of a possible path forward.
        touisteur 8mo ago
        BTW if you're interested in the concept of upgrading a sensor without retooling the RF part, and the impact of 'just' putting new COTS racked server hardware and engineering man-hours to get a 'new' sensor with new capabilities, have a look at Julien Plante's work on NenuFAR (which isn't like the SKA at all :-) : https://cnrs.hal.science/USN/obspm-04273804v1 https://cnrs.hal.science/USN/obspm-04273804v1 . Damien Gratadour, his PhD supervisor is an amazing technologist, dedicated to improving astronomy instruments, and I was very lucky to work with him and his team... the things the French can string together with small teams and thin budgets...
        jacquesm 9mo ago
- Etheryte 9mo ago
  This makes no sense though? Given the Nyquist theorem, simply increasing sampling frequency past a certain step doesn't change the outcome.
  - labcomputer 9mo ago
    Actually, it does. You can decimate the higher sample rate to increase dynamic range and S/N ratio. Also, for direct down conversion, you can get better mirror frequency rejection by oversampling and filtering in software.
    - Etheryte 9mo ago
      None of this changes the actual real amount of data you have at the end of the day though after all is said and done, that's what I mean, so long as you don't botch it and capture too little. In computing terms, the amount of real data in a compressed archive and the uncompressed original is the same, even if the file size is larger for the latter.
  - touisteur 8mo ago
    Sorry, not sure I follow from what I said (explaining how much data sensors produce) to 'increasing the sampling frequency' ? You're usually sampling at larger width to then put specifically taylored pass-band filter and removing aliasing effects and then downsampling. This is a classic signal acquisition pattern : https://dsp.stackexchange.com/questions/63359/obtain-i-q-components-from-a-real-signal-on-the-fly-hilbert-transform-or-digit https://dsp.stackexchange.com/questions/63359/obtain-i-q-com...
dylan604 9mo ago
Are you deliberately obtuse to the play on words of an array being used from a programmer's use of the word in contrast to an array of antennas?
- JamesTRexx 8mo ago
  If you have to ask, you know the answer. :-p

dylan604 9mo ago

I'm thinking that you'd need to print out the array to be able to properly measure this. So we'd need to decide on the print itself. Do we use A4 paper, old school green bar? What size font is used? We'd also need to decide on the contents of each element in the array. Let's say they are floats with a set limit of precision. So an A4 is 210mm x 297mm. 1km / 210mm = 4761.9 1km/297mm = 3367.0. If we go with 4bytes per float, that's 4761.9 * 3367.0 * 4 = 64,133,269.2 bytes => 61.16 MB It's also before coffee, so my logic might not be right for basic math yet